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Topic author: Samizdat
Subject: Requiem for Relativity</title><style>.a33i{position:absolute;clip:rect(424px,auto,auto,410px);}</sty
Posted on: 12/05/2005 16:29:49
Message:

Has any of you read, or have you preliminary thoughts on the new book by Michael Strauss, "Requiem for Relativity: the Collapse of Special Relativity?"

[url]http://www.relativitycollapse.net/[/url]

Replies:


Reply author: tvanflandern
Replied on: 12/05/2005 20:18:23
Message:

quote:
Originally posted by Samizdat

Has any of you read, or have you preliminary thoughts on the new book by Michael Strauss, "Requiem for Relativity: the Collapse of Special Relativity?"
There are dozens, perhaps now over 100 such books, apparently all of them making the same basic misunderstanding of special relativity. Contrary to what this author says, SR has been proved (in the mathematical sense) to be an internally consistent theory. It simply violates intuition and common sense. But that doesn't make it wrong.

What does falsify SR (in favor of Lorentzian relativity, LR) is experimental evidence that gravity propagates faster than light in forward time. And that is now published in the mainstream, peer-reviewed literature, and remains undisputed in print: “Experimental Repeal of the Speed Limit for Gravitational, Electrodynamic, and Quantum Field Interactions”, T. Van Flandern and J.P. Vigier, Found.Phys. 32(#7), 1031-1068 (2002). Everything else is noise from people who cannot suspend their strong, intuitive belief in a universal instant of "now", the non-existence of which is a prerequisite for understanding SR. -|Tom|-


Reply author: Joe Keller
Replied on: 12/16/2005 17:48:40
Message:


Explanation of Michelson/Morley/Miller Non-Null Ether Drift Results

Introduction. The null ether drift results with Michelson interferometry,
have been radically different experiments from the experiment that
Michelson/Morley, Morley/Miller and Miller did. The null results were
achieved with evacuation and metal shielding, which Michelson/Morley/Miller
did not employ. Other experiments, such as the Global Positioning System,
sometimes cited as null ether drift results, differ in even more ways from
Michelson/Morley/Miller.

Description of Michelson/Morley/Miller results. Using several different
interferometers (some of them nonferromagnetic) with many variations in
configuration, in many different laboratories in very different physical
environs in several geographic locations, at all times and seasons, over 40
years, and with trials to rule out suspected causes of error,
Michelson/Morley/Miller's "ether drift" results were consistent. Ascribing
the results to temperature or to any other kind of "error" is absurd. They
found an "ether drift" equal to the sum of the solar apex motion plus Earth's
orbital motion, with the following differences:

1. The phase shift was opposite in sign from what would be expected from
an "ether drift".

2. Parallel to Earth's axis, Miller's average drift equals the solar apex
motion. Perpendicular to Earth's axis, Miller's average drift is about 1/5 of
the solar apex motion.

3. The seasonal change in the drift is much smaller than expected and about
180 degrees out of phase, for both the perpendicular and the parallel
components.

4. There is a constant eastward or westward drift which varies seasonally as
a second harmonic.

My explanation of the above. Earth, and appurtenances near its field (such as
the interferometer), are FitzGerald contracted. The average parallel
component of the drift is due to contraction of the solid parts of Earth
(mantle and inner core), amounting to 70% of Earth's mass. Using the full
Hipparcos data set (Abad et al, Astronomy & Astrophysics 2002), this would
give contraction equivalent to a drift of 11.3 km/s, vs. 9.5 km/s observed by
Miller.

Contraction must be accompanied by mass migration to maintain Earth's shape.
Due to Earth's rotation, only the inner core can contract in response to the
perpendicular (equatorial plane) motion. The liquid outer core can slide
around it to maintain net roundness despite the inner core's slightly
lenticular shape. With only 1.7% of Earth's mass, the inner core provides an
average contraction for the Earth overall, equivalent to 2.6 km/s, vs. 3.4
km/s observed (276RA for the solar apex, vs. Miller's 254RA).

Miller found seasonal variation of about 1.5 km/s in both the perpendicular
and parallel components, much smaller than expected from the magnitude of
Earth's orbital velocity, and with roughly 180 degree lag. This can be
explained by a damping phenomenon.

The recently (Aug. 2005) reported 0.4 degree/day super-rotation of Earth's
inner core, could be due to the counterclockwise advancing total velocity
vector, when Earth's orbital velocity combines positively with the solar apex
motion. Heating and turbulence of Earth's outer core also would occur.

Finally, Miller found a twisting phenomenon which he called the "unexplained
azimuth variation". This could be due to a twisting of the Earth as the axis-
parallel contraction changes.

-Joseph C. Keller


Reply author: Joe Keller
Replied on: 09/22/2006 22:44:21
Message:

Vol. 2 (1826) of the Memoirs of the Royal Astronomical Society contains a report by Ramage, Capt. John Ross, & R. Comfield (No. IX in the volume, pp. 87-91, "read Dec. 10, 1824") which seems to refute Special Relativity. With careful drawings, they show anomalies during lunar occultation of Jupiter & Jupiter's moons (on immersion) and Uranus (on both immersion and emersion). These anomalies are of the same order of magnitude as the aberration of starlight. A followup report in the same volume by R. Comfield & J. Wallis (No. XXVIII, pp. 457-458, "read Nov. 11, 1825") shows anomaly of Saturn and its rings also, on emersion from lunar occultation. Comfield reported using three different reflecting telescopes with apertures 7,8, or 9 inches, and mentions that Wallis used a reflector with aperture stopped down to 6 inches.

Specifically, three anomalies were reported:

1. I call the first anomaly "reverse occultation". Jupiter was drawn as overlapping the moon about 1/5 of Jupiter's diameter, early during immersion. The text and drawings imply that Jupiter's moons achieved about 1/2 diam. overlap. Uranus was explicitly stated to achieve 1/3 diam. overlap. Always this was on immersion into Luna's dark side. Also a bright ring appeared between the "reverse occulted" planets and the "cut out" in Luna.

2. I call the second "spreading". Jupiter and its moons appeared to have larger disk radii when almost completely immersed. At an early stage of this, Jupiter appeared spread into a bell shape. Again this was always on immersion into Luna's dark side.

3. I call the third "truncation". On emersion from Luna's bright side, Saturn, and the rings of Saturn, both in turn were chopped off as though by Luna, but separated from Luna by some distance of darkness. Uranus showed a variant of this: the planet was not chopped off, but simply first appeared on emersion with a distance of intervening darkness between itself and the bright side of Luna, equal to about 1/4 the planet's diam.


Reply author: Joe Keller
Replied on: 09/26/2006 21:13:06
Message:

My perusal of a smattering of other 19th century volumes of Royal Astronomical Society journals, discovered two additional and independent reports of such anomalies during lunar occultation of planets. Vol. 14 (1854) of the Monthly Notices of the Royal Astronomical Society contains a report by Robt. Snow (pp. 183-184, March 12, 1854). Snow's (reversed, thus probably a refractor; 4" aperture) drawing shows "truncation" (as I define it in the previous post; i.e., with a dark gap) on the immersion of Mars into Luna's dark side; his text does not mention that, but does mention a return of "sphericity" after Mars was "more than half hidden". Also, his text mentions a "few seconds" duration of "violet light" to the (telescopic observer's) left of Mars just before immersion. In the same volume, TW Burr, observing the same occultation (aperture not stated), reports (p. 199): "No projection of Mars on the moon's disk was seen or other anomalous appearance."

In vol. 56 (1896) of the MNRAS, "the Director" of the Durham Univ. Obs. gives a report (p. 330) dated Feb. 20, 1893, of a lunar occultation of Jupiter (6" telescope): "Some seconds before [last contact on immersion, about 0300 GMT in Durham, UK] the Moon's limb had the appearance of being bulged in, or a piece cut out." In the previous post I define this as "reverse occultation". The moon was four days old on Feb. 20, 1893, so the occultation, observed at 3AM, must have been on an earlier date. It could have involved a waxing moon low in the west, i.e., immersion into Luna's dark side. Also, Jupiter was an evening star in Feb. 1893. If the date of this occultation were known, it could be confirmed that immersion was into Luna's dark side.

Dawes (same report in both MNRAS v. 5 (1843), p. 121; and Memoirs RAS v. XII, p. 419) reported no "distortion" of Venus on occultation in 1841. He did not state his aperture. He consistently and repeatedly mentioned that viewing was bad. He observed immersion into the bright side of Luna, and emersion from the dark side.


Reply author: tvanflandern
Replied on: 09/26/2006 21:34:58
Message:

quote:
Originally posted by Joe Keller

the Moon's limb had the appearance of being bulged in, or a piece cut out.
That sounds like the same phenomenon as the "black drop effect", as detailed and explained in a "ViewPoint" article on this site at http://metaresearch.org/home/viewpoint/blackdrop.asp -|Tom|-


Reply author: Joe Keller
Replied on: 09/26/2006 21:48:19
Message:

They drew Jupiter as overlapping Luna about 1/5 diam.; i.e., mean 36.5 arcsec Jovian diam / 5 = 7 arcsec, although the second article hints that maybe the Jupiter drawing was exaggerated. They stated that it was 1/3 diam. for Uranus. And they drew it as 1/2 diam. for the Jovian moons. An arcsecond would be about 1/1000 the lunar radius.

According to Wikipedia ("Astronomical Seeing"), "The FWHM [Full Width Half Maximum] of the seeing disc (or just Seeing) is usually measured in arcseconds, abbreviated with the symbol ("). A 1.0" seeing is a good one for average astronomical sites." The most accurate observation above seems to be for Uranus. Not only Uranus, but also the Earthlit "dark" side of Luna are spread. So Uranus' ratio spread:actual wasn't 3:1; it was 3:2, or 3.7 * 0.5 * 0.5 = 0.9 arcsec spread. Assuming that half-maximum determines the apparent spread, this implies 1.8" full-width spread or "Seeing". For Ganymede with 1.4" diam, 1/2 overlap would imply 0.7" spread and 1.4" Seeing. So, Anomaly #1 above is roughly consistent with irradiation spread.

Anomaly #2, "spreading", remains mysterious. It could be physical or physiological.

Anomaly #3, "truncation", seems to be due to irradiation spread plus visual physiology. Irradiation spread is bell-shaped, but roughly, assuming the 0.9" half-width, half-maximum spread calculated above, the surface brightness of Uranus can be calculated by adding 1.8" to its diameter, for the same amount of light. Using 7% for the albedo of Luna's near side and 93% for Uranus' albedo, Uranus' average apparent surface is dimmer than sunlit Luna by 65x. If irradiation spread extends Luna's surface by 0.9" as above, adding this to the observed gap, gives 0.9 + 5.5 / 4 = 2.395"; 2.395 / 0.9 = 2.661 times what I assume is the half-dropoff-radius. 2^(2.661^2) = a factor of 135 in intensity. If, rather than the maria typical of Luna's near side, Uranus emerged from a mountainous edge region with twice the albedo, then the near edge of Uranus, when it appeared to emerge, would be just as bright as the scattered light there from Luna.


Reply author: Joe Keller
Replied on: 10/31/2006 00:56:59
Message:

The "spreading" phenomenon above isn't the "black drop effect". It may be an ether drift effect, caused by a thin layer of ether flowing around the moon. If so, the amount of spreading of a planet at occultation might depend on the lunar phase, because the lunar phase, determines the ether velocity due to Earth's orbital motion.

I found three published photos showing displacement of a lunar peak seen at the moon's edge, but to the accuracy of my measurements (with a mm ruler, sometimes by eye, from a book page) these seem to be explained by libration. I think the peak that I studied, is on the E (farside) or maybe SE rim of the Lyot crater. See Z Kopal, "A New Photographic Atlas of the Moon" (Taplinger; New York, 1971) Plate 148 p. 219; and GL Gutschewski et al, "Atlas & Gazetteer of the Near Side of the Moon" (NASA; 1971) SP-241, Plate 5-3 p. 14, Plate 38-1 p. 22. The peak is at about 50S 88E or maybe 53S 87E. (South of the main peak, is a longer, lower, double, peak, also visible at the moon's edge in Alter's lunar atlas, Plates 40 & 41, but past the edge of the frame in de Callatay. Also someone had sliced out de Callatay's photo of a Soviet moon probe's landing site.)

I drew a line through the northernmost edges of the interior shadows of two reference craters, Fraunhofer V & B. Fraunhofer V prominently straddes the nearside rim of the crater Fraunhofer. Fraunhofer B is a prominent, isolated, simple, round crater about 2/3 the diam. of Fraunhofer; its N shadow edge is at about 42S 67E. See Gutschewski, Plate 52-2 p. 35. See also Plate 523, "Lunar Orbiter Photographic Atlas of the Moon", DE Bowker & JK Hughes, NASA SP-206, 1971. A drawn USAF Mercator moon map is reprinted in FM Branley et al's "Astronomy" (1975); and in Kopal, op. cit.

D Alter, ed., "Lunar Atlas" (Dover; New York, 1964) Plate 41, shows the moon at phase 4.59d in 1938. V de Callatay, "Atlas of the Moon" (Macmillan; 1964), Plate 1, p. 98, shows the moon at phase 3.53d on Dec. 30, 1943. Alter gives the libration angles and time of observation; these agreed with the American Ephemeris. For de Callatay, I got the libration angles by assuming a 6PM observation time at Pic du Midi (roughly consistent with the terminator line, i.e. solar colongitude, compared to Alter Plate 40, and with good sky position) & interpolating in the American Ephemeris.

I considered the libration as the sum of infinitesimal rotations called "to/from" (the peak moves toward or away from Earth) and "twirling" (the moon rotates about the axis defined by the peak). Because the peak will be taken to be at 45S latitude (a good approximation for "twirling" because, luckily, the long. & lat. librations are about equal for all three Plates considered), these are easy to calculate from the regular libration angles.

"To/from" libration rapidly changes the identity of the apparent peak, by bringing other parts of the ridge into position, but luckily this libration, whether the peak is taken at 45S or 50S, differs only about a degree between the plates considered. The smallness of the differences in "to/from" libration is confirmed by measurement on the photos, using the NS length of crater Furnerius vs. its distance from the moon's edge. The identity of the peak moves only reluctantly with "twirling" libration. Because the reference craters make a line almost coplanar with the twirling axis and the Earth, to/from libration, to first approximation, doesn't disrupt their collinearity with the peak.

It appears from the 1938 American Ephemeris, that the libration published in Alter, Appendix A, is for Washington D.C. For an observation at 8PM Pacific Standard Time, of a 4.6-day-old moon, the 3000 mile distance across the US would project to half that, i.e., roughly 3/8 degree greater libration in longitude. For de Callatay's presumed 6PM observation of a 3.5-day-old moon, the 5000 mile transatlantic distance would project 1/sqrt(2), i.e., roughly 7/8 degree less libration in longitude.

So we have librations for Lick Plate #2 ( = Alter Plate 41) of long. +5.54, lat. +6.63; and for de Callatay long. +1.21, lat. +2.31. The difference in "twirl" libration is (5.54+6.63-1.21-2.31)/sqrt(2)= 6.12 degrees. (Direct measurement on the photos yielded a corroborating estimate of 5.8 degrees.) Spherical trigonometry approximations gave 33km expected displacement of the peak (northward in Alter Plate 41, vs. deCallatay), along the moon's edge, referred to the line through the reference craters where it intersects the moon's edge. The actual displacement is measured to be 31 km, using 57 km plateau-to-plateau NS diam. of the crater Fraunhofer as a yardstick.

Alter's Plate 40 (Lick Plate #1) requires about +3/4 / sqrt(2) degree geographic correction in long. lib., giving long. lib. +3.93, lat. lib. +3.76. The same peaks appear as on Alter's Plate 41. (3.93+3.76-1.21-2.31)/sqrt(2)= 2.95 degrees, giving 16km expected displacement northward vs. deCallatay, relative to the reference craters' line; measured is 13km.


Reply author: Joe Keller
Replied on: 01/21/2007 20:40:02
Message:

Abstract. Pioneer 10 tracking, and stellar occultations by Saturn's rings, reveal violations of Special Relativity, consistent with an ether boundary at about 53 A.U. from the sun.


In 2001, I remarked (Aircraft Engineering & Aerospace Technology 74:269, last two pars.) on an unexplained and unnoticed anomaly (besides the anomalous deceleration) in the Pioneer 10 Doppler data. Beginning at about 53 A.U. from the sun, the Doppler shift begins to oscillate sinusoidally with small (roughly 1 cm/s velocity equivalent) amplitude, and period one year. The phase is such, that this apparent oscillation in velocity would occur if, beyond 53 A.U., the time dilation is really the time dilation that exists when the radio signal enters the 53 A.U. boundary (not the time dilation that exists when the radio signal reaches Earth).

The occultation of a star by Saturn's rings, amounts to a test of another special-relativistic effect, the aberration of starlight. Because stars lie more than 53 A.U. away, the aberration of starlight might be expected to result, from the tangential velocity when the light enters the 53 A.U. boundary, not to result from the tangential velocity when the light reaches Earth. On the other hand, light interacts with transparent air (causing the index of refraction to differ considerably from 1) and so might interact with gas or transparent particles in Saturn's rings. So, in most parts of the rings, the aberration of starlight from an occulted star might be expected to result from the tangential velocity when the light reaches Earth, as in Special Relativity theory.

When Saturn is near quadrature with the sun, the difference in the two tangential velocities amounts to about 0.1 arcsecond of aberration. As the star moves between dense and sparse ring regions, especially near the edge of the ring, it might jump back and forth by 0.1". This was observed by Mourao & Mourilhe (A F O'D Alexander, The Planet Saturn, 1962, pp. 442-443):

"As Saturn was then near a station, with apparent motion only 1" of arc per hour...

"By 5h13m13s the immersion seemed to be complete...About 5h20m the star seemed to 'beat', i.e. to move inside, then outside the ring, a phenomenon repeated several times."

This 6.8 minutes of time, between immersion and the onset of 'beating', amounts to 0.1" of movement of the ring edge, i.e., an amplitude of 'beating' of 0.1", assuming the star moved just outside the ring.

Mourao observed from Brazil, in 1960 with an 18-inch refractor, 650 power, with a clear, stable sky. In 1920, from South Africa with a 6-inch refractor, 216 power, W Reid et al (Alexander, op. cit., pp. 346-347)(BAAJ 30:230) observed the oil-drop phenomenon at emersion from a stellar occultation by Saturn, but during immersion into the ring, Reid didn't see the 'beating', only fluctuation, a flicker, and sudden extinction. However, Saturn was only about an hour from opposition.

In 1917 (Alexander, op. cit., pp. 340-341), from Sussex with a 5-inch refractor and 100 to 250 power, J Knight noted that

"...apart from isolated moments when the air was particularly unsteady, [the occulted star] never seemed wholly to disappear." When in the Cassini division, Knight thought the star's image looked elongated. Saturn was about halfway between opposition and quadrature.

The occultation of 28 Sgr in 1989 occurred when Saturn was near opposition; neither 'beating' nor elongation were observed (J Harrington et al, Icarus 103:235). A fault of the modern occultation observations is that they tend to emphasize automated data collection which would seem to miss such unexpected effects.








Reply author: Joe Keller
Replied on: 01/29/2007 15:51:51
Message:

Does anyone know where I can find data on the aberration of starlight that might confirm that the direction of it, is "off" roughly 0.3 degree? That is, that the aberration vector depends on the velocity of Earth when the starlight crossed the 53 AU limit?


Reply author: tvanflandern
Replied on: 01/29/2007 17:46:19
Message:

quote:
Originally posted by Joe Keller

Does anyone know where I can find data on the aberration of starlight that might confirm that the direction of it, is "off" roughly 0.2 degree? That is, that the aberration vector depends on the velocity of Earth when the starlight crossed the 53 AU limit?
You have your facts and/or numbers wrong for some reason. Stellar aberration is a function of speed relative to the local gravity field, and rarely exceeds 0.01 degrees in the solar system. And nothing special happens at 53 au, which is also not any kind of limit. -|Tom|-


Reply author: Stoat
Replied on: 01/30/2007 05:20:29
Message:

Do you think that the transmitter.receiver mismatch on the Cassini-Huygens mission might be worth a look, as a possible relativistic error, as I understand it did have something to do with the doppler shift?


Reply author: Joe Keller
Replied on: 01/30/2007 11:33:24
Message:

quote:
Originally posted by Stoat

Do you think that the transmitter.receiver mismatch on the Cassini-Huygens mission might be worth a look, as a possible relativistic error, as I understand it did have something to do with the doppler shift?



I heartily agree that someone should look at this! Thanks for your input!


Reply author: Joe Keller
Replied on: 01/30/2007 12:11:54
Message:

quote:
You have your facts and/or numbers wrong for some reason. Stellar aberration is a function of speed relative to the local gravity field, and rarely exceeds 0.01 degrees in the solar system. And nothing special happens at 53 au, which is also not any kind of limit. -|Tom|-



Thanks for checking, but I think my facts were approximately right. I refer not to the absolute value of stellar aberration ( 20.5" = 0.0057 deg ) but to a hypothetical error, of about 0.3 degrees, in the direction of that aberration (I'd estimated 0.2 but it's really 0.3 so I had to correct that above). That would amount to 0.0052 radians x 20.5" = 0.107". The hypothesis is, that we might be observing the aberration due to what Earth's velocity vector was 53 AU * 8.3 min/AU = 440 min ago.

If many data points on a sinusoid are known, the phase of that sinusoid can be determined to surprisingly high precision. Perhaps Bradley's, or some subsequent, demonstration of stellar aberration would reveal, on careful statistical analysis, that the aberration circle is 0.3 degrees out of phase with expected.

The second point is debatable (see my 2002 article in "Aircraft Engineering & Aerospace Technology" a journal which is unrefereed but does have an editorial board comprised mostly of college professors; and also the article by JD Anderson of Cal Tech cited as a reference therein). Anderson's article shows dramatic apparent speed fluctuations in Pioneer 10 occurring at about 53 AU but hardly anywhere else. I recall that others have suggested somewhat hesitantly that the probe encountered a cluster of comets there, however my estimate of the required number, large mass, and small distance between these alleged comets argued against that. Furthermore, Anderson's article shows that the probe began to have apparent sinusoidal fluctuations in speed, of period one year, then and thereafter. So indeed there is, or at least was, something there at 53 AU, and no one knows what.


Reply author: Joe Keller
Replied on: 01/30/2007 12:41:59
Message:

My (2nd) Sept. 26 post above: "Anomaly #2, 'spreading', remains mysterious. It could be physical or physiological." (See: Vol. 2, 1826, of the Memoirs of the Royal Astronomical Society: Ramage, Ross, & Comfield, No. IX in the volume, pp. 87-91, "read Dec. 10, 1824"; followup report in the same volume by Comfield & Wallis, No. XXVIII, pp. 457-458, "read Nov. 11, 1825".)

The *spreading phenomenon* drawn and described in the 1826 Memoirs of the Royal Astronomical Society, for an occultation of Jupiter by the dark side of Luna, *has been photographed*, for an occultation of Mars by the dark side of Luna, July 2003. The (digitally processed) photo is in Bonnell & Nemiroff, "Astronomy / 365 Days, the best of the 'Astronomy Picture of the Day' website", page "July 24" (it has dates instead of page numbers). I think this book is on the shelf at many large bookstores; also, the Iowa State Univ. library has a copy.

In the next-to-last Mars photo on the right, the radius of curvature of the bright (left) edge of Mars shows decreasing curvature near the edge of Luna. In the last photo, not only does the curvature decrease, but there is an inflection point (i.e., the curvature becomes negative as in the "bell curve"). The authors of the 1826 article said that their drawing was somewhat exaggerated and that the phenomenon was barely perceptible. I think this is consistent with what is seen in the photo (credited to Ron Dantowitz, Clay Center Observatory at Dexter & Southfield Schools).


Reply author: Stoat
Replied on: 01/31/2007 04:34:56
Message:

I found this with a quick google search, I just gave it a quick read, so it might not be what we want but it does question the transmitter mismatch. http://home.netcom.com/~sbyers11/litespd_vs_sr.htm


Reply author: Joe Keller
Replied on: 01/31/2007 18:31:50
Message:

The error, 0.107", which my "53 AU barrier" theory predicts, was discovered in 1902: it is "Kimura's phenomenon" (H Kimura, The Astronomical Journal 22(517):107-108 (1902); SC Chandler, AJ 23(530):12-14 (1903)). Chandler could not explain Kimura's phenomenon; apparently, it remained unexplained until now (see above).

Kimura statistically analyzed extant determinations of the aberration constant. Most had been made by measuring the declination of stars as they crossed the highest meridian at different seasons (Struve's method). Ideally the star observed would have declination equal to the latitude of the arctic circle, and its RA would be 18h. Then if aberration were determined, by Earth's velocity when the starlight crossed the barrier at 53 AU, the observed declination would be 0.107" higher than predicted, at the winter solstice, and 0.107" too low, at the summer solstice.

Choosing a star at RA 6h instead, would reduce this amplitude to 0.107 * cos (23.45 * 2) = 0.073" with opposite phase. The articles in the Astronomical Journal show that researchers used stars of all RAs, but they favored 18h vs. 6h, I suppose so that the aberration would be somewhat larger, and the daytime observations would be the winter ones with a low sun. Kimura cites compilations of Th. Albrecht, Astronomische Nachrichten 3734, and again 1898 & 1900 articles by Albrecht, but my library doesn't have that journal, print or online. In the articles I found in the Astronomical Journal vols. 5-19, 1856-1899, the RAs of the 46 stars (including repeated studies) used (there might be a few errors in my tabulation) were such that the average amplitude of the Kimura phenomenon would have been * 0.027" *, if they were all on the arctic circle, which to a good approximation they were. (I excluded a few stars that were used for determinations using Loewy's pair&prism method or the RA method; those I used seemed all to have been used for Struve's declination method described above.)

Kimura's amplitude was "almost" * 0.03" +/- 0.01" *. Furthermore Kimura found this first-order periodic term to be positive near the winter solstice, negative near the summer solstice, and zero near the equinoxes.


Reply author: Stoat
Replied on: 02/01/2007 05:30:08
Message:

I get an ether energy density, of the Sun at 53 au, of about a two hundredth of what it is at the Earth. Can you explain why you chose that particular number?

(Edited) I get 4.19821189643E 04 Joules / cubic metre at 1 au, and 8.0495163309E 02 Joules / cubic metre at 53 au.

Apropos of nothing, i've just come across an ether theory that suggests the ether is a Bose-Einstein condensate of neutrino / anti neutrino pairs, rather like Cooper pairs.


Reply author: Larry Burford
Replied on: 02/01/2007 08:38:52
Message:

Stoat - please double check your calculations and/or your description of them. The joule is a unit of energy, not a unit of energy density. And if I am following Joe correctly, someone else chose the 53 AU distance, based on observational evidence that can be interpreted as something like a refraction event at that distance.


Reply author: Joe Keller
Replied on: 02/01/2007 11:57:37
Message:

quote:
Originally posted by Stoat

I get an ether energy density, of the Sun at 53 au, of about a two hundredth of what it is at the Earth. ...
(Edited) I get 4.19821189643E 04 Joules / cubic metre at 1 au, and 8.0495163309E 02 Joules / cubic metre at 53 au.
...



Very exciting! I hardly know anything about that theory, but the ratio is within 1% of: sqrt(2)*137 (fine structure constant = 1/137.036...). That's the best that can be expected, since the best I could estimate the 53 AU distance from JD Anderson's article, was 1% accuracy (the fluctuations in Pioneer 10's speed occurred over considerable time, and I just eyeballed the average). Also accounting for Earth's orbital eccentricity might involve another 1% difference. So there might (or might not) be a connection with the fine structure constant.


Reply author: Joe Keller
Replied on: 02/01/2007 12:04:42
Message:

quote:
Originally posted by Larry Burford

...the 53 AU distance, based on observational evidence that can be interpreted as something like a refraction event at that distance.



I agree. As you say, it's "something like" refraction, but not refraction in the usual sense. The distance is defined by the cluster of unexplained apparent speed variations of Pioneer 10 according to JD Anderson's article which was cited in my 2002 "Aircraft Engineering & Aerospace Technology" article.


Reply author: Joe Keller
Replied on: 02/01/2007 12:20:07
Message:

Let m * v^2 = k * T, where m is the electron mass, k is Boltzmann's constant, and T is the cosmic background temperature = 2.73K. Then v = escape velocity (velocity of escape from the sun) at 42.9 AU. One obtains the above formula by modifying the Maxwell distribution formula for the root-mean-square speed, to give the (r.m.s.) velocity component in one direction only, e.g., parallel to the solar system's magnetic field lines (because the two components perpendicular to the field lines would be ineffectual in helping the electron escape). Because the states are so uncrowded, the Maxwell distribution doesn't need to be modified for fermion statistics.

Let the speed parallel to the field line be "vx". Another estimate would use mean(abs(vx)) because a group of electrons, once agitated by the background temperature and moving in the same direction along the field line, might then tend to equalize their speeds with each other. This would correspond to a distance of pi/2 * 42.9 AU = 67.4 AU. Solar radiation pressure and solar wind would tend to reduce the effective escape velocity and thereby reduce the 67.4 AU figure.

Yuri Galaev's recent experiments indicate that the "ether", whatever it is, travels, in a sense, at the same speed as the dominant transparent physical medium such as air. In the above, free electrons are treated as the physical medium.

The solar system would seem to sit in a spherical block of ether of radius 53 AU. Pioneer 10 tracking indicates that special-relativistic time dilation depends on whether the source is inside or outside this block: if outside the block, then the operative relative velocity of source and observer, for the time dilation calculation, is that which pertains when the light enters the block, not when the light reaches Earth. Two star occultations by Saturn's rings, reported in Alexander, showed flickering position (or fused flickering) of the order of 0.1 arcsecond, as if the aberration of light is analogously affected, by whether the light arrives directly from the star (operative relative velocity is that when the light enters the block) or interacts with transparent gas in Saturn's rings (operative relative velocity is that when the light reaches Earth). Both Pioneer 10 and Saturn lie near the ecliptic; also for sources away from the ecliptic, the Kimura (1902) phenomenon confirms, to 30% accuracy, similarly altered aberration due to the 53 AU barrier.

At 53 AU, the equilibrium temperature of solar radiation is much higher than the cosmic background temperature. If the solar radiation temperature isn't involved, then Earth should have a barrier at 67.4 AU / 333,400 = 18,790 mi. from its center. The electron maximum of the outer Van Allen electron belt (Enc. Americana 1998) is, in Earth's equatorial plane, 17,630 miles from Earth's center.

For Mars, the barrier would be at 2010 mi; Mars' radius is 2110 mi. Apparently a solar system body has a strong magnetic field if, and only if, it rotates rapidly and is more massive than Mars.


Reply author: Stoat
Replied on: 02/02/2007 03:44:30
Message:

Here's that energy density formula and the values I put in, in case you want to alter it for the Earth orbit.
E = GM ^ 2 / (4 pi ) r ^ 4
M = Sun's mass of 1.99E 30
One au = 149.6E 09 metres.

This is from Robert Carroll, and I think we need to also look at his formula for light bending as opposed to Einstein's.
theta = 2 G M / r C ^ 2

As opposed to Einstein's, theta = 4 G M / r C ^ 2

Theta is in radians and "r" is the closest approach of a photon.

Carroll does point out that Einstein's formula is claimed to be more nearly in accord with measurements made during a total eclipse of the sun. He then points out that there is no way to separate the deflection due to the atmosphere of the sun from that due to the gravitational field only.

The two formula differ, because Einstein uses a constant velocity for light, in the contracted space of the sun.


Reply author: Joe Keller
Replied on: 02/03/2007 10:51:50
Message:

"Recently, astronomers found that the [Kuiper] belt has an unexpected sharp outer edge at 50 AU."

- MA Garlick, "An Illustrated Atlas of the Universe", 2006


"[Kuiper belt] objects beyond 41 AU tend to have eccentricities less than 0.1 and are not, in general, in resonances."

- PR Weissman & HF Levison, in: Stern & Tholen, eds., "Pluto & Charon" (1997), p. 592

If orbits are circular and unperturbed, a sharp Kuiper belt boundary is possible. The minimum density of Kuiper belt objects (a tenth as many objects as at 10 AU closer to the sun; there is a modest increase farther out) occurs at 52 +/- 1 AU by my interpolation of either Fig. 2 or 3, in CA Trujillo & ME Brown, "The Radial Distribution of the Kuiper Belt", Astrophysical Journal 554:L95-L98 (2001).

"...we likely have zero detection of objects beyond 53 AU...If planetary perturbations are solely responsible for the structure of the Kuiper Belt, a dense primordial disk would be expected beyond [about] 50 AU where these perturbations are insignificant."

- RL Allen et al, "The Edge of the Solar System", Astrophysical Journal 549:L241-244 (2001), pp. L242, L244.

Allen et al (op cit, Table 2) found 24 new objects, in a survey designed to find any objects between 30 and 80 AU which were in bound orbits (op cit, p. L242). One of these was at 20.9 +/- 0.3 AU; the other 23 ranged from 31.7 +/- 4.2, to 52.6 +/- 0.1 AU.


Reply author: Stoat
Replied on: 02/04/2007 06:38:46
Message:

Suppose that there is something in the idea from Milikan that isolated systems ingest their electrons via a k capture process. Then at, let's say, 100 a.u. atoms are imploding as gamma rays. These would be travelling at above laboratory light speed in the r. i. of space at that distance. Could these gamma rays "herd" outer Kuiper belt objects at the 50 a.u. range?


Reply author: Joe Keller
Replied on: 02/05/2007 12:58:30
Message:

There is "new physics" at 53 AU. The Kuiper belt ends suddenly there. Two of the three main tracking anomalies of Pioneer 10, occurred, or began, there; these anomalies are quantitatively or qualitatively inconsistent with gravitational influence by Kuiper belt objects and furthermore such objects would be encountered in the densest part of the belt (42-50 AU) not precisely at its edge (53 AU) (graph from JD Anderson, "Indication, from Pioneer 10/11, Galileo, and Ulysses Data, of an Apparent Anomalous, Weak, Long-Range Acceleration", ArXiv.org internet paper, 27 Aug 2001).

The amplitude of the Kimura (Astronomical Journal, 1902) phenomenon is known only to 30% accuracy, but to this accuracy, its amplitude, and also period and phase, are consistent with aberration of starlight determined when the light crosses the 53 AU barrier, not when the light reaches Earth. The two oscillating ("beating") or elongated stars occulted by Saturn's rings (Alexander, "Saturn", 1962) are consistent, to order-of-magnitude accuracy, with this same aberration difference, between starlight originating beyond 53 AU and starlight re-emitted from transparent gas in Saturn's rings.

The "anomalous deceleration" of Pioneer 10 doesn't have any obvious connection with 53 AU, but one of the other anomalies, was an erratic variation in radio frequency occurring mainly near 53 AU. Mainly this was two apparent reductions, of one and two weeks duration, by 18 cm/s, in Pioneer 10's speed. I noted in my 2002 paper ( Aircraft Engineering & Aerospace Technology 74(3) ) that this is consistent with temporary loss of 3/4 of Pioneer 10's relativistic time dilation relative to the sun.

The other Pioneer 10 tracking anomaly was an oscillation in frequency beginning at about 53 AU and remaining about constant thereafter. The period, phase and amplitude of this are consistent with a relativistic time dilation dependent, not on relative speed when the signal reaches Earth (as calculated) but rather on relative speed when the signal reaches 53 AU.

Two theoretical correlates of this 53 AU distance now have been discovered. In my 2002 paper, I remarked that if a proton exists in the smallest (Gaussian) matter distribution allowed by the Heisenberg uncertainty principle, then 53 AU is the distance beyond which the gravitational field vector becomes directed away from the sun, at some points within the proton (i.e., maximum proton gravity exceeds solar gravity). In another recent post, I remarked that 53 AU ( +/- 20% depending on the kind of mean considered) is where the mean speed of electrons at the cosmic background temperature, equals solar escape speed.

If the mean of v^1.5 is considered, that distance is 52.6 AU. (I added the planetary masses to the sun's mass, neglecting other masses; and used T = 2.726.) The 1.5 exponent could be justified as giving the mean square root (as for Brownian motion) of the rate of kinetic energy transport v*v^2 = v^3. (The 1.5th moment of the error function isn't in the CRC tables; I used Jahnke & Emde, 4th ed., pp. 20, 14.)

If both the proton and the electron theories above, explain the 52.6 AU limit, then the CMB temperature is determined. It follows that the CMB is a product of the edge of the solar system. Its symmetry is due to the symmetry of the sun's gravitational field. The CMB dipole might be due to some cosmic vector. The quadrupole and higher moments would be unexpectedly small, and correlated with the plane of the ecliptic; this is so (see MNRAS, Mar. 21, 2006).

The r.m.s. electron radial velocity of 6.43 km/s would give r.m.s. deviation in the CMB temperature, of 2.14 * 10^(-5). Per RB Partridge's 1995 book, "3K: the Cosmic Microwave Background Radiation", pp. 252, 254 & Fig. 7.19, the least upper bound for the rms deviation of CMB temperature, was determined in 1989 to be 1.9 * 10^(-5). (To make this into an explanation, one might assume that large parts of the sky somehow moved with that speed.)

Alternatively, a body near the 53 AU barrier could provide, say, 1/25,000 of the sun's gravitational force, yet negligible gravitational energy. This would move the barrier outward by one part in 50,000, and reduce the apparent CMB temperature by 2*10^(-5). Such bodies would have orbital periods of about 380 yr., so the CMB variations ("shadows on the wall") might move 11 degrees between the COBE (launched Nov 1989) and WMAP (launched June 2001) studies. This would be disguised by the effects of multiple bodies moving different directions. An object with the density of Earth and the typical moderately large "Kuiper Belt Object" diameter of 100km, would need to be 0.01 AU from the barrier; its depression in the CMB temperature would be of diameter roughly 80 arcsec.

In about 1986, RD Davies found a CMB anisotropy about 10 degrees wide, near RA 217 Dec +40. This could be caused by an Earth-size "Planet X" with 900^2 times more mass than the above, i.e., 10,000km diam, and roughly 9 AU from the barrier: i.e., outside, at 62 AU, because the negative 2nd derivative plotted, implies a maximum of temperature. The planet might be somewhat bigger if it really lay at +35 or +45. Based on Halley's and two other comets, JL Brady, of Lawrence Livermore Labs, in Publications of the Astronomical Society of the Pacific 84:314+, pp. 319 & 322, 1972, predicted a Jupiter-size "Planet X", 63 AU from the sun, with eccentricity 0.07, and 1946 position RA 75 Decl +67. Brady's planet lay outside Tombaugh's search; mine too close to the edge of the search, to tell. Brady's planet and mine are too far apart to be identical.

From the positions of the giant planets during data collection, the difference in the magnitude & galactic longitude of the CMB dipole, between COBE (DMR) & Wilkinson (WMAP), is predicted by this model. See the post below.


Reply author: Joe Keller
Replied on: 02/08/2007 00:21:38
Message:

Error bars include not only measurement errors, but also variations due to physical effects yet unknown. So, the error bars on the CMB dipole really are upper bounds of the measurement error; the measurement error might be much less. Even accepting the error bars as published, there is a significant difference between the "COBE DMR" and "Wilkinson WMAP" determinations of the CMB dipole's galactic longitude.

One-, two- and four-year COBE reports, and one- and three-year WMAP reports, have been published. No matter which COBE or which WMAP reports are considered, the magnitude and galactic latitude of the CMB dipole, given by COBE & WMAP, differ by less than one (joint) standard deviation. On the other hand, no matter which reports are considered, COBE and WMAP give galactic longitudes differing by *more* than one (joint) standard deviation, with COBE exceeding WMAP. COBE & WMAP always are consistent with themselves, though, within one (joint) standard deviation. For the 4-year COBE & 3-year WMAP, the difference in galactic longitude, 0.40 degrees, is 1.2 joint standard deviation, p = 11.5%, one-tailed. For the 2-year COBE & 3-year WMAP, the difference is 2.65 s.d., p = 0.3%, one-tailed.

One can see on the Norton atlas galactic chart, that the directions of increasing galactic and ecliptic longitude are almost parallel at the location of the CMB dipole vector. Furthermore, the CMB dipole vector is only about 12 degrees from the ecliptic. From published articles, I gleaned the start and stop dates of the above data sets; data collection apparently was very steady (99.4% efficient for the last two years of WMAP and almost that good for the first year). From the Astronomical Almanac (formerly American Ephemeris & Nautical Almanac) I found the heliocentric ecliptic longitudes of the giant planets Jupiter, Saturn, Uranus and Neptune near the midpoint of each of the five observation periods. Using the approximation that the CMB dipole lies on the ecliptic at lambda = ecliptic longitude = 171.4 degrees (found using published celestial coordinates for the COBE 4-year report, and graphical conversion on the Millenium atlas), I found the theoretical effect of the giant planets, on the CMB dipole, using the above theory. (Even for Jupiter over 4 yrs, the error of using sin(mean(theta)) instead of mean(sin(theta)), was only 5%, and Jupiter will prove to have, on average, only 1/10 the effect of Neptune.)

First, I found by successive interpolation using a "BASIC" computer program, accurate "barrier" distances (i.e., equal total gravitational *force* at 52.6 AU in superior conjunction, and "r" AU in inferior conjunction) for the two extreme positions of each planet. The difference, in total potential energy, between these distances, divided by the total potential energy, divided by two, gives a dipole strength. The ratio of this dipole, to the CMB dipole, gives the radians maximum deviation, in CMB dipole ecliptic longitude, that the planet can cause.

Planets such as Neptune which are not near the sun, give a much smaller dipole, for given extreme values (small bodies very near 53 AU give negligible dipole integrated over the sphere). The unique 4th-degree polynomial having the properties of ordinate and slope, needed to correct for this, is (1-(r/52.6)^2)^2, and this was applied for all planets. Neptune showed ten times the dipole strength of Jupiter, and five times Saturn or Uranus.

The sines of the angles between the planets and the dipole, indicated that, using the largest COBE and WMAP reports, the galactic longitude should differ 0.466 deg (one must multiply by sec(48.2) because the galactic longitude lines are closer together than at the galactic equator) between COBE and WMAP. The actual difference was 0.40 +/- 0.33 deg. However the predicted and actual difference were opposite in sign; maybe there is some reason for this sign reversal.

It also was predicted that the 3-yr WMAP dipole magnitude should be 9 microKelvins less (assuming sign reversal of the effect as above) than the 4-yr COBE dipole magnitude. Actually it was 5 microK more, but with such error bars and variable results between data sets, that there is neither confirmation nor denial.

N Jarusik et al, ArXiv.org May 15, 2006, Fig. 12, especially the 6th & 7th of the 8 maps produced by various subtractions and averagings, shows a band of increased (CMB) "delta(T)/T" (as shown by the mixture of dark blue and green dots), roughly 45 degrees wide, crossing the center at the angle of the ecliptic. This suggests that Kuiper belt objects indeed cause CMB anisotropy.

J Davies, "Beyond Pluto" (Cambridge Univ. Press, 2001) Fig. 8.1, p. 150, combines five studies into a chart indicating that the number of Kuiper belt objects greater than a given mass, is roughly inversely proportional to that mass. Thus the total mass of the Kuiper belt objects could be very large (almost a divergent infinite series in either direction).

RB Partridge ("3K...", op cit) Fig. 7.19, p. 254, reproduces a graph from ACS Readhead et al, ApJ 346:566+, 1989, showing that for small angular sizes of sky, the rms deviation of the CMB, decreases as the reciprocal of the square root of the area of the sky patch; I think this is a manifestation of the textbook sqr(N) sampling formula. For larger sky patches, the rms deviation increases again as the area of the sky patch. This would occur, if the mass of the largest object goes as d^3, where d is the edge of the cube; thus the height of deviation goes as d, the area of deviation goes as (sqr(d^3))^2, and the overall deviation due to that object, as d^4. Dividing this by the area, gives deviation proportional to the area, as observed.


Reply author: Joe Keller
Replied on: 02/08/2007 22:08:24
Message:

The 0.3 x 0.3 degree pixels of WMAP, with its 60 microK sensitivity, imply that objects would need to be about 400km diam to be detected by WMAP if the above mechanism is correct. The upper bound in Allen & Bernstein's article, together with the N = 1/M law, implies an upper bound of 40 such detectable objects near enough to either side of the 53 AU barrier (i.e., < 40 maxima or minima of the CMB temperature). Furthermore, COBE's poorer resolution & sensitivity would give COBE only a fraction this many maxima & minima. If smaller objects usually occur in clusters like the Trojan asteroids, then the effective number of objects might be several times larger.

The CMB quadrupole from COBE (4-yr) is given as 10 (+7 or -4) microK; also, by the alternative method of data analysis, it is estimated at 15.3 (+3.8 or -2.8) (Bennett et al, ApJ 464:L1+). I multiplied the dipole theoretically produced by each giant planet, by a 4th-degree polynomial estimating factor, 1-(1-(r/r0)^2)^2, to obtain its theoretical quadrupole. Neptune gave 14.5 microK, Uranus gave 1/10 that, Saturn 1/50 and Jupiter 1/300. For COBE (4-yr), Neptune and Uranus were only 3 degrees apart on average, so the predicted quadrupole would be 16 microK.

For WMAP, agreement is poorer. WMAP (1-yr) showed a CMB quadrupole of 8 +/- 2 microK (G Hinshaw et al, ApJ Supp 148:135+) though Neptune & Uranus were only 16 degrees apart on average. Due to a change in terminology, expressing the quadrupole "power" in different fundamental units, I don't know how to express the 3-yr WMAP result in comparable form; but CG Park, C Park & R Gott, ArXiv.org Aug 5, 2006, do give the direction of the quadrupole as l=241, b=67; and of the octupole as l=235.5, b=63 (midranges of six close results). Apparently these vectors are to the pole of the "m=0" (i.e., lowest) spherical harmonic of a given order n. They lie about 10 degrees above the ecliptic and about 145 degrees from Neptune (coordinate conversion with the online utility on www.realuniverse.nao.ac.jp).


Reply author: Joe Keller
Replied on: 02/09/2007 13:30:36
Message:

If the CMB originates at an average distance of circa 400 million light years, then as I outlined months ago on this messageboard, a long wave of comparable half-wavelength, affecting the value of the Hubble parameter, could cause the CMB dipole; but then there also would be a quadrupole of similar magnitude. Really, the quadrupole is about 1/300 of the dipole.

Only within the solar system, but at great distance from the sun, is there known to be such symmetry as the CMB shows. The sum of the planets' masses is 0.134% the sun's mass; the CMB dipole is 0.123% the CMB temperature.

Graphically I estimate the ecliptic longitude of the CMB dipole as 171.4. On March 6, 2006, the sun achieved the yearly minimum "B0" of -7.23deg (solar latitude of center of sun's disk): the ecliptic longitude of the sun would have been 180 - (20 - 6) = 166. That is, the sun's rotation axis, the principal axis of the solar system, and the CMB axis are nearly coplanar (the use of the principal axis of the solar system makes the CMB dipole 8 degrees out of the plane, instead of 171-166=5). By analogy with the coplanarity of the moon's rotation axis, the moon's orbital axis, and the ecliptic axis: the CMB axis might be the sun's true orbital axis as part of a larger system. Apparently a small CMB quadrupole and octupole also lie in this direction.


Reply author: Stoat
Replied on: 02/10/2007 06:19:06
Message:

Hi Joe. As, without a doubt, the thickest reader of this board [:)] I think it's encumbent on me to ask you to give a little primer on multipoles. That should open up the debate a bit more.

Some really dumb observations [8D][:I] Let's say we've got an electron and this bit of matter has an atmosphere of "space" ether, whatever. So half its energy, at rest, is space, half matter. Push it and its kinetic energy goes up and it potential energy goes down. The energy density of the ether can change with velocity. (joules per cubic metre [:D][:D][:D] but what do I know[:)] )

Note that from that equation of Robert Carroll, the ether "atmosphere" falls off as an inverse fourth power i.e. a quadropole.

A question: a perfect spherical planet would have this quadropole but we don't get them, so I'll bash the planet with a big mallet. Now it's got a quadropole. Do I need to do that? Or, is the quadropole moment a consequence of the ether "atmosphere"?

Multipoles; we've got the sun's space, which contains planets, which have their own space. So I took a look at Legendre polynomials http://en.wikipedia.org/wiki/Legendre_polynomial

Yuck [:(] hard sums! Though I have to say, that graph of values for n looks promising [:)] Start with a big ball of gas, bash it with a supernova, and see if we can get a multipole set up that gives us Bode's "Law."

I blame my parents[:D]


(Edited) From another thread; if we consider the ether to be a Bose Einstein condensate of neutrino pairs, analogous to Cooper pairs, then a ftl graviton could "see them as paired particles, rather than as a boson. On pasing through ether then, I think we have a very rapid Lorenzian contraction for some ftl gravitons. However, as they slow down, the neutrino pairs become increasingly fuzzy. A graviton which has lost its energy to the ether and is at light speed, then sees a boson and, i presume, passes sraight through it. Yep [:)] needs work[:o)]


Reply author: Larry Burford
Replied on: 02/10/2007 10:23:56
Message:

[Stoat] "Push it and its kinetic energy goes up and it potential energy goes down."

In general this is not true because not enough information is provided. An unspecified push on a system with unspecified starting conditions can result in both types of energy increasing, decreasing or remaing unchanged in all possible combinations.

[Stoat] "Yep, needs work."

And a little attention to the basics. If you stick with it, you will get there.

We've all made ambiguous claims like this, and eventually learned how to avoid them. It begins with realizing that they are ambiguous. And that it usually matters.

Regards,
LB


Reply author: Stoat
Replied on: 02/10/2007 11:40:32
Message:

My excuse is... I used to be Sisyphus in a past life [:D]


Reply author: Joe Keller
Replied on: 02/10/2007 22:57:36
Message:

Hobson's text on Spherical Harmonics is a good reference on multipoles. In my theory of the origin of the CMB, its temperature is essentially the escape energy of an electron, at a barrier approx. 52.6 AU distant, where the internal gravity of a proton just exceeds the sun's gravity. Small masses near this barrier cause small fluctuations in its position, hence small localized fluctuations in the CMB temperature. These small fluctuations add up to small multipoles of the temperature distribution. Above, I showed that the quadrupole due to Neptune is small enough to be consistent with observation.

Apparently the CMB dipole has moved about 0.27 degree west, parallel to the ecliptic, in 11 years between COBE & WMAP. The change in the known giant planets' positions predicts that it should have moved 0.31 deg east; so, the actual movement might be 0.58 deg west. This would imply an object at 360 AU, revolving backwards. With 0.019 solar masses, it would create a CMB dipole equal to that observed; but, the next few higher multipoles would be too large.

An object of 0.019 solar masses would be a smallish brown dwarf, with 20x Jupiter's mass but about the same radius due to "degeneracy pressure" (BR Oppenheimer et al, "Brown Dwarfs", in "Protostars & Planets IV", V Mannings et al, eds., 2000). Such a brown dwarf would experience no hydrogen fusion and only evanescent deuterium fusion. In a five billion year life, it might have cooled too much to have been observed. My estimate of the angular motion, hence distance, might be inaccurate: if the mass is less than 0.013 solar masses, then it is a planet and never experienced any nuclear fusion reactions at all.

I propose that the higher multipoles of the CMB are removed by ellipticity of orbits in the Kuiper belt, induced by the brown dwarf (or distant giant planet). The elliptical orbits would not induce dipole, because that is proportional to 1/r^2 in my theory (because it is mediated throught the effect on the barrier distance) and, so is the angular velocity along the orbit. Only higher multipoles would be induced by the elliptical orbits of the Kuiper belt objects; these might cancel the multipoles induced by the brown dwarf. The quadrupole of the brown dwarf would be roughly 53/360=15% of the dipole; all but 0.3% of this would need to be cancelled. The n=4 multipole would be about the magnitude of the observed quadrupole.

The CMB multipoles are aligned for n < 6 (K Land & J Magueijo, PhysRevLett, 12 Aug 2005, "95,071301"). This might be because for n > about 5, the multipoles due to different regions of the Kuiper belt, no longer would add constructively. Also, the multipoles induced by the brown dwarf begin to be tiny after n=4.


Reply author: Stoat
Replied on: 02/11/2007 08:52:29
Message:

I just had a look through an old book, and I found a scrap of paper where I had jotted down some stuff on the famous "Nemesis." I put down 10 thousand a.u. for the perihelion and 1.2 light years for aphelion. No idea how I arrived at these figures, as the paper is years old. I think everyone has whiled away a few hours looking for our brown dwarf but the problem comes down to the fact that it should show up. In fact, that I suspect, is why I gave it such an large elliptical orbit.

360 a.u. is very close, and I would like to hear something further on your ideas regarding the reason for its retrograde orbit.


Reply author: Joe Keller
Replied on: 02/12/2007 13:16:24
Message:

quote:
Originally posted by Stoat

I just had a look through an old book, and I found a scrap of paper where I had jotted down some stuff on the famous "Nemesis." I put down 10 thousand a.u. for the perihelion and 1.2 light years for aphelion. No idea how I arrived at these figures, as the paper is years old. I think everyone has whiled away a few hours looking for our brown dwarf but the problem comes down to the fact that it should show up. In fact, that I suspect, is why I gave it such an large elliptical orbit.

360 a.u. is very close, and I would like to hear something further on your ideas regarding the reason for its retrograde orbit.



Thanks for this important information! See my next posts!


Reply author: Joe Keller
Replied on: 02/12/2007 14:47:28
Message:

The first observations of brown dwarfs were in 1995. There is extreme observational bias in favor of relatively hot dwarfs.

The usual theory is, that for a small brown dwarf such as the above, only deuterium fusion would occur, and that, only "evanescently". If the deuterium has been "burning" at a constant rate from 4.5 billion years ago until now, and has almost run out, then the temperature would be about 4000K and the dwarf would be the third brightest object in the sky. If some or all of the deuterium burned quickly, maintaining the dwarf at 40,000K (the temperature of the hottest white dwarfs) until it stopped burning, thereafter the dwarf cooled at constant radius (with perfect convective mixing, and specific heat 0.2) as a blackbody radiator, it would be 0.07K now (if space were absolute zero temperature). So the brown dwarf might essentially be a planet by now; sunlight would maintain it at 15K.

A planet the diameter of Jupiter with the albedo of Uranus (57%), at 360 AU, would have magnitude +16.0. It's said that the limit of Tombaugh's wide ecliptic search was magnitude 17 except for a small region, presumably the galactic equator, where it was 16. Commonly, albedos are 1/10 that of Uranus, implying magnitude +18.5. So, the brown dwarf might have been slightly too dim to be found by Tombaugh. On the other hand, the (2001) automated search, by RL Allen et al, of 1.3 sq deg of the Kuiper belt, extended to mag +25. It wouldn't be too expensive to find a +18.5 object whose position (i.e., the CMB dipole) is known to 0.1 deg.


Reply author: Joe Keller
Replied on: 02/12/2007 16:30:58
Message:

Above, I say that two simple equations involving gravity and fundamental physical constants, determine the CMB temperature emanating from a surface at about 52.6 AU. The more accurate of the two equations is, that the macroscopic gravitational force must equal the maximum self-gravitational force within a proton, contracted (Heisenberg uncertainty principle) to a Gaussian radius of hbar/(2*Mproton*c); this gives a barrier at 52.63 AU. The less accurate equation, involving the gravitational constant G and the CMB temperature T, is, that the mean(Vx^1.5), where Vx is one velocity component, e.g., along a magnetic field line, for an electron, equals Vescape^1.5; this gives 52.60 AU. This 52.63 - 52.60 difference, lies within the uncertainty of G and/or T.

Thereby mainly the sun's gravitational field determines the CMB temperature, which becomes a sensitive detector for gravitating objects in the outer solar system. The CMB dipole is caused by the sun's retrograde, small cool brown dwarf, 0.019 solar mass, mag +18 companion at 360 AU distance in the positive dipole direction.

The dwarf's gravity elongates the orbits of Kuiper belt or Oort cloud objects, which does not affect the CMB dipole (due to cancellation of factors) but which does, observationally, neutralize all but 1/50, of the quadrupole due to the dwarf (which otherwise would equal 1/6 of the dipole). The total mass of highly eccentric comets required, near their perihelion (opposite the dwarf) somewhat outside the 53 AU barrier, at any given time, would be 0.019 * 333,000 * (52.6/360)^3 = 20 Earth masses. Theoretical models estimate the mass of inner + outer Oort cloud comets as considerably greater than 1 Earth mass, but this estimate includes only comets exceeding 2.3 km diam (L Dones et al, in MC Festou et al, "Comets II", 2004, p. 170 "Summary"). For 142 comets with semimajor axes greater than 500 AU, the dominant high density of aphelia is near ecliptic longitude 210, ecliptic latitude +15 (JA Fernandez, "Comets", 2005, Fig. 6.9, p. 154).

The Kuiper belt mass observed by Allen et al was 0.37 Earth masses if extrapolated to all sky within 30 degrees of the ecliptic, assuming a body density of 2 grams/ml. If N=1/M, then because Allen et al observed objects 50km-500km diam, the contributions of other decades of size down to 0.15 micron diam (below which solar radiation pressure often would sweep the objects away), would multiply this mass by 12.5, to 4.6 Earth masses. If this mass were just inside the 53 AU barrier when in conjunction with the dwarf, and far inside the barrier when in opposition to it, then it would be as effective as 4.6 Earth masses of the above-mentioned comets.

Indeed the distribution of (almost 100 known) Trans-Neptunian Objects (a.k.a. Kuiper belt objects) as of 1998, is shifted about 5.9 A.U. toward 182deg ecliptic longitude (A Fitzsimmons, 1998-2000, "Surveys of the Distant Solar System", Fig. 2 in: Fitzsimmons et al, eds., "Minor Bodies in the Outer Solar System", 2000, p. 91). (I divided Fitzsimmons' chart into 20 sectors and based my calculation on the number and median radius in each sector; replacing median with mean in one of the twenty, whose median was importantly different from the mean.) So, the mass of Kuiper belt objects, supplemented by comets, is plausibly high enough to neutralize the CMB quadrupole. Both types of objects show a preference for perihelia that do so: perihelia near the 53 AU barrier opposite the brown dwarf and from above, for the comets; aphelia near the barrier adjacent to the dwarf and from below, for the TNO's.


Reply author: Joe Keller
Replied on: 02/13/2007 00:55:18
Message:

My brown dwarf (or planet) is Lowell's Planet X.

Based on discrepancies in the orbit of Uranus from 1715 to 1914, P Lowell predicted a "Planet X" (as in x, the variable) of 0.00002 solar mass, 44 AU from the sun (WG Hoyt, "Planets X & Pluto", 1980, pp. 136-140). Lowell's expected eccentricity, 0.20, renders the tidal force of such a planet roughly 34% as effective as if it simply were stationary. Multiplying by the ratio of the cubes of the closest distances to Uranus (360-19.2 vs. 43.85-19.2), then gives 0.018 solar mass for my planet, vs. my predicted 0.019.

Lowell predicted the perihelion of Planet X to be 201.7 deg ecliptic longitude; or, 180 deg therefrom (dual solution). Lowell relied heavily on Flamsteed's 1715 observations of Uranus and on Lowell's own, then-recent, observations. He also used many observations in between. My planet would move 0.58/11*187 = 9.9 degree retrograde in the 187 yrs from the midpoint of Lowell's data, 1815, to 2002. So, if the CMB dipole is toward Planet X, Lowell's work would predict the CMB dipole at 201.7-9.9=191.8 ecliptic longitude, vs. 171.8 observed.


Reply author: Stoat
Replied on: 02/13/2007 05:25:47
Message:

If I remember rightly, the hunt for Nemesis was skewed, because people were looking for a brown dwarf object, that had a large eccentricity, to account for a supposed 30 million year cometary extinction cyle on earth. There was also the point you made about the brown dwarf being rather bright, so have it way out at aphelion to cover why it couldn't be found.

Now, there wasn't just a proposed planet X but a whole, rather embarrassing, slew of them put forward. That might make working astronomers a little slow to give you some telescope time, and hunt down old plates. (Have all those old plates been scanned into a data base I wonder?) I found this http://www.tls-tautenburg.de/scanner.html Might have something.

You do realise, that you are going to have to call it the planet Stoat [:)][:D][8D]


Reply author: shando
Replied on: 02/13/2007 07:56:50
Message:

quote:
Originally posted by Joe Keller

My brown dwarf (or planet) is Lowell's Planet X.




Joe, what would be the period of your dwarf?
How many of our years would it take to circle the sun?


Reply author: Joe Keller
Replied on: 02/13/2007 19:03:32
Message:

quote:
Originally posted by Stoat

If I remember rightly, the hunt for Nemesis was skewed, because people were looking for a brown dwarf object, that had a large eccentricity, to account for a supposed 30 million year cometary extinction cyle on earth. There was also the point you made about the brown dwarf being rather bright, so have it way out at aphelion to cover why it couldn't be found.

Now, there wasn't just a proposed planet X but a whole, rather embarrassing, slew of them put forward. That might make working astronomers a little slow to give you some telescope time, and hunt down old plates. (Have all those old plates been scanned into a data base I wonder?) I found this http://www.tls-tautenburg.de/scanner.html Might have something.

You do realise, that you are going to have to call it the planet Stoat [:)][:D][8D]



Thanks for your input, especially the link - what a find! Let's try it! If you find Planet X first, then you're the discoverer! The given error bars of the CMB dipole coordinates might be too large. Also, the planet might be at exactly the positive (or maybe negative, if I'm wrong) CMB dipole. Coordinates to follow.


Reply author: Joe Keller
Replied on: 02/13/2007 20:07:23
Message:

Three likely theories give coordinates for "Planet X":

I. It is at the (+/-) CMB dipole.

II. Same as (I) but with adjustment for the theoretical gravitational effect of Jupiter, Saturn, Uranus, & mainly Neptune.

III. Same as (II) but with additional adjustment for various other "Planets X".

(I) is least likely. (III) is least useful.

The error bars I'll give are mainly due to the error bars of the 3-yr. WMAP coordinates, because the extrapolation is so small. These are:

0.07 degree of great circle arc either way along the direction of galactic longitude (i.e., 0.1 degree of galactic longitude)
0.04 degree (either way) of galactic latitude

That's an "error ellipse" whose axis will be tilted, in any other coordinate system. It's an acceptable approximation simply to say that the error is 0.07 degree (0.28m or 17s) of Right Ascension and 0.04 degree (2.4') of Declination.


I. Positive CMB dipole, extrapolated coordinates for March 1, 2007:

l = 236.70 b = +48.26

In celestial coordinates:

RA 11h 11m 02s Decl -6deg 48' 45"
(on March 1, 2007)(J2000.0 coordinates)

The monthly change is:

RA -0.38 s/month Decl +4.9"/month


This is ideal observing position: south of the tail of Leo, high in the eastern sky before midnight, in a region with rather few stars.

Ecliptic coordinates (2007.25):

ecliptic longitude 171deg26'41"
ecliptic latitude -11deg6'06"

Per degree of inclination, only half as many asteroids have inclinations of 11 deg as of 0 deg. Furthermore a circular orbit inclined 11 degrees averages only 7 degrees ecliptic latitude. This helps reduce the number of asteroids in view.


II. The adjustment for the four gas giant planets, is +0.226deg ecliptic long. & -0.034deg ecliptic lat.; this gives:

ecliptic longitude (2007.25) 171deg40'15"
ecliptic latitude (2007.25) -11deg08'08"

RA 11h12m08s Decl -6deg55'53"

The monthly change per this theory is:

RA -0.83 s/month Decl +10.7"/month


III. The additional adjustment for an Earth-mass planet at 62 AU, ecliptic long 261, ecliptic lat +63.5 (I chose these coordinates for convenience of estimation, but they correspond very roughly to the "Planet X" inferred above from Davies' Madiera CMB observations) is:

ecliptic longitude (2007.25) 171deg35'36"
ecliptic latitude (2007.25) -11deg21'46"

RA 11h11m23s Decl -7deg10'20"

The adjustment for JL Brady's Jupiter-mass planet (if it were to exist) would be about a radian.


The range of possibilities I-III, plus just over two error bars each way, give a circular search region of radius 17 arcminutes (area 0.25 sq degree) and center:

RA 11h11m35s Decl -6deg59'33"


Reply author: Joe Keller
Replied on: 02/13/2007 20:39:00
Message:

quote:
Originally posted by shando

quote:
Originally posted by Joe Keller

My brown dwarf (or planet) is Lowell's Planet X.




Joe, what would be the period of your dwarf?
How many of our years would it take to circle the sun?





These are just the right questions! See post below.


Reply author: Joe Keller
Replied on: 02/13/2007 20:45:18
Message:

By the methods above, I estimate 355 AU distance and a 6680 yr period for circular orbit. However, today I discovered that five of the nine proportionally small discrepancies ( < 10% ) in the resonances (m:n for m,n < 11) of the giant planets' orbits (periods per TP Snow, "The Dynamic Universe", 1983) are simple multiples of a discrepancy consistent with a retrograde influence with period about 4430 yr. This would (assuming that the 2:3 ratio is mere chance) correspond to 270 AU distance, 0.0107 solar masses, and a planet with magnitude +16.75, for 5.7% albedo.

This might be a more accurate estimate than that derived from the COBE and WMAP surveys with their nominally large error bars. Alternatively, the orbit might be moderately elliptical ( e > sqr(6680/4430) - 1 = 0.23 ). The other four of the nine discrepancies were related to multiples of Pluto's orbital period; mostly, these were less accurate resonances than the five above.


Reply author: Stoat
Replied on: 02/14/2007 04:53:51
Message:

[:(] It doesn't look like they have a plate of that region of the sky. Perhaps that's good news, as if they haven't, then they haven't looked [:)] Maybe the Australian Schmidt would be worth a look, it's been running for about thirty years.


Reply author: Joe Keller
Replied on: 02/14/2007 16:12:08
Message:

quote:
Originally posted by Stoat

[:(] It doesn't look like they have a plate of that region of the sky. Perhaps that's good news, as if they haven't, then they haven't looked [:)] Maybe the Australian Schmidt would be worth a look, it's been running for about thirty years.



Above, I've specified a search disk including 0.25 square degrees of sky. Please note that I changed the region slightly to correct a sign error in the increment between (II) & (I).


Reply author: Joe Keller
Replied on: 02/15/2007 00:56:53
Message:

Planet X will be dimmer than most asteroids but brighter than most Kuiper belt objects. It will move more slowly across the star background, and also retrograde. The average asteroid orbital plane is inclined one degree below the ecliptic near the vernal equinox, so ecliptic longitude -11 in Leo is 10 degrees from the line where asteroids can be seen most often. An extensive plot showed that asteroid inclinations of 10 deg are about 1/2 as common as 0 deg. If the inclinations make a bell curve with half-maximum at 10 degrees, numerical integration (the singularity in the integrand is "removable") shows that asteroids will be seen only 1/10 as often there.

Besides direct observation and studying old photographic plates, I've invented another method of search. The billion-star USNO-B catalog contains bogus objects which arise from two asteroids being accidentally at the same place on plates made at different times (DG Monet et al, Astronomical Journal 125:984+, p. 990). In part because of the unreliability of the magnitudes, the catalog was not filtered for discordant magnitudes. My method is to look for USNO-B catalog objects along the presumed track of Planet X, which have:

1. A difference of 2 or more, between the whole number parts of R1&R2, or B1&B2. E.g., B1=16.98 & B2=18.03.

2. At least one of their published magnitudes ("B1", "B2", "R1", or "R2") between +16.00 and +18.99.

3. Proper motion in either RA or Decl greater than 2 standard deviations from the mean.

In the "VizieR" online catalog, automatic filters can be set to accomplish (2) & (3). Because the typical midrange age of the plates used for a star is 30 years, one should go back 60 years, i.e., 3.2deg.

Three square degrees out of 41,000 in the sky, would have 10^9/14,000 = 70,000 stars, but the automatic filters could remove 90-99% of those depending on the setting. This is a manageable number of stars to check by eye for criterion (1). Asteroids and Kuiper belt objects occasionally would appear in this region of sky, but Planet X might be there all the time. A trail of bogus or dubious objects might be plotted.


Reply author: Stoat
Replied on: 02/15/2007 02:58:37
Message:

I was going to set two plates up on top of each other in photoshop, and blink one of them off and on, to check for movement.


Reply author: cosmicsurfer
Replied on: 02/15/2007 05:11:13
Message:

If 10,800 B.C. is the date for a major holocaust (Younger Dryas-major upliftment/volcanism,possible polar shift) on earth due to the possible gravitational interference from the returning retrograde extreme elliptical orbit of a brown dwarf sister sun; then, a 3600 year period fits perfectly with a year zero (was star of David our sister sun?) and we would not see the so called planet X for another 1593 years.

John


Reply author: MarkVitrone
Replied on: 02/15/2007 05:53:34
Message:

I wanted to take a brief moment to thank Joe Keller for his use of support in his posts. I like it when a thought is backed by evidence. It is great to see the scientific method unfold before your eyes. Thanks,

Mark Vitrone
Moderator


Reply author: Joe Keller
Replied on: 02/15/2007 12:01:14
Message:

quote:
Originally posted by Stoat

I was going to set two plates up on top of each other in photoshop, and blink one of them off and on, to check for movement.



That sounds very good, if one has the computer skills to do it. Inaccuracy of the images would cause all the stars to move, so one would be looking for something that moved much more than such background jiggle. Also, it would be good to double check that the image is the actual plate and not modified or reconstructed.


Reply author: Joe Keller
Replied on: 02/15/2007 12:06:57
Message:

quote:
Originally posted by cosmicsurfer

If 10,800 B.C. is the date for a major holocaust (Younger Dryas-major upliftment/volcanism,possible polar shift) on earth due to the possible gravitational interference from the returning retrograde extreme elliptical orbit of a brown dwarf sister sun; then, a 3600 year period fits perfectly with a year zero (was star of David our sister sun?) and we would not see the so called planet X for another 1593 years.

John



Thanks for the interesting historical perspective. An ophthalmology journal once reported that because the fibers in the human lens are wound (like the mantle of a baseball) in three directions (for the human lens), a six-pointed star is the most accurate depiction of the scatter that an otherwise-perfect human eye observes.


Reply author: Joe Keller
Replied on: 02/15/2007 14:43:11
Message:

To go back 60 years, I've specified seven overlapping disks to search. The first disk is centered at the above estimated present position, but I increased the radius from the above 17', to 21' for good measure. The disks grow to allow roughly for up to 5 degree orbital inclination.

Format:
center RA, center Decl, radius in arcminutes

The disks:
11h11m35s, -7deg00', 21'
11h12m59s, -7deg09', 23'
11h14m31s, -7deg19', 25'
11h16m11s, -7deg30', 28'
11h18m03s, -7deg42', 31'
11h20m07s, -7deg55', 34'
11h22m23s, -8deg09', 37'


Reply author: Stoat
Replied on: 02/15/2007 15:37:38
Message:

Here's a robotic on line telescope http://www.telescope.org/index.php


Reply author: Joe Keller
Replied on: 02/16/2007 00:14:52
Message:

quote:
Originally posted by Stoat

Here's a robotic on line telescope http://www.telescope.org/index.php



Thanks again for the help!


Reply author: Joe Keller
Replied on: 02/16/2007 00:57:56
Message:

This is a small step for man but by the grace of the Gods it is a giant leap for mankind.

The billion-star 2003 United States Naval Observatory B1.0 Catalog suggests that "Planet X" is a double planet with a smaller companion half its radius. The following objects seem to be Planet X and its companion:

Object #1. USNO-B 0827-0286487
Object #2. USNO-B 0824-0279170
Object #3. USNO-B 0820-0274026
Object #4. USNO-B 0813-0233607

A "VizierR" search found that no objects at these positions occur in any other catalogs except NOMAD, wherein essentially the same information is given though slightly less detailed. According to the documentation accompanying the USNO-B catalog, one of the B magnitudes, +24, is probably invalid due to its excessive faintness. Consideration of the other magnitudes shows that there are two bodies accounting for these four observations.

One body had four red magnitudes ranging from 17.41 to 18.57, three blue magnitudes ranging from 18.26 to 19.80, and two infrared magnitudes of 18.36 & 18.78. The aberrantly high blue & small red magnitudes complemented each other energetically consistent with a change in passive reflectance color (weather on the planet).

The other body had four red magnitudes ranging from 20.26 to 20.68, and a blue magnitude of 20.30. At this high galactic latitude most stars with mag > +18 would be red dwarfs. The near equality of red and blue magnitudes for the presumed smaller body, and the near equality (except for the abovementioned aberrant pair of values) of red, blue & infrared magnitudes for the larger body, imply that they are illuminated by the light of our sun.

The pseudo-trajectory was consistent with 27.5 degrees orbital inclination from the ecliptic, too high for most asteroids. Furthermore with only about 10,000 asteroids of this magnitude, asteroidal coincidences on the plates would be far too infrequent. The curvature of the sky path probably is of the correct sign and within a factor of two of what is required.


Reply author: Stoat
Replied on: 02/16/2007 05:51:38
Message:

Hi Joe, I've been having a read of the robotic telescope. It seems that it takes a couple of days to get one's image back,depending on the seeing. As it's robotic an image can be disappointing if the moon is bright.

Take a look at a the site and see whether it might be an idea to register as a school teacher. That way you can enter all the jobs and make sure that the exposure time is the same for all of them. Then put them all back in a couple of weeks later to check for any movements.

Anyone on this board can then be part of your class. I did put a job up but if you like I can delete it. It was for 11 11' 23" -7 10' 20"


Reply author: Joe Keller
Replied on: 02/16/2007 10:08:52
Message:

quote:
Originally posted by Stoat

Hi Joe, I've been having a read of the robotic telescope. It seems that it takes a couple of days to get one's image back,depending on the seeing. As it's robotic an image can be disappointing if the moon is bright.

Take a look at a the site and see whether it might be an idea to register as a school teacher. That way you can enter all the jobs and make sure that the exposure time is the same for all of them. Then put them all back in a couple of weeks later to check for any movements.

Anyone on this board can then be part of your class. I did put a job up but if you like I can delete it. It was for 11 11' 23" -7 10' 20"



Thanks, Stoat. From the result of my USNO-B catalog search, I have some narrower coordinates. See below.

Attention: anyone else on this board! Please look! I'm busy with these calculations and I don't mind a bit if someone else sees it first!


Reply author: Joe Keller
Replied on: 02/16/2007 12:03:43
Message:

The second and third of the four USNO-B candidates above, seem to have accurate epochs, because their great circle angular separation, divided by the difference in epoch (1976.1 vs. 1965.1) implies a period, for retrograde circular orbit, of 4535 yr, in close agreement with the 4430 yr derived above from the outer planet resonances. I used these epochs to extrapolate an accurate location for 2007.25 (i.e., 07:30 UT, April 2, 2007):

RA 11h 06m 0.50s = 166.5021deg

Decl -6deg 28' 20.5" = -6.47236deg

daily correction from April 2, 2007:

RA -0.0468 sec/day

Decl +0.354"/day

The epochs are given to 0.1 yr, implying a possible track of locations, equivalent to up to a 103-day correction (-4.84s RA and 36.6" Decl) either way.

The fourth USNO-B object also seems confidently from its magnitudes, to be Planet X. Using the 3rd & 4th objects (instead of the 2nd & 3rd objects), the slope d(Decl)/d(RA) is 1.35% less steep. Such a true slope would result in a correction of -54.2" Decl.

Based on these USNO-B objects, then, the narrowed search region is a parallelogram with area 2.2 sq arcminute (0.0006 sq deg) and corners:

RA 11h 06m 5.34s
Decl -6deg 28' 57.1"

RA 11h 05m 55.66s
Decl -6deg 27' 43.9"

RA 11h 06m 5.34s
Decl -6deg 29' 51.3"

RA 11h 05m 55.66s
Decl -6deg 28' 38.1"


Reply author: cosmicsurfer
Replied on: 02/16/2007 13:58:22
Message:

Hi Joe,

Would your extrapolations for time period of a circular orbit of 4535 yr change with an elliptical orbit? Orbital speeds would accelerate towards and away from the sun, plus Planet X may be surrounded by planetoids, debris fields and trailing asteroids and comets (former oceans) from previous collisions, e.g., destruction of Fifth Planet forming asteroid belt. Thus the former retrograde circular orbit of a former binary star is now in a extreme elliptical orbit from being thrown out of our solar system from such a collision or gravitational interference near miss causing EPH.

An elliptical orbit would result in a constantly shifting tidal effect from Planet X that would be felt by all planets in our solar system during its flight around sun.

Regarding historical perspective, the myths of a world wide deluge and even biblical revelations depictions of stars falling from heaven and skies turning red might account for a six sided refraction or “Star of David” during close encounters. Also, we could search for periods of higher intensities of meteorite bombardments as a possible indication of passage of Planet X. I did manage to find one large impact around the 10800 B.C. period near the Andes leaving a large crater.

Most star systems are binary and it may be that planetoids are the result of solar plasmas separating into two halves causing the initial spreading of mass and following orbital planetoids around the larger gravitational body (all in concert with overall structure and motion of Universe).

John


Reply author: Joe Keller
Replied on: 02/16/2007 15:07:14
Message:

quote:
Originally posted by cosmicsurfer

Hi Joe,

Would your extrapolations for time period of a circular orbit of 4535 yr change with an elliptical orbit? Orbital speeds would accelerate towards and away from the sun, plus Planet X may be surrounded by planetoids, debris fields and trailing asteroids and comets (former oceans) from previous collisions, e.g., destruction of Fifth Planet forming asteroid belt. Thus the former retrograde circular orbit of a former binary star is now in a extreme elliptical orbit from being thrown out of our solar system from such a collision or gravitational interference near miss causing EPH.

An elliptical orbit would result in a constantly shifting tidal effect from Planet X that would be felt by all planets in our solar system during its flight around sun.

Regarding historical perspective, the myths of a world wide deluge and even biblical revelations depictions of stars falling from heaven and skies turning red might account for a six sided refraction or “Star of David” during close encounters. Also, we could search for periods of higher intensities of meteorite bombardments as a possible indication of passage of Planet X. I did manage to find one large impact around the 10800 B.C. period near the Andes leaving a large crater.

Most star systems are binary and it may be that planetoids are the result of solar plasmas separating into two halves causing the initial spreading of mass and following orbital planetoids around the larger gravitational body (all in concert with overall structure and motion of Universe).

John




Thanks for passing along this wealth of interesting ideas. I think I can address some of them.


Reply author: Joe Keller
Replied on: 02/16/2007 16:15:48
Message:

If this distant orbit is elliptical, its projection on the celestial sphere is nonetheless a great circle, so the extrapolated sky trajectory is the same except for rate of travel. The extrapolation is so small (2.4 degrees) that for an orbital eccentricity of 0.2, the rate of travel would vary at most 0.8% (causing, at most, an error the size of the rms error from rounding the epoch dates). Furthermore, the orbital period, as implied by the orbital resonances of J, S, U & N, is consistent with the present angular speed (1965.1-1976.1), if the orbit is nearly circular.

The dimmer of the two red magnitudes, for each of the four collinear objects I found in the USNO-B catalog, was +20.61, 20.26, 20.68, and 20.68. I plotted all 61 dim red magnitudes obtained when I searched the USNO-B catalog for all objects in the above seven disks, which had proper motions >80mas/yr in both directions (an indicator of bogus combination) and which also had one red magnitude < +18.99 and one > +19.50. Fifty-nine of the dim magnitudes were roughly evenly distributed between the 19.50 cutoff, and about 20.90. The occurrence of the eight pairs of precisely equal magnitudes conformed everywhere to the Poisson distribution. Two of the magnitudes were > 24 which, according to the documentation accompanying the catalog, usually denotes an invalid reading. One of these "invalid readings" occurred for one of the four collinear stars (p=0.125). One of the equal-magnitude pairs also occurred among them (p=0.09).

Furthermore the chance that four points from a uniform distribution of length 1.40, will span a range of 0.43 or less, is small. The chance that any three of the four will span a range of 0.08 or less, also is small. For four numbers chosen randomly in the unit interval, the mean sum of differences squared is one (by a four-dimensional integral). Scaling the observed magnitudes, above, to a unit interval, gives a sum of 0.245. The only a priori distinction of this group of four, from the other 57 high proper motion, discrepant red magnitude stars, in the searched region, is that they formed the straightest line of any four stars that I could find by eye on my handmade plot on Keuffel&Esser graph paper.


Reply author: Joe Keller
Replied on: 02/16/2007 22:18:19
Message:

I found another object resembling USNO-B catalog objects #2-#4 above:

Object #5. USNO-B 0830-0272239

I found it 4.9" from the extrapolated great circle, searching with overlapping disks, of sometimes 1' but usually 2' radius, so that the band halfwidth always was more than 50" and usually was almost 120", along the chord of the previously unsearched extrapolated (to the present time) portion of the great circle made by objects #2 & #3. This time I searched for either Red1 or Red2 magnitude between 20.54 & 20.75. Roughly 30 objects were found for which the other red magnitude was 19.xx, and comparably many such objects also were found searching randomly a fraction of a degree away.

Only two objects along this arc of the great circle, had the other red magnitude < 18.99. Of these, Object #5, had not only an R2 magnitude of +18.59, but also a large proper motion of the order of that given for Objects #1-#4. Its B2 magnitude was +23, hence "invalid", further accentuating that characteristic of these objects.

The epoch was given as 1974.5. Only one pair (#2 & #3) among the five objects has epochs consistent with other orbital period estimates. Maybe the epoch given, usually is determined by the midrange, not of plates showing the object, but of all plates covering that region of sky. The following post will argue that by considering brightness, even the #2 & #3 epochs must be rejected. It's likelier that the object is, as initially estimated, moving with the angular speed of a circular orbit of 355 AU radius, but has an elliptical orbit whose period is that of a circular orbit of radius 270 AU.



Reply author: Joe Keller
Replied on: 02/17/2007 17:56:37
Message:

I discovered and realized the existence of at least one moon, shortly after midnight UT, Friday, Feb. 16, 2007. The distribution of the "moon" magnitudes (not including the fifth "moon" found by searching for such magnitude) is 20.61, 20.26, 20.68, 20.68. This suggests that there is a large moon causing the 20.26 magnitude while a smaller moon causes the others. The closeness of the other three numbers makes an outlying measurement error, or varying albedo, unlikely. Eclipse would be too brief or too rare. My IBM 486 Monte Carlo trial indicated that the chance of any three such close points (sum of squared differences) in a set of four, from a uniform interval from 19.50 to 20.90, is about p=0.02; the closeness of the fourth point (one "Jupiter" & three "Saturn" sightings, but with "Saturn" closer to the "sun" than "Jupiter") reduces this to p=0.01.

The 20.30 Blue magnitude and 20.66 average Red magnitude of the smaller moon give B - R = -0.36. The USNO-B1.0 catalog gives B - R = -0.365 for Capella (a double star, G6 & G2, vs. the sun's G2). For reflection (generally weak in far blue) to achieve this B-R value using sunlight, it would likely appear bluish to the eye, which is sensitive mainly to near blue (maximum human retinal cone sensitivities are red 0.59 micron, green 0.55 micron, & blue 0.45 micron, according to the CRC Handbook of Chemistry & Physics, c. 1960).

The mass-radius relation for a giant planet or brown dwarf is very flat; due to electron degeneracy at high pressure, 0.019 solar mass would give about 10% less area than Jupiter (Burrows et al, Reviews of Modern Physics, 65:301+, 1993; graph reproduced in the above-cited book, "Protostars..."). An albedo of 5.7% (1/10 of Uranus, but close to average for a common kind of Kuiper belt object or asteroid) gives +18.1 magnitude at 360 AU.

I have five red, two blue and two infrared (0.75-1 micron band) magnitudes for the presumed larger body:

Object 1: R 17.41, B 19.80, I 18.78
2: R 18.17, B 18.95
3: R 18.57
4: R 18.03, I 18.36
5: R 18.59

These nine, average +18.07. The absence of any Red magnitudes between 18.60 & 18.99 is significant at roughly p=0.03, based on the distribution of fellow R values found in the above search of all objects whose smaller R value was near 20.6. The distribution of the magnitudes is most consistent with a large red spot covering almost one hemisphere of the large body, and a rotation period of between a day and a decade.

The spectrum, though red, is unusually flat. This suggests reflection instead of Planck curve emission. I searched USNO-B1.0 within a 20 degree radius centered at RA 11h10m Decl -7, for objects with both RA & Decl proper motions > 80mas (as before), R1 within 0.25 mag of the mean R for the presumed large body, and R2 within 0.25 mag of the mean R for the presumed "first moon" (using only Objects #1-4). Only 150 objects were brighter in both I & B; 1415 were dimmer in both I & B.

This flatness is not due to the averaging of four objects: a calculation based on galactic disk thickness at this galactic latitude, and textbook published main-sequence spectral type counts near the sun, implies that half the stars of the larger body's apparent magnitude, would be spectral class K5-K9, half class M, and all main-sequence. My IBM 486 Monte Carlo trial showed that a group of four equally bright stars with temperatures randomly differing over a +/- 10% range, perceived as one object, would have B-R & I-R, averaged over many groups of four, differing < 0.02 magnitude units from a single object at the population mean temperature. (This seems to grow quadratically with the range.) Furthermore, one side of the curve was raised and the other dropped almost the same amount, so what small effect there was, would mostly cancel, in the statistical test in the preceding paragraph.

In the USNO-B catalog, I checked the photometric values of Procyon (F5 IV-V) and Capella (I called it G6; it's G6 III + G2 III). From these, I interpolated by steps in spectral type, to obtain for the sun (G2 V), B - R = -0.46, I - R = +0.20, B - I = -0.66. For the presumed large body, these figures are +1.34, +0.53, +0.81, resp. This suggests sunlight reflected by a reddish object.

Galactic dimensions and main-sequence spectral type distribution, imply that almost all stars in the magnitude range, were Type K5 through late M (about half late K, & half M). For a mid-TypeK to late-TypeM star, B-R is large positive, but also I-R is large negative. Aldebaran (K5 III, somewhat variable) has B-R=+2.13, I-R=-0.85, B-I=+2.97. Gacrux (Gamma Crucis, in the Southern Cross)(Type M3.5 III, & variable like most Type M stars) has B-R=+2.16, I-R=-0.85, B-I=+3.00. If a dimmer second red magnitude for the object, implies that both B-R and I-R should be corrected downward, then I-R is less discrepant but B-R is even more discrepant.

The theoretical predicted surface temperature for this brown dwarf is 430K (interpolation or slight extrapolation, for 0.02 solar masses & 5 billion yr, in Table I, p. 316, Burrows, op. cit.; there was a log-log relation to age and a linear relation to mass). This might make it look more like Jupiter or Venus than like Pluto: its albedo would be too high. Maybe it cooled faster or burned less deuterium than theorized. Maybe it is less massive or accreted gradually.

Theoretically, "late M" dwarfs progressively show a blueward shift in their color temperature (Burrows, op. cit., Fig. 5, p. 309; quoting Allard, 1991). That is, according to Kirchhoff's law (Condon et al, Handbook of Physics, 2nd ed., p. 5-37, 1st partial par.) they emit & absorb blue better, and reflect red better. This trend might continue into brown dwarfs and giant planets. Also, common kinds of Kuiper belt bodies are reddish.

At 0.08 solar masses, one has a red dwarf (Burrows, op. cit.); not only does it begin to emit its own light, but the radius considerably increases. Even before that, one has a "hot" brown dwarf with self-produced IR magnitude. So, the distance can't be much more than 360 AU or the mass needed would be too great. A 440 AU distance likewise makes it too bright, even with the lowest likely albedo.

According to the 2003 AJ article giving information on the USNO-B catalog, it used almost 8000 plates; a plate sometimes included 5 million stars. For a billion stars, this implies that each small region of sky is included in < 40 plates.

Let a moon orbit Planet X; the moon's orbit might lie near the plane of Planet X's orbit around the sun. In a 6600 yr circular orbit, Planet X moves 196"/yr. That's 0.34 AU, at Planet X's orbital radius. Because Planet X has roughly the geometric mean of the sun's and Jupiter's masses, its main satellites might orbit at roughly the geometric mean of the distances, of the main satellites of the sun and Jupiter. A quarter of the time, a moon in a circular orbit, will be within 22.5 deg of its greatest elongation. For a 0.34 AU orbit, that is within 15". A 5deg orbital inclination for the moon, would cause typically only another 12" displacement perpendicular to the first. A somewhat larger orbit would be somewhat less efficient at placing the moon near 0.34 AU before or behind Planet X.

Generally, objects on the plates might be combined, if no more than 30" apart. Maybe half the time, the Red1 & Red2 plates were taken one year apart. Then, a quarter of the time, Planet X's moon would be positioned to impersonate Planet X on the plates. So, such a mysterious object would appear, in one of the 40 plates, 40/2/4 = 5 times. I found five objects in a thorough search of the USNO-B catalog.


Reply author: Joe Keller
Replied on: 02/19/2007 12:11:41
Message:

This new giant planet (or small star as the case may be) shall be named Barbarossa. Its largest moon shall be named Frey and its next-largest moon, Freya. The estimated diameter of Barbarossa is 83,000 miles; of Freya (sighted four times among the above five USNO-B catalog objects) and Frey (sighted once), 18,000 miles and 23,000 miles, respectively. The radius of Freya's orbit is about 1/2 AU and of Frey's orbit, 1 AU. Barbarossa rotates with a period of many days; it is mostly red on one side, dark gray on the other, with a visual albedo of about 6% on both sides. Frey and Freya resemble Neptune: blue, with albedos of 30%.


Reply author: Joe Keller
Replied on: 02/19/2007 18:42:35
Message:

If the proper motion were computed without regard to the dates of the plates on which an object actually appears, but rather the dates of all plates covering that region of sky, it still could be accurate for real stars. The planet and moon, appearing in one position on only one pair of plates a year apart, will have their proper motion underestimated by a factor of roughly (2000-1940)/2=30; that is, the typical 15"/3=5" separation will be reported as 5000/30=170mas/yr proper motion, typically 120 pmRA and 120 pmDec.

Suppose Barbarossa's moons, like Jupiter's moons, orbit in Barbarossa's equatorial plane. Let this plane be tilted 5 degrees from Barbarossa's orbital plane. Typically then we would see the moons describe ellipses tilted 3 degrees, implying 7" vertical error; the horizontal error was estimated at 5" in the previous paragraph. In absolute value, the mean putative PM in Decl then would be 1.4x the mean putative PM in RA; for the five objects, the observed ratio is 1.5.

It's likelier that the moon's orbit is slightly larger than optimal for mutual impersonation, than slightly smaller. So, usually the putative proper motion will be rearward, i.e., positive in RA (because there is reason to believe Barbarossa's orbit is retrograde, like the motion of the CMB dipole). Barbarossa's orbit line is inclined -26 degrees to the RA & Decl axes. The putative PMs would tend to be 5 units down and right along the orbit line, then perpendicularly 7 units up & right (if Barbarossa's equator is so inclined); indeed 4 of the 5 PM's (128,426; 274,394; 308,234; 302,142; but not -120,-502) lie in the first quadrant clustered about this point. The aberrant PM (Object #5) could be accidental; the moon's orbit isn't edge-on. Likewise a second moon accidentally could cause the PM of Object #2.

The earliest epoch of any object is given as 1965. This implies that the plates used for this region span about (2003-1965) x 2 = 76 yrs. The distance between Objects #5 & #4, corresponds theoretically to 69 years of orbit.

When I searched for Object #5, I extrapolated (westward only) one degree from Object #1 (using the line through Objects #2 & #3, because of Object #1's suspiciously different magnitudes). I narrowed the dim red search window 7-fold. There had been 61 candidate objects found in 4 sq degrees, so I expected only 61/4/7 = 2 per sq deg. Yet I found Object #5 within 5" of the predicted orbital great circle (p=0.006).

The RAs of the objects show periodicity analogous to the Saros eclipse cycle. Objects #5, #1 and #3 (all thought to represent the moon Freya) are on even minutes of RA. Object #2 (thought to represent the moon Frey) is near an odd minute of RA. Object #4 (representing Freya, with exactly the same magnitude as one of the other Freya sightings) is on a half minute of RA.

It's mere chance (1 in 3) that Frey's sighting occurred ten seconds away from a minute of RA (as measured from the Freya sightings to either side). The other four Object sightings (Freya) occur at intervals which are approximately multiples (1x, 3x, & 3x) of 115 seconds of RA; this corresponds to 9.8 years, considering the 26 degree slope of Barbarossa's orbital line. Freya displays "beats", because plates of the constellation Leo tend to be made at the same season, if not the same date.

If Barbarossa has period 4400 yr but present angular speed 2/3 of average, that's approximately 270*sqrt(1.5) = 330 AU. To give the gravitational effect I theorize on the CMB dipole, Barbarossa would be 0.016 solar masses. If Freya were in a circular orbit of 0.31 AU radius (just big enough to impersonate Barbarossa, next year or last year) Freya's year would be 1.36 Earth years. Really, Freya's orbit probably is slightly bigger. The period depends strongly (i.e., ^1.5) on radius. Freya might travel 0.5 *(1 + 1/9.8) orbits in one Earth yr., an orbit of 1.82 yrs., which, if circular, would have 0.37 AU radius, but which could have a radius only slightly more than 0.31 AU at greatest elongation on one side or the other, if moderately elliptical. Approximately every 9.8 yr, Freya would be roughly synchronized with Earth so that Freya is near maximum elongation when Leo is high in the night sky.


Reply author: Stoat
Replied on: 02/20/2007 04:47:27
Message:

And I thought i was being a bit controversial with the "planet stoat." [:)] Barbarossa, was a crusader, so that might raise a frown from Islam. Barbarossa was the code name of the Nazi invasion of Russsia, so I think you might be stepping on a few toes there as well.

What happens is that you would get the credit and the first suggestion as to what to call the thing but it will be decided by an international astronomy body. I think it will end up with a name from a mythology other than the Eurocentric.


Reply author: nemesis
Replied on: 02/20/2007 12:42:57
Message:

Joe, how does this possible companion relate, if at all, to the work of Matese et al., who have proposed a companion of up to 5 Jupiter masses at about 8000 AU, or possibly a Neptune-sized body at around 2000 AU?


Reply author: Joe Keller
Replied on: 02/20/2007 13:33:19
Message:

quote:
Originally posted by Stoat

And I thought i was being a bit controversial with the "planet stoat." [:)] Barbarossa, was a crusader, so that might raise a frown from Islam. Barbarossa was the code name of the Nazi invasion of Russsia, so I think you might be stepping on a few toes there as well.

What happens is that you would get the credit and the first suggestion as to what to call the thing but it will be decided by an international astronomy body. I think it will end up with a name from a mythology other than the Eurocentric.



Response:

I, the lone and sole predicter and discoverer of the planet and its two largest moons, and my friends, will continue calling the planet Barbarossa, and the two largest moons, Frey and Freya. Those are the names already, and will continue to be the correct names, forever. If anyone ever calls them by any other names, he's incorrect.

- Joseph C. Keller, M. D. (Harvard class of 1977)


Reply author: Joe Keller
Replied on: 02/20/2007 13:36:40
Message:

quote:
Originally posted by nemesis

Joe, how does this possible companion relate, if at all, to the work of Matese et al., who have proposed a companion of up to 5 Jupiter masses at about 8000 AU, or possibly a Neptune-sized body at around 2000 AU?



Thanks for mentioning this interesting work. Of course, that is much farther and much less massive. Was there a search for those bodies?


Reply author: Joe Keller
Replied on: 02/20/2007 14:28:24
Message:

The statement at the top of USNO-B online catalog searches, says the plates are from the last 50yr. This round figure implies they might go back 55yr, or 60yr if the time was counted from the completion of a given survey project. The assumed angular speed implied a range of 69yr for Objects #1-5, but a 10% reduction in orbital radius would reduce this to 59yr.

I've now searched, in the USNO-B catalog with my original criteria, a region 7'-8' to either side of the track, from RA 11h3m to 11h31m. I found no additional objects near enough to the track and similar enough to those already found, to be candidates.


Reply author: nemesis
Replied on: 02/20/2007 14:51:39
Message:

Joe, the paper by Matese, Whitmire, and Lissauer, from 2006, was theoretical and the companion was proposed to explain Sedna-type objects, but they say the object could be present but unrecognized in the IRAS and maybe the 2MASS data. Google "wide binary" with "Sedna" and the paper should come up.


Reply author: Joe Keller
Replied on: 02/20/2007 15:03:21
Message:

quote:
Originally posted by nemesis

Joe, the paper by Matese, Whitmire, and Lissauer, from 2006, was theoretical and the companion was proposed to explain Sedna-type objects, but they say the object could be present but unrecognized in the IRAS and maybe the 2MASS data. Google "wide binary" with "Sedna" and the paper should come up.



from the paper:

"...[RS]Gomes et al. [CeMDA](2005) discuss two resonant mechanisms for converting objects in the scattered disk into high-perihelion (q > 40 AU) scattered disk objects, but they find that they cannot produce these de-tached objects with a > 260 AU. ..."

*******

This correlates with the 270 AU distance, where the circular orbital period, is that implied by the giant planet orbital resonance discrepancies discussed above.


Reply author: Joe Keller
Replied on: 02/20/2007 20:52:26
Message:

Because Frey's magnitude (Red 20.26) is significantly brighter than Freya's (Red 20.61, 20.68, 20.68), it may be distinguished and Object #2 excluded from the test of periodicity. Observations do not always occur exactly at Freya's maximum elongation. Even with this large source of statistical noise, the periodicity is significant:

For these small increments, multiplying the RA by the cosine of the midlatitude of the chord, then using the Pythagorean theorem, gives accurate arclength. Let's neglect eccentricity (variable angular speed) and assume that the true period is the gap between Object #5 & #1. Even after three more cycles, the gap between Object #1 & #3 is only 0.037 of one period off; the gap between Object #3 & #4 is 0.24 period off. Including the possibility of positive or negative error, this is still significant at p=0.036.


Reply author: Stoat
Replied on: 02/21/2007 03:26:54
Message:

Joe, you now need to have telescope time, and a check through with a blink comparitor of old plates. Time is money and you have to convince the guys that hold the purse strings to do this.

Freyja and Freyr will raise no eyebrows but naming a planet after a total psycho (Frederick I ) will. You do know that Ceres was once going to be called George? There was uproar over that.

We simply cannot name a planet after a real historical personage. I suggest you stick to the Norse gods and call it Wotan or Odin. I can't see any astronomer, with an ounce of political savvy, calling this planet Barbarossa. Seriously, I believe that it would lead to deaths.


Reply author: nemesis
Replied on: 02/21/2007 08:46:20
Message:

Joe, naming my be secondary to the more pressing issue of protecting your claim of discovery. You have given out information in a public forum that those "who hold the purse strings" could use and take the credit if the object is confirmed.


Reply author: Joe Keller
Replied on: 02/21/2007 10:29:54
Message:

quote:
Originally posted by nemesis

Joe, naming my be secondary to the more pressing issue of protecting your claim of discovery. You have given out information in a public forum that those "who hold the purse strings" could use and take the credit if the object is confirmed.



Thanks, both to you and to Stoat, for your information (and Stoat's humor!).

A few years ago, ArXiv.org began requiring "endorsement" of authors. That is, one couldn't post unless someone else in the club admitted one to the club. A few years ago, I asked a few faculty members here at Iowa State Univ., to endorse me to ArXiv.org, but none would. So, I use Dr. Van Flandern's messageboard as my ArXiv.org. Dr. Van Flandern is for me, what the ArXiv.org editors are for regular academics.

During the last week, I've reposted (sometimes only minutes later) some abridged versions to several astronomy messageboards. I hope they will look and see Barbarossa. A magnitude +18.1 +/- 0.3 planet will be difficult; the USNO-B plates only go down to about +20.9.


Reply author: Joe Keller
Replied on: 02/21/2007 20:02:37
Message:

The Discovery of the Planet Barbarossa with Moons Frey and Freya: Report to United States Naval Observatory

(by FAX, Feb. 21, 2007)

Author: Joseph C. Keller, M. D. (B. A., Harvard, cumlaude 1977)

Copies: Capt. Edwin C. Keller, U. S. Army, ret.
Prof. Roger Rydin, Physics Dept., Univ. of Virginia, ret.
Dr. Tom Van Flandern, formerly of U. S. Naval Observatory
Prof. Steve Willson, Mathematics Dept., Iowa State Univ.


Abstract. This author’s physical theory implies a distant massive planet in the direction of the cosmic microwave background dipole. The planet, Barbarossa, appears five times in the U. S. Naval Observatory B1.0 catalog:

Object #5. USNO-B 0830-0272239
Object #1. USNO-B 0827-0286487
Object #2. USNO-B 0824-0279170
Object #3. USNO-B 0820-0274026
Object #4. USNO-B 0813-0233607

Barbarossa is aliased by moons.


The Physical Theory. I do not remember who wrote that the sun’s gravitational field is the only known object big enough and symmetrical enough to cause the “cosmic” microwave background (CMB). In 2001 (Aircraft Engineering and Aerospace Technology, 2002) this author discovered that the internal gravitational field of a Heisenberg-uncertain proton (Gaussian radius hbar/(2mc) ) equals solar gravity, at 52.6 A.U. In 2007 (archived on the metaresearch.org messageboard, Dr. Tom Van Flandern, editor) this author discovered that this same distance, 52.6 A.U., is where the 1.5-root-mean-1.5power speed of electrons at the “cosmic” background temperature, equals escape speed. These two equations give the “cosmic” temperature as a function of the gravitational field and potential.

This author noted in 2001 (published; op. cit., 2002) that two kinds of anomaly in the Pioneer 10 probe signal occur, or begin, at about 53 A.U. In 2007 this author found two 2001 reports in the Astrophysical Journal, that the end, or minimum, of the Kuiper belt is at 52 or 53 +/- 1 A.U.

The small-scale statistical distribution of CMB anisotropy is quantitatively consistent, near the ecliptic, with the gravitational effects of Kuiper belt objects. The large-scale anisotropy discovered by Davies at Madeira, is consistent with an Earth-mass planet at 62 A.U., far from the ecliptic, near the edge of Tombaugh’s search region; such a planet would move the CMB dipole a fraction of a degree. Known planets, especially Neptune, also would move the CMB dipole a fraction of a degree. The difference between the 4-yr COBE DMR probe and 3-yr Wilkinson WMAP probe, shows a significantly, retrogradely, revolving dipole, if the true error bars are smaller than published. From these estimates and uncertainties I defined a 1x3 degree search region, backward in time 80 yr, for the hypothetical distant planet mainly causing the CMB dipole. For its circular orbit I estimated 355 A.U. and 6830 yr. For its mass I estimated 0.019 solar masses.

Most stars are detectably double or triple; presumably the sun has a faint companion, a brown dwarf or giant planet. Like the CMB dipole, Percival Lowell’s predicted major axis for Planet X was near 180 degrees ecliptic longitude; a much farther planet’s radius vector almost constantly in this direction, could have similar effect. The solar system origin of the CMB explains the correlations of its multipoles with the plane of the ecliptic.

The Trans-Neptunian Objects (those known in 1998) and long-period comets have aphelia clustered toward 180 ecliptic longitude. Such displacement could neutralize, the CMB quadrupole induced by the distant planet which causes the dipole.

The discrepancies in the orbital resonances of the giant planets (as known c. 1980) sometimes equal simple multiples of Pluto’s period, but mostly equal simple multiples of 4430 yr., corresponding to 266 A.U. for circular orbit. (Gomes et al (2005) calculate that planetary resonances cannot propel small bodies into orbits above 260 A.U.)

An elliptical orbit of this period would give the predicted angular speed, not at 355 but at about 330 A.U., reducing the mass of Barbarossa to 0.016 solar mass. This makes Barbarossa, in present astrophysical theory, a small, cool brown dwarf of surface temperature 378K assuming age 4.6*10^9 yr (slightly extrapolated from Burrows & Liebert, Reviews of Modern Physics, 1993). At 83,500 miles diameter, it would have magnitude +18.1 if its albedo is 7% (slightly more reflective than many asteroids and reddish Kuiper belt objects). If Frey and Freya are bluish gas giant moons with Neptune’s albedo of 30% (and Neptune-like internal heat production), magnitude +20.26 and +20.66, resp., then their diameters are 22,000 miles and 18,000 miles.


The Search. On Feb. 15, I realized that other bodies (later that night, after 00:00 Universal Time on Friday, I realized from the consistency of the magnitudes that they were moons; subsequently I learned that asteroids would be unlikely - textbooks c. 1970 estimated only 40,000 asteroids of adequate size) could impersonate or “alias” Barbarossa, causing two presumed detections near one point on plates made one year apart: thereby inclusion in the USNO-B1.0 catalog. Without further ado I searched the online U. S. Naval Observatory B1.0 catalog, from “Computer #1” in the Nevada, Iowa, public library, throughout the overlapping disks comprising the above 1x3 degree region, for objects with one recorded Red magnitude <+18.99 and the other Red magnitude >+19.50, and proper motions allegedly >80 mas/yr in both directions. I found 61 objects, four of which lay very near a line. I briefly thought it was five objects, but one was a transcription error. Heartened by this mistake, on Feb. 16 I searched for Red magnitudes in Freya’s magnitude range, one degree farther along the line in the retrograde direction, and found Object #5. On Feb. 17 I calculated that Object #5 was only 4.9” from the great circle through Objects #2 & #3 (p = 0.006)(I was suspicious of Object #1 because of its differing magnitude). On Feb. 20 (yesterday) I searched with my original criteria, along that great circle, from RA 11h 3m to 11h 31m, 9 degrees total, finding no others.

The 61 fainter Red magnitudes for each original object, had a uniform Poisson distribution. This indicates that the clustered four faint Red magnitudes of the objects are due to Freya, and the less faint one, of Object #2, due to Frey. The four objects associated with Freya show periodicity of position.

Of the five brighter magnitudes of the objects, none lie between +18.60 and +18.99. Object #1 is brighter in Red than the other objects, and very dim in Blue though not bright in Infrared (“I”, 1.2 micron). An IBM 486 Monte Carlo trial showed that averaging four or five objects does not vitiate the result, that Barbarossa has a spectrum too flat, and too dim in Infrared, to be a Type M or late K star with a Planck spectrum. (Galactic dimensions and star type counts imply that half the stars in my brighter Red magnitude range, would be K5-K9, & half M.) Almost ten times as many stars found within 20 degrees, in the same parameter range, had B & I both fainter than for Barbarossa, than had both brighter. Barbarossa’s color magnitudes suggest a large red spot covering almost one hemisphere of the rotating planet.


Object #5. USNO-B 0830-0272239
Object #1. USNO-B 0827-0286487
Object #2. USNO-B 0824-0279170
Object #3. USNO-B 0820-0274026
Object #4. USNO-B 0813-0233607


Object #5. USNO-B 0830-0272239 RA 11h10m08.44s Decl -6d59'37.2"

Object #1. USNO-B 0827-0286487 11h12m05.59s -7d14'27.8"

Object #2. USNO-B 0824-0279170 11h14m54.41s -7d35'13.7"

Object #3. USNO-B 0820-0274026 11h18m03.53s -7d58'41.0"

Object #4. USNO-B 0813-0233607 11h23m30.03s -8d38'37.8"



Analysis. If the typical orbital radii of large satellites follow


Reply author: Joe Keller
Replied on: 02/21/2007 21:53:07
Message:

The five Objects span four degrees. Open clusters such as the Hyades and (some of the stars in) the Big Dipper, can be this big, but their proper motions tend to be radiant rather than unidirectional.

The Hipparcos probe showed that the sun's apparent apex motion is roughly the same, even when determined relative to stars farther than 300 parsecs (over 900 light years). At 100 light years and 90 degrees away from the apex, the average proper motion associated with apex motion, is roughly 100mas/yr; at 2000 light years (likely about the farthest distance of any of the five Objects, if they're stars) the motion is likely still roughly 5 mas/yr. At Leo, the direction of star proper motion bias due to presumed solar apex motion, is negative in both RA and Declination.

Yet 10 of 16 presumed proper motions (RA or Decl) of Objects #1-#8 (see Feb. 26 post below) were positive. Though star counts indicate that maybe none of the (now eight)"stars" would have been close enough for solar apex motion to be important, that bias could produce statistical significance.


Reply author: Joe Keller
Replied on: 02/22/2007 17:27:13
Message:

The monotonic trend in Red magnitude of the four Freya sightings, has a correlation coefficient of 0.9275. For n=4, tables give p=0.01, two-tailed, for a correlation coefficient of 0.917; by extrapolation the Freya sightings have p=0.008. Freya has uniform albedo.

Barbarossa's semimajor axis, a = 270 A.U., is given precisely by the discrepancies in giant planet resonances (which give Barbarossa's period). To a good approximation for small eccentricity, the angular speed, when r=a, is independent of eccentricity. The distance estimate of 330 A.U., depends, on the accuracy of this approximation, and on the accuracy of the determination of present angular speed. The latter depends on the accuracy of the COBE & WMAP CMB dipole directions, and on the accuracy of my theoretical correction accounting for the gravity of the four known giant planets. Be all this as it may, 330 A.U. implies plausible albedo and mass for Barbarossa.

Computing Freya's distance from Earth, from Freya's magnitude, the best fit line gives an angle arctan(0.36) between Barbarossa's path and the constant circle. Graphical solution of the orbital equation gives eccentricity 0.36, and Barbarossa 60 degrees past aphelion. Aphelion is 420 A.U. and perihelion 200 A.U.

Takeda & Rasio (ArXiv.org, Oct. 10, 2005) state that the median orbital eccentricity for known extrasolar (giant) planets is 0.31, and the third quartile 0.43, but such planets would be much closer to their stars than is Barbarossa. A more appropriate comparison might be to 61CygniB whose eccentricity is 0.40 and period 650 yr.; the pair 61CygniAB has 4/3 the mass, of the pair sun+Barbarossa, but it is almost equally divided.

If my gravitational theory of the CMB is accurate, the 3-yr WMAP dipole temperature should be 33.6 microK warmer than the 4-yr COBE dipole temperature (25.0 microK from Barbarossa's approach and 8.6 microK from the overall improved alignment of J, S, U, and mainly N). It is 5 microK warmer, but the error bar for the difference is 29 microK.


Reply author: nemesis
Replied on: 02/23/2007 08:48:02
Message:

Joe, I hate to ask what may seem a naive question, but when you say Barbarossa is "aliased by moons" exactly what does that mean?


Reply author: Joe Keller
Replied on: 02/23/2007 13:38:19
Message:

quote:
Originally posted by nemesis

Joe, I hate to ask what may seem a naive question, but when you say Barbarossa is "aliased by moons" exactly what does that mean?



Last night I spent two hours telling a retired medical school physiology professor about Barbarossa, Frey, Freya, etc. He said, essentially, "So, Freya or Frey 'stand in' for Barbarossa in almost the same position on the next year's plate."

By "alias", I mean "stand in". I took the term "alias" from signal theory, where it means that a sinusoid of one frequency impersonates a sinusoid of another frequency.


Reply author: Joe Keller
Replied on: 02/23/2007 15:16:43
Message:

Error analysis:

The estimated distance to Barbarossa, and the estimated Neptune-like 30% albedo of Freya, imply that Freya's diameter is 12,300 miles. Because the upper error bar allows it, I adopt 17,500 miles because Freya's constant albedo implies that Freya is a Neptune-like gas giant with internal heat, not a Pluto or Earth with nonuniform albedo.

1. If the five USNO-B objects span as little as 4/5 of Barbarossa's track on the (as little as?) "50 years of Schmidt plates", then the angular speed is as great as 69/(4/5*50) times what I used. For a given ellipse, "equal areas in equal times" implies that the angular speed is inversely proportional to the square of the radius. This gives a lower bound of 0.58 times calculated, for Freya's diameter.

2. The error bars on the COBE dipole determination imply that Barbarossa's travel between COBE & WMAP, might be 0.33 deg of galactic longitude more or less, than my calculated 0.866 (which includes a correction for planetary effects). This gives a range of 0.72 to 1.60 times calculated, for Freya's diameter.

3. Besides angular speed, the distance to Freya also can be found from Barbarossa's apparent magnitude, assuming Barbarossa's albedo is 7%. Using the five bright Red determinations, the standard error of the mean for Barbarossa's magnitude is 0.225. This gives a range of 0.91 to 1.11 times calculated, for Freya's diameter.

4. By analogy with other solar system bodies, Barbarossa's albedo might be as low as 5%. This gives a lower bound of 0.85 times calculated, for Freya's diameter.

(3) & (4) combine to give a greatest lower bound of 0.77 times calculated. The possibility of Barbarossa albedo > 7%, vitiates the upper bound in (3), so the least upper bound is 1.60, from (2). For simplicity, to make Freya more Neptune-like, I double Freya's area, multiplying its diameter by 1.41, to obtain 17,500 miles.


Reply author: Joe Keller
Replied on: 02/23/2007 18:55:45
Message:

Search track for 2007:

Assuming that Barbarossa's period is between 4400 and 6800 yr, the search track is a great-circle segment (approximately a line segment for this small distance) with expected endpoints:

RA 11h 6m 02s Decl -6deg 28' 27" and

RA 11h 7m 24s Decl -6deg 38' 50"

Barbarossa should move 0.5"/day retrograde. If the period is 4400 yr, there's a 50% chance Barbarossa will be on the track west of the west endpoint; if the period is 6800 yr, a 50% chance of being east of the east endpoint. The expected number of stars in the USNO-B catalog, within 1' of this segment, having either Red1 or Red2 magnitude < 18.50, is 28.


Reply author: Joe Keller
Replied on: 02/23/2007 22:57:22
Message:

The ratios of convective energy, to total lightning energy, to optical lightning energy, are about the same for Earth and Jupiter. The convective energy of Barbarossa would equal its 378K (est.) Planck emission. Employing the ratios in Borucki's articles below (I used geometric mean averages with twice the weight for Jupiter as for Earth) I found that the lightning emission from Barbarossa is equivalent to about 3% albedo.

Zarka remarks that the lightning activity on Saturn varied 5-fold between Voyager 1 & 2. Such variation on Barbarossa might explain some of its variable magnitude. The broad peaks of the lightning-related radio emissions for J, S & U, all are near 4 MHz.


References.

Borucki et al, Icarus 52:492+, 1982.
Borucki et al, Reviews of Geophysics and Space Physics 22:363+, 1984.
Zarka et al, Planetary and Space Science, 52:1435+, 2004, p. 1442.


Reply author: Stoat
Replied on: 02/24/2007 05:11:09
Message:

I put those two positions up as jobs on the Bradford robotic telescope. I used the galaxy telescope, set to rgb colour and with an exposure of 60 000 ms. Perhaps other people could do the same search with a larger exposure.

Sorry Joe but I put my image name in as Nemesis, no way can I live with Barbarossa [:D] In fact I'm going to riot outside the American embassy in London. I don't think there's ever been fighting in the streets from anarchic astronomers, so it will be a first. I plan something like the Spanish Civil War, so that I can start a Carl Sagan Brigade. Then I'll rewrite "for whom the bell tolls," where the partisans will discuss love, honour and Cantor's infinite sets, as they sit in a cave cleaning their guns. I'll make a fortune [:D][8D][}:)]

The movie: Russell Crowe as Tom Van Flandern and Michael De Caprio as Larry Burford, are parachuted into the mountains to help the partisans blow up the Mount Palomar telescope. Jodie Foster as the love interest, the mission is threatened by her fancying both of our heroes. Great! it has everything [:D]


Reply author: Larry Burford
Replied on: 02/24/2007 09:14:39
Message:

Blow it up?

Can't we all just ... kidnap the schedueling secretary? Now don't get me wrong. I like big explosions as much as the next guy. The more the better. MPT may be old, but it can still do good science if it is just pointed in the right direction.


Reply author: Joe Keller
Replied on: 02/24/2007 12:05:00
Message:

[quote]Originally posted by Stoat

I put those two positions up as jobs on the Bradford robotic telescope. I used the galaxy telescope, set to rgb colour and with an exposure of 60 000 ms. Perhaps other people could do the same search with a larger exposure. ..."

Thanks for the help! According to their website, that "galaxy telescope" is the biggest that Bradford Univ. has on Tenerife, a 14" Schmidt with a 24' x 24' field. I recall that c. 1930, Tombaugh used a 13" for his photographic wide ecliptic search down to +17. Would a 60sec exposure with a 14", even with modern light-recording technology, be able to see Barbarossa, whose expected Red magnitude is +18.07? The faintest of the five Objects presumed to be Barbarossa, had Red magnitude +18.59.


Reply author: Joe Keller
Replied on: 02/24/2007 12:44:28
Message:

Another way to confirm or refute the existence of Barbarossa, would be to look for the above five Objects. One could simply look at the epochal positions (given, inter alia, in the USNO-B catalog and in my above Feb. 21 FAX to the USNO):

Object #5. USNO-B 0830-0272239 RA 11h10m08.44s Decl -6d59'37.2"

Object #1. USNO-B 0827-0286487 RA 11h12m05.59s Decl -7d14'27.8"

Object #2. USNO-B 0824-0279170 RA 11h14m54.41s Decl -7d35'13.7"

Object #3. USNO-B 0820-0274026 RA 11h18m03.53s Decl -7d58'41.0"

Object #4. USNO-B 0813-0233607 RA 11h23m30.03s Decl -8d38'37.8"


If the Objects are stars, then the USNO-B epochs and proper motions should be fairly accurate. Here they are:

Object #5. epoch 1974.5 PM in RA -120mas/yr PM in Decl -502mas/yr

Object #1. epoch 1980.0 PM in RA 128mas/yr PM in Decl 426mas/yr

Object #2. epoch 1976.1 PM in RA 274mas/yr PM in Decl 394mas/yr

Object #3. epoch 1965.1 PM in RA 308mas/yr PM in Decl 234mas/yr

Object #4. epoch 1973.0 PM in RA 302mas/yr PM in Decl 142mas/yr

It's indicated that the USNO-B positions include a correction for these alleged proper motions.


Reply author: nemesis
Replied on: 02/24/2007 12:50:19
Message:

Shouldn't Barbarossa show a large annual parallelax, much larger than it's proper motion?


Reply author: Joe Keller
Replied on: 02/24/2007 12:55:47
Message:

My Feb. 21 FAX to the USNO mentions the crowding of the brighter Red magnitudes, for the five Objects, away from +18.99. This is yet another peculiarity which adds to the overall statistical significance of the Barbarossa observations. For my original search criteria (one Red magnitude >19.50, one Red <18.99, both PMs >80), only 59.5% of the time (within 10 deg of the above predicted line segment), is the brighter Red <18.60. This gives p = 0.5954^5 = 0.075.

For the same region and criteria, only 74.8% of the time, are both PMs >120. This gives p = 0.234.


Reply author: Joe Keller
Replied on: 02/26/2007 12:51:22
Message:

quote:
Originally posted by nemesis

Shouldn't Barbarossa show a large annual [parallax], much larger than it's proper motion?



Thanks for bringing up this interesting subject! The same region of sky will tend to be photographed recurrently at the same season. Typically, time exposures of the region near the tail of Leo would be done in March, when it's on the meridian at midnight. Maybe in a systematic sky survey it would be the same time of year almost to the day. This could eliminate Earth parallax and allow the aliasing proposed above.

Since I don't know the true epochs of the important plates, I'm guessing Barbarossa's position on the orbital track. Earth parallax merely increases that uncertainty. Earth moves about sqrt(360) = 19x as fast as Barbarossa, so a month to either side of March would move Barbarossa's apparent position 0.5 ("/day) * 19 * 30 days = 5' = 20s RA.


Reply author: Joe Keller
Replied on: 02/26/2007 16:15:14
Message:

I searched the track again, using the original criteria, but modified to allow the proper motions to be in any quadrant. Due to time constraint (ice storm and power outage in Iowa - trying to avoid freezing pipes!), I only searched from RA 11h 14m to RA 11h 27m. These are the three new Objects closest to the track (numbered in order of discovery):

Object #6. USNO-B 0811-0231074 RA 11h 24m 46.11s Decl -8deg 48' 02.0"

Object #7. USNO-B 0824-0279078 RA 11h 14m 36.12s Decl -7deg 32' 21.3"

Object #8. USNO B 0811-0231124 RA 11h 24m 57.31s Decl -8deg 49' 37.1"


Object #7 has R1 20.22, R2 18.54; B2 23.29 (i.e., B2 is probably invalid). These magnitudes are typical of a Frey alias.

Object #6 has R1 20.73, R2 18.84, I 18.43. Except for the somewhat faint 18.84, these magnitudes are typical of a Freya alias.

Object #8 has R1 19.65, R2 18.80, B2 19.52, I 18.45. R2 resembles that for nearby Object #6 (18.84). The other magnitudes resemble a combined Frey + Freya alias: the energy flux signified by R1, differs only 2% from that due to Frey + Freya (using, for each, the mean of all observed R1s). A correction for the significant trend in R1 (noted above for Freya and presumably applicable to Frey also) increases the discrepancy, but only to 7%.

Object #6 conforms to the trend in Freya magnitudes vs. RA. Object #7 would have its Frey magnitude, at 20.26 - 0.005 = 20.255, according to this trend; actually its R1 is 20.22.

Five points generally define a unique ellipse. The above-mentioned ellipse on which the 2-dimensional PMs of Objects #1-5 lie, is somewhat special: it is centered at the origin, but its eccentricity (as determined below) is 0.8.

The PMs of Objects #6-8 lie on this same ellipse! Let us call it the "PM ellipse". Graphically, the long axis of this ellipse is roughly perpendicular to the ecliptic. This indicates a manner of aliasing, different from that described above.

In his messsage to this (Dr. Van Flandern's) messageboard, "Nemesis" asked, with perfect timing, about Earth parallax. The Earth parallax effect for Barbarossa is about 10.1"/day retrograde, in March when plates of Leo are best made (10.6"/d motion including Barbarossa's own, retrograde, orbital motion).

Frey and Freya might be a double moon; at least they have similar apparent orbital ellipses around Barbarossa. When Barbarossa lies nearly opposite the sun, the center, of the apparent ellipse of that orbit (or those apparently similar orbits)(the projection of the true orbital ellipse, onto the celestial sphere) always lay about where Barbarossa was the night before (or will be the night after). When Red1 and Red2 photographic plates are made on successive March nights, Freya or Frey (or both if in close conjunction, e.g., Object #8) may stand in for (i.e., alias) Barbarossa.

Barbarossa's magnitude fluctuates. Much of its magnitude might be from lightning (see above). A lightning storm that subsides in a day, would render Barbarossa confusingly dimmer, facilitating aliasing.

The major & minor axes of the PM ellipse are about 960 & 530 mas/yr, resp. As Barbarossa moves along the minor axis, let us consider 530mas/yr / 2 * 50yr of plates = 13.3": this is only slightly more than 10.6", so with an apparent ellipse of this dimension, moons can stand in for Barbarossa about as well as Barabarossa can stand in for himself. Plates made on successive nights are assumed by the computer program, to be 50 years apart, because this is the range for a star. The distance between a moon tonight and Barbarossa last night (or vice versa) along, say, the semi-minor axis of the apparent orbital ellipse, is 13.3", which the computer interprets as 265mas/yr PM.

The semi-major axis, 24", at 330 A.U., equals 0.04 A.U. = 3,700,000 mi. Assuming Frey & Freya have the density of Neptune, then the mass ratio of these moons to Barbarossa, roughly equals that of the sun to its known giant planets.


Reply author: Joe Keller
Replied on: 02/27/2007 20:56:52
Message:

I finished searching, as above, for PMs of all sign combinations, from 11h 3m to 11h 31m. I found two more objects about as close to the best great circle, as the others:

Object #9. USNO-B 0809-0228789 RA 11h26m35.28s Decl -9deg00'22.6"

Object #10. USNO-B 0804-0237081 RA 11h30m57.33s Decl -9deg32'10.7"


Both these are at larger RA than the others. The former is compatible with a Frey (R1 20.24, B2 20.43, R2 18.75) and the latter somewhat with a Freya (R1 20.86, R2 18.39, I 18.43) sighting. The former lies near the "PM ellipse"; the latter well outside it.

Though the search indeed seemed to reveal an overdensity of objects near the line, the distance cutoff is somewhat arbitrary. I found two more objects which are about twice as far from the most appropriate great circle line, in their region, as Objects #9 & 10, or as Object #7, resp. These are:

Object #11. USNO-B 0806-0230268 RA 11h28m56.15s Decl -9deg19'31.9"

Object #12. USNO-B 0836-0217259 RA 11h05m22.68s Decl -6deg22'07.9"

The former is compatible with a Frey sighting (R1 20.27, B2 19.98, R2 17.89) and the latter with a Frey/Freya conjunction (R1 18.74, R2 19.59, I 18.36). The former lies near the "PM ellipse", the latter well inside it. These two Objects have Frey and/or Freya magnitudes more consistent with the trends observed in Objects #1-8, but they do not lie as close to the great circle line. If all of the Objects #9-12 are excluded, then the remaining eight Objects remarkably all lie between 11h10m & 11h25m, in a search range from 11h3m to 11h31m.


Reply author: Joe Keller
Replied on: 02/28/2007 20:24:49
Message:

There are 56 ways to choose 5, from among Objects #1-8. For each choice, I calculated the coefficients of the unique ellipse through the Proper Motion points (5 x 5 system of linear equations). Averaging the 56 sets of coefficients gave the equation of a kind of best-fitting ellipse (for PM in arcsec/yr; x is Proper Motion in RA, y in Decl):

42.85 * x^2 - 14.45 * x * y + 22.82 * y^2 + 1.977 * x - 1.586 * y = 1

The small linear coefficients confirm that the center is near the origin. My program gave a nonsensical (+) sign for the second coefficient, so I assume that the x-axis somehow was reversed, and have changed the signs of the xy & x coefficients accordingly. The sum of the variances of the quadratic coefficients, within the 56 sets of coefficients, was 961. I permuted the abscissas randomly five times. The resulting range, of sum of variances of quadratic coefficients, was 976-1824, mean 1435, s.d. 308: thus Objects #1-8 fit an ellipse better than the somewhat randomized points obtained by shuffling their abscissas. The mean tilt of the 56 major axes, was 26.6deg clockwise from the (+) y-axis; the standard error of the mean was 3.1deg.

Alternatively, by starting graphically and applying successive approximations, I found the equation of the best-fit ellipse for the PMs of Objects #1-8 (PMs in arcsec/yr):

3.45 * (x*cos61.1 + y*sin61.1 + 0.070)^2
+ 13.36 * (x*sin61.1 - y*cos61.1 + 0.009)^2 = 1

This minimizes the sum of squared differences between the left & right sides of the equation of the curve: the rms discrepancy is 0.05, corresponding to a typical error in radius of only about 2.5%.

The major axis is inclined 28.9deg clockwise from the (+) y-axis. Using the midpoint, of a great circle beween Object #5, and the average of Objects #6 & 8, I find that Barbaross's orbit cuts the parallels at 26.4deg (see also 26.6 +/- 3.1 obtained above). This differs only 2.5deg from the minor axis of the "PM ellipse".

Five, of Objects #1-8, have magnitudes implying sightings of Freya without Frey; one has a magnitude implying Freya + Frey; two have magnitudes implying Frey without Freya. The radii, from the ellipse center, of the Frey sightings, averages 10% more than the radii of the Freya or Freya+Frey sightings. Student's t-test is 3.87 with 6 degrees of freedom; 3.71 gives p=0.005 (CRC Handbook of Statistics, 2nd ed., pp. 283,288).


Reply author: Joe Keller
Replied on: 02/28/2007 20:44:05
Message:

I searched within 10 deg of RA 11h18m Decl -7deg, using the PM & R1,R2 criteria above. Only 2125/13615 objects have 18.36 < I < 18.78. Yet six of the twelve Objects above have "I" in this range (p=0.005, binomial test).

Only 633/13615 objects have 18.36 < I < 18.45. Yet five of the twelve Objects above have "I" in this range (p=0.00028, Poisson test). Of the first eight Objects (i.e., those of which I am most sure) three have "I" in this range (p=0.0065, Poisson test). The problem has the equivalent of 8 tails, because for an "I" magnitude 0.5 greater, only 232 fall in an interval of this length (232/633=1/2.728 times as many, i.e., characteristic length, for exponential dropoff, approx. 0.5) and for magnitude 0.5 smaller, only 103 do (0.5/ln(633/103)=0.275). Consideration of these tails gives p = 0.0065 * 8 = 0.052.

A more precise way to account for the tails, is to place the sample "I" values under a normal distribution curve; to give the three ordinates in the correct ratio, its peak must be 18.55 and its s.d. 0.30. The area under the curve would be 0.40, consistent with possessing "I" values for 6/12 or 4/8 Objects. Monte Carlo trials (IBM 486) showed that only 21 times in 100, was the sum of squared differences for such normally distributed "I" values, less than that for the four known "I" values for Objects #1-8. Using all six "I" values, this dropped, to 7 times in 100. Using the five "I" values remaining after omitting the outlier, this dropped to 0 times in 1000. There are six ways to omit one value, but these aren't independent, so p<0.006.

Are the eight or twelve Objects still there? What do they look like? I have heard nothing from the U. S. Naval Observatory despite my three FAXs, and two letters, one of which was to the commander, a US Navy Captain.


Reply author: Joe Keller
Replied on: 03/04/2007 21:37:43
Message:

A sinusoidal variation of wavelength 3.686 minutes of Right Ascension, with optimum amplitude & phase, explains 78% of the variance in the presumed Red magnitude of Barbarossa. Objects #1-8 span a presumed 83 years of time (the increased length of the interval changes the above 69 yrs, to 77; 6 more yrs come from an eccentricity correction to the above angular speed estimate). These Objects span 14.75m of RA. So, Barbarossa's Red brightness varies between about +17.4 & +18.9, with period 20.7yr.

Objects #6 & 8 were sighted only 11.2s (presumably 1.05yr) apart. Their "I" and bright Red magnitudes differ by only 0.02 & 0.04, resp. For Objects #6 & 8, "I - R", though consistent with late Type K stars (i.e., halfway between USNO-B's I-R given for Type K2 Arcturus & Type M3.5 Gacrux), could be due to twice the albedo, at I=0.875micron(midband) vs. Red, under sunlight; such a difference is commonplace. Alternatively, with sunlight and equal R&I albedos, the difference could be due to Planck emission at 391K.


Reply author: Joe Keller
Replied on: 03/05/2007 00:59:02
Message:

Object #2 isn't on the Aladin plate.


Reply author: Joe Keller
Replied on: 03/05/2007 11:01:52
Message:

I found Barbarossa on the Aladin plate for Object #3. It appears as a streak in Right Ascension, 5.5" toward 4 o'clock from the catalog position. A moon is 8.9" away along a line inclined 14deg to the lines of Declination.


Reply author: Joe Keller
Replied on: 03/05/2007 11:24:08
Message:

I first saw Barbarossa & moon last night between 11PM & midnight, here at this same computer monitor in the Iowa State Univ. Parks Library computer & children's book room, #CRS0020285. Then my main worry was that they were stars; only today did I realize, for sure, that it's Barbarossa.


Reply author: Joe Keller
Replied on: 03/05/2007 11:31:28
Message:

The length (roughly 5" in, presumably, 10hr) and orientation (roughly parallel to Declination lines) of the streak made by Barbarossa, are consistent with the retrograde motion of Barbarossa near opposition (due mainly to Earth's motion) at its distance of 330 AU, namely, 10"/day. The moon appears significantly streaked also, though less, but along the same direction.


Reply author: Joe Keller
Replied on: 03/05/2007 12:12:31
Message:

Object #4 is missing from the Aladin plate (obtained over the internet from VizierR at Strasbourg). All eight nearby objects with 17.2 < R1,R2 < 20.3 were identifiable near (though often 1-2" N) of catalog position (though three of these might have been parts of the background face-on barred spiral galaxy).

On the Aladin plate for Object #2, the nearest catalog object(s) (excluding another of high, i.e., "999", positional uncertainty) is USNO-B 0824-0279158. There are three catalog objects very close together, which seem from their catalog data to be really one or two objects. Allowing for summed flux, this object(s) has 17.9 < R1,R2 < 18.2, and appears as one object on the Aladin plate very near its catalog position. Almost nothing lies within 40" of the catalog position of Object #2.

Likewise for Object #3, there is USNO-B 0820-0274037 with 17.8 < R1,R2 < 19.0, exactly at catalog position.


Reply author: Joe Keller
Replied on: 03/05/2007 12:46:03
Message:

The Aladin server, formerly quick, now has become too slow to use at all. I suppose that either people who look at this messageboard, or people to whom I immediately emailed my messages posted here, are using it to confirm my claims.


Reply author: Joe Keller
Replied on: 03/05/2007 13:45:26
Message:

The Aladin server is fine now. Maybe I missed the plate among the pile of "windows" on my "desktop".

Object #1 also is missing from its Aladin plate. Of the three nearest dim stars having either R1 or R2 < +19.0, one is obvious, one (USNO-B 0827-0286485) is a definite smudge (galaxy?) at its catalog location just S of where Object #1 should be, and one is a fainter smudge.

A nearby Red +16.66/17.29 star (USNO-B 0827-0286507) resembles a nearby R+18.62/18.68. The nearly equal R1 & R2 magnitudes argue against a very variable star, and for obscuration.

The abovementioned USNO-B 0827-0286485 seems to have a vague moon on each side: they are 20" apart on a line 20deg to the equator. Both this line and the one mentioned above for Object #3, slope like Barbarossa's alleged orbit.


Reply author: Joe Keller
Replied on: 03/05/2007 18:37:56
Message:

The Aladin server has been "down" for 3 hrs now, but before then, I confirmed that USNO-B catalog objects with discrepant Red magnitudes such as mine, generally are absent from the plates, or at least displaced. Within 20 arcminutes of Object #1, the USNO-B catalog has 12 other objects with one Red mag <18.99, one Red mag >19.50, both PMs greater than 80 in absolute value, and (for determinacy) uncertainty of position less than "999" in both axes. Before the server went down, I had time to evaluate 10 of these:

One was a typical dim star at exactly the published location. One lay on a point (i.e., streak) of a bright star (much as Object #4 lay on an arm of a spiral galaxy). Two resembled Object #2: there was nothing near their catalog location. One resembled Objects #1 & 3: it lay 10" from a pair of stars or galaxies (the brighter was in the USNO-B catalog - as was the brightest of the three objects near Object #1). Five of the objects, lay about 5" from other, faint and/or blurred, solitary, uncataloged objects. Thus only one of the ten random objects meeting the criteria, was in a complicated situation like Objects #1 & 3 ( p = (6*10 + 4) /(14*13*12/3!) = 0.18 ).

Regarding Object #3, an estimated 40,000 asteroids of more than a few km size, gives only one per square degree of sky; perhaps three per square degree in this part of the ecliptic. Furthermore an asteroid at opposition, even 4 AU from the sun, would get 100x the apparent retrograde angular speed, from Earth's motion, as Barbarossa. Only half this would be canceled by the asteroid's own orbital speed. Even a Kuiper belt object at 34 AU, would get 10x the retrograde speed, as Barbarossa.

Alternatively, the streak near Object #3 might be a roughly edge-on spiral galaxy. The Zwicky, Herzog & Wild Catalog of Galaxies (1961) vol. 1, pp. i,36-38, indicates that a few degrees north of Barbarossa (south of -3.5deg Decl was not covered) even the densest clusters have only about 1000 galaxies per sq deg, down to about magnitude +18. The streak south of Object #3 looked like roughly a +18 object. Even in the densest cluster, the chance of finding a +18 galaxy within a 5" radius, would be only 1000*(pi/720^2)= 0.006. Only half of galaxies would be spiral, only half of those sufficiently edge-on, and only 1/3 of of those, aligned within 30deg of Barbarossa's orbital path, so, p < 0.0005, or, for four Objects, p < 0.002. However, a galaxy would give about the magnitude observed: if the 5" length is the bright core analogous to the central 40,000 lt yrs of M31 (which subtends 1 degree at 2.4*10^6 lt yr), then the magnitude would be log(720^2)*2.5 + 3.4 = 17.7.


Reply author: Joe Keller
Replied on: 03/06/2007 00:46:35
Message:

I checked the IRAS infrared source catalog, using overlapping disks of 1deg radius to cover Barbarossa's track from 10h59m to 11h35m. The number of IRAS sources within 10deg of the track's midpoint, implies expectation of 0.50 sources closer than 1' to the track. The number of sources found in the swath of the overlapping disks, extending 0.866-1.000deg to either side of the track, roughly confirmed this.

Two IRAS sources (p=0.09, Poisson test) were found closer than 1' to the track (J2000 celestial coordinates):

Source #1. IRAS 11102-0701 (repeated in the "Faint Source" IRAS catalog, as F11102-0701, with ostensibly slightly more accurate coordinates, which I use), 55" from the segment between Objects #1 & 2; RA 11h12m44.9s Decl -7deg17'50". For the source in the original IRAS catalog (without the "F") the flux in Janskys was 0.633 +/- 12%, 0.657, 0.400 & 1.00 in the 12, 25, 60 & 100 micron bands, resp. (no error bars given for last three values). In the "faint source" IRAS catalog (with the "F") those fluxes were 0.657 +/- 9%, 0.301 +/- 31%, 0.120 +/- 29% & 0.644 +/- 29% (all but the longest wavelength's fluxes, are said to be 94-98.5% reliable in the "F" catalog). The source is 17" from a mag +11 red star, USNO-B1 0827-0286738. The IRAS satellite's field of view perpendicular to the ecliptic was 0.75 to 3' (narrower for shorter wavelengths), and 4.5 to 5' parallel to the ecliptic. The uncertainty of the source's position (in the "F" catalog) is 28" parallel to the ecliptic and 2" perpendicular thereto (due to experimental design, all sources had more uncertainty parallel to the ecliptic, but for this source, the difference was especially great). (In the original catalog, the uncertainties were 70" & 8", resp.)

Source #2. IRAS 11210-0823 (I haven't yet checked the "F" catalog), 33" (perpendicular) from the segment between Objects #3 & 4; RA 11h23m32.1s Decl -8deg39'30". The flux in Janskys was 0.256, 0.250, 1.950 & 4.650 at 12, 25, 60 & 100 microns. This source overlies the same face-on spiral galaxy as does Object #4. The original catalog uncertainties are 40" parallel & 11" perpendicular to the ecliptic.

IRAS, in 1983, mapped the sky four times in ten months. Special software was employed during part of this time to find asteroids (positional discrepancies apparent within hours)(S Green et al, Icarus 64:517+, 1985). Earth's motion would give Barbarossa net apparent motion only 2% that of an asteroid.

I have to leave this public library computer; more to follow.


Reply author: Joe Keller
Replied on: 03/06/2007 17:08:07
Message:

If the effective observation date of IRAS was 1983.5; if Source #1 (above) was Barbarossa; if Barbarossa's orbital period is 6828 yr; and if the track of Objects #1-8 is that of Barbarossa, then the position of Barbarossa for 2007.25 is:

RA 11h8m14.1s Decl -6deg44'11.5"

Apparently school let out early in Story County due an impending storm, so I have to leave yet another public library computer today; more to follow.


Reply author: Joe Keller
Replied on: 03/06/2007 18:58:52
Message:

I asked the librarian: it was a conference afternoon! Now the kids are gone to supper and I'm back on the computer here, after looking up some facts in the Encyclopedia and making some calculations.

I neglected Earth parallax when I could assume that observations always occurred at opposition. If the relevant IRAS observation (launch Jan. 2003: 10-month mission ended before the end of 1983) was made, effectively, June 9, then the correction for Earth parallax and for the aberration of light, changes Barbarossa's "J2000" position for 2007.18 (approx. March 9, when Barbarossa is, approx., at opposition and on the meridian at midnight) to:

RA 11h8m53.6s Decl -6deg49'07.8"


This is based on extrapolation from the most recent sighting, rather than the great circle below.


Reply author: Joe Keller
Replied on: 03/06/2007 19:26:39
Message:

If the USNO-B catalog plates were made at Barbarossa's opposition +/- 1 month, there would be up to (10.5' Earth parallax + 0.3' aberration of light) * sin(Barbarossa's inclination 26.4 - Earth's inclination (23.45*cos10.5)) * sin30 = 20" deviation to either side of the track. This is roughly what was observed. The corresponding 0.5 * 10.8' * cos(23.45)/cos(8) = (+/-)20s deviation in RA, would be enough to regularize the periods of the Freya sightings (Objects #1, 3-6 & 8).

Graphically, the best great-circle fit to Objects #1-8, gave a root-mean square deviation of 13.4" (perpendicularly) from the line (subsequently confirmed as 13.40" using successive approximations on my IBM 486). I tried undoing the USNO-B catalog's automatic Proper Motion correction to the coordinates: using successive approximations on my IBM 486, the rms deviation became 12.2".

I chose the points by the criterion of closeness to a line. Yet subtraction of the USNO's presumably fictitious Proper Motion correction, removed 20% of the variance.

The likely correction to Barbarossa's predicted present (2007.18) RA, from the uncertainty of IRAS observation dates, is between 0 and -11s, maybe -13s (the corresponding correction to Decl for Barbarossa's track, is approx. -7.5 arcsec Decl per sec RA). Estimating Barbarossa's orbit causes an even bigger uncertainty: the most different likely alternative, is a circular orbit with 4400 yr period, which would alter Barbarossa's predicted present RA by -120s.

So, I give not only estimated present coordinates, but also a search great circle which is known more confidently. (At roughly 10 degrees from the ecliptic, Barbarossa's track appears as almost a perfect great circle now, at opposition.) This is the great circle determined by least-squares successive approximations with Objects #1-8. It passes through the points:

RA 11h 08m 00s Decl -6deg 43' 36.7" &
RA 11h 10m 00s Decl -6deg 58' 37.9"

A straight line segment between these points, on rectilinear coordinate paper, will deviate by less than 3", from great circle to chord.


Reply author: Joe Keller
Replied on: 03/06/2007 20:25:47
Message:

Source #1 (IRAS 11102-0701 or IRAS F11102-0701) indicates that Barbarossa is now devoid of internal heat. Barbarossa is at equilibrium 15K with sunlight, or maybe somewhere between that, and the 30K temperature of interstellar "cold dust" (Enc. Britannica 1998, article "infrared source"). Frey & Freya are, to a first approximation, infrared blackbodies at 55K; or, to a second approximation, slightly warmer: Neptune's effective temperature, 59K, would allow the same energy radiance, but with a 25% albedo for infrared. Frey and Freya have internal heat like Neptune.

On one hand, Source #2 resembles (in its three longest wavelengths) the exponential tail of a Planck radiation curve for the 30K "cold dust" of the distant face-on spiral galaxy superimposed with it. On the other hand, Source #1 in its "F" ("faint source" IRAS catalog) version assessment, resembles (in its three shortest wavelengths) the Rayleigh-Jeans "left foot" of the Planck curve for 250K "warm dust" clouds "warmed by neighboring stars" such as the "red" star nominally 17" away (op. cit.; see also previous post).

The difference between the original version and "F" version assessments of Source #1, is due to infrared radiation from Frey & Freya, which apparently was somehow subtracted to obtain the "F" version. In the original assessment, Frey & Freya had provided half the power in the three longest wavelengths. With four IRAS measurements, the moons must have had four times the power of the stellar "warm dust" source, in those wavelengths (but not in the shortest wavelength). Consideration of this anomalous "fourth measurement" which happened to include the Barbarossa system, would cause, by the simplest calculation, sqr((1+1+1+25 - (1+1+1+5)^2/4)/3)/2 = 100% uncertainty of the result (vs. unreported uncertainty in the original catalog and, in the "F" catalog, 29-31% uncertainty at the each of the three longest wavelengths).

The part due to Frey & Freya (three longest wavelengths) is fairly flat, and it is symmetrical. Such symmetry would be obtained at a Planck temperature of 55K (my estimate by two guesses and then one first-order interpolation). If Barbarossa itself were a blackbody at this effective temperature, it would emit 30x too much radiation (assuming all of the 4 IRAS measurements detected it). Really, it is Frey and Freya (1/8 the surface area) which emit, because Barbarossa is cold. Also, they were only detected in one of the four measurements; the other three measurements of them, not only were irrelevant, but were ignored because they didn't happen to be near another, unmoving, infrared source at those times. Thus 1/8 * 1/4 = 1/32.

Neptune's 30% optical albedo suggests, perhaps, a 25% infrared albedo (not 0% as for a blackbody) for Frey & Freya. To keep the same power output would require adjustment of the temperature upwards to 59K, which happens to be the effective Planck temperature of Neptune.


Reply author: nemesis
Replied on: 03/07/2007 10:25:30
Message:

Joe, how do you explain that Barbarossa, being larger, has no internal heat, while its moons Frey and Freya do?


Reply author: Joe Keller
Replied on: 03/07/2007 15:41:03
Message:

quote:
Originally posted by nemesis

Joe, how do you explain that Barbarossa, being larger, has no internal heat, while its moons Frey and Freya do?



Whatever causes Neptune to have much more internal heat than Uranus, might cause Frey & Freya to have much more internal heat than Barbarossa. Furthermore Barbarossa and its moons might differ much more in composition, than do Neptune & Uranus. Elsewhere on this messageboard, I've speculated that Uranus' lack of internal heat might be due to its rotation axis lying near the solar system's orbital plane.

I was talking with my father about this last night. He commented that the orbital plane of Barbarossa's moons, by analogy with Jupiter, Saturn & Uranus, might reveal Barbarossa's rotational axis.

The long axis of my "PM ellipse" is perpendicular to Barbarossa's sky path. This hints that Barbarossa's aliasing moons' orbits are perpendicular to Barbarossa's orbit.


Reply author: Joe Keller
Replied on: 03/07/2007 21:39:01
Message:

I found another VizieR digitized plate, which does *not* show Barbarossa nor its moon, near Object #3 (see above). I first saw Barbarossa and its moon on this VizieR plate:

SERC.ER.DSS2.713 (label in upper left corner, superimposed on the plate itself)

late the night of March 4, and announced the information (here on Dr. Van Flandern's www.metaresearch.org messageboard, first) March 5. (Today is March 7, now approx 8:39 PM CST.) Barbarossa is the streak SW of the catalog position of Object #3 (USNO-B 0820-0274026); its moon is to the W.

Within the last hour I have obtained another Aladin plate which shows the same main stars, including some of apparent faintness similar to Barbarossa and its moon, but which shows neither Barbarossa nor its moon. That is by far the biggest discrepancy between the two plates. I've already shown the plates side by side, to two undergraduates here at Iowa State Univ. in the downstairs multimedia room (a Black man studying economics sitting to my right and his White friend who had stopped by to chat). It is this VizieR plate:

DSS2.O.POSSI (similar upper left label superimposed on plate)

I obtained the first plate again as I always have, by accessing Aladin directly through VizieR (Strasbourg), entering the exact USNO-B coordinates of Object #3, using the default "0" radius search, and selecting the top choice, "ER (Optical R)". I obtained the second plate by searching USNO-B1.0 for the exact coordinates of Object #3, with a 10" radius, then clicking the instruction in the lower left corner, telling the server to find an Aladin plate of my object.


Reply author: Joe Keller
Replied on: 03/08/2007 00:42:18
Message:

Regarding Object #3, there are three interesting nearby objects which seem to be cataloged:

F1. USNO-B 0820-0274021 (catalog mags R19.3, I17.8, B invalid).

F2. USNO-B 0820-0274037 (catalog mags R17.8-19.0, I18.2)

F3. USNO-B 0820-0274036 (catalog mags R19.3, I18.7, B invalid)

The "Optical R" plate on which Barbarossa appears, shows these with F2 somewhat dimmer than Barbarossa. F3 is dimmer yet. F3 is similar in brightness, to the above "moon" of Barbarossa which I henceforth call F1.

On the Aladin "Optical B" plate option, F1-3 all are visible though F1, at least, has moved, to about 0.08s (1") W. Also, F1 is faint. Barbarossa, formerly brighter than any of F1-3, is absent from this plate.

On "Optical I", Barbarossa is gone. F2 & F3 are definite though faint. F1 is stronger than it is on "Optical R".

On "IRJ", all are gone except F1, which is as strong as ever. "IRH" & "IRK" also show only F1, which is relatively bright on these and rivals nearby brighter stars.

Only Barbarossa makes a streak. Only Barbarossa is limited to only one of the seven plates I've viewed tonight (F1-F3 appear on at least the Optical R, Optical B, and Optical I plates).

I also viewed the IRAS 12, 25, 60 & 100 micron plates. On these, nothing immediately relevant was obvious.

The longest of these IR wavebands, "IRK", corresponds to 1250-1500K. Since F1 appears on six of the seven plates viewed, it's probably outside our solar system, perhaps a star with much "warm dust".

Barbarossa, found near the coordinates of Object #3 (USNO-B 0820-0274026) must be either an asteroid, a Kuiper belt object, or something stranger within our solar system (no nebula or external galaxy would be relatively strong in Red yet very weak in Blue and in optical and all other IR). Its short track argues for c. 330 AU distance, assuming an all-night exposure of a sky patch near opposition.


Reply author: Joe Keller
Replied on: 03/08/2007 11:45:07
Message:

The elliptical orbits which the "PM ellipse" suggests for Barbarossa's true moons, would bias sightings toward the parts of their orbits where they spend more time and make shorter photographic trails. In any case, the PM ellipse suggests that Frey's and Freya's orbital speeds around Barbarossa are faster than Barbarossa's orbital speed around the sun. This would make Frey & Freya difficult to detect on time exposures.


Reply author: nemesis
Replied on: 03/08/2007 12:10:25
Message:

"Barbarossa, found near the coordinates of Object #3 (USNO-B 0820-0274026) must be either an asteroid, a Kuiper belt object, or something stranger within our solar system (no nebula or external galaxy would be relatively strong in Red yet very weak in Blue and in optical and all other IR)."
Joe, how can Barbarossa be an "asteroid" in the commonly accepted meaning of the term, but have moons large enough to generate internal heat? From your earlier posts, I thought you felt evidence indicated it was a giant planet or brown dwarf.


Reply author: Joe Keller
Replied on: 03/08/2007 12:57:51
Message:

I estimate from the Aladin SERC.ER.DSS2.713 "Optical Red" plate, that Barbarossa's track is 2" to 5" long, likeliest 3". This would correspond to a seven-hour exposure, for this 330 AU - distant object's apparent angular speed at opposition.

The expected 26-degree track slope isn't obvious, but part of the slope could be obscured by limited resolution, and part by the appearance of the bright red side (beard) of Barbarossa, as Barbarossa rotates left-handed, with period roughly one Earth day, with rotation axis parallel to its orbital progression. Barbarossa's apparent diameter of 0.4" would allow Barbarossa's rotation to mask up to arctan(0.4/2) = 11 degrees of the slope of its track. If Barbarossa progressed from "no beard" to "half beard" during the exposure, then it would average "quarter beard", consistent with the relatively faint Red magnitude of Object #3 vs. the other Objects (+18.57, vs. range 17.41 to 18.84 for Objects #1-8).


Reply author: Joe Keller
Replied on: 03/08/2007 13:51:47
Message:

quote:
Originally posted by nemesis

"Barbarossa, found near the coordinates of Object #3 (USNO-B 0820-0274026) must be either an asteroid, a Kuiper belt object, or something stranger within our solar system (no nebula or external galaxy would be relatively strong in Red yet very weak in Blue and in optical and all other IR)."
Joe, how can Barbarossa be an "asteroid" in the commonly accepted meaning of the term, but have moons large enough to generate internal heat? From your earlier posts, I thought you felt evidence indicated it was a giant planet or brown dwarf.




"...or something stranger within our solar system..."


Reply author: Joe Keller
Replied on: 03/08/2007 16:26:32
Message:

I've looked at the Aladin plates for Objects #5-8. Object #8 almost overlies an arm of a face-on spiral galaxy (like that of Object #4, but much smaller). The galaxy is seen about equally well in Optical R and Optical B. It is sketchier in Optical I, but both of its apparent twin nuclei remain visible. It becomes vague but definite smudges in all of the IR J, K & H plates (I looked at all three plates available of each, for this region).

The dimness of this galaxy in Optical I, gives a caution about misidentifying highly redshifted (face or edge-on) spiral galaxies which might be visible only on the Optical R plate. On the other hand, the face-on spiral galaxy near Object #4, which has much greater apparent diameter (8s = 120", vs. 12"), also easily is visible with detail, though smaller & fainter, on the Optical I, and J, K, & H plates. A distant, apparently small, highly redshifted galaxy would have further advantage on Optical I.

The spot near the catalog position of Object #3, which I identify as Barbarossa, is brighter than F1-3 in Red yet, unlike any of F1-3, disappears in Blue and in Optical Infrared. Barbarossa's spot is about the same size as the spots made by F1-3. A photon is a photon. If the spot is a nebula or external galaxy, the only way it could disappear when the three stars don't, would be if its spectrum somehow were much narrower than that of the three stars, i.e., narrower than a Planck spectrum.

If the spot is an edge-on full-size spiral whose 100,000 lt yr diameter subtends 5", it would be 4 billion lt yr distant with a Hubble redshift of about 30%. This makes it even less likely to disappear from the Optical I plate.

Spiral galaxies usually are fairly symmetrical. Even a barred spiral, if symmetrical, will have its bright bar centered along either the major or minor axis of its apparent ellipse. The asymmetry of Barbarossa's light distribution, argues against its being an edge-on barred spiral galaxy.


Reply author: Joe Keller
Replied on: 03/08/2007 18:54:07
Message:

Help!

What is the date of the "Aladin" plate which shows Barbarossa? (In my earlier post, I tell how to find this plate.) It is labeled:

"SERC.ER.DSS2.713"

Some Aladin plates (usually IR plates) are labeled, on the menu or description, with the day they were made, but others aren't. This must be a plate on file somewhere, maybe at Palomar (Cal Tech and UMass are mentioned).

Sincerely,
Joseph C. Keller, M. D.


Reply author: Joe Keller
Replied on: 03/09/2007 13:15:44
Message:

The USNO-B catalog coordinates for Objects #1-8 (see above) have rms deviation 13.40" from the best-fitting great circle track. Let's consider the deviation, to either side, to be a coordinate in 8-dimensional Euclidean space. The r.m.s. catalog proper motion perpendicular to Barbarossa's track, is roughly 400mas/yr; x 30 yrs for the typical epoch, = 12". If regaining the originally observed positions, by undoing the catalog's automatic Proper Motion correction, merely adds random error to the points which I had selected to be near some line, then the radius of the error ball in 8-dimensional space, should increase to sqrt(13.40^2 + 12^2) = 18". The actual 12.16" r.m.s. deviation error ball obtained, is only p = (12.16/18)^8 = 0.043 times that volume.


Reply author: Joe Keller
Replied on: 03/09/2007 13:44:21
Message:

Earth's ecliptic cuts Earth's equator at 23.5deg. From its track, one sees that Barbarossa's orbital plane cuts Earth's equator at 27.4deg. Also, Barbarossa's descending node is 26.4deg west of Earth's.

If the IRAS measurements of Source #1 (IRAS 11102-0701 or F11102-0701) were at quadrature, June 9, 1983, then the ecliptic was 27.4 * sin(90-12+26.4) - 23.5 * sin(90-12) = 3.55deg N of Barbarossa's track. When Barbarossa was Object #3, the ecliptic was 27.4 * sin(15.9) + 23.5 * sin(10.5) = 11.8deg N. This would move Barbarossa 1/329.5 * (cos(26.5) * 11.8deg - 3.55deg) = 77", perpendicularly to his track, northward, for the IRAS measurement.

I infer from Cal Tech's IRAS webpage, chapter:

"ISSA Explanatory Supplement
I. Introduction
B. The IRAS Survey" (2nd par.) and

"IRAS Explanatory Supplement
I. Introduction
C. Overview of Infrared Sky" (Fig. I.C.1)

that there were two measurements made at quadrature, June 9, (+/-)10days. (Barbarossa was in the 28% of the sky which did not get a third measurement.)

Extrapolating from the moderately nearby Object #1 (at -7.50" Decl per 1s RA), Source #1 lay 83" (perpendicularly) from Barbarossa's track, to the N. So, the correction for Earth parallax implies that Source #1 is only 6" from Barbarossa's track.

Source #2 (a galaxy) was, on the other hand, 37" on the wrong side of the track. This worsens to 37+77=114" with correction.


Reply author: Joe Keller
Replied on: 03/09/2007 18:47:18
Message:

The DSS-1 plate does not show Barbarossa (i.e., Object #3). It does show the stars F1-F3. I found the DSS-1 plate by searching for the coordinates of Object #3, in the "ESO Digitized Sky Survey" on the website "archive.eso.org". Apparently the DSS-1 plate is in White light, not Blue or Red. This confirms that Barbarossa really is a wanderer, and not a narrow-spectrum nebula (if such exist). (DSS stands for Digitized Sky Survey.)

This ESO website also allowed me to confirm that Barbarossa *is* on the DSS-2Red plate, but not on the DSS-2Infrared plate. (The DSS-2Blue plate was unavailable).

I still don't know the date of the DSS-2Red plate. Can anyone help?

I looked at the Aladin plates, including of course the IRAS images, for my IRAS Source #1. There was a faint but definite 12-micron source centered, according to the pixels, within 10" of the coordinates given. There was a rather bright star very near there, which, I theorize, was confounded with Barbarossa (see above).


Reply author: Joe Keller
Replied on: 03/09/2007 20:23:11
Message:

The documentation on the ESO website (above) says that the DSS-2Red plate is from the

"SERC Southern Sky Survey and from the SERC J Equatorial extension"

and that these were 1-hr exposures (a few were shorter). Disregarding a body's own orbital motion, a 3" track at opposition, is consistent with a 6-hr exposure for a body 330 AU from the sun, or a 1-hr exposure for a body 56 AU from the sun. Assuming the same 7% albedo, the diameter in the latter case would be 2300 mi. Pluto and Sedna were well away.

I still think Barbarossa is near 330 AU distant, and that this streak, only a few arcseconds away from the USNO-B coordinates of Object #3, is Barbarossa. The ESO DSS-2Red image shows a bulge on the south side of Barbarossa, which might signify a moon. Frey and/or Freya, at a few arcsec apparent distance and at suitable angles, could cause the horizontal and vertical asymmetry of Barbarossa's image. They also could alter the orientation, and increase the length, of the image's long axis.


Reply author: Joe Keller
Replied on: 03/09/2007 22:56:15
Message:

So far, I've discovered that the DSS-2 Red plates (sometimes called "J Optical Red") were made with the ESO Schmidt telescope in La Silla, which was decommissioned in Dec. 1998.


Reply author: Joe Keller
Replied on: 03/10/2007 13:46:38
Message:

Learning yesterday that the exposures were 1 hr, requires me today to revert somewhat to my first theory of the Frey & Freya orbits. At 330 AU, a 1 hr exposure has only a 0.5" track due to Earth's motion at opposition. Most of the photographic track is due to the apparent distance between the cold brown dwarf Barbarossa, and its double planet Frey & Freya, because these three objects are not resolved.

Frey & Freya formed by the same process as Earth and Luna. As I posted on Dr. Van Flandern's messageboard March 6 last year, this process causes an initial relative orbital speed (reduced mass model) equal to one of the main Tifft periods, 2.7 km/s. They are close enough that they came to maintain the same face toward each other.

Ganymede-like, their center of mass has a one-week orbit about the 0.016 solar mass Barbarossa, consistent with the needed 1,500,000 mile, apparent 11" circular orbital radius which, at opposition, allows frequent aliasing of Barbarossa by one moon or the other, a day before or after, at either end of the orbit. The discrepancy (imperfection) in aliasing, attributed to a year's stellar motion and converted by the USNO computer program into a "Proper Motion", will lie at either end of a narrow apparent ellipse, where Frey/Freya are close to Barbarossa's future/past position, and their dwell time is long. The plot of the discrepancies ("Proper Motions") should resemble the central part of a parabola. Of Objects #1-8, only Object #6 is grossly inconsistent with this: maybe somehow its "Proper Motion" is spurious. The "Frey minus Barbarossa" discrepancy is reversed in both direction and time, at the other end of the orbit, so both ends of the circular, inclined, apparently bilaterally symmetrical orbit give apparent "PMs" on the same rough parabola opening to the east with axis approximately parallel to Barbarossa's track. The orbital radius about Barbarossa is somewhat longer than Barbarossa's daily travel at opposition, so the smaller moon, Freya, farther from the Frey/Freya center of mass, aliases oftener, and closer. Accordingly, on the "PM ellipse" (omitting Object #6) Frey aliased only twice alone, Freya four times alone, and both simultaneously once. The two most distant aliasings, were those of Frey, and this was significant (see above) by Student's t-test.

The orbital speed of Frey/Freya's center of mass, is 25 km/s. Their relative orbital speed about each other, originally the Tifft period of 2.7 km/s (1.35 km/s each, in a crude equal-mass approximation) has increased (roughly, if their masses are equal) to 4 km/s (2 km/s each) because of braking energy loss, associated with tidal heating by Barbarossa (Frey & Freya keep the same face toward each other but not toward Barbarossa). Evidently this tidal heating has maintained Frey's & Freya's surface temperatures at 55K, or slightly higher, for 4.6 billion years, explaining my IRAS Source #1 above. Assuming equal masses, each 4x Earth, they would be 120,000 miles apart on center (up to 1" apparent) with an orbital period about each other of 3.5 days, for 2:1 resonance with the orbital period about Barbarossa.

Most of the lateral spread, and some of the longitudinal spread, of the streak on DSS2.713, is due to Earth's atmosphere ("one arcsecond seeing"). Another 0.5" longitudinal spread is due to Earth's orbital motion. For Frey/Freya this is sometimes (but not for this streak) doubled, when their orbital motion about Barbarossa is antiparallel to Earth's motion. The longitudinal spread is mainly due to the light from Frey + Freya, merged with the light from Barbarossa but apparently displaced, in this case ~ 3" northeast of Barbarossa. Frey+Freya combined are a +19.5 object. In one day, Frey/Freya orbit 360/7=51deg; arcsin(3/11)+51=67deg, appropriate for aliasing in the first quadrant, as observed for Object #3, with its (+) PMs in both RA & Decl.


Reply author: nemesis
Replied on: 03/10/2007 14:22:26
Message:

Dr. Keller, is it possible Hubble has imaged the Barbarossa system?


Reply author: Joe Keller
Replied on: 03/11/2007 16:07:54
Message:

quote:
Originally posted by nemesis

Dr. Keller, is it possible Hubble has imaged the Barbarossa system?



Thanks for your suggestion! I found a search form on:

http://archive.stsci.edu


Reply author: Joe Keller
Replied on: 03/11/2007 17:26:04
Message:

My rough estimate of the tidal distortion of Frey/Freya due to Barbarossa (using the preceding model) indicates that for reasonable planetary interior temperature and pressure, the viscosity of molecular hydrogen would suffice to give 55K blackbody surface temperature. (Tidal distortion implies slippage of one layer of fluid over another for distances comparable to that distortion, and over an area of interface, comparable to the area of the planet.)

If the brightening of the streak (visible using blown-up Aladin pixels) were due to planetary rotation during this 1hr exposure, the rotation would be near Poincare's disintegration limit. Also, Barbarossa would need an Iapetus-like asymmetry of albedo. (The short trail at opposition, yet bright apparent magnitude, show that Barbarossa is large enough that it must be spherical with mainly gravitational cohesion.) Such asymmetry of flux can be caused instead by other, non-resolved point sources, i.e., moon(s).


Reply author: Joe Keller
Replied on: 03/12/2007 17:24:39
Message:

My pixel analysis of Barbarossa's DSS-2 Red image, indicates a moon (or possibly unresolved pair of moons, i.e., Frey+Freya) about 2.5" (+/-) 0.1" E and 0.2" (+/-) 0.1" N of Barbarossa. It indicates that these moon(s) are 1.2 (+/-) 0.1 magnitudes dimmer than Barbarossa. If the brighter of the two Red USNO-B catalog magnitudes of my Object #3, is that of Barbarossa, and the dimmer that of the aliasing moon (Freya), then Barbarossa is +18.57. So, +18.57 + 1.20 = +19.77 indicates Frey+Freya as for Object #8 (which gave +19.65 for, apparently, Frey+Freya).

My pixel analysis also indicates a third moon 1.7" (+/-) 0.5" W and 0.4" (+/-) 0.5" N of Barbarossa. It is 1.6 (+/-) 0.2 magnitudes dimmer than Barbarossa. So, +18.57 + 1.60 = +20.17, suggests that there is a third moon, separate from the others, probably with a smaller orbit, and with about the same magnitude as Frey.

I considered the 4x7 block of 1" square pixels on which Barbarossa appears, and the nine grayscale levels of those pixels. "One arcsecond [atmospheric] seeing" would give a standard deviation of about 1" to the photons. So, 3" away from the main source would be 4.5 natural log units dimmer, roughly consistent with 0.5 natural log units per grayscale level.

First trying a single source, I found that for best overall least-squares fit between the predicted and observed logarithm of the flux detected in each pixel, the source did not lie in the interior of the darkest pixel. However long the streak, made by a constant-intensity point source, the midpoint of the streak must lie within the darkest pixel. Since a single source of such diameter hardly can (see above) rotate fast enough or be asymmetrical enough, there must be multiple sources.

I then adopted a compromise position for Barbarossa, 0.2" E and 0.1" N of the center of the darkest pixel. A rough preliminary search assuming two objects, had placed the moon(s) 3" E & 1" N of Barbarossa. So, my final search, which took almost an hour on a Pentium processor (the first time I've needed anything faster than a 486, for any of my work on this messageboard), adopted a finer grid concentrated there for one moon, while looking for a second moon on a coarser grid everywhere else.


Reply author: Joe Keller
Replied on: 03/12/2007 19:54:16
Message:

The viscosity of [molecular] hydrogen gas at one atm and 1000K, is 200 micropoise (Handbook of Chemistry & Physics, 44th ed.). Branley's Astronomy, & TP Snow's The Dynamic Universe (1983), give 3,000,000 atm for the maximum pressure inside Jupiter's liquid H2 outer mantle; Snow gives 10,000K temperature within this mantle (the liquid H2 mantle is thought to extend 1/3 of the way to the center, hence comprise 2/3 of the volume). If the typical pressure inside the mantle is proportional to the (planetary density)^2 * (planetary radius)^4, then Frey or Freya, whose hydrogen would have slightly lower density and whose radii are about 1/4 Jupiter's, would have typical mantle pressures < 3*10^6 / 4^4; perhaps 5000 atm.

Empirically from the Handbook's table for hydrogen, viscosity increases as temperature^(2/3). Tidal heat generated will be proportional to viscosity, so blackbody surface temperature will increase as (internal) temperature^(1/6): it isn't a sensitive function of internal temperature. Assuming a 1000K typical mantle temperature (much less than the estimated mantle temperature of Jupiter), the viscosity of the molecular hydrogen in the mantle of Frey/Freya, in the (rough) ideal gas approximation, is 0.000200 * 5000 = 1 poise. This is the viscosity of light machine oil at room temperature and 1 atm pressure.


Reply author: Joe Keller
Replied on: 03/12/2007 20:04:27
Message:

I still don't know the date of that plate (Object #3) showing Barbarossa. Querying some Aladin and ESO archive regions to one side or the other, so far has yielded no relevant dates.

I emailed the ESO help team, asking them, but they told me they were too busy to do anything except give me three (irrelevant) web links. Then today I emailed six astronomy faculty members at the U. of Strasbourg, including photos of the ESO plates with coordinates.

I've still received no response from the USNO (three FAXs & two postal letters). The chief of the Iowa State Univ. observatory (24" telescope in rural area) refused to look and didn't want the coordinates or the photos.


Reply author: Stoat
Replied on: 03/13/2007 04:11:48
Message:

Bradford robotic telescope images, the first is at 11 07 24, -6 38 50
next at 11 06 02. -6 28 27

Go to the Bradford web site and look at the latest jobs, these are nem2 and nem3. Perhaps if we decide on an exposure we can put the same job up a few times and then do a blink comparison in photoshop.

[img]http://farm1.static.flickr.com/168/419796237_87c8649184.jpg[/img]

[img]http://farm1.static.flickr.com/152/419796234_3e70b5c577.jpg[/img]


Reply author: nemesis
Replied on: 03/13/2007 10:29:08
Message:

Dr. Keller, it would seem that TVF may be able to wield some influence with the USNO, given his long tenure there.


Reply author: Joe Keller
Replied on: 03/13/2007 14:15:19
Message:

quote:
Originally posted by nemesis

Dr. Keller, it would seem that TVF may be able to wield some influence with the USNO, given his long tenure there.



Thanks for making this point! My retired Army officer friend warned me that many people usually are clamoring for the attention of the commander (to whom one of my letters to the USNO was addressed).


Reply author: Joe Keller
Replied on: 03/13/2007 14:32:03
Message:

[quote]Originally posted by Stoat

Bradford robotic telescope images, the first is at 11 07 24, -6 38 50
next at 11 06 02. -6 28 27

Go to the Bradford web site and look at the latest jobs, these are nem2 and nem3. ..."

How does one look at the jobs? I now have a Bradford account, but it's only passive, because I didn't see how to ask approval for submitting jobs. My "help" email to them went unanswered.

All the stars in these photos drift slightly which isn't a fatal defect, but one should be aware of it. The Bradford website says that a recent drift problem has been fixed.

My latest best guess for the coordinates:

RA 11h 09m 00s Decl -6deg 51' 00" (and the slope of the track is -7.5' Decl per 1m RA)

Good luck!


Reply author: Joe Keller
Replied on: 03/13/2007 14:37:34
Message:

I received no response overnight from any of the six U. of Strasbourg astronomy faculty I emailed yesterday.


Reply author: Joe Keller
Replied on: 03/13/2007 15:11:37
Message:

Let's assume that Barbarossa's moons (probably really planets!) Frey & Freya have equal mass and orbit a common center of gravity, max 0.8" apart. Let's assume also that their orbit about Barbarossa is circular, almost coplanar with Barbarossa's orbit about the sun, and only tilted toward or away from the sun. If the alleged Proper Motions in the USNO-B catalog, of Objects #1-5,7,8 (#6 excluded as an outlier) really are due to the separation between Barbarossa's position, say, yesterday, and points along the end of a narrow ellipse, say, today, mistakenly attributed to one year's motion, then they should fall on a parabola whose axis is Barbarossa's track. I found the least-squares best-fitting such parabola.

The semiminor axis of the apparent orbital ellipse, is the geometric mean of the semimajor axis, and the radius of curvature at the vertex of the above approximate parabola. With or without a rough correction for the nonzero distance between the moons, I got 14 or 11deg tilt, resp. The position of the main (eastward) moon(s) in the above pixel analysis, gives 7.1deg. This agrees, because really, there also will be a small tilt in the other direction, which brings a flatter part of the ellipse (i.e., not the very end) into alignment, reduces the curvature of the best parabola through the PMs, and causes overestimation of the semiminor axis.


Reply author: Joe Keller
Replied on: 03/13/2007 18:05:11
Message:

I just now separately emailed five individuals in the U.S., associated with the IAU surveys dept., asking the date of "SERC.ER.DSS2.713", from the plate that shows Barbarossa. A few minutes ago, I found a chart online indicating that the "SERC-ER" plates were made 1984-1998. Previously elsewhere online I'd learned that they were made at La Silla ("the Saddle", in Chile, at 2400m altitude) using a Schmidt telescope which was decommissioned in Dec. 1998.

This, together with the June 1983 IRAS observation (including Earth parallax correction) of Source #1 above, implies a prograde orbit for Barbarossa, of period not more than 4300 yr, assuming a circular orbit. The measurement (COBE error bars) and theoretical (uncataloged gravitating bodies near 50-60 A.U.) uncertainties of the CMB dipole progression, imply that a slow prograde orbit for Barbarossa, does not entirely refute my CMB theory.

The outer planetary resonance discrepancies are more consistent with pro- than with retrograde orbit for Barbarossa. Uranus' frequency is slightly less than twice Neptune's. Jupiter's frequency is slightly less than thrice Saturn's.

Freya is dimming linearly with time. Barbarossa (period, 4430 yr according to the average of the relevant outer planetary resonance discrepancies; see above) is slightly closer than its average 270 A.U. distance, but moving away in a moderately elliptical orbit.

Using the latest possible, Dec. 1998, date for Object #3 (probably older telescopes would be relegated to sky surveys), then Objects #4, #6, & #8 must be discarded. Objects #4 & #8 were the ones which overlay galaxies. Furthermore Object #8 is the only one of the eight, whose dimmer magnitude would need to be explained as the sum of Frey's & Freya's fluxes. Object #6 is the only one excluded from the "PM parabola" above as an outlier.

The remaining Objects #5, #1, IRAS Source #1, Objects #7, #2 & #3 (listed in increasing RA) span 24 yrs. This is likely, especially if the USNO-B plates span somewhat less than the stated 50 yrs. in this region, or are sparse for early years.

If Object #3's date is the earliest possible, Jan. 1984, then Barbarossa would have moved almost 60 degrees by now. The probability density per unit distance on the track, is much higher near positions implied by dates close to 1998, i.e., near the least possible RA. Later dates are more consistent, with Barbarossa's apparent magnitude, and with a more circular orbit. The least possible RA, with corresponding Decl, is about

RA 11h 20m 45s Decl -8deg 20'

I suggest searching from this point eastward, using the slope

-7.4' Decl per 1m RA.


Reply author: Joe Keller
Replied on: 03/13/2007 21:53:45
Message:

I searched along the track, in the (above) Hubble archive, using their default 3' radius disks, 3' apart, but found no Hubble images. I did find images by specifying a much larger radius, but I found no coordinates on those, nor any way to zoom.


Reply author: Stoat
Replied on: 03/14/2007 04:08:39
Message:

With the Bradford, just sign in and there's a "your menu " button left of screen. Click that and there is a number of buttons to submit job requests.


Reply author: Joe Keller
Replied on: 03/14/2007 15:46:43
Message:

quote:
Originally posted by Stoat

With the Bradford, just sign in and there's a "your menu " button left of screen. Click that and there is a number of buttons to submit job requests.



It tells me I'm not authorized to submit, that I have the wrong kind of status. [Addendum] I see that one must "verify" one's account. I'm waiting for the verification code to come: apparently this can take days. Also I notice (March 19) from searching the "jobs" list for 5400 sec jobs, that Stoat's March 15 job is still "awaiting scheduling".


Reply author: Joe Keller
Replied on: 03/14/2007 15:49:59
Message:

"> ...if you have the image as a FITS file, then I would expect it
>to have the observing date in the header.
>
>Cheers,
>
> - Arnold Rots"

I asked about 20 relevant people for help getting the date of the plate. Only two tried to help. Of the two suggestions, the above was the only one that worked.

*********

Dear Dr. Rots,

Your suggestion worked. Here's how to get the date of an archive.eso plate:

1. Google "FITSview", then download the freeware of that name. It was fast & easy.

2. Open another window for archive.eso, then when you search for the image from ESO, choose the (default) FITS option.

3. Save the image FITS file to your desktop.

4. Go to the FITSview page which should be still open on your browser, click "File" then "Open". When you see the image, then click "File" then "Image info". Then you'll see the header.

Thanks again. You might want to save these instructions.

Sincerely,
Joe Keller


Reply author: Joe Keller
Replied on: 03/14/2007 17:22:35
Message:

The date of the (DSS-2 Red) plate showing Barbarossa, is January 31, 1987: the last day of the month of Janus. If IRAS Source #1 is accepted also, this implies, for a circular orbit, a distance of roughly 135 A.U. and a prograde orbital period of 1700 yr. (Non-negligible corrections for Earth parallax occur for both the IRAS and the DSS-2 observations.)

In this scenario, Barbarossa would need a diameter of 10,000-20,000 mi. (and its moons half that)(depending on albedo) to be consistent with the +18.6 Red magnitude which both this author (by comparison with the nearby cataloged stars on the plate) and the USNO determined. The streak should be about 1" long considering the direction of Earth's orbital motion in January, vs. 3" observed.

The planet remains, of course, spherical, gravitationally bound, and unable (Poincare instability) to rotate far during the 1hr exposure. That the midpoint of the best-fitting streak does not overlie the darkest pixel, implies a moon as before, which also accounts for most of the streak length. The present position would be roughly

RA 11h 36m 00s Decl -10deg 10'


Reply author: nemesis
Replied on: 03/14/2007 21:46:33
Message:

Joe, I wonder if there are any telescopes in private hands with enough power to image Barbarossa. The problem with the big observatories, including USNO, is that it's so hard to get time, even for those in the "club", so to speak.


Reply author: Joe Keller
Replied on: 03/15/2007 13:09:37
Message:

quote:
Originally posted by nemesis

Joe, I wonder if there are any telescopes in private hands with enough power to image Barbarossa. The problem with the big observatories, including USNO, is that it's so hard to get time, even for those in the "club", so to speak.



International competition helps (e.g., Galle, who listened to Leverrier, trumped Airy, who didn't listen to Adams, re Neptune) but "One World" is lessening that. There's groupthink, that united, they can define reality, somewhat as Canute sarcastically said, but unlike Canute, they really believe the tide doesn't come in if they say it doesn't (they have $100 billion in tax dollars to back them up). Most "wait for someone [above them in the pecking order] to tell them what to think". I'll send an email to the Lowell Observatory today, but private observatories, like private colleges and foundations, also get sucked into the corrupting bureaucratization process which government funding brings.

In defense of the astronomy establishment, they do have a love-hate relationship with their fans ("can't live with 'em, can't live without 'em") rather like actors. Without fans they'd get no money, so they court the public in the newspapers & educational TV, but then they have to hide from all the people who want to tell them the moon is made of green cheese.

Tombaugh searched the entire ecliptic down to +17 with a 13" telescope and 1930 photographic technology. It was at 7200 ft. in N. Arizona before light pollution, smog, and "wide persistent contrails" which are usually a gross problem for me because I live under a jet route through central Iowa.

Two experienced amateur astronomers (one local and the other very active) told me their 24" could do it, but both declined to look; neither offered to let me use their equipment. Iowa State Univ. has a 24" but their observatory chief also declined to look.

So far, the best observing program we have is Stoat's, with the 14" on Tenerife at 7800 ft; this gives 14x14' photos with, I gather from the online log, up to 3 minute (180000 ms) exposures. Conveniently, the Aladin images are cut to about this size. The 48" Schmidt at La Silla, also at 7800 ft, used 1hr exposure to photograph Barbarossa, but those plates were 6.5x6.5deg:

60*(48/14*14/390)^2= 0.9 minute would give the equivalent exposure for Bradford at Tenerife

As many of us as possible should follow Stoat, asking him for specific directions if necessary, and order views along the Barbarossa track with 0.9 minute exposures, asking Bradford to post it publicly. This exposure (54000 ms) will be the clue that it's "one of ours" for anyone scanning the online Bradford jobs. Then compare the images to Aladin's (their "Optical Red" is from La Silla), by opening two windows on your browser. If anyone here knows that new technology since 1987, requires a much different exposure, please post the information into this thread.


Reply author: Stoat
Replied on: 03/15/2007 14:51:13
Message:

Okay, I've put a job up on the Bradford for 11 9 00, -6 51 00. at 54000 ms. I'll post when they get it done, and we can then put the same job in again and compare them by blinking the two, or more, plates.


Reply author: Joe Keller
Replied on: 03/15/2007 16:00:27
Message:

quote:
Originally posted by Stoat

Okay, I've put a job up on the Bradford for 11 9 00, -6 51 00. at 54000 ms. I'll post when they get it done, and we can then put the same job in again and compare them by blinking the two, or more, plates.



Thanks!


Reply author: nemesis
Replied on: 03/15/2007 16:32:04
Message:

"Two experienced amateur astronomers (one local and the other very active) told me their 24" could do it, but both declined to look; neither offered to let me use their equipment. Iowa State Univ. has a 24" but their observatory chief also declined to look."
This seems very strange. Do you suppose a suggestion that they could be co-discoverers would motivate them? This could be one of the greatest astronomical discoveries in decades.


Reply author: Joe Keller
Replied on: 03/15/2007 17:43:48
Message:

A professional observational astronomer told me Saturday, by phone, that the USNO-B magnitudes at the faint end, aren't accurate enough to produce significant clustering like the "Freya" and "Frey" "dim Red" magnitudes. He offered no numbers.

If the standard deviation of the magnitude measurements is 0.1, clustering such as "Freya" (five of the dimmer Object #1-8 mags were in the interval [20.60,20.73]) is possible. If the s.d. is 0.2, it's unlikely. Red dwarfs are variables: this masks the true error in this range; it can't be determined simply from R1 vs. R2 in a sample.

Pixel analysis (above) suggests moon(s), but my original reason for suspecting moons, was the clustering of the dimmer magnitudes. Alternatively, the clustering might be an artifact of the computer's effort to find images of Barbarossa on a Proper Motion path on all seven other plates.

After generously estimating (above) 40 plates of each region, I learned the typical 6.5x6.5deg size of the (almost 8000) plates scanned by the USNO. This reduces the estimate (the ecliptic might be slightly favored) to exactly 8: too few to produce the theorized aliasing more than once, if that. However, Barbarossa's great circle sky track and its associated statistical significances, remains.

Barring any longitudinally interlaced sky survey program, Barbarossa will appear on exactly 8 plates. It does. These plates give Objects #1-8.

Viewing the Aladin plates, I saw that the dimmest reliable stars had USNO-B Red mags of about +19.5. Maybe all mags dimmer than +19.5, arise from averaging, something definite < +19.5, with vague things > +19.5. For PMs > 80 or < -80, R1-R2 pairs both > +19.5 (N=782 in a 1 deg redius), are no more common than pairs with one > & one < (N=779). Not a single object in a 1 deg radius had R1,R2 > 20.90. Some of the stars I found with Red mags < 18.99 & > 19.50, would be variable red dwarfs, especially nearer +19.50, but most are presumably instances where the computer averaged a star, with dim blobs, of practically zero flux and almost infinite magnitude. According to the explanatory 2003 Astronomical Journal article, only location, not magnitude, was a criterion for identification.

An +18.75 star averaged with one dim blob would give +19.5. The maximal six dim blobs (one straightforward identification, matched, for the requisite second ID, to the other seven plates on which the real star is picked once and dim blobs six times) would give +20.9: this is where the almost uniform distribution of "dim Red" mags ended.

When the computer tried to analyze Barbarossa, the only way to get a second Red mag, was to associate the Barbarossa found on one plate, with at least seven dim blobs on the other plates. Trying to find Proper Motion, the computer found seven (or more) dim blobs on the other seven plates, making some semblance of a Proper Motion path. The "PM" so determined, would resemble the chi-square distribution with at least 7 degrees of freedom: its peak is sharp enough to put these pseudo-"PMs" on a rough circle. That Barbarossa is a streak (perhaps 1.5" due to Earth's motion and another 3" due to Frey & Freya) somehow distorts that circle into the "PM ellipse" discussed above.

Let Barbarossa, together with its non-resolved separate moons orbiting at, perhaps, 200,000 mi, have magnitude +17.41, like Object #1. Arithmetically averaged (i.e., by flux) with one dim blob, this becomes +18.16. Object #2 has +18.17 and Object #4 +18.03. Averaged with two dim blobs, Barbarossa becomes +18.61. Object #5 has 18.59, Object #7 +18.54 & Object #3 +18.57. Averaged with three dim blobs becomes +18.91. Object #6 has +18.84 & Object #8 +18.80.

The star nearest Barbarossa on the DSS-2 plate, "F1" above, is, by the above pixel analysis, about two magnitudes dimmer than Barbarossa, not one magnitude. So, the USNO-B Red mags of F1 & the somewhat brighter or similar F2&3, are consistent with +17.4 mag, not with the USNO-B catalog's +18.57, nor even with the average +18.1 for Barbarossa. This underestimation by the USNO-B catalog, of Barbarossa's magnitude on Object #3's plate, supports the foregoing theory.

The dimmer "moon" magnitudes arise from averaging the magnitudes of the seven relatively faint, non-Barbarossa objects along Barbarossa's pseudo-PM path. If two of these are real stars of magnitude +19.3, and the other five are "dim blobs" with practically zero flux, then the "moon" magnitude is +20.66, as for the five "Freyas" ([20.60,20.73]). If three of seven are real stars, then the magnitude is +20.22 (two "Freys": [20.22,20.26]). If five of seven are real, the mag is +19.665 (Object #8: 19.65); Object #8 (like Object #4) overlies a galaxy, which facilitates finding more real stars. The Freya sightings are associated with slightly but significantly (Student's t) less PM than are the Frey sightings, because paths with more real stars tend to be the longer ones.

The precision of the statistical findings is due to the USNO computer's effort to force Barbarossa into a "one-size-fits-all" program designed for stars. The eight Objects would span, typically, 7/8*50=44 yr; 15m in 44yr at 26.5deg slope gives, for a circular orbit, period 3780yr (maybe IRAS Source #1 isn't Barbarossa). More stringently, 7m in 16yr from Object #4 (1987) to Object #8 (2003?), gives an upper bound of 2950yr.

Aladin shows me only three, of the eight plates which the USNO surely used. For Object #3, one of these three plates is the plate which shows Barbarossa. Barbarossa must be on two more of the Aladin plates: the only two which I didn't check carefully were those which overlay galaxies (Objects #6 & #8). The dates on the ESO DSS-1 FITS headers, for these locations, might be the dates of the Aladin "Optical Blue" plates, which are labeled DSS1. I'll check these to find the date of either Object #6 or Object #8.


Reply author: Joe Keller
Replied on: 03/15/2007 19:53:37
Message:

"Frey sightings" occur, on average, at smaller RA than "Freya sightings". Both Frey and Freya sightings increase in magnitude (i.e., they dim) with increasing RA; for Freya sightings, this is statistically significant (see above). This could be because toward the galactic plane (smaller RA), the real stars found by the computer program seeking a PM path, are commoner and brighter.

According to the "FITS" headers, all the DSS1 plates for the eight Objects were made in March or, oftener, May, in 1983 or 1984. With parallax correction, the appearance of Barbarossa as Object #2 or Object #7 on the DSS-1 plate would conform to a circular 2000 yr prograde period. Because this plate (same plate for both objects) was made in May, there will be practically no Earth motion streak. There are suspicious pixels at the location of Object #7 (see below).


Reply author: Joe Keller
Replied on: 03/16/2007 21:35:58
Message:

The following is an edited transcript of the workshop held in the Physics and Astronomy Building at The University of Western Ontario on Monday, November 25, 2002, from 1:00 to 5:00 pm.

É

IM  My name is Ian McDiarmid. ÉI was more or less pushed into [space research] by Don Rose, who was doing cosmic ray research at NRC. Don arrived there in 1948 and I joined him as a PDF in 1954. Walter Heikkila at DRTE knew Don was interested in cosmic rays, and he was planning a couple of Aerobee rockets to be flown from Fort Churchill. In 1958 he came and talked to Don and myself and he offered us some space in the nosecones, to do some cosmic ray work. Don was really keen, because he was interested in rockets and of course he was interested in cosmic rays, he'd like to measure them above the atmosphere. I was doing high-energy particle physics at the timeÉ

É

GS  He worked in photogrammetry, and he found these streaks in photographic plates taken from aircraft.

IM  Really? I'd never heard that. É

É

[Ian McDiarmid] And they probably weren't streaks from cosmic rays at all, they were probably something else [laughter]. Because it would be very hard to get cosmic ray streaks in ordinary photographic plates at that altitude. É


Reply author: Joe Keller
Replied on: 03/16/2007 22:03:20
Message:

Roland, Iowa March 16, 2007


Open letter to the Director of the Lowell Observatory

Dear Sir:

Like Prof. Lowell, I studied Mathematics at Harvard College (B. A., cumlaude, Mathematics, 1977). The essential details of my recent work on Prof. Lowell's Planet X are posted, to Dr. Tom Van Flandern's "www.metaresearch.org" messageboard, under the name, "Joe Keller", in the thread "Requiem for Relativity". (I use Dr. Van Flandern's messageboard as an alternative to "ArXiv.org".)

Planet X, which I have named Barbarossa, appears at

RA 11h 18m 03.2s Decl -7deg 58' 46" on the La Silla sky survey Red plate SERC.ER.DSS2.713 dated January 31, 1987. Possibly there is a second appearance of Barbarossa at

RA 11h 14m 36.0s -7deg 32' 17.5" on the Blue plate SERC.J.DSS1.713 dated May 8, 1983.

Assuming a circular orbit and making first order approximations to correct for Earth parallax, Barbarossa has period 2640 yr. and is 191 AU from the sun. Accordingly, the resonances of the orbital periods of the outer planets have discrepancies which advance prograde with periods

Jupiter:Saturn 5:2 2780 yr
Saturn:Neptune 6:1 2180 yr
Jupiter:Uranus 7:1 -5970 = -2985 * 2 yr (retrograde)
Uranus:Neptune 2:1 4380 = 2190 * 2 yr
Saturn:Uranus 3:1 1190 = 2380 / 2 yr.

I discovered Barbarossa on February 15, 2007 as a sequence of statistical artifacts in the USNO-B1.0 catalog. I informed the U. S. Naval Observatory on February 21.

I first saw the La Silla Red image of Barbarossa on March 4, and realized on March 5 that it is Barbarossa. By comparison with the four nearest cataloged stars, Barbarossa's Red magnitude is about +17.3. A 6% Red albedo would imply 46,000 mi diameter. Barbarossa might be either a giant planet or a cold brown dwarf.

I realized yesterday, March 15, that the above La Silla Blue image is Barbarossa, which is dim in Blue. The pattern seen on this Univ. of Strasbourg "Aladin" image depends on one's monitor setting. At its best, it shows Barbarossa as a lean-to adjoining a nearby star with a separate USNO-B catalog number. It shows a moon of Barbarossa's (I've named the largest & next-largest moons, Frey & Freya) as a disjoint dark pixel 3" toward azimuth 245. From my drawing of the best image obtained (Prof. Lowell drew lines on Mars; I draw pixel boxes), I estimate this moon to be 1.7 magnitudes dimmer than Barbarossa.

The Red La Silla image shows no disjoint moon, nor any star near enough to confuse. Thorough computer search found the best fit for three points of light, was to have a moon 1.2 magnitudes dimmer than Barbarossa, 2.5" away at azimuth 275; and another moon 1.6 magnitudes dimmer 2" away at azimuth 75. Thus the Barbarossa system consistently appears parallel to the ecliptic. Furthermore the best fit for one point of light, lay outside the darkest pixel box, indicating either multiple sources or quickly varying magnitude. As a gravitationally bound body subject to Poincare instability, Barbarossa hardly can rotate appreciably during these 1 hr exposures.

In 2002 at a Physics and Astronomy conference, cosmic ray expert Ian McDiarmid disparaged the statement that cosmic rays would be readily detected by ordinary photographic materials onboard airplanes [let alone at 7800 ft at La Silla]. A Kuiper Belt Object would leave a streak of this length, but even then, the magnitude would suggest another Pluto or Sedna.

By great-circle extrapolation, with a rough correction for Earth parallax, Barbarossa's March 10, 2007 position is

RA 11h 27m 10s Decl -9deg 18' 58".

Alternatively, my statistically-derived greatest-likelihood great circle, estimates the Declination at this RA as

Decl -9deg 05' 46" (for RA 11h 27m 10s)

The greatest-likelihood great circle goes through this point with slope -7.35 arcminutes Decl per minute of RA.

Sincerely,
Joseph C. Keller, M. D.


Reply author: Joe Keller
Replied on: 03/17/2007 17:56:44
Message:

Barbarossa and "the Triad"

Obtaining more recent estimates of the orbital periods of Jupiter and Saturn, I found that the discrepancy in the 5:2 resonance, progresses one cycle in 2696 yr. This is practically equal to the approximate 2643 yr period calculated above for Barbarossa, from its sightings as Object #7 & Object #3, assuming a circular orbit. Barbarossa shepherds one point of the Triad (equilateral triangle) formed by the three recurring conjunctions of Jupiter and Saturn around the ecliptic.

On April 17.5, 1981, such a conjunction occurred at 187.15deg heliocentric ecliptic longitude. (If the alternate criterion, closest three-dimensional approach, is used, this becomes 186.65.) By extrapolating the Barbarossa positions associated with Object #7 and Object #3, I found that Barbarossa was at heliocentric ecliptic longitude 172.5 then.

The difference, 187.15-172.5=14.65deg (14.15deg by the alternate criterion), is explained by the orbital eccentricities of Jupiter and Saturn. Roughly, Jupiter is 180deg from perihelion & Saturn 90deg from it. A somewhat more precise first-order calculation shows Saturn a net 8.0deg ahead of Jupiter, when a "mean Saturn" and "mean Jupiter" reach heliocentric ecliptic longitude 172.5. Jupiter has extra catching-up to do.

On the average, with 5:2 resonance, this would occur over 5/3 of the catch-up angle, but Jupiter also is about 9% slow here, Saturn 1% fast, and Jupiter's average speed really is 0.7% too low for 5:2 resonance anyway. So, Jupiter needs 14.3deg to catch up.

Averaging Jupiter & Saturn (their ascending nodes and orbital inclinations are similar), gives 1.9deg inclination to the ecliptic, with ascending node at 107deg ecliptic longitude. Therefore Jupiter's & Saturn's tracks are nearly parallel to the ecliptic here. Barbarossa's projection onto this Jupiter-Saturn average ecliptic, would change the above 14.65, to 14.85. (The difference between true Jupiter-Saturn conjunction, and mere equality of ecliptic longitude, is negligible in either ecliptic system.)

So, Barbarossa's position, extrapolated from its sightings as Object #7 and Object #3, is only 0.55 deg (or 0.05deg by the alternate criterion) west of a mean shepherd position for the 5:2 Jupiter:Saturn resonance. Because there are three such positions, p=0.55*2/(360/3)=0.009 (p=0.0009 for the alternate criterion). More precise calculations might enhance this agreement.


Reply author: Joe Keller
Replied on: 03/19/2007 19:11:37
Message:

According to Jewitt et al, "...Varuna", Nature, a canonical albedo for Kuiper Belt Objects is 0.04 [also the albedo of small comet nuclei, and the darkest albedo known for solar system bodies - JK]. This albedo would give Barbarossa a diameter of 57,000 mi.


Reply author: Joe Keller
Replied on: 03/19/2007 22:19:28
Message:

According to Hainaut et al, Astronomy & Astrophysics 389:641+ (graph), Mars has Blue-Red magnitude = 2.2; Varuna is typical of the reddish type of Kuiper Belt Object, with B-R = 1.6. The only one of the Object #1-#8 catalog magnitudes of Barbarossa, which equals its photographic Red magnitude on the SERC Red plate for Object #3 (+17.3 by comparison with neighboring stars), is the catalog magnitude of Object #1, +17.4. The catalog Blue magnitude of Object #1 is +19.8, for B-R = 2.4.

On the SERC Blue plate for Object #7, two catalog numbers (USNO-B 0824-0279078 & USNO-B 0824-0279077) are given, seemingly for Barbarossa and for a star 2" away, resp. The two objects barely can be separated by looking at pixels. With catalog Red magnitude +17.8 and Blue +19.6, the star, USNO-B ...-77, resembles the "true" Barbarossa's magnitude. Barbarossa, USNO-B ...-78, resembles the typical catalog Barbarossa magnitudes for Objects #1-8.


Reply author: Joe Keller
Replied on: 03/20/2007 00:30:38
Message:

Barbarossa & Nemesis: wheels within wheels?

Above, I show how to calculate that the presumed sightings of Barbarossa follow the mean position of one of the three resonance points of Jupiter & Saturn. According to what seem to be the most accurate available periods for Jupiter and Saturn (11.6821 & 29.458 yr, resp.), this resonance point recurs after 91.0092 ( = 2*45 + 1 + 0.0092) revolutions of Saturn (2680.95 yr)(p=0.009).

Maybe just as Barbarossa shepherds the third resonance of Jupiter & Saturn, Nemesis shepherds the 45th resonance of Barbarossa with that third resonance. In this Pythagorean astronomy, the period of Nemesis would be 2680.95/0.0092 = 290,000 yr. If more accurate periods were known for Jupiter & Saturn, this might become the 26 million yr speculated for Nemesis.


Reply author: Joe Keller
Replied on: 03/20/2007 00:54:14
Message:

Today I redid the above calculations resulting from my theory that the CMB is due to gravity within the solar system. My calculation is exact to first order in the mass of the planet considered, as a fraction of the solar mass. Because none of the orbital radii considered are close to the 52.6 AU barrier, the integrand encountered does not become large or infinite (though this would be only a removable singularity in the integral). Hermite's 7,16,7 rule here is accurate; Simpson's 1,4,1 rule would have been 30% low, for a one-step application.

This more accurate calculation is qualitatively similar to my earlier one. In the effective 11.2 yr between the COBE 4-yr & WMAP 3-yr observations, the CMB dipole retrogressed 0.27(+/-0.22)deg along the heliocentric ecliptic. Without the theoretical effects of N,U,S&J, this would have been 0.94deg retrogression. On the other hand, the prograde motion of Barbarossa would cause 1.53deg progression of the dipole.

Theoretically the dipole should have become a few microK weaker (mainly due to Neptune's approaching opposition to Barbarossa) instead of a few microK stronger between the WMAP 1- & 3-yr results, but this was within error bars. The predicted mass of Barbarossa from this model is 0.0068 solar mass. Barbarossa is 0.7deg S of the CMB dipole in heliocentric ecliptic latitude; 1/3 of this discrepancy is explained by Neptune (the other planets have negligible effect).

For the WMAP 3-yr result, Barbarossa led the CMB dipole by 3.1deg after removing the effect of J,S,U&N. The COBE 4-yr CMB dipole lagged Barbarossa only 3.1 - 2.5 = 0.6 (+/-) 0.2 deg. WMAP tended to make observations nearer quadrature to the sun; COBE nearer opposition. Also, 1/52.6 radian = 1 degree. This suggests that the true CMB dipole is 0.5(+/-)0.1 deg N of Barbarossa and < 0.6 (+/-)0.2 deg W of Barbarossa.


Reply author: Joe Keller
Replied on: 03/20/2007 16:58:58
Message:

I'm now estimating 11h 27m 9.6s, -9deg 12' 36" (this is the average of two differently derived estimates 7' above and below this, which I gave to Lowell Observatory) for March 10. To correct for the date, subtract 1.1s RA and add 7.5" Decl (i.e., less negative, toward equator) per day.

The correction is pretty accurate until April 10. A 14' field of view is barely big enough to include my two about equally likely estimates, so there's a 50% chance of being in the field, 25% chance of being above & 25% chance of below.


Reply author: Stoat
Replied on: 03/21/2007 07:45:55
Message:

Okay, I've put tha job onto the Bradford as nem4, dumped the last nem4 job. I hate to harp on about it but these two people in your state with telescopes, that won't look for this object. Try them again but this time say that to your friends it will have the name you gave it but to the world it will be what the international astronomy body decide what it's to be called.

In my local pub nobody cares much about the speed of gravity but they did care, very much, about the name of a possible brown dwarf in the neighbourhood. They asked if it was legal for the discoverer to name the thing. I said, I didn't believe it was. Perhaps the people with telescopes are wary of the name you've given the object. Honestly, the press conference would end up in an uproar, and that would steal their kudos.

If an unprincipled mercenary dog, such as myself [:D] discovered it, I'd call it the planet "Coca Cola." That way I'd be sure of being passed a huge cheque under the table, by a company that shan't be named. Of course I wouldn't do that [:I] No, I'd e - mail every large multinational, to get them to bid [8D] The planet "Nike" sounds good, I could claim it as being from mythology [:D] Then there was a little known Greek god called "Kentucky fried chicken." [:)][:)][}:)][;)]


Reply author: Joe Keller
Replied on: 03/21/2007 18:13:12
Message:

quote:
Originally posted by Stoat

Bradford robotic telescope images, the first is at 11 07 24, -6 38 50
next at 11 06 02. -6 28 27

Go to the Bradford web site and look at the latest jobs, these are nem2 and nem3. Perhaps if we decide on an exposure we can put the same job up a few times and then do a blink comparison in photoshop.

[img]http://farm1.static.flickr.com/168/419796237_87c8649184.jpg[/img]

[img]http://farm1.static.flickr.com/152/419796234_3e70b5c577.jpg[/img]



I'm unable to match these photos with the Aladin plates centered at the same coordinates, even if I switch the labels or move the Aladin coords a few arcmin. Nor is there a very good match with the Millenium Star Atlas.

If the labels were switched, the bright star at the bottom edge of the lower photo, corresponds to the 8th magnitude star near the SE corner of the degree grid on p. 801 of the Millenium Star Atlas. The bright star toward the bottom right of the top photo, corresponds to the 9th mag star near the center of the same degree grid in the atlas. The centers of these photos then would have to be displaced one to four arcminutes in each direction.

The Hipparcos cataloged stars were observed ~100x, between mid-1989 & mid-1993. From the Millenium catalog census, apparently inclusion of 11th mag stars was only partial, and a few 12th mag were included.


Reply author: Joe Keller
Replied on: 03/21/2007 20:39:11
Message:

The Tautenburg plates (link originally found by Stoat above)(80" Schmidt, world's biggest, near Jena) include two, #3855 & #8559, which lie on the track I've been discussing, but according to my newest estimate of the position and speed, neither plate should show Barbarossa.


Reply author: Joe Keller
Replied on: 03/22/2007 00:57:36
Message:

The difference between the CMB dipole, and Barbarossa's position, might be explained by non-Lambertian emission at the 52.6 AU surface. This phenomenon is well-known for medical X-ray screens:

"The spectral distribution of the X-rays proved not to be important. ...

"The X-ray angle of incidence on the screen's surface was also found to be unimportant. ...

"...total light flux actually emitted from the screen is systematically less than that calculated according to Lambert's law, varying from 8% for very thin screens to 25% or even more for screens of medium and large surface densities."

- Giakoumakis & Miliotis, Physics of Medicine & Biology 30:21+, 1985.

The WMAP scan geometry is much less symmetrical than COBE's. A deviation of the CMB dipole of the order of Earth's orbital eccentricity could ensue.


Reply author: Stoat
Replied on: 03/22/2007 04:28:46
Message:

Hi Joe, I just checked that I hadn't mixed up the images when I uploaded them to flickr.com They are in the right order, and I can't see how an automated telescope could mix up plates. Actually, setting up a computerised telescope makes it easier to calibrate. Errors could creep in when the telescope is looking close to the horizon, due to the weight of the beast but these plates should be okay. Four minutes error sounds way too much to me.

(Edited) We should have no problems if we compare plates made on the same telescope but perhaps Tom has come across the possible problems of trying to compare plates made on two telescopes.

Perhaps we could use a named star to get an idea of what the Bradford can see. Hip55890 is at
RA 11 27.277'
Dec -8 52.136'
RA proper motion -0.0472
Dec motion 0.0068
Magnitude 8.31


Reply author: nemesis
Replied on: 03/22/2007 12:35:53
Message:

My suggestion for a name for the planet would be Niflheim, the Norse name for the dark realm of cold, ice, and mist. This would also be in keeping with the mythological naming tradition.


Reply author: nonneta
Replied on: 03/22/2007 15:06:07
Message:

Just out of curiosity, how is this thread related to the thread title, which is "Requiem for Relativity"? Maybe the current discussion should be given its own thread with a more representative title.


Reply author: Larry Burford
Replied on: 03/22/2007 16:11:50
Message:

On line discussions are much like face to face discusions. The often move off on tangents that have a life of their own. You would have to go back and start from the beginning to see just how this particular one came into being.

We have a number of participants who use our resources for their own purposes. Without shame or embarassment, aparently. (Many of them are kooky, even to us.)

Some of them we cut off.

Some we don't.

Our reasons vary, but they are confidential.

===

We make an occasional attempt to get them to tie their stuff (either as support or as refutation) into our stuff. Sometimes they try.

===

As you can see, moderation is fairly loose here (but the rules are not especially objective, and are subject to change). The only thing that will always get you into trouble (right now) is shooting at a messenger. Off topic comments (our definition, not yours) are another.

Messages are always fair game.

LB


Reply author: Joe Keller
Replied on: 03/22/2007 19:07:25
Message:

quote:
Originally posted by Stoat

Hi Joe, I just checked that I hadn't mixed up the images when I uploaded them to flickr.com They are in the right order, and I can't see how an automated telescope could mix up plates. Actually, setting up a computerised telescope makes it easier to calibrate. Errors could creep in when the telescope is looking close to the horizon, due to the weight of the beast but these plates should be okay. Four minutes error sounds way too much to me.

(Edited) We should have no problems if we compare plates made on the same telescope but perhaps Tom has come across the possible problems of trying to compare plates made on two telescopes.

Perhaps we could use a named star to get an idea of what the Bradford can see. Hip55890 is at
RA 11 27.277'
Dec -8 52.136'
RA proper motion -0.0472
Dec motion 0.0068
Magnitude 8.31



These are good ideas. However, I had no problem correlating stars on POSSI (Palomar) and SERC (La Silla) 48" Schmidt plates, or on La Silla Red vs. Blue plates, or even Optical IR (0.75-1.00 micron) plates.


Reply author: Joe Keller
Replied on: 03/22/2007 19:23:31
Message:

quote:
Originally posted by nonneta

Just out of curiosity, how is this thread related to the thread title, which is "Requiem for Relativity"? Maybe the current discussion should be given its own thread with a more representative title.



Thanks for your input. The Cosmic Microwave Background often is cited as the main evidence for the Big Bang, and moreover for the orthodox version of General Relativistic cosmology. So, if the CMB dipole were proven to be almost perfectly aligned with a massive solar system body, a psychological door would be opened for questioning other facets of orthodox Relativity theory.


Reply author: Joe Keller
Replied on: 03/22/2007 19:51:24
Message:

A good description of the WMAP scan geometry is on the WMAP website. A good description of the COBE scan geometry is in Smoot et al, Astrophysical Journal 360:685+, p. 686.

From the COBE description, it seems that to first order in Earth's orbital eccentricity, a nearby non-Lambertian emission shell for the CMB would not affect the apparent dipole. However, the article admits that there was enough deviation from the idealized description, that sometimes part of Earth appeared over the edge of the sunshield. So, an effect on the apparent dipole might be seen due to the COBE scan geometry. It likely would be smaller, but of the same sign as, that due to the WMAP geometry.

WMAP, basically, scanned the hemisphere away from the sun. A non-Lambertian emission surface at 52.6 AU (e.g., a sum of rainbows) would give an apparent cross dipole of strength

0.5/52.6^2*(90/theta)^2*8/pi*epsilon

where epsilon is Earth's orbital eccentricity and theta is the angular radius of the "rainbow" (the rainbow is a model of extreme non-Lambertian emission).

By the method of Newton's "Optics", Book I, Part 2, Prop. 9, Problem 4 (pp. 446-447, Britannica Great Books edition) one sees that the index of refraction which gives the largest deflection (39.4deg) for light transmitted through a droplet, is sqrt(2). The light reflected in this droplet gives a rainbow radius of theta = 31.6deg.

Using this as a model or guide, one finds that Earth's orbital eccentricity causes WMAP's apparent CMB dipole to lag 3.0 degrees. The actual lag found above is 3.1 degrees. COBE's smaller lag (0.6 deg) might be due to a lesser amount of the same phenomenon.


Reply author: Joe Keller
Replied on: 03/22/2007 20:32:12
Message:

quote:
Originally posted by Larry Burford

On line discussions are much like face to face discusions. The often move off on tangents that have a life of their own. You would have to go back and start from the beginning to see just how this particular one came into being.

We have a number of participants who use our resources for their own purposes. Without shame or embarassment, aparently. (Many of them are kooky, even to us.)

Some of them we cut off.

Some we don't.

Our reasons vary, but they are confidential.

===

We make an occasional attempt to get them to tie their stuff (either as support or as refutation) into our stuff. Sometimes they try.

===

As you can see, moderation is fairly loose here (but the rules are not especially objective, and are subject to change). The only thing that will always get you into trouble (right now) is shooting at a messenger. Off topic comments (our definition, not yours) are another.

Messages are always fair game.

LB



I have something to add to the above incisive, valuable comments: please consider the message, not the messenger. Almost never will one find a person who is a competent heretic in exactly one's own heretical area of choice and no other.


Reply author: Stoat
Replied on: 03/23/2007 04:52:30
Message:

Hi Joe, one thing that the image problem could be, is that the images put up here are simple jpg's. You can download them as FIT files from the Bradford. jpg files will only be at 72 dpi, so they won't fit onto the much larger fit files. Do yo think it could be down to that?


Reply author: Joe Keller
Replied on: 03/23/2007 16:22:46
Message:

quote:
Originally posted by Stoat

Hi Joe, one thing that the image problem could be, is that the images put up here are simple jpg's. You can download them as FIT files from the Bradford. jpg files will only be at 72 dpi, so they won't fit onto the much larger fit files. Do yo think it could be down to that?



When I saved the ESO versions of my Object #3 area, as JPG instead of FITS files, they were a little blurrier, but it still was easy to identify the main landmark stars.


Reply author: Stoat
Replied on: 03/23/2007 17:47:33
Message:

I just downloaded one of the images as a fits file. My viewer didn't want to know, so i downloaded the other older fits file, not for colour images. One of them opened and after adjusting the intensity I got about three times as many stars. So it looks as though it's best to use the black and white image format on job requests. Should I put up the images again? Or not bother? One good thing I found, my fits viewer has a blink animator to it.


Reply author: Joe Keller
Replied on: 03/23/2007 21:01:05
Message:

quote:
Originally posted by Stoat

I just downloaded one of the images as a fits file. My viewer didn't want to know, so i downloaded the other older fits file, not for colour images. One of them opened and after adjusting the intensity I got about three times as many stars. So it looks as though it's best to use the black and white image format on job requests. Should I put up the images again? Or not bother? One good thing I found, my fits viewer has a blink animator to it.



Please put up the new improved (B&W) images! With three times as many stars, it should be easy to confirm the region. Also, it will let others know what to expect, even though I no longer think that region is the best bet for finding Barbarossa. It's also good that you have a blink animator now.


Reply author: Joe Keller
Replied on: 03/23/2007 22:57:27
Message:

JL Brady determined the mass of "Planet X" (a different "Planet X" revealed by ephemeris discrepancies of Halley's, and other comets) as a multiple of Pluto's mass (Brady, Pub. Astron. Soc. Pacific 84:314+, 1972, Sec. II). Brady's Fourier analysis showed, an influence with the period of Pluto, and another influence 300x bigger with a period corresponding to a circular orbit at 65 AU. When Brady wrote, Pluto was thought to have 0.10 Earth mass (Baker's "Astronomy", 9th ed., 1971). Noting that the influence also should be proportional to the inverse square of the major axis, Brady estimated, in Sec. II, that Planet X should be 1000x heavier than Pluto, i.e., 100 Earth masses according to the presumed mass of Pluto at that time.

Also, the mass should be proportional to the cosecant of the orbital inclination (which Brady eventually estimated at 120 deg, vs. 17 deg for Pluto). This would change Brady's Sec. II estimate, to 300x heavier than Pluto (keeping the same number of significant digits as Brady). The mass of Pluto has been revised drastically downward, to 0.0020 Earth masses (Astronomical Almanac, 2004, p. E88; 2007 ed. is the same). So, Brady's Sec. II estimate becomes 300*0.002 = 0.6 Earth masses, not 100.

In later sections of his article, Brady made computer integrations of gravitational effects, and found 300 Earth masses for Planet X, implying 1 Earth mass for Pluto. The resolution of this paradox, is that the perturbations of Halley's comet, though proportional to the masses of Pluto and Planet X, are not due to gravity "as we know it" (a phrase I borrow from actor DeForest Kelly). The perturbation is 500x that due to gravity, perhaps through the mediation of phenomena at the 52.6 AU barrier, to which Pluto (inside) and Brady's Planet X (outside) make comparably close approaches.

If Brady's Planet X were as heavy as Jupiter, it would have sqrt(65/5.2) = 3.5x Jupiter's angular momentum and would "wag the dog" (the phrase originally appeared in Von Arnum's "Frauelein Schmidt", 1907, was used by F. Scott Fitzgerald in 1935, & was the title of a 1997 movie) of orbital precession in the solar system (Goldreich & Ward, PASP 84:737+). At 1/500 Jupiter's mass, it has 0.007x Jupiter's angular momentum. Not only is its force on the other planets 500x less (which by itself would increase the disruption time from Goldreich & Ward's 1 million yr, to 500 million yr) but it is Planet X which precesses fastest, thereby canceling by averaging, what little disruptive effect Planet X has.

Though Brady predicted magnitude +13 to 14, Klemola & Harlan (PASP 84:736) found nothing in Brady's predicted region, down to mag +15 or, mostly, +17 or 18 (sensitivity varied). The actual 500x smaller mass of Brady's Planet X equates to 4.5 magnitude units, i.e., +17.5 to 18.5. The difference between Pluto's 30% albedo (Astronomical Almanac, 2004) and the canonical Kuiper belt albedo of 4-8%, gives another 1.4 to 2.2 mags, for +18.9 to +20.7. Thus Klemola & Harlan's observational failure also is obviated. (Brady checked that his Planet X was outside Tombaugh's search, i.e. the Lowell Proper Motion Survey, which was down only to +16 or 17 anyway.)

A 500x smaller mass for Brady's Planet X reduces the hypothetical sum of squared residuals due to its effect on the known outer planets, 250,000x. This thoroughly obviates the findings of Seidelmann et al, PASP 84:858.

According to Brady's orbit (Brady, op. cit., Sec. IV last par., & Fig. 5) Planet X now would be near ecliptic longitude 0, ecliptic latitude +52. Using its 0.6 Earth mass, which the modern estimate of Pluto's mass implies, and its "x1" = 65/52.6 = 1.24 (vs. "x2" = 191/52.6 = 3.63 for Barbarossa) Brady's Planet X moves the CMB dipole

0.6/(0.0068*333,000) * f(x1)/f(x2) radian * sin52 = 0.58 degree north,

where f(x)= -2*(1- 3*x^2)/(1-x)^2/(1+x)^2

is the Hermite one-step approximation, for x > 1 (and not too near 1), for an integral arising in my calculation of the CMB dipole caused by a massive body. (The approximation for 0 < x < 1, x not too near 1, is

f(x) = -4*x^3/(1-x)^2/(1+x)^2 .)

Together with the contribution of Neptune ( +0.0243 * 11.8 deg * sin(311-173); conveniently, Neptune was at its descending node) and Uranus ( +0.00323 * 11.8 deg * sin(329-173) ) (the contributions of Jupiter & Saturn are small), this explains 0.58 + 0.21 + 0.03 = 0.82deg of discrepancy between Barbarossa's and the CMB dipole's ecliptic latitudes (observed discrepancy: 0.7deg).

Also, a negligibly larger predicted mass for Barbarossa, results from Brady's Planet X. The rational expression f shows that the induced CMB dipole decreases as inverse distance squared, rendering interstellar effects negligible.


Reply author: Stoat
Replied on: 03/24/2007 07:34:41
Message:

Oops!


Reply author: Stoat
Replied on: 03/24/2007 07:36:59
Message:

[quote]Originally posted by Stoat

Here's the nem3 image, which is at RA 11 07 24 Dec -6 38 50

[img]http://farm1.static.flickr.com/155/432260927_b426006da6.jpg[/img]


Reply author: Joe Keller
Replied on: 03/24/2007 13:42:45
Message:

quote:
Originally posted by Stoat

[quote]Originally posted by Stoat

Here's the nem3 image, which is at RA 11 07 24 Dec -6 38 50

[img]http://farm1.static.flickr.com/155/432260927_b426006da6.jpg[/img]



This is the previous version of this coordinate shot, turned upside down. I'm not laughing: this is only the third photo in the world ever made in the search for Barbarossa, and things will go wrong. The important thing is that it's corrected. I still can't correlate it with the Millenium catalog, however. Part of the reason might be that the Millenium catalog typically shows only one or two stars per Bradford frame in this region.


Reply author: Stoat
Replied on: 03/24/2007 14:42:23
Message:

It doesn't matter. When we get two shots the same from the Bradford, we can animate them to look for movement.

We look at upside down pictures of the moon without any problems [8D]


Reply author: Joe Keller
Replied on: 03/24/2007 15:17:06
Message:

quote:
Originally posted by Stoat

It doesn't matter. When we get two shots the same from the Bradford, we can animate them to look for movement.

We look at upside down pictures of the moon without any problems [8D]



Good point! Carry on!


Reply author: Stoat
Replied on: 03/24/2007 17:48:32
Message:

I think we should be thinking about going there[:D] An ion rocket engine will give us some good data on aether density, maybe a few more relativity experiments as well as that. What's the next launch window for this brute? As it would have to be sent round the houses to get up to speed. I would hate to watch the launch, then pop my clogs before it gets there[:(][8D]


Reply author: Joe Keller
Replied on: 03/24/2007 18:13:11
Message:

Telescope aiming directions: large distant planet Barbarossa in our solar system

Here are aiming directions for March 30, but they're good for the
next 10 days, because this predicted magnitude +17.3 reddish
planet, "Barbarossa", is so far away (190 AU), that it doesn't move
very fast at opposition:

1. Find the 5th magnitude star 14epsilon Crater (the constellation
Crater is S of Leo).

2. Using a finder scope, move slightly more than 1.0 degree NE by N,
to find two unnamed stars, one 7th & one 8th mag, which make
the "crossbar" of the "T" (a very short crossbar) with the star 14epsilon
Crater.

3. Move 0.7 degree straight N of the middle of the "crossbar" in #2.
The 10th mag star there is only 3' NW from my "best" calculated
position for "Barbarossa". My "next best" calculated position
(different method with alternative data) is 11' N of this 10th mag
star.

4. If your field of view is well over 14', aim 4' N of the 10th mag
star in #3. If your field of view is *much* bigger than 14', simply
aim at the 10th mag star, if you wish.

Sincerely,
Joseph C. Keller, M. D.


Reply author: Stoat
Replied on: 03/25/2007 07:26:16
Message:

Here's the general lay of the land, 14 epsilon crater is marked and a little left of centre. If you like I can put up a bigger map from flickr but that will make the page wider and make it awkward to read posts. Maybe Joe, download it and mark up the image.

[img]http://farm1.static.flickr.com/186/433413616_4da49c8626.jpg[/img]


Reply author: Joe Keller
Replied on: 03/26/2007 22:04:29
Message:

Thanks for the map! It's excellent. The 7th & 8th mag stars I mention are visible on it. Simply go 0.7 deg N from their midpoint.


Reply author: Joe Keller
Replied on: 03/26/2007 22:06:54
Message:

To: Joan Genebriera, Tacande Observatory, Canary Islands
From: Joseph C. Keller, M. D.
Re: search for planet Barbarossa, your second photo made 49 hr ago.

Congratulations, you found it!

Two or three bodies forming a line roughly parallel to the ecliptic (i.e., also to Barbarossa's presumed orbit) spread over about 6", of summed magnitude (by comparison with nearby USNO-B catalog stars) +18 to 18.5, are found in the lower right corner near

RA 11h 26m 22.2s Decl -09deg 04' 59"

This object is not in the USNO-B catalog. Nearby cataloged objects include

USNO-B 0809-0228748
USNO-B 0809-0228758
USNO-B 0809-0228763
USNO-B 0809-0228757.

It does not appear on the Aladin Optical Red, Optical Blue, nor Optical Infrared images. I note that the exposure was short. The presence of two or three points of light aligned with the ecliptic, argues against an asteroid or detection artifact.

The object is 30s W of its estimated RA (correcting for Earth parallax) which corresponds to a 2800 yr period for circular orbit. It is 2.4' ecliptic latitude S of its predicted track (by the second prediction method, i.e., that by which this photo was directed).

Sincerely,
Joseph C. Keller, M. D.
March 26, 2007

>From: Joan Genebriera <PLEIADES@telefonica.net>
>To: josephkeller100@hotmail.com
>Subject: New planet ?
>Date: Sun, 25 Mar 2007 15:33:47 +0100
...


Reply author: Joe Keller
Replied on: 03/26/2007 23:32:13
Message:

Joan Genebriera is an amateur astronomer from Barcelona, Spain, now working in the Canary Is. The photo was taken with a 16" Cassegrain telescope from 2500ft elevation at 29deg N latitude. The low elevation and equatorial latitude argue against a cosmic ray artifact, though presumably electronic sensors, not photographic film, were used.


Reply author: Joe Keller
Replied on: 03/27/2007 15:20:22
Message:

I've emailed Joan Genebriera's photo of Barbarossa, to "Stoat", but I'm unable to cut & paste it into this window. If you would like to have the photo, please email me or Stoat. He's been successful posting astronomical photos here; he might be able to post this photo of Barbarossa.


Reply author: Joe Keller
Replied on: 03/27/2007 15:35:44
Message:

In the 0.7" resolution photo by Joan Genebriera (electronic, not a scan of film) Barbarossa is north of the date "25" printed in the lower right corner. Barbarossa & its apparent moon are nearer the edge than the bottom. Together with the moon, and three stars to the NW, they make a line of five points of light.


Reply author: Stoat
Replied on: 03/28/2007 04:28:43
Message:

Here's that image, I reduced its size, so that it wouldn't create problems with the text window. Has Joan put up the fits file on the web?

[img]http://farm1.static.flickr.com/147/437374242_2f3d96cdf5_o.jpg[/img]

[img]http://farm1.static.flickr.com/155/437383449_9b5367b226_o.jpg[/img]


Reply author: nemesis
Replied on: 03/28/2007 09:24:38
Message:

Stoat, in the blowup I assume Barbarossa and its moon are directly over the "25" about 1/3 of the way from top to bottom?


Reply author: Joe Keller
Replied on: 03/28/2007 19:42:08
Message:

Letter re: identification, of Barbarossa candidate, as asteroid

Dear Ms. Genebriera & Mr. Riley [Steve Riley, amateur astronomer, USA],

I compared all of Mr. Riley's photo to its counterpart, Ms. Genebriera's photo "Barbarossa_3" (i.e., the photo with the more northerly coordinates, on which I saw the candidate object). Also I compared them to the SERC DSS2 (Red filter) image in the Aladin archive. Both photos showed excellent correspondence to the archive image.

Other than the candidate object on Ms. Genebriera's photo, I saw nothing on either photo that did not match the DSS2 image (except for obvious very slight defects). In particular, I saw nothing retrograde thereof on Mr. Riley's photo.

I spoke lengthily yesterday with *********, to whom I had emailed Ms. Genebriera's photo. His opinion was that the candidate object was *not* a cosmic ray artifact.

So, it might have been an asteroid on Ms. Genebriera's photo. An asteroid would be out of the field of view of Mr. Riley's photo. A trans-Neptunian object, even as close as 30 AU from the sun (which would give 5 arcminutes motion in 3.2 days now, near opposition) would have been inside the field of view (centered on the candidate object).

Thank you both for your assistance. This initial negative result neither proves nor disproves the existence of a distant planet shepherding a point of the 5:2 Jupiter:Saturn resonance. The 1987 SERC image I discovered, is consistent with such a planet.

I'll forward to both of you, any important information I acquire in the future about this. Meanwhile, if either of you take more photos along the ecliptic in this area, I will give my full attention to their analysis.

Sincerely,

Joseph C. Keller, M. D.


Reply author: Joe Keller
Replied on: 03/28/2007 19:50:03
Message:

quote:
Originally posted by nemesis

Stoat, in the blowup I assume Barbarossa and its moon are directly over the "25" about 1/3 of the way from top to bottom?



[I'll answer since I'm online. - JK] Nemesis, you're correct. Also, in the top picture, it's 1/4 of the way up from the bottom.

It didn't show up in Steve Riley's verification photo. Almost all the stars, even much dimmer ones, did, so now I think it's an asteroid, maybe one with irregular shape or albedo, hence the double appearance.

- Joe Keller


Reply author: Joe Keller
Replied on: 03/29/2007 00:23:12
Message:

The theoretical mass of Barbarossa, 0.0068 solar mass, and distance from the sun, 191 AU, cause gravitational tidal force which comprises part of the anomalous acceleration of Pioneer 10 & 11. O. Oelsen, Astronomy & Astrophysics 463:393+, 2007, Table 1, p. 395, gives, for each of three time intervals for Pioneer 10 & one time interval for Pioneer 11, three different estimates of the anomalous acceleration by different investigators. The standard deviation of these three investigations (considered as a meta-analysis) roughly given by half the range, varies from 1% to 4%, depending on which of the four time intervals is considered. Oelsen remarks that measurement error is believed to be small compared to the differences (~ 10%) observed between time intervals generally.

Using probe distances and heliocentric ecliptic coordinates from "lewes.gsfc.nasa.gov" via "nssdc.gsfc.nasa.gov", and a slide rule, I exactly calculated the sunward tidal acceleration due to Barbarossa, for the midpoint of each interval (by tidal, I mean the difference between that at the probe and that at the sun). Subtracting this from the mean of the three values given by Oelsen for each interval, makes these four measurements of the Pioneer anomaly accurately conform to a simple rule:

Rule: [see post of April 24 below for more accurate rule. - Joe Keller]

...


JD Anderson et al, 1998, also quantified a sunward anomalous acceleration for the Galileo and Ulysses probes. For these probes, sunward was grossly different from Earthward or along the velocity vector or along the spin axis. Anderson's estimates, though rough, confirm that the acceleration is sunward, not along those other directions. Anderson's 1998 figures for these nearby probes, fit 7.0 considerably better than 3.5. However, recent authors suggest that Anderson's corrections, more important in the nearby solar system, were inaccurate, and that the true figures might be smaller or zero.


Reply author: Stoat
Replied on: 03/29/2007 01:50:06
Message:

Hi Joe, does this mean that we've lost a newly found asteroid? Now I believe that there are astronomers who do nothing else but look for asteroids. Don't they have to be reported? This might be a useful argument to get people to point telescopes at the region of sky where the planet may be.

These look like the people to see about reporting asteroids. There's also links to some telescope arrchives. I tried to use one but it just didn't recognise that I had put in the coordinates. That might be down to my using an apple mac.

[:(] I can't put the link up for some reason. Do a google for the, minor planet center.


Reply author: Joe Keller
Replied on: 03/29/2007 14:09:56
Message:

quote:
Originally posted by Stoat

Hi Joe, does this mean that we've lost a newly found asteroid? Now I believe that there are astronomers who do nothing else but look for asteroids. Don't they have to be reported? This might be a useful argument to get people to point telescopes at the region of sky where the planet may be.

These look like the people to see about reporting asteroids. There's also links to some telescope arrchives. I tried to use one but it just didn't recognise that I had put in the coordinates. That might be down to my using an apple mac.

[:(] I can't put the link up ror some reason. Do a google for the, minor planet center.



This asteroid was photographed on a ten-minute exposure, so, the length of the track is consistent with the distance to the orbit of, say, Ceres. (This is inaccurate because not only is the length of the track vague, but increased Earth parallax for nearer asteroids is somewhat compensated by their faster revolution.)

This asteroid might have bright spots on its surface. Assuming it has the relatively high albedo of Ceres (9%), it would be 3km across based on its magnitude, +18.3 by comparison with nearby cataloged stars.


Reply author: Joe Keller
Replied on: 03/30/2007 00:55:53
Message:

I found another disappearing dot, again on a scanned archival Red plate (absent from two other archival Red scanned plates, also one archival Blue and one archival Optical Infrared). Its Red magnitude is +17.8 by comparison with a nearby star with stable catalog magnitudes. Its coordinates are

RA 11 03 12.4 Decl -5 58 09

This is a POSS I Palomar plate from 1954.154. The position is quite consistent with the track, the period and the 1987 archival object (est. Red mag +17.3) discussed above.

Tonight's coordinates, based on great-circle extrapolation & corrected for Earth parallax are

RA 11 25 49.4 Decl -9 03 02.


Reply author: Stoat
Replied on: 03/30/2007 04:28:08
Message:

Can you say what its ascending node is, and its inclination to the ecliptic, Joe? I thought I'd add it to my planets in the program "Starry Night Back Yard" You might want to donload a demo copy of that and see if it lets you add a planet in demo mode. Setting up a planet is pretty easy as it's all done by sliders, and it does some of the maths for you, like working out the planet's year. About 2600 for this one at 191 AU.


Reply author: cosmicsurfer
Replied on: 03/30/2007 15:13:02
Message:

Hi Joe,

Sure enjoying your posts! Wondering about pioneer anomaly, mass accelerations, and the phenomenon of time delayed echo’s of images. Your astute calculations indicated the pioneer anomaly was caused by a zone of differential acceleration towards gravitational center of Sun. Boundary zones of resonance around our solar system may exist where FTL accelerations are more concentrated as they enter this scale. If all mass acceleration is the result of a collapsing higher scale FTL fields (we cannot see because these fields exist outside of our spectrum view plane), then could shells of concentrated resonant electromagnetic fields surrounding solar systems and galaxies result in lensed time delayed echoes of images revealing light wave splitting of image from these boundary zones?

Just how images bounce back and forth on there way through the Elysium or Aether may reveal where these boundary zones exist. A slightly skewed field may bounce the light waves in such a way that reveal a time delayed portion of the spectrum resulting in double images. Space is certainly not empty and from my perspective if you were to look at space from a higher scale viewpoint and were able to take a picture (beyond the frequencies of light) of Universe, all mass in this scale would appear as empty holes/space and then the negative space would reveal bands of extreme frequency energies as they collapsed into this scale. The viewer suddenly would realize that everything we presently see as Universe was the result of this FTL interactions with lower frequency light waves and matter. So, that 90% of the Universe rather then being Dark Matter is now higher scale higher frequency wave forms collapsing into this lower frequency electromagnetic spectrum of light creating mass and accelerations. Just some thoughts….

John


Reply author: Stoat
Replied on: 03/31/2007 07:45:44
Message:

Hi John, it's something I've been thinking about too. if we say that half of the energy of matter at rest is this "atmospere" of aether and that this falls off as an inverse fourth power. Then, I think that this extremely rigid substance has a slight curve to it. Of course we don't know the ratio of light speed to gravity speed but I like the idea that of it being in the region of pi times 10 billion. A few constants actually start to get into the right ball park with that ratio.

Put matter in motion and it tries to to get to equilibrium, it rebalances its aether energy, so that it's no longer half and half. It rotates faster. A planet going near light speed will not flatten but speed up its rotation.

On this brown dwalf. I don't think that it was formed from the sun in a fission process. Much more likely that it was an early break up of the cosmic cloud. So, lets say that two aether "bubbles" form. Where they meet, do we get an almost flat disk of aether energy density, which would have a massive surface area but little volume. Also thiis surface area would more than likely have fractal properties. Boundaries have different surface areas for different frequencies. Such a disk, at the 54 Au mark"?" would have interesting properties.

Bode's Law; well it's not a law but we should expect that there is something to do with resonances going on here. Suppose we say that, in order for a solar system like ours to happen, we need a brown dwarf to direct the Sun's fission of planetary mass objects. No brown dwalf and the sun will still fission off planets but they will be closer in to the parent.


Reply author: cosmicsurfer
Replied on: 03/31/2007 15:04:15
Message:

Hi Stoat, excellent post yes I agree most likely "break up of the cosmic cloud" is a typical process mimicking overall motions of Universe. The original plasma begins to rotate as a dipole and two electromagnetic fields produce spin at top and bottom in reverse directions. This spin causes the plasma to divide in half or variations there of and that is how you gain two opposing rotational stars and the splitting of “cosmic cloud” mass causes a trailing of plasmas following the greater gravitational center or larger sun. So that the binary star system is the normal causation for planetary structure not a fission process, which may also be a later function of the gravity differences between centers.

Accretion is constant from higher scales feeding like a spring that is attached to every atom. I see all matter as a condensate of the slowing down of this higher frequency constant connection from the higher scales. To include as you correctly understood both boundary resonant conditions and fractals. Galaxies are like gemstones in their brilliance and repetitive fractals that appear to fall into an infinite center (galaxies really move in two directions from collapsing fields inwards to outwards motion of mass formed in creational zones, and resonance boundary zones).

As you said, the inverse square rule certainly does apply but maybe what we are not seeing is this “fall off” zones near boundaries are literally not of this scale high frequencies concentration regions that are the result of a shock wave resonating off of the centers of gravity. So that the collapsing field “hits a wall” of resonating boundary zones and falls off from that point out into space around mass subject to the inverse square rule (of combined gravitational centers of mass).

I have really enjoyed the great science involved in your guys search for our brown dwarf sister sun. I would think that there will be one large trailing component found of comets and other trailing destruction of asteroids. Most star systems are binary because that is the common shape of Universe. Two opposing fields in reverse rotation and that is why I will say that until we understand that nothing can exist period with out time going in two directions (as it is attempting to collapse back to zero time) we will still be flailing around in the water without a paddle: Trying to fabricate all kinds of work arounds to describe the unusual conditions of motion that we observe from false big bangs, to dark matter that does not exist. Thanks to TVF first bringing forward the idea of (multiple scales, FTL CG impacts) high frequency higher scale attributes of Universe, we can now see that all mass is generated as a component of this constant streaming of FTL collapsing fields around all mass.

john



Reply author: Joe Keller
Replied on: 03/31/2007 21:54:32
Message:

quote:
Originally posted by Stoat

Can you say what its ascending node is, and its inclination to the ecliptic, Joe? I thought I'd add it to my planets in the program "Starry Night Back Yard" You might want to donload a demo copy of that and see if it lets you add a planet in demo mode. Setting up a planet is pretty easy as it's all done by sliders, and it does some of the maths for you, like working out the planet's year. About 2600 for this one at 191 AU.



Thanks for the information. As I recall, I had estimated that Barbarossa crosses the celestial equator, descending in Declination, at RA 10h 14m. I estimated the angle of Barbarossa's orbit to the equator, as 27.5 deg, and Barbarossa's inclination to the ecliptic (the usual meaning of inclination), as about 16 deg.


Reply author: Joe Keller
Replied on: 03/31/2007 21:59:59
Message:

quote:
Originally posted by cosmicsurfer

Hi Joe,

Sure enjoying your posts! Wondering about pioneer anomaly, mass accelerations, and the phenomenon of time delayed echo�s of images. Your astute calculations indicated the pioneer anomaly was caused by a zone of differential acceleration towards gravitational center of Sun. Boundary zones of resonance around our solar system may exist where FTL accelerations are more concentrated as they enter this scale. If all mass acceleration is the result of a collapsing higher scale FTL fields (we cannot see because these fields exist outside of our spectrum view plane), then could shells of concentrated resonant electromagnetic fields surrounding solar systems and galaxies result in lensed time delayed echoes of images revealing light wave splitting of image from these boundary zones?

Just how images bounce back and forth on there way through the Elysium or Aether may reveal where these boundary zones exist. A slightly skewed field may bounce the light waves in such a way that reveal a time delayed portion of the spectrum resulting in double images. Space is certainly not empty and from my perspective if you were to look at space from a higher scale viewpoint and were able to take a picture (beyond the frequencies of light) of Universe, all mass in this scale would appear as empty holes/space and then the negative space would reveal bands of extreme frequency energies as they collapsed into this scale. The viewer suddenly would realize that everything we presently see as Universe was the result of this FTL interactions with lower frequency light waves and matter. So, that 90% of the Universe rather then being Dark Matter is now higher scale higher frequency wave forms collapsing into this lower frequency electromagnetic spectrum of light creating mass and accelerations. Just some thoughts�.

John




Thanks for the compliment! These are very valuable thoughts.


Reply author: Joe Keller
Replied on: 03/31/2007 22:47:11
Message:

I found Barbarossa on a third archive plate. There is no longer any doubt of Barbarossa's reality nor of Barbarossa's position. Using the two most credible of these "disappearing dots" (1987 & 1997) Barbarossa's period, assuming circular orbit, matches the progression of the imperfect 5:2 Jupiter:Saturn resonance, to 3% accuracy. As in earlier estimates, Barbarossa aligns in longitude, with one of the five (mean, corrected for Jupiter & Saturn eccentricity) resonance points, to 0.4 or maybe 0.1 deg error.

Corrected for Earth parallax, the four points (there are two competing dots on the 1954 plate scan) lie nearly on a great circle. The change in angular speed corresponds to orbital eccentricity of at least 0.1, or at least 0.25, if one or the other of the 1954 points is used in addition to the 1987 & 1997 points. The Red comparison-based magnitudes of three of the objects are all +17.6 +/- 0.3; one of the 1954 objects is about Red +18.3.

The plates (from online 1.0"-resolution scanned versions) are:

1. POSS-I E (a.k.a. POSS-I Red)(exposures for this series varied from 2400 to 4200 sec) Plate XE671, February 25, 1954, epoch 1954.154.

First dot: RA 11h 03m 12.4s Decl -5deg 58' 09"

I determined the Red magnitude as +17.6 by comparison with the USNO-B Red1 magnitude (Red1 was chiefly determined from plates of this series) of a nearby star with stable magnitude. I found this dot, March 29. If it is Barbarossa, then Barbarossa's eccentricity must be at least 0.25, assuming the validity of the 1987 & 1997 dots.


Second dot: RA 11h 02m 25.16s Decl -5deg 56' 11.3"

By comparison with nearby stars, this dot's Red mag is about +18.3. I found this dot, March 28. It is the brightest of a "flying circus" of five disappearing dots spread over about 1'. Together with the 1987 & 1997 dots, it would imply an eccentricity of at least 0.1.


2. SERC-ER (a.k.a. SERC2 Red)(exposure 3600 sec), Plate 713, January 31, 1987, epoch 1987.08215.

RA 11h 18m 03.18s Decl -7deg 58' 46.1"

Because this sky survey was only one of three used to determine the USNO-B Red2 magnitudes, I determined the magnitude of Barbarossa on this plate, by comparison with both the R1 & R2 magnitudes of four nearby stars, finding +17.3. I saw Barbarossa on this plate, March 4, and realized on March 5 that what I saw, was Barbarossa.


3. SERC-I (a.k.a. Optical Infrared)(exposure 5400 sec), Plate IS713(A438), March 3, 1997, epoch 1997.16711.

RA 11h 22m 16.77s Decl -8deg 29' 30.9"

I determined Barbarossa's Infrared magnitude as +18.1 by comparison with two nearby stars. Though the authors of the USNO-B catalog warn that it is a relatively inaccurate source for magnitudes of bright stars, I used the USNO-B's I-R value for Capella, 0.2, to correct the sunlit Barbarossa's Red magnitude to +17.9. I found this dot March 31. I've found no other Optical Infrared plate online with which to prove the disappearance of this dot. Instead, I found that it is absent from both the SERC Red and MASS IR J,K,H plate scan series, indicating, if not disappearance, then an aberrantly narrow spectrum.


The 1987-1997 track implies a 2775 year period for circular orbit. Recent values of Jupiter's and Saturn's periods indicate that their 5:2 resonance progesses with a period of 2696 yr.

Corrected for April 1 Earth parallax, Barbarossa's geocentric coordinates tonight, assuming a circular orbit through the 1987 & 1997 objects, are:

RA 11h 26m 30.9s Decl -9deg 00' 11"

The position might be 7' W to 1.5' E of this, if one or the other of the 1954 dots is used for prediction instead of, or in addition to, the 1997 dot. Last night Steve Riley imaged an approx. mag. +17.3 dot which tonight will be 3' NW of these coordinates (only 1' above the predicted track), if Steve indeed imaged Barbarossa.

Barbarossa's estimated apparent diameter is 0.9". Barbarossa's estimated retrograde motion is 0.7"/hr.

Sincerely,
Joseph C. Keller, M. D.


Reply author: Stoat
Replied on: 04/01/2007 05:07:15
Message:

Here's the latest image from the Bradford. at 11 27 9.60 RA
-9 12 36 Decl.

(edited) I took a look at the faq page at the Bradford and it says that sometimes the image is upside down because of the mounting. The image is flipped horizontally by default, So if the stars are upside down, then one rotates the image rather than flipping it vertical. So I changed the image to a rotated 180 degree image, rather than the flipped one I'd put up at first.

[img]http://farm1.static.flickr.com/206/441804821_5d5988a54a.jpg[/img]


Reply author: Joe Keller
Replied on: 04/01/2007 21:05:26
Message:

quote:
Originally posted by Stoat

Here's the latest image from the Bradford. at 11 27 9.60 RA
-9 12 36 Decl.

(edited) I took a look at the faq page at the Bradford and it says that sometimes the image is upside down because of the mounting. The image is flipped horizontally by default, So if the stars are upside down, then one rotates the image rather than flipping it vertical. So I changed the image to a rotated 180 degree image, rather than the flipped one I'd put up at first.

[img]http://farm1.static.flickr.com/206/441804821_5d5988a54a.jpg[/img]





Thanks! So far, though, even by mentally rotating it, I can't correlate this photo with the same-size (13 or 14' square) Aladin Optical Red image ostensibly centered at those coordinates. Earlier today I didn't have very much trouble mentally rotating another photo someone sent me. I've tried two or three times and also sent an email to the webmaster, but I can't get any response from Bradford about getting an authorization to submit jobs myself.


Reply author: Stoat
Replied on: 04/02/2007 03:00:47
Message:

Hi Joe, why don't you just join again?

Now, on the question of those images from the Bradford. in one place it says that the galaxy camera take images half the size of the full moon but in the hardware section of information, it says 24 arc minutes per side. I think that the bright star, right of centre, lower quadrant, is the star 14 epsilon Crater. So, 24' per side looks right.

I sent them an e mai as well, about the rotation of images. In computer animation there is a problem with cameras. These virtual cameras take the y axis as "up." If they are given a target to follow and the target passes over the top of them, then the y axis drops down to the zero zx plain. When that happens, the y axis can flip round. It's called gimbal lock. It can be fixed by giving the camera a second target that the camera has to try and follow at the same time. It's a software problem that perhaps they haven't sussed out yet.

Anyway, I've put that job back up and hope they take only a few days to do it.

Your fits viewer should have a stack command somewhere. Open two fits files and then stack them. They will be scaled properly, you then have to move one to line up the stars. There should also be a command to animate the stack, which will show any blinking of stars, which indictates movement.

The viewer i got was MicroObservatoryImageMac2.0 which can be got for a pc.


Reply author: Stoat
Replied on: 04/02/2007 06:24:46
Message:

Just had a hunt for a fit viewer that will open the 3d fit files from Bradford. Found this one that's available for pc's as well as macs. It will take a while to get the hang of it, as the docs aren't that good but with a little bit play I got a pretty good image up.

http://heasarc.nasa.gov/lheasoft/ftools/fv/fv_download.html

I did a check with starry sky back yard, and one can see that there's a hexagon of stars, with 14 epsilon Crater bottom right corner , that fits the Bradford image. No need to rotate it at all. Now, is our brown dwarf still about a degree up and a little to the right of 14 eps?


Reply author: Stoat
Replied on: 04/02/2007 11:51:33
Message:

"Barbarossa's estimated apparent diameter is 0.9". Barbarossa's estimated retrograde motion is 0.7"/hr."

Not with you there Joe, I get about 0.05"/hr. A good few months to get a decent blink going but we should be able to get a blink for its planets' orbits


Reply author: Stoat
Replied on: 04/03/2007 04:10:38
Message:

Nem 5 is up now on the Bradford. Same coordinates as the last one but not as good an image as there's some wash from the moon in the image. I think as the Bradford does jobs quite quickly, once the weather improves, I'lll wait a few weeks before putting the job up again.

[:I] oops, I seem to have transposed the minutes and seconds of nem 5. No great problem [:D] practice in moving the images over eah other.

Well, just goes to prove that even we omnipotent folk can make mistakes. I once created a stone that I couldn't lift [;)][8D]

That link to the fits file viewer. I just found out it links to nasa's skyview database. A few seconds to download files.


Reply author: Stoat
Replied on: 04/04/2007 04:10:55
Message:

Here's a few images from the sky survey of the last coordinates given. I didn't alter the last two as i don't know how a clean up would effect data reading.

[img]http://farm1.static.flickr.com/228/445869332_cf65685c98_o.jpg[/img]

Close up of the first image. White dot in centre, with a green dot above and to the left. If this is a star, then our mystery planet is almost on top of it. Could be hard to spot but that might help with our blink over a month or so.
[img]http://farm1.static.flickr.com/207/445902391_24c5322304_o.jpg[/img]


Reply author: Stoat
Replied on: 04/04/2007 04:42:16
Message:

Here's the header for the first image.

SIMPLE = T / Written by IDL: Wed Apr 4 03:44:17 2007
BITPIX = -32 / Bits per pixel.
NAXIS = 2 / Number of dimensions
NAXIS1 = 300 / Length of x axis.
NAXIS2 = 300 / Length of y axis.
CTYPE1 = 'RA---TAN' / X-axis type
CTYPE2 = 'DEC--TAN' / Y-axis type
CRVAL1 = 171.6329 / Reference pixel value
CRVAL2 = -9.0335 / Reference pixel value
CRPIX1 = 150.500 / Reference pixel
CRPIX2 = 150.500 / Reference pixel
CDELT1 = -0.00047222222 / Degrees/pixel
CDELT2 = 0.00047222222 / Degrees/pixel
CROTA1 = 0.00000 / Rotation in degrees.
EQUINOX = 2000.00 /Equinox of coordinates
COMMENT
COMMENT This file was produced by the SkyView survey analysis system from
COMMENT available astronomical surveys. The data are formatted
COMMENT as a simple two-dimensional FITS image with the same units as
COMMENT the orginal survey. A single ASCII table extension may be present
COMMENT which describes catalog objects found within the field of view.
COMMENT Copies of relevant copyright notices are included in this file.
COMMENT
COMMENT Questions should be directed to:
COMMENT
COMMENT lmm@skyview.gsfc.nasa.gov
COMMENT or
COMMENT mcglynn@grossc.gsfc.nasa.gov
COMMENT
COMMENT SkyView
COMMENT Code 660.2
COMMENT Goddard Space Flight Center, Greenbelt, MD 20771
COMMENT 301-286-7780
COMMENT
COMMENT SkyView is supported by NASA ADP grant NAS 5-32068.
COMMENT
SURVEY = 'Digitized Sky Survey'
ORIGIN = 'CASB -- STScI' /Origin of FITS image
COMMENT Note these values refer to Southern plates.
TELESCOP= 'UK Schmidt (new optics)' /Telescope where plate taken
SITELONG= '+149:03:42.00' /Longitude of Observatory
SITELAT = '-31:16:24.00' /Latitute of Observatory
SCANIMG = 'CASB -- STScI ' /Name of original scan
COMMENT Properties of original survey:
COMMENT Provenance - Data taken by Royal Observatory of Edinburgh and AAO,
COMMENT Compression and
COMMENT distribution by Space Telescope Science Institute.
COMMENT Copyright - Space Telescope Science Institute and
COMMENT AAO, UK-PPART and ROE, restrictions on data transmissions
COMMENT prior to June, 1995.
COMMENT Frequency- 600 THz (J band image)
COMMENT Pixel Scale - 1.7".
COMMENT Pixel Units - Pixel values are given as scaled densities.
COMMENT Resolution - Depends on plate. Typically 2" or better.
COMMENT Coordinate system - Equatorial
COMMENT Projection - Schmidt
COMMENT Equinox - 2000
COMMENT Epoch - 1973-1994
END


Reply author: Joe Keller
Replied on: 04/05/2007 00:00:58
Message:

Dear Stoat:

Above posts noted and appreciated!

- Joe Keller


Reply author: Joe Keller
Replied on: 04/05/2007 00:11:36
Message:

Apparently a 16" telescope, even with excellent modern digital equipment and technique, is barely enough for this job. I've been studying such photos. A star with USNO-B catalog Red mag +19.6 +/- 0.1 was a "can't miss" in the photo. Yet another star, on the Aladin scan image, with catalog Red mag +18.1 +/- 0.2, in the same field, was absent except for a slight pixel overdensity impossible to recognize unless one knew where to look. Some of this inconsistency seems to be due to inconsistency of the Schmidt sky survey plates and catalog mags, too, but I think much of it is due to the 16" being near its performance limit.


Reply author: Joe Keller
Replied on: 04/05/2007 00:57:15
Message:

Searching online scans of four overlapping photographic plates from different sky surveys, I've found 21 disappearing dots near Barbarossa's theoretical track, which are plain and starlike on one plate yet absent from multiple others. (The Red plates are compared, inter alia, to at least one other modern Red plate. The Optical Infrared plate is compared, inter alia, to a plate at the wavelength immediately to either side, but not to another of the same wavelength - which I lack.)

The disappearing dots generally are somewhat smaller in diameter and subtly sharper at the edges than stars of similar magnitude. Depending on the response curve of the detector, it is theoretically possible for a dim planet to make a smaller disk than a dim star of the same bolometric magnitude (the planet is brighter than the star far from the center, but the star is brighter than the planet near the center: the star might be able to expose the detector when the planet fails to do so at all).

Thirteen of the disappearing dots are (by comparison to catalog stars) of Red magnitude +19.3 to +20.4; another was about +19 and another +18.8. These 15 dim dots are in position consistent with inner or outer moons of Barbarossa. Each of the three Red plates examined, had at least two dim dots consistent with an inner moon (by the graphing method described below) inside the orbits of Frey & Freya. By magnitude and position, three of them may well be the same inner moon.

Six of the disappearing dots are of magnitude +17.3 through +18.3. So, the distribution of the magnitudes is bimodal. Four of these six bright dots are noticeably oblong, consistent with the 0.7" theoretical one-hour track (the length of the track mainly is due to Earth's motion, not Barbarossa's)(plate exposures were as much as 75 min and maybe as little as 40 min). Three of these oblong ones are sloped roughly 0 degrees and one roughly 45 degrees (SE to NW). The 1.0" pixelization process would tend to introduce luck into the actual apparent direction of the track, but these directions are reassuringly near the theoretical 23 deg.

I converted the plate positions of the six bright dots, to heliocentric celestial coordinates, then graphed them relative to the presumed position of Barbarossa. This revealed that two of the dots are Barbarossa (Red mag +18.0, 18.3), two are Frey (Red mag +17.6, 17.3), and two Freya (Red mag +17.9, 17.9). Sketching the orbital ellipses showed that Frey's orbit is 0.7 AU and Freya's 2.0 AU. Their orbital periods are consistent with Barbarossa having 0.010 solar mass. Both orbits are circular, inclined 5.5deg to our line of sight, and tilted 16.5deg to Barbarossa's track.

Two of these six dots appeared on the 1954 POSS-I Red plate; one of them, presumably Barbarossa, was surrounded by four dimmer disappearing dots ranging from 25" (hence at least ten times the Roche limit even for a moon with only the density of Saturn) to 2' distant. Two of the six dots appeared on the 1997 SERC Optical Infrared plate; I chose as Barbarossa the one which best fit the rest of the theory (the one mentioned in my earlier post, turned out to be Freya). Two appeared on the 1987 SERC Red plate, which, despite its superiority to the 1954 plate, showed no very nearby moons for either; so, I called one of these Frey and the other Freya. Because I found no Barbarossa on the 1987 plate, there I used an interpolated position for Barbarossa, on my graph.

The choices I made, give a period of 2685 yr for Barbarossa, vs. the Saturn:Jupiter 5:2 resonance progression period of 2696 yr. This suggests low eccentricity and an orbital radius of 193 AU (I have used 191 AU in my geocentric-heliocentric corrections). The geocentric coordinates for Barbarossa for 08:01 UT April 5, 2007, would be

RA 11 26 36 Decl -8 58 01

The magnitude of Barbarossa is consistent with 1% visual albedo. Approximately 1% albedo at 5500 Angstroms, in particular, and throughout most of the visual spectrum, is predicted by the main theoretical model of "roaster" ("Class IV") giant planets, which amount to borderline brown dwarfs ( Sudarsky et al, Astrophysical Journal 538(2):885+). At 0.0068 solar mass, Barbarossa would exceed the 0.006 solar mass limit for early lithium burning, so could be just barely a brown dwarf rather than a giant planet. The main theoretical model of brown dwarfs is uncertain in its temperature prediction for Barbarossa, because of Barbarossa's uncertain age, and the strong dependence of log(T) on mass at low masses (A. Burrows et al, RevModPhys 65:301+). One must extrapolate from the corner of Burrows' table to find T=320K for Barbarossa assuming 0.01 solar masses and 4.6 billion yr age. Due to inaccuracy from extrapolation, or in the mass or age of Barbarossa, or in Burrows' model itself, Barbarossa could well in reality be a roaster. The lack of strong infrared sources in the area, on the other hand, suggests that Barbarossa is colder than 100K. Maybe an inaccuracy in Sudarsky et al's albedo model allows Barbarossa to have 1% albedo at cold temperature.

Of twelve expected appearances of Barbarossa+Frey+Freya on four plates, I found at most six. Because a single plate never has more than two bright disappearing dots, maybe Freya is one of the small moons, while Barbarossa and Frey both are giant and orbit each other: this requires only eight appearances of bright dots. The six known dots are roughly consistent with a circular mutual orbit nearly in the plane of the ecliptic. If in 1986 the Barbarossa system were eclipsed by an outer solar system dust cloud causing it to lose one magnitude, then some pair of dim dots from that plate could be Barbarossa & Frey. This would amount to complete detection of Barbarossa & Frey, with detection of a much smaller Freya and miscellaneous minor moons also.


Reply author: Stoat
Replied on: 04/05/2007 05:03:53
Message:

[:I] sorry about that Joe, I worked out the orbital motion of the dwarf but forgot about the earth moving.

On the imaging of telescopes, what we really need is the application of hdri technology. This is used mainly in the 3d community at the moment but is making its way slowly into the photography realm. It stands for high dynamic range image.

An image set is taken with 20 f stops, then this is merged into one image. let's say we want a picture of the inside of an old church. One f stop will give us detail in the lights but lose detail in the darks. Another f stop will give us dark details but wash out the lights. hdri takes care of that. It will have to be used for imaging any terrestial planets round suns, so i'm sure it's being looked at. The main man for this work is here. Maybe he can be asked to give some thought to the problem. http://www.debevec.org/

I've stuck up those new coordinates on the Bradford anyway. You never know, we might get lucky.

Still trying to work out that FITS viewer. It seems that one can rotate the fits file but I just haven't managed to do it yet.


Reply author: Stoat
Replied on: 04/06/2007 04:14:53
Message:

[:)] The Bradford did that job straight away. (not rotated yet)

01:40 on Friday 6 April 2007 (00:40:18 UTC)

[img]http://farm1.static.flickr.com/200/448095296_bcb890fdcd.jpg[/img]


Reply author: Stoat
Replied on: 04/06/2007 09:24:38
Message:

This might be nothng but i was monkeying about with some of tools in my fits program and got this image. This is very close to the centre spot of the plate. The tool does something called a top hat histogram. The light from the star is bright in the centre and has a marked fall off at the edges. Stars look like little blobs of pixels but this one is a bit odd.

[img]http://farm1.static.flickr.com/228/448282408_2268db4442_o.jpg[/img]


Reply author: Stoat
Replied on: 04/07/2007 09:28:32
Message:

Got another job done by the Bradford, nem7. No obvious signs of any movement when I blinked the two fits files. What I thought might be, could also be just junk pixels.


Reply author: Joe Keller
Replied on: 04/07/2007 22:19:04
Message:

Six of the seven "disappearing dots" I've found (on online archive sky survey plate scans) with magnitudes about +18.8 or brighter, fit a mildly elliptical two-body mutual orbit with major axis at least 1.95 AU. The eccentricity is between 0.19 (absolute lower bound) and about 0.38 (better than 67% confidence upper bound). The mutual orbital plane is inclined about 10.7deg to Barbarossa's orbital plane. The mass ratio of these two, Barbarossa and Frey, is 5:1. The period is 42 yr. The implied mass of Barbarossa is between 0.0051 solar mass for 0.19 eccentricity and 0.0080 solar mass for 0.38 eccentricity. The orbital period of the center-of-mass about the sun is 2850 yr, assuming a circular orbit.

The magnitudes of Barbarossa are about +17.3, 17.9 & 18.0; of Frey, +17.6, 18.3 & 18.8. The dimmest Frey magnitude occurs, near maximum elongation, but much nearer to a point at which Frey's orbit intersects that plane (through the center of mass of Barbarossa & Frey) which is parallel to the principal plane of the solar system.

There might be a dust belt there. Neither Frey nor Barbarossa could be found on the March 1986 plate. Then, the line through Barbarossa & Frey, as seen from Earth, was theoretically only 1.8deg from the plane which contained the center of mass and was parallel to the principal plane of the solar system. Now, 21 yrs = 0.50 orbit later, the same situation holds.

In this model, the geocentric coordinates for April 5 (for the next few days the usual minus 1s RA & +6" Decl per day correction applies) are:

Barbarossa: RA 11 26 07.5 Decl -8 59 53.5

Frey: RA 11 27 35.5 Decl -9 10 47.5


Reply author: Stoat
Replied on: 04/08/2007 04:19:05
Message:

Hi Joe, on one blink composition I thought I had what looked like a rotation of two objects but it's at the point where the pixels could be noise.

I think we should see if anyone has a super duper fits viewer, as those two images from the Bradford, need to be looked at for a variety of settings. On a home pc it takes ages but I imagine that any observatory has software to find the best level mix for the two plates.

Anyway, I've put the new RA dec up but I've upped the exposure and changed to a neutral density filter, rather than rgb. See if that helps any.

There's a lot of people reading this thread, if some were to get a fits viewer and download the two Bradford files, nem6 and nem7 we can speed up the search.


Reply author: Joe Keller
Replied on: 04/09/2007 21:28:41
Message:

I now have a pair of relatively dim starlike "disappearing dots" on the March 1986 "UK Red" plate which are consistent with the above Barbarossa/Frey orbit. I don't find dust dimming of the background stars in the area, but the area is sparse in stars so a narrow dust belt in Barbarossa's system can't be ruled out.

Also today I found two bright starlike disappearing dots on a nearby SERC-I (Optical Infrared) plate from February 1995 (the March 1997 SERC-I plate overlaps this area; these dots do not appear on it). One of these dots, when paired with the 7th bright dot above, is consistent with a 5 AU orbit for Freya.

Numerical integration shows that Simpson's rule would have been more accurate than Hermite's rule above (in that integration, the two rules differ by a factor of 5/7). The accurate prediction, based on the gravitational CMB dipole theory, for the mass of Barbarossa + Frey + Freya plus all the other moons and the entire Barbarossa system, is 0.0109 solar masses (perhaps Barbarossa 0.0080, Frey 0.0016, Freya as much as 0.0013). This still makes the net Pioneer 10/11 acceleration more regular than before.


Reply author: Joe Keller
Replied on: 04/09/2007 21:29:49
Message:

quote:
Originally posted by Stoat

Hi Joe, on one blink composition I thought I had what looked like a rotation of two objects but it's at the point where the pixels could be noise.

I think we should see if anyone has a super duper fits viewer, as those two images from the Bradford, need to be looked at for a variety of settings. On a home pc it takes ages but I imagine that any observatory has software to find the best level mix for the two plates.

Anyway, I've put the new RA dec up but I've upped the exposure and changed to a neutral density filter, rather than rgb. See if that helps any.

There's a lot of people reading this thread, if some were to get a fits viewer and download the two Bradford files, nem6 and nem7 we can speed up the search.



Thanks for the help!


Reply author: Stoat
Replied on: 04/10/2007 09:25:19
Message:

I went looking for any other robotic telescopes, due to the Bradford's web site being down at the moment. There's a small one from the Univ of Iowa, there's a 2 metre reflector from the Univ of Liverpool, and there's two 2 metre ones from the Faulkes foundation. One in Australia and one in Hawaii.

Interesting that, I once suggested that if aliens have arrived here, they'd set up Google as a study aid [8D][:D] This guy Faulkes is an amateur astronomer, who just buys 2 metre telescopes out of his pocket money, he's one of the founders of Google, so he must have a shilling or two to rub together.

It might be an idea to set up something here at meta research, to use the Faulkes telescopes. The guy wants to build a network of them for schools' use.

Nem 8 is now up as a finished job at the Bradford. With a bit of luck the next one nem9, will be up shortly as well. Do you think I should go for a full 3 minutes exposure?


Reply author: Stoat
Replied on: 04/11/2007 03:53:58
Message:

Here's the image of nem8. I've rotated this one 180 degrees. I've just disdovered that one can't download the jobs as fits files, unless they are ones own jobs. So Joe, do you want me to e mail them to you? Next ones I think I'll drop the neutral density filters.

Here's an address for Dill Faulkes. it might be worth contacting him through this site. I think he would see the pr benefits of finding a solar systtem brown dwarf, to his robotic telescope project, http://www.faulkes.com/

[img]http://farm1.static.flickr.com/207/455063436_5fda70e48c_o.jpg[/img]


Reply author: Joe Keller
Replied on: 04/12/2007 00:18:22
Message:

I continue to work on arranging the "disappearing dots" into a model of Barbarossa's system. Herein I present my latest, most accurate (correcting my transcription error in the coordinates of one object, and increasing the accuracy of my adjustments for Earth parallax), most plausible and most conservative model to date.

Three dots are plausibly Barbarossa. These are (using the names of convenience I've assigned them as I've worked):

A2. POSS-I (Red) plate date 1954.154, geocentric position RA 11 02 25.16 Decl -5 56 11.3

C. SERC (Red) plate 1987.08215, RA 11 18 03.18 Decl -7 58 46.1

D. SERC-I (Optical Infrared) plate 1997.16711, RA 11 22 16.77 Decl -8 29 30.9.

The position of C differs 115 arcsec from its expected, parallax-corrected, great-circle interpolated position between A2 & D. Though this deviation is, I think, about 10x bigger than the errors inherent in my model or in my calculations, the only component of the deviation big enough to demand explanation, is that perpendicular to Barbarossa's path (i.e., the path from A2 to D).

Two more dots are consistent with the moon Frey in a near-circular approx. 1.4 AU, 22-yr orbit inclined only a few degrees to Barbarossa's orbital plane:

A. POSS-I (Red) plate date 1954.154, geocentric position RA 11 03 12.4 Decl -5 58 09

D2. SERC-I (Optical Infrared) plate 1997.16711, RA 11 22 32.9 Decl -8 26 56.

I've found yet three more dots, two of which could be consistent with the moon Freya in a near-circular 2 AU orbit moderately inclined to Frey's:

B8. UK-Red plate approx. date 1986.199, geocentric position RA 11 14 58 Decl -7 42 20

or

C5. SERC (Red) plate 1987.08215, RA 11 16 04.4 Decl -7 47 51 ;

and

E2. SERC-I (Optical Infrared) plate approx. date 1995.140, approx. RA 11 19 43 Decl -8 06 50.

These orbits imply about 0.0054 solar mass for Barbarossa and much lower mass for Frey and Freya. The total system mass thus could be close to the earlier predicted 0.0068 solar mass which smoothed the net Pioneer 10/11 accelerations.


Reply author: Joe Keller
Replied on: 04/12/2007 00:20:02
Message:

Thanks again, Stoat, for the above!


Reply author: Stoat
Replied on: 04/12/2007 03:48:55
Message:

Nem 10 is done and it looks a lot better with no filters and 3 minutes exposure 03:29 on Thursday 12 April 2007 (02:29:32 UTC)

Should I send the fits to you, or should we try again with two new plates at those settings and new coordinates?

[img]http://farm1.static.flickr.com/172/456357954_5b9c036c93_o.jpg[/img]


Reply author: Joe Keller
Replied on: 04/13/2007 00:13:13
Message:

[quote]Originally posted by Stoat

Nem 10 is done and it looks a lot better with no filters and 3 minutes exposure 03:29 on Thursday 12 April 2007 (02:29:32 UTC)

Should I send the fits to you, or should we try again with two new plates at those settings and new coordinates?


I hope you can find time, to keep taking these Bradford Tenerife pictures at likely coordinates (see below, for latest developments). This can lead to finding that "moving dot" we need.

Yes, although posting the images here is good, emailing the files to me too would be a plus. I know how to download "FITS view" freeware, so I can look at them thoroughly from my computer in the library.


Reply author: Joe Keller
Replied on: 04/13/2007 00:55:31
Message:

Here's the latest "dot theory". These three dots are Barbarossa:

A2. POSS-I (Red) plate date 1954.154, geocentric position RA 11 02 25.16 Decl -5 56 11.3

C3. SERC (Red) plate 1987.08215, RA 11 18 37.6 Decl -7 54 09.5

D. SERC-I (Optical Infrared) plate 1997.16711, RA 11 22 16.77 Decl -8 29 30.9.

These three dots are Frey:

A. POSS-I (Red) plate date 1954.154, geocentric position RA 11 03 12.4 Decl -5 58 09

C. SERC (Red) plate 1987.08215, RA 11 18 03.18 Decl -7 58 46.1

D2. SERC-I (Optical Infrared) plate 1997.16711, RA 11 22 32.9 Decl -8 26 56.

(Some of the other dots are Freya. There are some disappearing dots on the 1986 and 1995 plates which could be these bodies too.)

I assumed that A2 & D are Barbarossa, then making my most accurate correction for Earth parallax, interpolated the expected position for Barbarossa on the 1987 plate. Both C and C3 are a small distance away from that position.

Then I drew lines between C3 & C, A2 & A, D & D2. If these are Barbarossa & Frey in mutual orbit, the center of mass should be displaced at a constant rate. This is best seen by graphing all six bodies on the same sheet, each body relative to the presumed Barbarossa of the pair for its epoch. Generally there will be one mass ratio which makes the centers of mass collinear.

However, when the centers of mass became collinear, they also assumed the correct distance ratio, i.e., constant speed, to within 2% accuracy. (I refer to the residual small speed remaining after the speed from A2 to D is deducted.) This is a very precise and unlikely result. The implied period for circular orbit around the sun was 2847 yr (vs 2688 yr for the J:S resonance progression). Furthermore the mass ratio which gave this precisely constant-velocity center of mass, was 1:1. The conditioning of the graphical solution was such that a 1.2 :1 ratio either way might occur, but certainly not 1.5 :1. Alpha Centauri A & B are said to have a 1.2 :1 ratio, as do Earth & Venus.

The mutual orbit cannot be perfectly circular, because no ellipse centered on the center of mass, fit the points. Slight displacement of the ellipse center (if a noncircular orbital ellipse is tilted, the center of mass generally is not even a focus) allows an infinitude of ellipses. I chose one such that was especially easy to calculate, and found constant angular speed between A, C & D, within 10%; distance between Barbarossa & Frey, 0.7 AU; inclination 18 deg; tilt to Barbarossa's solar orbit, 25 deg; tilt to orbital plane, 30.5 deg; combined mass of Barbarossa & Frey, 0.0036 solar masses.

The trajectory of the presumed center of mass of Barbarossa & Frey, is so constant that Freya likely would have to be of much smaller mass than Frey, or much more distant from Barbarossa. Alternatively, let 1954 be t=0, 1997 be t=1. The midpoint of the interval A-C then is t=3/8 and the midpoint of C-D is t=7/8. If Freya were at conjunction (near our line of sight to the center of mass of Barbarossa & Frey) at t=5/8, then to a first-order approximation the acceleration due to Freya would be zero in the plane of the celestial sphere.

In this model, Barbarossa & Frey are always within 15 arcminutes of the more recent of my various predicted positions. So, the best I have to offer now, is to keep looking within 15' of those coordinates.


Reply author: Joe Keller
Replied on: 04/13/2007 16:22:51
Message:

[quote]Originally posted by Stoat

Nem 10 is done and it looks a lot better with no filters and 3 minutes exposure 03:29 on Thursday 12 April 2007 (02:29:32 UTC)

This photo Number 10 in your Bradford Tenerife series in the search for Barbarossa, is a good one and was instantly recognizable vs. the SERC Red image. I wasn't able to magnify it, but at the magnification present on this messageboard, I found two objects which are not on the SERC Red image, nor in the USNO-B catalog. Their approximate coordinates and comparison magnitudes are:

11 25 31.9 -8 55 40 mag +18.1 and
11 25 38.2 -9 01 24 mag +18.4.

At this low magnification, both seem somewhat smaller and sharper than stars of the same magnitude, but this isn't obviously extreme enough to disqualify them. My present model indicates that they are too far from the predicted center of gravity of Barbarossa & Frey to be either, but either could be Freya even if my model is accurate. I would recommend another photo including these positions to check for reoccurrence with slight movement.

Near the limit of detection, the photo definitely showed a star with USNO-B catalog Red magnitudes +18.2 & 18.5, but, at least at this magnification, did not show another star with catalog Red magnitudes +17.0 & 16.9. The latter star (or object) appeared on both the SERC Red and POSS-I Red images. So, sometimes +18.2 is visible and sometimes +17.0 isn't. I encountered the same inconsistency in images by a skilled, careful amateur astronomer using a 16" telescope in the southwestern US.

The brightest comparison Red magnitude of any object presumed to be Barbarossa, Frey or Freya, on scanned sky survey plates, is +17.3. So presumably all three astronomers I know of who have looked for these bodies so far - J. Genebriera with 16" at Tacande Observatory, S. Riley with 16" at Buena Vista Observatory, and Stoat with 14" at Bradford Observatory on Tenerife - are working at a detection level which might or might not miss the bodies on any given photo.


Reply author: Joe Keller
Replied on: 04/14/2007 12:40:37
Message:

Stoat: Thanks for posting the (rotated) blowup of Number 10 on the other thread. I found the locations of the two suspicious dots. The more southern dot is small and faint, barely visible. By comparison with Aladin, I can revise its coordinates slightly to

11 25 38.6 -9 01 21.

I easily could see the more northern dot, but it's far smaller and more intense than any stars. I think experts call this a "hot pixel". I'll look at this blowup more thoroughly for anything starlike that doesn't match, say, the SERC Red image on Aladin.


Reply author: Stoat
Replied on: 04/14/2007 18:25:52
Message:

Okay Joe, I'll do the same with nem 11 and post that. It's only a day later and as I said, it has some crud that must have got onto the mirror but it's no great problem. Is a "hot pixel" some sort of noise then? If it is, then it shouldn't appear at all on the second image.

I'll post that image tomorrow as it's late here in the u.k.


Reply author: Stoat
Replied on: 04/15/2007 06:50:18
Message:

I put the second image up Joe. What's your take on the kind of "weather" we might expect from a brown dwarf? Will it radiate fairly smoothly , or have periodic mad falir ups? Could it have electromagnetic storms that are braking Frey, or Freya's orbit, somwhat like Jupiter and Io?

If we have a collapsing cloud of hydrogen and chunks fission off, then they will still have flux pipes connecting them. Then the new sun reaches a stage where the pipes can't find deep enough anchor points in the new sun's surface.

A double sun system, where one cloud of gas is just too small to produce anything but a brown dwarf. Should have a tear drop shaped collapsing cloud, it fissions off some planets of its own but the rather "slushy" surface of the dwarf can retain its flux pipes. Ah [:I] just a thought.


Reply author: Stoat
Replied on: 04/17/2007 09:09:49
Message:

Hi Joe, have you looked at the last post by cosmic surfer on this board. Looks like it's worth a read, and there's some good animations of the issue in the link he gives. http://metaresearch.org/msgboard/topic.asp?TOPIC_ID=708&whichpage=8

This notion of a shear region at about 50 AU sounds very interesting.

(edited) Just been reading that web site. It sounds like Nemesis they are thinking about. It's thought to be at about 1000 A.U. and at 17hr 45 minutes and declination –22 degrees.

They do have some serious money up for anyone that can advance the bianry sun theory.


Reply author: nemesis
Replied on: 04/17/2007 16:23:19
Message:

This was discussed rather extensively last summer under "Origin of the Solar System" in the thread "Is the Sun a Binary"? It may be worthwhile to go back and read that.


Reply author: Stoat
Replied on: 04/18/2007 04:20:58
Message:

I've put the new job up on the Bradford, depends on the weather but it should be up by Friday.

Hi Nemesis, I've readdthe thread and most of what that web site says. I think they've got the cart before the horse on this. Their precession notion, which is wrong, drives the maths for the size of the missing binary partner. Now, adding a brown dwarf to the solar system is going to shift the centre of gravity of the system but with Joe's dwarf it's not going to be by very much. The sun does wobble a bit round this point, and as we are changing that wobble, that will have an effect upon the Earth's precession. It's a radius cubed effect, so it will not be any great alteration.

It looks like the money is up only for someone who discovers their solar partner, and not for anyone who finds a brown dwarf, that doesn't cause this curved motion of the sun. Bummer!


Reply author: nemesis
Replied on: 04/18/2007 08:54:52
Message:

Barbarossa can't be the same as as the hypothesized Nemesis in any case, which has been proposed to have a period of up to millions of years.


Reply author: Joe Keller
Replied on: 04/19/2007 20:50:35
Message:

I've spent much of my time this week accurately confirming that J. Genebriera did indeed image Frey on March 25, 2007 (00:42 UT) and that S. Riley did indeed image Barbarossa on April 1, 2007 (07:39 UT). The J2000 coordinates of Genebriera's object are

11 26 22.2 -9 4 59

and of Riley's

11 26 28.0 -8 58 14.5.

The center of gravity slowed only 0.48% between the second and third segments, i.e., [Objects C3 & C 1987, Objects D & D2 1997] and [Objects D & D2 1997, Riley & Genbriera Objects 2007]. This corresponds to about 2s RA. The direction changed only 0.0048 radian (i.e., 0.48%) between the two segments. These deviations are smaller than likely would arise from observation bias: I searched entire 15x15' square images for "disappearing dots" and parts of adjoining squares also, rarely finding more than one or two starlike "disappearing dots" per square.

The direction changed 1.06% between the first and second segments, i.e., [Objects A2 & A 1954, Objects C3 & C 1987] and [Objects C3 & C 1987, Objects D & D2 1997]. The center of gravity slowed 3.6%.

The correction for Earth parallax was made by interpolating the sun's position according to old volumes of the American Ephemeris & Astronomical Almanac in the Iowa State Univ. library. For 1954 I had to use the formula therein to convert to J2000 coordinates. On an IBM486 computer I wrote a "BASIC" program to find the rectangular coordinates of every object precisely. I adjusted the objects' distance from the sun (presumed the same for all) so that the angle subtended between 1954 and 2007 equalled that for a body with a slightly elliptical orbit of period 2688 yr (my best estimate of the period of progression of the 5:2 Jupiter:Saturn resonance) when at said distance from the sun, which was 198.5 AU (corresponding to 2834 yr for circular orbit).

Then I adjusted the mass ratio of Barbarossa (i.e., A2, C3, D & Riley) and Frey (i.e., A, C, D2 & Genebriera) to 0.63:0.37, at which the torsion of the great circle was about constant: that is, the (small) break between the first (32.9 yr) and second (10.1 yr) segments was about twice the break between the second (10.1 yr) and last (10.1 yr) segments.

Corrections for proper motion of reference stars, between the 1987 date of the SERC-Red reference plate, and 2007 or 1954, all were negligible (1997 had its own Aladin reference plate). The aberration of light from the objects is negligible because it affects the reference stars as well; the aberration of sunlight is negligible because the sun's position is needed only for the small Earth parallax correction. The correction for the different distances of Barbarossa vs. Frey from the sun, is negligible.

The decreasing angular speed could be due to the influence of a small distant moon. The moon(s) seem to orbit Barbarossa in a plane near that of Barbarossa's orbit, so any torsion of the great circle would be relatively small.

The strange shapes of both Genebriera's and Riley's objects might be due to rings like Saturn's. The Roche limits for these bodies, a likely distance for rings, would be 1-2" depending on densities. The bodies' diameters should be about 0.8". A barely detectable planetary disk could make a spot smaller than that of a star of equal bolometric magnitude.


Reply author: Stoat
Replied on: 04/20/2007 04:31:47
Message:

Hi Joe, in the image thread I put up, simply to allow bigger images, I asked what you thought of that odd little smudge. A galaxy maybe? It struck me that on a pc monitor it might not even show up so here's a blow up of the region and a further blow up of the smudge, inside the blue oblong.

As I was waiting for the next Bradford image i thought I'd see if coud find anything odd,and this was the only thing.

[img]http://farm1.static.flickr.com/197/465924784_512b2f49a0_o.jpg[/img]


Reply author: Joe Keller
Replied on: 04/20/2007 17:17:47
Message:

quote:
Originally posted by Stoat

Hi Joe, in the image thread I put up, simply to allow bigger images, I asked what you thought of that odd little smudge. A galaxy maybe? It struck me that on a pc monitor it might not even show up so here's a blow up of the region and a further blow up of the smudge, inside the blue oblong.

As I was waiting for the next Bradford image i thought I'd see if coud find anything odd,and this was the only thing.

[img]http://farm1.static.flickr.com/197/465924784_512b2f49a0_o.jpg[/img]



Thanks for the blowup of the smudge. Maybe it's a star surrounded by nebulosity. The core is too bright, relative to the periphery, for an ordinary galaxy.

I'm glad you're looking at these photos intensively. The best bet is to look near the locations at which something showed up on Genebriera's & Riley's photos. A rough but adequate correction for date, through mid-May, is -0.8s RA, +5" Decl, for each day after March 29.


Reply author: Stoat
Replied on: 04/21/2007 05:33:55
Message:

Sorry about that Joe, if you take a look at the place on the image, all you see is a fairly flat grey oval of light. The human eye is not great at differentiating scales of grey, so I used levels to bring out the variations. The white dot in the middle is log scaled to hell. It has structure but its core is actually very dim indeed. Its angle isn't down to the telescope tracking, which makes the stars look like ovals.

It's interesting but the acid test will be if it moves. I'll have a look at it with the next image that comes in but I'm not expecting it to be anything. I just remarked on it because you mentioned the idea of an accretion disk.


Reply author: Joe Keller
Replied on: 04/21/2007 20:38:59
Message:

quote:
Originally posted by Stoat

Sorry about that Joe, if you take a look at the place on the image, all you see is a fairly flat grey oval of light. The human eye is not great at differentiating scales of grey, so I used levels to bring out the variations. The white dot in the middle is log scaled to hell. It has structure but its core is actually very dim indeed. Its angle isn't down to the telescope tracking, which makes the stars look like ovals.

It's interesting but the acid test will be if it moves. I'll have a look at it with the next image that comes in but I'm not expecting it to be anything. I just remarked on it because you mentioned the idea of an accretion disk.



Then indeed it would be well to recheck the object for movement. One of the objects I suspect as Barbarossa ("D" above) has an Optical Infrared glow around it.


Reply author: Stoat
Replied on: 04/23/2007 10:25:26
Message:

What are your thoughts on Sedna and 2003 UB313 then Joe? I note that the old rogue star has been blamed for the wildy eccentric orbits of these.


Reply author: Joe Keller
Replied on: 04/23/2007 22:05:13
Message:

quote:
Originally posted by Stoat

What are your thoughts on Sedna and 2003 UB313 then Joe? I note that the old rogue star has been blamed for the wildy eccentric orbits of these.



How eccentric are their orbits? Are they merely eccentric or do they have unexplained perturbations?


Reply author: Joe Keller
Replied on: 04/24/2007 00:06:39
Message:

"Dot theory" update: high significance.

With a more accurate calculation of Earth parallax, the above-mentioned alignment of presumed centers of gravity of A (1954), C (1987) & D (1997) objects became less perfect. I've been resorting to the IBM486 computer more, to avoid such inaccuracies.

Also, I've found two "disappearing dots" of interest on the B (1986) plate:

"B3" 11 16 51.55 -7 49 41.1

"B" 11 16 56.07 -7 55 14.3.

The following theory is hindered by my lack of success in finding accurately corresponding objects on the C plate. However A2 & A, B3 & B, and J. Genebriera's (March 25) & S. Riley's (April 1) objects (see above for coordinates and other details) seem to correspond to Barbarossa & Frey, resp.

I found accurate heliocentric coordinates for these six objects, using old ephemerides and the above-mentioned "BASIC" computer program. I assumed that all six objects have the same distance from the sun. This distance, 197.7283 AU, was adjusted, as above, so that the overall angular speed 1954-2007 (equivalent to a circular orbit with period 2811.866 yr) would equal that for an elliptical orbit with period 2688.000 yr.

The mass ratio of the contemporaneous pairs of objects was adjusted so that the two great-circle arcs, from center-of-mass A to center-of-mass B and from c.o.m. B to c.o.m. Genebriera/Riley, had exactly the same direction to within a millionth of a radian. The resulting mass ratio was 0.87710:0.12280. Generally there is a ratio which will cause the directions to be the same, because often an equation will have a real solution. Generally a different equation will not be simultaneously satisfied. Here, I found simultaneously that the c.o.m. angular speeds from 1954 to 1986 and from 1986 to 2007 became nearly equal, only a -0.0508% change (-0.001915% per yr). This is typical of the speed change expected in the lower range of possible solar orbital eccentricities.

I checked this graphically as above. By successive approximations at the computer, I found the geocentric coordinates (for G.'s & R.'s objects, coordinates always were adjusted to the midpoint of Genebriera's & Riley's observation times) which would put Genebriera's object on a perfect constant-speed heliocentric great circle with objects A2 & B3. Then I graphed all six objects on the same sheet, with these three points superimposed as the origins for objects of their epoch. I approximated the short great circles between contemporaneous points, as lines, and neglected the nonparallelism of longitude lines. (Also on my graph I neglected the difference between RA degrees & Decl degrees: this changes 0.7% between 6 & 9deg Decl.)

When I found the centers of gravity according to the above ratio, they were, as expected, collinear and showing constant speed. On the graph paper, the errors of the collinearity and of the constancy of speed both were about 1% of the graphed portion, which was in turn about 1% of the total motion; i.e., about 10^(-4) accuracy overall. Translating the Genebriera & Riley objects equally so that Genebriera's object also lay at the origin, I found by quick but generalized searching at the computer, the approximate unique ellipse centered at that origin and passing through the three presumed Frey objects (A, B, and Riley's). The ellipse was tipped 10deg NE-SW (i.e., 37deg to Barbarossa's orbit). The major and minor semiaxes were 0.25deg (0.86 AU) and 0.10deg, resp., giving i=22deg and a total tip for the Barbarossa-Frey system, of 43deg to Barbarossa's solar orbit.

The positions on the presumed orbital circle then were found. This was consistent with an orbital period of 8.3 yr (almost 4 revolutions, 1954-1986, & almost 2.5, 1986-2007) with only 3% discrepancy between the A-B and B-Riley arcs. The total mass Barbarossa+Frey would be 0.0094 solar mass (Barbarossa alone, 0.00825 solar mass).

The theoretical mass at this distance, needed to produced the CMB dipole, was calculated as above using precise 200- or 2000-step trapezoidal rule integration, and found to be 0.0116 solar mass. (Any additional bodies in the Barbarossa-Frey system, unless closely orbiting Barbarossa or Frey, or very distant, would need to have small mass because of the precise motion of the Barbarossa-Frey center of mass).

An inaccuracy in the tidal Pioneer 10/11 acceleration calculation above, was fixed, and the new parameters applied. The net sunward anomalous Pioneer 10/11 acceleration, after subtracting the tidal forces from the presumed 0.0116 solar mass Barbarossa system, at the four distances tabulated by O. Olsen (A&A, op. cit. 2007 above) becomes:

27 AU: 6.03*10^(-8) cm/s^2
45: 5.38* "
52: 5.68* "
63: 4.25* "

The best figure for the Hubble shift is 72 km/s/Mpc, which is equivalent to 7.0*10^(-8) cm/s^2. Assuming that near the sun, the net anomalous acceleration equals the Hubble-equivalent acceleration (the Galileo and Ulysses measurements of the anomalous acceleration lack the accuracy to confirm or refute this) the anomalous acceleration approximates a normal distribution with peak 7.0 and standard deviation 53 AU:


observed/predicted/difference (relative scale, i.e., 7.0*10^(-8) is 1.00)

0 AU: ~1.0+-0.3/1.00/?
27: 0.86/0.88/-0.02
45: 0.77/0.70/0.07
52: 0.81/0.61/0.20
63: 0.61/0.50/0.11

So, accounting for the tidal force of Barbarossa et al, the magnitude of the anomalous acceleration, as a function of radius, approximates a normal curve with standard deviation roughly 53 AU; with a small hump added near 53 AU.

Object A2 has four dimmer disappearing dots within about an arcminute (6,000,000 mi at 198 AU); two of these are a pair about 20" (2,000,000 mi) apart. Object B3 has probably one dimmer disappearing dot about 45" S. Barbarossa is a cold brown dwarf: these might be Barbarossa's inner planets. Frey is Barbarossa's Jupiter. Barbarossa's next-biggest satellite already has been named Freya.

Five additional more-or-less starlike disappearing dots have been found (on my "C", 1987 Red, and "D" 1997 Optical IR) sky survey images, which seem bright enough to be Barbarossa, Frey or Freya and are close enough to fall on the same sheet of graph paper on which I graphed the Barbarossa-Frey orbit. The three telescopes used so far to search prospectively for Barbarossa or Frey have been the 16" telescope of J. Genebriera (Tacande Observatory, Tenerife, Canary Is.), the 11" of S. Riley (Buena Vista Observatory, California, USA) and the 14" of R. Turner (Bradford Observatory, Tenerife, Canary Is.). Comparison with sky surveys shows that the sky survey dots I seek to relocate are near, if not beyond, the detection limits of these telescopes. Genebriera and Riley have achieved, so far, one detection apiece (Barbarossa and Frey, resp., according to this latest theory).


Reply author: Stoat
Replied on: 04/24/2007 10:49:15
Message:

Sedna (the inuit goddess of war, I think) goes from 76 to 990 AU.

UB313 is now titled Eris (A Greek goddess) I don't know its eccentricity but the drawing up on the web site for it suggests a large eccentricity as well.

I don't know about you but I'm always wary when the explanation of something is a rogue passing sun. I take that as meaning, "we haven't a clue why."

Data on the two objects can be got from here. http://ssd.jpl.nasa.gov/horizons.cgi

I got image 12 back from the Bradford, should I put it up? As it looks like it will miss the revised position of old Barb. Maybe new jobs put up with a red filter to bring out the dwarf?


Reply author: Bill_Smith
Replied on: 04/25/2007 01:18:59
Message:

You do not want to use a filter. If the target is at or near the detection limit of the scope adding a filter is just going to make it disappear altogether.

Are these images being calibrated? The last one posted appeared to have a lot of vignetting so it appears it wasn't flat field corrected at least. Without proper calibration the images are going to have a lot of 'noise' making detection of objects unreliable.

Cheers

Bill


Reply author: Stoat
Replied on: 04/25/2007 04:48:47
Message:

Hi Bill, won't a red filter lighten the reds and dim the blues? Obviously it will mean upping the exposure to compensate for the filter but we're just after movement over two plates. I think it will just come down to the fact that the Bradford isn't big enough to do the job.

(edited)
03 21 36.36 +06 18 17.5 Sedna
01 38 19.71 -04 52 36.5 Eris

So these two are at perigee about 180 degrees from Barb. They are both about Pluto mass, and there must be more of them, which are way out in the boonies at the present time.


Reply author: Bill_Smith
Replied on: 04/25/2007 08:44:28
Message:

Hi Bob,
No, filters will cut light to the CCD. A Red filter will allow red light through and block the other wavelengths so a red object will appear brighter than a blue object but you have still cut the amount of light getting to the CCD. Whenever dealing with anything close to your detection limit, you do not want to use anything that will limit the amount of light your CCD will receive.

Cheers

Bill


Reply author: Bill_Smith
Replied on: 04/25/2007 08:51:23
Message:

Bob,

Given the poor racking in NEM10 and 12, will Bradford allow you to select guiding? If not then you might be better off using shorter integrations and then stack the results.

Cheers

Bill


Reply author: Joe Keller
Replied on: 04/25/2007 15:16:21
Message:

quote:
Originally posted by Stoat

Sedna (the inuit goddess of war, I think) goes from 76 to 990 AU.

UB313 is now titled Eris (A Greek goddess) I don't know its eccentricity but the drawing up on the web site for it suggests a large eccentricity as well.

I don't know about you but I'm always wary when the explanation of something is a rogue passing sun. I take that as meaning, "we haven't a clue why."

Data on the two objects can be got from here. http://ssd.jpl.nasa.gov/horizons.cgi

I got image 12 back from the Bradford, should I put it up? As it looks like it will miss the revised position of old Barb. Maybe new jobs put up with a red filter to bring out the dwarf?





Thanks for the info & link. I'm also wary of the rogue sun idea, especially because elsewhere on this site I've theorized that even what little interstellar red- & blueshift there seems to be, might be mostly non-Doppler.

If convenient, please do put the image up. Somebody else might find something before I do. The Bradford photos have improved dramatically and I have high hopes for them. It's OK if the coordinates are a little off because the still-uncertain Barbarossa-Frey orbit introduces several arcminutes uncertainty.


Reply author: Joe Keller
Replied on: 04/25/2007 15:28:12
Message:

quote:
Originally posted by Bill_Smith

You do not want to use a filter. If the target is at or near the detection limit of the scope adding a filter is just going to make it disappear altogether.

Are these images being calibrated? The last one posted appeared to have a lot of vignetting so it appears it wasn't flat field corrected at least. Without proper calibration the images are going to have a lot of 'noise' making detection of objects unreliable.

Cheers

Bill



Dear Bill,

Thanks for your expert input. I mostly look at Red sky survey plates because these are the commonest, and have the best resolution and sensitivity.

Stars are rather sparse in this part of the sky. It hasn't been too laborious to rule them out by comparison with plate images.

- Joseph C. Keller


Reply author: Joe Keller
Replied on: 04/25/2007 15:38:37
Message:

Now, I have more reason to think the Bradford Observatory might be big enough to do the job: Steve Riley read my last major post and emailed me that he's not using a 16", he's using an 11"! This is with maximum personal care and attention, though.

correction 6/24/08: 8", not 11"


Reply author: Stoat
Replied on: 04/25/2007 16:14:19
Message:

On the Sedna problem. The only thing I can think of at the moment, is the idea that masses clump much much earlier than we thought. A gas cloud that splits into two unequal portions. One becomes our sun, the other a brown dwarf. This thing is about a light year across and there are other clouds in the cosmic nursery which are quite close. If matter starts to clump at this very early stage, it might explain why Sedna and co have such large eccentricities. They could be nearly as old as the cosmic cloud from which they are born. In a sense they get left behind when the planets move closer into the parent sun. They are left in these daft orbits as the cosmic nursery breaks up. The problem would then be to explain how an event such as a nova, can cause particles int eh cloud to collapse into fairly large masses so quickly.


Reply author: Joe Keller
Replied on: 04/25/2007 16:30:31
Message:

"Dot update"

One of my five additional nearby starlike disappearing Sky Survey dots, is

"C11" (Jan 31 1987, SERC Red) Strasbourg "Aladin" RA 11 18 00.41 Decl -8 01 57.7

Assuming this is Frey and that my previous post is valid, I find with my IBM 486 "BASIC" program above, that Barbarossa should be on this 1987 plate at

11 18 03.006 -7 56 29.4.

There is a magnitude +19 star at

11 18 02.80 -7 56 21.8.

This star seems brighter relative to its neighbors, on this 1987 La Silla sky survey vs. the 1986 "UK Red" sky survey. Its R1 & R2 USNO-B catalog magnitudes are nearly equal, but the documentation doesn't reveal the relative influence of the two or three plates used to compute each magnitude. So, it might be a nonresolved conjunction, or (less likely) occultation.

My computerized Earth parallax correction uses the sun's apparent position. Barbarossa orbits the center of gravity of the solar system: including Jupiter & Saturn. Saturn was near quadrature with Barbarossa then, so its influence changed little between 1986 & 1987. On the other hand, Jupiter was nearly opposite Barbarossa. Jupiter's 30 degree motion in one year would subtract about 0.15s RA from Barbarossa's predicted apparent position in 1987, vs. 1986. So, only the predicted and observed Declinations of the star are significantly discrepant.

My very first discovered sky survey "Barbarossa" object, "C", lies 145" exactly S of the predicted position of Barbarossa. My pixel analysis of this object indicated that it is at least two different unresolved objects (the center of luminosity doesn't lie in the brightest pixel). Object C might be an unresolved conjunction of two of Barbarossa's inner planets. (Such "inner planets" appear on the A and B plates.) If so, then these planet(s) would need 6% the mass of Barbarossa, to move the center of gravity of the non-Frey, inner, portion of Barbarossa's system, 145*0.06/1.06 = 8" S. This would place the predicted Barbarossa at the coordinates of the observed star.


Reply author: Stoat
Replied on: 04/28/2007 06:50:29
Message:

Talking to Bill via e mail about brown dwarfs, he mentioned that he thought that Barbarossa was already taken. i did a check on the JPL Horizons site and it has indeed been taken by an asteroid.

Sooo, I think it should be called the planet "Billgates." Billgates was a Greek mythological god, the god of dodgy operating systems [8D][:D] That should be worth a few backhanders from a man who shall remain nameless [8D]


Reply author: Bill_Smith
Replied on: 04/29/2007 18:29:36
Message:

Hi Joe,

I should point out that there are no disappearing/moving objects on Bobs NEM10 and 12 images. The disappearing dots are artifacts. The Minimum FWHM of every stellar object in the images was 5.1" indicating poor seeing and/or poor focus and the targets you identified had a FWHM of 1.3". Given the image resolution was 1.48"/pixel this indicates hot pixels.

Cheers

Bill


Reply author: Bill_Smith
Replied on: 04/29/2007 18:49:47
Message:

Let Bill Gates know you want to name such a potentially important object after him and he may fund all your search efforts. I think he's got a few dollars to spare.


Reply author: Stoat
Replied on: 05/01/2007 07:29:13
Message:

[:)] We'd have to pronounce the name as Billgatease, in order to sneak it past the powers that be. They'd be angry once the realised they'd been duped but the money would be in the bank account by then [8D]

I was thinking about Sedna and Eris. if the sun condensed out of a pure gas cloud, then its diameter at the start would be about 14 light years. That's saying that we have about one atom per cubic centimeter. So, add some dust. The ratio for clouds near us is about 1 in a hundred. Not much change in the radius then. So, let's say that in the early stage of star formation we have a build up of dust particles. Further let's say that when two atoms combine, their aether "atmosphere" increases in extent and this breaks some of the angular momentum of the cloud. Something like watching the bubbles in the centre of your coffee cup merging into one.

If we have Sedna sized chunks of stuff form early on, then their orbits would be effected by the nearest new star to ours in the nursery. Then the whole thing pulls in towards what we see today.


Reply author: Joe Keller
Replied on: 05/01/2007 17:11:40
Message:

quote:
Originally posted by Bill_Smith

Hi Joe,

I should point out that there are no disappearing/moving objects on Bobs NEM10 and 12 images. The disappearing dots are artifacts. The Minimum FWHM of every stellar object in the images was 5.1" indicating poor seeing and/or poor focus and the targets you identified had a FWHM of 1.3". Given the image resolution was 1.48"/pixel this indicates hot pixels.

Cheers

Bill



Dear Bill,

Thanks for letting me know what you found on these Bradford Observatory photos.

- Joe Keller


Reply author: Joe Keller
Replied on: 05/01/2007 17:13:09
Message:

quote:
Originally posted by Bill_Smith

Hi Joe,

I should point out that there are no disappearing/moving objects on Bobs NEM10 and 12 images. The disappearing dots are artifacts. The Minimum FWHM of every stellar object in the images was 5.1" indicating poor seeing and/or poor focus and the targets you identified had a FWHM of 1.3". Given the image resolution was 1.48"/pixel this indicates hot pixels.

Cheers

Bill



Dear Bill,

Thanks for letting me know what you found on these Bradford Observatory photos.

- Joe Keller


Reply author: Joe Keller
Replied on: 05/01/2007 18:09:08
Message:

Steve Riley (Buena Vista Observatory, California) photographed what I call his "Object #1" at approx. 07:39 UT on April 1, 2007. By comparison with the 1987 La Silla archive image (on which it is absent), its position (J2000 geocentric coordinates) is

RA 11 26 24.6 Decl -8 57 48.5.

Riley has a new photo which shows what I call his "Object #2" likewise not found on the 1987 archive image. It was taken (midrange of contributing exposures) 05:41 UT on April 24, 2007. Its coordinates (again by comparison with 1987) are

RA 11 26 09.58 Decl -8 55 56.0.

(This second point is corrected for the proper motion of the reference stars but the first point isn't. That's alright, because the relevant stellar PMs are uncertain and of marginal size in both cases, especially the first.)

Using the ephemeris position of the sun at these times, in my above-mentioned computer program, I found that the heliocentric xyz position of these objects differs as though they are the same object which has moved 0.1 AU prograde in an orbit inclined 34 degrees (and moving south) to the celestial equator. Barbarossa's orbit would move it 0.03 AU prograde and is inclined 27 degrees.

There is little remaining discrepancy in the x coordinate. Two-thirds of the remaining discrepancy in the y coordinate is removed by employing my above-mentioned model of the Barbarossa-Frey orbit, but this model does not remove the z coordinate discrepancy. If I do not insist on a circular actual orbit for Frey about Barbarossa, I can move the apparent semimajor axis 37 degrees, until it is parallel to Barbarossa's solar orbit: then, the predicted x, y, and z coordinates of Riley's Object #2 all become roughly what is observed.

Riley uses an 11" [correction 6/24/08: 8"] telescope. In Riley's photo I found two different +19.3 stars with stable catalog magnitudes. One of these stars was obvious, and one amounted to a barely discernible pixel overdensity. So, I think the position of Riley's Object #2 is more important than its appearance, which is faint and abnormally small.

Most of the apparent motion between Riley's Objects #1 & #2, is due to Earth parallax. The daily change in Earth parallax is shrinking rapidly, but for the next few days, linear interpolation will predict the geocentric coordinates accurately enough to find Riley's Object, which I think is Frey. When Riley's Object (Frey) is found a third time, then quadratic interpolation can be used to predict the geocentric coordinates.


Reply author: Stoat
Replied on: 05/03/2007 03:25:33
Message:

I tried a clustercam shot of RA 11 26 09.58 Decl -8 55 56.0
with the Bradford. No joy [:(] It takes a 3 degree image, and the exposure can be set to one minute maximum. I got an image with only one star on it [:I] They are all dim stars in that bit of sky, so I assume that exposure is way too low.


Reply author: Joe Keller
Replied on: 05/03/2007 22:22:38
Message:

quote:
Originally posted by Stoat

I tried a clustercam shot of RA 11 26 09.58 Decl -8 55 56.0
with the Bradford. No joy [:(] It takes a 3 degree image, and the exposure can be set to one minute maximum. I got an image with only one star on it [:I] They are all dim stars in that bit of sky, so I assume that exposure is way too low.



Thanks, and don't give up! Here's something that would help me and everyone else: on the other thread where you've been posting the Bradford photos, it would be helpful to amend those posts so that each photo has, right next to it, the time of exposure (the Universal Time of the midpoint of the exposure, would be ideal) and also the J2000 celestial coordinates of the midpoint of the image (this can be obtained by comparing the photo to the online "Aladin" image). (Other information is relatively unimportant and could be omitted to avoid clutter.)


Reply author: Joe Keller
Replied on: 05/07/2007 19:32:31
Message:

I've looked retrospectively at some of the photos that have been taken. By chance, the presumed coordinates of Frey are included in these, much oftener than those of Barbarossa. I now see four likely detections of Frey. With my numbering system, they are:

Steve Riley's Object #1. April 1, 2007, 07:39 UT.
RA 11 26 24.60 Decl -8 57 48.5.
Though this object is not entirely starlike, and several even less starlike objects occur nearby, its reality is corroborated by its agreement, when teamed with J. Genebriera's Object #1 (Barbarossa), with Barbarossa/Frey Objects A2 & A on the 1954 Palomar plate scan and with Objects B3 & B on the 1986 UK-Australia plate scan. (See above for details.)

Joan Genebriera's Object #2. April 2, 2007, 23:54 UT.
RA 11 26 23.10 Decl -8 57 40.0.
This amounts to a heart-shaped pixel overdensity several arcseconds across. It does not appear on a followup photo taken by Genebriera only a few minutes later.

Robert Turner's Object #1. April 12, 2007, 02:30 UT.
RA 11 26 16.46 Decl -8 56 57.9.
No more impressive than the preceding object, this amounts to a hazy vertical streak parallel to a vertical black streak. Several barely-visible spots of light are nearby. On this photo by Turner through the Bradford Observatory 14" at Tenerife, the nearby star USNO-B 0810-0228763, Red1 & 2 magnitudes +17.6 & +18.1, is moderately easy to see, but another nearby star, with R2 +18.14, is, though starlike in size & shape, moderately difficult even with high magnification. Several nearby stars with magnitudes +18.5 to +19 are more-or-less impossible to see; however, an R2 +19.05 star shows a very faint though starlike exposure pattern.

S. Riley's Object #2. April 24, 2007, 05:41 UT.
RA 11 26 09.58 Decl -8 55 56.0.
This object is starlike except that it is abnormally small. Occasionally catalog stars are likewise small, at this magnitude near the detection limit.

Interpolation between Riley's Objects #1 & #2, linear with quadratic correction terms, shows that Turner's Object #1 is only 5" from expected position. Then the short distances allow plain linear interpolation between Riley's #1 and Turner's #1, which finds that Genebriera's #2 is only 4" from expected.

In all these photos, here near the detection limit, the intensity of stars' exposure patterns varies drastically despite equal catalog magnitude. Also, some stars show abnormal exposure patterns. I think that the positional consistency of these objects, such as they are, warrants the use of larger telescopes to take photos near Frey's predicted position extrapolated from these. This needs to be done soon, because the constellation Crater (S of Leo) is nearing the sun.

Technique for predicting position during the next two weeks:

1. Plot RA vs. time for the four Objects above, on one sheet of graph paper, and Decl vs. time on another sheet.

2. Use a long string to draw a circular arc through the four points on each sheet.

3. Read the RA & Decl off the two graphs.

This should be accurate enough to allow fields of view much smaller than 15'.

If Barbarossa & Frey have more rock & metal than assumed in published models of giant planets, then they will have smaller diameters and fainter magnitudes than predicted by those models. Also, nearby satellites and/or rings would tend to produce aberrant exposure patterns.


Reply author: Stoat
Replied on: 05/08/2007 07:26:13
Message:

Looks like it's time to try some begging letters to this guy Faulkes and the Liverpool two metre perhaps. As they say in my part of the world, "shy bairns get nowt!"

Another group that might be worth chasing up on would be the small band of supernova chasers. Amateurs but they tend to have very good kit.

I'd go for the short but sweet approach. I wouldn't mention Frey at all. No more than four or five paragraphs. Where it is, how long you need on the scope, and the offer to send fits files from the smaller telescopes.

The last paragraph would be the hard one to write. It has to suggest that for a small outlay, the results could be astonishing, without being too gushing about it.


Reply author: Joe Keller
Replied on: 05/08/2007 11:52:01
Message:

quote:
Originally posted by Stoat

Looks like it's time to try some begging letters to this guy Faulkes and the Liverpool teo metre perhaps. As they say in my part of the world, "shy bairns get nowt!"

Another group that might be worth chasing up on would be the small band of supernova chasers. Amateurs but they tend to have very good kit.

I'd go for the short but sweet approach. I wouldn't mention Frey at all. No more than four or five paragraphs. Where it is, how long you need on the scope, and the offer to send fits files from the smaller telescopes.

The last paragraph woul be the hard one to write. It has to suggest that for a small outlay, the results could be astonishing, without being too gushing about it.



Dear "Stoat",

Thanks for the ideas. Yesterday I emailed 21 serious amateurs on a list who had their own observatory domes. I think two small colleges were among them. I discriminated in favor of those with scopes at least 12" and latitudes +35 or below. The largest scope I noticed among them was 24". Several were in the S. hemisphere.

- Joe Keller


Reply author: Joe Keller
Replied on: 05/15/2007 16:18:30
Message:

Open letter to the Secretary of the Navy (U. S. A.)


To: the Hon. Donald C. Winter, Ph.D.
Secretary of the Navy
Washington, D. C., USA

From: Joseph C. Keller, M. D.
POB 9122
Ames, Iowa 50014 USA

Date: May 15, 2007

Subject: planetary discovery

Dear Secretary Winter:

Please order someone with a big telescope to look for planets near these coordinates:

RA 11h 26m Decl -8.9 deg (J2000)

At 00:42 on March 25, 2007, UT, amateur astronomer Joan Genebriera of Barcelona, Spain, using a 16” telescope and electronic camera at Tacande Observatory on Tenerife in the Canary Is., aiming her telescope at coordinates I provided, recorded objects near

RA 11 26 22.2 Decl -9 04 59 and
RA 11 26 31.8 Decl -9 00 11

These objects seem to be identical with objects found on the online scan of the1954 (48” Schmidt camera) Palomar sky survey plate “POSS-I Red” at these coordinates:

RA 11 02 25.2 Decl -5 56 11 and
RA 11 03 12.4 Decl -5 58 09

and the scan of the 1986 (also 48” Schmidt) Australian sky survey plate “UK Red” at:

RA 11 16 51.8 Decl -7 49 40 and
RA 11 16 56.1 Decl -7 55 14

Assuming a mass ratio of 0.865: 0.135, and a distance, from the sun, of 197.75 Astronomical Units, the center of gravity seems to be in orbit around the sun. The eccentricity of the orbit is < 0.009 with 90% confidence. (The eccentricity of Neptune’s orbit is 0.009.) Though uncommon, double star orbits of this size and eccentricity are known. Known double stars are observationally biased toward systems of more equal mass.

The apparent period of the orbit equals the period of advancement of the 5:2 Jupiter:Saturn resonance, to the accuracy to which the latter is known. Correction for the eccentricities of Jupiter’s and Saturn’s orbits shows that my newly discovered objects lie, in projection, within a fraction of a degree of one of the five mean Jupiter:Saturn resonance points.

The objects lie only two degrees prograde from the positive Cosmic Microwave Background dipole. The dipole lies on the objects’ orbit to within a fraction of a degree, the accuracy to which the dipole is known. This is consistent with a new, gravitational theory of the CMB, which would revolutionize our understanding of the relationship between gravity and electricity.

For reasonable masses, correction for the tidal gravity of the objects, reduces the variation of the Pioneer Anomalous Acceleration. The net Anomalous Acceleration becomes fairly smoothly decreasing with distance from the sun.

I think that despite the likely 0.01 solar mass for the combined objects, they fail to disrupt the solar system, because small shifts in the orbital planes of the known planets, counter the torque. Depending on composition, the objects might have smaller diameters than predicted by published giant planet theory.

The first photos taken in the search for these objects were by amateur astronomer Robert Turner of England, using the 14” telescope and electronic camera of the Bradford College Observatory, also on Tenerife, Canary Is. Amateur astronomer Steve Riley of California, USA, took many photos using an 11” telescope and electronic camera at Buena Vista Observatory in California. The objects are near the detection limit for all these observers. Known stars of magnitude equal to the objects, often are absent or distorted in the photos. Despite these problems, Turner’s later photos, Genebriera’s, and especially Riley’s, show several alternative or additional detections of the objects.

Sincerely,



Joseph C. Keller, M. D.


Reply author: Joe Keller
Replied on: 05/16/2007 17:19:57
Message:

Barbarossa and Cassini's third law.

Cassini's third law is widely applicable (Colombo, Astronomical Journal, 1966). Using the Barbarossa-Frey center of mass from 1954 and 1986, I found (epoch 2000.0):

ascending node: 283.688 deg ecliptic long.
inclination: 12.162 deg
inclination to mean (angular momentum average) orbital plane of J,S,U,N: 13.75 deg

The tilt of the sun's axis, relative to the planets' orbits, is about 30 deg away from what would be needed, for the sun's equator, the mean (see above) orbital plane of the planets, and Barbarossa's orbital plane, to obey Cassini's third law. The tilt of Saturn's orbit, relative to Jupiter's, also is about 30 degrees away from what would be needed, for Saturn's, Jupiter's and Barbarossa's orbits to obey Cassini's third law: Jupiter's ascending node is 101 deg ecliptic longitude, Saturn's 114, and Barbarossa's descending node 103.7. On the other hand, Cassini's third law applies fairly accurately to the trio of Barbarossa's orbit, Jupiter's orbit, and the mean plane of everyone else's angular momentum combined (basically Saturn + Sun). The five biggest non-Earth planets have ascending nodes ranging from 74 to 132 (Mercury's is 48 and Mars' 50.)


Barbarossa manifests the Tifft period.

"The problem in working with short [redshift] periods...is not measurement, it is physical intuition. ...it seems inconceivable that fluctuations within systems would not mask periodic order on a very small scale. Despite the near universality of this feeling, the data appear to speak otherwise."

- WG Tifft, Astrophysical Journal 468:491+, 1996, p. 506

In the Perseus and Cancer regions (i.e., in much of the sky) Tifft found that galactic redshifts tend to cluster around multiples of 2.8817 km/s. Including a small correction for the masses of the planets, Barbarossa's nearly circular orbital speed is 2.10 km/s. This corresponds to an escape speed, at Barbarossa's distance from the sun, of 2.97 km/s. Maybe binary companions such as Barbarossa tend to form at a distance consistent with Tifft's redshift period. As I've noted elsewhere on Dr. Van Flandern's messageboard, the Tifft periods also manifest themselves within the Milky Way galaxy and within the solar system.


Barbarossa is not a planet; it is a binary companion.

My estimate of Barbarossa's mass, is near the theoretical threshold for transient lithium burning, therefore near the boundary between star and planet, as defined by the presence, at any time, of nuclear fusion. Another definition of planet, could be based on the mechanism of formation. Binary star companions farther than 40 AU, from primaries resembling our sun, show little correlation between their orbital planes and the equatorial planes of the primaries (A. Hale, Astronomical Journal 107:306+, 1994). Barbarossa's orbit is about 14deg out of the principal plane of the solar system and 14+7=21deg out of the sun's equatorial plane ( p = (0.37 radian)^2 / 4 = 0.03 ).

When a lone planet orbits a star in a binary system, "Kozai's integral", (1-eccentricity^2)*(cos(inclination))^2, is invariant (M Holman et al, Nature 386:254+, 1997). For a lone planet with a circular orbit initially inclined < 39 deg ( p = (0.68 radian)^2 / 4 = 0.12 ) to the companion star's orbit, the planet's orbit merely precesses. (If the planet's orbit is initially eccentric, the 39 must be replaced by a somewhat smaller angle.) With an inclination > 39 deg, even an initially circular orbit eventually will oscillate between small and large eccentricity, and correspondingly large and small inclination, keeping Kozai's integral invariant. Barbarossa's low mass and near-circular orbit would lengthen the timescale, but not reduce the size, of this oscillation. It is Barbarossa's rather small inclination to the orbits of the planets, that allows the eccentricities of the planets to remain small.


Reply author: cosmicsurfer
Replied on: 05/17/2007 15:30:57
Message:

Hi Joe,

I agree with your statement from above post:

"...This is consistent with a new, gravitational theory of the CMB, which would revolutionize our understanding of the relationship between gravity and electricity."

Wondering if you have any source documents or links pertaining to this research area?

My own views are a modification of MM theory in that the Graviton/Antigraviton is a charged particle of higher frequency above the speed of light spectrum and is in itself the driving force behind all mass fluctuations as shown in collider experiments revealing three trillion times per second flipping between states of matter and antimatter in sub B Mesons.

Thanks, John


Reply author: Joe Keller
Replied on: 05/18/2007 15:22:11
Message:

quote:
Originally posted by cosmicsurfer

Hi Joe,

I agree with your statement from above post:

"...This is consistent with a new, gravitational theory of the CMB, which would revolutionize our understanding of the relationship between gravity and electricity."

Wondering if you have any source documents or links pertaining to this research area?

My own views are a modification of MM theory in that the Graviton/Antigraviton is a charged particle of higher frequency above the speed of light spectrum and is in itself the driving force behind all mass fluctuations as shown in collider experiments revealing three trillion times per second flipping between states of matter and antimatter in sub B Mesons.

Thanks, John




Dear Cosmicsurfer (John),

Thanks for the timely comment! See next post!

- Joe Keller


Reply author: Joe Keller
Replied on: 05/18/2007 17:56:49
Message:

Throughout almost all of the solar system out to about 52.6 AU, the direction of the gravitational field is approximately constant over nuclear distances. The gravitational field in Australia is opposite that in England, but the gravitational field is practically the same, in adjacent nucleons.

At 52.6 AU, the sun's gravity becomes equal to the strongest gravity achievable within a proton (the Heisenberg uncertainty principle limits the mass density, achievable with DeBroglie waves). Outside 52.6 AU, the direction of gravity varies up to 180 degrees, within an atomic nucleus; inside 52.6 AU, the direction of gravity is never more than 90 degrees from sunward. This is a quantized change. Almost anywhere inside 52.6 AU, a space-crystal can exist such that all gravity dipoles within the crystal are aligned parallel to the crystal axis, and also within 90 degrees of parallel to the microscopic gravitational field. Outside 52.6 AU, some gravity dipoles (in crystals near protons) must align antiparallel with the crystal axis in order to be within 90 degrees of parallel to the microscopic gravitational field. The energy penalty for this might correspond to the CMB temperature.

A probe could be sent to the region of space between Earth and Sun, 160,000 mi from Earth, where Earth's gravity cancels the Sun's, to within one part in 2500, thereby simulating the 52.6 AU barrier. This region would be about 65 mi thick and 130 mi wide. At sunrise/set this "agravitational region" would be 1.4 deg from the sun, on-center, and 2.6 arcminutes diam. (Never look at or near the sun with or through anything at any time, not even sunset! Let experts make photographs, then look at the photographs; or, project the sun on grey cardboard and look at the cardboard.)

Wikipedia has accurate information on the discovery of Vulcan by Lescarbault & LeVerrier. More complete information is found in the book, "In Search of Planet Vulcan" by Baum & Sheehan, 1997. The original English source is Rev. Prof. Challis, Proc. Camb. Phil. Soc. 1:219-222, March 12, 1860 (cites Comptes Rendus January 2, 1860, p. 40). The transit time observed for "Vulcan" differs only 2.67% from that expected for projections (see next par.) from the center of the above "agravitational region". This would seem to be within observer error; Vulcan's chord gave only a 37 degree sector. So the transit time was about right; though at 48N latitude, the phenomenon should have missed the sun by 1.0 degree, on-center. Instead it apparently was Liais' serendipitous simultaneous negative observation, from Rio de Janeiro, which missed.

In the preceding paragraph, I assume that the "agravitational region" is a sidereally nonrotating transparent optical structure. For some reason, it blocks wavefronts propagating through it in a certain narrow cone of directions. Lescarbault estimated Vulcan's apparent diameter as 1/4 Mercury's. Thus the angular diameter of this black cone ("Vulcan") happens to be in about the same ratio to the distance from the "region" to Earth, as the angular diameter of the sun is to the distance from the "region" to the sun. Relative to the sun, the region's optical structure rotates retrograde. This causes the black cone to appear as a dot moving prograde as if the region were moving around the sun at the same speed as Earth, not one part in 500 slower. This apparent motion, which is in the plane of the ecliptic, causes about 1/3 of the dot's apparent motion.

The other 2/3 of the dot's apparent motion is due to Earth's rotation, and lies in Earth's equatorial plane. This results in a 15 degree inclination to the ecliptic, tolerably close to Lescarbault's estimate, 6.3 +/- 1 degree (both ascending).

LeVerrier, too, calculated an orbital inclination and ascending node for Vulcan. His results differ from Lescarbault's because, at least according to one anonymous internet source, LeVerrier combined Lescarbault's data with data from five other similar sightings gleaned by searching the literature and eliminating sightings that likely were sunspots, or which were Mercury or Venus. LeVerrier found:

Vulcan (LeVerrier) inclination to ecliptic: 12# 10'
(Barbarossa/Frey (Keller) inclination: 12# 9.7')

Vulcan (LeVerrier) ascending node: 12# 59'
(Barbarossa/Frey (Keller) asc. node: 283# 41' = 13# 41' - 90#)

Maybe both Vulcan and Barbarossa are manifestations of the same real phenomenon which is not a planet. Of five sky surveys I have seached for Barbarossa/Frey, the two surveys on which I have found both of them, were the two surveys on which they were closest to opposition: the survey dates were February 25, 1954, and March 15, 1986. The only prospective photo showing both Barbarossa & Frey, also is the first good-quality photo taken, and the closest one to opposition: its date is March 25, 2007.


Reply author: Bill_Smith
Replied on: 05/18/2007 22:00:43
Message:

quote:
Originally posted by Joe Keller

Barbarossa and Cassini's third law.

Barbarossa manifests the Tifft period.

"The problem in working with short [redshift] periods...is not measurement, it is physical intuition. ...it seems inconceivable that fluctuations within systems would not mask periodic order on a very small scale. Despite the near universality of this feeling, the data appear to speak otherwise."

- WG Tifft, Astrophysical Journal 468:491+, 1996, p. 506



Hi Joe,

I don't pretend to know jack about theory - I just observe and apply logic. So my logic says that the above is a bit of a stretch. 1. it applies to galactic redshift, that is the acceleration away from us, not their actual velocity in space, and 2. He's talking "multiples of" rather than "a value of".

As a side note, what did you actually ask those amateurs to do? I'm looking at your open letter to the MOD(N) and it doesn't appear to be a very practical request. You might get a better response by asking to observe a region of sky within a certain timeframe to a given detection limit!

Also, talking from a experienced observers perspective, looking at current images and comparing them to historic images is only useful to a certain extent, particularly if there was something on the old image that is not on the new image. It is near impossible to rule out asteroids or even space junk that may have been captured on the old image (and they can mimic both fast moving NEO's and very slow moving objects at all brightness levels requiring observations over a number of days to rule them out).

Cheers

Bill


Reply author: Bill_Smith
Replied on: 05/18/2007 22:47:16
Message:

quote:
Originally posted by Joe Keller
A probe could be sent to the region of space between Earth and Sun, 160,000 mi from Earth, where Earth's gravity cancels the Sun's, to within one part in 2500, thereby simulating the 52.6 AU barrier. This region would be about 65 mi thick and 130 mi wide. At sunrise/set this "agravitational region" would be 1.4 deg from the sun, on-center, and 2.6 arcminutes diam. (Never look at or near the sun with or through anything at any time, not even sunset! Let experts make photographs, then look at the photographs; or, project the sun on grey cardboard and look at the cardboard.)



Hi Joe,

Isn't this the L1 point of the Lagrange points? According to the reference we already have 4 probes/satellites at the L1.

Cheers

Bill

http://en.wikipedia.org/wiki/Lagrangian_point


Reply author: Joe Keller
Replied on: 05/19/2007 00:00:58
Message:

quote:
Originally posted by Bill_Smith

quote:
Originally posted by Joe Keller
A probe could be sent to the region of space between Earth and Sun, 160,000 mi from Earth, where Earth's gravity cancels the Sun's, to within one part in 2500, thereby simulating the 52.6 AU barrier. This region would be about 65 mi thick and 130 mi wide. At sunrise/set this "agravitational region" would be 1.4 deg from the sun, on-center, and 2.6 arcminutes diam. (Never look at or near the sun with or through anything at any time, not even sunset! Let experts make photographs, then look at the photographs; or, project the sun on grey cardboard and look at the cardboard.)



Hi Joe,

Isn't this the L1 point of the Lagrange points? According to the reference we already have 4 probes/satellites at the L1.

Cheers

Bill

http://en.wikipedia.org/wiki/Lagrangian_point



Dear Bill,

Thanks for your input! One of the Lagrangian points, is where the sun has enough gravity left over, even after subtracting Earth's, to swing a satellite in an orbit slightly smaller than Earth's but with period one year. What I discussed, is a different point at which Earth cancels approximately all the sun's gravity. That's much closer to Earth, than the Lagrangian point is.

Regarding magnitudes: Uranus is +5.7 at opposition and almost 20 AU out. Barbarossa is almost 200 AU out. The brightness goes as the fourth power of distance because the inverse square law holds both ways (it's farther away, and also the sun is dimmer there). Four powers of 10, is 10 magnitudes. So, if Barbarossa were just like Uranus, it would be about +15.7 (more precisely, +15.9).

Uranus has a rather high albedo, 66% according to a 1983 college textbook. Albedos of 4 to 8% often have been used in recent years as "canonical" albedos in academic journal articles about comet nuclei and Kuiper Belt Objects. Also, a recent article theoretically estimated the albedo of one type of borderline brown dwarf, as 1%! So, it would be fair to use 6.6%, i.e., about 1/10 the albedo of Uranus: +15.9+2.5=+18.4. Using 1% albedo would give +20.45.

The objects I've found and suspected of being Barbarossa, have comparison albedos (these are necessarily somewhat inaccurate) ranging from +17.3 to +20, but mainly +18 to +19. Theoretically, Barbarossa (assuming 1 to 10 Jupiter masses) should be the size of Jupiter only if Barbarossa has a H/He composition like the sun or Jupiter. The same article predicts that if a medium-weight brown dwarf isn't made of hydrogen (even if it's helium or oxygen), it has 1/2 to 1/3 the diameter of the hydrogen version. Of course, this is only theory. That's why observation is so important.

- Joe Keller


Reply author: Bill_Smith
Replied on: 05/19/2007 00:17:21
Message:

Looks like the forum has gone a little haywire with duplication.

About the magnitudes, so this is why the object is at or near the limit of Bobs images. I think I quoted mag 18.5 to Bob when I measured them so it leaves a large gap between 18.5 and 20.

If the object is a brown dwarf would you be relying on albedo only?

Cheers

Bill


Reply author: Stoat
Replied on: 05/19/2007 04:49:27
Message:

Hi Joe, a couple of questions. On the Tifft period, are you saying that this 2.1 km/sec is a statistical effect due to our solar system revolving round the new centre of gavity, due to the brown dwarf being there?

If the cloud that created our sun and its companion dwarf split very early, rather than fissioned off from the collapsing protostar later , then it will have taken a higher percentage of the heavier elements with it. If it happened that way here, it would suggest that other brown dwarf companions, round stars, are similar. Jupiter mass but smaller radius. That would knock down the albedo but so could the core's convection of heavier elements through the dwalf's atmosphere. The surface layers being more dense, drops the albedo further. Is this roughly what you're saying?

The Vulcan thing. "Maybe both Vulcan and Barbarossa are manifestations of the same real phenomenon which is not a planet." This has me totally stumped. What do you think it might be then? I tend to think that the whole vulcan affair was part of the new planet mania sweeping the world at the time, and not real.


Reply author: Joe Keller
Replied on: 05/19/2007 13:30:10
Message:

quote:
Originally posted by Stoat

Hi Joe, a couple of questions. On the Tifft period, are you saying that this 2.1 km/sec is a statistical effect due to our solar system revolving round the new centre of gravity, due to the brown dwarf being there?

If the cloud that created our sun and its companion dwarf split very early, rather than fissioned off from the collapsing protostar later , then it will have taken a higher percentage of the heavier elements with it. If it happened that way here, it would suggest that other brown dwarf companions, round stars, are similar. Jupiter mass but smaller radius. That would knock down the albedo but so could the core's convection of heavier elements through the dwarf's atmosphere. The surface layers being more dense, drops the albedo further. Is this roughly what you're saying?

The Vulcan thing. "Maybe both Vulcan and Barbarossa are manifestations of the same real phenomenon which is not a planet." This has me totally stumped. What do you think it might be then? I tend to think that the whole vulcan affair was part of the new planet mania sweeping the world at the time, and not real.



Hi Stoat!

I think the Tifft periods are a fundamental property of space or matter, which tends to favor certain redshifts and energies within our solar system and everywhere else.

"...[the brown dwarf companion] will have taken a higher percentage of the heavier elements with it. If it happened that way here, it would suggest that other brown dwarf companions, round stars, are similar. Jupiter mass but smaller radius. That would knock down the albedo but so could the core's convection of heavier elements through the dwarf's atmosphere. The surface layers being more dense, drops the albedo further." - Robert Turner

It's been calculated in journal articles, that moderately warm, borderline brown dwarfs have heavier elements like sodium in their atmospheres, which reduce the albedo to 1%. Again, this is very theoretical. No one really knows what Wien's-law surface temperature is needed for this. All we know is that it might happen.

My reading yesterday increased my respect for the skill of Lescarbault & LeVerrier. I think Lescarbault and some of the others saw something real but it wasn't a sunspot or a rock. The duration of the phenomenon was consistent with the duration of a pseudo-transit of the "agravitational region". Luna should have a small "agrav region" of its own which usually transits Luna at full moon; this might explain some of the lights seen on Luna. Jupiter should have a big agrav region which often eclipses Jupiter when Jupiter's opposition occurs not too far from the ecliptic.

- Joe Keller


Reply author: Joe Keller
Replied on: 05/19/2007 19:48:57
Message:

Regarding the likelihood of a brown dwarf companion, the most informative journal article I found today, is Gizis et al, Astrophysical Journal 551(2):L163+, 2000:

"Our evidence [is] that [4 to 32% of][spectral type] F [to] M0 dwarfs [e.g., the sun is a type G dwarf, i.e., not giant - JK] have...[brown dwarf] companions at [separations >1000 AU, i.e., considered "widely separated"]...

"The fraction [of sun-like stars having brown dwarf companions] in the range 100-1000 AU cannot yet be constrained, but...*may* be as common as wide companions..."


Gizis et al sliced and diced the data various ways and came up with similar numbers. Other articles made weaker but not contradictory statements. Another encouraging article, with Gizis' colleague JD Kirkpatrick as primary author (Astronomical Journal 121(6):3235+, 2001) reports the discovery of two type G stars, each with a widely separated brown dwarf companion.

From the foregoing, I can say that the sun had a big chance of having a brown dwarf companion beyond 100 AU (other articles report that sun-like stars seldom have brown dwarf companions inside 40 AU). Old systems like ours tend to have old, cold brown dwarfs. My post above, discussing magnitudes, shows that an old, cold brown dwarf at Barbarossa's distance, could be as bright as +13.5, but with typical trans-Neptunian albedo and Jupiter size would be +16. Theories of diameter and albedo indicate that it might be as dim as +20. So, the failure of other astronomers to detect the sun's brown dwarf companion, Barbarossa, does not much decrease the likelihood of Barbarossa's existence. Furthermore the likelihood that astronomers would have detected a cold brown dwarf beyond 1000 AU (which could only be as bright as +20.5) is almost zero.

Often brown dwarfs have brown dwarf companions. These usually have small orbits. So, for Barbarossa to have a Jupiter-like planet (or micro-dwarf) Frey, orbiting at about 1 AU, is not surprising.

The sun rotates somewhat slower than average for a star of its type. Therefore it is more, not less, likely to have a companion which removed angular momentum from its formative cloud. Mere planets are commonplace. Furthermore, recent theoretical articles conclude that planets alone are not enough. A companion forming in a relatively circular orbit might have been more effective at removing angular momentum. If the sun's Oort cloud is unusually dense, then Barbarossa's orbit might have been circularized by interaction with the Oort cloud. An especially old brown dwarf might be likelier to have its orbit circularized (by means other than the tidal mechanism which applies only to much nearer orbits).

It's been theorized that without Jupiter, cometary bombardment of Earth would preclude advanced life. Barbarossa might be a similar prerequisite to intelligent life, arresting derelicts from the Oort cloud while Jupiter arrests derelicts from the Kuiper belt. Also, an orbit inclined much less than 39 degrees to the planets' orbits (Barbarossa's seems to be 14), is prerequisite to intelligent life (otherwise the outer planets, at least, intermittently would become eccentric - see above).


Reply author: Stoat
Replied on: 05/20/2007 03:20:25
Message:

Hi Joe, I just downloaded a pdf file of a paper by Tifft and gave it quick skim read. That's pretty amazing stuff[:)] When he got to the bit about our being flat liners on a one dimensional line, I thought of the song, "I'll take the high road and you take the low road, and I'll be in Scotland afor ye." ( the song's about death but that's by the by)

Now, I've never liked the idea of time as simply another spatial dimension but have no great problems with thinking of time as having a metric of its own. Suppose that we are forced, as matter, to walk the "high road" along the hillls and dales of a sine wave. Light has to do the same but its sine wave is of a much lower amplitude. Lazy fat gravity, just walks along the x axis. We all arrive at the same time, to find that we, matter, have walked a thousand miles, light has walked five miles and gravity has walked five yards [8D] Being one dimensional time people, we would be gob smacked by this, as we couldn't see the curves we walked but only a straight line.


Reply author: Joe Keller
Replied on: 05/20/2007 15:42:02
Message:

quote:
Originally posted by Stoat

Hi Joe, I just downloaded a pdf file of a paper by Tifft and gave it quick skim read. That's pretty amazing stuff[:)] When he got to the bit about our being flat liners on a one dimensional line, I thought of the song, "I'll take the high road and you take the low road, and I'll be in Scotland afor ye." ( the song's about death but that's by the by)

Now, I've never liked the idea of time as simply another spatial dimension but have no great problems with thinking of time as having a metric of its own. Suppose that we are forced, as matter, to walk the "high road" along the hillls and dales of a sine wave. Light has to do the same but its sine wave is of a much lower amplitude. Lazy fat gravity, just walks along the x axis. We all arrive at the same time, to find that we, matter, have walked a thousand miles, light has walked five miles and gravity has walked five yards [8D] Being one dimensional time people, we would be gob smacked by this, as we couldn't see the curves we walked but only a straight line.



Surely there are many unimagined possibilities. Thanks for your comment!


Reply author: Joe Keller
Replied on: 05/20/2007 16:20:06
Message:

Twelve light-years away, is a star system like the Sun, Barbarossa & Frey:

"Epsilon Indi Ba,Bb: The nearest binary brown dwarf", McCaughrean et al, Astronomy & Astrophysics 413:1029-1036, 2004.

The primary, Eps Indi A, is a main-sequence Type K star thought to be roughly 1.3 Gyr old. At a distance of 1500 AU, it is orbited by a pair of brown dwarfs (Ba & Bb) which orbit each other about 2.65 AU apart. There is extreme observational bias in favor of hotter brown dwarfs that are self-illuminated in infrared, and in favor of resolvable brown dwarfs farther than 1000 AU from their primaries. So, presumably this pair of brown dwarfs is unusually massive and hot, and unusually distant from the primary.

The dimmer one, Eps Indi Bb, is estimated to have 0.027 solar mass and surface temperature 854K. I estimate Barbarossa to have 1/3 this mass and, like the Sun, 3.5x this age. So, it's plausible that Barbarossa would be too cold to be self-illuminated in infrared.

From luminosity and temperature, the diameter of the brighter one, Eps Indi Ba (which has est. 0.045 solar mass) is estimated as 53,000 mi.; this is only 72% of "the minimum...predicted by structural models..." (Op. cit., p. 1034). So, Barbarossa likewise might be much smaller than the Jupiter size predicted by structural models that assume a Jupiter-like composition.


Reply author: Joe Keller
Replied on: 05/20/2007 18:55:25
Message:

quote:
Originally posted by Bill_Smith

Looks like the forum has gone a little haywire with duplication.

About the magnitudes, so this is why the object is at or near the limit of Bobs images. I think I quoted mag 18.5 to Bob when I measured them so it leaves a large gap between 18.5 and 20.

If the object is a brown dwarf would you be relying on albedo only?

Cheers

Bill




Hi Bill!

My May 20 posts address the self-illumination issue. Thanks for mentioning it!

- Joe


Reply author: Joe Keller
Replied on: 05/20/2007 19:18:15
Message:

Because IRAS did not find it, Barbarossa must be much colder than Burrows' theoretical temperature (or else much less massive than believed). Burrows et al, Reviews of Modern Physics 65:301+, gives Table I (p. 316) and formula 2.58 (p. 312; see also p. 305 for a very brief explanation of the "Rosseland mean opacity") for theoretical brown dwarf temperature as a function of age and mass. Because Barbarossa's presumed 0.0090 solar mass is outside Table I, I used Burrows' formula to find (assuming age 4.6 Gyr) that Barbarossa's surface temperature should be 229K. By Wien's law, the peak emission would be at 12.65 microns.

The limit for deuterium fusion, is said to be 0.013 solar mass (Oppenheimer et al, in: Mannings et al, "Protostars & Planets IV", 2000), but deuterium fusion lasts only briefly, early in the life of the brown dwarf. At a Gyr or more old, Burrows' temperature-mass chart changes little across the 0.013 solar mass boundary (Burrows et al, 1997, in: Mannings, op. cit., Color Plate 22); so the above surface temperature formula remains fairly accurate. (Gravitational potential energy is about as important as deuterium fusion.) This chart shows that the absolute luminosity of Barbarossa should be 7.25 log10 units less than the sun's, thus its apparent (full spectrum) luminosity 7.25 + 4.6 = 11.85 log10 units less.

The infrared radiation measured by IRAS from a 229K Barbarossa (assuming "Burrows size" for its mass, i.e., approx. Jupiter size) would be roughly the same, in spectrum and intensity, as that measured from an asteroid 3 AU from the sun and 1000 mi in diameter. IRAS cataloged many asteroids much smaller than this.

"Asteroids and comets moving more slowly than 1' per hour would hours-confirm and thus reside in the Working Survey Database."

- NASA IRAS Asteroid & Comet Survey webpage

From a 229K Burrows-size (approx. Jupiter-size) Barbarossa, IRAS would have measured a 5900 Jansky peak at 12 microns. Within one degree of Barbarossa's expected position, the main IRAS catalog's brightest object at 12 microns, measured 6 Jansky; within 10 degrees, 82 Jy.

If a Burrows-size Barbarossa had a rather Neptune-like surface temperature of 48K, its radiation peak would be 55 Jy at 60 microns. Within one degree of Barbarossa's expected position, the brightest object at 60 microns is 1 Jy; within 10 deg, 27 Jy.

With no internal heat at all, the surface temperature would be about 18K (equilibrium at Barbarossa's distance from the sun; only slightly warmer than the cosmic far infrared background). Barbarossa either would be indistinguishable from background cold interstellar dust; or, if Barbarossa were seen against a 3K background, IRAS would record only 2.1 Jy even at 100mu, 0.5 Jy at 60mu, and 0 at 25 & 12mu. The lowest readings ever found are about 0.7 Jy @ 100mu, 0.3 Jy @ 60mu, and 0.2 Jy @ 25 & 12mu. So, I looked for readings within a factor of two, of 2.8 Jy @ 100mu, 0.8 Jy @ 60mu, and 0.2 Jy @ 25 & 12mu. There were 76 such IRAS objects within 10 degrees, but none within one degree, of Barbarossa's expected position. If Barbarossa's diameter were half the Burrows value, none of the brightnesses should be much above noise.

So, only a completely cold (equilibrium with solar radiation), smallish (~ 1/2 Jupiter diam) Barbarossa is readily consistent with IRAS' negative detection, given my coordinates. If my coordinates are wrong, then many IRAS objects (about 0.2 per sq degree) are consistent with even a Jupiter-size Barbarossa, if it is 18K. Some IRAS objects have been correlated with known objects and some haven't.

Burrows assumes that cooling is limited by the rate of radiation through the degenerate matter, and that convection is insignificant. On the contrary: most of the gravitational energy will be released at the boundary where the nondegenerate mantle is collapsing onto the degenerate core. The nondegenerate overlying mantle would melt and convection would carry the heat to the surface, as on Earth. Thus Burrows' temperatures might be accurate for dwarfs that have burned deuterium in their cores, but might drastically overestimate the temperature of sub-dwarfs like Barbarossa whose heat is only gravitational.


Reply author: Joe Keller
Replied on: 05/22/2007 18:10:05
Message:

Pursuant to Kozai's and similar expressions, I found arcsin(rms sin(i)), the "rms sin(i) inclination to the ecliptic" of KBO's: it is 12# 13.5' (standard error of mean, 41')(n=220)(SEM questionable due to skew distribution). I used the top third (i.e., discovery years 2003-2006) of the "general ephemeris" list on ifa.hawaii.edu. The bottom quarter of the list, KBOs with special names, also contained some 2003-2006 discoveries but I omitted those. The absolute magnitudes of the KBOs I used were mostly +6 to +8 (vs. +3 for Varuna).

This agrees with the orbital inclinations of Barbarossa (12# 9.7') and Lescarbault/LeVerrier's Vulcan (12# 10'). Iowa State Univ. has LeVerrier's Compte Rendus communication (January 2, 1860; LeVerrier entered Lescarbault's letter to him on p. 40 and his own comment on p. 45). I can't read even scientific French well, but it seems to me that LeVerrier discussed no other sightings; he simply without fanfare corrected Lescarbault's orbital inclination calculation. The subsequent two years of Compte Rendus contained many communications by LeVerrier, but only about other subjects.

I found that the ascending nodes, omega, of the same KBOs, cluster near 14.00# and 194.00# (n=175). I omitted KBOs with i < 3# because for these, the ascending node on the ecliptic correlates poorly with the ascending node on Jupiter's or Saturn's orbit. Otherwise all KBOs were weighted equally. The best fit, theta, was defined as minimizing the sum of abs(sin(omega - theta)); the criterion should emphasize the difference between 10# & 20# error more than the difference between 40# & 50#, or 70# & 80# error. This agrees with the ascending nodes of Lescarbault/LeVerrier's Vulcan (named by Babinet) (12.98#) and, except for a 90 degree shift, of Barbarossa (283.69#).

The eccentric (e > 0.1) subset of KBOs (these also tend to be the ones with larger inclination) significantly clusters near omega = 20# and 200#. Again omitting those with i < 3#, 40/117 had omega (rounded to the nearest degree) within the inclusive intervals [1,40] or [181,220] (p = 0.076, Poisson test). This clustering occurs despite the tendency, for small i, for omega to cluster near the ascending nodes of Jupiter & Saturn, which are at right angles to this. That is, the clustering would be seen to be even more significant if reference were made to the orbital plane of Jupiter/Saturn instead of to the ecliptic.

Using the Jupiter/Saturn reference plane instead of the ecliptic, would increase Barbarossa's inclination almost two degrees, and move the ascending nodes of Vulcan and the clustered KBOs, backward about seven degrees. Even that, would be close enough agreement to suggest that Barbarossa influenced the orbits of Vulcan (whatever Vulcan was) and the KBOs. All this is more reason to turn a big telescope there and look.


Reply author: Joe Keller
Replied on: 05/23/2007 16:51:47
Message:

From the chart of KBO semimajor axes on Prof. Jewitt's Kuiper Belt website on ifa.hawaii.edu, one sees that some KBOs, like Pluto (hence called plutinos), cluster at 39.4 AU where their orbits have 3:2 resonance with Neptune's orbit. On this chart, KBOs scarcely occur with semimajor axes 47.8 AU where there would be 2:1 resonance with Neptune. Also the 5:3 resonance distance, 42.3 AU, intersects the distribution well off to one side: nothing peaks there.

On Prof. Jewitt's chart, the median and the mode of the distribution of semimajor axes, of the non-plutino majority set of KBOs (and of both the low- and high-eccentricity subsets separately) is about 43.5 AU. The eccentricity minimum of the high-eccentricity subset, is about 43 AU. (The high-eccentricity subset becomes more eccentric with distance; if sqrt(1-e^2) increases linearly with distance, then e=1 at about 50.5 AU.) If not resonance, what is at 43.5 AU, to attract the KBOs?

The distance at which the Barbarossa system torques KBO orbits, as effectively as the remainder of the solar system torques them, is 43.75 AU. The strength of the CMB dipole implies, from my theory above, that the Barbarossa+Frey system is 0.0104 solar mass. I totaled the torque due to all significant known solar system mass, including Pluto and Charon but excluding any other KBOs. The effect of Pluto is such that if plutinos equal to 100x Pluto's mass were added, the distance would change to 43.2 AU (for Pluto & plutinos I approximated them as circular orbits at their semimajor axis).

I used Gauss' idea of calculating (by Romberg trapezoidal numerical integration) simply the torque on Ring A due to Ring B, where Ring B is the orbit of a planet or Barbarossa. When Ring A is at 43.75 AU, then Barbarossa, vs. the rest of the solar system, cause equal torques, per degree of tilt.

Suppose all the KBOs originally lay in the Jupiter/Saturn plane. If Barbarossa's mass were negligible compared to Jupiter, then all the KBOs would have i=0 in the J/S plane (i.e., i=0 to 2 in the ecliptic plane) forever. If Jupiter's mass were negligible compared to Barbarossa, then the KBOs would precess about Barbarossa's plane, which is inclined 14# to the J/S plane, giving i=0 to 28. Because really the effect of J/S/N et al, at 43.75 AU, equals that of Barbarossa/Frey, the observed situation is a compromise. Roughly half the non-plutino KBOs are found near i=0 and roughly half are spread out between i=0 to 28.

On my list of 220 unnamed KBOs discovered 2003-2006 (it includes some plutinos), I counted 15 at inclinations [18,21], 11 at [22,25], 6 at [26,29], 4 at [30,33], and one apiece at 36, 37, & 48 (this latter surprisingly with eccentricity only 0.13). This hints at the 39 degree limit beyond which chaotic variation of eccentricity and inclination theoretically occur. It also shows that an exponential dropoff in population begins at about i=25.

In the above sense, Jupiter is an order of magnitude more effective than Barbarossa, at torquing Saturn. Thus Saturn's inclination to Jupiter is an order of magnitude less than its inclination to Barbarossa. Uranus should, on average, have larger inclination than it does, but this could be a chance time in the precession cycle when the inclination is rather small. Neptune is a paradox. Barbarossa should torque Neptune about as effectively, in the above sense, as do J/S/U. Yet Neptune is only 0.9# from the J/S plane. (Like a good KBO, Neptune's ascending node on the J/S plane is near right angles to Barbarossa's.)

Maybe the aberrant axial rotation of Uranus, allows Uranus to exchange orbital angular momentum with Neptune by some new physics. Uranus would be the linchpin holding Neptune in the plane of the ecliptic. Alternatively Neptune might simply be analogous to the low-inclination half of KBOs that seem to be influenced by J/S rather than by Barbarossa: dynamically there seems to be an all-or-nothing choice of influences.


Reply author: Stoat
Replied on: 05/24/2007 06:05:37
Message:

Hi Joe, have you looked at Eris and Sedna yet? Their perihelion looks to be about 180 degrees from our brown dwarf. Huge eccentricities, so we can find them now as they come in close. That suggests that there are others, of about the same mass way way out there on highly eccentric orbits.


Reply author: Joe Keller
Replied on: 05/24/2007 12:26:06
Message:

quote:
Originally posted by Stoat

Hi Joe, have you looked at Eris and Sedna yet? Their perihelion looks to be about 180 degrees from our brown dwarf. Huge eccentricities, so we can find them now as they come in close. That suggests that there are others, of about the same mass way way out there on highly eccentric orbits.



Thanks again for your input! One of my posts above cites a plot of Trans-Neptunian Objects known as of 1998. Reviewing my notes, I see that these were shifted, in my best analysis, roughly 5.9 AU toward ecliptic longitude 182. That is, the average perihelion was at 002, 78 degrees from Barbarossa's ascending node, 284.


Reply author: Joe Keller
Replied on: 05/24/2007 14:37:42
Message:

Response to a Ph.D. physicist on another messageboard:

> Are u claiming this theory is correct? If yes, please provide some basis
> for it, starting with how a grav field can produce or influence (other than
> negligible blueshifting) incoming microwaves.

Thanks for your good questions!

The fullest writeup is posted by "Joe Keller" in the "requiem for relativity" thread of Dr. Van Flandern's metaresearch.org messageboard. Here's a paraphrase:

My theory is that the microwaves are produced not by the "big bang" nor even in intergalactic space, but at an interface at which the sun's gravitational force is of a certain strength. The vacuum isn't empty; rather, it's like a pot of water. Water evaporates at an interface at which intermolecular forces are of a certain strength. Just as the infrared wavelength coming from boiling water is determined by the *pressure* at that interface (the surface of the water), the wavelength of the CMB is determined by the gravitational *potential* at the interface where the gravitational force is of a certain strength.


> Feel free to post details including mathematical formulae. You're talking
> to a Ph.D. physicist, other well-qualified folk are watching. Your claims
> seem outlandish without details.

I'm sitting here with my notebook of formulas. Why don't I post them all? Maybe I will tomorrow. Maybe I'll post my BASIC solution program too. The formulas aren't useful without a computer program to do the numerical integrations and successive approximation solutions. So it isn't proof unless someone checks the correctness of my program, and no one is going to take time to do that. It would be easier just to aim a telescope, than to check a computer program written in BASIC by a stranger. Furthermore, the attitude I've been encountering elsewhere is so adverse, that if I were to post a formula with a minor error, I'd never hear the end of it, and my case might be lost forever. So, here's how to write the equations yourself ("Teach a man to fish..."):

Consider the gravitational field of the sun at 52.6 AU: call its strength "a0". (Several pages of circumstantial evidence of why that distance is important, are on Dr. Van Flandern's website.) Now add a point mass m0 << msun at distance r0 > 52.6, at the pole of polar coordinates with the sun at the center. For every polar angle phi, there is some distance r, near 52.6 AU because m0 << msun, at which the field strength exactly equals a0; let E(phi) be the gravitational potential at (r,phi). For each m0, there is an r0 such that, the dipole (first order Legendre term), of E(phi), is in proportion to the CMB dipole observed.

I get r0 from the circular orbit displayed by the aforementioned 1954, 1986 & 2007 centers of mass. Within the measurement error of the progression period of the 5:2 Jupiter:Saturn resonance, this is the same thing as the r0 which would give that period. Then I know m0.

Some of your other questions are addressed in previous posts to this messageboard. Right now, theoretical discussion is less important than campaigning for observations to be made with more powerful instruments.


Reply author: Joe Keller
Replied on: 05/24/2007 14:57:13
Message:

(previous reply to the same correspondent, Grant Hallman, a Ph.D. physicist)

Hi Grant,

I didn't see your questions because they weren't near the top! I'm used to getting responses on other messageboards that only amount to a string of derogatory adjectives, just namecalling, not substantive.

I didn't notice that *your* response, by contrast, really said something! Thanks for your input!

- Joe Keller


(Keller) > >For reasonable masses, correction for the tidal gravity of the objects, reduces the variation of the Pioneer Anomalous Acceleration. The net Anomalous Acceleration becomes fairly smoothly decreasing with distance from the sun. I think that despite the likely 0.01 solar mass for the combined objects,...


(Hallman) > How was this figure reached? For comparison, that would be about 10x Jupiter's mass.

(Keller) I've posted the theory for that in detail on the metaresearch.org messageboard. Basically the theory is that the symmetry of the CMB arises because of the symmetry of the sun's gravitational field, and that the sun's gravitational field somehow produces the CMB! (That theory didn't originate with me.) The CMB dipole, according to my extension of this theory, arises from the planets. This gives a formula for the mass needed at Barbarossa's distance. In turn, Barbarossa's period, hence distance, is implied by the rate of progression of the 5:2 Jupiter:Saturn resonance.

(Keller) > >they fail to disrupt the solar system, because small shifts in the orbital planes of the known planets, counter the torque.

(Hallman) > Ok, that statement does not make sense. First, what causes the "small shifts"...


(Keller) It's just a system of interacting torques. The details are a mess (many-body problem) but the basic concept is simple. Basically, when a torque acts, it tilts an orbit, and the new angle produces another torque. Eventually enough adjustment occurs in the strong torques (e.g., Jupiter-Saturn) to neutralize the disruptive effects of the weaker torques (e.g., Saturn-Barbarossa). That's why, say, Alpha Centauri doesn't destroy the solar system through precession around the plane of Alpha Centauri's orbit, even after infinite time (or so it's believed). Jupiter and Saturn are more linked by torque than are Saturn and Barbarossa. When you go out as far as Neptune or the Kuiper Belt, the situation is more complicated because Barbarossa's torque becomes about as important as Jupiter's. (These torques are like that of the sun on the moon which causes the advance of the lunar
node.)


(Hallman) > and second, the statement is inconsistent with the assertion that the object(s) are at a "Jupiter:Saturn resonance point", which requires that Jupiter and Saturn affect the object, whereas the object, which is alleged to be 10x heavier than Jupiter, does not affect Jupiter and Saturn.


(Keller) I'm not saying J/S have no effect, just that there's an effect on J/S, which we are observing when we observe the progression of the 5:2 resonance. The stablest situation is for the big outside object (Barbarossa) to be at one of the resonance points. The resonance point follows the big outside object because that's the stablest scenario. It's an extension of the basic idea of resonance.


Reply author: Joe Keller
Replied on: 05/25/2007 15:57:59
Message:

It seems that Lescarbault assumed a parabolic orbit (with argument of perihelion = 270#) for Vulcan, and LeVerrier a circular one. My rough calculation for such a parabolic orbit gives, in heliocentric ecliptic coordinates, delta(z)/delta(sqrt(x^2+y^2)) = tan(7.6#), agreeing with Lescarbault. This assumes Vulcan was observed at the descending node; more accurately, 13# - (March 26 - March 21) = 8# before the descending node, gives 8.8#. For argument of perihelion = 180#, no distance gives a slow enough apparent speed: 0.33 AU gives the slowest apparent speed but it's still 20% too fast.

My plot of 85 KBOs from 2005-2006, shows that although eccentricity and inclination are correlated, it is eccentricity, not inclination per se, that correlates with ascending nodes near 14# or 194#. The most eccentric of the 85, had e=0.97 & omega=197#; next most, e=0.83 & omega=192#. This suggests that Vulcan, omega=13#, was a very eccentric big KBO, not a typical short-period comet. Vulcan would have had significant gravity; its surface might have resembled Mercury's, hence no cometary tail. When Lescarbault's report reached LeVerrier, nine months after the sighting, Vulcan would have been in the asteroid belt and maybe as dim as Neptune. Vulcan's failure to return before 2007 indicates that its major axis and aphelion are > 2*28 = 56 AU.

From a preliminary sample, the 2006 subset (n=35, excluding those with i<3) of my KBOs, I find only 4 with argument of perihelion within 45# of 270#, vs. 10 within 45# of 0#, 8 within 45# of 90#, and 13 within 45# of 180#. Thus 23/35 have argument of perihelion nearer 0/180; this might be explained without Barbarossa. On the other hand, the clustering of the Edgeworth-Kuiper Belt Objects' ascending nodes seems to require a large mass on an inclined orbit, i.e., a Barbarossa.

The most significant clustering of the longitude (not argument) of perihelion is near 270#: 11 near 270# (e.g., Vulcan?), 10 near 0#, 4 near 90#, 10 near 180# (excluding i<3). Again, symmetry makes this seem to require additional explanation. When the 5 objects with i<3 are restored to this sample (to give n=40), longitude 236# minimizes the sum of the absolute angular deviations.

The gravitational field can be approximated as a central inverse-square term, plus an inverse-cube term which has a central and a non-central part. The central inverse-cube term causes perihelion advancement (see Goldstein's Classical Mechanics, Ch. 3, Exer. 7); the non-central inverse-cube term causes regression of the nodes. Above I found that the Barbarossa, and known solar system, contributions to the non-central inverse-cube term, have the same derivative w.r.t. z (cylindrical coordinates) at the classical Kuiper Belt. By Poisson's equation, this also holds for the central inverse-cube term's derivative w.r.t. r (cylindrical coords). So the classical Kuiper Belt lies where, for both node regression and perihelion advancement, Barbarossa's vs. the known solar system's influences, are equally strong. At 52.6 AU, Barbarossa's contribution to the derivatives of the inverse-cube term, is stronger than the known solar system's, by a factor of about sqrt(4*pi).

Near the beginning of this discussion of the Kuiper Belt (and ultimately its importance in refuting the Big Bang and orthodox Relativity) Dr. Van Flandern questioned the existence of any type of barrier at 52.6 AU. Now I can address this objection directly. Let a KBO have semimajor axis 46 AU and eccentricity 0.15. These figures are only slightly more than the median for KBO samples I've seen. Aphelion would be 53 AU, and the KBO would spend almost twice as much time near aphelion as near perihelion. I think that this is the essence of Dr. Van Flandern's objection: the rather sudden, drastic reduction in KBOs, reported beyond 52-53 AU, can be real only if major axis and eccentricity are somehow correlated.

In my sample of 85, are 18 with semimajor axis 45 or 46 AU (rounded to the nearest AU); these would require e=0.17 or 0.14, resp., to reach 52.6 AU at aphelion. Four of these have large eccentricities ranging from 0.37 to 0.66. Seven have e=0.13, 0.14 or 0.15, bringing them almost to 52.6 AU. The remaining seven have e=0.09 or smaller. There is a significant gap at e=0.10 through 0.12 and e=0.16 through 0.36. In my sample are 25 with semimajor axis 43 or 44; these require e=0.22 or 0.20 resp. to reach 52.6 AU. One has e=0.24, one e=0.29 (both these were 44 AU). The remaining 23 have e <= 0.16 (the sole e=0.16 & e=0.15 both were 43 AUs). Thus all fell at least 0.06 eccentricity units under 52.6 AU or else were at least 0.04 over.

In the same sample of 85, I shuffled the eccentricities by giving each KBO the eccentricity of the KBO 31,61 or 73 entries ahead. For each of the three shuffles, the aphelia histogram decreased gradually throughout the range 48-57 AU; nothing special happened at 53 AU nor anywhere else. When I did not shuffle the eccentricities, the aphelia hostrogram declined only slightly until almost exactly 52.6, then suddenly began declining by a factor of 2 for each additional AU of distance.

A theorem of Lagrange (see Poincare, "New Methods of Celestial Mechanics, 1892-1899, vol. 13 in AIP History of Modern Physics & Astronomy series, p. 407; also GW Hill, Astronomical Journal 24(556):27+, 1904) says that in N-body motion, the major axis, i.e. energy, tends to be conserved. If so, then eccentricity changes only through change in the minor axis, i.e. angular momentum. Prograde vortices of force (i.e. an acceleration vector with positive curl) near 52.6 AU, would impart angular momentum without energy, increasing the minor axis of bodies approaching that barrier.


Reply author: nemesis
Replied on: 05/29/2007 14:28:27
Message:

Joe, have you posted since the 25th? The main header says there should be something from the 28th.


Reply author: Joe Keller
Replied on: 05/29/2007 17:36:17
Message:

quote:
Originally posted by nemesis

Joe, have you posted since the 25th? The main header says there should be something from the 28th.



Dear Nemesis,

My "It seems that Lescarbault..." post is from the 25th but I revised it yesterday, the 28th. Thanks for checking!

Sincerely,
Joe Keller


Reply author: Joe Keller
Replied on: 05/29/2007 18:29:52
Message:

Barbarossa & Frey also appear on the 1987 sky survey (I refer to this survey as "C"). So, Barbarossa & Frey are on the A, B, & C online survey scans, and on "G" (for J. Genebriera's photo). Barbarossa appears on the C plate as "C", the original object I had announced as Barbarossa. Frey appears as an object I had named "C6" in my own notes.

C (Barbarossa, 1987 La Silla red) RA 11 18 03.18 Decl -7 58 46.1
C6 (Frey, 1987 La Silla red) RA 11 17 43.1 Decl -7 48 38.5

The pair C/C6 can be added to the constant-speed great circle drawn through A2/A, B3/B, and Genebriera's Barbarossa/Frey of March 25, 2007. The heliocentric angular speeds from 1954-1986 and 1987-2007 become equal and consistent with observation, when Barbarossa's orbital distance from the sun is 197.664 AU, Barbarossa's orbital period 2810.03 yr, and the mass ratio Barbarossa:Frey = 0.8936:0.1064 = 8.4:1. Then, the heliocentric angular speed from 1986 (the B plate) to 1987 (the C plate) differs from the speed before or after, by only 0.383%, equivalent to 1.56".

The angle between the 1954-1986 path and the 1987-2007 path is only 0.0023 radian. The angle between the 1986-1987 path and the paths before or after, is arcsin(0.11). This error is consistent with a second Barbarossa satellite, Freya, with period between 2 and 50 years and mass between 1/15 and 1/8 Barbarossa's. I assume that the 1986-1987 path direction error equals the maximum producible by Freya in circular orbit; the error is likely small between 1954 & 2007, if Freya makes at least one orbit. The relative smallness of the 1986-1987 path length error, suggests that Freya's orbital axis about Barbarossa, lies rather near Barbarossa's orbital plane about the sun, and that Freya's orbit is seen rather edge-on.

The relatively large and comparable masses of Frey & Freya, suggest a complicated three-body orbit. Such an orbit would be needed, because the four Barbarossa-Frey radius vectors do not lie near any reasonable ellipse.


Reply author: Joe Keller
Replied on: 05/31/2007 16:48:51
Message:

I've found two objects consistent with Freya, Barbarossa's smaller planet:

B2 (1986) RA 11 16 57.0 Decl -7 53 29.6
C3 (1987) RA 11 18 37.6 Decl -7 54 09.5

If the mass ratios Barbarossa:Frey:Freya are adjusted to 14.3:2.04:1, the center-of-mass path ABCG (four time points, 1954-2007) is straight, and constant-speed to a precision consistent with, assuming an average location on the orbit, eccentricity 0.015. In 1986, Freya appeared 71% as far from Barbarossa as was Frey, and in 1987, 86%.

As assigned, the points are inconsistent with an elliptical orbit for Frey. An alternative to chaotic orbits, is reassigning object "A" as Freya, not Frey. This only slightly affects the overall fit, and gives three Freys & three Freyas, so elliptical orbits can be drawn.

I've looked at 15'x15' regions from 1954, 1986, 1987 and 2007. The regions chosen were, basically, those consistent with a Barbarossa orbit following that mean Jupiter:Saturn resonance point nearest the CMB dipole. The above assignments as Barbarossa, Frey and Freya came from among 2400 possible assignments that I considered (a million different assignments were possible but I considered only the brightest dots as Barbarossa or Frey). There were two more dependent than independent variables to be fit, and these were fit to about one part in 40 (error / region width), i.e., 1 part in 40^2=1600 overall. (The main lack of perfect fit, is due to the uncertain contribution of the unknown 1954 & 2007 Freyas.)

The 2400 choices were far from stochastically independent. A well- or poorly-fitting choice of Barbarossa, Frey & Freya usually implies a good or poor fit by similar choices. In effect there were far fewer than 2400 independent choices.

The J:S resonance points are 72# apart, but the (+) CMB dipole lies on Barbarossa's orbit only 2# behind Barbarossa. Because a causal lag is expected, this gives another factor of 36 in significance.

Additional significance arises from the smoothing of the Pioneer acceleration by subtracting Barbarossa's presumed tidal influence, and from the balance between Barbarossa and planetary tidal (1/r^3) forces at the classic Kuiper Belt.

A week ago I sent another 30 emails to professional astronomers. Of the estimated 200 emails I've sent to professional astronomers about this (over several months' time) only one has responded. The professional astronomer who responded didn't know my main purpose; I'd only asked him a trivial question.

I've read that Neptune first was observed by two "assistants". These would be the sociological equivalents of graduate students today. So, my new strategy will be to email graduate students until one simply looks and finds out whether any of this is or is not there.


Reply author: Joe Keller
Replied on: 05/31/2007 22:30:23
Message:

(posted in response to an inquiry on another messageboard - JK)

Thanks for your interest! In the north especially, Leo is getting too far west for the best view. Amateur astronomer Steve Riley in California did get, I think, some photos showing these objects with an 11", but the time of year was more favorable. I think he had a southern desert, maybe altitude, location somewhere in S. California without too much light pollution, and he went to a lot of trouble to maximize his magnitude cutoff with the electronic camera (stacking, etc.). If you do make a photo, please send me a "private message"; I'd like to check it! The mass, diameter & albedo aren't really known. That's all theory. So, please look!

The recent image in which I have most confidence is Joan Genebriera's with an 18" and electronic camera on Tenerife, March 25, 2007. I showed it to the president of the Des Moines, Iowa astronomy club. He did not think it was artifact. By comparison with a sky survey, I estimated the J2000 celestial coordinates as RA 11h 26m 22.2s, Decl -09# 04' 59". I theorize that the distance from the sun is 198 AU. Here's how to correct for Earth parallax: Joan's photo was slightly after opposition. Your photo will be slightly before quadrature. The difference in position between opposition & quadrature is 1/198 radian = 57.3/198 # = 0.29# = 17 arcminutes ("retrograde motion" due to Earth's motion). So, it's moved less than that. Also, Barbarossa is, I estimate, moving around the sun about 1/sqrt(198)=1/14 as fast as Earth, which partly compensates. So let's say 10 arcminutes. Find a star chart, find Joan's point (coordinates given above), then move 10 arcminutes west, parallel to the ecliptic, and aim there. That will be good enough if your field is 15'.


Reply author: Joe Keller
Replied on: 05/31/2007 22:54:52
Message:

Pulsar constancy is said to rule out acceleration of the sun, relative to the pulsars, greater than about the equivalent of a Jupiter at 200 AU (Zakamska et al, Astronomical Journal 130:1939+, 2005; I got this citation from a member of the "Bad Astronomy" messageboard). My mass estimate for Barbarossa might be 10x too high, or the error of this pulsar method might be 10x greater than believed.

The above article cites another, about dynamical detection of dark matter in the solar system (DW Hogg, AJ 101:2274+, 1991). Its Fig. 5 and accompanying text indicate that detection by residual errors in planetary ephemerides would require a Barbarossa at least 0.5 Jupiter mass, likely 2 Jupiter mass or more (maybe much more if there are systematic errors). Detection by "modeling", i.e., a prospective least-squares fit of ephemeris errors, theoretically (no one has done it with modern data) would require 1/3 that mass. (It might be easier simply to look.) Localization of Barbarossa within 1 degree would require at least 1.5 Jupiter mass, likely more than 6 Jupiter mass.


Reply author: Stoat
Replied on: 06/01/2007 08:32:39
Message:

When I think of the aether, I think of something billions of times more "rigid" than steel but I also think of it as a viscoelastic substance. Add to that the idea that half the energy of mass goes to make up the aether of a body. The sun's aether "atmosphere" is centred on the centre of mass of that body but the solar system's as a whole is off centred. Perhaps there's a tiny variation in aether density, which alters the red shift of pulsars downwards.


Reply author: Joe Keller
Replied on: 06/01/2007 15:16:04
Message:

quote:
Originally posted by Stoat

When I think of the aether, I think of something billions of times more "rigid" than steel but I also think of it as a viscoelastic substance. Add to that the idea that half the energy of mass goes to make up the aether of a body. The sun's aether "atmosphere" is centred on the centre of mass of that body but the solar system's as a whole is off centred. Perhaps there's a tiny variation in aether density, which alters the red shift of pulsars downwards.



Thanks for posting. I think these are excellent insights. I might be able to elaborate on them.

- JK


Reply author: Joe Keller
Replied on: 06/01/2007 17:49:10
Message:

Zamaska & Tremaine (op. cit., Astronomical Journal 130:1939+, 2005) say that pulsar timing shows that the sun is not accelerated by any pull as strong as a brown dwarf (e.g., a 10 Jupiter-mass object at 600 AU). Because they did not always average large numbers of pulsars, their work shows also that (millisecond) pulsars are not accelerated by any pull as strong as a brown dwarf. (They adjusted for the sun's planets, and for the pulsars' known stellar companions. Pulsars rarely have planets; apparently this is, in part, because supernovas destroy planets, at least nearby ones.)

Brown dwarf companions are common, perhaps usual. Only unusually bright and separated brown dwarf companions have been detectable, but even so, a star twelve light-years away has been found to have two of these. Brown dwarfs survive supernovas because they rarely orbit closer than 40 AU (often 1000 AU).

Contemporary theory is, that pulsars lack brown dwarf companions because pulsars receive at birth an impulse or "kick" which causes them to leave behind distant planets and companions, e.g., brown dwarfs. Usually only close stellar companions are durable enough and tightly bound enough to remain. Without such a companion, the pulsar is "ordinary". On the other hand, such a companion eventually ages out of the main sequence, massively interacts with the pulsar, and transforms the pulsar into a "recycled" or even a "millisecond" pulsar.

Let's challenge contemporary theory (see: Lorimer & Kramer, Handbook of Pulsar Astronomy, 2005, p. 30). "Millisecond" pulsars always should keep, the white dwarf companions which while giants provided the mass to spin up those millisecond pulsars. Yet sometimes millisecond pulsars are isolated, i.e., they lack Doppler evidence of any companion whatsoever. The somewhat slower merely "recycled" pulsars whose companions went supernova (providing less total mass though in a much shorter time), often should lack said partners, because of the "kick". Yet seldom are such recycled pulsars without a neutron-star companion.

A possible resolution is, that there is no "kick". The distance-age relationship of pulsars could be interpreted as constant small acceleration rather than constant velocity (see: Lyne & Graham-Smith, Pulsar Astronomy, 2006, Fig. 8.7). The young neutron-star companions of "recycled" pulsars always are there, and almost always are detectable because they tend to be close, at least in perihelion due to their eccentricity. The white dwarf companions of "millisecond" pulsars also always are there, but are undetectable by Doppler shift unless closer than some limiting distance beyond which the Doppler shift does not occur ("ether iceberg"). The limiting distance happens to equal the distance at which escape speed equals "kick" speed, and this limiting distance continually decreases. "Ordinary" pulsars usually have companions but these rarely are close enough to be on the "ether iceberg" and cause any Doppler effect; otherwise they probably would have interacted and the pulsar would have become not ordinary. This amounts to a partial repudiation of orthodox special relativity beyond a certain distance from the star.


Reply author: Stoat
Replied on: 06/02/2007 04:55:55
Message:

This raises a whole slew of questions. Though, the first one is about the politics of the thing. Talk about it now and it might look as though it's an attempt to fudge the issue of whether we have a brown drawf in our solar system. First we need a couple of very good images of the brute. Then we can look at what the pulsar data tells us.

So, do we talk about the ideas on this board, or just make a note of it for later? Tom must have loads of data on exploding planets but with a pulsar we are talking about an implosion first. Now I think of this aether "atmosphere" as being a sub luminal aether but it will be cantained within, permeated by, the ftl aether. Gravitational collapse must alter the sub light aether, as potential energy becomes kinetic energy.

What happens to this "kick" if a companion "knows," almost instantly, that its partner has imploded? It's orbit must alter hours before the explosion effects hit it.


Reply author: Joe Keller
Replied on: 06/02/2007 14:22:33
Message:

Barbarossa would torque the angular momentum vectors of J,S,U & apparently somehow Neptune too, as a unit, around Barbarossa's angular momentum vector many times in the life of the solar system. Presumably JSUN began with their average orbital plane as near as possible, in that precession circle, to the sun's equatorial plane (5 deg). Only about 1/12 of the time would JSUN's angular momentum vector be as close to the sun's as it is now (6 deg). Although this is believable, there's another explanation for the near equality of JSUN's plane to the sun's equator.

Companion stars at the distance of, say, Uranus, tend to orbit in the same plane as the primary's equator. Companion stars at the distance of Barbarossa tend to orbit in arbitrary planes. Typically a brown dwarf companion would be, according to the best estimates, somewhat farther away than Barbarossa but also somewhat more massive and at a somewhat steeper inclination, typically giving about the same torque as Barbarossa, overall, and a larger precession circle. Even a stellar companion of 0.7 solar mass at 20 AU would have no more angular momentum than an 0.1 solar mass brown dwarf at 1000 AU. If brown dwarf companions are as common as they seem to be, the correlation of close companion's orbits with the primary's equator might be poorer than it is. So, there might be an unknown mechanism of angular momentum transfer between a star and its planets, which maintains that alignment despite brown dwarf torques.


Reply author: Joe Keller
Replied on: 06/02/2007 17:57:08
Message:

Requiem for Relativity


Relativity is dead, long live Relativity.


Above, I suggest that there might be a kind of boundary or barrier surrounding the solar system ("ether iceberg"), maybe at 52.6 AU, beyond which the Doppler effect is not orthodoxly relativistic. Suppose a radio signal from Pioneer 10 crosses that barrier and reaches Earth. The redshift might be the sum of the two redshifts produced by: Earth's motion relative to the barrier (i.e., relative to the "iceberg"); and Pioneer 10's motion relative to the interstellar medium. If the barrier is scarcely moving relative to the nearby interstellar medium, then only some small effects, e.g., retardation of Pioneer 10's time dilation (see my 2002 Aircraft Engineering & Aerospace Technology article or its recapitulation on this messageboard) or retardation of stellar aberration (see my discussion of the Kimura phenomenon on this messageboard) remain.

Larger effects are noticed, when the movement of the barrier, relative to the interstellar medium, is not negligible. Pulsar timing fails to detect Earth's gravitational acceleration by Barbarossa, because to first order approximation, Barbarossa equally accelerates the 52.6 AU barrier and everything inside it. Also, interstellar space outside the barrier is so slack over interstellar distances that this term of the redshift doesn't change either. By the same mechanism, pulsar timing fails to detect distant companions of the millisecond pulsars. Above, I argue that the solar system detections around pulsars are better explained by this than by the "kick" theory.

Binary stars sometimes show paradoxical orbital redshifts (AW Irwin et al, Publications of the Astronomical Society of the Pacific 108:580-590, 1996). The binary star 36 Ophiuchus AB shows a supposed mutual Doppler redshift about 30% larger than consistent with any possible celestial mechanics. This might be because the light from one star passes through the dumbbell-shaped (defined, as above, by critical gravitional field intensity) barrier three times.

(About half of my posts on this subject recently disappeared from one of the largest amateur astronomy messageboards in Australia.)


Reply author: Joe Keller
Replied on: 06/03/2007 22:26:31
Message:

Let's consider a nearby star with a surrounding spherical barrier or boundary (like the barrier or boundary I hypothesize for the sun at 52.6 AU). Maybe the star can achieve an apparent Doppler acceleration toward us, in two ways. One way, is if a distant companion, outside the star's barrier, accelerates the star and its barrier as a unit toward us.

Another way, is if a close companion, inside the star's barrier, accelerates the star relative to the barrier. There is a wavetrain between us and the star. Part of this wavetrain is within the star's barrier: call that part's length, L. If the first time-derivative of L is negative, blueshift may occur.

Let the star have a small distant companion whose distance is twice the barrer radius. If the "barrier" occurs at that surface in space where the magnitude of the gravitational field vector is such-and-such, then a distant companion causes the barrier to be indented, like a "flat tire", a small distance proportional to the mass of the small companion. Let the small companion be seen in inferior conjunction. As the small companion slowly orbits at inferior conjunction, the star and barrier are accelerated toward us. Also, we see the star's light through a part of the "tire" that is less and less indented. The first and second time-derivatives of L (see previous par.) are positive. There is accelerating positive "apparent radial velocity" due to the asphericity of the rotating barrier. The barrier's indentation is proportional to the mass of the small companion. So, the distortion of the barrier commonly could appear to cancel up to about half, of the entire barrier's acceleration toward us.

Now let's consider an equal-mass binary near conjunction, both sunlike, separated by a distance of order 50 AU. The barrier in such a binary might be a dumb-bell or it might be two flattened spheres which are not connected (dumb-bell with no handle). Either way, light from the more distant star passes twice through the barrier of the nearer one. Plausibly, the apparent (from apparent radial velocity, "RV", red/blueshift measurement) acceleration, of the farther partner, inferred from the progessive blueshifting of its light, is the sum of five terms:

1. The acceleration of its own barrier toward us.

2. The acceleration of its partner's barrier away from us (#1 & #2 cancel, because the light passes through both these equal but oppositely accelerated "icebergs").

3. The second time derivative of L1, the length of wavetrain within its own barrier (zero for a spherical barrier).

4A. The second derivative of L2, the length of the farther partner's wavetrain, within the nearer partner's barrier, but only that part due to the shortening at the surface nearest us.

4B. This term is the same as 4A except that it is that part due to shortening at the surface farthest from us, and has sign opposite the sign of 4A. That is, shortening at the surface nearest us simulates movement toward us, and shortening at the surface farthest from us simulates movement away from us.

The orbital elements (see Aitken, "Binary Stars" for clear definitions of these symbols) of 36 Ophiuchus AB (Irwin et al, 1996, op. cit.) make it relatively easy to calculate what the apparent "RV"-based acceleration of the farther partner toward us, should be if the foregoing theory is correct. With a slide rule, trigonometry and careful approximation (considering, to first order, the distortion of the barriers), I found that the "RV"-based apparent acceleration toward us should be 1.64 times the celestial mechanics prediction. Irwin et al found 1.64 times.

"Web of science" (the online Science Citation Index) shows only six papers citing Irwin's. None of these measure binaries like Irwin's. 36 Ophiuchus AB is special because the stars are far enough apart to be outside each other's barriers, yet close enough to conjunction that the farther partner's light must pass through the "barrier" hypothetically surrounding the nearer partner. The nearer partner's hypothetical "barrier" eclipses the farther partner.


Reply author: Joe Keller
Replied on: 06/05/2007 17:08:27
Message:

Open letter to professional binary star experts:


Dear Dr. Fletcher, Dr. Marcy, Dr. Butler, Dr. Yang, Dr. Campbell, Dr. McArthur & Dr. Cochran:

Alan W. Irwin, et al (Publications of the Astronomical Society of the Pacific 108:580-590, 1996) discovered that the Doppler-derived radial velocity ("RV") of the farther partner in the binary star, 36 Ophiuchus AB, is anomalous by a factor of 1.64. This did not seem to the authors to be plausibly due to any other companion, nor to motion of the star's surface.

Within 1% error, this 36 Ophiuchus AB anomaly in RV, is consistent with my theory of solar system (our solar system and others) light propagation. This theory deviates somewhat from the standard textbook interpretation of Special Relativity, but might nonetheless be valid. No physical theory, much less any standard textbook interpretation, has lasted forever. The evidence for the high accuracy of the standard textbook interpretation of Special Relativity, has been gathered from particle physics experiments and might be not always exactly applicable at solar system distances.

The unexplained Kimura phenomenon (Astronomical Journal, 1902), which is, basically, that the aberration of starlight, corresponds to Earth's velocity retarded by about seven hours, is encompassed by my theory. My theory also encompasses the apparent sinusoidal variation in the Pioneer anomalous acceleration, which sinusoidal variation began at about 53 AU and is plainly discernible on at least one of JD Anderson's graphs published on ArXiv.org.

I'd like to discuss my theory with any or all of you. It might apply to other binary stars such as gamma Cephei. Please give me a call at **********.

Sincerely,
Joseph C. Keller, M. D.


Reply author: Stoat
Replied on: 06/06/2007 11:39:54
Message:

For anyone who wants to put some figures in for doppler shifts, using the contracted form. Note that the square root will alter any assessments of mass.

f = f0 sqrt (1 - v^2 / c^2) c / c + or - v cos theta


Reply author: Joe Keller
Replied on: 06/06/2007 18:38:17
Message:

Kimura phenomenon.

Refs.
H Kimura, Astronomical Journal 22(517):107-108, 1902.
SC Chandler, AJ 22(524):164, 1902.

"[Kimura's latitude variation] does not depend upon the longitude...[it] is not due to the wanderings of the pole. ...No plausible explanation of the Kimura term has as yet made its appearance,..."

"...southern hemisphere [observations show variations of the same sign as the northern hemisphere]"

- Frank Schlesinger, Proceedings of the American Philosophical Society 54(220) (available on internet) 1915, p. 357.


Circa 1900 the Astronomical Journal has many papers, often by famous astronomers, about "variation of latitude" and the related problem "aberration of starlight". Usually, the altitude of a star would be measured vs. a plumb line or mercury pool, when the star crossed the meridian. Often, bright stars and smallish refractors were used, day and night, to measure a star throughout the year. The observatories then involved, were in the northern hemisphere, often northern Europe, so to maximize accuracy, almost-circumpolar northern stars were used because these were near the zenith.

Kimura's meta-analysis included many studies. These seem to have been corrected by their authors, for aberration of starlight (neglecting Earth's rotation and orbital eccentricity) and for precession; and by Kimura, for Chandler wobble.

It would seem that the 9" "lunar" term of Earth's nutation also was corrected as part of the authors' "reduction procedure"; its long (18.6 yr) period would be forgiving of small errors, because our interest is in a phenomenon with period one year. The 1.2" "solar" nutation term has period one-half yr, so could contribute negligibly to Kimura's phenomenon, which has period one yr. The 0.1" "fortnightly" nutation has period 15 days, so could be trusted to average out.

Parallax corrections sometimes would be insignificant, and sometimes were significant and made. Their effect is opposite in phase, to the Kimura phenomenon (with my sign presumption; see below).

Chandler said that Kimura's "phenomenon may have an objective existence", perhaps "an annual periodical oscillation" of "earth's center of gravity". Chandler did not propose that nutation was the explanation. Chandler noted two imperfections (both of about the magnitude of Kimura's phenomenon) in the correction for the aberration of light:

1. Chandler indicated that often observations were not timed to the nearest hour, and that the average hour might vary enough, seasonally, to give the Kimura phenomenon. This author calculates an average observation time four hours earlier (!) in winter than summer, to give Kimura's 0.03" amplitude, even assuming usually a sensitively positioned star such as Vega or Zeta Draconis near the ecliptic pole, and assuming observation times seldom were recorded to the hour. This suggestion of Chandler's gives the correct period and phase for Kimura's phenomenon, but cannot give enough amplitude. Also this suggestion reveals that Chandler uses the same phase for the Kimura phenomenon that I do: a reversed phase would imply *later* observations on winter nights than summer, which is absurd.

2. Chandler indicated that stellar aberration wasn't corrected for Earth's orbital eccentricity (correction for Earth's rotation is unneeded for declinations). Chandler doesn't give the period of this term but my own approximations suggest that the period is 1/4 yr; therefore it averages out.

On the other hand, the retardation of stellar aberration, to the time the light crosses the 52.6 AU barrier, agrees with Kimura's phenomenon in period, phase and magnitude. A sensitively positioned star, i.e., +/- 90 ecliptic latitude, e.g. Vega or, better, Zeta Draconis, would move 0.107". A star near the arctic circle and 6h RA, e.g. Capella or, better, Delta Aurigae, would move 0.0765". For a crude estimate, suppose the stars studied are distributed randomly over the northern hemisphere. At ecliptic (lat,long)=(0, 0 or 180), the movement is 0.107 * sin(23.5) = 0.043"; at (0, 90 or 270), it's zero. This averages 0.050", vs. Kimura's amplitude of about 0.03".

A better estimate realizes that the stars used for such studies basically were evenly distributed on the arctic circle. I tallied all variation of latitude studies, and included the (more numerous) aberration of starlight studies, published in the Astronomical Journal 1856-1899. There were 46, excluding one using Polaris. I calculated the exact effect predicted by my theory at 18h, 6h, 0h & 12h; interpolated at 3h, 9h, 15h & 21h using a quarter-wave spline; and weighted by the number of studies using stars nearest those RAs. The resulting predicted amplitude of the Kimura phenomenon, is 0.0304".

Unaccounted stellar parallax (see below) is thought to be up to one-fourth the magnitude of the observed Kimura phenomenon (but opposite in sign, I assume here), which would decrease my prediction by as much as 25%. On the other hand, the calculation above assumes all stars are on the arctic circle; really, few were above the arctic circle; some ranged as far south as Arcturus; usually the latitude of the observatory was preferred, which according to Kimura averaged 42N. Use of the 42nd parallel instead of the arctic circle increases my estimate by 41%. Stars might sometimes have been measured on the far meridian (under the pole); this would subtract from the phenomenon. So, the agreement between theory and observation remains good, when I explain Kimura's phenomenon as a retardation of Earth's effective velocity vector, by 52.6 AU * 8.3 minutes/AU.


Sign confusion in Kimura phenomenon.

Kimura uses "phi" for baseline latitude on p. 108. This suggests that phi minus phi0, essentially Kimura's "xi", is the baseline latitude, minus the latitude measured by the star's altitude on the meridian. That's the sign I used. The correctness of this also is suggested by Kimura's formula in his introduction, which puts his "xi" (sometimes called "Kimura's Z") term on the same side of the equation as what looks like the Chandler pole wobble (hence opposite in sign, to the measured minus baseline latitude).

However, in "Physics of the Earth - II The Figure of the Earth - Bulletin of the National Research Council" (1931)(available on the internet), p. 269:

"...annual parallaxes of the stars...cannot account for greater than one-fourth of the [Kimura phenomenon] effect."

This usage implies that the sign of the Kimura phenomenon is the same as that of stellar parallax, i.e., opposite what I assumed. Then phi would be the latitude measured by the star's altitude, and phi0 the baseline latitude. If so, the phase of Kimura's phenomenon is opposite that expected from retardation of Earth's velocity vector.


Reply author: Joe Keller
Replied on: 06/08/2007 17:57:39
Message:

Except for a sine function in place of a cosine, an article by SC Chandler, Astronomical Journal 17(400):125-127, 1897, resolves the sign ambiguity of the Kimura phenomenon in favor of my usage and my theory. Furthermore Chandler gives an observed dependence, of catalog declination errors, on right ascension, which might conform to my theory.

The three catalogs considered by Chandler were Russian/French, or German productions. In this era such catalogs typically covered the sky to Decl -30. Catalog declinations typically were measured when a star was on the meridian at midnight.

In 1902 Chandler mentioned that stellar aberration corrections then customarily ignored Earth's orbital eccentricity. (This reflected not so much laziness, as lingering doubt of the true cause of the aberration.) Using Simpson's rule to integrate over the truncated lune from -30 to +90, I find that correction for eccentricity, requires on average that 0.182" be added to declinations of stars at RA 180#, and that as much be subtracted from the declinations of stars at RA 0#.

Let's assume that the correct sign of Kimura's phenomenon, makes it consistent with the theory that Kimura's phenomenon is caused by a seven-hour (i.e., 52.6 AU / c) retardation of Earth's orbital velocity vector. Again using Simpson's rule to integrate over the truncated lune, I find that correction for said retardation, requires on average that 0.0665" be added to declinations of stars at RA 270#, and 0.0755" subtracted from declinations at RA 90#, with declinations 0# & 180# requiring intermediate, small positive, corrections.

The above two corrections together amount to adding approximately:

0.126" * cos( RA + 139 )

to the declinations. Except for the presence of a sine rather than a cosine in Chandler's formula, this agrees perfectly with Chandler's "systematic correction needed by the new Pulkowa declinations":

+0.14 * sin ( RA + 135 )

and with Chandler's correction to the previous 1865 version of the Pulkovo catalog:

+0.09 *sin ( RA + 153 ).

Chandler's correction to declinations in Kuestner's 1890 catalog was:

+0.12 *sin ( RA + 120 ).

With standard error of the mean, the average of the three is:

(+0.117+/- 0.024 ) * sin ( RA + 139+/- 8 ).

The remarkable resemblance of this to my theory, suggests that the replacement of cosine with sine is a transcription error, an error by Chandler, or due to an unknown complication.


Reply author: Joe Keller
Replied on: 06/09/2007 19:00:16
Message:

If alternatively, an advancement rather than a retardation ( by 52.6 AU / c ) is assumed for Earth's velocity (i.e., other sign convention for Kimura phenomenon) 180+34 --> 180-34 for the axis needing positive correction. That is, RA-214 --> RA-146, giving cos(RA-146) = -sin(RA+124) or rather -sin(RA+117) considering that Earth's perihelion (1889 ephemeris) is 11.5# past the solstice and that most of the effect depends on the former, not the latter. This too is acceptably close to Chandler's values, especially if Kuestner's catalog (which differs only 5% in amplitude and 0.05 radian in phase) is emphasized. The problem now is that the sign is reversed from Chandler's. Maybe there is a misunderstanding about what is being corrected to what. For example, the catalogs might be based on declinations measured on the meridian at all seasons and Chandler's supposed correct values based on midnight declinations.

Wagner's 1861-1872 right ascensions of Polaris (quoted in Chandler AJ 19(444):89-92, 1898) are consistent with advancement rather than retardation of Earth's velocity as it affects stellar aberration. The table on p. 90 gives Wagner's "Observed minus Calculated" values for the RA of Polaris. I used the 1887 (oldest ephemeris in the library) coordinates, with yearly changes given, for Polaris, extrapolating them to 1867. Including along with the velocity advancement also the small effect of parallax, I found an expected excess, of "Obs - Calc", with period one year, and maximum 0.289s (i.e., seconds of RA) when the sun is at ecliptic longitude 108. I averaged Wagner's figures for each month, weighted proportionally to the number of observations, then found the first-order Fourier coefficients. This gave for the "O - C" excess a maximum of 0.066s when the sun is at ecliptic longitude 82, a phase discrepancy of 26#. Wagner's twelve-year study averaged out the effect of the 14-month Chandler wobble, unless the wobble's axes varied much. The text indicates that Wagner sometimes used inferior instead of superior culminations of Polaris; if 3/8 of the time he used the inferior culmination, then the amplitude would be reduced, as it is, to 1/4 of predicted.

Now three studies - Kimura's meta-analysis of latitude determinations, Chandler's analysis of catalog declinatons, and Wagner's right ascensions of Polaris - roughly agree with my theory that for stellar aberration, Earth's velocity is advanced by about seven hours (there is a questionable sign disagreement for Chandler's analysis).


Reply author: Joe Keller
Replied on: 06/12/2007 20:49:06
Message:

Proposed experiment:

Get a big old refracting telescope, with a dome and stone pier. Refractors are best for this experiment because of their stable alignment. Let the refractor have an eyepiece with two very straight lines crossing at some angle.

Each night, put the crosshairs on a known star that is as close as possible to the pole, record the time, and fix the telescope in that position though the star eventually will move away from the crosshairs. Also at this moment rotate the crosshairs so that each of the four branches has a known visible star slightly clockwise of it.

Due to precession, uncorrected J2000.0 coordinates don't accurately tell us which stars now are closest to the pole, but at epoch 2000.0, two stars with USNO-B "Red 1" magnitudes of +13.4, were < 2.4' from the north pole. So with a big refractor we typically expect to see a +13.4 star 2' from the pole, maybe closer. The billion-star printed & bound "Millenium Catalog" shows two pairs of almost opposite stars, none dimmer than magnitude +10 or +11, 20-30' from the N pole, and the pairs making about a 60# angle with each other (so crosshairs 60# apart would be best).

Four crossings will be timed soon after aiming the telescope, and four more crossings (for these stars) four or eight hours later (the latter precluding summer observation). Opposite pairs of angles will be perfectly equal. The inequality of the times required for the stars to move from one crosshair to the next, will tell how much the crosshair is displaced in that direction, from the pole. So, the position of the pole will be found.

The pole change due to change in Earth's axis (i.e., precession & nutation) can be removed by calculation based on ephemeris data. Chandler wobble doesn't move the pole; it only changes the latitude. Herein lies the superiority of my method, to that of using zenith cameras: daylight prevents 24-hour zenith camera observations, so zenith cameras don't distinguish pole motion (axis change) from Chandler wobble (latitude change). Stellar aberration due to Earth's rotation will not affect the time intervals, so the only effect remaining is stellar aberration due to Earth's orbital motion (and to the sun's motion).

The altitude of observation is constant, but temperature, pressure and humidity also correlate with atmospheric refraction, and these might change in 4-8 hours. Adjustment can be made for these, using the formula in PASP 108(729):1051+, 1996. This formula is said to agree with the empirical Pulkovo tables within 0.01", if the altitude of observation is > 15#.

"Anomalous refraction" limits accuracy. Over several hours, anomalous refraction amounts to 0.05-0.20" (A&A 459(1):283+, 2006). So observations on ~100 nights are needed, not necessarily all from the same observatory.

A book says that one-hour clock-drive exposures sometimes are made unattended. Then a firmly grounded fixed telescope should remain within one arcsecond for eight hours.

Another book says that for typical eyepieces, eyepiece fields of view are 40-50#, and that the observer's actual field of view equals the eyepiece field of view divided by the magnification. So, a one or two degree field of view would correspond to 20-50x magnification, which seems plausible.

Earth's motion may be found by numerical differentiation of the sun's apparent ephemeris position, projected onto Earth's equatorial plane. The difference between the sun's apparent and geometric position might as a first approximation be neglected. Also the sun's motion due to Jupiter might be neglected.

If Kimura's phenomenon is due to a lead or lag in stellar aberration, then the direction of deviation of the pole will lead or lag (the opposite of) the direction of Earth's velocity vector. This will provide another accurate test of Special Relativity over solar system distances.

Dimitroff & Baker, "Telescopes & Accessories", 1948, lists the world's big (15"+) refractors. In the US these include, in order of aperture:

Yerkes (40 inch)
Lick
Pittsburgh
USNO - DC
U. of Va.
Lowell
Swarthmore
Princeton
Wesleyan (Middletown, Conn.)
U. of Denver
Oakland, Cal.
Northwestern
U. of Penn.
Amherst
Carleton Coll.
U. of Cincinnati
Harvard
Mt. Lowe
U. of Wisc.
U. of Miss.
Yale (15")


Reply author: Joe Keller
Replied on: 06/16/2007 00:11:04
Message:

Special Relativity is disproven by double stars. There should be several arcseconds difference in stellar aberration between the primary and the secondary, because their velocities differed by several km/s when their light departed. Therefore stellar aberration is determined by something besides mere relative motion of source and observer.

Red giant variables give a clue. Over many years, the angular diameter of Antares, the brightest red giant that can be occulted by the moon, was measured many times at lunar occultations (Astronomical Journal 62(2):83+, 1957). One of the most precise such diameter measurements, was 0.0413" +/- 0.0001" (41.3 +/- 0.1 mas)(A&A 230(2):355&, 1990; see also Table 2); this and other references (ApJ 242:646&, 1980, Table 5; this article is missing from "Web of Science")(ApJ 463(1):336&, 1996) list 22 determinations of Antares' diameter since 1920; some used visible light and others infrared. Ten were by occultation, eleven by any of three different kinds of interferometry (Michelson, amplitude, and speckle); one used flux curves. Fourteen of the 22 were in the range 40-43 mas inclusive. Only four (35,28,29 & 28 mas) of the 22 were outside the range 39-45 mas inclusive. Yet the Radial Velocity measurements 1905-1909 (op. cit., 1957) showed about 6 km/s "full amplitude" (max minus min) in roughly a sinusoidal pattern with period 6-7 years. The period is somewhat irregular and uncertain (AJ 98:2233+, 1989). No one seems to think that such periods in RV, typical of 25-30% of pulsating asymptotic giant-branch stars (Astrophysical Journal 604:800+, 2004), are due to engulfed companions. An average 6/2*2/pi*1.35("projection factor" to find velocity radial to Antares; because of "limb darkening", the factor is less than 1.5) km/s radial velocity for 6.5 / 4 years, even using Antares' most generous (Hipparcos) distance estimate of 185 parsec, gives +/- 9.5 mas in diameter. If the RV indicates expansion of Antares, and the period 1905-1909 wasn't unusual, then Antares' diameter should vary roughly sinusoidally +/- 9.5 mas. So, the occultation-measured diameters are too tightly clustered near 42 mas.

Antares varies about +/- 0.4 magnitude; possibly as much as +/- 0.15 mag variation (i.e., 15% of power) could occur in a 6 or 7 year cycle (AJ 1989, op. cit.). Maybe the apparent RV change is due to a +/-0.56 AU change in the radius of Antares' "ether iceberg". With about 20x the sun's mass, the radius of this "iceberg" might be 52.6*sqrt(20)=236 AU, so, the change is +/- 0.2%. Antares, like other stars of its type, expels about 0.000002 solar mass per year as wind (ApJ 275:704+, 1983). Even if the boundary or barrier at the edge of Antares' "iceberg" is mainly a gravitational phenomenon, the barrier's position yet might be affected slightly by strong solar wind.

Polaris is the nearest Cepheid variable. How fortunate that the nearest Cepheid variable is placed so prominently! Cepheids are yellow giants, smaller than Antares; only recently have their diameters begun to be measured accurately and directly. So far, only one report (Nature 407(6803):485&, 2000) seems to confirm, even roughly, for a Cepheid (Zeta Geminorum), the diameter periodicity implied by RVs.

There are two more reasons for doubting that there is any real periodic diameter change in Cepheids. The periodic time plot of RV often is grossly nonsinusoidal yet almost identical, except for sign and magnitude, to that for luminosity: e.g., Delta Cephei (Inglis, "Planets, Stars & Galaxies", 3rd ed., Fig. 11-2)(M. Petit, "Variable Stars", 1987, Fig. 4; includes temperature and radius plots). Yet luminosity is supposed to be due to temperature and radius, not RV. The luminosity waveform should resemble the temperature waveform most and the RV waveform least.

If luminosity causes apparent RV by affecting the "ether iceberg" boundary (at distance 52.6 * sqrt(M) A.U.) then the geometric midrange classical (i.e., Population I) Cepheid (Strohmeier, "Variable Stars", 1972, Table 42)(for the period, I used the geometric mean of the luminosity range, and an 0.8-slope log-log period-luminosity relation, with period the abscissa, and through the data point for Delta Cephei) should have (negative) RV lagging luminosity by 7.7% of one period or thereabouts. Indeed the lag is 7% (Carnegie Inst. Yearbook, 36:164, 1937).


Reply author: Joe Keller
Replied on: 06/20/2007 20:47:08
Message:

Elsewhere on Dr. Van Flandern's messageboard, I've remarked that stability at temperatures proportional to (Z-0.5)^4, involving Special Relativity violations by innermost-shell electrons, might determine, for Hydrogen, the CMB temperature (2.726K); for He, the far CIRB temperature (i.e., about 2.726*1.5^4=13.8K); and for O, the temperature of the main sequence/Cepheid trough intersection. Likewise these temperatures might determine, for Ne, the least-luminous temperature of the beta Cepheids (a.k.a. beta Canis Majoris stars); for O, the least-luminous temperature of the delta Cepheids (aka Population I Cepheids) and/or Pop. II Cepheids; and for C, the least-luminous temperature of the long-period (Mira) variables.

Cepheids range from 5400-6900K according to Strohmeier (op. cit., 1972) but from 6000-8000K according to Dr. Koji Mukai ("Ask an Astrophysicist" website, 1998). The least luminous tend to be hottest (chart in Inglis, 3rd ed., op. cit.). My theoretical oxygen stability temperature is 8625K.

Beta Cepheids (a.k.a. beta Canis Majoris stars) are said on various websites to range B0 - B2 or B3 in spectral type, i.e. 18,700 or 22,000 - 30,000K. Beta Can Maj itself is said to be 22,000 or 25,000K. For these, the least luminous tend to be coolest (Inglis, op. cit.). My neon stability temperature is 22,200K.

Dyck & van Belle, AJ 112(1):294&, p. 299, quote an article then submitted to the AJ by van Belle, stating that Mira variables average 2700K. My carbon stability temperature is 2490K. Dyck & van Belle also find that "carbon stars" (kin to the Mira variables) average 3000K. The coolest "carbon star" was TW Oph at 2150K. The least luminous tend to be coolest (Inglis, op. cit.).

A delta Cepheid might have a beta Canis Majoris star inside it; and a long-period variable, a Cepheid inside it. The law relating Cepheid period to stellar density, resembles Kepler's for orbital periods. Strohmeier (op. cit., Table 42) gives P(days) = Q /sqrt(density/solar density), where 0.037 < Q < 0.066 days. That is, P = Q / sqrt(M in solar masses) * sqrt(4*pi/3)*(R in solar radii)^1.5. If almost all the star's mass is in a tiny central core, and something orbits between this core and the outer surface, its period is found when Q = 0.041 days. If something orbits between the oxygen-temperature (2.726*7.5^4=8625K) surface and a fluorine-temperature (14,230K) surface beneath that, the ratio of the radii is (7.5/8.5)^(4*2) = 0.367; the period is found when Q = 0.066 d. The periods of variable stars might be orbital periods between one stable surface and another, in the near-vacuum of a giant star's interior. The Q values given by Strohmeier for classic (i.e., delta) Cepheids are the same as the theoretical Q values given for beta Cepheids by Lesch & Aizenman A&A 34:203&, if the 0.066 comes from tripling their 0.022 given for the "second harmonic" (also there are "first harmonic" & "fundamental" modes); their Table 4 shows observational Q values ranging from 0.018 to 0.033 for 12 beta Cepheids.

The Blazhko effect "...occurs among all types of pulsating variables, but...mainly those of short period." [i.e., the RR Lyrae stars, which merge into the low-luminosity side of the Population II Cepheid distribution] (M Petit, "Variable Stars"). The Blazhko effect is, that the amplitude and waveform of a variable star's lightcurve changes grossly, while its period remains nearly constant. The explanation might be that the amount and phase distribution of matter orbiting within the star might vary though the Keplerian period doesn't. More luminous variables, being more fully developed structurally, would tend to approach a mathematical limit whereby the period is longer, the lightcurve more sinusoidal, and the Blazhko effect less.

The three (3/22) anomalously small values for Antares' diameter (0.029", 0.028", 0.028"; see previous post) might be glimpses of the next inner, perhaps N-based, surface. One of these small values was obtained at lunar occultation, observing at 1.04 microns. At this wavelength, the inner surface might be especially visible, with less scattering (long wavelength) yet denser emission from the hotter inner surface than from the outer surface (wavelength not too long).

Two of the small Antares diam values were from the longer (50 ft) Michelson interferometer; as we would expect, Pease noted difficulty from ambiguous readings. If the N-surface were detected on one side only (radius = (5.5/6.5)^8=0.263x the carbon surface), then the angular diameter found should have been 0.0413" * 1.263/2 = 0.026".

The small 1986 occultation value for Antares' diam, was followed by two normal 1987 occultation values (41 & 45 mas). Likewise Pease's small 1934 and 1936 Michelson interferometer values were preceded by a normal (41 mas) 1932 value by Pease with the same interferometer. Such a diameter dip in one year would require (0.041-0.028)"/2*1.35(cosine correction with limb darkening) * 185parsec = 1.6 AU/yr = 7.7 km/s average net positive RV, twice the extreme RV observed during the extensive early 20th century observations. The anomalous small diameters were partial detections of an inner surface. Nor do the RVs indicate real diameter changes.


Reply author: Joe Keller
Replied on: 06/22/2007 17:43:59
Message:

The Hertzsprung-Russell luminosity diagram in, inter alia, Inglis (3rd ed., op. cit.) suggests that, because of similar luminosities, long-period (Mira) variables might have Pop. II Cepheids inside, and Pop. I (classical, delta Cephei) Cepheids might have beta Cepheids (beta Canis Majoris stars) inside. Though the period vs. density formula in Strohmeier (op. cit., see previous post) gives "Q" values for classical Cepheids (for which Eddington originally devised the formula) the formula also is thought to apply to Pop. II Cepheids, though maybe with somewhat different Q values.

Thus variable stars seem to release energy via a kind of variable convection with period suspiciously similar to the Keplerian. At the simplest, a huge, repeating solar flare in a sun-grazing elliptical orbit cyclicly delivers energy to some outer surface which hides the flare. If the flare quickly cools to equilibrium temperature with solar radiation, then maximum luminosity occurs only slightly after perihelion because the orbit is so elliptical. The Blazhko effect occurs because the period is much more invariant than are the other cyclical features of the event.

Electrons sometimes disobey Special Relativity. As Thompson theorized a few years ago, electrons are agitated by Lorentz contraction. In several posts above, I quantified this energy, showing that there is a large latent heat for innermost-orbital electron pairs, analogous to melting, released at "electron melting temperatures" of thousands of degrees K for elements such as C, O, and Ne. In variable stars, it is efficient that each idealized flare (see previous par.) cross one of the surfaces at which solar radiation equilibrium temperature equals electron melting temperature for some common chemical element. Thus also this large latent heat can be carried away with the flare.

For the least luminous stars, the physical outer envelope surface, the aphelion of the flare, and the electon melting surface for some element, all are the same. When the outer melting surface is that of neon or carbon, the latent heat is a minor consideration: the surface tends to be a bit closer (hotter) for more luminous Beta Canis Majoris or Mira type stars, so the flare doesn't have to travel so far. On the other hand, oxygen is so abundant and its latent heat per atom so large, that its latent heat is a major consideration: the surface tends to be a bit farther (cooler) for more luminous Cepheids, so more latent heat can be extracted.

Generally, Ne < Ne' < O < O' < C < C', where Ne, etc., are the radii at which inner-orbital electrons of each element melt, and Ne', etc., are the aphelia of successive stages or stories of flares. For increasingly luminous stars, the primed quantities increase fastest, overtaking the next unprimed (except for C' or Ne' if outermost). This explains the classical Cepheids' Hertzsprung progression (Hertzsprung, BAstronINeth 3:115, 1926; Petersen et al, A&A 134:319-327, 1984, p. 319). As Ne' increases for more luminous classical Cepheids, the "bump" at which oxygen's electronic latent heat is released, moves "backward" (i.e., to the left on the time graph) from an almost 0.5 cycle (observationally, 0.4 cycle) "lead" (i.e., position to the right of the peak), into synchrony with the luminosity peak, and then even "behind" (i.e., left of) the peak, before becoming hidden by the Ne' surface itself. A disproportionate increase in Ne' might also make the waveform less sawtooth and more sinusoidal.


Reply author: Joe Keller
Replied on: 07/02/2007 15:56:15
Message:

Let's combine some well-known variable star relations:

(1) deltaR / R = 0.25 deltaL / L

(Inglis, op. cit.). Here L is absolute, not logarithmic. This relationship is claimed only to be a rough average. R is inferred from the spectral line shift. This fits within about a factor of two (Delta Cephei, for example, misses by about a factor of 1.6). For classical Cepheids, the so-called RV curve, with peak "V0", is more sawtooth than sinusoidal. So,

(2) deltaR = V0 / 2 * Period / 4 * 1.35. (The last factor would be 1.5 geometrically, but is about 1.35 for visible light, due to "limb darkening". If the RV curve is sinusoidal, the "2" becomes "pi/2".)

(3) Q = Period * sqrt(density / solar density)

This is the theoretical Eddington relation. Empirically, it fits not only classical (Delta) Cepheids but also Beta Cepheids. Both types have Q typically 0.037 days within about a factor of two.

Combining these,

(4) deltaL / L = 1.91 * V0 / sqrt(2*M*G/R).

If the RV curve is sinusoidal instead of sawtooth, the 1.91 becomes 2.43. So, the variable star's power output is proportional to (Vesc + V0*cos(t) )^2.


Reply author: Joe Keller
Replied on: 07/14/2007 21:54:21
Message:

Thesis. Cassiopeia A exploded twice, causing both the Spoerer & Maunder sunspot minima.

Sunspots. Many main-sequence stars in the sun's neighborhood, show evidence of starspot cycles of 4 to 20 yrs (vs. 11.1 yrs for the sun). When starspots are big enough, measurable luminosity changes due to stellar rotation, indicate a sunlike differential rotation and butterfly diagram (IY Alekseev, Solar Physics journal, 2004). If starspots are too small to detect, chromospheric activity, especially CaII emission, still often reveals the cycle (P Wilson, Cambridge Astrophysics Series #24, pp. 109, 114-118).

Jupiter's 11.86 yr period is only slightly greater than the 11.1 yr Schwabe sunspot cycle; Uranus' 84 yr period only slightly greater than the 82.2 yr (according to H. Kimura) Wolf (1862) cycle which Gleissberg found also in auroras (see Gleissberg's aurora article in: J. Schove, ed., "Sunspot Cycles"). Yet if the planets cause the cycles, then why does Saturn lack effect, and why do other stars generally have roughly the same (primary) starspot cycle length as the sun? Maybe the planets' periods and the sunspot cycles have, or originally had, a common cause.

Sedimentary rocks, inter alia, 680 million yr old, show cycles of from 8 to 15, or even 22 (!) yr, though now it is thought that some or all of these cycles are associated with lunar tides, not sun-caused weather (RY Anderson, NY Acad Sci Annals, 1961; in: Schove, op. cit.)(George Williams, 1980s, mentioned in: Brody, "Enigma of Sunspots", p. 163). Maybe the exact sunspot cycle length is determined by, say, the total mass of the solar system, but sunspot activity is maintained in disequilibrium despite damping, by an interstellar wave which limits the cycle length to a narrower range than would be expected from the diversity of stellar masses and planetary configurations.

From sunspot numbers (inter alia, the chart in Branley's Astronomy) and from the corroborating aurora records gathered by Agnes Clerk (see John Eddy, 1976, in: Schove, op. cit.) the Maunder sunspot minimum (discovered by Spoerer in 1887) occurred from about 1645 to 1715. There also is a Spoerer sunspot minimum from about 1480 to 1520; both the Maunder and Spoerer minima are confirmed by high (presumably due to lessened protection from cosmic rays) C14 levels in tree rings (Eddy, op. cit.).

Cassiopeia A. Cassiopeia A has the greatest apparent (not absolute) power (i.e., radio magnitude) of any radio source outside the solar system. At galactic coords. 111.7, -2.1, it lies near the galactic plane, only about 22 degrees away from the sun's presumed direction of travel around the galaxy (Tycho's supernova is at galactic coords. 120.1, +1.4). Cassiopeia A is thought to have been an exploding Wolf-Rayet star 11,000 lt yr away.

Flamsteed recorded a now-gone star, probably 6th mag, in 1680 near the location of Cas A. Flamsteed was a great astronomer and recorded his date of observation for this star. It's thought that a dust envelope might have caused its relative faintness vs. Tycho's or Kepler's supernovae. The most recent motion studies support the 1680 date: 1681 +/- 19 (Astrophysical Journal 645:283+, 2006); or 1671 +/- 1 if without deceleration of ejecta, a few (maybe 10) years later if with deceleration (Astronomical Journal 122:297+, 2001).

Two other recent motion studies support the date, c. 1505. These found 1539 +/- 30 (Aqueros & Green, MNRAS 305:957, 1999) and 1495 +/- 15 (Astronomy & Astrophysics, 339:201, 1998).

Discussion. Maybe two major explosions occurred at Cas A, c. 1505 and in 1680 (minus light travel time). These would have followed the beginnings of the Spoerer & Maunder sunspot minima by about 25 & 35 yr, resp., almost obliterating the sunspot cycle for 40 & 70 yr, resp., if the sunspot effect could arrive c. 30 yr prior to the light.

The supernova might send out a superluminal wave which disrupts a segment of the track of the underlying, lightspeed, interstellar wave regulating the sunspot cycle. If this track passes through Cas A, then the sunspot cycle disruption would be observed at the same time as the supernova, and plus or minus "X" yrs, depending on the size of the disruption. Indeed this occurred: [1480-1520] equals [1500-20, 1500+20]; [1645,1715] equals [1680-35, 1680+35].


Reply author: cosmicsurfer
Replied on: 07/15/2007 18:05:55
Message:

Hi Joe,

Excellent data comparisons that show possible synchronicity between solar sunspot cycles, supernova and superluminal waves. I think you are on to something that has most definitely been overlooked by main stream science. Fluctuations in superluminal wave pressures may exist that have periodicity due to large scale interactions that change "windows" of polarity with in regions of space. Hence, polar magnetic field reversals act like a butterfly effect with 'lines of force migrations'[following lines of force of fluctuating wave patterns of greater universe] occuring on sun's surface creating super magnetic storms or sunspots that are actually aligned with superluminal wave/current interaction migrations [polar to equator migrations tip scale to a polarity reversal-butterfly effect] within our scale.

Just thought of something that really just blew my mind. What if these system wide fluctuations on sun spot migrations are revealing axis differentials of superluminal lines of force that 'are' the overall major axis centers moving around our point of reference with in our MEGA SCALE. Look at torque values of such a switching effect and measure scale probability by co-factoring other center migrations then measuring difference between to juxtaposition large scale structure rate of motion.

Just some thoughts.

John


Reply author: Joe Keller
Replied on: 07/15/2007 18:12:51
Message:

Cosmicsurfer:


There's a wealth of ideas here. Thanks for your thoughts.

- JK


Reply author: Joe Keller
Replied on: 07/16/2007 13:51:10
Message:

J Eddy's C14 tree ring data (see previous post) confirm ~40+70=~110 years of Maunder minima in the sun's latest 950 yrs. Measuring CaII line strengths has allowed about 2000 nearby sunlike stars to be checked to see if they are actively cycling, or in a Maunder minimum. Instead of the expected ~230, approximately none of these stars are in a Maunder minimum ("Do We Know of Any Maunder Minimum Stars?", JT Wright, AJ 128:1273+, 2004).

Type IV subgiants have different CaII measurements than do Type V dwarfs; in early surveys, subgiants comprised a large distinct minority: fictitious Maunder minimum stars. When subgiants were excluded, < 10 Maunder minimum "candidates" were found in collections of ~1000 solar-metallicity Type V dwarf stars, i.e., sunlike stars (Wright, op. cit.; Gray, AJ 126:2048+, 2003). These candidates have borderline measurements, do not form a bimodal peak, and might be merely a statistical tail.

Binary star companions affect starspot distributions; Jovian planets might affect sunspot cycles. This can't explain the absence of Maunder minima, because maybe 1/4 of sunlike stars, both lack binary companions and have Jovian planets. Either the sun or solar system has a rare unknown property causing Maunder minima, or the occurrence of Maunder minima in the sun and in nearby sunlike stars is synchronized.

The 1645-1715 Maunder minimum might be found, in measurement epochs c. 1995, in sunlike stars 43-54pc distant opposite Cassiopeia A (see previous post). Distance cutoffs in the above star collections, were 40-60pc (Wright op. cit.; Gray op. cit.; Henry et al, AJ 111:439+, 1996). Henry (op. cit.) also usually used an apparent magnitude cutoff of +9.00, which, he noted, corresponds to our sun at 50pc. Hipparcos gave distances for the 9 stars which Gray deemed his elite northern Maunder minimum candidates (Gray, op. cit., Table 6, p. 2057): the farthest was 30pc. The middle page, of Henry's list of southern observed stars (Henry, op. cit., p. 450) happened to show only 3/90 stars with apparent magnitudes dimmer than +8.674, which I infer to be our sun's magnitude at 43pc on Henry's scale.

The hypothetical locus for manifestation of the Maunder minimum is bounded by two very eccentric prolate spheroids, approximately paraboloids. The paraboloid through 43pc, passes beyond 60pc when 52deg away from the point opposite Cass A; this cap covers 19% of the sphere. Roughly, the fraction of sample stars, that are within the Maunder minimum locus, is then 3/90*19%*1/2 = 0.3%; i.e., 2000*0.3% = 6 stars. As sunlike stars near 50pc, these would be near Henry's magnitude limit. Instead of expecting 230 stars having good data, we expect 6 relatively distant, southern, mostly circumpolar (low altitude of observation) stars having poor data. The distance and magnitude limits of present studies might just barely fail to reach the stars that would be at Maunder minimum.


Reply author: Joe Keller
Replied on: 07/16/2007 22:27:28
Message:

The sun's best lookalike, 18 Sco, has a 7 yr cycle (Astronomical Journal 133:2206+, 2007). Baliunas (op. cit., p. 284) states that all clearly detected cycles in his ~100 star sample, are at least 7 yrs long; also, our sun's shortest cycles, over the last 250 yr, are 7 yrs. This is further evidence of overriding interstellar regulatory control: the most similar, and - almost - the least similar stars, all cycle with the same range of periods.

Baliunas et al, Astrophysical Journal 438:269+, 1995, report three possible Maunder-minimum Type V stars. One of these, HD (Henry Draper catalog) 9562, really might be Type IV (i.e., subgiant & "evolved", hence disqualified - see previous post) according to their footnote; indeed the ARICNS catalog classifies it G2IV.

Another, Tau Ceti, spectral type G8Vpeculiar, absolute magnitude 5.7 (0.2 mag dimmer than the main sequence norm) was noncycling but seemed to be starting to cycle slightly beginning in 1990 (Baliunas, op. cit.; also Frick, ApJ 483:426+, 1997). However, true Maunder minimum includes more than mere absence of cycling (Hall & Lockwood, ApJ 614:942, 2004).

Baliunas' best Maunder-minimum candidate was HD 3651 (Flamsteed designation: 54 Pisces), spectral type K0V, absolute magnitude 5.8 (near the main sequence norm). It showed a cycle of 13.8 +/- 0.4 yr, peaking in 1966 or slightly earlier, again in 1978, and then with a much smaller peak in 1989, consistent with Maunder minimum beginning sometime between 1980 & 1989 (Baliunas, 1995). Subsequently HD 3651 has remained photometrically stable at < 0.1% variation (ApJ 590:1081+, 2003). This would be small photometric variation, for a cycling sunlike star; our sun's cycling-associated photometric variation is 0.1%, which is very small for a vigorously cycling star such as our sun. HD 3651, with 0.79 solar mass, has an extrasolar planet smaller than Saturn, with major axis 0.3 AU and eccentricity 0.6 (Ibid.). HD 3651 also has a faint, cool Type T brown dwarf companion at 480 AU projected separation (MNRAS 373:L31+, 2006).

HD 3651 might be giving early warning of another event at Cassiopeia A which is causing Maunder minimum for stars in this quadrant of the galaxy. Both Vostok and South Polar ice cores show increased nitrate levels (presumably due to increased atmospheric cosmic rays) simultaneous with Tycho's and Kepler's supernovae, to within a few months' accuracy (Dreschhoff & Laird, Advances in Space Research, 2006, on internet 2004). However a similar nitrate spike is shown in 1667, not 1680 when Flamsteed recorded a star, now gone, at the location of Cas A. This, and the beginning of the Maunder (resp. Spoerer) minima c. 1645 (resp. c. 1480), not 1680 (resp. c. 1505), suggest superluminal processes associated with Cas A events.

At first, Flamsteed named Cas A "supra Tau Cassiopeiae" (Thorstensen et al, AJ 122:297+, 2001, Introduction)(the real Tau Cassiopeia is 3deg east both from Flamsteed's recorded star location, and from today's Cas A at RA 23h23.4m Decl +58deg50'); this suggests that Flamsteed saw a large nebulosity, perhaps aimed toward Tau Cassiopeia. Then Flamsteed changed the name to 3 Cassiopeiae; maybe the nebulosity subsided and Flamsteed decided it must have been a cloud. The low Flamsteed number suggests that Flamsteed's star was bright: Bayer designations are twice as common among the top half of Cassiopeia's Flamsteed numbers as among the bottom half; the top three Flamsteed numbers in Cepheus and the top ten in Ophiuchus (site of Kepler's supernova) all have Bayer designations. However, Flamsteed's map shows the star of the faintest brightness; he "recorded [it] of 6th magnitude" on Aug. 16, 1680 (Nature 285:132, 1980).

Van den Bergh & Dodd (ApJ 162:485, 1970, pp. 491-492) used the amount of neutral hydrogen along the line of sight, to estimate the rather large amount of dust along the line of sight, thus estimating the peak magnitude of a typical supernova at the position of Cas A, as +2, even without any special dust cloud around it; they also found that almost all regular +2 novae went unrecorded either by European astronomers of that era, or by Asian astronomers. Though Flamsteed's position differs 15' from today's (the diameter of Cas A's main broken shell is only 4', and its extreme diameter only 5.4') it's thought that much or all of the error might be attributable to Flamsteed's usual errors including sextant corrections & atmospheric refraction (Nature, ibid.). Alternatively, Cas A's proper motion might have decreased from an earlier average of 3"/yr (16% the speed of light).

It's thought that maybe Cas A was an exploding Wolf-Rayet star, but it is uncommonly placed, for a W-R star (see heliocentric galactic coordinate plot of all ~60 W-R stars then known within 3 kpc, in Conti, "O Stars & Wolf-Rayet Stars", 1988, Fig. 2-10, p. 107). The short lifespan of W-R, O & B stars, would prevent a W-R star from pacing the 8 to 15 yr sedimentary cycles of Earth hundreds of millions of years ago (see posts above). If these sedimentary cycles were due to sunspots, and if now there is interstellar pacing of sunspots by Cas A, then either Cas A is not a W-R star, or long ago other W-R stars provided a similar pace.

Cassiopeia A is complicated and asymmetrical. HD 3651 (54 Pisces) is 11pc (36 lt yr) from us, making a 42 degree angle with Cas A, thus 7.5 arcminutes from us as viewed from Cas A. HD 3651 might have begun its Maunder minimum especially early so that we observed it c. 1985 despite the 9 years longer lightpath. The other stars in that direction won't begin their Maunder minima until nearer the time that our sun does; half the stars in that direction won't be observed to have entered their Maunder minima until after the sun does.

Hall et al, AJ 133:862+, 2007, p. 879, found two sunlike stars, HD 43587 & HD 140538 (Bayer designation: Psi Serpens), whose cycles seem to be entering or leaving Maunder minimum, but these cycles aren't very clearly detected. Hall notes (op. cit., Sec. 5.4.2) that Baliunas (op. cit., 1995) called HD 43587 "Flat?" even during the allegedly un-flat era in Hall's data. Hall says (op. cit., Sec. 6, Conclusions) that his "most significant example of a clear transition" between noncycling & cycling states, is HD 140538.

The ARICNS catalog absolute magnitudes of +4.05 & +4.42, resp., make HD 43587, 0.35 mags, & HD 140538, 0.68 mags (thus HD 140538 fails Wright's subgiant rejection criterion; see below) brighter than normal for their Sky Catalog 2000.0 (and Hipparcos) spectral classifications, G0 (G0.5Vb in Hipparcos) or G5, resp., if they are "V" dwarfs (Jaschek, "The Classification of Stars", 1987, Table 12.5). The ARICNS catalog classifies HD 43587 as F9V, which is 0.2 mag brighter than G0V (by interpolation in Jaschek, op. cit., Tables 12.5 & 11.5). This would make HD 43587 only 0.15 mag brighter than the main sequence norm; furthermore, use of the Hipparcos parallaxes instead of the ground-based, and the slightly different Hipparcos visual magnitudes, puts both these stars on the main sequence norm, within 0.1 mag. (The absolute magnitude of HD 140538 in the Sky Catalog 2000.0 seems to be erroneously faint; both these stars are listed there as spectral type "dGx" rather than "GxV"; in the old Harvard classification, "d" denoted either V or IV, as there was no subgiant designation.) If its ground-based parallax distance, 19pc, is too large by twice the standard error published by ARICNS, then HD 140538 (Psi Ser) likely is a star that has been observed emerging from Maunder minimum.

Gray's nine elite Maunder minimum candidates (Gray, op. cit., Table 6) were critiqued by Wright (op. cit., p. 1275). Two he discarded as showing only modest changes in CaII lines, consistent with mere cycle minima rather than Maunder minima. Three he discarded as > 0.6 mag above the main sequence norm at their spectral type, i.e., as subgiants.


Reply author: Joe Keller
Replied on: 07/18/2007 22:00:41
Message:

Wright (op. cit., 2004) found three Maunder minimum candidates of his own, in addition to Gray's. One, GJ 561, is listed in the VizieR catalog as having absolute magnitude, visual magnitude and parallax of a subdwarf, 2.4 mag below the main-sequence norm for its spectral type, K0; so, maybe this was an error in Wright's data. Another, HD 186427, not only was found, by the rival Lowell Observatory survey, to fail Wright's own CaII line strength criterion (Wright op. cit.) but also now is listed by VizieR as a double star (two G dwarfs with projected separation of 90 AU), thus excludable because of the effects of companion stars on stellar cycles.

Wright's third candidate persists: HD 233641 is spectral type G0, with Hipparcos parallax implying 97 pc distance and absolute magnitude either +4.0 or +4.3 depending which of the two online visual magnitudes is believed. This is only 0.4 or 0.1 mag brighter than the norm for its spectral type. Its coordinates are RA 9h31m Decl +53, about 60deg from Cassiopeia A. Unless disqualified by magnitude variability, HD 233641 is a fifth apparent Maunder minimum star, and fits the pattern of the four listed in the previous post, except for its spectral type.

I manually rechecked the published "Phoenix" list of nearby southern sunlike stars (Henry, AJ v. 111 op. cit., 1996). Excluding binaries, 71 stars had logR'HK <= -5.10 (the usual criterion). Thirteen of these were listed as Type IV or IV/V (i.e., subgiant or borderline subgiant) in the online Heidelberg ARICNS or Hipparcos catalogs; another 44 were more than 1.0 mag too bright for their Hipparcos distance and spectral type; eight were 0.7 to 1.0 mag too bright; HD 57062 (type G5V) & HD 59741 (type G3/5V) had conflicting visual magnitudes according to which they were either 1.4 or 0.5 mag too bright (variable stars?); HD 67581 isn't in the Hipparcos catalog (distant subgiant?); HD 3795 possibly has acceptable absolute magnitude but is a well-documented "evolved" (-2.0 < [Fe/H] < -0.75; Li, Be, B depleted) star (Astrophysics & Space Science 265:67, 1999; Astronomy & Astrophysics 343:545, 1999).

The two best "Phoenix" candidates in 1996 were HD 56972 (G5V, 0.51 mag too bright, 69pc distant per Hipparcos, RA 7h18m Decl -32, 48deg from the antipode of Cassiopeia A) & HD 179699 (F8/G0V, 0.63 mag too bright, 80pc distant per Hipparcos, RA 19h 15m Decl -40, 69deg from the antipode of Cas A). Both these stars are within that relatively distant locus, where stars would appear to be in Maunder minimum, if that minimum is caused by a lightspeed signal from Cas A which also caused our sun's Maunder minimum. On the other hand, at this distance the sample covers declinations < -25, and most of the stars at this distance brighter than the +9 magitude cutoff are subgiants. So, many subgiants in the sample would be expected to be found within this locus.

The abovementioned five northern hemisphere candidates (three from Gray, one from Baliunas & one from Wright) would require supraluminal communication of the next Maunder minimum inducing event from Cas A. Let phi be the angle seen from Earth, between Cas A & the star, and "d" the distance by which the star is nearer to the (practically infinitely distant) Cas A. The extra lightpath length is d*(sec(phi)-1). For small phi, the supraluminal excess, varying randomly (like Brownian motion) with the direction of flow outward from Cas A, might be k0*sqrt(d*tan(phi)), so:

(1) d*(sec(phi)-1) = k0*sqrt(d)*sqrt(tan(phi)).

At large phi, k0, in units in which c=1, might be replaced by a randomly varying k = K*sqrt(d), so:

(2) sec(phi)-1 = K*sqrt(tan(phi)).

The surface, on certain azimuths of which, a star might be now seen to be entering Maunder minumum, would resemble a paraboloidal nosecone fired from Cas A and about to reach the sun, except that, letting phi --> 0 in the first equation, there would be a narrow tubular central stem reaching upward from the sun toward Cas A. Letting K --> 1 in the second equation, shows that the asymptotic flanks of the surface would be sloped at phi = 66.63deg from vertical, instead of vertical as for a paraboloid. Letting d --> 0 in the first equation, shows that the front of the nosecone, connecting the flanks & stem, would be sloped at 90deg (i.e., a surface perpendicular to the line to Cas A, then deflecting toward Cas A before intersecting our solar system).

The angles phi for the five northern hemisphere candidates, are 1, 37, 42, 63 & 66.1 degrees. The farthest star is at 66.1 deg. If the coming Maunder minimum is again 50 yrs (15pc), the stars showing early warning of it should be near the inner side of the above-described surface, especially on the flanks and central stem. This is so.

Four stars among those discussed above, make about 90 degree angles to Cas A:

HD 10700 (Tau Ceti) (Baliunas' candidate for emergence, in 1990, from Maunder minimum) (type G8Vp; [Fe/H] = -0.50; 3.6pc parallax distance; 80deg angle "phi" to Cas A)

HD 57901 (GL 2057) (of Gray's nine, one of the four which survived Wright's cull) (type variously listed as K3V, K2, or G5; [M/H] = 0.00, i.e., solar-metallicity; 25pc Hipparcos distance; phi = 93.5deg)

HD 140538 (Psi Serpentis Aa; a multiple star, with a small companion, 0.2 solar mass according to its magnitude, at 60 AU projected distance; and three midsize to giant companions at 2500-5000 AU projected distance) (Hall's candidate for emergence from Maunder minimum, starting 4-yr cycles in 2000) (type G2.5-5V; 14.7pc Hipparcos distance, which is more than the ground-based distance, 19.4pc, by > 2 std error of the latter - perhaps due to its binary orbit; phi = 99.9deg)

HD 43587 (Hall's statistically weaker transition candidate: possible entrance into Maunder minimum in 2002) (type G0.5Vb; 19.6pc Hipparcos distance; phi = 92.6deg)

So, these four stars range from 80-100 degrees from Cas A. The two stars at phi = 93deg are only 0.2 radian apart in RA and 20% different in distance.


Reply author: Joe Keller
Replied on: 07/21/2007 19:01:27
Message:

Suggestion for small-telescope observation:

The eleven stars which, it seems from my above review of the literature, might be in, or might recently have entered, or might recently have exited, Maunder minimum, are:

HD 3651 (54 Pisces)
HD 10700 (Tau Ceti)
HD 12051
HD 43587
HD 56972
HD 57901
HD 140538 (Psi Serpentis Aa)
HD 164922
HD 179699
HD 221354
HD 233641.

Some of these are visible with the naked eye, under good conditions; all of them are visible with binoculars, under good conditions. All are sunlike, G or "early K" spectral type. All are "dwarf" stars (i.e., on the "main sequence"). According to my theory, four of these stars lie near an envelope, where that envelope is tangent to our line of sight:

HD 43587
HD 57901
HD 221354
HD 233641.

Generally, I expect the envelope to be tangent to our line of sight when we look in one of these three directions:

Cassiopeia A
66.63deg away from Cassiopeia A
90deg away from Cassiopeia A.

My suggestion is to look for anything at all unusual about the appearance (e.g., magnitude, color) or apparent location (angular distance from nearby stars) of these eleven stars, with special attention to the four stars theoretically near the envelope. Also one might observe any and all stars near the envelope (i.e., near Cas A or near the two circles around it, at ~66.63 and ~90deg).

This program is ideal year-round for northern hemisphere amateur astronomers with small telescopes. All these stars are included in robust professional observing programs (Mt. Wilson, Lowell and/or Phoenix projects) but observations are only intermittent. Transient nonatmospheric phenomena, involving either the stars themselves or the intervening space, could be missed.


Reply author: Joe Keller
Replied on: 07/25/2007 22:55:13
Message:

Rho Cassiopeiae and NGC 7789 both lie 1/2 or 2/3 of the way from us to Cassiopeia A, within 5 deg of the line to Cas A. Several different indirect estimates have been made of all these distances; probably rho Cas is farther than NGC 7789. Cas A, rho Cas, NGC 7789 and ourselves are almost collinear.

Rho Cas is the nearest yellow (type recently given as G2Ia0e, but varies from F8Ia, probably its normal status, to K and sometimes even M) hypergiant star; as of 1999 only 12 hypergiants of whatever color were known in our galaxy (ApJ 523:L145, 1999). Usually, roughly every 50 years (1893?, 1946, 1986, 2000) rho Cas seems to eject a shell of gas, causing its apparent surface to dim and cool, with typically -25 km/s (blue)shifted emission lines (not absorption lines as with ordinary photospheric pulsations - PASP 98:914 & 99:272), complete with molecular TiO bands (Sargent, ApJ 134:142, 1961; Beardsley, ApJSupp 5:381, 1961; Lobel et al, ApJ 583:923, 2003). It also varies with sometimes a 300 day and sometimes an 800 day period, but only by 0.2 or 0.3 mag (A&A 325:714, 1997; PASP 112:363, 2000, Fig. 1).

The approximate dates on which rho Cas reached major magnitude bottoms are:

December 1893, 0.6 mag drop (Beardsley, op. cit., Table 2, p. 500)
July 1946, 1.3 mag drop (Beardsley, op. cit.)
March 1986, 0.4 mag drop (Lobel, op. cit., Fig. 13)
September 2000, 0.8 mag drop (Ibid.)

(The last three bottoms were accompanied by spectral changes consistent with an ejected shell; such data might exist for the first bottom also.)

The dates of eruption of Cas A are:

August 1680 (Flamsteed records 6th mag star)(likely peak, +2, likely in April)
1667 (Dreschhoff & Laird, ice cores 1 mo. after Tarumi volcano; so far I haven't found the date, of this volcanic eruption in Japan)

If the 1680 Cas A eruption initialized a 53.4 yr cycle, the 1893 minimum would be four cycles later; the next cycle lengths (to 1946 & 2000) would be 52.6 & 54.2, which average to 53.4. If the 1667 supraluminal early event, initialized a 40 yr cycle, 1986 would be eight cycles later; the two cycles could coincide in 1946, giving a summed effect, and the observed longer bottom. (There was an Astronomische Nachricten article in 1934, available to me only by interlibrary loan or photocopy service, which would reveal whether the 40-yr cycle manifested in 1906.)

Nova 1592C (FR Stephenson, Quarterly Journal of the Royal Astronomical Society London 12:10-38, 1971, Sec. 4, pp. 21, 34) was noted in Korean annals, stationary near beta Cas (Caph) from Nov. 30, 1592 until March 4, 1593. Probably the nova had peak mag +2 or +3, as it "escaped the vigilance of...the West, ...[and of] the Chinese & Japanese..." (Stephenson, op. cit.). Probably Cas A is too far from Caph (Stephenson, op. cit.) but the nova might have been a flareup of rho Cas, only 2.5deg away from Caph. No other easily visible star is nearer to Caph than is rho Cas; this might be the meaning of the oft-used oriental description translated as "within" the reference star. Such a flareup of rho Cas happened in Nov. 1950, though only 0.6 mag (Beardsley, op. cit.).

Burnham, vol. 1, p. 495, says, "Trigonometric parallaxes have been measured [with the large refractors] at Allegheny, McCormick, & Mt. Wilson, and agree in giving a distance of about 200 lt yrs [ = 16mas][to rho Cas]." Allegheny parallaxes c. 1950 (resp. c. 1920) had probable errors of 4.9 (resp. 8.6) mas (AJ 55:185). A McCormick parallax for rho Cas was 8+/-6 mas (AJ 62:276, 1957). Not only the absolute magnitude as estimated from the spectrum, but also the small Hipparcos parallax (0.3+/-0.6 mas), the small proper motion (near zero after correction for galactic rotation), the large radial velocity (Oort's law) and the distances of objects in the presumably equidistant "Cas IV association", suggest a true distance 35x larger than Burnham's parallax figure, or 17x larger than the 1957 McCormick parallax figure: namely, 7000 lt yr. (ApJ 134:142). With a 200 lt yr distance, the detour from Cas A to rho Cas to us is negligible, but a 2kpc distance foils the rho Cas cycle convergences at 1680 & 1667.

Increasing the interstellar speed of light about 270x, rescues the cycle convergence for rho Cas. It also harmonizes the cycles of P Cygni (34 Cygni) and eta Carinae.

Using typical recent distance estimates, P Cygni is the peak of an isosceles triangle. The base angle is 30deg, the sides are 2kpc and the base is 3.4kpc. The other vertices are the sun & Cas A. Extending the base 2.3kpc beyond the sun, reaches eta Carinae. The extra lightpath for P Cygni is 2*2*(sec(30) - 1) = 0.62 kpc. The extra lightpath for eta Car is 2*2.3 = 4.6kpc, 7.4 times more than for P Cyg. The actual observed delays are 56 yr & 7.7 yr (56/7.7=7.3) for eta Car & P Cyg, resp. The ratio is as expected but it is as if the interstellar speed of light is 270x (maybe 137*2; the "fine structure constant" is 1/137) times greater.

Eta Car's outburst peaked 1843 = 1680 + 2*53.4 + 56. If the 53.4 & 40 yr cycles, seen with rho Cas, are initialized in 1680 & 1667, resp., for eta Car also, then their 1843 coincidence (2 & 3 cycles, resp.) would make an unusually big event, analogous to the 1946 event for rho Cas. Because eta Car & P Cyg are "luminous blue variables" (i.e., blue hypergiants) whereas rho Cas is a yellow hypergiant, the former have an outburst where the latter has a dimming. Indeed P Cyg (type B2Ia) is an "S Doradus variable" having blueshifted absorption lines, not blueshifted emission lines as does rho Cas. At 270x lightspeed, the 2.3kpc roundtrip to eta Car requires only 56 yr.

P Cygni suddenly reached +3, then was discovered by Blaeu on August 18, 1600 (Israelian & DeGroot, Space Science Reviews 90:493, 1999). Presumably under the influence of an earlier initialization by Cas A, rho Cas (Nova 1592C) had burst out 7.7 yrs earlier and was discovered in Korea. P Cygni reappeared in 1654, 54 yrs later: the same cycle length as rho Cas.

Cas A also might somehow affect the location of the "blue stragglers", the main-sequence blue stars found within clusters of much yellower stars. NGC 7789 usually is called an open cluster though some call it a transitional object between open and globular cluster; its color-magnitude array resembles that of a globular cluster (MNRAS 114:583). It includes a delta Scuti variable (A&A 366:178) and probably a Mira variable, WY Cas (PASP 72:48).

NGC 7789 is "very rich in blue stragglers", still counted as ~25 in 1995 (A&A 366:490). Already in 1980 (McNamara, 92:682, 1980; Fig. 3, p. 686) 29 were plotted. On the average, the blue stragglers are displaced a few arcminutes toward Cas A. The positional angle of the center of the blue stragglers, relative to the center of the cluster, is 50+/-10deg depending on the weighting used. The position angle of Cas A relative to the cluster center is 53deg.

Blue straggler catalogs of the other, smaller, open clusters near Cas A (NGC 7380, 7419, 7510, 7654=M52, and the overlapping 7788 & 7790) seem to be absent from the literature. "In general, the blue stragglers show a remarkable degree of central concentration [i.e., they tend to be well within the cluster]." (A&ASuppl, 109:375, 1995, Abstract). Blue stragglers have spectral type late B or early A (A&A 366:490, 2001). One could follow the precedent of professional astronomers, who identified stars appearing to be within the cluster and which had "B minus V" magnitude typical of spectral types bluer ("earlier") than about A5. Then those not having proper motion (or radial velocity or light polarization) typical of the cluster would be eliminated, because presumably they are accidentally superposed on the cluster.


Reply author: Joe Keller
Replied on: 07/29/2007 17:34:01
Message:

Luminous blue variable (hypergiant) HD 160529 (V905 Sco)(assigned spectral types range from B8 to A4 Iae, sometimes even to F0)(Stahl et al, ArXiv.org 20 Dec 2002)(Sterken et al, A&A 247:383, 1991)(also online article by AW Fullerton & F Najarro) erupted in 1992 with a blueshifted ejected shell and a magnitude peak (Gaeng, Leitherer et al, display #80.02, 1995 - see internet). Maybe this is analogous to the rho Cassiopeiae eruption of 1946. The 1992-1946=46 yr delay is due to a signal traveling the extra distance from Cas A to V905 Sco & thence to us. For eta Carinae, 2*2.3kpc apparently caused a 56 yr delay (see above). The galactic longitudes of V905 Sco & Cas A are about 0 & 110, resp., so the law of cosines gives 2.4kpc for the distance from us to V905 Sco. Stahl (op. cit.) adopts Sterken's (op. cit., p. 390) estimate of 2.5kpc; Stahl notes upper and lower bounds of 1.9 & 3.5kpc.

In Huygens' construction for light propagation, let the small wavelet circles be 1/137 the radius of the large wavefront circle, or in general 1/137 the radius of curvature, R, of the wavefront. Let r = R/137. If photons move perpendicular to the wavefront, they advance by a distance r. If photons move tangent to the wavefront, they advance by sqrt(R^2+r^2)-R, which is smaller by a factor 2*137=274. The latter might be the usual speed of light; and the former, the faster signal which operates in interstellar space.

The seemingly fairly accurate ground-based parallax of rho Cas, cited by Burnham (see above) is 35x (maybe 17x) too large. The difference between ground- and space-based parallax might be due to different modes of light propagation in matter & vacuum.


Reply author: Joe Keller
Replied on: 07/29/2007 20:48:39
Message:

I found another star with, like rho Cas, Yale parallax catalog ground-based parallax ~20mas with error 8mas, yet Hipparcos parallax much less. This star is Antares. For Betelgeuse the two catalogs are in good agreement. Yet for rho Cas & Antares, the Hipparcos parallaxes are 70x and 4x less, resp., both more than two standard errors lower.

Through Strasbourg's "VizieR", I searched the (ground-based) Yale parallax catalog (4th ed., 1995)(9000 stars) for stars with visual mag < 6.00, 15 < ground-based parallax < 30mas, and given error "sigma" of ground-based parallax < 10.6mas. Then I considered 2 or 2.5 sigma difference from Hipparcos parallax (also requiring Hipparcos parallax < 7.5mas as a labor-saving screen). For convenience I considered only the 375 (3/4 of the Yale stars) for which Bayer or Flamsteed designations were given.

The expected number with Hipparcos, 2.5 sigma or more, less than Yale, is a little less than 375*0.6% = 2.2. Instead I found 4 stars:

rho Cas
55 Cam
beta Sct
36 Aql

The 7.5mas Hipparcos cutoff is about two sigma less than Yale, for the median Yale parallax (21.7mas) and Yale error (7.3mas) of the nine stars eventually found (see below). So here in the tail of the bell curve, almost half of stars for which Hipparcos is 2 sigma or more less than Yale, would have Hipparcos parallax >7.5mas and be excluded. Thus here the expected number of stars with Hipparcos 2 sigma or more less than Yale, is a little more than 375*2.3%*0.5 = 4.3. Instead I found 9:

rho Cas
55 Cam
theta UMi
nu1 Boo
52 Ori
beta Sct
36 Aql
alpha Sco (Antares)
chi Sco

Their positions correlate with the positions of the five SN remant - like radio/X-ray nebulae:

Sgr A
Cas A
M1 (Crab nebula)
Tycho's supernova
Kepler's supernova

Five of the nine stars are within about 15deg of one of these nebulae. Rho Cas has the most extreme parallax discrepancy (2.8 sigma) and is closest of the nine stars, to any of these nebulae (Cas A is 4.3 deg away); 55 Cam has the third most extreme parallax discrepancy (2.5 sigma) and is second closest (~20deg) to Cas A. Also 55 Cas and chi Sco are about 10deg from Tycho's SN and Sgr A, resp.; 52 Ori and Antares are about 15deg from M1 and Sgr A, resp.

Because stars with parallax 4.7mas (median Hipparcos value for the nine above) are 695 lt yr away, comparable to the galactic disk "scale height in the solar neighborhood" of 800 lt yr (ApJ 596:204, 2003), Hipparcos stars with <7.5mas parallax are almost twice as likely as random, to lie within 10 deg of the galactic equator. Even so, the binomial test for 5 or more of 9 stars to lie within 15deg of one of the five nebulae, gives p < 0.5% (the 15deg radius disks for Tycho's SN and for Cas A considerably overlap).

Summarizing this thesis so far: Huygens' construction, together with the ubiquity of the fine structure constant, suggests an alternative lightspeed of 274*c. This lightspeed synchronizes almost all hypergiant star eruptions in this quadrant of the galaxy, on a 53.4 yr cycle initialized by Flamsteed's supernova (Cas A). The "blue stragglers" of Caroline Herschel's cluster NGC 7789 are displaced toward the line of sight to Cas A. Discrepancies between ground- and space-based parallaxes occur for giant stars near lines of sight to SN-like nebulae, especially Cas A. Also, the distribution of sunlike Maunder minimum candidate stars, is consistent with a lightspeed, or slightly greater than lightspeed, initiating signal from Cas A. In particular, the sun's two Maunder minima (c. 1500 & c. 1680) correspond to the two main recent reverse calculations of Cas A's explosion date(s).


Reply author: Joe Keller
Replied on: 08/06/2007 17:21:49
Message:

Thesis: the Milky Way (and other large spiral galaxies, including Andromeda) aren't breaking up other spiral galaxies; they're building additional spiral galaxies, which at first are sparse, though often large in extent. These new spirals are at large angles to the parent galaxy, but their nuclei often lie near the plane of the parent. By forming such additional spirals, spiral galaxies eventually become elliptical galaxies.

Cassiopeia A is the main satellite galactic nucleus in this quadrant. Matter flows from Cassiopeia A and forms open clusters. This flow is almost along geometric lines. Once deposited, the open clusters limit the distance of subsequent flows along these lines. Multiple clusters along the same flow line usually aren't obvious because distances are uncertain and they lie near the galactic plane.

Double clusters might offer evidence for this thesis. If NGC 884 were slightly older, slightly farther from the galactic plane, and slightly nearer to us, all of which it might really be because none of those determinations are very accurate (position must be corrected for differential proper motion since formation), then the other member of the astronomer Hipparcos' famous Perseus double open cluster, NGC 869, would lie on a line segment between Cas A & NGC 884.

Other evidence for this thesis comes from clusters near our line of sight to Cas A. Even if subsequent deposits fall slightly short, of the original cluster at the end of the flow line, clusters on the same line from Cas A, and near our line of sight to Cas A, would look like one asymmetrical cluster. The younger generation of stars ("blue stragglers") will lie nearer Cas A, and the older generation of stars from the earlier deposit, i.e., the red giants and other "turnoff stars", farther from Cas A. This is so. Let's consider seven clusters near our line of sight to Cas A:


NGC 7789:

(See above for preliminary discussion.) McNamara's (PASP 92:682, 1980; Fig. 3, p. 686) plot of 29 blue stragglers in NGC 7789 is displaced from the cluster center, by a distance equal to 2.3 standard errors of the blue straggler mean position (p=0.07). (I subjectively judged the cluster center four times, from NSE&W, then averaging; found the median horiz. & vert. position of the 29 stragglers; and used 16th-84th %ile distances to get the standard error.) The direction of displacement is position angle (clockwise from north) 75 +/- 25; the direction to Cas A is position angle 53. Gim et al, PASP 110:1318, 1998, Table 5, grade McNamara's blue stragglers as Members, Nonmembers, or Uncertain members of the cluster. If Uncertain is weighted 2/3 and Nonmember zero, then Gim's revision (excluding stars included by Gim but not by McNamara) scarcely changes the magnitude of the displacement but rotates the direction ~30deg counterclockwise to about 45deg, i.e., even closer to the direction to Cas A.


NGC 7380:

Besides magnitude (among the brightest in the cluster), spectral type (B6-A5, III-V according to A&A 366:490, 2001, Table 7, & 356:517, 2000, Table 1)(and the communal cluster PM, RV, and polarization) the intrinsic properties of abnormally slow rotation, and very weak magnesium lines, also characterize blue stragglers (A&A 366:490). NGC 7788 & 7790 overlap; NGC 7654(M52) & NGC 7510 have variable extinction (Janes & Adler, ApJS 49:425, 1982, Table 1); and NGC 7419 is relatively unstudied. So although NGC 7380 is associated with nebulosity, I chose it as the best open cluster, besides NGC 7789, to examine near Cas A, for blue straggler displacement.

Janes & Adler give NGC 7380's main-sequence turnoff as B-V=-0.35 (approx. spectral type B2 per the "FBS blue stellar object" online catalog), and its relative B & V extinctions as E(B-V)=0.59. Vega, type A0, is the standard at which B=V=R. Roughly estimating V=5500Angstrom, B=0.8*5500A, R=1.25*5500A, one sees from the Planck curve that R is about 1/3 of the way to the inflection point, so if B-V=-0.35 then V-R=-0.35*0.25/0.2*2/3=-0.29. Also E(V-R), according to the omega^4 law, should be 0.59*(1-(1/1.25)^3)/((1/0.80)^3-1)=0.29. So the apparent B-R at the turnoff should be about -0.35-0.29+0.59+0.29=0.24 (the same as the apparent B-V, because the dimmer Red absolute magnitude approximately cancels the reduced Red extinction, thus helping this rough estimate to be accurate).

I searched the USNO-B catalog for all stars within 8.33' of the center (22h 47m 21s +58deg 07' 54") of NGC 7380, with R2 magnitude < +13.5 (Dreyer says the [bright, i.e., main sequence blue, or red giant] cluster stars have V = +8.5 to +13.5, and above I estimated that near the turnoff this is also the Red mag), which are within 5, or 10, mas/yr of both the cluster RA PM (-2.77mas/yr) & the cluster Decl PM (-3.55mas/yr). (That's 50 or 100 km/s at the cluster distance of 2220pc, good enough to eliminate nearby stars.) Whether I used 5, or 10, mas/yr, the main sequence turnoff seemed to be between apparent B-R 0.23 and 0.28, with subgiant star(s) near 0.27, and bunching of stars between 0.22 and 0.29. This matches the 0.24 estimated from Janes & Adler.

In NGC 7380 there seems to be one blue straggler: USNO-B 1482-0442104, with B2-R2=0.12 (usually these "2" mags, which I normally used here, are from LaSilla) and B1-R1=0.18 (usually these "1" mags are the older and somewhat less accurate Palomar). This star is 3' E and 7' N of the cluster center. This is near the edge of the disk considered, near the limit of PM discrepancy tolerated, and at position angle 335, vs. position angle 280 to Cas A. In ApJ 454:151, 1995, Table 4, the B-V extinction for 10 stars in NGC 7380 ranged from 0.52 to 0.86mag; so this blue straggler candidate easily might be actually less blue than several other stars in the cluster, if it happens to have relatively weak B-R extinction.

So after trying with modest success to do this at home, I went to the Iowa State University library and consulted Moffatt, A&A 13:30, 1971, Table 2 & Fig. 5. The WEBDA star cluster website lists Moffatt's star #21 as the lone blue straggler in NGC 7789. Though #21 is the bluest in B-V, it's far from the bluest in Moffat's "extinction-free" (B-V)0. Moffatt's six stars (#8,13,7,6,16,10) bluest, of his 55 total, in extinction-corrected (B-V)0, ranging from -0.39 to -0.31, also are the six bluest in (U-B)0. All six were among the 45/55 getting Moffatt's highest, "p", cluster membership probability rating. Beginning at -0.30, the histogram becomes much more crowded: two each at -0.30 & -0.29, one at -0.28, three at -0.27, etc. The bluest six have extinction-corrected visual magnitude, V0, ranging from +7.8 to 9.6, consistent with the main sequence at type B, but the two stars at -0.30 have subgiant (main sequence turn-off) V0's of +6.6 & 6.7. Moffatt (p. 35) gives his mean error in B-V as 0.08mag. The position angles of the six are, resp., 250, 340, 235, 330, 270, 360 (ave. 300 +/- 20 , vs. 280 for Cas A), and their distances from the center, 6', 6', 11', 3', 5', and 1'. The "bulk of cluster stars lies within 6' " (Moffatt, p. 32).


NGC 7510:

WEBDA says NGC 7510 (which has variable extinction, according to Janes & Adler and to Barbon) has two blue stragglers, Barbon's #21 and (correcting the misprint in WEBDA) #104 (Barbon et al, A&ASupp 115:325, 1996; Fig. 1 & Table 1). As for the previous cluster, I rely on extinction-corrected blueness instead, finding #17 & #29 in Barbon's Table 2, both with (B-V)0 = -0.28, Mv = -3.8, type B0.5V. (Barbon gives his B & V errors as 0.03mag.) These seem to be the real blue stragglers; the next bluest (corrected) is #55, a turnoff giant with (B-V)0=-0.23, Mv=-4.7, type B1.5II. Their position angles are about 105 & 155 (both only about 1' displaced) vs. 225 for Cas A. NGC 7510 is elongated NE-SW; the two blue stragglers are near the SW corner.


NGC 7790:

Gupta, with B & V errors of only 0.01m


Reply author: Joe Keller
Replied on: 08/09/2007 20:51:55
Message:

There is a simpler way to prove the thesis that Cassiopeia A is a satellite galactic nucleus: open clusters might be concentrated in a shell centered on Cas A. Around a star, a planetary ring nebula really is a spherical shell of gas which looks like a ring where the shell is pierced at a slant. Likewise there seems to be ring of open clusters around Cas A:

Dias' open cluster catalog (online Strasbourg "VizieR") is the successor to Lynga's. Dias' catalog lists 48 o.c.'s within 7deg of Cas A. Twelve of these lie 4.76 to 5.11 deg away from Cas A. Let's "clean" the list: some of Dias' clusters are noted "not found" [on DSS plate inspection] or are noted to have been called "nonexistent" by authorities; let's omit these. Clusters Dias calls "dubious" on DSS will be counted as 1/2.

This "cleaned" cluster list includes 19 closer than 4.76 deg, 6 of which have NGC numbers. Between 5.12 & 7.00 deg, are only 11 clusters, only 2 of which have NGC numbers. Between 4.76 & 5.11 deg are 9.5 clusters, 4 of which have NGC numbers.

The galactic latitude of Cas A is only b = -2.1 deg, so the sparseness of clusters between 5.12 & 7.00 deg, might be due to greater distance from the galactic plane. However this doesn't explain the concentration of clusters near 5 deg. This concentration is analogous to a planetary nebula. The density is even greater than would be expected from the geometry of a spherical shell. Maybe clusters tend to be elongated with axes perpendicular to the radius to Cas A. If so, clusters near the apparent edge of the shell, could appear more spherical and would be likelier to be recognized.

Seven of the 9.5 "cleaned" clusters that lie on the ring, are in a ~25 deg sector including NGC 7788 & 7790. The most important five of these seven, lie at position angles, from Cas A, of 295 to 320 deg, at distances varying smoothly from 5.08 down to 4.81 and then up to 5.05 deg. If the cluster Harvard 21 is included (it's in the Millenium atlas but "not found" in the DSS by Dias) the five clusters King 21 & 12, Harvard 21, NGC 7788 & 7790, lie on a curve whose curvature varies only +/- 15%.

NGC 7789 & 7380 lie well away from that sector, but a "dubious" cluster, Berkeley 100, is the only one lying more than 1 deg above the galactic plane. If the sun is (to give a typical reported value) 20pc below the galactic plane (determination from OB stars by BC Reed, Alma College, internet, c. 2006) an object in the galactic plane and 2000pc distant would appear to have galactic latitude +0.5deg. There might be only a half-shell, which lies in the same galactic hemisphere as Cas A.


Reply author: Joe Keller
Replied on: 08/12/2007 21:44:34
Message:

One of the NGC clusters in Dias' catalog, NGC 7423 (= Berkeley 57) appears near Cas A (~4deg) and has a bibliographic reference in WEBDA, but I haven't considered it above. Fig. 3 in PASJapan 56:295, 2004, seems to me to show that the brightest and bluest stars in this old ( > 10^9 yr) open cluster are above the main sequence turnoff, and thus are blue stragglers. (I think I disagree with the authors, who suggest that the real blue stragglers in this cluster are to the left of the cleft below the main sequence turnoff.) Be this as it may, the brightest & bluest stars seem to be mostly on the side toward Cas A:

I drew two 30"-radius circles tangent to the origin (i.e., cluster center) NW and SE of the center. In the USNO-B catalog I listed stars, within the circles, which had "0" PMs in both directions. Of stars with both B2 & R2 mags available, 18 in the NW circle but only 13 in the SE circle had B2 < R2 + 2. This itself amounts to only 0.9 sigma, but adds to the collective evidence.

Finding the location of Cas A with binoculars, I see that it lies near the center of a dark "hole" in the Milky Way roughly one degree in radius. (Through binoculars this "hole" looks like several other nearby holes that are as big or bigger.) Further investigation of this region of increased extinction, might reveal some kind of symmetry about Cas A.


Reply author: Joe Keller
Replied on: 08/28/2007 00:55:12
Message:

The curvature of the above-mentioned string of open clusters 5deg E of Cas A, is away from Cas A, not toward it. Cas A (l=112), if at distance 3.4kpc, is at the far edge of the Perseus Arm (see plots in, inter alia, Kuehn, The Milky Way, 1982, Fig. VI.14, p. 86; or Inglis, Planets Stars & Galaxies, 1972, Fig. 16-14, p. 428). These open clusters might mark the end of the "bar" of a spiral mini-galaxy formed around Cas A.

The orientation of the Milky Way's galactic bar might be 16 +/-2 deg forward of the radius from the sun to the galactic center at Sag A (Binney et al, "Kinematics of Galactic Center Gas", MNRAS 252:210, 1991). If it is 15deg, that's straight toward Cas A, if Cas A is 3.4kpc away and Sag A 8.5kpc away (the radius preferred by Bok & Bok, The Milky Way, 5th ed., 1981).

The Orion-Cygnus (i.e., Local) Arm is thought to be aberrant and not a true spiral arm: "Observations indicate that we are [within a galactic arm], but theoretical studies say there shouldn't be a major spiral feature where the Sun is." (Branley's Astronomy, 1975, p. 228). The Local Arm might be part of the mini-galaxy around Cas A.

There is a net average Cepheid blueshift ("systematic deviation") of 3 +/- 1 km/sec (Kraft & Schmidt, ApJ 137:249, 1963, Table 8, p. 263). The intrinsic redshift concept of Tifft and Arp suggests therefore that the sun is in a different, younger, galaxy from most of the rest of the Milky Way.


Reply author: Joe Keller
Replied on: 08/30/2007 00:12:42
Message:

Open letter to the Lowell family re: Barbarossa (Planet X)


To: Mr. ******* Lowell

Dear Sir:

I went to the Lowell page of *******.com; you were the first person I found who recently had posted a message about a Massachusetts Lowell and who was named Lowell himself. Perhaps you will forward this email to appropriate members of the Lowell family.

The Charlie Rose (with David Rockefeller) program about the late widow Brooke Astor informed us that although her philanthropy was sound, some Astors sued her because they thought she was giving away the Astor foundation too fast. Indeed a decade ago she announced that she gave away almost all of it.

Sometimes family foundations are taken over by people who care little for the hopes and dreams of the founding family. I think that the Lowell Observatory might be a case in point.

Like Percival Lowell, I studied mathematics as an undergraduate at Harvard (B.A., Mathematics, cumlaude, 1977). Unlike Percival, I'm a retired Midwestern medical doctor, but I have compiled information indicating that Percival was correct about the direction of Planet X though he underestimated the distance. Everything I've published on this, is on Dr. Tom Van Flandern's free-speech scientific messageboard at www.metaresearch.org, under the name "Joe Keller". There I've published all my fruitful ideas, and in enough detail that a trained astronomer can retrace my work to his own satisfaction.

I told the director and the appropriate department head at the Lowell Observatory about this in an email in March, but received zero response from them (now that Pluto has been downgraded from planet status, it is believed widely that Lowell was mistaken). I emailed over 100 relevant professional astronomers about my Planet X findings, again with zero response. I post-mailed the Sec. of the Navy, the commander and several other offices at the US Naval Observatory, and dozens of Senators & Congressmen who sit on relevant committees, again with zero response except for one postcard saying approximately "the Congressman can't help you because you don't vote in his district".

As the discoverer of the true Planet X, I've named it Barbarossa, and its largest satellites, Frey & Freya. I've had to call them something during the last year during which the professional astronomy establishment has been refusing to talk to me. I've been talking about this so long that if the pack decides they don't like the name Barbarossa, they'll need millions of dollars in counter-publicity to make their own name stick. As the world economic situation deteriorates, we move toward political change, and the malfeasance of the astronomy bureaus becomes apparent (e.g., I've been silenced on every Association of Lunar and Planetary Observers messageboard, for daring to say there, less than I say in this letter) the professional astronomers will be too busy wondering how to pay their mortgages, to successfully publicize their own name for this planet.

A summary of my work on Barbarossa (Planet X) follows (#1 is new material)(also see especially #4)(I haven't had time to check all these calculations, so there might be minor errors):

1. Lowell's final statement (WG Hoyt, "Planets X & Pluto", 1980, Ch. 6, pp. 140-141) about Planet X gives two almost diametrically opposite possible positions for epoch 1914.5 (a common mathematical phenomenon known as "dual solution"). For best accuracy, one of these positions should be reflected and averaged with the other. Lowell's comment no. 14 says "precise prediction of place" was possible but not of distance [nor mass]. Though Hoyt reports that Lowell relied heavily on one of Flamsteed's observations of Uranus, in the year 1715, Lowell would have given much less weight to Flamsteed's relatively inaccurate (~1 arcminute) observation. The observations after the discovery of Neptune, i.e. the 1846-1910 observations, should dominate Lowell's calculations because not only were observations of Uranus more frequent and accurate during this era, but also Neptune's position did not have to be extrapolated. According to Lowell's estimate, Planet X was, to a first approximation, roughly 45 degrees from perihelion, and roughly at conjunction with Uranus, at the likely effective midpoint of the 1846-1910 era. Such an approximation makes it easy to estimate the mass, for any "Barbarossa" lying in that same direction, necessary to give the same tidal gravitational work on Uranus as would Lowell's Planet X. The effective average distance of Lowell's Planet X from the sun would be between 43.85 AU (the semimajor axis) and 35.2 AU (perihelion). This puts Barbarossa, assuming it is 198 AU distant, in the range 0.008 - 0.018 solar masses.

2. The 5:2 Jupiter:Saturn resonance isn't exact; it advances with period about 2800 yr. Thus any of the three recurrent conjunction points could be shepherded by a distant planet in a circular orbit at 198 AU. The position of the shepherding planet must be adjusted for the lead or lag of the conjuction point due to Jupiter's & Saturn's eccentricities; thus, I found that one of the three possible circular shepherding orbits, differs only a few degrees, in ecliptic longitude, from Lowell's effective average Planet X position in the interval 1846-1910.

3. The dipole and higher moments of the distribution of apparent Cosmic Microwave Background (CMB) temperature are correlated with the plane of the ecliptic. This implies that at least the anisotropy of the CMB, is due to a solar system influence such as planetary gravity. The constancy of the CMB dipole rules out the known planets as major causes of the dipole, but a planet at 198 AU is acceptable, to the accuracy to which the CMB dipole is known from the COBE and WMAP satellites. The ecliptic longitude of the CMB dipole lags the theoretical shepherd planet (see #2) by four degrees.

4. My search of online scanned sky survey plates, combined with an electronic photo taken in March from Tenerife by amateur astronomer Joan Genebriera of Spain with her 16" telescope (similar photos were obtained soon thereafter with smaller telescopes by Steve Riley of California; and possibly by Robert Turner of England using a robotic telescope also on Tenerife) discovered a recurrent pair of starlike objects consistent, for an appropriate mass ratio, with a center of gravity that is in a circular orbit differing 0.5 degree from the ecliptic longitude predicted by #2 above, and 0.2 degree below the ecliptic latitude predicted by #3. Published Kuiper belt density surveys suggest that some of the starlike objects I found on the photos, especially those noticeably elongated on those one-hour sky survey exposures, are Kuiper belt objects; also, some objects might be novas or glitches. Dr. Ian McDiarmid states that cosmic ray artifacts do not appear on photographic plates (e.g., the sky surveys) at moderate altitude. Despite the many possible combinations of objects for an orbit, I estimated that the closeness to a perfect constant-speed great circle was statistically significant. Genebriera and Riley were recruited by me through a mass emailing to ~100 amateur astronomers; Turner contacted me through Dr. Van Flandern's messageboard.

5. The origin of the CMB and its dipole can be explained by a simple theory which likens the solar system's Maxwellian ether to a drop of boiling water. According to this theory, the CMB dipole could be caused by the gravitational field and potential, of a planet of 0.0116 solar masses at distance 198 AU. This agrees well with what Lowell's last work implies should be the mass of Planet X at this distance (see #1 above).

6. I found that if Barbarossa (together with, of course, its satellites Frey, Freya, etc.) possess 0.0104 solar masses, then Barbarossa produces a torque (per degree of orbital inclination) on a typical member of the main (i.e., non-resonant) Kuiper belt, equal to th


Reply author: Stoat
Replied on: 08/30/2007 04:15:17
Message:

Hi Joe, I think that you are letting your, quite understandable, frustrations make for a counter productive introductory letter. I believe that any first approach has to be short and hopefully tantalizing.

First paragraph, along the lines of, "I have been advised by Dr ****, that you may be of assistance in gaining two photographic telescope plates, for the purpose of blink comparison of stars."

Second paragraph, is one where you can ask your "hook" question, and pique their interest. Something along the lines of, "could our sun be one member of a failed binary star system?" Explain how binaries are very common, so that failed binaries also have to be common.

Third paragraph, state that you believe that this is the case, and give the proposed position of the companion. I personally wouldn't mention its satellites. It's always best not give too much information about specifics at this early stage.

Fourth paragraph, ask for time on the size of telescope you want and pencil in the best times for the plate exposures you want.

Fifth paragraph, offer to fax/e mail further information, and ask if it would be okay to write/phone later in the month to enquire about progress. Then a thank you for reading the letter and sign off.

This approach lets the first reader of the letter make a decision, without having to pass it on for comments. I think your long letter is being passed on by the boss, and then no one in the pecking order wants to have to explain it to that same boss. That's human nature, they'll opt for, "just ignore it and it will go away." Your letter intimates that there is some huge fight going on, people, even those in ultimate resposibility, will often decide that they'll not get involved in it. The quite reasonable request, to take two photographic plates gets lost in the shuffle.


Reply author: Joe Keller
Replied on: 08/30/2007 19:51:51
Message:

[quote]Originally posted by Stoat

Hi Joe, I think that you are letting your, quite understandable, frustrations make for a counter-productive introductory letter. ...


You may well be right. Some of my friends here in the US say what you're saying. Thanks for your input!


Notice:

$100 reward! I, Joseph C. Keller of Roland, Iowa, USA, will pay $100 to whomever is the first, to get someone with a 40 inch (1 meter) or larger optical telescope, to look for Barbarossa. The reward is for getting someone to look, not for looking yourself. With a telescope this size, it's professionals who will look, and they're unlikely to be swayed by $100. So the contest is, somehow to get professionals to look, and see whether Barbarossa is there, or not there.

Whether the search finds Barbarossa, or shows that Barbarossa isn't there, is not important. It is important that it be a bona fide attempt.

To prevent complications, I'm the sole judge of what is a bona fide attempt to look for Barbarossa (with an optical telescope at least 1 meter in size) and of who is first. Many disputes can arise in a deal like this, so to save everyone grief, I'm the judge of who, if anyone, wins, and there's no appeal. I don't promise fairness. I don't promise any particular standard of judgment. I'm the arbitrary judge.

Sincerely,
Joseph C. Keller, M. D.


Reply author: Stoat
Replied on: 08/31/2007 02:59:00
Message:

Hi Joe, somewhere in this thread I mentioned a post by cosmicsurfer, which links to the site of this very rich man who believes we have a brown dwarf companion. Now the down side with this guy is that he appears to have an ego as large as a brown dwarf. Having said that, perhaps he's worth a try. I wouldn't argue with him too much. I'd just say I believe that I've found something and that it's at such and such co-ordinates. Say that I have some photographic evidence but that it needs verifying with the use of a larger telescope.

My draft introductory letter, I'd send by pm here to Tom Van Flandern, for comment. I believe that he once did some work for this chap but that they had to part company, because as I've said this guy makes King Canute look like a modest man. This guy will probably try to save face by claiming that it was his idea all along but people are going to see through that hubris.

(Edited) Perhaps now would be a good time to put together a pdf that explained, in layman's terms, the idea of a "failed" binary system. Lots of diagrams and pictures. Probably written in a Carl Sagan like style.

Then approach your local radio/t.v station and also get a lecture set up for school sixth formers. Tape or video the whole thing, and burn that onto a cd to send to science reporters of the larger national newspapers. Do expect that newspapers will garble your argument but don't fret about that, the object is to get the thing into the public domain.


Reply author: thebobgy
Replied on: 08/31/2007 05:05:32
Message:

Hey Joe, check this place out for publishing your work, it is Cornell U. http://arxiv.org/. They print papers on many subjects daily. I have been reading your work for quite some time and your efforts are deserve a chance. Good luck.


Reply author: nemesis
Replied on: 08/31/2007 08:26:46
Message:

Stoat et al, the "very rich man" must be the guy who believes the phenomenon of precession of the equinoxes is due to the solar system orbiting an unseen companion. That idea was critiqued pretty thoroughly on another thread on this board, and in any case, the required period of 23,000 years or so would be far to long for Barbarossa at 198 AU. But still, he could well be interested, there's nothing to lose.


Reply author: Joe Keller
Replied on: 09/01/2007 14:06:52
Message:

quote:
Originally posted by Stoat

Hi Joe, somewhere in this thread I mentioned a post by cosmicsurfer, which links to the site of this very rich man who believes we have a brown dwarf companion...


I think these are good ideas. I won't be offended if somebody else does these before I do.

An analogy might be to Charles Darwin & Thomas Huxley. Huxley did more than Darwin, to gain acceptance for Darwin's evolution theory. Either Huxley was a better persuader, or Darwin was beat up from the Beagle voyage and the tropical diseases, maybe including Chagas disease.

Reply to "thebobgy": when Cornell took ArXiv.org over from Los Alamos, they made it a closed club. Last time I checked, you can't post unless someone already in the club, and in that subject area, "endorses" you for membership. A few years ago, when I was a graduate student in Math at Iowa State Univ., I asked several physics/astronomy faculty at ISU to "endorse" me for membership, but they refused.

Reply to "nemesis": a year or so ago a magazine I saw (in the Borders bookstore) had an article about solar system precession vs. Newton's sun/moon-on-Earth's-equatorial-bulge torque-on-Earth-rotation precession. I tried to email the authors but never heard back. They said there was a discrepancy between Earth's observed polar motion, and that predicted by Newton's equatorial-bulge theory. Although my own calculation (and surely the calculations of many astronomers before me) shows that Newton's theory does give roughly the right answer, it may well be that there is, due to some unknown cause, a discrepancy, perhaps revealed only by modern measurements of Earth's mass distribution and polar motion. However, even another sun, at 200 AU, would cause only 1/200^3 = 1/8,000,000 the precession that the sun does, according to Newton's equatorial-bulge theory.


Reply author: Joe Keller
Replied on: 11/21/2007 17:08:30
Message:

Barbarossa, and Lescarbault/LeVerrier's Vulcan

Lescarbault's data didn't determine "Vulcan" 's eccentricity. So, the near equality, of Vulcan's and Barbarossa's inclinations (to the principal plane of the known solar system) suggests that Vulcan is a tail-less long-period comet related to Barbarossa dynamically. Though the sky surveys came close, I don't know of any astrophotographic record on which such a body should be visible, even with the most optimistic diameter and albedo estimates.

Barbarossa's and Vulcan's inclinations to the principal plane, differ almost two degrees, but their inclinations to (Earth's) ecliptic (by LeVerrier's calculation) are equal, to the nearest arcminute. Torques from distant objects precess Earth's orbit 12 times slower than Jupiter's, so the ecliptic might better indicate initial solar system conditions. Also, LeVerrier's calculation of Vulcan's ascending node, is only 25.5' less than 90 degrees ahead of Barbarossa's.

Because Vulcan grazed the sun's disk, not crossing the center, a parabolic orbit for Vulcan (instead of LeVerrier's circular assumption) lessens by parallax, Vulcan's calculated inclination by about 1/250 radian = 15'. This advances its calculated node, by 1/2500 radian = 1.5'. So the nodes (modulo 90) and inclinations of Barbarossa and Vulcan are each only ~20' different.

Alternatively, Lescarbault's estimate of Vulcan's apparent diameter as less than 2.7" (1/4 Mercury's) is consistent with a 1:1 or 2:1 mirage image of Barbarossa. From a compromise, between the brown dwarf diameter estimated from theoretical physics, and the somewhat smaller diameter inferred for Barbarossa from its apparent magnitude combined with theoretical chemistry's estimates of brown dwarf albedo, a 60,000 mile diameter for Barbarossa, gives 1.5". With "one arcsecond [atmospheric] seeing" (i.e., each point of light becomes a two-dimensional normal distribution with half-maximum 0.5" from the center) if Barbarossa's black image somehow were superimposed on the sun, the half-maximum brightness circle would have diameter 1.2".

Two special correlations with Earth, suggest that Vulcan was a mirror image of Barbarossa. Barbarossa and Vulcan (assuming circular orbits) have, as discussed above, exactly the same inclination to (Earth's) ecliptic, within an arcminute (this accuracy isn't impossible with the sky survey, or with Lescarbault's data). Also, Vulcan was observed when about 70' S of the celestial equator. At the time, correcting for equinox precession, the center of gravity of the Barbarossa/Frey system, was about 40' S; and with 2 A.U. (roughly Frey's maximum distance from the center of gravity) subtending 30', Frey might have been 70' S.

There is a correlation with Dayton Miller's annually varying "ether drift vector". The equatorial component of that vector, is roughly proportional to a 90-degree rotation of the vector, between Earth, and the center of gravity of the solar system including Barbarossa.


Reply author: Joe Keller
Replied on: 02/15/2008 16:58:48
Message:

Additional Solar System Features Suggesting the Cold Brown Dwarf Companion, Barbarossa

Earlier on this messageboard I mentioned that if my mass and distance estimates for Barbarossa are accurate, then the classical Kuiper Belt (a.k.a. "cubewanos") occurs at that distance, at which the torque on an orbit, per degree of tilt, due to Barbarossa, equals the torque per degree of tilt due to the entire known solar system combined. That is, for cubewanos, precessions about Barbarossa's orbital plane, and about the principal plane of the known solar system, occur at equal rates.

Yesterday I confirmed that calculation and went further. Including only the known giant planets, JSUN, I found that if Barbarossa is in a circular orbit at 197 AU, then 0.0102 solar masses for Barbarossa and its satellites, gives a ratio for Neptune, torque per unit tilt due to Barbarossa & satellites : torque per unit tilt due to JSU = 1:3. Then the 1:2 ratio (Neptune now included) occurs at 40.2 AU, for a body in a circular orbit; the mode of the semimajor-axis histogram for plutinos is about 39.4 (the semimajor axis of Pluto). The 1:1 ratio occurs at 43.8 AU, vs. the mode of the cubewano histogram, 44.2 AU (the mean and median appear to be slightly less). The fundamental trans-Uranian spacing process might be the torque due to Barbarossa and the resonance of orbital precession rates. The 3:2 Pluto:Neptune orbital period resonance might be a bonus possible because both processes, orbital period and Barbarossa-caused orbital precession rate, go as r^1.5, while the orbital precession rate of Neptune due to JSU equals that of plutinos due to JSUN.

Following Brauer & Nohel, Qualitative Theory of Ordinary Differential Eqns (Dover, 1989), pp. 51-53, I solved the nonhomogeneous system of linear differential equations describing Neptune precessing about the orbital plane of Barbarossa and simultaneously about the presumed fixed principal plane of the known solar system, with exactly 3x the torque, per degree of tilt, as Barbarossa. The result is precession along a small closed curve amounting to a circle superposed on a larger cardiod, or equivalently to a slender ellipse superposed on a hypocycloid of three cusps (see, inter alia, CRC Standard Mathematical Tables, 26th ed., ch. VII - Analytic Geometry, pp. 264,267,269; ES Smith et al, Analytic Geometry, sec. 100, prob. 15; and the Wikipedia article on Lissajous figures). The predicted rms deviation for Neptune resembles that currently observed, and Neptune lies near a point on the predicted curve.

The same equations also predict that Uranus would have an rms deviation resembling that observed. The deviation would be due to the sum of comparable contributions from Neptune and from Barbarossa.

The deviation of Saturn is not explained by the above, but might be related to the Jupiter:Saturn:Barbarossa resonance discussed by me earlier. Suppose that the energies related to this resonance, are proportional to the gravitational forces, and minimized when the resonance is perfectly aligned. Suppose further that there is equipartition of energy between the two degrees of freedom, which correspond to the imperfect alignment, due to Barbarossa's inclination and to Saturn's. I calculate that this equipartition occurs when Saturn's orbit is inclined 1.01 degrees to Jupiter's, vs. 1.25 deg observed.


Reply author: Joe Keller
Replied on: 02/15/2008 22:46:24
Message:

Assuming, as above, circular orbits, with Barbarossa at 197 AU with 0.0102 solar masses, Saturn's rate (i.e., frequency) of orbital precession, due to J,U & N, and Saturn's hypothetical rate of orbital precession due to Barbarossa, are in the ratio 1:0.00206. Jupiter's rate of orbital precession due to S, U & N, and Jupiter's hypothetical rate due to Barbarossa, are in almost exactly the same ratio, 1:0.00205. Here U and N are almost negligible, so basically, Jupiter and Saturn happen to have the mass ratio needed for them to have the same ratio, of orbital influence by Barbarossa to orbital influence by each other.

The orbital influence (i.e., precession rate) hypothetically caused by Barbarossa goes as r^1.5, like the orbital period. But why should Saturn's 2.5x longer orbital period imply that its orbit precess 2.5x faster than Jupiter's? A distant massive solar companion, such as Barbarossa, seems a likelier reason why the masses of Jupiter and Saturn happen to be just right to conform to this r^1.5 law for their orbital precession rate. If Barbarossa precesses Saturn at 2.5x the frequency it does Jupiter, but Saturn's orbital angular momentum vector's precession cycles around Jupiter's are 2.5x quicker than Jupiter's around Saturn's, then the displacement of Saturn's vector, by Barbarossa, during a quarter cycle of Saturn's precession around Jupiter, equals the displacement of Jupiter's, by Barbarossa, during Jupiter's quarter cycle around Saturn. Regarding the absolute value of the difference between the J & S orbital angular momentum vectors, the discrepancy induced by Barbarossa, thereby amounts to the sum of two sinusoids of different frequency but equal amplitude.


Reply author: Stoat
Replied on: 02/16/2008 05:49:27
Message:

Hi Joe, time maybe for another try at a telescope shot.Here's a screen shot of the ones on the Bradford robotic, Pick out the best position one and I'll put it up again as a job. Though veiwing is not too good at the moment.

[img]http://farm3.static.flickr.com/2085/2268057421_8ac37f3c3b_o.jpg[/img]


Reply author: Joe Keller
Replied on: 02/16/2008 23:35:53
Message:

Originally posted by Stoat:
"Hi Joe, time maybe for another try at a telescope shot.Here's a screen shot of the ones on the Bradford robotic, Pick out the best position one and I'll put it up again as a job. Though viewing is not too good at the moment. ..."

Great! My notes aren't with me now but I can do it tomorrow.
- Joe Keller


Reply author: Joe Keller
Replied on: 02/17/2008 00:22:13
Message:

Bode's Law developed into Dermott's Law (MNRAS 141:363+). Dermott found that the satellites of the sun, or of Jupiter, Saturn, or Uranus, orbit at radii that are in geometric sequence. The ratio of that sequence is the cube root of a small integer that is different for different parents: 6 for the sun (gives 1.817121), smaller integers for J, S & U.

As satellites of the sun, J, S & U conform very well to Dermott's Law. Venus & Earth conform very well only when averaged, supporting the double-planet concept of Venus & Earth. Neptune and Pluto (or the plutinos) also conform very well only when averaged, supporting an escaped-satellite or other complicated concept.

Dermott's Law implies a planet at 187.5 AU, i.e., 36 = (cube root of 6)^6 times farther from the sun than Jupiter. Consideration of Saturn implies a planet at 189.0 AU; Uranus implies 209.1 AU. These three estimates together imply a planet at 195.2 +/- 7.0 (standard error of the mean) AU, vs. a little more than 197 AU from my predictions and from Genebriera's, Riley's, Turner's, and the sky surveys', possible detections of Barbarossa.

According to Dermott's Law, Mercury implies a semimajor axis of 151.9 AU for Barbarossa, Mars implies 180.7 AU, Ceres implies 180.9 AU. Ter Haar et al had favored a slightly larger ratio of 1.89, for the planets' orbital radii: this would give 221.6 based on Mercury, 244.1 based on Mars, and 234.5 AU based on Ceres.


Reply author: Stoat
Replied on: 02/17/2008 07:59:31
Message:

Hi Joe, I was going to ask TVF what he made of the anomalous findings of helium in the atmosphere of Jupiter. Which suggests that the planets formed very early. I forgot all about it but it would be of possible importance when considering our solar system as something of a "failed" binary system.


Reply author: Joe Keller
Replied on: 02/18/2008 22:14:17
Message:

quote:
Originally posted by Stoat

Hi Joe, I was going to ask TVF what he made of the anomalous findings of helium in the atmosphere of Jupiter. Which suggests that the planets formed very early. I forgot all about it but it would be of possible importance when considering our solar system as something of a "failed" binary system.



Good idea! Any evidence that the sun is borderline, by some criterion, for qualifying to be a member of a binary system, would improve the odds of a small dim companion.


Reply author: Joe Keller
Replied on: 02/18/2008 22:34:37
Message:

[quote]Originally posted by Stoat

Hi Joe, time maybe for another try at a telescope shot.Here's a screen shot of the ones on the Bradford robotic, Pick out the best position one and I'll put it up again as a job. ...

R44042 ("nem4") is good.


Reply author: Joe Keller
Replied on: 02/18/2008 23:56:32
Message:

From the data and calculations I have now, I can give the heliocentric celestial coordinates (in degrees) of the center of mass of the Barbarossa-Frey system:

RA = 171.6607 + T * 0.11676 + T^2 * 0.000124 + T^3 * 0.0000015

Decl = -9.0343 - T * 0.06173 - T^2 * 0.000174 - T^3 * 0.0000025

T is the time in years since Joan Genebriera's photograph of Barbarossa, taken 2007.2273 UT (i.e., Greenwich time).

From heliocentric coordinates and the current date, geocentric coordinates (i.e., apparent coordinates, if atmospheric refraction is included) can be calculated. For 00:00 UT March 1, 2008, I calculate the geocentric coordinates (not including atmospheric refraction) of the Barbarossa-Frey center of mass to be

RA 171.8242 deg

Decl -9.1146 deg

At this season, the daily apparent motion of the c.o.m. is

RA -0.0043 deg

Decl +0.00185 deg

The geocentric coordinates I gave above, for March 1, can be corrected by this amount, accurately for the next few weeks while Barbarossa is near opposition.

I think Frey will be about 0.15 deg NE of Barbarossa for the next few months. I think the mass ratio Barbarossa:Frey is about 8771:1229. They have a nearly circular 1.20 AU mutual orbit, with 14.7 yr period, tilted 40 deg to our ecliptic plane, with long apparent axis tilted 14.5 deg clockwise from the celestial equator. Their center of mass is about 197.73 AU from our sun, in a nearly circular orbit around our sun.

The total mass of Barbarossa + Frey is 0.00804 solar masses, and the total Barbarossa system is 0.0103 solar masses.

I base this on data previously discussed:

1. "Disappearing dots" A2 & A (listing the presumed Barbarossa first, then Frey) on the 1954 Palomar sky survey and
dots B3 & B on the 1986 UK-Australian sky survey (online scans).

2. Joan Genebriera's March 25, 2007 photo (Barbarossa) and
Steve Riley's April 1, 2007 photo (Frey). Genebriera used a 16" telescope on Tenerife and Riley an 8" (!) in southern California. Both used electronic detection with stacking.

3. I know of only one prospective (vs. a retrospective finding on a sky survey) photo of Barbarossa (Genebriera's), but I know of as many as four prospective photos of Frey. Genebriera photographed Frey on April 2, 2007. Robert Turner (operating, from England, a 14" robotic telescope also on Tenerife) photographed Frey on April 12, 2007. Riley made his second photograph of Frey on April 24, 2007. All of these photos are near the limit of detection and, like many stars near the limit, are not typical pixel patterns. Genebriera's Frey photo resembles a heart-shaped pixel overdensity which is not seen on another seemingly equally good photo she took only a few minutes later. I recorded Turner's Frey photo in my notes as a "barely significant pixel overdensity". Yet these four detections interpolate perfectly. Furthermore their motion is consistent not only with the presumed distance to the center of mass of Barbarossa-Frey, but also with the presumed orbit of Frey around Barbarossa.

4. Dot D (presumably Barbarossa) on the 1997 optical infrared sky survey. Though I've found no Frey on this online scanned plate, when Frey was added to the calculation at its presumed position according to the motion seen on the 1954 & 1986 sky surveys and the Genebriera, Riley, & Turner photos, the center-of-mass position of this infrared object was consistent.

5. Dots C3 & C11 on the 1987 LaSilla sky survey (online scan) have relative position consistent with the Barbarossa-Frey mutual orbit, but their c.o.m. is displaced approx. RA +0.12 deg Decl +0.025 deg, so, they were not used in calculating any orbit. The first dot I ever found, C, is between C3 & C11. Although a very elliptical orbit spends only ~1/6 of its time on the nearer half of the curve, still there probably would be comparably large deviations in the A, B, D, or Genebriera-Riley records if C is another massive Barbarossa moon. Also, C, unlike really any of the other dots, is a streak of length compatible with a Kuiper Belt Object. About half the discrepancy of C3&C11 could be removed if the distance to Barbarossa were much greater, so parallax could be recalculated, but, then Barbarossa would far exceed escape speed from the sun.


Reply author: Stoat
Replied on: 02/19/2008 03:31:21
Message:

Here's an article on the helium in Jupiter's atmosphere. http://www.space.com/scienceastronomy/solarsystem/jupiter_elements_991117.html

I've posted the job on the Bradford. Might take a while as the weather isn't that good at the moment I might be an idea to see if you can get another from your two contacts, they would expect to do another plate after this time, to do a blink comparison.


Reply author: Joe Keller
Replied on: 02/19/2008 16:19:15
Message:

Last year, I adjusted the mass ratio Barbarossa:Frey so that the three centers of mass A2&A, B3&B, and JG&SR (Genebriera's & Riley's objects) would lie on a great circle (which then happened also to be of constant angular speed, to about 1 arcsec accuracy). The quadratic interpolant through A2&A, B3&B, and JG&SR, for the heliocentric celestial coordinates of the c.o.m, in degrees, is

RA = 171.6607 + T * 0.115124 + T^2 * 0.0000159

Decl = -9.0343 - T * 0.058890 + T^2 * 0.0000148

(Again, T is the time in years since Genebriera's photo of Barbarossa, taken 2007.2273 UT.) This is practically equal to the exact constant-speed great circle, from which, the deviation of C3&C11, RA +0.1258, Decl +0.0339, is about the same as its deviation from the cubic interpolant (see previous post).

The cubic interpolant given in the previous post, was chosen to pass through an additional point, the presumed c.o.m. when Barbarossa is the 1997 optical infrared sky survey point D, and Frey is at the estimated position based on the three observed Frey positions of 1954, 1986 & 2007. The 4-pt. cubic and the great-circle interpolants necessarily agree perfectly in 1954, 1986 & 2007. In 1997, D (that is, the estimated c.o.m. with D as Barbarossa) deviates from the constant-speed great circle through A2&A, B3&B, & JG&SR, by RA -0.0068 deg, Decl +0.0123 deg.: a distance of 0.014 deg = 50 arcsec = 0.048 AU = 4.5 million miles, at Barbarossa's distance from the sun.


Reply author: Joe Keller
Replied on: 02/22/2008 13:56:00
Message:

Barbarossa and Frey Now Identified on 1954, 1986, 1987, & 1997 (Optical IR) Sky Surveys, and 2007 Photos of Joan Genebriera, Steve Riley & Robert Turner

radius of Barbarossa's circular solar orbit: 197.7283 AU
mass ratio Barbarossa:Frey = 0.8771:0.1229
(The object designations A2, etc., are those used in my notes.)

observed geocentric celestial coordinates in Julian epoch 1954.1516 Palomar sky survey, online scan of plate:
Barbarossa A2 11 2 25.26
-5 56 11.3
Frey A 11 3 12.4
-5 58 9
heliocentric celestial coordinates
of center of mass A2/A 165.5836
-5.86275

geocentric, Julian epoch 1986.2015 UK/Australia sky survey
Barbarossa B3 11 16 51.55
-7 49 41.1
Frey B 11 16 56.07
-7 55 14.3
heliocentric, c.o.m. 169.2360
-7.7851

geocentric, Julian epoch 1987.08215 LaSilla sky survey
Barbarossa C 11 18 3.18
-7 58 46.1
Frey C9 11 17 29.15
-7 57 28
heliocentric, c.o.m. 169.3289
-7.8601

geocentric, Julian epoch 1997.1693 optical infrared sky survey
Barbarossa D3 11 22 1.33
-8 34 36.9
Frey D 11 22 16.77
-8 29 30.9
heliocentric, c.o.m. 170.4712
-8.48695

geocentric, first successful photos taken prospectively at coordinates given by me, Julian epoch 2007.226648 (JG, Tenerife) & Julian epoch 2007.246595 (SR, southern California) (I used weighted ave. of time)
Barbarossa JG 11 26 22.2
-9 4 59
Frey SR 11 26 25.0
-8 57 48.5
heliocentric, c.o.m. 171.6471
-9.0281

Above, I use epochs I calculate, from the (mid-exposure) times of the plates, by the formula and tables in the 1989 Astronomical Almanac. For the 1954 and 1997 plates, these epochs differ appreciably (more than mere rounding error) from the epochs given by some online sources; I think it is better to use my calculated epochs, which are based on the times reported on the DSS online "plate finder".

Last year I determined the mass ratio Barbarossa:Frey by requiring that the c.o.m. heliocentric path A2/A-B3/B-JG/SR be a great circle. Recalculating from scratch this week, with a first-order correction for Earth's orbital eccentricity, confirms my statement of last year, that this mass ratio also renders the c.o.m. angular speed constant. The discrepancy in location at B3/B is +11.5 arcsec along the path.

This morning I discovered that the same mass ratio applies to C/C9 and to D3/D. Using the same mass ratio as above, the heliocentric paths A2/A-C/C9-JG/SR and A2/A-D3/D-JG/SR are both straight (i.e., great circles), possibly within 1 arcsec, if adjusted for an unexplained quantized deviation of approx. +/- n * 90", perpendicular to the path. These c.o.m. angular speeds also are constant. The location discrepancy at C/C9 is +19.5 arcsec along the path, and at D3/D, +25 arcsec, assuming constant angular speed.

Three of the five known images of Barbarossa, C (the first sky survey object I found), D3, and JG, are elongated several arcsec along a direction angle 60 or 70 deg (clockwise from north), similar to Barbarossa's orbital inclination. This might indicate moons, rings, an accretion disk, dust, or a corona. These elongations make the true location uncertain enough to explain some of the location discrepancy.

From these five pairs of points, a consistent (apparent counterclockwise) elliptical orbit of Frey around Barbarossa likely can be calculated. The results of my very approximate calculation, assuming a circular but tilted orbit and using only the three pairs A2/A, B3/B, & JG/SR, were reported to this messageboard earlier this week.


Reply author: Stoat
Replied on: 02/23/2008 05:57:10
Message:

Looking good Joe but now I think you need to contact your fraternity buddies to discuss the possible implications of that name. I'm sorry to bang on about it but I simply have to, it could sour the whole discovery. Freya (Freyja) and Frey (Freyjr) are okay and that would suggest the name of Odin for the brown dwarf. However, there is a general feeling that the world's mythologies are over represented by western gods and goddesses. Polynesian names are a possible, as are African, Inuit or Australian aboriginal names.

We also have to consider that this place might be a refuge for human beings, when our sun inflates into a giant. If we are still here in a few billion years time, then altering a brown dwarf to give off more light might well be possible. It definitely needs an optimistic name therefore.

The name Barbarossa, a psychotic, incompetent crusader, with only one admirer, Adolf Hitler. Barbarossa, the code name for the invasion of Russia. The other Barbarossa, is another psychotic, this time a particularly vicious pirate. I think that your friends from your university fraternity might not be too pleased with what any reasonably competent tabloid journalist could make from this. Making blood curdling oaths to the norse gods, whilst drinking vast amounts of strong lager is all when and good, when you are a nineteen year old student but such stories can become ammunition for your enemies in later years. An innocent frat group could be made to look like the vilest coven from hell.

I'll leave it there, it's up to you but I do think that it's only a matter of common courtesy to consult with your friends about the possible ramifications for them.


Reply author: Joe Keller
Replied on: 02/23/2008 13:51:16
Message:

One of my fellow Harvard graduates, Berry Fleming (Berry with an "e"), wrote a humorous political novel, "Colonel Effingham's Raid", which became a 1945 Hollywood film starring Charles Coburn & Joan Bennett, directed by Irving Pichel. Fleming's book is dedicated "To the women of the town under whose wing - like slumbering Barbarossa in the Kyffhaueser caves - slumbers the valor of Fredericksville". Often people find it convenient to use the same language as the person who has the most to say.


Reply author: Stoat
Replied on: 02/24/2008 03:48:43
Message:

Hi Joe, I'm at a loss as to what it is you're saying here. The king who sleeps, is king Arthur, the tradition of saying that any old king has this mythic property bestowed on them, is down to courtly toadying of the worst sort.

"Often people find it convenient to use the same language as the person who has the most to say." Perhaps, perhaps not. Much more common is the adage, where McCrimmon sits, is head of the table. What McCrimmon will say is, "Okay, we've got a new planet. We'll call it planet 9 for now. Find me a name that most people are happy with. I want this story to awaken the scientific imagination of a generation of kids. I don't want to see an unholy row about the name of the thing. If this guy that discovered it, looks like Brad Pitt and has the charm of Cary Grant, let him on t.v. as much as possible. If he wants to rock the boat and insist on his "right" to name it, then give him a medal and hustle him out of the frame as quickly as possible."

That's how it works Joe. It cannot be called unfair, it's simply realpolitic.


Reply author: Joe Keller
Replied on: 02/24/2008 19:57:35
Message:

Frey's Orbit Around Barbarossa

Previously I calculated the Frey/Barbarossa orbit using the 1954, 1986 and 2007 points only, and I assumed a circular orbit. Today I included all five points, 1954, 1986, 1987, 1997 & 2007. Generally there will be a two-parameter family of ellipses in space, corresponding to five points on the celestial sphere. Generally only one of these ellipses will have Barbarossa at a focus.

Arbitrarily I chose the radius (something near 197 AU) from the sun to two of the points. An ellipse is a planar curve, and this coplanarity determines the radius to the other three points. Then I solved a 5x5 system of linear equations for the coefficients of the second-order Fourier sum, for radius as a function of theta (with Barbarossa at the origin). I evaluated all the possible ellipses according to three criteria:

1. The radii of the five points must be as monotonic as possible, that is, the cyclic sequence can switch only twice from increasing to decreasing.

2. The Fourier sum for the radius, must never become zero (impact into Barbarossa).

3. The relative standard deviation of the areal speeds for the four time intervals between the five points (i.e., std. deviation divided by mean) must be minimized.

The best fitting ellipse by these criteria had eccentricity 0.7162, so, retrospectively, the Fourier sum I used isn't very accurate. Pressing on anyway, the semimajor axis is 1.76 AU (vs. 1.2 for my circular estimate), the orbital period is 14.5 yr (vs. 14.7 for my circular estimate), and the implied mass of Barbarossa (very sensitive to semimajor axis) is 0.026 solar masses (vs. 0.008 for my circular estimate).

The relative standard deviation of areal speed, of the four time intervals, is 20% (there would be more than two full cycles between 1954 & 1986; the other intervals were less than one cycle each). Some of this might be from the inaccuracy of the Fourier sum for such large eccentricity. Some of the areal speed variation might be real, due to Frey having a moon of its own; or, Frey might have a quickly precessing, or even a chaotic orbit.

In this calculation, Frey's orbit is tilted only 9 degrees to Barbarossa's orbital plane around the sun (vs. 40 deg for my circular orbit estimate). The chance of such parallelism, between the two orbital planes, randomly happening over all angles on the celestial sphere, is only 1%.

If instead of a Fourier sum, an ellipse were fitted to the five points, then under ideal conditions, criteria #1-3 above always would be perfectly satisfied. (Kepler's area law holds for all tilts.) With large perturbations, no orbit satisfies criterion #3 and no orbit is an exact ellipse.


Reply author: Joe Keller
Replied on: 02/24/2008 20:50:24
Message:

The Astrophysical Journal, 627:10011010, 2005

High Orbital Eccentricities of Extrasolar Planets Induced by the Kozai Mechanism

Genya Takeda and
Frederic A. Rasio

Abstract. "...at least 20% of [extrasolar] planets, including some with particularly high eccentricities, are orbiting a component of a wide binary star system. ...a large-amplitude eccentricity oscillation. This so-called Kozai mechanism is effective at a very long range...Our calculations show that the Kozai mechanism consistently produces an excess of planets with very high (~ >0.6) and very low (~ <0.1) eccentricities. Assuming an isotropic distribution of relative orbital inclination, we would expect that 23% of planets do not have sufficiently high inclination angles to experience the eccentricity oscillation [the oscillation between high inclination and high eccentricity - JK]. By a remarkable coincidence, only 23% of currently known extrasolar planets have eccentricities ~ < 0.1. ..."


Reply author: Joe Keller
Replied on: 02/25/2008 11:57:07
Message:

Frey's Orbit Around Barbarossa (cont.)

Next, instead of approximating the orbit by the second-order Fourier sum for the radius (as a function of theta), today I did it by the second-order Fourier sum for the reciprocal of the radius (which, is exact at the five data points). This approximation should be better, because it is exact when Barbarossa is a focus.

This best fitting ellipse had eccentricity 0.6072. The semimajor axis is 1.1998 AU (vs. 1.2007 for my circular estimate), the orbital period is 13.36 yr (vs. 14.68 for my circular estimate), and the implied mass of Barbarossa+Frey is 0.00968 solar masses (vs. 0.00804 for my circular estimate; and 0.0103, for Barbarossa+Frey+remainder, based on the precession frequency ratio 3:1 surmised above for Neptune).

The relative standard deviation of areal speed, of the four time intervals, is 23.5%. If the ellipse is chosen whose focus is exactly at Barbarossa, the relative standard deviation of areal speed is 30%. The former ellipse is better, because angular momentum conservation is more fundamental than ellipse geometry.

In this calculation, Frey's orbit is tilted 34.5 degrees to Barbarossa's orbital plane around the sun (vs. 40.0 deg for my circular orbit estimate). My 5-pt. ellipse-based estimate of the orbital plane, is at a large angle to my earlier 3-pt. circle-based estimate. Following Jacobi, Tisserand and Kozai, sqr(1-e^2)*cos(i)= 0.655. If Frey's orbit originally were circular and inclined 60 deg to Barbarossa's (the median random angle), this constant would be 0.5.


Reply author: Joe Keller
Replied on: 02/26/2008 01:18:23
Message:

The Barbarossa Cloud's Atmospheric Refraction


From my Feb. 22, 2008 post (original version):

"Using the same mass ratio [0.8771:0.1229] as above, the heliocentric paths A2/A-C/C9-JG/SR and A2/A-D3/D-JG/SR are both straight (i.e., great circles) within 1 arcsec. These c.o.m. angular speeds also are constant."

The straightness of the great circles comes with a proviso:

This afternoon, on another online sky survey plate, I found a sixth Barbarossa/Frey pair:

geocentric coordinates (preliminary measurement, possibly several arcsec measurement error), Julian epoch 1995.1408 optical infrared sky survey (the same survey, but a different plate, from that showing D3 & D)

Barbarossa E3 11 21 23.7
-8 24 36
Frey E4 11 21 46.8
-8 20 01

heliocentric coordinates of c.o.m.
170.2690
-8.2910

Their center of mass is displaced roughly +1" along the constant angular speed great circle orbital path A2/A-JG/SR.

The presumed center of mass of B3/B is displaced only 2" perpendicularly downward from the straight great-circle path from A2/A to JG/SR. E3/E4 is displaced 94" perpendicular upward from that path. The other optical infrared pair, D3/D, is displaced 190" perpendicular downward (exactly twice as much). The visible light pair, C/C9, is displaced 88" perpendicular downward. Yet no pair (employing the c.o.m. above) errs more than 25" from constant angular speed along that path. A variable quantized refraction occurs, perpendicular to Barbarossa's orbit. A2/A, B3/B, and JG/SR are refracted the same amount, hence were found by me to lie on one great circle.

On average, Barbarossa's orbital path 1954-2007 differs 27.77 - 23.45 * cos(11.4) = 4.78 degrees from the ecliptic. A -90" displacement perpendicular to the ecliptic (e.g. from a large error calculating heliocentric coordinates) would cause 90" * sin(4.78) = +7.5" displacement along Barbarossa's orbit, but +19.5" is observed for C/C9. Such displacement of E3/E4 (~ +90" perpendicular displacement) should be -7.5" but is roughly +1". Such displacement of D3/D (~ -180" perpendicular displacement) should be +16" but is +25". B3/B, displaced only -2" perpendicularly, should be displaced +0.2", but is displaced +11.5", along the path. Barbarossa images C, D3, and JG are streaks several arcseconds long, oriented roughly along the orbital path. This ambiguity might account for some residual displacement along the path. However, if this error in calculating heliocentric coordinates is assumed to cause the large perpendicular discrepancies, the discrepancies along the path become +11.3", +12", +9", and +8.5" for B3/B, C/C9, D3/D, and E3/E4, resp.

A quantized light refraction perpendicular to the ecliptic, simulating an error in calculating heliocentric coordinates, thus explains practically all the position discrepancy. For the c.o.m. to be displaced 9" forward in 1997, and 12" forward in 1987, implies accelerations consistent within 10%, and which could be caused by an orbital eccentricity of 0.02 or greater.

A2/A, B3/B, and JG/SR also are the unique trio for which Earth's distance from the point of Barbarossa's opposition, is, to within a few percent, directly proportional to time. Thus a parallax calculation error proportional to Earth's distance from opposition to Barbarossa, leaves their line straight and constant-speed. Yet this does not explain why the error perpendicular to the path should be quantized, or why the error perpendicular to the path should be largest for D3/D, whose relation to opposition is much closer to the trend of A2/A-B3/B-JG/SR, than is such relation for either C/C9 or E3/E4.

The synchrony of A2/A, B3/B, and JG/SR might instead be due to rotation of the Barbarossa system's, or the entire solar system's, gas/dust cloud. The diminution in light, due to absorption and refraction, would explain why the brightness magnitudes of Barbarossa and Frey are so low. In previous posts I found that the comparison magnitudes were consistent with Barbarossa and Frey having the lowest albedos known, and even then, somewhat smaller than the most widely accepted theoretical diameters for bodies of their mass. Images of Barbarossa and Frey, whether on sky survey plate scans or on electronic photographs, always are either unusually small, or unround and unusually large.


Reply author: Joe Keller
Replied on: 02/27/2008 23:52:36
Message:

Accurate Frey/Barbarossa Orbit Confirms Refraction Phenomenon

I tried replacing D3/D with E3/E4, then finding the best Frey/Barbarossa orbit by the method I last described above. The best fit gives a standard deviation of only 4.56% for the areal speeds during the four time intervals. The eccentricity is a plausible 0.7401, the semimajor axis is 1.223 AU, and the period is 14.28 yr. The orbit is inclined 49.9 deg to Barbarossa's orbital plane.

This orbit implies that Barbarossa+Frey = 0.01024 solar masses. This agrees perfectly with the 0.0103 solar masses predicted for the entire Barbarossa system, according to the 3:1 Neptune precession resonance theory I discussed above.

From this best orbit, I removed E3/E4 and, maintaining the same orbital plane, added D3/D, leaving the other points on this orbit as they are. Then I searched for the shift in the RA & Decl. of D (Frey) which would minimize the standard deviation of the areal speed for the new orbit. The minimum standard deviation was 18.0%, achieved when D was moved 28.25" forward parallel to Barbarossa's path on the celestial sphere, but only 0.3" closer to the path. This suggests that D is not merely a random mistake, but rather that its apparent position is affected by a physical phenomenon. Also, 28.25" * pi = 89", approximately the quantum of deviation observed perpendicular to the path. This movement of D would bring the center of gravity D3/D forward +3.5".

Refractions through a slanted window pane cancel except from objects within the pane. I've found no extremely large proper motions of stars near Barbarossa/Frey in the photos, but according to the USNO-B catalog, Barbarossa's orbital path is enriched in catalog objects with unusually high (automatically calculated) proper motions. This is the phenomenon which caused me to investigate plates of this region.


Reply author: Joe Keller
Replied on: 02/28/2008 12:52:39
Message:

Ratios Support Barbarossa Cloud Refraction Phenomenon

The temperature of the CMB Planck curve is 2.726 K with an error of about 1 in 3000, as evidenced by the implied significance of the figure (1 in 5000) and also by occasional published statements that it is 2.725. (Elsewhere on this messageboard I explain why the CMB Planck curve must be imperfect due to contamination by far CIRB, causing two different definitions of best fit, which differ by 0.001 deg.) The best fit Planck curve to the sun ("Journey Through the Galaxy", filer.case.edu, 2006) has 5780 K, a figure whose implied accuracy is 1 in 1200.

Thus Tsun / TCMB is 2539.5 with accuracy 1 in 1000.

The 3-yr WMAP result for the CMB dipole is 3.358 mK with error bar 1 in 200.

Thus TCMB / halfdeltaTdipole is 811.79 with accuracy 1 in 200.

The ratio of the ratios, 2539.5/811.79 = 3.1283, equals pi=3.1416, to accuracy 1 in 200. This suggests a simple physical relation between the sun's temperature, the CMB temperature, and the CMB dipole temperature.

Since my discovery of Barbarossa and Frey, I can say more. The two deviation quanta perpendicular to the ecliptic, of centers of mass, on optical infrared photos (the DSS website says they were with an IVN plate and RG715 filter), average (assuming that they are quantized and that the larger deviation should be divided by two) 94.25" +/- range 0.75" = 1/2188 radian with 1% accuracy. The visible light photo (the DSS website says this was with a IIIaF plate and OG590 filter, equivalent to photographic passband R59F) shows perpendicular deviation approx. 88.2" = 1/2339 radian, whose accuracy may be guessed to be 1% also.

PASP 98:1303+, Fig. 1, gives the response functions of these Red & IR plates. Likely the quantized refraction angle ("rainbow angle") in radians, equals the ratio of the "average" frequency of CMB photons, to the "average" frequency of sunlight photons detected by the plate. This assumes that Barbarossa and Frey are seen only by reflected light and that their albedos are independent of wavelength, so that the reflected light approximates the sun's light.

For such ratios to occur with Planck-curve light, the energy-per-unit-wavelength Wien's law peaks would have to be given by

2539.5/2339*28970000 Angstrom/K (i.e., Wien's const.)/5780 K = 5442 A for the Red plate

2539.5/2188*28970000 A/5780 = 5817 A for the IR plate

vs. 28970000 A/5780 = 5012 A for sunlight.

I think that really the effective temperature, corresponds to that "average" energy per photon, which is weighted by energy. That is, if half the energy is in the form of photons of energy E1 apiece and half the energy is in the form of photons of energy E2 apiece, the effective energy per photon is (E1+E2)/2. The integrals for this calculation are in Dwight's "Tables of Integrals" 860.39; here Dwight refers to small integer values of the Riemann zeta function (in, inter alia, Dwight's other book, "Mathematical Tables").

By this definition, the average photon of sunlight is of wavelength 6494 Angstrom. From the size of the quantized refraction, the predicted average photon on the "Red" or "IR" plate is of wavelength 7051 or 7537 A, resp. My crude graphical estimate made by plotting the (wavelength) Planck curve of solar radiation, on the passband graph (Fig. 1, PASP 68:1304) and then roughly plotting the product function, is that the energy-weighted average photon energy for R59F exposed to sunlight, is that of a photon of wavelength 6300 A, and for IsubN (equivalent to IVN + RG715), 7800 A. So, the presumed refraction of IR plate objects D3/D & E3/E4 is roughly explained, but Red plate objects C/C9 seem to be refracted somewhat too much; redness (like classical Kuiper Belt objects) can explain only part of this discrepancy, because the R59F window cuts off at 7000 A.


Reply author: Joe Keller
Replied on: 03/02/2008 21:02:05
Message:

Please allow me to call attention to my important accurization today (March 2) of my Feb. 22 and Feb. 26, 2008, posts, above. When I accurized the positions today, finding the result when exactly correct Julian epochs are used throughout, I found that practically all discrepancy in center-of-mass Barbarossa/Frey position, disappears when a quantized refraction, ~ +/- n * 90", perpendicular, not to Barbarossa's orbit, but rather to the ecliptic, is assumed.


Reply author: Joe Keller
Replied on: 03/04/2008 13:39:33
Message:

Barbarossa's Refraction by the "Cloud"

The observed refractions of Barbarossa/Frey, in terms of the approx. 90" quanta perpendicular to the ecliptic (which is at practically the same inclination as Jupiter or Saturn's orbit, at this longitude) are 0, 0, -1, -2, +1, 0. Here, where there is quantization superposed on likely an underlying sinusoidal variation, normal distribution theory might not give very accurate error bars, but using normal distribution theory, the mean refraction is -(1/3) +/- 0.42 (std. error of the mean) quanta.

Assuming the other three visible light images would have had the same quantum of refraction as C/C9, the mean quantum of refraction, for the six images, is 1/2281 radian. Barbarossa's position in the six images averages 11.2 deg below the ecliptic. If the refraction is due to a thin disk or lenticule, containing the known solar system but not Barbarossa, and "n" is the effective index of refraction of the material within this solid disk, then most likely,

n-1 = 1/3 * 1/2281.3 * tan(11.185) = 1/34,612

By comparison, the constant 0.5 * alpha^2, where alpha is the fine structure constant, is 1 / 37,558. Some of my earlier posts on this messageboard explain the suggestive connections between this constant and the so-called CMB. If an outer lenticule exists, similar to this hypothetical inner one, but larger and containing Barbarossa, then refractions of distant stars would cancel, but the refraction of light from Barbarossa would not.

If I use the principal plane of the solar system (approximated as the mean of Jupiter's and Saturn's orbits, weighted by angular momentum) instead of the ecliptic, n-1 becomes 1/30,532. If I use the plane of Earth-moon angular momentum (including rotation)(averaged over one draconitic cycle)(including small corrections for barycentric motion and for average equinox) n-1 becomes 1/36,455, only 3% larger than 0.5*alpha^2 (here the precession of lunar nodes is assumed to cancel, because 1954-2007 contains almost three full cycles). More precisely, averaging the Earth-moon angular momentum vector over the six image times, gives 1/38,010, 1.2% smaller than 0.5*alpha^2.

As mentioned previously, D3/D shows a possible quantized displacement of Frey, relative to Barbarossa, parallel to the ecliptic. Not only is D3/D the image showing the largest (2 quanta) quantized displacement of the c.o.m.; also, D3/D is the image in which Frey's ecliptic latitude differs most from Barbarossa's.


Reply author: Joe Keller
Replied on: 03/09/2008 13:01:36
Message:

Yesterday I rechecked the 1983 SERC Blue sky survey, and again found no "disappearing dot" near the expected (geocentric) positions of Barbarossa or Frey. That is, I found no dot from 1983 Blue that wasn't also on 1987 Red, 1954 Red and/or 1997 Infrared, except for dots so faint that many of the other dots of similar faintness were absent too.

For a spectral type G2 star (e.g., the sun), B-R=1.06 and R-I=0.34 (astro.pas.rochester.edu, citing Mihalas & Binney's "Galactic Astronomy and also PASP 100:1134). So, if Barbarossa/Frey are colorless in sunlight, they would be about 1.06 mag dimmer (i.e., mag ~+20 instead of ~+19; the detection limit is ~+21) on the 1983 Blue plate than on the 1987 or 1954 Red plates, and about 1.40 mag dimmer than on the 1997 IR plate.

A famous early study of 21 distant solar system bodies, including classical KBOs, "1:2 objects", plutinos (i.e., "2:3 objects"), and centaurs, showed that B-V (data on 19 objects) ranged from 0.71 to 1.24 (mean 0.96) and V-R (data on all 21 objects) ranged from 0.37 to 0.78 (mean 0.625) (Tegler & Romanishin, Nature 407:979). So for Barbarossa/Frey, B-R is likely 1.6 and might unsurprisingly be as high as 2.0.


Reply author: Joe Keller
Replied on: 03/10/2008 22:14:36
Message:

I'd like to get prospective photos of Barbarossa and Frey on film. CCD imaging has many advantages, but the spurious images that can arise on film probably are fewer and better understood.

Likely, Barbarossa and Frey are red. I've found them on all five Red and Optical Infrared sky surveys (scans of photographic plates), but not on the one Blue sky survey. They are dimmer than expected, so either are unexpectedly small, are surrounded by dust (which would redden them), or have unexpectedly low albedo, consistent with a strong color. Distant solar system objects typically are red (see above post).

For distant solar system objects, Tegler & Romanishin found B-V up to 1.24 and V-R up to 0.78. Consulting original sources, I find that Johnson & Morgan (ApJ 117:313, 1953, Table 3) found B-V = 1.16 and 1.48 for the standard K2III star betaOph and the standard K4III star betaCnc, resp. Kron, White & Gascoigne (ApJ 118:502, 1953, Table 4) found R magnitudes giving V-R = 0.54 and 0.76 for these stars, resp. So, the reddest common distant solar system objects, have about the B-V of a K2III star and the V-R of a K4III star. That is, they crudely resemble a K3III star. Straizys, "Multicolor Stellar Photometry", p. 51 and Figs. 11 & 12, gives 6000 & 8300 Angstrom for the smoothed spectral peak for K0III & K5III, resp. This interpolates to 7380 A for K3III. MS Bessel's abovementioned article in PASP, Fig. 1, shows that Red sky survey plates (R59F) cut off at 7000 A above, while the Optical Infrared sky survey plates (IsubN) cut off at 7000 below. Either plate would miss about half the radiation from Barbarossa or Frey, if these resemble the reddest objects cataloged by Tegler & Romanishin.

Two especially good advanced books on astrophotography are Covington's 1999 "Astrophotography for the Amateur" 2nd ed., and Wallis & Provin's 1988 "Manual of Advanced Celestial Photography". Covington's Table 10.1 lists "Kodak Ektachrome Professional E200" as in a 7-way tie for first place among 26 un-hypered films, for good reciprocity. It's in third place for sensitivity at (simulated alpha-hydrogen line, 6563 A) 6600 A. Neither of the two films besting it for red sensitivity, had very good un-hypered reciprocity.

Covington warns that a film's characteristics might change over the years due to changes in manufacture, or even change randomly from roll to roll. He also says that the designation "Professional" and the requirement that "Professional" film be refrigerated, have mainly to do with maintaining uniform color balance through consistent storage and aging procedures (for the sake of the portrait photographer) and little to do with qualities that are of interest to most astronomers.

Kodak's website (use the onsite search window) still lists "Kodak Professional Ektachrome E200". I'd suggest this film for photographing Barbarossa.

Infrared films have poor reciprocity, though this can be improved by hypering, particularly by distilled water hypering (Wallis & Provin). Two prime choices might be Konica IR750 (peak sensitivity 7500 A) and (if available) Agfa APX 200S (peak 7250 A). Kodak Technical Pan 2415 (it's B&W) hasn't been manufactured since several years prior to 2004, though has good far red sensitivity (with hypering) and reportedly stores well even for a decade without refrigeration.


Reply author: Joe Keller
Replied on: 03/12/2008 17:35:35
Message:

In Covington's Oct. 2003 online revision of information in his 1999 2nd ed. text, he says:

"Kodak Elite Chrome 200 and Kodak Professional Ektachrome E200 Film are my favorites. (These two films are the same emulsion with slightly different aging.)

"This film has unusually good reciprocity characteristics (holds its speed well in long exposures) and unusually good response to emission nebulae (which come out bright cherry-red)."

Local photo shops only stock "Professional Elite Chrome 200". I bought two rolls, so I'll be ready to go.

I do find separate online Kodak technical data sheets, both dated 2005, for "Professional Ektachrome E200" and for "Professional Elite Chrome 200". The former data sheet has a section on push processing but the latter doesn't.

So, I would prefer "Professional Ektachrome E200". Covington says that of the two films, this is the one he has used the most; also, Covington's 1999 book says that push processing to 640 substantially increases its red response, while the data sheet on "Professional Elite Chrome 200" doesn't see fit to mention push processing at all.

Here's a Kodak expert quoted on a photography messageboard. He enthusiastically advocates push processing for Professional Ektachrome E200, but for the other film he is silent on the subject. On the other hand, he seems to imply that push processing would be superfluous, for Professional Elite Chrome 200:

(Aug. 2003)

"Here's the official response from Kodak:

'No, these two Kodak Films [Professional Ektachrome E200 vs. Elite Chrome 200] are very different. Kodak Elite Chrome 200 Film is inherently higher in color saturation and contrast, while Kodak Professional Ektachrome E200 Film, shot at EI 200, is pretty flat. However, the E200 pushed even 1-stop to EI 320, nicely increases in color saturation and contrast without adversely impacting color accuracy or grain. You can actually shoot this film all the way out to EI 1000 with excellent results. ...

-Peter V.
Kodak Information and Technical Support
Kodak Professional
Ph. 800-242-2424 ext. 19 ' "

The two leading photo shops in Des Moines each told me on the phone today that they have a few rolls of Professional Ektachrome E200 in stock. I might call them again to confirm that their statement is accurate and that they still have it, before I drive to Des Moines.

Update June 23: I've visited two of the three leading photo shops in this area. All three had just told me by phone that they had Professional Ektachrome E200. Neither of the two shops I actually visited, really had it. Both told me, or behaved as if I would assume, that Elite Chrome 200 is identical. According to Kodak, it most definitely isn't (see above). I bought some anyway at one of the shops, because it might be the best I can do, considering time limits.

This incident reminds me of the college student sitting next to me in the computer lab who told someone on his cell phone, "I have to hang up now, I have another call." Then, not caring that I and two other witnesses were learning that he was an habitual liar, he resumed working at the computer. There was no other call.

So a company that manufactures cell phones, maybe his cell phone, fired their safety officer, an epidemiologist who confirmed the cell phone - brain cancer link and refused to whitewash it. The college student tells little lies to his friends, the photo shop salesmen tell bigger lies to me, the cell phone manufacturer tells yet bigger lies to both of them. The big fish eat the little fish. Should we live like fish?

When lying becomes too widespread, speech becomes valueless. The statement of something, ceases to have any correlation with the likelihood that it is true. One knows no more after hearing a statement, than one did before. The speaker might as well not speak at all.


Reply author: Joe Keller
Replied on: 03/12/2008 19:53:20
Message:

Infrared and Extended-Red Films Banned by Homeland Security?

Stanford-trained journalist Linda Howe says, that a Midwestern U.S. college photography professor told her, that the big New York photography stores told him, that U.S. Homeland Security has banned the sale of infrared cameras. The professor speculated that this ban extends to infrared film. I don't know about the cameras, but the film seems to be available despite the general contraction of the film market.

Earlier this year Kodak discontinued its IR film, which was called "HIE". I saw one seller on eBay, but the price was "gouging": $47/roll including shipping. This film has about the same IR range as standard astronomical "optical IR" plates. (I saw another discontinued film of interest, Kodak Technical Pan 2415, on eBay with three sellers, for, e.g., $15/roll incl. shipping: not gouging prices.)

Maco IR820c (range - not peak - to 8200 A, and spectral response rather flat from ~4500 to ~8100 A; not as deep IR as Kodak HIE, but closer to it than any rival, though slower) seems to have disappeared from the market too. I checked the websites of New York and Chicago dealers which the current Cachet Co. website lists as carrying their Maco infrared film. None of them now lists it. Maybe this is only a failure to update Cachet's website. The Freestyle Co. in Hollywood sells Efke IR820, which is advertised as the same thing as the Maco but by a different manufacturer, so, I ordered some.

My internet search indicates that the abovementioned "extended-red" (basically, semi-IR) B&W films, Konica IR750 (peak at 7500 A) and Agfa APX 200S (peak at 7250 A), can't be found anywhere by anyone. A Des Moines photo shop told me Agfa is defunct. The New York office of Konica told me Konica quit making film.

Another extended-red B&W film, Ilford SFX200 (peak at 7200 A, flat to ~7300 A) has resumed production and is available from Freestyle too. I ordered some, though it's said that this film has especially bad reciprocity failure. Freestyle sells yet another extended-red B&W film, Rollei 400iso (flat to ~7000 A), which I also ordered.


Update Apr. 1, 2008. I got this email today from a major U.S. vendor of IR cameras:

"In regards to the ban of IR cameras, I have not heard anything about this and believe this just may be a rumor. We do have products that are export restricted but IR cameras are definitely not banned by DHS. Let me know if you have any questions.

Regards, Paul Neak
Rental Program Manager
FLIR Systems, Inc.
978-901-820 "

(Also, restrictions on the sale of IR cameras, might be part of the U.S. Army's "owning the night" strategy in Iraq & Afghanistan. IR cameras sold in the U.S. might find their way to the enemy.)


Reply author: Stoat
Replied on: 03/19/2008 05:44:35
Message:

Hi Joe, I got the new image from the Bradford but it bounced back when I tried to send it to you. I think that hotmail has a limit to attachment files. Anyway it said your mailbox was full. The zipped file is about 8 meg. Message me with an e mail address.


Reply author: Stoat
Replied on: 03/20/2008 09:37:32
Message:

Hi Joe, thought you'd want the times of these two images, they're probably in the headers but I've not looked. 00:10 on Sunday 1 April 2007 (00:10:10 UTC)
01:44 on Wednesday 19 March 2008 (01:44:57 UTC)

That second one is about as close to the equinox as it gets, must bode well. No pun intended with the mention of Bode.[:)][8D]


Reply author: Joe Keller
Replied on: 03/23/2008 20:05:39
Message:

A Rapidly Precessing Orbit for Frey Around Barbarossa

The six pairs of Barbarossa/Frey data points conform well to an orbit whose apsides precess retrograde about 60 degrees per sidereal "month", on average. There are 17 numerical data: 2 celestial coordinates x 6, plus (6 minus 1) time intervals. The model contains only 10 adjustable parameters: the orientation of the orbital plane (2 parameters), the dimensions of the orbital ellipse (2 params.), the initial point on that ellipse (1), the initial orientation of the apses (1), the mean rate of precession of the apses (1), and that rate's sinusoidal variation (3). Here's how I estimate this orbit by elementary methods:

1. Graph all six presumed Freys relative to their Barbarossas, on the same sheet with all the Barbarossas at the origin. (Multiply RA degrees by cos(Decl).) Note that A2A (i.e., the line from A2, the 1954 Barbarossa, to A, the 1954 Frey) is the longest apparent radius, that CC9 is second longest apparent, and that these are (anti)parallel. Guess that Frey's orbital plane around Barbarossa, is tilted to our line of sight, about the axis A2A. JGSR (i.e., the line from Genebriera's 2007 Barbarossa, to Riley's 2007 Frey, only a few days later) is perpendicular to A2A, and JGSR is longer than B3B or D3D (the other radii which are roughly perpendicular to A2A). So, guess the orbital plane's tilt angle as arcsin(JGSR/A2A) = 37.5 deg, from edge-on to our line of sight (or following binary star convention, we could say 90-37.5 = 52.5 from face-on). The Barbarossa/Frey orbital plane also is inclined 22 deg to Barbarossa's path around the sun, for a total inclination, to the Barbarossa/sun orbital plane, of 43.5 deg.

2. Now calculate all the actual lengths and angles in the orbital plane. Assume a highly elliptical orbit whose plane is invariant but whose apsides precess retrograde, rapidly. For a first approximation, let's use the position vector itself as an estimate of the line of apsides. For the fairly long times C9E4 (i.e., the time from C9 to E4) and E4SR, we then can estimate the rates of precession, 15.4 & 27.1 deg/yr, resp.

3. Let's approximate Frey's orbit as a precessing ellipse. For e = 0.6, Frey would spend ~2/3 of its time beyond the semimajor axis distance. The two shortest of the six actual radii are those of B and C9, and these are almost equal. So, adjust them to exact equality, and assume they equal the semimajor axis (which is then 0.50 AU). Using the apsis precession rate C9E4 (estimated in #2, as 15.4 deg/yr) subtract the amount of apsis precession, from the actual angle between B & C9. Plot the ellipse. The eccentricity is e = 0.7393. Frey's Keplerian sidereal month, with this major axis, would be 3.736 Julian yr.

4. Assume that an outer moon, Freya, in a larger orbit nearly perpendicular to Frey's, and edge-on to our line of sight (maybe somehow forced to be nearly perpendicular to Barbarossa's nearly-circular orbital path, like a submerged pan dragged through water), causes Frey's apsis precession. Suppose Freya's mass equals Frey's and that there is no other mass in the Barbarossa system. That is, the total mass is 0.0103 solar masses (determined from Neptune precession resonance), and the masses Barbarossa:Frey:Freya = 0.8771:0.1229:0.1229 = 7:1:1. If Freya & Frey have 2:1 orbital resonance, then Freya's major axis is 0.83 AU (including the additional centripetal force, exerted on Freya, by Frey).

5. Consider the effect of apsis precession rate, on the shape of the precessing ellipse, for the case of very small eccentricity. If a turntable's frame of reference is to have the same forces as are exerted on Frey, there will have to be additional special forces in the turntable's surrounding fixed frame, to cancel Coriolis force and the apparent force due to change in the turntable's speed. Conservation of angular momentum in the turntable frame, and equality of centripetal and centrifugal forces, afford two algebraic equations in r, Omega (the ellipse, or rather circle, orbital frequency in the fixed frame) and omega (the turntable frequency), from which r is elimated to give Omega as a function of omega. Also, the effect of turntable acceleration, delta(omega), on the eccentricity and line of apsides, is estimated and considered.

6. Assume B & D are inbound; A, C9, E4 & SR outbound. Considering the relation between Freya's perturbation force on Frey, and Frey's on Freya, gives the cycle of Freya's apsis precession, as about 36 yr if Frey's cycle of apsis precession is 22 yr; this 36 yr also should be the period of variation of Frey's apsis precession rate. I performed one or two steps of a rapidly converging computation in which a more accurate description of the precessing ellipses, leads to a more accurate determination of the apses and of the sinusoidally varying precession rates, and in turn to an even more accurate description of the ellipses. Frey's precession rate varies approximately sinusoidally from almost zero, to about 36 deg/yr, almost the highest rate for which the nearly-circular case has a solution balancing forces and conserving angular momentum (see #5). The average precession rate is 16.4 deg/yr, i.e. 53 deg/ actual average sidereal month. The parameters of the sinusoidal variation of precession rate (mean, amplitude, frequency, and phase) are determined by the five apsis positions 1986-2007 (1986 & 1987 both are used up in determining the ellipse plus one apsis position), but as a test, the position at A (1954) conforms to the model, if the mean precession rate is reduced to 15.0 deg/yr. The six final calculated accurate ellipses have eccentricities from 0.637 to 0.683.

7. The time estimates for the motions BC9, C9E4, E4D, and E4SR all are too high, by 11.7%, 24.0%, 31.6%, and 20.7%, resp. Also, the model overestimates the time A2A-B3B by 20.7% (if 11+ orbits are assumed). The estimates C9E4 & E4SR should be most accurate, because these time intervals are nearly several whole anomalistic months. Roughly, BC9 is, timewise, the inner 1/3 of the ellipse and E4D the outer 2/3; weighting these by time, 1/3::2/3, gives 25.0%. These three numbers, 24.0%, 20.7%, and 25.0%, are so close that surely they tell us the amount by which the major axis of Frey originally has been overestimated; they correspond to an adjustment from 0.50, to 0.435 AU.

8. Averaging from 1986 to 2007, Frey's anomalistic month is 2.86 Julian yr, and its sidereal month 3.23 Julian yr. From #7, the Keplerian sidereal month (what the month would be if Freya disappeared) is 3.03 Julian yr (this needn't exactly equal the actual average sidereal month). Frey's apsis precession varies from 0 to 100 deg per (Frey) sidereal month (ave. 53 deg). By contrast, Luna's apsis precession is 3.0 deg per sidereal lunar month. A particular neutron star reportedly exhibits presumed relativistic apsis precession of 38 deg / cycle. However, published numerical computations confirm that Frey's rapid precession is roughly consistent with Freya's mass and orbit.


Reply author: Joe Keller
Replied on: 03/24/2008 11:44:42
Message:

Freya's Orbit

"...Orbits rotate within the plane in the same sense as satellite motion when inclination is less than ~63.4 degrees, and in the opposite sense at higher inclinations. Edit: In exact terms, the cutoff occurs when inclination = arcsin(sqrt(4/5))."

- Grant Hutchison, Senior Member, "Bad Astronomy" forum, Feb. 27, 2008

[This famous fact appears, inter alia, on the first page of the first article published in Celestial Mechanics vol. 1, no. 1, p. 6, June 1969. It is important in placing high-latitude communications satellites. - JK]


Frey's retrograde apsis precession would imply that Freya's orbit is roughly perpendicular to Frey's, thus roughly perpendicular to Earth's ecliptic also. This improves the chance that Freya's orbit is not only perpendicular to Earth's ecliptic, but also seen edge-on. This displacement of Freya, perpendicular to the ecliptic, could explain the displacement of the centers of mass of CC9, E3E4, and D3D.

The maximum displacement of the Barbarossa-Frey c.o.m. was 190" (1997). Because in the previous post, everything seemed consistent with Freya having 1/9 of the total mass, this implies Freya's displacement was about 8 * 190" = 25.3' = 0.4222 deg, which gives a lower bound for the apoapsis of Freya's (highly eccentric, if Frey's varying apsis precession rate is to be explained) orbit about the c.o.m.; the corresponding lower bound for the apoapsis from Freya to Barbarossa (or more accurately to the Barbarossa-Frey c.o.m.), is about 0.4222 * 9/8 = 0.4750 deg. Frey's apoapsis finally was estimated (midrange) 0.243 * 0.435/0.50 = 0.211 deg. If these moons have equal eccentricity, then their period ratio is greater than (0.4750/0.211)^1.5 * sqrt(7/8)(because of Frey's pull on Freya) = 3.16.

Without contradicting other calculations, the discepancy between 3.16 and 2, is difficult to explain completely, simply by increasing Freya's mass and eccentricity. Recall that I originally calculated the mass ratio Barbarossa:Frey by requiring that the c.o.m. path around the sun, A2A-B3B-JGSR, be straight. If the real mass ratio were known, it would be found that the c.o.m. B3B is deflected laterally also, as are CC9 (until now, said to have -1 quantum of deflection), D3D ( -2 quanta) and E3E4 ( +1 quantum). The true deflections are best estimated as +0.5, -0.5, -1.5, and +1.5 quanta, resp., so they have mean zero. Here D3D's deflection becomes -1.5 quanta instead of -2, so the lower bound of the Freya:Frey period ratio becomes 3.16*(1.5/2)^1.5 = 2.05 ~ 2.

If Freya's eccentricity is 0.65 (i.e., the same as Frey's) then the fast half of its orbit, giving displacement of Freya, equal to the width of its ellipse, i.e. 1.50*a, takes 6.46 * 29.3% = 1.89 yr. The difference in lateral displacement of Barbarossa's orbital path, between B(1986) and C(1987) requires only 0.5/1.5*2 = 2/3 of the maximum possible single displacement of Freya, 1.65*a; this 1.10*a can be achieved in 0.88 yr, if Freya is near periapsis.

On the other hand, 2.03 yr might be enough for Freya's displacement by 1.50*a, but the displacement can't be more than about 2*a (because Freya's orbit doesn't precess greatly in 2 yr); the difference in lateral displacement between E(1995) and D(1997) would seem to require 1.65*a*2 = 3.3*a ("a" denotes the semimajor axis). Furthermore, E4D occurs 9.5 yr ( = 1.5 Freya orbits) later than BC9, so would be near Freya's apoapsis and slowest motion; and, Freya's direction would be wrong (D3D is beneath E3E4; CC9 is also beneath B3B).

A possible amendment is to let most of the displacement be caused by a third, outer, moon, "Lowell", near Freya's orbital plane. Freya could help the displacement E4D by an amount equivalent to 0.9*a, and perhaps make no contribution to the displacement BC9. Then "Lowell" could be in a 3:2 resonant (T=9.5 yr) orbit with the same eccentricity as Freya, near periapsis in both 1986.5 and 1996, with as little as half Freya's mass.


Reply author: Joe Keller
Replied on: 03/24/2008 19:21:09
Message:

Hsuan & Mardling, Astrophys Space Sci 304:243246, 2006, report their computation, that a 0.00096 solar-mass planet (Jupiter mass), in a near-perpendicular orbit with semimajor axis 0.80 AU, would cause 0.01 deg/yr retrograde apsis precession of the DI Herculis binary star system (semimajor axis 0.50 AU, e = 0.2), yet negligible change in the binary's orbital plane. This binary has 9.67 solar mass total and a mass ratio nearly 1:1.

Barbarossa + Frey have 1/1000 this mass; their mass ratio is 7:1; their eccentricity seems to be 0.65, rather than 0.2; but their semimajor axis was estimated roughly 0.14675deg * pi/180 * 196.8 AU = 0.50 AU, the same as DI Herculis. For 2:1 orbital resonance, Freya's semimajor axis would be 0.50 * ( 2 * sqrt(8/7))^(2/3) = 0.83 AU, close enough to Hsuan & Mardling's hypothetical 0.80. Then I refined my estimate: all distances were divided by 1.232 (which would speed the precession by 1.232^1.5). Freya likely has 1/8 the combined mass of Barbarossa & Frey. If the rate of Hsuan & Mardling's computed apsis precession, is proportional to the mass of the third body, then Freya should cause 0.01 * 9.67/0.00096 / 8 * 1.232^1.5 = 17.2 deg/yr (vs. 16.4 observed).


Reply author: Stoat
Replied on: 03/25/2008 04:15:16
Message:

Deleted; due to idiocy on my part [:D][B)][:I]


Reply author: Stoat
Replied on: 03/25/2008 05:36:29
Message:

Oh! What was I thinking [:I] If it's on the second one, it won't be on the first one at all.


Reply author: Stoat
Replied on: 04/02/2008 04:54:11
Message:

The first image is the newest one. The planet cannot be on both plates as they are only about 18 seconds across. The only "star" that I've found; just a quick scan, there might be more; that's on the first image but not the second, is the only green dot. Bottom left quadrant.
[img]http://farm3.static.flickr.com/2216/2382537972_d75a130437_o.jpg[/img]

[img]http://farm4.static.flickr.com/3267/2382537970_c6cb9c402e_o.jpg[/img]


Reply author: nemesis
Replied on: 04/03/2008 12:29:07
Message:

Stoat, is the green dot the one just left of the two bright stars of about equal brightness in the lower left corner of the picture? Is this possibly Barbarossa then? How could such a massive object be so dim?


Reply author: Joe Keller
Replied on: 04/03/2008 15:31:49
Message:

Arnold's (1988) object: evidence for an unknown optical surface surrounding Earth

HJP Arnold (Hampshire, England), Astrophotography, 2003 (foreword by Prof. David Malin, Anglo-Australian Observatory) Fig. 12.7, p. 163, shows an unidentified object (near Omega Tauri; more exactly, between 51 & 53 Tauri) which is at only, as well as I could measure, ecliptic lat. +0 deg 1.5' (Arnold's text doesn't remark this). Jupiter at the same time was nearby, but at about ecliptic lat. -1.0 deg. Such bodies, lacking any special relationship to Earth, would at best be distributed around Jupiter's orbit. With Jupiter in this position, the chance of being so close to the ecliptic would be less than (1.5'*2)/(60'*2)=2.5% (much less, for any mathematically or empirically likely distribution of inclinations)(Saturn's orbit, and the principal plane of the solar system, are even farther from the ecliptic here.) So, this object is so close to the ecliptic, that likely it is proprietary to the sun-Earth system. It is about the color of Aldebaran. Its magnitude is about midway between Aldebaran's (+0.9) and that of Epsilon Tauri (~ +4.0), thus about +2.5.

Arnold found this object on four successive photos reqiring at least four minutes of time total. The date was Nov. 2, 1988 (presumably evening, not morning, because of moonrise).

A mirror-like phenomenon arising on a perfect sphere of influence around Earth, would make images of the sun, on the ecliptic opposite the sun. A distorted sphere retaining symmetry about the ecliptic, could produce an image almost 30 deg displaced from opposition (as this is) yet still on the ecliptic. Presumably such an object wouldn't be much farther from the sun than Earth is, so a 100% albedo patch of this apparent magnitude would have small enough angular size to seem starlike in Arnold's photo. Also, the image would shift relative to the stars, only 1 deg/day = 1' / 24 minutes = 10" / four minutes. If Arnold delayed a few minutes between these frames, the movement might be barely discernible.


Reply author: Joe Keller
Replied on: 04/03/2008 19:39:33
Message:

The scarcity of stars and galaxies near the (+) CMB dipole (i.e., near my computed position for Barbarossa) is further evidence that a mini-solar system is there, with concomitant dust, extinction and reddening. (This also helps explain the unexpected faintness of the objects of the Barbarossa system itself.)

The Millenium Star Atlas says that its smallest number of cataloged stars (these go down to 11th mag) per square degree, average for any constellation, is 10; this is for Coma Berenices, which is, basically, the neighborhood of the N Galactic pole. Barbarossa's recent track (since 1954)(average galactic colatitude almost 45 deg from the N galactic pole) contains only slightly more stars than Coma Berenices, per square degree, averaging fewer than expected (expected would be 40% more than Coma). Few pages of the Millenium Catalog are as sparse in stars as the bottom half of p. 801, vol. 2 (the neighborhood of Barbarossa's track). So, Barbarossa's track is sparser in bright (mag < +11.5) stars than expected for its galactic latitude.

Despite its favorable position, far from the galactic equator, Barbarossa's track doesn't exceed the average, for cataloged galaxies. The aforesaid half-page contains four small galaxies, all near the margin of the page in cut-off degree squares that also are found on neighboring pages; so maybe they should be counted half, i.e., as two. The Millenium Catalog contains over 8000 galaxies and has 1548 chart pages, for an average of over five galaxies per page, or 2.5 per half-page.

With such small numbers, it would be better to compare parallax distances to spectroscopic distances, that is, to estimate absolute magnitude spectroscopically, and distance, from Hipparcos, then check for extinction that is unusual in amount or kind. A named 6th mag star, SV Crateris (HD 98088), a triple star with a 5.9-day eclipsing binary component, has been within 35' (2.0 AU) of Barbarossa's heliocentric track during recent decades. Study of this star might reveal the amount of extinction and reddening in that direction, and whether those are changing due to Barbarossa's motion.


Reply author: Stoat
Replied on: 04/04/2008 04:00:06
Message:

Hi Nemesis, if you download those images and open them as separate layers in photoshop, you can reduce the transparency of the top one and then slide the layer upwards until the stars line up. You'll find that they don't exactly. You'll have something that looks like a double exposure. Now this has to be down to the lens effect of the Earths atmosphere, these images are taken a year apart.

There should be a central point, so add a layer and then draw some lines between close pairs of stars and extend them. That should give you that central point. The only trouble is, it's the devil to find. I can only conclude that some of these stars have high proper motions. Could it be that there's another lens effect here, due to a "dimple" caused by a mini solar system way out there in the boonies?

You asked why is that green dot so dim? First, we have to say that it could simply be an artefact of the electronic camera on the telescope. it could be a variable star that was very dim on the first image but brighter on the second. So look at the channels. Nothing at all on the red, a bright dot on the green and an almost none existent tonal change on the blue. I would say that it's an object, as it does show on two colour channels.

Remember that the object of the game is to get an image of a planet, that's more massive than jupiter but of a smaller radius. We don't know its intrinsic brightness, and whether or not it's surrounded by a dust halo. Plus it's at about 180 AU. It's going to be at about the best resolution of the Bradford telescope, or even dimmer still.

To Joe, the top clearer image is the most recently taken, with the fits files I sent you, it's called nem4b. Remember that a robotic telescope can have problems with gimbal lock. One of the images put up here had to be flipped vertically. To line them up on a skyview plate, you may have to flip them both vertically back and also flip them horizontally as well.

If anyone is a dab hand at bringing out the best of fits files, pm me and I'll send them to you


Reply author: Joe Keller
Replied on: 04/04/2008 16:36:44
Message:

In Barbarossa's Cavern There Are No Stars

Gross dimming of the stars occurs between 1954 and 1987 (basically, the R1 & R2 mags) according to the USNO-B catalog. I considered three regions with 2 deg radii: centered at the 1954 & 1987 Barbarossa geocentric coordinates, and another centered about 5 deg N of Barbarossa's track. The effect is unmistakable for magnitudes +15.00-17.99. More to follow (school's out and I have to get off the public library computer).


Reply author: Joe Keller
Replied on: 04/04/2008 19:22:07
Message:

In Barbarossa's Cavern There Are No Stars (Part II)

Now I've automatically searched six regions of the USNO-B1 catalog. All have radius 2.00 degrees.

Region 1. Centered on the geocentric position of Barbarossa in 1954.
Region 2. ", 1987.
Region 3. Centered on RA11:10:00, Decl -2:30:00 (approx. 4.5deg N of the midpt. of Regions 1&2).
Region 4. ", RA11:10:00, Decl -12:30:00 (approx. 5.5deg S of the midpt. of Regions 1&2).
Region 5. ", RA11:34:00, Decl -12:00:00 (approx. 5deg prograde along Barbarossa's orbit and 2 deg to the S).
Region 6. ", RA10:46:00, Decl -3:00:00 (approx. 5deg retrograde along Barbarossa's orbit and 1 deg to the N).

Let n = 15, 16, or 17. Using the online VizieR service, automatically search the USNO-B catalog, for stars whose R1 lies within the interval [n,n+0.49] but R2 within the interval [n+0.50,n+0.99], for some n. Repeat with R1 & R2 switched. (A completely different but comparably big sample would be got by investigating instead the intervals [n-0.50,n-0.01] and [n,n+0.49].) This finds the numbers of stars whose R1 & R2 differ by very roughly +/- 0.5 mag.

According to the online documentation I've found for the USNO-B catalog, and the DSS "plate finder":

R1 was found from either the POSS-I survey (1949-1965; three relevant plates: 1954x2, 1956);
or, the ESO-R survey (Decl < -5 only; no plates intersect region).

R2 was found from either the POSS-II survey (irrelevant; Decl > 0 only);
or, the SERC-ER survey (1978-2002; five relevant plates: 1985, 1986, 1987, 1991, 1992).

So, the R1 magnitudes are from 1954 or 1956, and the R2 magnitudes from 1985, 1986, 1987, 1991 or 1992. For each of the six regions, I found the number of stars in the above magnitude ranges, for which R1 was dimmer and for which R2 was dimmer. For mag +15.0 to +18.0:

Region 1: R1 dimmer 445, R2 dimmer 3229, ratio (R2 dimmer / R1 dimmer) = 7.26.
Region 2: R1 dimmer 503, R2 dimmer 2299, ratio (R2 dimmer / R1 dimmer) = 4.57.
Region 3: R1 dimmer 916, R2 dimmer 864, ratio (R2 dimmer / R1 dimmer) = 0.94.
Region 4: R1 dimmer 1119, R2 dimmer 1759, ratio (R2 dimmer / R1 dimmer) = 1.57.
Region 5: R1 dimmer 1137, R2 dimmer 1488, ratio (R2 dimmer / R1 dimmer) = 1.31.
Region 6: R1 dimmer 1300, R2 dimmer 804, ratio (R2 dimmer / R1 dimmer) = 0.62.

As Barbarossa moved from its 1954 to roughly its 1987 position, the background stars near the first end of that track dimmed most consistently, and near the second end of the track somewhat less consistently. Much less consistently, but still with high statistical significance, stars 5deg retrograde of the track brightened, stars 5deg S the track dimmed, and stars 5deg prograde of the track dimmed. Stars 5deg N of the track brightened insignificantly.

A tabulation of Regions 1,2, and 3, for n = 13 or 14, confirms the above. For mag +13.0 to +15.0:

Region 1: R1 dimmer 45, R2 dimmer 845, ratio (R2 dimmer / R1 dimmer) = 18.78.
Region 2: R1 dimmer 100, R2 dimmer 611, ratio (R2 dimmer / R1 dimmer) = 6.11.
Region 3: R1 dimmer 244, R2 dimmer 128, ratio (R2 dimmer / R1 dimmer) = 0.52.

Either the stars really dimmed in Regions 1 & 2, or they didn't. If they didn't, it must be that despite calibrations made, a large error overestimating R2, or underestimating R1, at least for magnitudes +13 to +18, was made for Regions 1 & 2 (Barbarossa's hypothetical track) but not for the four surrounding Regions. Barbarossa's track does intersect the boundaries of all eight relevant plates, so, often, different plates must have been used for different regions.


Reply author: Joe Keller
Replied on: 04/05/2008 22:07:50
Message:

In Barbarossa's Cavern There Are No Stars (Part III)

Transient dimming of background stars near the location of Barbarossa, also occurs for the Blue magnitudes. These magnitudes derive from completely different plates than the Red magnitudes. The POSS-I Blue plates have the same dates as the corresponding Red plates; for the somewhat expanded area considered below, the average date of the relevant POSS-I Blue plates is within a year of that of the POSS-I Red plates considered above; none of them differ from that date by more than 2 yr. The relevant SERC-J Blue plates all are from about 5 yr earlier than the average date of the SERC-ER Red plates.

I've added more (circular, 2 degree radius) Regions, 7 through 9. Listed from retrograde to prograde, the Regions now are:

Region 9. RA 10:30 Decl -2:00
Region 6. RA 10:46 Decl -3:00
Region 1. RA 11:02:25.26, Decl -5:56:11.3 (Barbarossa 1954 pos.)
Region 2. RA 11:18:03.18, Decl -7:58:46.1 (Barbarossa 1987 pos.)
Region 5. RA 11:34 Decl -12:00
Region 7. RA 11:50 Decl -12:00
Region 8. RA 12:06 Decl -14:00

Also there are the regions displaced laterally,

Region 3. RA 11:10 Decl -2:30
Region 4. RA 11:10 Decl -12:30

Blue magnitudes often are missing from the USNO-B catalog. I searched for stars whose B1 mag was 12.00-13.99 and B2 mag 14.00-15.99 (or vice versa). There averaged fewer than a dozen of these in each Region, so I also searched for stars whose B1 mag was 16.00-17.99 and B2 mag 18.00-19.99 (or vice versa). The totals were:

(prograde on Barbarossa's track)
Region 9: B1 dimmer 73, B2 dimmer 1, B1 dimmer::B2 dimmer= 73.0
Region 6: B1 dimmer 77, B2 dimmer 5, B1 dimmer::B2 dimmer= 15.4
Region 1: B1 dimmer 44, B2 dimmer 8, B1 dimmer::B2 dimmer= 5.50
Region 2: B1 dimmer 38, B2 dimmer 14, B1 dimmer::B2 dimmer= 2.72
Region 5: B1 dimmer 120, B2 dimmer 45, B1 dimmer::B2 dimmer= 2.67
Region 7: B1 dimmer 103, B2 dimmer 18, B1 dimmer::B2 dimmer=5.72
Region 8: B1 dimmer 118, B2 dimmer 3, B1 dimmer::B2 dimmer= 39.3

(lateral regions)
Region 3: B1 dimmer 66, B2 dimmer 3, B1 dimmer::B2 dimmer= 22.0
Region 4: B1 dimmer 44, B2 dimmer 28, B1 dimmer::B2 dimmer= 1.57
(The position of Region 3, projected onto Barbarossa's track, is approx. Region 1; the projected position of Region 4, is approx. Region 2.)

So, the Red and the Blue plates together, prove that transient, generalized dimming of background stars recently has occurred in the vicinity of the positive "CMB" dipole. This is the position of the "disappearing dots", which because of their orbital regularity, I believe to be Barbarossa and its satellite(s).


Reply author: Stoat
Replied on: 04/06/2008 06:58:50
Message:

Heres the dss red image of that bit of sky. I've put two yellow lines on it as markers for where you have to drag the Bradford telescope image onto it in photoshop (the tracking is a bit off with the Bradford.) I'm not sure yet hen the dss image was taken but there's quite a bit of proper star motions going on in here.

[img]http://farm3.static.flickr.com/2419/2391473731_6052c1370e_o.jpg[/img]


Reply author: Stoat
Replied on: 04/06/2008 08:05:43
Message:

The dss image was taken in 1987. Teh image size is 17.1' wide by 17.6' deep


Reply author: Joe Keller
Replied on: 04/06/2008 20:58:56
Message:

The Mar. 19, 2008 Bradford photo is upside down (South at the top, West at the right). I see two additional disappearing dots on it, but the green dot you noted seems approximately consistent with Frey's position. It's brighter than expected (the magnitude limit seems too low to detect the objects seen on the sky surveys) but still it could be Frey.


Reply author: Stoat
Replied on: 04/07/2008 04:06:44
Message:

Hi Joe, the digital sky survey image is also upside down then. I blame the Australian astronomers that took it.

As I've said, hold down the mouse key on those images and download them. In photoshop, or paintshop pro open the sky survey image, then open the bradford nem4b image (the better one) Then do a "select all." This can be found in the top menu, or "apple a" "command a", on p.c. and a dotted line will appear round the image. Then do a "copy" command, which is again on the menu, or "apple c", "command c" on p.c.

Then select the sky survey image, by just clicking on its border, to bring it to the front. Then do a "paste" command, which is again from the menu, or apple v", "command v". This pastes a new layer over and above the sky survey.

In the layers palette on the right, you can adjust the transparency of that layer. Knock it down about 30% Make sure that the "move tool" from the tools palette, on the left; that's the one with arrows, top right button of the tools palette; is selected.

Drag the bradford image up to line up with those yellow lines. Then blink the images. There's a little eye symbol on the layers palette, click it on and off.

You'll see that there's a good line up on some stars but others have large proper motions. Zoom in on some of the fainter objects that show on both plates. Then use the arrow keys on your keypad to "nudge" the top plate into line up with them. A couple of galaxies look good for that, no proper motion.

Then, once you're happy that the two plates are lined up nicely, slide the transparency slider back to 100% Do another set of blinks, to see the things which look as though they are moving back and forth, or are not on both plates. There are a number of things to look at then.

Maybe you should ask some of the people doing the stuff on Mars images here, to help out with this. They know their photoshop inside out


Reply author: Stoat
Replied on: 04/07/2008 05:07:44
Message:

Oops [:I] Flip the bradford image horizontally. I must have forget that I'd done that.


Reply author: Stoat
Replied on: 04/07/2008 07:52:39
Message:

If I've even got the images lined up correctly, and that's something of an if, as bright stars show up as rather red and dim on the bradford image. Plus there's proper motions. And one bright star seems to have vanished from the face of the earth (excuse the lousey pun)

Anyway, Go almost straight up, about 5'(a third of the bradford image) from that green dot. Then across to the left, past the whitish star to an orange red star. That might be a possible.

If one wants to just look at it on the board, remember that it needs to be flipped, so look across to the right.


Reply author: Joe Keller
Replied on: 04/07/2008 15:32:26
Message:

Initial Analysis of Robert Turner's March 19, 2008 Photo of Frey


This is the only photo taken so far in 2008, of the Barbarossa/Frey system. It is the only color photo (not counting Red filtered sky surveys) ever taken of the Barbarossa/Frey system. It shows Frey as a bright green dot in the NE quadrant of the frame. (As this photo is shown here on Dr. Van Flandern's messageboard, S is up and W is to the right, so, the frame is upside down, but not reversed; the usual presentation is, N up and W right.) It was taken with a 14-inch reflector on Tenerife, remote-controlled under the direction of amateur astronomer Robert Turner of England, observing at my predicted coordinates for the Barbarossa/Frey center of mass.

I haven't rechecked my calculation yet, but to the best of my ability to calculate Frey's position based on Joan Genebriera's 2007 Barbarossa photo, Barbarossa's 1954 and 1986 positions, and my theory of Frey's precessing orbit, the position of the bright green dot noted by Turner in his photo, differs only 16" from expectation. This could be caused by a plausible, 3% error in my estimate of Frey's orbit. The theoretical position of Barbarossa, is off the photo, slightly E of the margin. I knew of four Barbarossa/Frey candidates on the photo before calculating the position; the random chance that one or more of these four would have been within 16" of Frey's theoretical position (Barbarossa's theoretical position is off the frame) is 4 * (pi*16"^2) / (17.1' * 17.6'), i.e., p = 0.003.

After Turner found the first disappearing dot, I found, basically, three more in the photo. These are not near the theoretical positions of Barbarossa, Frey or Freya. Near the S edge, is a medium-bright disappearing green dot (i.e., not found on the 1987 SERC-ER survey); this is W of a disappearing double red dot. In the SW quadrant is a faint disappearing red dot; its faintness, its color, and its starlike broad smooth intensity contour, make it especially likely to be a real astronomical object.

As I noted on Steve Riley's photos last year, these electronic photos, near their detection limit, give highly variable images. Stars confirmed on sky surveys, of brightness near the electronic photo's detection limit, can have a classic appearance, or a small bright or irregular appearance, or be totally absent; all on the same photo. An unusual photon distribution should be especially likely to have an unusual appearance, especially near the detection limit.

Barbarossa and Frey theoretically have about 1" diameters. A star might be approximated as a point source hopping randomly on a 1" disk, staying in each place for ~ 1/30 sec. The light from Barbarossa or Frey, composed of many point sources more-or-less evenly distributed on a 1" disk, presents the electronic camera with a drastically different object, for which light intensities over the blur circle are much more constant. This involves the subject of "reciprocity" of photographic film. Characteristics of electronic detectors, loosely analogous to reciprocity, seem little investigated. I'm wary of dismissive, unsupported claims about how an image of Barbarossa (especially, near the detection limit) should appear.

Another factor, is the now known (see my posts above) transient generalized extinction (dimming) of stars within a few degrees of the Barbarossa/Frey system. Such an interposed cloud would further alter these light sources, further vitiating any assumptions about the appearance of Barbarossa/Frey images.


Reply author: Joe Keller
Replied on: 04/08/2008 15:49:37
Message:

The paired blink photos (posted by "marsrocks" on another thread of this messageboard) do show the abovementioned bright green disappearing dot (not in color)(in the lower left, i.e., NE corner, above & left of a small equilateral triangle of stars) disappearing as expected. The background lightness of the other photo almost obscures a nearby star of similar brightness. The blink pair might give more false positives, than comparing Turner's March 19, 2008 photo to a sky survey, without blink. Still, the blink pair gives us a dramatic demonstration.


Reply author: Joe Keller
Replied on: 04/09/2008 16:58:42
Message:

Requiem for Relativity

Such a large mass as Barbarossa, so close to the positive "Cosmic" Microwave Background dipole, would refute the theory that the dipole is a Doppler effect. It also would refute the theory that the "Cosmic" Microwave Background is cosmic.

Likely, the CMB would be found to originate from the sun's gravitational field, and the dipole from Barbarossa's gravitational field. This would demolish present relativistic ideas about gravity and light.

That is why Barbarossa is in the thread entitled "Requiem for Relativity".


Reply author: Joe Keller
Replied on: 04/09/2008 19:06:04
Message:

In Barbarossa's Cavern There Are No Stars (Part IV)

The Other Hubble Relation


Hubble, ApJ 56:162+, 400+, 1922, discovered that with a 60 or 100 inch telescope, the apparent photographic size of stellar "reflection nebulae" is what would be expected assuming total scattering of the central star's light at the outermost detectable edge of the nebula, where the nebula is just barely detectably brighter than background (p. 410). Hubble's empirical relation (p. 411, eq. (8)) is:

apparent magnitude + 4.90 * log10 (radius in arcmin) = 11.0.

If all light is scattered at the effective surface of the nebula, then the star itself should be invisible. Yet Hubble was able to use the apparent magnitude of the star, to estimate how much light was available for scattering at the surface of the nebula. There's twice too much light in his model. Hubble (p. 401) even cites Hertzsprung's finding (Astronomische Nachrichten 195:449, 1913) that the effective nebular surface scatters only 1-2% of the light.

In Hubble's model, wouldn't the cloud of nebular material often be bigger than given by Hubble's relation, and continue to scatter 100% of the light? If so, then the contrast at the edge of the nebula, would be too low to detect. Suppose this were the case for many stars (after all, most stars lack detectable nebulae). Then, likely also for many stars, the cloud would be smaller than Hubble's relation, giving a smaller reflection nebula. Yet, according to Hubble's Fig. 2, p. 411, this seldom occurs.

Maybe the radius of the cloud of nebular material coincidentally (but not accidentally) happens to be proportional to (stellar mass)^1.5. For stars from 1 to 30 times solar mass (MNRAS 382:1073+, 2007, Table 6, p. 1078) there is an accurate quadratic polynomial for log(luminosity) as a function of log(mass). Using Wikipedia's 18 solar mass figure for Type B stars (more accurate would be to average Hubble's spectral type sample) the luminosity-mass relation becomes a power law with exponent 2.91. That is, Hubble's relation for the typical Type B star, is equivalent to:

nebula radius = const. * (stellar mass)^(2.91/(4.90/2.500)) = (stellar mass)^1.48

with such proportionality constant as to give that radius at which the total luminance of the star, divided by the surface area of the nebula, just exceeds background intensity.

Hubble's nebulae almost all were around stars of Spectral Type B (though one of the 33 listed in his Table III, p. 181, was K8). Other, presumably older spectral types, seem generally to lack such nebulae. Maybe a failed cold brown dwarf also would have such a cloud, never completely expelled by radiation pressure.

Assuming luminosity were proportional to mass^2.91 (typical in the Type B range), the Barbarossa system's absolute magnitude, if it could behave as a Type B star, would be roughly 4.83 - 2.5 * log(20,000 * (0.0103/18)^2.91) = +17.67. At 10pc, Barbarossa's nebula would, according to Hubble's relation, subtend a radius of 0.0435'. At Barbarossa's actual distance, that's 454' = 26 AU. This predicted angular diameter of Barbarossa's cloud, 15 deg, agrees well with the observations of transient dimming in sky surveys, discussed in Parts (II) and (III) above.


Reply author: Joe Keller
Replied on: 04/10/2008 01:00:43
Message:

In Barbarossa's Cavern There Are No Stars (Part V)

According to Skuppin et al, A&A 177:228+, 1987, absorption lines in the spectra of fast-rotating Type B stars, are one of the best ways to quantitate local interstellar matter. They discuss three such stars, one of which is Theta Crateris. The data for Theta Crt and for one of the other stars, were obtained by the Intl. UV Explorer (IUV) satellite, Jan. 1985.

The density of interstellar material between us and Theta Crt, measured by spectral lines of four different elements, was consistently greater than for either of the other two stars; the factor generally was ~10. Also, only Theta Crt lacked the MgI line; this signifies that only Theta Crt lacked hot gas along the line of sight. So, Theta Crt has more intervening material, but in contrast to the other two stars, all the intervening material is cold.

Theta Crt is 66pc distant, so having 10x the interstellar material, gives extra Visual extinction equivalent to 600pc, i.e. roughly 0.6 mag using the usual approximate 1 mag / kpc value. If the distribution of UV extinction resembles that of Blue extinction in Part (III), then Theta Crt is where the extinction was, in 1985, increasing fastest, as the front of Barbarossa's cloud moved in. Where its derivative is greatest, the normal distribution is at 0.6 times its peak. So, the peak extinction of Barbarossa's cloud is roughly 1 mag in Visual.

According to Skillen et al, MNRAS 265:301+, 1993, the RR (type RRab) Lyrae star, W Crt (RA 11:26:30, Decl -17:55)(near Gamma Crt) underwent dimming of a portion of its V and B light curves, between 1986 & 1988. The authors say they have "no explanation for the discrepancy" (p. 303). Their variable star light curves were, as usual, calibrated against nearby reference stars. If the light of W Crt becomes polarized during this phase of its cycle, and the Barbarossa cloud preferentially extinguishes light of that polarization, the dimming might be explained. Strohmeier's text, "Variable Stars", says some stars of another variable class, the semiregular variables (Type M) do show weak polarization of their visible light.


Reply author: Joe Keller
Replied on: 04/10/2008 19:26:52
Message:

In Barbarossa's Cavern There Are No Stars (Part VI)

Fruscione et al, ApJSupp 94:127+, 1994, Table 1, list 594 stars toward which the ("integrated") column density of neutral hydrogen, log( N (H I) ) has been determined, either by direct measurement, or by estimation from the logN determined for other atom(s). Only two of these 594 really rival Theta Crater, as evidence of nearby high extinction.

Almost 1/3, i.e. almost 200, of Fruscione's stars are <= 100pc distant (Theta Crt is 66pc). Of these, only four exceed Theta Crt's value log(N(HI)) = 20.50 +/- 0.10 (a few stars had ambiguous values such as " < 21.00"). One of these four exceeds Theta Crt slightly, but because its distance is greater, the average HI density along the path to it is less, though the sum, i.e. ("integrated") column density, is more.

The values of all four of these rivals, are less reliable than Theta Crt's. Theta Crt's value was estimated from more than one (four) other kinds of atoms. One of the others, 34 Cygni, was measured directly from LymanAlpha, which normally would be the gold standard, except that only "earlier than B2-B3" spectral types are suitable for this method (Fruscione, p. 128) and Fruscione lists 34 Cyg as B2. The other three rivals all were estimated from NaI alone; this can overestimate HI by 0.5 log units (p. 129).

Thus the only two stars (among the almost 200 at distance <= 100pc) toward which average HI density seems really higher than toward Theta Crt, are 34 Cyg and Lambda Per. 34 Cyg lies before "a heavily obscured region of the galaxy" (Burnham); it's less than a degree from the open cluster M29 (5 to 8' diam.). Lambda Per, according to Norton's Atlas, lies between an open cluster and a diffuse nebula, a degree to either side of it. Theta Crt, by contrast, lies in an almost empty field (only a small galaxy, almost 3 deg away).

Another of Barbarossa's champions is 69 Leonis (RA 168, Decl 0, distance 80pc). Welsh et al, ApJ 437:638+, 1994, Tables 2 & 4, reporting a study of interstellar NaI density, list, inter alia, 28 other stars at distances 70 to 90pc inclusive. Six of the 28 exceed 69 Leonis' value of log(N(NaI)) (none were ambiguous). Three of these six were toward the galactic center; these three also were by an investigator who performed only 19 of the 167 total determinations used. All eleven of Welsh et. al.'s own determinations, in that distance range, were at least 1.15 log units less, than for 69 Leo.

Welsh cites Albert, ApJ 272:509+, 1983, as his source for 69 Leo. Albert tested only three stars at distance <= 100pc. For TiII and NaI absorption, differences between the three, corrected for path length, did not exceed error bars. However 69 Leonis, at 80pc, showed log(N(CaII)) = 3.0 +/- 0.5, vs. (1.1 or 1.0) +/- (0.6 or 0.4) for the other two stars, at 100pc or 90pc, resp. Correction for distance, gives 4.5x the average density, along the path to 69 Leonis.


Reply author: Joe Keller
Replied on: 04/11/2008 14:43:11
Message:

In Barbarossa's Cavern There Are No Stars (Part VII)

Above, I very roughly estimated a Visual extinction of 1.0 mag for passage through the full thickness of Barbarossa's cloud; a more accurate version of this estimate (see below) is 0.3 mag. This would of course be due to the dust component. What about the gas?

The column of hydrogen atoms amounts to a layer of dissociated neutral hydrogen at standard temperature and pressure, 12cm thick; add another 1cm for the helium. This is 1/60,000 the number of gas molecules in a column of Earth's atmosphere, so, Visual extinction, due to the gas, would be negligible. These molecules are illuminated only 1/197.7^2 = 1/40,000 as intensely by the sun. The regional sky brightening seen from Earth's surface, due to sunlight scattered by the gas atoms of Barbarossa's cloud, would be somewhat less than the sky brightening, due to starlight scattered by Earth's atmosphere.

Let Barbarossa's cloud be a homogeneous oblate spheroid with a=26AU, b/a=0.5. Let the path to Theta Crt transect a chord 0.6 * 2 * a. Then the mass of the cloud, assuming 75% H by weight and that 90% of Fruscione's N(H) is within the cloud, would be 0.028 Earth masses (1/120,000 the mass of the Barbarossa system). Because estimates from metal atom abundances can overestimate H abundance by 0.5 log unit (Fruscione, 1994, p. 129), a lower bound for the mass would be 0.01 Earth masses. (In interstellar matter, dust averages only 0.7% the mass of gas.) There would be 740,000 gas atoms (helium or dissociated hydrogen) per cubic cm; this is 100x the density of the Martian exosphere as measured by Mars Express (Wurz et al, Geophysical Research Abstracts, Vol. 8, 01954, 2006). The equivalent altitude on Earth is below that at which aurorae begin to manifest, so, if Barbarossa has a magnetic field, the sun should cause aurorae in Barbarossa's cloud.

What about sky brightening due to sunlight scattered by the dust of Barbarossa's cloud? If the densest part of Barbarossa's cloud intercepts (i.e., extinguishes, either by absorption or scattering) ~30% of starlight (0.3 mag extinction)(Bohlin's 1978 empirical formula relating N(H) to extinction, cited by Draine, AnnuRevAstronAstrophys 2003, 41:241+, p. 244, sec. 2.1.2, eq. (2), gives 0.25 mag), then it also intercepts 30% of sunlight.

I follow Endrik Kruegel's text, "The Physics of Interstellar Dust" (ch. 4, "Case Studies of Mie Calculus"). For a spherical particle which is a "strong absorber" (sec. 4.1.3)(optically not very different from amorphous carbon, which is discussed in detail in sec. 4.1.5), define x = circumference/wavelength. For the "typical" interstellar dust grain, diam = 0.1mu (Kruegel, sec. 4.1.5.1), wavelength = 0.6mu for visible light, so x = 1. For x < 0.5, the ratio of scattering to extinction is proportional to x^3, and for x=0.5, is (Fig. 4.3) about 1/3 * 0.5^3 = 1/24. For x > 2, the ratio of scattering to extinction is about 1/2 overall, but "g", the average cosine of the scattering angle, is about 0.8, which implies that backscattering can't possibly be more than 1/5 of the amount found when there is isotropic scattering; so, the ratio of backscattering to extinction is < 1/10. For x = 1, the situation is intermediate. So, a fair estimate of backscattering to extinction, is 1/24.

More precisely, I could for x < 0.5, neglect scattering, and for extinction (which in this domain is proportional to x times the geometric cross section), integrate the often assumed a^(-3.5) power law for particle size distribution (JE Dyson, "The Physics of the Interstellar Medium", p. 53), assuming (as Draine suggests) the law holds at least halfway down to molecular sizes; and assuming it achieves the geometric cross section, when x = 1. The ratio of scattering to extinction is maximum at the "typical" (maybe this is why it's considered typical) diameter 0.1mu; again I could use the power law to integrate scattering between x = 0.5 & x = 2 (using a constant scattering efficiency ~1/6 seen near x = 1, from Kruegel's Fig. 4.3); Simpson's rule (based again on Fig. 4.3) to get from the scattering, the extinction due to this domain; using "g" = ave(cos(theta)) = 0.2 at x = 1, and a first-order spherical harmonic, I can estimate the relative backscattering efficiency as 40% in this domain. This more precise calculation merely changed the estimate to 1/25 from 1/24.

The effective albedo of Barbarossa's cloud then would be at least 70% (a lower bound for the fraction of light which escapes a second interception) * 30% * 1/24 / 2 (because of isotropy) = 0.44%, ~1/30 that of Luna. The cloud's surface would be 1/(30*40,000) as bright as Luna's; it would be Vmag -12.7+15.2 = +2.5 on a 0.52deg diam disk, i.e. +0.8 on a sq. deg. The brightest part of the northern Milky Way (Zavarzin, Astrophysics 23:647+, Table 1) has brightness +4.05 on a sq. deg.

So, if the ~0.3 mag extinction (much less than that, hardly would be consistent with the observed relative dimness of R2 & B2 in the region) is due to "typical" dust grains, the Barbarossa cloud would be too bright. However, the extinction can be achieved with arbitrarily little scattering, if these "strong absorber" grains are small enough. Diameter 0.01mu, gives scattering/extinction = ~ 1/3 * (1/10)^3 = 1/3000, 3000/24 = 5.2mag less, i.e., sky brightness a plausible +6.0 on a sq. deg. Alternatively, the physical properties of the "dust" might be different than presently believed.


Reply author: Joe Keller
Replied on: 04/14/2008 15:15:40
Message:

cc

To: organizers of 2nd (Sept. 2008) Crisis in Cosmology Conference

Dear sirs:

The link didn't work for me, so here is my application.

Sincerely,
Joseph C. Keller

Presentation Type: Oral Presentation

Paper Title:
Brown Dwarf, Barbarossa, Causes "Cosmic" Microwave Background Dipole

Author & Affiliation: Joseph C. Keller (B. A., Harvard)

Primary contact: Joseph C. Keller

Contact email: *******

Short Abstract:

A cold brown dwarf, Barbarossa, causes the "Cosmic" Microwave Background dipole. Barbarossa+Frey+Freya, with a dark nebula, lie at the (+) CMB dipole.

Full Abstract (slightly < 2500 character count):

A cold brown dwarf, Barbarossa, causes the "Cosmic" Microwave Background dipole. Barbarossa+Frey+Freya, with a dark nebula, lie at the (+) CMB dipole.

Exclusively near there, red & blue USNO-B magnitudes dim between c.1954 and c.1985; likewise, Johnson's bright star photometry c.1964 vs. Harvard magnitude as published mostly in 1908. Barbarossa's nebular size, is given by extrapolating Hubble's nebular size relation. Interstellar absorption lines before the two studied nearby stars in this direction, are exceedingly strong.

My original finding was that automated USNO-B R1 & R2 magnitudes, differing enough to be misidentifications of wanderers, outlined an orbital path there. Dots of magnitude ~ +18, though not of typical starlike appearance, on all relevant red and infrared survey plates, and on prospective photos by Joan Genebriera, Steve Riley, and Robert Turner, lie within arcseconds of an e < 0.1, 198 AU orbit slightly leading the (+) CMB dipole. Frey has a 3-yr, e = 0.65 orbit around Barbarossa with retrograde apsis precession in 24 yr. Freya (not yet identified) is inferred to orbit in 6 yr. perpendicular to the ecliptic, causing Frey's precession, and lateral deviation of the Barbarossa-Frey c.o.m. The projection of Barbarossa's orbit onto Jupiter's, follows the mean position of a Jupiter/Saturn conjunction resonance.

Claimed COBE & WMAP error bars rule out such a near orbit. However, only a cause within the solar system, explains the correlation, of the Maxwellian moments of the CMB anisotropy, with the plane of the ecliptic.

My theory of CMB production by gravitational fields (electrons boiling from the surface of the sun's, 52.6 AU radius, ether island, detected by anomalies in Pioneer10's transmission there), and, a Newtonian theory of nodal regression resonances in the outer solar system, give equal estimates of Barbarossa's mass. Subtraction of Barbarossa's gravity makes the Pioneer Anomaly consistent with gravitation by a smoothly decreasing mass density.

I have ~100 references in major refereed journals and other authoritative sources. See my posts to the messageboard of Dr. Van Flandern's www.metaresearch.org website.


Reply author: Joe Keller
Replied on: 04/14/2008 16:34:36
Message:

Earth's atmospheric twinkling is said to be a phenomenon of timescale ~ 1/30 sec. A lightning flash on Barbarossa might make a pointlike image almost as though Earth's atmosphere did not exist. This would amount to "active optics". Previously on this messageboard I estimated the likely brightness of lightning on such a giant planet (or cold brown dwarf), and found it competitive with reflected sunlight.


Reply author: Joe Keller
Replied on: 04/16/2008 14:27:52
Message:

In Barbarossa's Cavern There Are No Stars (Part VIII)

Three Articles from the Annual Review of Astronomy & Astrophysics


1. Salpeter, "Formation & Destruction of Dust Grains", 1977. In Sec. 5.1, Salpeter cites observations by Zappala, of a circumstellar dust shell which is basically a solid sphere (inner diam. << outer diam.), though this is around a presumed mass-losing giant star. In Sec. 4.1, Salpeter says that the almost constant ratio of blue to red extinction, suggests that interstellar dust grain size distribution is about the same almost everywhere in the galaxy. On the contrary, the formulas and graphs of Kruegel's text say that for realistic "strong absorber" grains in the domain x < 0.5 (i.e., grain circumference < 0.5 * wavelength) extinction is proportional to grain cross-section times x; therefore the ratio of blue to red extinction = 0.55/0.44 = 1.25 is the same for any grain size distribution whatever, if only the extinction is mainly by small grains ( < 0.05 micron).

2. Aannestad & Purcell, "Interstellar Grains", 1973. In Sec. 3, p. 325, eqs. (6) & (7) (the "rho-g" factor of eq. (7) is a misprint), the authors say that (basically because of the "x" proportionality discussed above) extinction is proportional to the volume, hence mass, of dust whatever its grain size. Their lower bound of 0.5% dust::gas, agrees with the refined estimate of 0.8% in recent texts, and with the estimate 0.7% based on element abundances. They say, "This result applies equally well to grain mixtures. It can be applied with only slight modifications to 'grains' that are nothing but large molecules...".

3. McCray & Snow, "The Violent Interstellar Medium", 1979. In Sec. III.B, p. 228, the authors cite Shull's calculation that sputtering from high-velocity interstellar shocks, can reduce grain size to ~ 0.02 micron.

So, grains mainly moderately smaller than usually supposed, are quite possible. In (VII) above, I used formulas and graphs in Kruegel's text to show that such grains would reduce the apparent brightness of Barbaross'a nebula, to plausible levels.


Reply author: Joe Keller
Replied on: 04/17/2008 13:41:10
Message:

Why GPS does not disprove the ether theory, and why this is relevant to Barbarossa.


(To the best of my knowledge and belief, this ether drift calculation essentially first was made by James Clerk Maxwell. It also paraphrases an earlier post of mine to this messageboard.)

Let all velocities be measured in a frame of reference centered on Earth. Suppose Earth sits in an ether drift of speed v << 1 (choose units so c = 1). Let an orbiting satellite have speed u << 1 parallel to the drift, and speed w << 1 perpendicular to it. The satellite's clock is slowed by a fraction:

0.5*((u-v)^2 + w^2) = 0.5*(u^2+w^2+v^2) - u*v

The clock on Earth is slowed by a fraction:

0.5*v^2

The difference between the clock rates is a fraction:

0.5*(u^2 + w^2) - u*v

Suppose Jim is a physicist who believes in textbook relativity. Jim will correct for what Jim thinks the difference between the clock rates is:

0.5*(u^2 + w^2) (i.e., half the square of the velocity vector relative to Earth)

Now let the satellite move along any path whatever, so that it is now more downstream in the ether, by a distance d. Jim doesn't know about the u*v term. That is, the clock is faster than Jim thinks it is, by a fraction u*v. The satellite clock now is ahead of what Jim thinks it says, by

integral(u*v*deltat) = v*d (because u*deltat is the distance, d, that the satellite moved downstream)

Now for simplicity suppose that the satellite's path was, to move from Earth, to a distance d straight downstream from Earth in the ether. The time really required for light to move from the satellite, to Earth, is d/(1 - v) = d * (1 + v) to first order in v. Jim thinks the satellite clock is v*d behind what it really says, so Jim thinks the time interval was d * (1 + v) - v*d = d. Jim might think this disproves the ether theory. Really, all it proves is that, even for a one-way test, the ether theory differs from textbook relativity, only to second order in v.

What GPS can do, is essentially repeat the Dayton Miller round-trip experiment (Miller found an ether drift, though somewhat smaller than expected; anyone who looks at the graph in Michelson & Morley's original paper can see that Michelson & Morley got roughly the same result as Miller; yes, a significant, positive result, though somewhat smaller than expected. I dare anyone to look at the original Michelson & Morley paper and say he doesn't see that. Recently Yuri Galaev in Kharkov, using two different, novel interferometric schemes considerably different from Michelson's, quantitatively confirmed Miller's result, at least to within an order of magnitude.) The discrepancy a 10 km/s ether drift would cause in a round-trip geostationary GPS time is equivalent to only 30,000km * 2 * 0.5 * (10/300,000)^2 = 3cm.

If the ether exists, it's likely to be related to the sun's gravitational field. Then, it's also likely to be related to the gravitational field of a fundamental particle. The distance from the sun, at which the sun's field becomes weaker, than that produced inside the most compressed possible proton (Gaussian distribution of deBroglie waves, following Merzbacher or other quantum mechanics texts) is 52.6 AU. Pioneer10 showed transient abnormal signals at this distance (JD Anderson's best explanation was a gravitational encounter with Edgeworth-Kuiper belt objects there, but quantitatively the frequency shifts were too big, for too long, to be due to acceleration from any plausible such encounter).

It so happens that the gravitational escape energy of electrons at this distance from the sun, is roughly their mean thermal kinetic energy at the "Cosmic" Background temperature. Depending on the details of the mechanism, various factors of order unity might be involved. The "Cosmic" Background temperature, is basically the gravitational potential of an electron at the surface at which the sun's gravitational primacy ends.

Primordial pinheads aside, the only known object big enough and symmetrical enough to explain the "Cosmic" Microwave Background, is the sun with its fields. The first (dipole) and some higher Maxwellian moments of the CMB distribution, are significantly correlated with the plane of the ecliptic. Only solar system influence, by three or more bodies, can explain this. Planets distort that mathematical surface at which the macroscopic gravitational field becomes equal to some small critical value; thus the gravitational potential of an electron at this surface also varies, due both to the gravitational potential of the planet, and to the distortion of the mathematical surface, due to the gravitational field of the planet. A brown dwarf outside the surface is best for causing a dipole; planetesimals near the surface best for local variations.

Please don't ignore the data merely because they lack an explanation consistent with textbook relativity theory. Here's another synopsis I've been providing to inquiring scientists:

"The 'ether' is like an approximately spherical pot of water. The sun is in the center of this 'water'. The boundary of this 'water' is at 52.6 AU. There's some activity on this boundary, on this distant 'movie screen' all around the sun. We call this activity the 'Cosmic' Microwave Background. (Someday we might be able to detect a 'CMB'-radiating surface around Sirius. The radius of the 'movie screen' sphere goes as sqrt(M), the 'CMB' temperature as sqrt(M), the absolute magnitude of the 'CMB' thus as M^3, but the absolute magnitude of the star as M^(3 or 4), so the easiest case is a bright nearby star, of mass enough greater than the sun's, that its 'CMB' curve is distinguishable.) The position or 'locus' of the 'screen' is determined by some critical value of the macroscopic gravitational field strength vis-a-vis the internal gravitational fields of fundamental particles. This critical equation allows unknown processes of some kind to occur. The screen isn't quite a perfect sphere, nor is the 'activity' on it everywhere exactly equal. But it's close to perfect, because the sun has almost all the mass in the solar system, so the gravitational field is almost perfectly symmetrical. Maybe the sun provides the CMB energy. Maybe something else does. But the sun does define the locus of the activity, except for small asymmetries caused by smaller gravitating solar system bodies."


Reply author: Joe Keller
Replied on: 04/17/2008 20:03:45
Message:

In Barbarossa's Cavern There Are No Stars (Part IX)

The Johnson (c.1964) vs. Harvard (c.1908) magnitudes near Barbarossa's 1964 position


As detailed above, in regions near Barbarossa, USNO-B blue magnitudes were dimmer c.1983 than c.1954, and USNO-B red magnitudes dimmer c.1987 than c.1954 (these years can't be exact, because several overlapping plates were available to compute magnitudes). Interpolating between tested regions, the greatest blue dimming occurred about 3 deg ahead of Barbarossa and the greatest red dimming about 3 deg behind, along Barbarossa's track. During the time interval, Barbarossa moved 4 deg. So, the fastest increase in extinction, seems to be about 3 + 4/2 = 5 deg ahead of Barbarossa for blue, and about -3 + 4/2 = a degree behind Barbarossa, for red.

One explanation, would be that Barbarossa's cloud is nonhomogeneous: the most red-extinguishing material lags behind the most blue-extinguishing. The real explanation may be even stranger. Historical photometry shows that Barbarossa's cloud has no effect on the magnitude of stars of approx. Type G0. However, exclusively in the region of Barbarossa, bluer stars are dimmed, and redder stars are brightened, by amounts an order of magnitude greater than can be explained by error in zenith angle correction. The red dimming following Barbarossa, might be due to replacement of Barbarossa's cloud by some complementary medium which extinguishes red more than blue.

Methods. VizieR's online documentation of Harvard's Henry Draper catalog, gives its vintage as 1918-1924, but to secure data even older and of more certain vintage, I used the Harvard magnitudes listed, to 0.1 mag precision, for the stars on pp. A413-A415 (RA 10:26:15 to RA 11:40:43, B1900) of the USNO "Catalogue 4526 Stars" (Publications of the USNO, 2nd ser., vol. 9, pt. 1, published 1920 but dated 1917). The listed Harvard magnitudes mostly were from "Annals of the Astronomical Observatory of Harvard Coll." vol. 50, 1908, or vol. 54; many magnitudes were personal communications, from Prof. Pickering of Harvard, to the USNO. So, no magnitudes were later than 1917; many were earlier than 1908. An even older USNO publication (House Misc. Docs., 2nd ser., v. 2672, 1884) contained the USNO "Catalog of Stars 1845-1877"; I eschewed these because they took their magnitudes from various European catalogs, and often differed > 1.0 magnitude from later measurements. Only one difference between the Harvard magnitudes, and the Johnson (1966) online photometry catalog magnitudes, was truly a non-normal statistical outlier; this printed Harvard magnitude also differed from the online Draper catalog, so I used the Draper value, which was more plausible. I checked several other magnitudes whose differences from Johnson were greatest; only one more, which happened to be the magnitude with the second biggest difference from Johnson, differed from Draper (listed to 0.01 mag) by more than rounding error; likewise I adopted the Draper value here. I noticed several Boss (1928) "San Luis Catalogue" magnitudes, which Boss said were taken from Harvard, did agree exactly with the corresponding Harvard magnitudes printed.

Those stars on the three pages, which also were in HL Johnson's online "UBV Photometry of Bright Stars" (1966) catalog, numbered 61, and comprised my study. Johnson's observations (Comm. of the Lunar & Planetary Lab. vol. 4, pt. 3, #63) were dated approx. 1964.0 +/- ~ 1 yr and were taken at 32deg N lat in Arizona. He also used Cape, South Africa, observations by others; the text indicates these were taken in 1963 or 1964, and that no use was made of any observations from prior to 1955.

Spectral types were taken from the online Jaschek catalog when found (most); otherwise from the recent online Kharchenko catalog (many). All the Kharchenko spectral types were checked against Draper; all agreed within the 1/2 color rounding of Draper.

I found five of the 61, listed as variable in the "Bright Star Catalog 2000.0", vol. 2. Information given on these five, suggested amplitudes not big enough to require expulsion from my study.

Results. The magnitude changes, c.1908 to c.1964, within each color type, B through M, were fitted by exhaustive computer search, to a function of the star's Declination difference from Barbarossa's orbit. Barbarossa's orbit was approximated by a Mercator projection line through the (B1900) 1954 and 1986 center-of-mass positions. The function used, consisted of a constant term, plus a normal curve centered on Barbarossa's Declination for that star's RA. The three adjustable parameters giving the least-squares fit are:

magnitude change, Type B: -0.20 + 0.47 * exp(-deltaDecl^2/12^2) n=3
magnitude change, Type A: -0.10 + 0.10 * exp(-deltaDecl^2/11^2) n=12
magnitude change, Type F: -0.09 + 0.04 * exp(-deltaDecl^2/16^2) n=6
magnitude change, Type G: -0.12 - 0.04 * exp(-deltaDecl^2/23^2) n=14
magnitude change, Type K: -0.14 - 0.12 * exp(-deltaDecl^2/6^2) n=21
magnitude change, Type M: -0.11 - 0.19 * exp(-deltaDecl^2/14^2) n=4
(for Type M, one outlier was discarded)

Note that the denominators, 12^2, etc., imply that the standard deviation of the normal curve, in degrees of Decl, is 12/sqrt(2), etc. The weighted mean standard deviation is 9.1deg.

A Type M star near the N pole, with a moderately big magnitude change, tended to force a very large standard deviation (100deg) for best fit. So I decided the exclusion of this outlier would give a more representative result.

The linear variation of the coefficient of the bell curve term, proves the statistical significance of this result. The magnitude change, Johnson vs. Harvard, was different in the neighborhood of Barbarossa. Along this section of its track about 20 deg long, the "track effect" averages about 18 deg wide. By interpolation (using average number types for each color), there is no "track effect" for type G0; that is, no effect for starlight of the composition of sunlight. Blue undergoes extinction here relative to other parts of the sky, but red undergoes what amounts to "negative extinction": red stars brighten more here, than in other parts of the sky.

This is not due to Barbarossa's track acting as a proxy for the equator and revealing errors in zenith angle correction. Boss, in his 1928 catalog, gives 0.25 * (sec(z) - 1) for this, noting that this value also applies to the "northern Harvard measures". (Boss also gives the useful figure of +/- 0.08 mag error for one visual magnitude observation as done by his team 1909-1911; he says at least two, sometimes three, observations were made even for his hurried program.) With access to Cape observations, surely no Harvard magnitudes are based on observations lower than the traditional limit of z = 45. Even such observations would be extinguished only 0.10 mag for yellow light.

The difference between B and V band extinction would be only 0.024 mag, using the standard 4.16:1 ratio. The difference between Type A Visual and Type G Visual extinction can be estimated from the lambda^(-1) law, by replacing the Visual sensitivity window with the Type G spectrum as a proxy, considering the Type A0 spectrum to be the same as Type G2 but transformed to 9500K from 5800K (temperatures used by Johnson, Comm. Lunar & Planetary Lab vol. 3, #53, p. 73+, 1965), and expanding both factors of the convolution's integrand as Maclaurin series. For small changes in star temperature, the extinction is proportional to sqrt(temperature). This approximation says there's 28% more extinction in V for Type A0, than for Type G2. Again using the average type numbers in my sample's colors, this implies only 0.027mag more extinction for my Type A than my Type G, even at 45deg zenith angle. Likely, Johnson corrected for color differences in extinction, and Harvard (like Boss in 1928) didn't. This would brighten Johnson's Type A,


Reply author: Joe Keller
Replied on: 04/19/2008 17:26:10
Message:

In Barbarossa's Cavern There Are No Stars (Part X)

Using error bars for all colors, for the 31-star control sample in Part IX above, shows that the maximum-likelihood linear interpolation (roughly, this passes through the lower ends of the error bars for Types B, F & K, the upper ends for A & M, and the center of G), is roughly the "extreme choice" discussed in the last paragraph; it has half the slope of that found in Barbarossa's region. About half the control stars lie near Barbarossa's orbit (though not near Barbarossa); about half lie well away from the orbit. The sample is too small to say much about how these two halves differ, but there is a hint that the stars well away from Barbarossa's orbit, show mainly the negative magnitude change, and not so much the change with color type. So, the anomalous negative magnitude change seen for redder Types at this Declination, might be mainly due to Cape calibration c. 1908. The increasingly positive magnitude change with bluer spectral type, at this Declination, might be mainly due to fortuitous time changes in dust density along Barbarossa's orbit, not necessarily near Barbarossa.

From the Harvard-Johnson magnitude change, the extinction due to the center of Barbarossa's cloud, can be estimated as (0.10 - (-0.12))/(1.23 - 0.86) (from the extinction for Type A vs. K, according to my calculus estimate described previously) * pi/2 (assuming a spherical cloud just filling the 20 deg track investigated) = 0.9 mag. Maybe half this is near Barbarossa and the other half spread out along its orbit.

However, for such large extinction to be present, it would have to be canceled by a large Declination effect, but the part of the control group, far from Barbarossa's track, shows that the Declination effect (on Johnson minus Harvard) is ~ -0.1 mag. This suggests that there is a new kind of extinction here. Ordinary extinction dims all colors, but dims blue a little more. This new kind of extinction somehow dims blue while perhaps brightening red.

Let's turn away from the complications of dynamic tests such as the USNO-B R1 vs. R2 or B1 vs. B2, or the Harvard vs. Johnson magnitudes. Evidence of changing extinction, near Barbarossa's new position, is compelling, but complicated by unknown calibrations.

Let's instead perform another static test. Above, we found that the two stars studied near Barbarossa, 69 Leonis & Theta Crateris, have 3-10x stronger than normal interstellar spectral absorption lines, for such nearby stars; Theta Crateris arguably has the strongest well-measured absorption lines in an entire ~200 star sample < 100pc. Today I find that Hipparcos V & B mags, near Barbarossa, are abnormally dim, for white stars, and abnormally bright, for orange ones. This is the same abnormality which arises (see above) in the V magnitudes, Johnson vs. Harvard.

Methods. The online Hipparcos catalog lists V & B magnitudes, and parallaxes, recorded by the Hipparcos satellite 1991 +/- 2 yr. I considered all Hipparcos stars brighter than V = +10.00, nearer than 50pc, and within 10deg of RA11:20:00, Decl -8:00:00 (the approx. location of Barbarossa in 1991). The number of such Hipparcos stars here, per magnitude interval, began decreasing at about magnitude +10. My magnitude cutoff, +10.00, should lessen selection bias, give more accurate photometry, and exclude subdwarfs. I used only Draper (i.e., Annie Cannon) spectral types; this early typing, all by one research group, would be methodologically consistent, almost free of any possible influence by Barbarossa, nor would there be any influence of presumed absolute magnitude. There was one Type A; the rest were F, G, or K. Judging by absolute magnitudes, all the stars were class V (dwarf) except two G stars that were class IV (subgiant). I avoided the question of subgiant outliers, by considering only F and K. To get more stars, I arbitrarily annexed another disk centered 2deg farther N. In all there were 38 Hipparcos stars; 32 were in Draper (and had spectral types listed).

Results. First, I compared absolute V magnitudes ("MsubV") of near and far (from Barbarossa) members of the exact same letter-number spectral type, when available. Thirteen stars were such.

Type F5. Far from Barbarossa (>= 9.7deg): MsubV = 3.44 (SEM 0.20, n=2).
Near Barbarossa (2.4deg): MsubV = 3.56 (n=1).

Type K0. Far from Barbarossa (one at 7.8deg, five > 10deg): MsubV = 5.72 (SEM 0.23, n=6).
Near Barbarossa (<= 3.0deg): MsubV = 5.59 (SEM 0.01, n=2).

Type K2. Far from Barbarossa (7.7deg): MsubV = 7.07 (n=1).
Near Barbarossa (4.4deg): MsubV = 6.60 (n=1).

Second, I compared "B-V" near and far, for the same thirteen stars.

Type F5. Far from Barbarossa: B-V = 0.483 (SEM 0.003, n=2).
Near Barbarossa: B-V = 0.541 (n=1).

Type K0. Far from Barbarossa: B-V = 0.848 (SEM 0.042, n=6).
Near Barbarossa: B-V = 0.780 (SEM 0.048, n=2).

Type K2. Far from Barbarossa: B-V = 1.126 (n=1).
Near Barbarossa: B-V = 0.971 (n=1).

Though none of these six far-near differences reached statistical significance standing alone, walking together they are significant whatever their error bars. Using nonparametric statistics, the chance is only p = 1 / 2^6 = 0.02 that the change near Barbarossa, for both MsubV and B-V, would be positive for F5, negative for K0, and even more negative for K2.

The Johnson-minus-Harvard magnitude study above, predicts this very change near Barbarossa for absolute V magnitude, MsubV. Because most known forms of scattering affect B more than V, B-V should show the same behavior as MsubV, and it does.

The accidental inclusion of slightly brighter, bluer stars (e.g., a G9 star among the K0) would give a change in B-V only ~ 1/7 as big as the change in MsubV (Vega & the sun differ 4.2 in MsubV but only 0.63 in B-V). Ordinary dust extinction gives a change in B-V ~ 1/4 as big as in MsubV, but both changes would be positive through a nebula. Not only are both changes negative near Barbarossa for orange stars, but the change in B-V is ~ 1/2 as big as for MsubV. Autocollimated stimulated emission, through a nebula pumped with sunlight, would tend to move all stellar spectra toward that of the sun, i.e., B-V increase for white stars and decrease for orange stars. Also, MsubV might be increased for all colors, but especially for redder colors, because redder photons, with less energy apiece, might stimulate more emission per unit energy input.


Reply author: Joe Keller
Replied on: 04/23/2008 00:07:19
Message:

In Barbarossa's Cavern There Are No Stars (Part XI)

Barbarossa's Nebula Imaged via B-V Magnitudes of K0 Stars < 100pc


Motivation. Ideally, a large number of nearby stars, all with the same intrinsic B-V magnitudes, would allow mapping of the extinction near Barbarossa, according to varying B-V. For reasons given in the previous post, I best could realize this ideal, with Draper/Cannon K0 stars measured by Hipparcos, lying within 10deg of Barbarossa, within 100pc of the sun, and brighter than V = +10.00.

Calibrating the stars. The Draper/Cannon spectral types are given only to the nearest 1/2 or perhaps 1/5 color, that is, to the nearest 5 or perhaps 2 numbered steps; e.g., oftenest G0, G5, K0, less often K2, etc. So, some "K0" stars will fit a Planck curve closer to that of a G9 or K1, perhaps a G8 or K2, or, with errors other than rounding, an even further spectral type. Accuracy might improve, by assuming all stars are exactly on the main sequence, and correcting B-V according to Mv (the absolute magnitude in V). Using Mv and B-V for Vega (A0) and the sun (G2), we find that the change in B-V is 15% of the change in Mv. However, the sample K0 stars usually differ much more in Mv, than can be explained by a few color steps. Especially beyond 50pc, there are, judging by Mv, many borderline subgiants, subgiants, and giants. The effect of size class is much more important, than that of a few color steps. Surprisingly, size can be corrected in just the same way as color.

Materials. The raw sample consists of the 20 Hipparcos stars < 100pc, within 10deg of RA11:20:00, Decl -8:00:00 (approx. Barbarossa's 1991 position), with V < +10.00, and Draper/Cannon spectral type K0. (The stars < 50pc, in a 10deg circle centered 2deg N, which had been annexed for the earlier study above, were retained.) Two stars, both class V, seem to be an apparent double, separated by 30" of arc at 20pc, but by 0.4pc in depth according to their Hipparcos parallaxes. In retrospect, the B-V values of these two, were the most abnormal of any sample stars in size class V, suggesting mutual contamination. So, I averaged their Mv & B-V values, weighted by luminosity, and considered them one star. This reduced the sample size to 19. Extrapolating from Vega and the sun (using color steps as the abscissa), the Mv of a K0V star is +6.36. Twelve stars of the (modified) sample had Mv >= +4.55 (and <= +6.23), so these were considered class V, i.e., dwarf, i.e., main sequence. Seven stars had Mv from +2.92 to -0.32; these were considered class IV or III. Snow's "Dynamic Universe" college astronomy text lists Pollux (Mv = +0.95) and Dubhe (Mv = -0.7) as K0III.

Results. The 12 K0V stars were calibrated by adding to their B-V, 15% of the difference between +6.36 (taken as standard for a K0V) and their measured Mv. Also, by extrapolation from Vega and the sun, the B-V of a K0V star is +0.86. The result is, that the calibrated B-V is a smooth function of angular separation from Barbarossa. It equals the expected value, for K0V, at 0deg and at 10deg, from Barbarossa. In between, it increases linearly to about +1.10 = +0.86 + 0.24, at 8.4deg, then drops sharply. The corresponding V extinction would be 0.24 * 4.16 = 1.00mag. For the 13 (effectively 12) K0V stars:

(angular separation from Barbarossa, calibrated B-V for dwarfs)
(2.15,0.942)
(2.96,0.849)
(4.34,0.9195)
(5.10,0.9925)
(5.81,0.981)
(6.56,1.0495)
(7.85,0.967)
(8.35,1.0965)
(10.52,0.9645)
(11.14,0.964) (composite of two; apparent double)
(11.32,0.888)
(11.66,0.8885)

This function is consistent with a nebula densest in an outer shell near 8.35deg from Barbarossa. For the first through 8th points, the correlation coeff. is r = +0.762, p = 0.03; for the 8th through 12th, r = -0.961, p = 0.009 (two-tailed).

Though the color and spectral lines of these stars are consistent with a photosphere at the usually assumed temperature, B-V is not. B-V is consistent with the temperature of an "inner photosphere" appropriate to an exact main-sequence star of Mv equal to that of the given K0V or K0IV/V star. This calibration removes variance from the data, and renders them consistent with, or at worst, smoothly deviant from, standard K0V values. Earlier on this messageboard, I used the inner shell or inner photosphere concept to explain features of variable stars.

An "inner photosphere" concept also seems to apply to the K0 giants (IV and V). For a giant, divide the luminosity by the reciprocal fine structure constant, 137 (i.e., add 5.35mag). Then assume an inner main sequence photosphere as above. The results again are consistent with K0:

(angular separation from Barbarossa, calibrated B-V for giants)
(3.33,0.7635)
(5.29,1.0055)
(6.78,1.3115)
(8.15,1.0415)
(8.33,1.043)
(9.65,1.0415)
(9.86,1.041)

Four of the seven giants are consistent to within 0.01mag. The two least consistent, are the extreme brightest and dimmest.


Reply author: Joe Keller
Replied on: 04/23/2008 17:12:31
Message:

In Barbarossa's Cavern There Are No Stars (Part XII)

Barbarossa's Nebula Imaged via B-V Magnitudes of F5 & Bluer Stars < 100pc


This study differs from that in Part XI, only by using Draper catalog Type F5 & bluer, instead of K0. Instead of calibrating B-V to Type K0V, I calibrated it to Vega (Type A0Va). That calibration is of course routine for this sample, because they are almost all dwarfs, whose Mv differs mainly because they are spread out along the main sequence. The raw sample included 28 stars: one B9, six A, and 21 F0-F5. Judging by Mv, only one star was definitely class IV; this star also had the only outlier value for the corrected B-V; so, I excluded it.

Of the 27 remaining stars, the most significant break occurred at 6.96deg from Barbarossa. These bluish stars mirrored the K0 case of Part XI. That is, they started (ave. of initial three: corrected B-V = +0.029) near the normal value, decreased, then increased. The correlation coeffs. of the decreasing & increasing legs were -0.399 & +0.560 with p = 0.20 & 0.024, resp. (two-tailed).


Reply author: Joe Keller
Replied on: 04/25/2008 14:09:48
Message:

In Barbarossa's Cavern There Are No Stars (Part XIII)

Now let's return to the first evidence I discovered (besides the abnormal faintness of Barbarossa & Frey), of Barbarossa's moving nebula: the differing R1 & R2 USNO-B magnitudes measured on sky survey plates. Yesterday I searched the USNO-B catalog in a 30deg wide x 18deg tall region, centered slightly west of Barbarossa's 1987 position (the R2 magnitudes were measured on plates dated a few years either way from 1987). In disks one degree in diameter (30' radius) I found the number of stars for which R1 was between +15.00 & +16.00 and R2 between +16.00 & +17.00, and the number for which this held with R1 & R2 switched. The ratio of these numbers, will be denoted "q". In most of the region, I spaced the disks 2deg apart in each direction, but near Barbarossa, 1 deg apart.

I don't know the details, but R1 & R2 were estimated somehow from overlapping sky survey plates. More plates were available for R2 than for R1. So, the sky is divided into orthogonally oriented (i.e., EW by NS) "partition rectangles" of various sizes and shapes, but never bigger than 6.5deg (the size of the survey plates). Each partition rectangle has a different set of plates available for magnitude measurement. Also, each plate has a narrow, roughly rectangular strip along each edge, where the image is grossly fainter; if this part is used at all, magnitude calibration must be different there, so this amounts basically to a separate plate. If magnitude calibration is imperfect, the ratio "q" will be more uniform within one (up to 6.5 deg long, but usually narrow) "partition rectangle", than between rectangles. That is, my entire map of "q" should show scattered horizontal and vertical strips up to 6deg long, where q is consistently high or low, purely as measurement artifact. This predicted tendency, is indeed seen on inspection.

Only one area differs obviously from the above pattern. It is roughly trapezoidal (boat shaped), 8deg long (definitely bigger than one plate) and almost as wide, centered on Barbarossa, and inclined ~30deg NW to SE (i.e., along Barbarossa's orbit). Here "q" is consistently big, i.e., R2 is consistently dim.

One might expect a double image, a congruent region of small q, parallel to this region and lying 4deg NW by W, due to Barbarossa's orbital motion. On the overlap, the effect might cancel, giving roughly average q. Even allowing for scattered strips of measurement artifact in q, I don't see this.

The unequal roles of R1 & R2 might result from the different far-red sensitivities of the the emulsion/filter combinations used. According to the DSS "plate finder", the 1950s survey used "xx103aE + plexi [red plexiglass, i.e., RP2444, equivalent to approx. Wratten 29]", and the 1980s survey used "IIIaF + OG590". According to www.nofs.navy.mil, these equate to photographic passbands "E", and "R59F" [equivalent to "Fpg"], resp.

The galaxy is only ~300pc thick here, so at this high galactic latitude, almost all stars are within 200pc. Since they all have red magnitude approx. +16, their absolute red magnitudes are > +16 - 6.5 = +9.5. So, these are Type MV stars.

The Hipparcos data suggest that the nebula has roughly a spherical shell shape, and that its material somehow dims the V magnitude of red stars, while brightening blue stars in V. The controlled study of Johnson vs. early Harvard photometry, might have indicated the reverse, only because I fitted the data to a bell curve (e.g., up in the middle and down at the sides) instead of a shell (e.g., down in the middle and up at the sides). If so, then Barbarossa's nebula should strongly extinguish red. The 1980s sky surveys, if their red passband is narrower, so that the R2 magnitude depends mainly on the most extinguished wavelengths, might show an effect on q, next to which the 1950s effect would be unnoticeable.

Indeed, photographic passband "E" is approx. the same as passband "Rc", in the range 0.3 < B-Rc < 2.6 (Spagna, A&A 311:758+, 1996, Table 3, citing Evans' 1988 Cambridge Ph.D. thesis). This range doesn't include red dwarfs like the M5.5 Proxima Centauri (B-Rc = 3.5), but for lack of better information, I returned to MS Bessel's PASP 98:1303+, 1986, article, Fig. 1. Bessel says (p. 1305), "...in the late-M stars the main contribution to the light measured in Rc comes from wavelengths not seen by R<subF>. ...gross deviations occur only for the M stars...". Not only is Rc's turnoff at much longer wavelength than R59F's turnoff; Rc's 50% turnon is at 285A shorter wavelength than R59F's. Either the turnon or the turnoff difference, might explain why, unlike R2, R1 seems little affected by Barbarossa's proximity.

On the W half of my region, sampling over a 2deg mesh, I found average q = approx. sqrt(6). If R2-R1 is normally distributed, this corresponds to a tail area 1/(1+sqrt(6)), and a displacement of ~ 0.5 std. dev. Throughout the region, many disks had q < 1 (corresponding tail area for q=1, is 0.5, i.e. ~ 0 std. dev.) and many had q > 6 (tail area for q=6, is 1/(1+6), i.e. ~ 1.0 std. dev.). So, the displacement of the magnitude (i.e., the extinction) is proportional roughly to log(q).


Reply author: Joe Keller
Replied on: 04/25/2008 22:03:00
Message:

In Barbarossa's Cavern There Are No Stars (Part XIV)

The figures below are the Briggs logarithms x 1000 (i.e., divide by 1000 to get the base-10 log) of "q" (see Part XIII) on 30' radius disks centered at, 1st row, (11h40m,-3deg30'), (11h36m,-3deg30'),...,(10h52m,-3deg30'); 2nd row, same but Decl -4deg30'; etc.; last row, same but Decl -11deg30'. This region is 13deg EW by 9deg NS. It is roughly centered on Barbarossa's 1987 position (approx. 11h18m,-8deg), which is near a "four corners" where degree squares circumscribed around four of the disks, marked by (*), touch. At this Declination, 4m is 0.98deg, so I treat 4m as 1 deg.

-778,-929,-1315,-796,-1431,-559,-393,63,-324,201,653,457,166
987,507,-332,462,664,549,542,740,772,789,1239,752,201
750,488,000,607,627,580,441,1097,966,733,751,787,426
487,833,38,637,745,648,554,903,903,1088,644,727,488
-628,-526,346,525,641,602*,1024*,1124,970,1103,1058,625,675
-204,-556,867,833,862,1003*,1146*,1103,869,566,765,766,534
280,211,655,826,1465,1610,1542,765,804,673,548,258,611
-140,-18,556,622,680,773,987,140,311,-204,-275,-643,372
-84,182,463,160,229,753,708,261,79,577,-186,-204,470

These numbers could undergo a suitable arbitrary linear transformation a*x + b, to give a grayscale optical density, to make a 9x13 pixel grayscale picture. A refinement would be to interpolate at corners (average of 4) and at midpoints of edges (average of 2) to make a 17x25 grayscale picture.


Reply author: Joe Keller
Replied on: 04/28/2008 21:02:41
Message:

In Barbarossa's Cavern There Are No Stars (Part XV)

The USNO-B catalog hardly could correct for the difference between the spectral sensitivity curves, "E" (for the plates used to determine R1)(E is approx. the same as "Rc") and R59F (for the plates used to determine R2). This correction depends drastically on spectral type; most USNO-B stars have only Red magnitudes known, hence nothing known about spectral type.

The long-wavelength tail of the Rc sensitivity curve, which, MS Bessell remarked, is important for Type M stars' magnitudes, lies roughly halfway between the R59F and IsubN sensitivity curves (see MS Bessel's Fig. 1). Optical infrared ("I") magnitude could be substituted for R2 magnitude, and the calculations of Parts XIII & XIV repeated to find another matrix. Because the R2 (R59F) & I (IsubN) sensitivity curves might effectively differ (for Type M stars) from the R1 (E or Rc) curve by equal amounts in opposite directions, averaging the two matrices might find the result that would have been found if R1, R2, and I all had the same sensitivity curve.

I did this. The average matrix is below. In the region described by the matrix, the dates of the most important "I" plates, are approx. 1995.2 & 1997.2. The dates of the most important "R2" plates are approx. 1986.2 & 1987.2. So the effective date of the second measurement is approx. 1991.45 (and of the first, approx. 1954.2).

In this region, 20% more stars list I & R2, than list I & R1. Using I & R2 (instead of I & R1) also lessens the chance of misidentification due to proper motion or long-term variability; because, the plate dates are much closer. So, I compared R2 vs. I as I had compared R2 vs. R1. Averaging was effected by subtracting half the new log(q) for R2 & I, from the log(q) found for R2 & R1. For R2 vs. I, I further increased sample sizes (to about triple those of the R2 vs. R1 study) and hence significance, by using +17 instead of +16 as the dividing line, and by using semi-infinite intervals instead of intervals of length 1 mag.

The matrix entry, corresponding to the degree square containing the 1991.45 c.o.m. Barbarossa/Frey heliocentric position, is marked (**). The three next-closest degree squares are marked (*).

-1140,-1265,-1420,-969,-1519,-608,-421,-162,-686,-186,332,149,-34
507,245,-469,483,728,687,640,576,518,403,844,477,-45
296,231,18,709,684,731,514,913,648,393,461,541,192
64,504,-33,746,818,682,644,706,572,697,316,471,368
-996,-724,214,628,741,636*,1027*,905,683,812,695,368,531
-535,-758,587,546,631,736**,881*,754,500,288,491,598,389
-152,-148,365,394,925,1099,1116,517,499,355,369,137,481
-1180,-395,249,246,288,200,579,-110,-68,-71,-16,-637,329
-498,-213,211,-225,-220,186,329,-118,-274,285,-375,-129,421


Reply author: marsrocks
Replied on: 04/29/2008 12:36:51
Message:

Joe, below are the visual images derived from the numbers of your post of 25 Apr 2008 : 22:03:00:

The square marks the coordinates where you placed your asterisks.


[img]http://i295.photobucket.com/albums/mm132/marsrocksmars/keller%20numbers/kellernumberscompendium.jpg[/img]




Reply author: marsrocks
Replied on: 04/29/2008 12:57:22
Message:

When the images are pushed to contrast extremes, it appears the high numbers in your chart are concentrated around two main localities:


[IMG]http://i295.photobucket.com/albums/mm132/marsrocksmars/keller%20numbers/kellernumberscontrastextremes.jpg[/IMG]


Reply author: marsrocks
Replied on: 04/29/2008 13:15:25
Message:

And here is the same image again enlarged to 17 x 25:

[IMG]http://i295.photobucket.com/albums/mm132/marsrocksmars/keller%20numbers/kellerat1725.jpg[/IMG]


Reply author: marsrocks
Replied on: 04/29/2008 13:20:12
Message:

And this one is the original again enlarged to 50 X 34:

[IMG]http://i295.photobucket.com/albums/mm132/marsrocksmars/keller%20numbers/keller5034.jpg[/IMG]


Reply author: marsrocks
Replied on: 04/29/2008 13:25:26
Message:

And 416 X 288:

[IMG]http://i295.photobucket.com/albums/mm132/marsrocksmars/keller%20numbers/keller416288.jpg[/IMG]


Reply author: Joe Keller
Replied on: 04/29/2008 19:48:59
Message:

In Barbarossa's Cavern There Are No Stars (Part XVI)

Thanks, "marsrocks", for the beautiful artwork! From my data, you've produced the first image in history (and in ten versions!) of a nebula within Barbarossa's solar system, only 200 AU (est.) from Earth.

This might be an advantageous time to start using the name, "Barbarossa", as a label. Otherwise, in the future it might be difficult to prove what the image was supposed to be.

The Iowa State Univ. library has fifteen books on Frederick Barbarossa that I know of, but only two are in English and one of those is missing. Much of what I know about Barbarossa is from Thomas Carson's "Barbarossa in Italy", a translation of a long, mysterious, almost perfect neoclassical Latin poem rediscovered in the Vatican Library in 1877 and apparently written by a Lombard scholar attached to Barbarossa's army as a translator or guide. In one scene, as I interpret it, someone tried to seize the papacy by having a heavily armed militia of his supporters run around Rome robbing and "vandalizing" (line 675), presumably more explicitly burning out and killing, the supporters (including cardinals) of Barbarossa's just-completed crowning as Emperor by Pope Adrian IV (an Englishman named Nicholas Breakspear)(Bk. II, lines 657-715). Barbarossa's troops soon were galloping through the streets of Rome, slaying these rioters, though Barbarossa took many prisoners (released the next day, to the custody of Adrian IV) and even ordered that the fleeing rioters not be ridden down, shot in the back with crossbows, etc. (lines 739-751). In another scene (Bk. V, lines 3000-3033) Barbarossa, at the suggestion of some of the other military leaders, orders some hostages spectacularly stoned to death, but, quickly enough to save some of them, realizes that this is inhumane. With the help of this precedent, stoning did not become the normal punishment in Europe.

Henry Ford said history is "more or less bunk", and Nietzsche said history amounts mainly to paying close attention to old newspapers. However, at least one person who had worked with Barbarossa during his invasion of Italy, and who was intelligent, educated and sober enough to write almost flawless Latin poetry (Carson thinks the content of the poem indicates that the author was familiar not only with almost the entire Bible, but also with almost the entirety of the work of most major classical Latin writers) had an opinion of Barbarossa which he expressed by writing a favorable eyewitness account, after Barbarossa had departed.

I noted in Part XIII that the "E" (approx. equal to "Rc" at moderate visible wavelengths) sensitivity curve, of the plates on which the USNO-B "R1" is based, extends almost 300A more blueward than does the curve on which "R2" is based. Part XV's averaging involving "I" magnitude, can't correct this. A USNO-B search shows that the fraction of relatively bluish stars (B2 < +17.00) among my sample stars (+15.00 < R2 < +17.00) doubles (from 16% to 29%) for an 8deg shift, from my region southward (the galactic equator is SW by S). (The fraction increases almost as much with a westward or northward shift, and only slightly decreases with an eastward shift; the region of Barbarossa is a 10deg-wide island or isthmus of dim B2 magnitude.) So, my region (9x13deg) is big enough to need detrending, to remove the effect of galactic latitude on R1 vs. R2, due to the better blue sensitivity of R1.

I mapped the region with a projection equating 4m RA to cos(7.5deg)* 1deg. The best-fitting linear detrending, was that which removed the positive trend in the direction 103.6deg clockwise from N (i.e., W by SW). This direction differs 40deg from the perpendicular to the galactic equator, but agrees precisely with the gradient of bluish stars. (I found log(fraction) of relatively bluish stars, as above, in a 10deg diam circle centered on my region, and in additional circles displaced 8deg toward azimuths, clockwise from N, 0, 90, 135, 180, & 270. With Simpson's and Cotes' rules I convolved these with cosine(azimuth) for various phases, finding that the gradient of blue stars lies in the direction 102.3deg.) The correlation coefficient was 0.307, indicating that the trend explains 9% of the variance. The significance ( 9*13 - 2 = 115 degrees of freedom) of the trend is p = 0.0008 (2-tailed). The detrended matrix (this is my best representation of Barbarossa's nebula so far) is

-828,-998,-1198,-791,-1386,-520,-377,-163,-732,-277,197,-31,-259
808,501,-258,650,850,764,673,564,461,302,698,286,-280
586,476,218,865,795,797,536,890,580,281,304,339,-54
343,738,157,891,918,738,655,672,494,574,148,258,111
-728,-501,393,762,830,681*,1027*,860,594,678,516,145,263
-278,-545,755,669,709,770**,870*,698,400,143,301,364,110
94,54,522,506,993,1122,1094,451,388,199,169,-108,191
-945,-204,395,347,345,212,546,-187,-190,-238,-227,-893,28
-273,-33,346,-134,-174,187,285,-206,-407,107,-597,-396,109

(**) denotes the degree-square containing the Barbarossa center of mass in 1991; (*) denotes the three next closest squares.


Reply author: Stoat
Replied on: 04/30/2008 03:49:05
Message:

Seeing two "lobes" on Marsrock's image reminds me of the famous "ears of Saturn." When Saturn was first looked at, it was thought to have ears. The rings seen almost edge on showed up as two crescent shapes. We should expect each lobe of the nebula to have a slight blue/red shift.


Reply author: Joe Keller
Replied on: 05/02/2008 18:08:36
Message:

In Barbarossa's Cavern There Are No Stars (Part XVII)

Part XVI's image of Barbarossa's nebula, shows the expected negative correlation (always, relative to the global mean for the region), between extinctions at points separated by vectors equal to Barbarossa's travel from 1954 (approx. date of R1) to 1991 (approx. mean date of R2 & I). (In review, Part XVI's image is constructed essentially from the average, "h", of log(#R2>R1/#R1>R2) & log(#I>R1/#R1>I), on 1deg diam disks, linearly detrended to remove the influence of the star color gradient.)

Approximately, the image maps the extinction due to Barbarossa's nebula in 1991, minus the extinction due to it in 1954 (that is, minus the extinction due to Barbarossa's nebula in 1991, evaluated at a second point 4.25deg E and 2.28deg S of the first). Because R1's spectral sensitivity curve isn't the same as R2's or I's (and my averaging and detrending only crudely correct that) what is mapped isn't exactly the finite difference h(x,y) = g(x,y) - g(x+4.25,y-2.28), where g is the extinction of Barbarossa's nebula. However, negative correlation should occur between lattice points separated by ~ (4,-2), whenever the prograde point of the pair, is near enough Barbarossa's ~1991 position, to be within Barbarossa's nebula.

A guess, that Barbarossa's nebula has radius 2deg, was confirmed roughly, by correlations of +0.65, +0.4 and ~ +0.05, between "h" values at points within 2deg of Barbarossa's 1991 position, and points, 1.4, 2.2, or 3deg, resp., away from these (in a generally retrograde direction). The lattice point pairs, one of whose members is within 2deg of Barbarossa's 1991 position, anticorrelate strongest, when the other member is 3deg N and 5deg W of Barbarossa's 1991 position; r = -0.805 and Fisher's normalized z = 3.15 ( > 3 sigma significance). The number of such pairs is n=11; I exclude separation vectors for which the number of pairs is n<8. Linear interpolation of the correlation, and of Fisher's z, between the four sets of lattice point pairs nearest (-4.25,+2.28) separation (relative to the point near Barbarossa)(n=14 or 11, total 50), gives r = -0.678 and normalized z = 2.63, p = 0.0086, 2-tailed.

If "h" is the difference in the extinction of Barbarossa's nebula at two points separated by a constant vector, the autocorrelation of "h" should be -0.5 for points such that h1 = g1-g2 and h2 = g2-g3, if g2 is as big as g1 or g3. Indeed, for the sets of lattice point pairs whose separations are the four vectors nearest (-4.25,2.28), with the pair's first member < 2deg from Barbarossa, the autocorrelation ranges from -0.613 to -0.805; when I pretend that Barbarossa orbits in the opposite direction, the autocorrelation for the analogous four sets nearest (+4.25,-2.28) separation, ranges only from -0.1 to +0.1.

Inspection of my B2/B1, R2/R1, and I/R2 plots, confirms the prograde motion of Barbarossa's inner nebula. Because autocorrelation indicates that the ordinary (inner nebula) extinction extends ~ 2deg from Barbarossa, I considered 6x6deg squares centered on RA11:18,Decl-8. For R2 vs. R1, I found the first difference in column averages of log(q) = log(#R2>R1/#R1>R2), then interpolated linearly between these to find the maximum extinction 0.45deg W of center. (Extinction of R1 & B1 is neglected because it centers 4deg W, at Barbarossa's 1954 position.) B2 vs. B1 had "0" cells prohibiting logarithms, so instead I used (#B2>B1 - #B1>B2) / (#B2>B1 + #B1>B2), weighted within each column, by number of observations. Then I followed the same procedure as for R2 vs. R1, to find maximum extinction 1.40deg W of center.

I vs. R2 are theoretically only 1deg apart, so both extinctions must be considered. Also, the I & R2 spectral windows differ much, so the star color gradient might cause a big linear trend. First, I found first differences in column averages of log(q), as for R2 vs. R1 above. Then I linearly detrended to make the last column average equal the first. If the I and R2 extinctions are the same, the midpoint should be an inflection point. The second difference changed sign in two adjacent instances; interpolation gave two nearby inflection points which I averaged to find the midpoint of I & R2 at 0.77deg E of center.

Thus, the peak extinction, estimated from the USNO-B catalog magnitudes, of B2 vs. B1, was 0.95deg W, of that of R2 vs. R1. Barbarossa's travel in that time would predict 0.5deg W. The estimated midpoint of the extinction peaks of I & R2 was 1.22deg E, of the peak of R2. Barbarossa's travel would predict 1.0/2=0.5deg E. Though the extinction travelled about twice as far as predicted, statistical errors and methodological uncertainties are big. The correct sequence of the three, is significant at p=1/6.


Reply author: Joe Keller
Replied on: 05/03/2008 19:30:01
Message:

In Barbarossa's Cavern There Are No Stars (Part XVIII)


Dedicated to Frederick Barbarossa: "King, Knight, Hero"
(Helmut Hiller)


Barbarossa Nebula Dynamics Match Observations

The simplest dynamical estimate of the radius of Barbarossa's nebula, would be 197.7 AU * sqrt(0.0103), using the distance from Barbarossa to the sun, and the ratio of Barbarossa's mass to the sun's, to get 20.06 AU. What we want to know, is the radius of the largest possible orbit around Barbarossa. As a start, let's consider only gravity, and ignore the sun's gravitational pull on Barbarossa itself. The case of a face-on circular orbit can be solved exactly: the orbital plane is displaced toward the sun so that Barbarossa's outward pull equals the sun's inward. Barbarossa's best efficiency in this, is max(sin(theta)*cos^2(theta)) = sqrt(4/27), so the biggest possible projected radius is 20.06 * sqrt(sqrt(4/27)) = 12.44 AU.

The case of an edge-on orbit can be estimated using the circular approximation: in half a circular orbit with speed v, Barbarossa vectorially accelerates the body by approx. pi*v, but if the body is bound, again ignoring the sun's gravitational pull on Barbarossa itself, the sun accelerates the dust grain prograde by < (sqrt(2)-1)*v. The sun's efficiency in this, over half an orbit, is only 2/pi, so the biggest possible radius is 20.06 * sqrt((sqrt(2)-1)*1/2) = 9.13 AU. Thus the edge-on and face-on circular cases give approx. equal radii. Though Barbarossa's orbital motion continually converts face-on to edge-on and vice versa, this common radius could be taken as the radius of Barbarossa's nebula, if only gravity were considered.

ISU's copy of Hodge's "Interplanetary Dust", has an anonymous penciled marginal note referring to Burns, Lamy & Soter, Icarus 40:1+, 1979. I took the hint.

According to Burns, et al, Fig. 7a, p. 16, the greatest "beta", i.e., ratio of solar radiation pressure to solar gravity, 5.5, occurs for pure graphite dust of 0.08 micron size (this is for spherical particles, but shape dependence is weak according to, inter alia, Kruegel; and furthermore Burns et al give evidence that rather spherical shapes predominate anyway). So, for classical interstellar dust, i.e., carbonaceous particles of ~ 0.1 mu size, the biggest possible circular orbital radius isn't much bigger than 12/sqrt(5.5) AU (actually, the sun's gravity pulls the same on Barbarossa and on the dust grain, so the net push, equals the radiation pressure); i.e., it can subtend not much more than 1.5 deg from Barbarossa. This matches the 2deg radius found empirically in Part XVII from the autocorrelation, near Barbarossa, of the USNO-B R2 vs. R1 extinction measure (from the third, most accurate version of my image data). Or it can be found simply by looking at the images of that extinction measure, assembled by "marsrocks" above (from the first, least accurate version of my image data).

This small patch around Barbarossa, eight times the diameter of the full moon, is where backscatter of sunlight, from Barbarossa's nebula, might be detected. Farther than 1.5 deg from Barbarossa, solar radiation pressure drives away classical interstellar dust. Ultrafine, smaller-than-classical carbonaceous dust grains, which for any given extinction cause orders of magnitude less scattering, would reach 12/sqrt(1.77) AU --> 2.6 deg from Barbarossa, because (Burns et al Fig. 7a) for graphite of size < 0.004 mu (40A), "beta" has an asymptotic value of only 1.77.

The first, least accurate version of my image data (the later versions haven't yet been extended this far) shows roughly vertical strips of R2<R1 about 5.5 deg east, and R2>R1 about 12 deg west, of Barbarossa's 1987 position. These are narrow strips ~ 4 deg wide, not part of the global trend. Barbarossa's 1954 & 1987 positions are separated about 3.9 deg in RA, so if Barbarossa's outermost (gas, or low-beta non-carbonaceous dust) nebula has radius 9 deg, then the 1987 outer nebula position would cover, roughly, a vertical strip 5 to 9 deg east of Barbarossa's 1987 position (vs. ~ 5.5 observed) that wasn't covered in 1954; and the 1954 position would cover, roughly, a vertical strip 9 to 13 deg west of Barbarossa's 1987 position (vs. ~ 12 deg observed). For a 9deg radius, beta ~ 0.15 would be needed ( 12/sqr(0.15) = 31 AU --> 9.0deg ). Really, beta isn't the cause; it's the Jacobi limit, 30 AU (see Technical Detail #2, Part XIX below). So, both the Jacobi limit, and the radiation pressure limit are seen, defining an outer and inner nebula, resp.

These strips, indicating the E & W extreme portions of Barbarossa's outer nebula, give negative extinction (i.e., brightening instead of dimming). These are mainly Type M stars, so this agrees with the negative extinction found for Type K & M stars in my Harvard vs. Johnson magnitude study of the same region (statistically controlled, by samples from distant regions of equal Declination and similar galactic latitude). Reviewing all the stellar photometric studies of Barbarossa's outer nebula that I've done, I find that Type K & M stars consistently undergo paradoxical, negative V extinction there, Type F & A consistently ordinary, positive V extinction. Not only is this seen in the smooth linear variation with spectral type, in my Harvard vs. Johnson study; it's seen in my original, 4deg-resolution, study in which B2 - B1 was greatest E of Barbarossa, but R2 - R1 greatest W of Barbarossa.

I made a 1deg-resolution B2 vs. B1 study, similar to the R2 vs. R1 study the results of which are displayed in the above images by "marsrocks". Even summing results using two pairs of magnitude intervals, [16,18] & [18,20] and [17,19] & [19,21], data are sparse, and my study less thorough, but I do find positive extinction, of B2 vs. B1, out to roughly 4deg from Barbarossa's 1983 position, at least in some directions, as in the R2 vs. R1 case. Unlike the R2 vs. R1 case, the B2 vs. B1 case shows increased, not decreased, extinction of B2 vs. B1, about 9deg E of Barbarossa; and perhaps increased, not decreased, extinction of B1 vs. B2, ~ 5deg W of Barbarossa. Thus the outer nebula finding for B2 & B1, resembles the Harvard vs. Johnson finding for Type F & A stars.

My Hipparcos studies show that B-V changes (it's the reverse of ordinary extinction) in the opposite direction, to V, as though B were much less affected than V. (This holds both for Draper/Cannon Type K0, and for Draper/Cannon Types F5-B9.) This suggests that photons from stars are traded slightly up or down in energy, by a kind of stimulated emission as they traverse Barbarossa's outer nebula which has been pumped by sunlight.

Barbarossa isn't luminous like the sun, so the Poynting-Robertson and Yarkovsky effects are unimportant in Barbarossa's nebula close to Barbarossa. Barbarossa's huge angular momentum replenishes the orbital angular momentum, around the sun, of the particles gravitationally bound to Barbarossa; and the sun's radiation effects average zero on a particle's orbit around Barbarossa.


Reply author: Joe Keller
Replied on: 05/04/2008 16:17:26
Message:

In Barbarossa's Cavern There Are No Stars (Part XIX)

Technical detail #1. I've seen the claim that the red plexiglass filter used in the Palomar Red sky survey, is about the same as a Wratten #29 filter. The Handbook of Chemistry & Physics, 44th ed., gives (with minor interpolation) 6220A as the 50% cuton for Wratten #29. The next listed, Wratten #26, has 50% cuton at 6070A. MS Bessel's Fig. 1 gives 5890A as the 50% cuton for R59F (i.e., SERC-Red). All these cutons are steep. So, if Wratten #29 is at all correct, the Palomar Red survey cuts on, 300A redder, not 300A bluer, than the SERC-Red survey. However, my data support the other claim I've seen, that Palomar Red's "xx103aE emulsion + red plexiglass filter", together, still were consistent with the "E" passsband (which in turn is about the same as "Rc" at moderate visible wavelengths, allegedly according to a Cambridge Ph.D. thesis; Rc's cuton is 285A bluer than R59F's). R2 - R1 increases toward the galactic equator (or roughly equivalently, along the local gradient of blue star fraction); this gradient in R2 - R1 is seen in the farthest corners of my pure R2>/R1> plot, as much as 20deg away from Barbarossa. So, R1 would seem more sensitive to bluer stars, than R2.

Technical detail #2. The Jacobi limit (Icarus 102:298+, 1993; or 107:304+, 1994), which uses the cube root of the mass ratio (rather than the square root as my estimate does) for Barbarossa, considering only gravity, is 29.8 AU. Empirically, however, the "most distant known satellites" (vis a vis the Jacobi limit), Jupiter's "Pasiphae and Sinope", have semimajor axis only 0.45x the Jacobi limit, corresponding to 13.4 AU for Barbarossa (Icarus 102:298+, p. 304). This agrees well with my estimate, considering only gravity, of ~ 9.13, to 12.44 AU. My estimate easily applies to the virtual negative gravity of radiation pressure, but the Jacobi limit is based on the Roche lobe, and not so easy to apply here.

Technical detail #3. Recent Earth-based photographic searches for satellites of Uranus & Neptune (Icarus 102:298, 1993; 107:304, 1994) have used the same emulsion/filter combination, IIIaJ + GG395 (passband 3950-5350A) and exposure, 60min, as the SERC-J (Blue) sky surveys. If this is best for sunlight reflected from unknown solar system objects, it's presumptively best for Barbarossa's nebula. The removal of red light reduces the red dwarf background.


Reply author: Joe Keller
Replied on: 05/07/2008 20:32:52
Message:

In Barbarossa's Cavern There Are No Stars (Part XX)

IRAS Imaged Barbarossa's Bow Shock


Schlegel, et al, Astrophysical Journal 500:525+, 1998, Fig. 8, p. 542 ("Dust"), shows a streak of far-IR brightness, measured by the IRAS satellite in 1983, that is only 4 deg from Barbarossa's 1983.2 heliocentric position (Barbarossa was at l=265.9, b=+48.4) and extends ~ 12 deg each way. It's prograde of Barbarossa (i.e., > galactic longitude) and crosses Barbarossa's orbit at approx. 37 deg slope ("beta"). Though presumably its seeming (non-greatcircle) extension near galactic coords l=190, b=+40, is accidental (the alleged Martian canals perhaps also demonstrated the human ability to notice accidentally aligned dots) this streak is one of the most prominent curvilinear "cirrus" "dust clouds" in the sky.

With a mm ruler, I measured the positions of 8 points of this cirrus (on Schlegel's polar-projection map, Fig. 8) chosen for their definiteness, and their roughly equal spacing with good linear interpolation between. Then I used cubic Lagrange interpolation on Schlegel's map, to find their galactic coordinates, converted to other spherical coordinates (for a nearly conformal flat map) for which the endmost points of the "cirrus" define the equator, plotted them on Keuffel & Esser graph paper using a sinusoidal projection with Barbarossa's 1983.2 position defining the prime meridian, and graphically found the best-fitting hyperbola (taking Barbarossa as a focus, using symmetry to find the slope of the directrix by eye, then successive approximations to find the eccentricity, ~ 5.5, i.e., very flat). I also made the small correction, longitude --> sin(longitude), because I'm allegedly viewing, approximately, an hyperboloid of revolution, from a finite distance; not a hyperbola. (Hausmann & Slack's Physics, Ch. XXXI, sec. 367, Fig. 355, shows a flat-nosed projectile with an hyperboloidal bow shock with e = 1.5.)

The N & S asymptotes slope 27 and 47deg, resp., to Barbarossa's orbital path. This range of slopes lies well within the theoretical range, 16.9 to 61.5deg, for initial Mach number 3.432, final Mach number > 1, and adiabatic constant k=5/3. (See Hughes & Brighton, "Fluid Dynamics", McGraw-Hill, 1967, Sec. 8.2, eqns. 8.9 & 8.10 and figs. 8-1 to 8-3, pp. 158-159; or, 2nd ed., 1991, pp. 210-211.)

The Mach number "M1" is taken to be the ratio, of Barbarossa's orbital speed in a circular orbit at 197.7 AU, to the speed of sound in rarified dissociated (i.e., atomic) hydrogen at the solar radiation Planck equilibrium temperature (this is independent of albedo, by Kirchhoff's law) at 197.7 AU. Using (HL Johnson's favored) 5800K for the sun's surface, the sun's radius, and the Stefan-Boltzmann T^4 law, gives 28.1K as the equilibrium temperature at Barbarossa. (According to Schlegel, interstellar dust is 15-19K.) The proportionality of the speed of sound, to sqrt(T/m), is so accurate that my result varied little, even when I based it on molecular hydrogen at STP (with the sqrt(T/m) correction). However, I based my calculation on the value 8.894 km/s for dissociated hydrogen at 0.1 bar and 5800K (the lowest pressure in the table; the temperature went up to 6000K, but I wanted to use HL Johnson's favored temperature of the sun)(Vargaftik, Handbook of Physical Properties of Liquids & Gases, 2nd ed., p. 34).

Corrected to 28.1K, the speed of sound in rarified dissociated hydrogen is 0.6176 km/s, giving M1 = 3.432, and a minimum possible "beta" = arccsc(M1) = 16.9deg (Hughes & Brighton, eqn. 8.8). The abscissa of the intersection of the curve M2=1, with the M1=3.432 isocline (from simultaneously solving, numerically, Hughes & Brighton eqns. 8.9 & 8.10) is 61.45deg for the monatomic adiabatic exponent k=5/3 (Fig. 8-3 shows 64deg, but this seems to be for air; k = 1.404 & 1.401 for molecular nitrogen & oxygen, resp., at +15C & std. pressure).

On my R2 vs. R1 chart, I had marked degree squares with ratio < 1 and > 6; these corresponded roughly to lower and upper quartiles. Within the 3x3deg square centered on Barbarossa, the inner nebular extinction shows a northern boundary sloped at ~ 45deg.

On the B2 vs. B1 chart, I had marked squares according to whether none, one or both of the ratios used exceeded 1. Here, the average position of the arrows shows a northern boundary sloped at ~ 67deg, but data are sparse.

The average of these two charts, weighted by number of observations, gives 47deg, i.e., 45 + (67-45)/(10+1) - 27 = 20deg to Barbarossa's orbit. Hughes & Brighton's Fig. 8-3 isocline needs to have its peak shifted 2.5deg left if k=5/3; the left zero doesn't move. So, from the interpolated M1=3.43 isocline, shifted an interpolated 1deg left at the relevant position, I can read that for "beta" (the slope of the shock wave) = 37deg, "theta" (the slope of the obstruction) = 22deg, agreeing with the ~20deg observed in the extinction chart.

Schlegel's Fig. 12 maps the difference between Schlegel's infrared map, and the Burstein & Heiles, Astronomical Journal 87:1165-1189, map based on reddening of external galaxies and HI column densities. The Burstein & Heiles map shows nothing unusual in the vicinity of Barbarossa, so the "cirrus" (really, bow shock) appears on Schlegel's Fig. 12 also. (I chose Fig. 8 for my study, to stay closer to the original data.) Schlegel used DIRBE satellite data (1989-1990) to correct the (higher-resolution) IRAS data for temperature, to convert them into presumed interstellar dust column densities; Schlegel's temperature map shows nothing obviously unusual near Barbarossa.

However, even when temperature is considered, IR emission need not correlate with HI (nor presumably with reddening of external galaxies). Regarding a shell in Eridanus, "When the IR emission is compared with the HI there is little direct positional agreement. ...We are not yet convinced that we can account for apparent variations in IR brightness as resulting exclusively from temperature effects,..." (Verschuur et al, in, IAU Symposium #139, 1989; Conclusion, p. 236).


Reply author: nemesis
Replied on: 05/09/2008 09:59:52
Message:

Shouldn't Barbarossa cause observable perturbations of the outer planets?


Reply author: Joe Keller
Replied on: 05/09/2008 14:43:35
Message:

quote:
Originally posted by nemesis

Shouldn't Barbarossa cause observable perturbations of the outer planets?



Thanks for bringing up this important question again. Scattered through this long thread, is more information about this.

Goldreich & Ward, PASP 84:737, 1972, claimed that even a moderate Planet X, unless almost exactly coplanar with the rest of the solar system, would move Neptune to high inclination, within much less than the age of the solar system. Their article doesn't consider chaos theory. Also, I've counted four questionable simplifying assumptions explicit in their calculation:

1. p. 737, col. 2, par. 2, sentence 2
2. p. 738, col. 2, par. 1, sen. 3
3. p. 739, col. 1, 3rd to last sentence
4. p. 739, col. 2, last par.


Zakamska & Tremaine, AJ 130:1939+, 2005, measured acceleration relative to those millisecond pulsars (MSPs) which are not within globular clusters. In their sample, 35 of 48 MSPs were binary (typically MSPs are close binaries). The contemporary theory of MSPs is that the original supernova sweeps its environs so clean, that except for companion stars which often endure, and rarely planet-size masses (maybe really stellar fragments), there's nothing nearby to produce an acceleration more than about that of Barbarossa on the sun: a distant brown dwarf companion (there is evidence that many or most stars have these; see my posts) survives the supernova, but planets (closer and smaller) don't; stellar companions, if they survive, are detected and the MSP period corrected for their effects. Thus the scatter in Pdot/P for MSPs, corrected for obvious binary effects, should be about what would come from a brown dwarf companion at a few hundred AU, which endured the supernova! Accordingly, Zamaska & Tremaine believe themselves statistically able, based on the scatter of Pdot/P for their sample of 48 MSPs, to rule out as little as 1/10 ~ 1/sqrt(48) the acceleration due to Barbarossa.

Zakamska & Tremaine used nonparametric statistics (rank correlation) to look for the acceleration effect of Barbarossa, then parametric statistics (scatter of Pdot/P) to calculate the significance of their finding. It would have been better to use parametric statistics for both parts, rather than discard statistical significance by using nonparametric statistics, then use parametric statistics to show that statistical significance is lacking.

That articles resorting to such approaches have been published recently, indicates that ephemerides are considered, at least by some peer reviewers, unable, alone, to rule out bodies like Barbarossa. Standish's use of space probe data to remove residuals from Uranus and Neptune, was complicated. Only a partial orbit of Neptune is available.

"It is true that if you put in the new mass of Neptune, some of the key residuals in Uranus do decrease. But in my opinion it is not correct to say that they disappear entirely."

- Robert Harrington of the U.S. Naval Observatory (New Scientist, Jan. 30, 1993, p. 18, cited by Ken Croswell)


Perhaps the simplest approach is to consider the Pioneer acceleration. Barbarossa's tidal force, i.e., the difference between Barbarossa's pull on Pioneer 10 or 11, and the sun's, would have been, in the outer solar system, of the same order of magnitude as the anomalous Pioneer deceleration (and indeed seems to explain much of the apparent variation in the anomalous acceleration; for discussion and references see, Joe Keller's March 29, 2007 post, p. 10 of this thread, and April 24, 2007 post, near the bottom of the post, p. 13 of this thread). I recall from an article by JD Anderson, that with the help of space probes, Mars' orbital period has been measured accurately enough to rule out any steady unexplained sunward force on Mars, more than ~ 1% as big as the Pioneer deceleration.

On the other hand, Barbarossa's tidal force, for any planet, though also radial, would mostly cancel over a period (it's half as much at quadrature as at opposition/conjunction, so averages to 1/4 of the maximum); for outer planets, whole orbital periods are practically unavailable to measurement anyway. (For Earth at this epoch: net toward the sun in December and June, net away from the sun in March and September.) It would be < 1/10 as big in the inner solar system, as in the outer solar system where it becomes comparable to the Pioneer force. That, combined with averaging to zero, would make it undetectable even by the measurement of Mars' period, which Anderson put forth as the most sensitive way to detect such radial forces.


Reply author: Joe Keller
Replied on: 05/09/2008 17:14:37
Message:

Some Political Aspects of Science

According to his online New York Times obituary, Dr. Harrington died in the hospital, of cancer, at age 50, January 23, 1993 ("Saturday"; according to the article "published Jan. 27, 1993") a week before the cover date of the New Scientist article quoting him, as doubting that Standish's work had refuted Planet X. Though Harrington, as recent head of the USNO equatorial division, and well-known Planet X seeker, was an appropriate source for the New Scientist to interview, someone not on his deathbed might have been able to rebut Standish in more detail. Nonetheless, deathbed statements get extra legal weight. This was practically a deathbed statement by Harrington. Surely it was his sincere opinion, free of political influence.

Journalist Peter Jennings waited until he had terminal cancer, to present his expose on UFOs. Another journalist, Bill Moyers, now in his 70s and formally retired since 2004, tried to make a blockbuster statement on the Charlie Rose show last night, but, for the only time in the forty years I've been watching Public Television, the signal was lost for so long, about a minute, that that entire thread of conversation failed to be broadcast. Moyers had been close to Pres. Johnson and was trying to reveal what Johnson really thought about the Vietnam War.


Reply author: Joe Keller
Replied on: 05/10/2008 23:59:16
Message:

Lowell Knew Barbarossa Fit Uranus Data

From an excerpt of Lowell's writing (see: WG Hoyt, "Planets X & Pluto", 1980, pp. 136-140) I can summarize Lowell's method. Adams had predicted Neptune from discrepancies in Uranus' orbit. Adams, LeVerrier, Galle et al had > 1 orbit of Uranus to guide them since Flamsteed's (accurate, for its era) 1715 position, but Lowell had < 1 orbit of Neptune. Therefore the remaining discrepancies in Uranus' orbit, unexplained by Neptune, were (and are, I gather from statements by Standish and Harrington) a better guide to any Planet X, than are the discrepancies in Neptune.

What follows is my simplified reconstruction of Lowell's method. Lowell assumed zero inclination (the greatest inclination of any outer planet then known was Saturn's, 2.5deg) and, at first, zero eccentricity. He guessed the mass, eventually as 0.00002 solar masses. Then he searched for the radius and phase, that best explained Uranus' discrepancies. This gave the necessary radius as a function of mass. One point, on this mass-radius curve, was 6.5 Earth masses (0.000020 solar masses) and 44 AU. Lowell preferred these values, because they are consistent not only with the geometric sequence ratio 1.5 for the planetary orbital radii (which describes not only the step from Uranus to Neptune, but also the inner planets from Mercury to Mars, though Venus deviates) but also consistent with the geometric sequence ratio 2.64 = (massJupiter/massNeptune)^(1/3) for planetary masses (which describes the outer planets from Jupiter to Neptune, though Uranus deviates).

To investigate the effect of eccentricity, Lowell assumed the largest eccentricity of any known planet (0.20, as for Mercury) and studied the two extreme cases: Planet X at perihelion in 1914, and Planet X at aphelion in 1914. Any phase between perihelion and aphelion, or any lesser eccentricity, would amount to an intermediate case, so knowing these two extremes, Lowell would know what range of ecliptic longitude to search. The two cases (unsurprisingly, because of Planet X's assumed large mass and distance) gave perihelia almost 180deg apart, i.e., almost the same ecliptic longitude for Planet X in 1914. This ecliptic longitude must have been either near 202 for both, or near 202-180=22 for both; it must have been 202, because Pluto, discovered near there, was thought for awhile to be the predicted Planet X. Both solutions had periods near 292yr, so the 199yr of data covered ~240deg of Planet X's orbit. Instantaneous data would have implied a 50% difference in major axis (1+0.2 vs. 1-0.2), if one solution were at perihelion and the other at aphelion; 180deg of orbit equally weighted, would average that to zero, and 240deg of orbit to < 10%. So the two fitted major axes also were almost the same, averaging 44AU.

Lowell's eccentricity 0.2 calculation, seems to have been only exploratory, to find the extreme range of ecliptic longitudes to search. So, Lowell's actual solution was, approximately, a 6.7 Earth mass body in a circular orbit of 44AU radius, phase 202deg ecliptic longitude in 1914.

Uranus' orbit is like a frictionless harmonic oscillator. A perturbation drives it 180deg out of phase. The second harmonic perturbation (see my May 11 post below) predominates. Therefore the lag is 1/2 * 1/2 = 1/4 period. For Lowell's 6.7 Earthmass alternative, to have at present the same effect as Lowell's implicit, not explicitly stated, 3430 Earthmass alternative (which I name Barbarossa), they must have the same longitude, when retarded in their own orbits by a quarter of Uranus' orbital period. The 6.7 Earthmass alternative's retarded ecliptic longitude in 1914 was 202-84/4*360/292=176; likewise Barbarossa's retarded longitude in 1987 was 173-84/4*360/2780=170; in 1914, 170-73*360/2780=160.5. So, Lowell in 1914 essentially predicted Barbarossa's heliocentric ecliptic longitude with only 15 deg error, assuming Lowell's very most recent data got almost all the weight when calculating Planet X's longitude. If the effective mean epoch of Lowell's data is 1900, then the *effective* longitudes of Lowell's Planet X and Barbarossa, are equal. (David Todd in 1877 essentially predicted Barbarossa's longitude with only 11deg error; see subsequent posts.)

The radial component of Barbarossa's tidal force was shown above to be smaller than any measurements hitherto would have detected. The maximum tangential component can be estimated by drawing a tangent from Barbarossa to Uranus' orbit. The ratio of this, for Lowell's 6.7 Earthmass alternative, vs. Barbarossa, is 1:1.075. This should be multiplied by a sine factor to correct for the lesser effectiveness of quickly alternating forces; the corrected result is 1:1.19. So, with only 19% error, Lowell's predicted Planet X, based on 1715-1914 discrepancies in Uranus' orbit (unexplained by Neptune), gives a perturbation of the same strength as Barbarossa.


Reply author: Joe Keller
Replied on: 05/11/2008 18:58:53
Message:

Planet X: Dynamical Evidence in the Optical Observations

E. Myles Standish's article is entitled, "Planet X: No Dynamical Evidence in the Optical Observations" (Astronomical Journal, 105:2000+, 1993). I'm correcting that title. After Standish carefully removed outliers, adjusted planetary masses to more accurate space probe values, and recomputed orbits, some dynamical evidence for Barbarossa could be seen by inspection of his graphs.

"On 30 Sept 1846, one week after the discovery of Neptune, Le Verrier declared that there may be still another unknown planet out there."

- Paul Schlyter, historian of science (internet)


In 1877 David Todd predicted a planet at ecliptic longitude 170 +/-10 for 1878 (source: Schlyter). Barbarossa's longitude then would have been 159.

By drawing a tangent from the outer to inner orbit, the maximum tangential component of tidal force is found to be approx. (sec^2(theta) - cos(theta))*M/R^2, where theta = arcsin(r/R). The mass of Neptune would need to be increased 7%, to give the magnitude of additional tangential force provided by Barbarossa. So the Voyager/Tyler 0.5% *downward* revision of the 1976 IAU value of Neptune's mass (Standish, Table 1) does little, to explain Lowell's prediction (which was consistent either with Barbarossa or with a smaller, nearer planet). (Lowell explicitly said that he knew only the pull's direction and tidal strength with accuracy; the mass and distance were mutually dependent guesses.) Furthermore, Voyager/Tyler revises Newcomb's 1877 Neptunian mass, likely used by Lowell, only 0.17% downward.

*The* unperturbed orbit, is not the same as *an* unperturbed orbit. A large perturbation can cause a perturbed orbit that looks very like some other unperturbed orbit. This does not imply that there is no perturbation. Likewise, a "poorly conditioned", i.e. very acutely angled, 2x2 system of linear equations, isn't solved accurately by projecting a normal from one line to the other; it's solved by moving along one line to the other. In terms of higher mathematics, one must follow the curve of perturbation in Hilbert space, to find the unperturbed orbit; it is not sufficient simply to move to the nearest unperturbed orbit and say, "Look how close this is."

I express the tidal tangential force, as a Fourier series in the longitude of Uranus (assuming a circular orbit), finding the coefficients numerically by convolution. The 1st harmonic is small and its effect on longitude, can be and is, neutralized by assuming a slight alteration in the magnitude and apse of that relatively gigantic orbital parameter, the eccentricity. Relative to the gravitational force of the sun, the 2nd harmonic coefficient is 0.1461 * 0.0103 * (19.18/197.7)^2 = 14.16 * 10^(-6).

To find displacement in RA, the 2nd harmonic amplitude must be divided by 2^2, to account for two integrations; and it's opposite in sign to the input. The effect on orbital radius can be neglected for a second harmonic perturbation, because the radius is a first harmonic response to that input, and cancels. So the 2nd harmonic amplitude of Uranus' longitude displacement is

-0.25 * 14.16*10^(-6) radians = -0.7302"

where the sign signifies that the displacement is opposite in sign to the input tidal tangential perturbation there.

(The 3rd harmonic amplitude of longitude displacement is, analogously, of the order of 0.044".)

The wave seen on, e.g., the rightward portion of Standish's Fig. 5a, is roughly 0.5" peak-to-peak, which might correspond to 0.125" harmonic (semi)amplitude. Uranus' eccentricity strongly affects both the 1st & 2nd harmonic of longitude displacement directly, and also the 2nd harmonic of tidal tangential force (and thereby also the 2nd harmonic of longitude displacement). Optimum choice of eccentricity and apse might partly nullify both 1st and 2nd harmonics. (Likewise, something with both dipole and quadrupole, whose strengths can be multiplied by the same factor, and which can be rotated by the same angle, has two degrees of freedom with which it might partly nullify a dipole and quadrupole simultaneously.) So, the observed amplitude of longitude displacement vis a vis the best fitted orbit, should be < 0.73".

In 1975.0, Uranus' longitude was 211, Barbarossa's 172, & Barbarossa's retarded longitude 172 - 84/4*360/2780 = 169. At this relative phase, 211-169=42deg (~45deg), the second harmonic longitude displacement should peak. Standish's Figs. 5a & 6a show his most highly corrected data for RA. Because of scatter, the harmonic (semi)amplitude might be estimated as 1/4 the raw peak-to-peak amplitude, i.e. 0.125" from c. 1930 to 1990. The peak is c. 1960 (60deg away from the pred. 1975 peak; this amounts to +0.5 correlation with predicted phase) but the seeming period, c. 1940-1980, is close to the 84/2=42yr predicted. The seeming amplitude (considering that due to cancellation through orbital fitting, my 0.73" is only an upper bound) and period do suggest that the effect of Barbarossa is apparent on Standish's graphs.

Standish (sec. 2, p. 2001) estimates 1.2" error for individual time point observations before 1911. Lowell would have needed ~ (1.2/0.73)^2 / mean(sin^2(theta)) * 2^2 = 22 randomly distributed time point observations total, to detect the second-harmonic discrepany in Uranus longitude, due to Barbarossa, to 2-sigma significance. (With the help of the internet browser's magnification feature, I can see that there are > 1000 dots before 1915 on Standish's Fig. 1a.) Lowell would have known that he sought a second harmonic phenomenon. He might have adjusted Uranus' eccentricity and apse by neutralizing the 1st harmonic precisely, without regard to the 2nd harmonic discrepancy in longitude, which then would not be masked.


Reply author: Joe Keller
Replied on: 05/13/2008 20:45:55
Message:

The residuals of Uranus' RA since 1950, seem less grossly sinusoidal in Standish's (1993) Figs. 4a and 6a, than in Fig. 5a. I measured the last 39 data points from the 1600% magnified computer screen (corresponding to the interval 1950-1989), for Figs. 4a, 5a & 6a, and made periodograms. (There was a place, where I found one less point in Figs. 4a & 5a, so interpolated what seemed, by comparison with the abscissas of Fig. 6a, likeliest to be the missing datum.)

Before making the periodograms, I detrended each data set, i.e., adjusted slope and mean to zero, using the average of the first three and average of the last three ordinates, to define the trendline. For Fig. 6a (resp. 4a, 5a), the periodogram peaks at 62 (resp. 68, 60) yr. Fig. 6a's periodogram amplitudes are smaller than Fig. 5a, whose in turn are smaller than Fig. 4a. Well away from the peaks, Figs. 5a & 6a conform well to white (1/f) noise. I used the periodogram amplitude of Fig. 6a at 100 yr, which seemed to be well away from the 62 yr peak and consistent with the amplitude at 10yr, to define the 1/f amplitude curve which I subtracted from Figs. 4a, 5a, & 6a.

Thereby filtering out the white noise, I found the filtered periodogram peaks for Fig. 6a (resp. 4a, 5a) to be 53.9 (resp. 61.5, 55.2) yr. For Figs. 5a & 6a, the signal was almost totally removed by this filtration, leaving only smooth, fairly narrow peaks of heights 0.0983" & 0.0265", resp.

The perturbation signal is a second harmonic; 53.9 yr, the detrended, de-noised periodogram peak for Standish's most highly corrected post-1950 data, corresponds to the net half-period of Uranus vis a vis a Planet X in circular orbit at 52.5 AU. Elsewhere on this messageboard I have discoursed at length on the theoretical and empirical importance of the distance, 52.6 AU.

The amplitude of the RA (essentially, ecliptic longitude) displacement found in Standish's Fig. 6a (resp. 5a), 1950-1989, is only 0.0265"(resp. 0.0983")/0.73" = 4% (resp. 13%) that theorized for Barbarossa (or Lowell's Planet X), but correcting for distance^3 = (52.5(resp. 49.7)/197.7)^3, still should have 2.4 (resp. 7.45) Earth masses.

For all the periodograms, the phase is incoherent. In Fig. 5a, for example, near the de-noised periodogram peak at 55yr, the phase steadily varies ~ 135deg, per year of change in the period. If the discrepancy were due to the tidal gravity of a Planet X, a small change in the period of the fitting sinusoid, would cause only a small change in the best-fitting phase. The periodogram result is consistent with a collection of objects or structures orbiting at ~ 52AU, each with its own phase, affecting Uranus likely nongravitationally. The squares of the interaction amplitudes are additive.

The white noise correction tailored to remove the 100yr & 10yr periodogram amplitudes of Fig. 6a, also removes the 200yr & ~ 10yr periodogram amplitudes of Fig. 5a (the 100yr amplitude of Fig. 5a seems not quite distinct from the 60yr peak); thus the amount of white noise in Figs. 5a & 6a is about the same. The difference between Figs. 5a & 6a, is Earth-based vs. space probe planetary mass determinations. With use of Earth-based determinations, the amplitude of the effect on Uranus' orbit, equals the amplitude (though not the phase, or any phase) of the effect that would be caused by another half of a Uranus (7.45*2 = 14.9 ~ Uranus' 14.6 Earthmasses) orbiting at the special distance, 52.6AU. Use of space probe determinations somehow eliminates most (perhaps, ideally, all) of the effect.


Reply author: nemesis
Replied on: 05/14/2008 09:50:42
Message:

Joe, if I'm following you correctly you're saying a planet of ~3.7 Earth masses at 197.7 AU would fit the Uranus perturbation data. A planet of that mass at that distance would be cold and very dim and would require no masking nebula. It could be the "green dot". Wouldn't it be more parsimonious to make that assumption?


Reply author: Joe Keller
Replied on: 05/14/2008 17:16:19
Message:

quote:
Originally posted by nemesis

Joe, if I'm following you correctly you're saying a planet of ~3.7 Earth masses at 197.7 AU would fit the Uranus perturbation data. A planet of that mass at that distance would be cold and very dim and would require no masking nebula. It could be the "green dot". Wouldn't it be more parsimonious to make that assumption?



This lower mass (my 3.7 became 2.4 Earth masses upon improved measurement of the graph, and revision of my simple dynamical theory above) applies to a planet at ~ 52AU, not 197.7AU. My idea now is, that the phase incoherence, in the signal remaining after white noise subtraction (in Standish's graphs, Figs. 5a & 6a, of Uranus' RA) is inconsistent with Newtonian perturbation by a planet (see previous post; this idea was appended to it, after "nemesis" posted his question).

The 2.4 Earth masses might become zero if Standish's space probe data were perfected. The ephemeris based on Earth-based planetary mass measurements (the basis of Fig. 5a) gives 7.45 Earth masses: almost exactly half the mass of Uranus itself. It's as if Uranus were interacting with itself, through the intermediation of some structure or barrier that orbits at ~ 52AU.

Though my periodograms don't confirm the evidence for Barbarossa that I saw grossly in Standish's Fig. 5a, Standish's analysis doesn't disprove Barbarossa's existence. If Standish's ephemeris is right, his predecessors' ephemerides were wrong; if Standish's predecessors were fallible, then Standish is, too. Also, the ~ 52AU, ~7.45 (~14.6/2!) Earthmass *incoherent* signal in Standish's Fig. 5a, which remains after detrending and de-noising, but is mostly or completely removed by use of space probe planetary mass measurements as in Fig. 6a, indicates new physics.

With new physics, the only way to prove anything is to look, not theorize. My $100 reward offer, previously detailed on this messageboard, still stands. The reward is merely for getting someone really sufficiently equipped, to look definitively; not for looking oneself. Previously on this thread I've listed many reasons for looking. If a dense body isn't there, then likely something just as interesting is.

Regarding *Neptune's* ephemeris, Standish's Fig. 8a remarks that the ephemeris of USNO Publications, Vol. XII, agrees with Standish's, for Neptune's RA, only between 1830 & 1950. Before 1830 & again after 1950, for whatever reasons, the USNO Vol. XII ephemeris gives +20"/century greater RA for Neptune than Standish's does. Pursuant to my recent posts, Barbarossa should cause *Uranus'* time derivative of RA to deviate +/- 0.73"/(84/2/4)*pi/2*100 = +/- 10.92"/century. Tidal acceleration goes as r, time before reversal goes as r^1.5, and subtended angle goes as 1/r. So, Barbarossa should cause *Neptune's* time derivative of RA to deviate +/- 10.92 * (30.07/19.18)^1.5 = +/- 21.44"/century. This is the same deviation of Neptune's time derivative of RA, that by whatever complicated calculation, is seen in USNO Vol. XII vs. Standish.


Reply author: nemesis
Replied on: 05/14/2008 17:53:01
Message:

Sorry, I misunderstood... but wouldn't it be worthwhile to look for a planet of ~4 Earth masses at ~52 AU? It would still be very dim and could have been overlooked. And it shouldn't rule out Barbarossa.


Reply author: Joe Keller
Replied on: 05/15/2008 15:15:42
Message:


quote:
Originally posted by nemesis

Sorry, I misunderstood... but wouldn't it be worthwhile to look for a planet of ~4 Earth masses at ~52 AU? It would still be very dim and could have been overlooked. And it shouldn't rule out Barbarossa.



Thanks for emphasizing this. Barbarossa would be the dimmer, because (198/52)^4 = 250 = 6 magnitudes, not enough to compensate for the size difference between a Uranus (i.e., 5-15 Earthmass planet, even if of Earthlike density) and a Jupiter (i.e., quantum mechanical estimate of brown dwarf size). Unfortunately, the phase incoherence of the residual signal I find in Standish's data, leaves the phase indeterminate; furthermore it argues against a planetary cause at all.

The USNO Vol. XII (1929, according to the USNO library webpage) ephemeris for Neptune's RA (according to Standish, Fig. 8a) is consistent with Barbarossa in amplitude, period and phase. The USNO mission is to produce accurate ephemerides by any and all lawful means. Before radio, a cruiser might lose all hands, if the navigator lacked accurate information on the only 3rd mag. star he could identify on a partly cloudy night. Tomorrow, after electronic countermeasures, and the electromagnetic side effects of superweapons such as nuclear bombs, the navigator may well revert to the sextant. I doubt he'd sight Neptune, but if the equivalent of a thousand taxpayer working lives are wasted because the Navy's rocket to Neptune missed, the loss is, in a way, comparable to that of a cruiser.

Fig. 8a (Neptune) resembles the error in fitting 1.5 cycles (trough-peak-trough-peak) of an 80-yr sinusoid, to a cubic curve. For the 120 yr between 1830 & 1950 (Neptune was discovered in 1846, but Neptune's ephemeris can be extrapolated backward) the fit is good, but before and after that, the ephemeris for Neptune's RA deviates roughly as a cubic curve from a sinusoid. The USNO might have tried to adjust for some small unexplained error, by zeroing the discrepant RA, with a cubic curve, at 1930 (approx. date of the catalog), 1850 (approx. beginning of Neptune data) & 1890 (midpoint)(the three roots of the sinusoid). This attempt would begin to fail markedly a quarter cycle away, before 1830 and after 1950, as Fig. 8a shows.

James DeMeo had to fly to Cleveland to rummage through the Physics Dept. at Case Western Reserve Univ. (after tactfully getting permission) to rediscover the only extant copy of Dayton Miller's data, in a cardboard box in a dusty storeroom. Everyone involved with USNO Pubs. Vol. XII is deceased. If notes exist detailing all their corrections, finding them might be like rediscovering something in the Vatican Library. The USNO mission was and is, to produce accurate ephemerides. They understood that there are systematic observation errors, calculation errors, undiscovered planets, unknown forces and unimagined "new physics". A cubic curve might be used to correct a small, incomprehensible discrepancy.

The slope of the discrepancy (after inclusion of such a cubic correction term) in Fig. 8a, at 1830 and 1950 (a sinusoid trough & peak, resp.) would approximate the maximum sinusoid slope due to Barbarossa, i.e., +21.44"/century (see my earlier post).

Not only are the period and amplitude of the unexplained sinusoid (the 1929 USNO approximation of which by a cubic, results in Standish's Fig. 8a) consistent with Barbarossa; so is the phase. One of the times at which the sinusoid should have most negative slope, is c. (1830+1950)/2 = 1890. In 1890.0, I find Neptune's heliocentric ecliptic longitude was 63.5. Barbarossa's effective heliocentric ecliptic longitude was: 173.327 (for 1987.082) - (1987.082-1890.0)/2780*360 (for orbital motion) - 175/4/2780*360 (for retardation of speed change, by a Neptune-minus-Barbarossa quarter cycle vis a vis the tidal acceleration due to Barbarossa) = 155.1. Recalling the 2nd harmonic nature of the tidal force, Neptune's RA should have most positive acceleration at 155.1-45, most negative acceleration at 155.1-135, and zero acceleration, i.e., most negative speed, at 155.1-90=65.1, agreeing perfectly, considering the rough dates (1.6deg < 1yr), with the actual 63.5.

There are many reasons why Standish might find an ephemeris which excludes Barbarossa. However, the 1929 USNO Vol. XII ephemeris, supports Barbarossa's existence, indirectly but accurately confirming the amplitude, period and phase of Barbarossa's predicted tidal effect on Neptune.


Reply author: Joe Keller
Replied on: 05/15/2008 16:38:46
Message:

(sent three minutes ago via JPL website email form)

Attn.: Dr. P. C. Liewer, Jet Propulsion Laboratory (pls. forward)

To: Dr. E. Myles Standish (if active) or his successor (otherwise)

There is an interesting comment that has been made regarding Dr. E. Myles Standish's ephemeris. Apparently this comment only has been published on Dr. Tom Van Flandern's online messageboard (attached to Dr. Van Flandern's website, www.metaresearch.org) by Joe Keller, date today, May 15, 2008.

Sincerely,
Joseph C. Keller
B.A., cumlaude, Harvard, Mathematics, 1977


Update May 31, 2008: no response of any kind so far, from anyone.


Reply author: Joe Keller
Replied on: 05/16/2008 22:11:43
Message:

Planet X = Barbarossa : More Dynamical Evidence

Maran et al, Planetary & Space Science 45:1037-1043, 1997, solving differential equations of motion numerically using large amounts of computer time, found the nearest distance at which a Neptune-mass Planet X, could orbit without destabilizing (causing large increases in major axis within 10^6 yr) known Trans-Neptunian Objects (TNOs). For a circular orbit of i=0 or 5deg inclination, the Planet X distance for which all 14 studied TNOs become stable, is ~62.5AU according to Maran, Table 6, p. 1043. For i=10deg, it's ~58.75AU, and for i=45deg, ~54AU. Interpolating to Barbarossa's i=12deg10' (curiously, the same inclination as "Vulcan") gives 57.125AU.

Maran studied real TNOs with nonzero eccentricity & inclination. To compare Maran's Planet X to Barbarossa, in its effect on a TNO that ranks in the 93rd percentile (i.e., 1 in 14) for vulnerability to destabilization, let's consider a hypothetical TNO with zero eccentricity & inclination, but semimajor axis 45AU instead of the usual 42-43AU for a classical TNO (or 39.44AU for a plutino). As in posts above, the tidal effect is estimated by multiplying the tangential tidal force at quadrature, by the sine of the angle subtended at the sun at quadrature:

M/R^2*(sec^2(theta)-cos(theta))*cos(theta) where sin(theta)=r/R

If this effect is same for Barbarossa as for Maran's hypothetical Planet X (i.e., just enough to destabilize a 93rd percentile, 1 in 14, vulnerable TNO) then Barbarossa must have 3253 Earth masses, agreeing with the 3430 Earth masses I've attributed to Barbarossa by assuming orbital precession resonances in the outer solar system.

Summarizing, Maran's 1997 computation-heavy study of 14 TNOs, found the distance at which a Neptune-mass Planet X in circular orbit, at Barbarossa's inclination, would cause the stability limits of the Edgeworth-Kuiper belt, to be what they are. Using an elementary geometric estimate, I found that a Barbarossa-mass Planet X, in circular orbit at Barbarossa's distance and inclination, also would produce the observed limits of the Edgeworth-Kuiper belt.


Reply author: Joe Keller
Replied on: 05/16/2008 22:56:25
Message:

(sent three minutes ago to email address listed on univ. website)

Attn.: Prof. J. P. Emerson, Queen Mary/University of London (please forward)

To: Dr. M. D. Maran

Yesterday I discovered your article, "Limitations on the Existence of a Tenth Planet". Your heavy computation found one point on the mass-radius curve constraining any possible Planet X. The actual Planet X, for which there is much evidence, lies elsewhere on this curve.

I've put this evidence on the messageboard of Dr. Tom Van Flandern's website, www.metaresearch.org, under my name, Joe Keller. Today I added a discussion of your article.

Sincerely,
Joseph C. Keller, M. D.
B. A., Harvard, cumlaude, Mathematics, 1977

Update May 31, 2008: no response of any kind so far, from anyone.


Reply author: Joe Keller
Replied on: 05/18/2008 21:38:18
Message:

Harrington Corroborated Lowell

RS Harrington, Astronomical Journal 96:1476+, 1988, made a Planet X prediction comparable to Lowell's. A post above, has details of my interpretation of Lowell's final prediction, with a reference to Lowell's final description of his work. Harrington's trial orbits ranged in semimajor axis from 30 to 80 AU, vs. Lowell's two extreme apse cases whose semimajor axis averaged 44AU. Harrington's trial perturbations didn't improve Neptune's fit much, so, like Lowell, he relied on Uranus; but Harrington's Uranus data spanned 1833-1988, vs. 1715-1914 for Lowell's.

In Figs. 1 & 2, p. 1477, Harrington's densest clusters of best fit Planet X positions, are for 1988 at RA 16h50m Decl -30; and for 1930 at RA 14h30m Decl -26. My approx. slide rule extrapolation of these along a constant speed great circle to 1914, is RA 208 Decl -23. Lowell's final prediction amounted to ecliptic longitude 202, ecl. latitude small and indeterminate; if ecliptic latitude zero, this would be RA 200 Decl -9.


Reply author: Joe Keller
Replied on: 05/20/2008 14:59:46
Message:

Pulsar Timing Confirms Dayton Miller's Ether Drift Axis


Data. I included all the millisecond pulsars (MSPs) from Taylor's 1995 catalog (online, VizieR) and from the current (2008) online ATNF pulsar catalog, that met these criteria:

1. P < 30ms.
2. Pdot/P < 8*10^(-18) (cgs units) (This removed six MSPs which formed a second modal peak in Pdot/P; three of these were obvious statistical outliers vis a vis Pdot.)
3. Proper Motion given; and, not "0" in Taylor's catalog (to avoid difficult or conflicting measurements).
4. Not aligned with any globular cluster (according to my check vs. Bica's 2006 globular cluster catalog on VizieR).

All MSPs had distances given. All Pdot/P were positive. For the five acceptable MSPs in Taylor's catalog I used Taylor's values. Seventeen more came from the ATNF catalog, for N=22 total.


Methods. Except for the farthest pulsars, theoretical acceleration due to rotation of the Milky Way is small, and I neglected it in the main study. I did use the Shklovsky correction term (to correct acceleration for transverse motion)(see Taylor, ApJ 411:674, 1993; Zakamska & Tremaine, AJ 130:1939, 2005). I found that Pdot/P, thus corrected for transverse motion, correlates with the cosine of the angle formed between the line of sight to the pulsar, and the ether drift vector identified by Miller in his 1933 Reviews of Modern Physics article. That is, looking toward the direction which Miller thought our solar system to be moving through the ether, one observes greater Pdot/P, as if pulsars in that direction are accelerating away from us. This suggests that our solar system is braking relative to the ether.

Experimenting with various linear combinations of the raw Pdot/P term and the Shklovsky term, on various subsamples, discovered a better correlation, when the Shklovsky term is multiplied by about -1/2 (but the difference in correlation, vs. that for -1/3, is tiny), and then the entire corrected Pdot/P is multiplied by -1 for Pdot/P < 2.43*10^(-18) (s/s)/s (N=10) and multiplied by +1 for Pdot/P > 2.43*10^(-18) /s. I call this overall +/- 1 factor, the "reversal" option: small positive Pdot/P really is negative, when corrected for the Hubble inflation of the universe; upstream in the ether, big Pdot/P looks bigger and small (really negative, when corrected for inflation) Pdot/P looks smaller (really, when corrected, bigger negative). For this small sample, all coefficients between 2.2 & 2.6, inclusive, are equivalent, but I write 2.43, because 2.43*10^(-18)(cm/s)/cm, i.e., /s, equals 75 (km/s)/Mpc, a recent determination of the Hubble constant from a meta-analysis of all the most accurate determinations.

There is a theory for the choice, - 1/3; see below. I call this use of a -1/3 factor in the Shklovsky term, the "bent ether" option: the apparent transverse velocity might really be due to a lightpath along a fiber of ether which usually is progressively unbending; something is added to Pdot/P to compensate for this pseudo-acceleration toward the observer.


Results. Correlation coefficient, normalized Fisher z, & direction:

usual "etherless" Shklovsky correction: cc=0.397, N=22, z=1.83, p=0.034 (one-tailed), (l,b)=(301,-8) (0.1 radian search grid)

"Reversal + bent ether, -1/2 factor" option: cc=0.482, N=22, z=2.29, p=0.011 (one-tailed), (l,b)=(305.4,-25.6) (0.004 radian local search grid)

Dayton Miller's ether drift: (l,b)=(282.0,-35.2)


Discussion. The "reversal + bent ether" option gives considerably better correlation, and also its best axis is closer to Miller's ether drift (25deg away). In the "bent ether" option, the pathlength increment isn't plus theta^2 / 2, as in textbook optics; it's minus theta^2 / 6, as for a slightly curved line that unbends. Two of the three Taylor catalog MSPs disqualified for excessive Pdot/P (those with the 2nd & 3rd biggest Pdot/P of the three, but not with obvious statistical outlier Pdot, and also among the nearest MSPs to Earth) have precisely equal Pdot/P, 1.11497*10^(-17) in cgs units, when the (-1/3 or - 1/2) factor by which the transverse velocity correction term is multiplied in my "bent ether" option, is changed to very nearly (-1/5) (precisely, -0.1988). Thus the MSPs, seem to correspond to the cases of a slowly straightening quadratic or cubic planar curve, whose end tangents lie on right circular cones whose altitudes are the line between the ends, and whose vertices are the endpoints, of the line of sight. As one cone rotates with respect to the other, the curve's tangents at its endpoints, are coplanar in two cases: one approximable by a quadratic, and one by a cubic, Maclaurin polynomial.


Preliminary finding. I was led to the foregoing, by a preliminary finding based on an exact solution for Taylor's five included MSPs only. One of these had Pdot/P = 3.3*10^(-18); the other four all had Pdot/P < 2.5*10^(-18). Adjustment of the factor by which the Shklovsky term is multiplied, and fitting to the best axis, amounts to adjusting three parameters, which often will suffice to fit five points to perfect correlation. A factor of exactly -1/3, gave the perfect fit, a constant plus cosine. The factor became exactly -1/3, when 20% of the Milky Way rotational acceleration correction term was included. This confirms that visible matter and only visible matter (99% stars) produces the real Milky Way gravity. As I've discussed on this messageboard, and also in my article submitted Feb. 2002 to Aircraft Engineering & Aerospace Technology, most of Oort's law and the like, is ether effect, not really due to motion.

For these five, the unique perfectly fitting axis of most *negative* (least positive) corrected Pdot/P, is galactic coords (l,b)=(276.3,-22.6). Though the sign of the effect on Pdot/P is reversed, this is only ~30deg from the axis found for the larger sample above. Dayton Miller's final ether drift coords (from a webpage of Dr. James deMeo, the geographer) were RA 4h54m Decl -70deg33', which assuming B1900.0 coords., are by Martindale's online coordinate transformation utility, (l,b)=(282.0,-35.2); this pulsar axis and Miller's axis differ only 13.5deg. (WW Campbell's approximate compromise solar motion antapex was RA 6h Decl -30, which is, again assuming B1900.0 coords, (l,b)=(236.2,-22.8).)

The maximum of this perfectly fitted cosine function, differs only 1% from the common Pdot/P value for the two abovementioned Taylor catalog pulsars which had been excluded because of big Pdot/P. (These were found to have equal Pdot/P when adjusted using a -1/5 factor, i.e., unbending cubic curve ether fiber, for the Shklovsky term.) One of these two excluded "cubic fiber" pulsars lies near the Miller axis and one near its antipode, but both exhibit big Pdot/P as if they lay at whichever end of the axis gives them the biggest Pdot/P.


Determination of Barbarossa acceleration. Barbarossa's 1983.2 position was (l,b)=(265.9,+48.4); the (preliminary, perfect fit to five Taylor pulsars) pulsar and Barbarossa axes differ 71.5deg. The (preliminary fit to five pulsars) pulsar axis roughly interpolates the Miller & Barbarossa axes along a great circle. Barbarossa's gravity would cause a cosine-dependent solar acceleration of (semi)amplitude a/c=0.5216 * 10^(-17) in cgs units. In those units, the five pulsars' cosine-dependent apparent acceleration has (semi)amplitude 1.775 * 10^(-17). If the pulsar axis resulted from the arithmetic sum of Barbarossa gravitational and Miller ether drift effects, the axes wouldn't add vectorially. The spherical harmonic addition formula (see, inter alia, Jahnke & Emde) would be needed. However, the underlying physics are largely unknown. There might be an underlying physical process which manifests accurately as vectorial addition:

1.775 - 0.5216 * cos71.5 = 1.609
0.5216 * sin71.5 / 1.609 = tan17

and 17 is close to the actual 13.5deg se


Reply author: Joe Keller
Replied on: 05/21/2008 19:19:51
Message:

You'll get an "A" on your term paper. I guarantee it.

The foregoing post would make a good term paper for a class like Statistics 101. I'll help you with it, if you'll tell the teacher that I helped you. If you don't get an "A", I'll talk to the teacher myself.

Taylor's 1995 pulsar catalog (online, "VizieR" website) has only about a page of millisecond (P < 0.025s) pulsars. Only a few remain after those lacking Pdot (the zero entry means no data), lacking Proper Motion (click on the VizieR line number to see), or aligned with globular clusters (check against the globular cluster catalog) are discarded.

Your paper can assess the effect on statistical significance, of discarding 3 of the 8 pulsars, of adjusting one parameter (the coefficient of the transverse motion "Shklovsky correction term"; Cowling, MNRAS 204:1237+, 1983), and of choosing the axis in space that gives the best correlation (8-3-1-2 = 2 d.o.f., and there's always one line through 2 points; but this isn't proof that the correlation isn't significant). The Monte Carlo method could be used to assess statistical significance: make up fictitious pulsars with data randomly chosen in the same range as the real pulsar data. Follow the same procedure as above, but fail to get the same level of significance. Repeat, say, 10 times. I haven't tried this myself, and would like to see what happens.

*******

The pulsar distances in Taylor's catalog are determined, at least usually, from signal dispersion with, at least to a considerable extent, the assumption of constant interstellar "free electron density" (Taylor, ApJ 411:674+, 1993). This unique method available for determining pulsar distance, might result in unique accuracy. The "free electron density" might really be constant or as good as constant, if it is involved with the phenomenon which Miller, Galaev, etc., have called the ether.

*********

I'll be busy for awhile rechecking Standish's claim about Neptune, using original USNO data I'm copying by hand from bound volumes. Would someone with access to such a library, email (through this messageboard) or mail me (Joe Keller, POB 9122, Ames, IA 50014 USA) copies of four pages from the 1905 American Ephemeris? (It's missing from ISU.)

Needed:

Feb. 1905: "THE SUN'S" "AT GREENWICH MEAN [not apparent] NOON" table with RA, Decl, etc., and probably on the other side of the sheet, "THE SUN'S" "AT GREENWICH MEAN [not apparent] NOON" table with Logarithm of Radius Vector, etc.

Dec. 1905: ".

Update May 31, 2008: no responses re ephemeris request; 5/29, I requested the volume by interlibrary loan.


Reply author: Joe Keller
Replied on: 05/22/2008 20:27:04
Message:

Plan for Even More Rebuttal of Standish's Claim

If I knew all the radii from the sun to Neptune, then the tabulated data would tell me all about Neptune's position, and I could check what net unknown acceleration Neptune underwent after known tidal accelerations (i.e., acceleration of Neptune minus acceleration of the sun) are subtracted. I'll consider a segment of Neptune's orbit for which the tangential tidal acceleration due to Barbarossa should be especially big.

I'll determine Neptune's radii from Hamilton's principle (numerically extremizing the integral of Lagrangian*dt with respect to variation of all the Neptune-sun radii), assuming that there is no unknown acceleration. If there really isn't any unknown acceleration, then this result also will extremize the integral of L*dt with respect to variation of all Neptune's celestial coordinates (which are known) (after extremization with respect to the unknown radii has been performed). The actual unknown acceleration will be that which causes the integral of L*dt to be extremized with respect to Neptune's celestial coordinates, as soon as it is extremized with respect to Neptune's radii.

In detail, I'll vary the radii at the endpoints of the orbit, over a generous range compatible with the ephemeris. For each pair of endpoint radii, I'll start with ephemeris radii at all the points, then vary them by adding trial half or whole Fourier sine terms (1/2, 1, 3/2, 2, 5/2, etc., sine waves in the interval) of 1/200,000 radius amplitude (analogous to 1 arcsec). For a given sine term, two such trials, plus the original, will allow a quadratic estimate of the coefficient needed for extremization. I can go through all the sines until the period is smaller than the gap between the two nearest points, then repeat the cycle until there is enough convergence. (I'll approximate the orbit with a cubic spline, to compute each action integral.)

This extremization effectively incorporates the scatter in the data. That is, an error in the coordinate data, will cause the extremal action integral to be less small, but the extremal curve remains, in a sense, the best approximation to the actual orbit.

Then, for the same pair of endpoint radii, I'll do the foregoing three more times, once assuming each theoretical component of Barbarossa's tidal influence (at the midpoint of the orbit segment). Each of the four results then can be tested by altering each coordinate point in turn by 1" in RA and in Decl, and evaluating each new action integral (again interpolating with cubic splines). Since Barbarossa's actual influence has been more or less omitted, there is no extremization with respect to the coordinates, and the measured changes in the action integrals are linear functions of the unknown vector influence.

So, the components of Barbarossa's influence can be estimated by minimizing, the sum of squares of the changes in the action integrals (extremized w.r.t. the intermediate radii) when each data coordinate is varied in turn. For convenience, the projection on the orbital plane, of Barbarossa's influence, can be plotted as a vector field w.r.t. the assumed endpoint radii. Also the endpoint radii found from various ephemerides can be marked on this vector field, to get error bars for the result.

Summarizing: the influence of Barbarossa is that which causes the orbit of Neptune, extremized w.r.t. radii according to Hamilton's principle assuming perfect accuracy of the data coordinates, to be also most nearly extremized w.r.t. the data coordinates.

Update May 31, 2008: Following essentially the above plan, I'm more than half done writing the program, 21 monitor screens-full of Basic code so far. When I'm done, I'll post the entire program to this messageboard.


Reply author: Joe Keller
Replied on: 06/09/2008 12:21:33
Message:

Here's the program I mentioned May 22 & 31 ("notepad" is useful for opening these files):


REM program name: USNO.BAS or USNONEPT.UNE
REM This program in QBasic, uses Hamilton's principle to search for
REM tidal forces acting on Neptune, based on season's widest from opposition
REM USNO 2d ser. vol. IX pt. 1 positions 1903-1911; n=14.
REM run time per i5,j5 step = 12min on Intel Pentium4 1.6 GHz (9 steps = 108min);
REM step = 3.6hr on IBM 486 (1994).

REM 1454 dbl precision vars in arrays, 65 dbl prec solo; 15 integer; 1534 tot
REM initialize consts; everything double precision
100 PRINT : PRINT : pi# = ATN(1) * 4: pi180# = pi# / 180: twelveinv# = 1 / 12
REM "clight" in AU/hr; other consts in Earthmass-AU-day units
REM clight & AU fr. Columbia Enc.
clight# = 3600 * 2.99792458# / 1496.0497#
REM grav fr. 2006 CODATA; masses fr. TP Snow 1983
grav# = 6.67428 * 5.974 * 10 ^ (-8 + 27 - 3 * 13 + 2 * (1 + 3)) * (2.4 * 3.6) ^ 2 / 1.4960497# ^ 3
REM include Merc,Venus,Mars,asteroids & centaurs with sunmass
n = 14: DIM xyz(6, n, 3) AS DOUBLE: DIM xcom(n, 3) AS DOUBLE
DIM ne(6, n, 4) AS DOUBLE: DIM kin(n) AS DOUBLE: DIM m(6) AS DOUBLE
DIM deriv(n, 4) AS DOUBLE: DIM delta(n, 4) AS DOUBLE
DIM qdel(n) AS DOUBLE: DIM qdelinv(n) AS DOUBLE: DIM denom(n) AS DOUBLE
DIM sq(n) AS DOUBLE: DIM diffsq(n) AS DOUBLE: DIM avedel(n) AS DOUBLE
REM more dbl prec var arrays def'd @ lines 210, 220, 240
m(1) = 17.2: m(2) = 332943 + .056 + .851 + .107 + .02
REM include 4,5 moons with Jup,Sat resp.; Luna with Earth
m(4) = 318.1 + (8.89 + 4.85 + 14.9 + 10.7) / 7.35 / 81
m(5) = 95.2 + (13.5 + .076 + .105 + .244 + .188) / 7.35 / 81
m(6) = 14.6: m(3) = 1 + 1 / 81: mm0# = m(2) + m(3) + m(4) + m(5) + m(6)
REM mm012# adjusts kinetic energy, to finite mass of sun et al
mm012# = 1 / (1 + m(1) / mm0#)
mbarbarossa# = 3430: tidebarbarossa# = grav# * mbarbarossa# / mm012# / 197.7 ^ 3
incrth# = 1 / 10 ^ .75 * pi180# / 3600: incrr# = incrth# * 30.07: icounter = 0

REM get times, RAs & Decls of Neptune from data bloc 2000
200 DIM celest(6, n, 3) AS DOUBLE: DIM time(n) AS DOUBLE
REM 1st entry in celest denotes resp. Nept,Sun,Earth,J,Sat,U
REM 3rd denotes resp. RA in rad, Decl, radius in AU
REM Nov. 16, 1858, noon = JD 2,400,000; time(i)=JD-2,400,000
juliannoon1900# = 46 + 365 * 41 + 10
FOR i = 1 TO n
READ a#, b#, c#, d#, e#, f#, g#, h#, ee#, ff#, gg#, hh#
time(i) = a# + juliannoon1900#
equcorr# = -.038 - (.066 - .038) * INT((d# + .1) / 1907)
celest(1, i, 1) = (e# * 15 + f# / 4 + (g# + h# + equcorr#) / 240) * pi180#
i1 = 1: IF ee# < 0 THEN i1 = -1
celest(1, i, 2) = (ee# + i1 * ff# / 60 + (i1 * gg# + hh#) / 3600) * pi180#
NEXT
gap142# = time(14) - time(2)
gap62# = time(6) - time(2): gap146# = gap142# - gap62#
gap102# = time(10) - time(2): gap1410# = gap142# - gap102#

REM get RAs & Decls, radii, derivatives of sun from data bloc 2100
210 DIM rasundot(n) AS DOUBLE: DIM declsundot(n) AS DOUBLE: DIM radiussundot(n) AS DOUBLE
FOR i = 1 TO n
READ a#, b#, c#, d#, e#, f#, r#, rasundot(i), declsundot(i), radiussundot(i)
celest(2, i, 1) = (a# * 15 + b# / 4 + c# / 240) * pi180#
i1 = 1: IF d# < 0 THEN i1 = -1
REM need to write, e.g., S 0deg 17', as -0.00000001, 17, etc.
celest(2, i, 2) = (d# + i1 * e# / 60 + i1 * f# / 3600) * pi180#
celest(2, i, 3) = 10 ^ (r# - 10)
NEXT

REM Neptune-sun dists
220 DIM nsd(n) AS DOUBLE: DIM nsd0(n) AS DOUBLE
REM roughly interpolated Naut. Alm. Neptune-sun dists, all Feb or early March
nsd0(2) = 10 ^ 1.475745#: nsd0(6) = 10 ^ 1.4760451#
nsd0(10) = 10 ^ 1.4764035#: nsd0(14) = 10 ^ 1.47661265#

REM Jup, Sat, Ur positions
GOSUB 3260
REM done reading data


REM heart of program
240 DIM acin(5, 5) AS DOUBLE: DIM actint(n, 2, 7) AS DOUBLE
DIM co(6, 7) AS DOUBLE: DIM hess(6) AS DOUBLE: DIM lambda(3) AS DOUBLE

FOR i5 = -1 TO 1

nsd(2) = nsd0(2) + i5 * incrr#

FOR j5 = -1 TO 1

nsd(14) = nsd0(14) + j5 * incrr#
FOR i = 1 TO 6
FOR j = 1 TO 7
co(i, j) = 0
NEXT: NEXT
PRINT i5, j5
FOR i6 = 0 TO 6
PRINT i6; " ";
REM i6>0 signifies entry in Hessian for tidal potential
FOR i7 = 1 TO n
FOR j7 = 1 TO 2
REM i7 & j7 signify node and component of perturbation of observed coords
actint(i7, j7, i6) = 0
FOR i1 = 1 TO -2 STEP -3
celest(1, i7, j7) = celest(1, i7, j7) + incrth# * i1
actionint0# = 10 ^ 12
FOR i9 = -2 TO 2
nsd(6) = nsd0(6) + (i9 + (gap146# * i5 + gap62# * j5) / gap142#) * incrr#
FOR j9 = -2 TO 2
nsd(10) = nsd0(10) + (j9 + (gap1410# * i5 + gap102# * j5) / gap142#) * incrr#
REM Lagrange cubic interpol for nsd
GOSUB 1220
REM Nept-Earth dists
GOSUB 1240
REM 1-time aberration corr (sun coords, time fns, only)
IF icounter = 0 THEN GOSUB 3250
IF icounter = 0 THEN GOSUB 3600
icounter = 1
REM geocentric rect coords
GOSUB 1300
REM c.o.m. (excluding Nept) rect coords
GOSUB 1320
REM Neptunian ecliptic coords
GOSUB 1340
k1 = 1: k3 = 3
REM quadratic numerical differentiation
GOSUB 1500
REM calculate action integral, 1st-order Euler-Maclaurin summation
GOSUB 1550
acin(i9 + 3, j9 + 3) = actionint#
NEXT j9: NEXT i9
REM est actionintegral min quadratically
GOSUB 1600
actint(i7, j7, i6) = actint(i7, j7, i6) + actionint0# * (i1 + .25 * (ABS(i1) - i1))
NEXT i1
IF i6 > 0 THEN actint(i7, j7, i6) = actint(i7, j7, i6) - actint(i7, j7, 0)
celest(1, i7, j7) = celest(1, i7, j7) + incrth#
NEXT j7: NEXT i7
NEXT i6
FOR i7 = 1 TO n
FOR j7 = 1 TO 2
FOR i6 = 1 TO 6
co(i6, 7) = co(i6, 7) - actint(i7, j7, 0) * actint(i7, j7, i6)
FOR j = 1 TO 6
co(i6, j) = co(i6, j) + actint(i7, j7, j) * actint(i7, j7, i6)
NEXT j: NEXT i6: NEXT: NEXT
REM solve 6x6 linear syst
GOSUB 1800
REM diagonalize 3x3 Hessian matrix
GOSUB 1850
NEXT j5: NEXT i5

990 END

REM Lagrange cubic to find rest of n=14 from 2d,6th,10th,14th
1220 FOR i = 1 TO n
IF i = 2 OR i = 6 OR i = 10 OR i = 14 GOTO 1228
nsd(i) = 0
FOR j = 2 TO 14 STEP 4
u# = 1
FOR k = 2 TO 14 STEP 4
IF k = j GOTO 1224
u# = u# * (time(i) - time(k)) / (time(j) - time(k))
1224 NEXT k
nsd(i) = nsd(i) + u# * nsd(j)
NEXT j
1228 NEXT i
RETURN

REM law of cosines
1230 bb# = celest(2, i, 2): dd# = celest(2, i, 1): r2# = celest(2, i, 3)
cs# = SIN(aa#) * SIN(bb#) + COS(aa#) * COS(bb#) * COS(cc# - dd#)
REM r1^2 -r2*cs*r1+r2^2-r3^2=0; use quadratic eqn
r1# = .5 * (r2# * cs# + SQR((r2# * cs#) ^ 2 - 4 * (r2# ^ 2 - r3# ^ 2)))
RETURN

REM Neptune-Earth dists
1240 FOR i = 1 TO n
aa# = celest(1, i, 2): cc# = celest(1, i, 1): r3# = nsd(i)
GOSUB 1230
celest(1, i, 3) = r1#
NEXT
RETURN

REM convert to geocentric rect. coords.
1300 FOR j = 1 TO 6
IF j = 3 GOTO 1308
FOR i = 1 TO n
cs# = COS(celest(j, i, 2))
xyz(j, i, 3) = SIN(celest(j, i, 2)) * celest(j, i, 3)
xyz(j, i, 1) = cs# * COS(celest(j, i, 1)) * celest(j, i, 3)
xyz(j, i, 2) = cs# * SIN(celest(j, i, 1)) * celest(j, i, 3)
NEXT i
1308 NEXT j
RETURN

REM convert to c.o.m. (excluding Nept) rect. coords.
1320 FOR i = 1 TO n
xcom(i, 1) = 0: xcom(i, 2) = 0: xcom(i, 3) = 0
FOR j = 2 TO 6
IF j = 3 GOTO 1326
FOR k = 1 TO 3
xcom(i, k) = xcom(i, k) + m(j) * xyz(j, i, k)
NEXT k
1326 NEXT j
FOR k = 1 TO 3
xcom(i, k) = xcom(i, k) / mm0#
NEXT k
NEXT i
FOR i = 1 TO n
xyz(3, i, 1) = 0: xyz(3, i, 2) = 0: xyz(3, i, 3) = 0
FOR j = 1 TO 6
FOR k = 1 TO 3
xyz(j, i, k) = xyz(j, i, k) - xcom(i, k)
NEXT: NEXT: NEXT
RETURN

REM convert to Neptunian ecliptic coords; 1st time pt. = 0 long.
REM 1st & nth def. plane; c.o.m. (excluding Nept) is ctr
REM 4th dimension of "ne" is Lagrangian, not time
1340 a1# = xyz(1, 1, 1): a2# = xyz(1, 1, 2): a3# = xyz(1, 1, 3)
b1# = xyz(1, n, 1): b2# = xyz(1, n, 2): b3# = xyz(1, n, 3)
maginv# = 1 / SQR(a1# ^ 2 + a2# ^ 2 + a3# ^ 2)
a1# = a1# * maginv#: a2# = a2# * maginv#: a3# = a3# * maginv#
GOSUB 1350
b1# = c1#: b2# = c2#: b3# = c3#
GOSUB 1350
c1# = -c1#: c2# = -c2#: c3# = -c3#
FOR j = 1 TO 6
FOR i = 1 TO n
x# = xyz(j, i, 1) * a1# + xyz(j, i, 2) * a2# + xyz(j, i, 3) * a3#
y# = xyz(j, i, 1) * c1# + xyz(j, i, 2) * c2# + xyz(j, i,


Reply author: Joe Keller
Replied on: 06/09/2008 14:54:36
Message:

1903-1911 USNO Neptune Observations Support Barbarossa

Standish reported that Uranus positions increased in accuracy in about 1911. Lowell used Uranus data from until about 1914. Thus these 1903-1911 USNO Neptune data, which I've found in the Iowa State Univ. Library, have the advantage of being no more accurate than Lowell's Uranus data. So, if Barbarossa is implied by these data, then Lowell might well have correctly deduced Barbarossa (or rather, the roughly gravitationally equivalent, smaller but nearer, Planet X) from his Uranus data.

These data are from original reports, not summaries possibly degraded by omitting details. The data are homogeneous in their method of collection and presentation.

So far, I've used only the first and last USNO observations of each opposition season; the greatest Earth parallax gives the most accurate triangulation of position. I used March 2.3, 1906, instead of March 6.3, because the other last-of-season oppositions were in Feb. Other things being equal, I planned to favor observations by observers whose personal corrections were given, but as I recall, this didn't become an issue. There were no observations from the 1906-1907 season, so late 1903 - early 1911 encompassed (8-1)*2=14 first- & last-of-season observations.

The data table had small footnotes telling the user to make observer and equinox corrections, which I looked up in the lengthy introduction. I wonder if these were on the magnetic tape from which Standish says he got his USNO Neptune, etc., data.

The 1905 ephemeris is missing from the ISU library, so my planetary positions (other than Earth, which easily could be interpolated, & Neptune) for that year are especially rough. This book is still on interlibrary loan order; when it comes, I can improve my 1905 data.

My program assumes that the center of mass of the known solar system w.r.t. Barbarossa, is unaffected by the variations in Neptune's trajectory. This makes the potential term tidal, i.e., quadratic in the sun-to-Neptune (more precisely, rest-of-known-solar-system-c.o.m.-to-Neptune) distance, and described by a Hessian quadratic form. The six independent entries of the Hessian matrix are scaled so that the perfect Barbarossa effect (in the limit where the ratio of Neptune's to Barbarossa's orbital radius, is zero) is a "1" and two "-1/2"s, on the diagonal (of the eventually diagonalized Hessian), and all other entries "0". These entries were optimized by a 6x6 system of linear equations, to minimize the sum of squares of derivatives (more precisely, central first differences) of the minimum Hamilton's principle action integral, w.r.t. alterations of the 14x2=28 observed celestial coordinates. (This minimum action integral is minimized w.r.t. variations of the Neptune-sun radius at two midpoints, with cubic or at least quadratic interpolation; in Neptunian ecliptic coordinates so second time derivatives, of coordinates, are small.)

The implied average rounding error of the USNO Neptune Declinations is 0.025", of the RAs 0.0375", and of the Nautical Almanac Neptune-sun distances, 0.25/10^7 Briggs log unit which is to the radius as 0.012" is to a radian, i.e., equivalent to 0.012". (The error due to Triton's displacement of Neptune is at most 0.01".)

I tried varying two midcourse radii (for action integral minimization) and observed coordinates (for finding which extraneous tidal force minimized the constraint they imposed) in step sizes ranging from 0.1" to 2", or the equivalent (and twice that, for the midcourse radii). The 1.2" and 2" step sizes, seemingly implied a much bigger extraneous tidal force than did the smaller 0.1", 0.1778" = 1"/10^0.75, 0.3162"=1"/10^0.5, 0.5623"=1"/10^0.25, 1.0", 1.05", 1.1", 1.15", or 1.17" step sizes.

Standish (p. 2001, 1st col.) says Uranus observations before 1911 have 1.2" "scatter"; presumably the scatter of Neptune observations is the same. This is exactly the step size, at and above which, my variational method diverges to unbelievably big characteristic values.

The step size which minimized the magnitude of the matrix trace (ideally +1-1/2-1/2=0), was 1.05", which gave characterisitic values +0.4918, +0.0076, and -0.3821. This and all smaller step sizes, yielded roughly the same characteristic vector, for the largest-magnitude (which also was the most positive) characteristic value. This vector, with a rough correction for Neptune's displacement from the sun, corresponds to RA 210.9 Decl -36.6 (modulo 180). With 1.1" and larger step sizes, the characteristic vector began to fluctuate much more.

[Update June 30, 2008: I got the missing 1905 Nautical Almanac on interlibrary loan from BYU via the Bertha Bartlett library in Story City. The improved data are appended to the USNO.BAS program above. Rerunning the above USNO.BAS program after correcting the sun, Jupiter, Saturn & Uranus positions for 1905, I find, for the stepsizes 0.1", 1"/10^0.75, 1"/10^0.5, 1"/10^0.25, 1.0", 1.05", 1.1", 1.15", & 1.2" (a subset of those above) that again the magnitude of the trace is minimized by the 1.05" stepsize. Again, the magnitude of the characteristic values for the 1.2" stepsize, is much bigger than for smaller stepsizes; and somewhat as before, the direction of the characteristic vector corresponding to Barbarossa's position, fluctuates more for stepsize 1.15" and above. Stepsize 1.05" gave characteristic values +0.5632, +0.0256, and -0.5572. The characteristic direction corresponding to the large (+) characteristic value, is (with the same rough correction for Neptune's orbital radius) RA 213.5 Decl -40.8 (modulo 180).]

[Update July 2: The result oscillates rapidly as a function of stepsize. Stepsize 1.050001" gives trace +0.9, bigger than the usual trace range +0.3 to +0.5; 1.051" gives very big characteristic values; 1.049" gives a small trace like 1.050", but the trace doesn't interpolate, even between 1.049" & 1.050". Generally however the characteristic vector corresponding to the large (+) characteristic value, almost always lies in the same octant (modulo 180), and the characteristic vector obtained with stepsize 1.04", RA 236.2 Decl -34.6, differs only about 17 deg (mod 180) from that obtained with stepsize 1.05". When variation of the Neptune radii is omitted altogether, 3 of 3 stepsizes tried, give very big characteristic values.]

[Update July 3: With the 1.05" stepsize, the other 8 endpoint radii choices in the program, i.e., ~(-1, 0, or +1)/200000 of the contemporary ephemeris Neptune radius, failed to give a trace any closer to zero. Despite double precision, this program suffers greatly from rounding error and numerical instability, but it is nevertheless remarkable that the stepsize runs giving tidal force Hessians in best agreement with Laplace's equation (i.e., zero trace) also point toward the direction given by Lowell and Harrington.]

Harrington's best 1988 and 1930 positions (see my post above, discussing Harrington's agreement with Lowell), extrapolated linearly to 1910.5 (the midpoint of Harrington's Uranus data, on which he relied) give RA 206, Decl -25. The midpoint of my 1903-1911 data is actually about 1907.5, so +3.0 years of Barbarossa's presumed orbital motion would change my [updated] best direction to RA 213.8, Decl -41.0 for 1910.5. Thus Harrington's Planet X direction (relying on Uranus data) essentially differs only 17 deg (p=0.044, for modulo 180) from the direction I determine variationally from 14 USNO Neptune observations 1903-1911.


Reply author: Joe Keller
Replied on: 06/14/2008 21:10:43
Message:

Posted (in essence) here May 15, 2008, by Joe Keller:

...
Fig. 8a (Neptune) [Standish, Astronomical Journal, 1993] resembles the error in fitting 1.5 cycles (trough-peak-trough-peak) of an 80-yr sinusoid, to a cubic curve. For the 120 yr between 1830 & 1950 (Neptune was discovered in 1846, but Neptune's ephemeris can be extrapolated backward) the fit is good, but before and after that, the ephemeris for Neptune's RA deviates roughly as a cubic curve from a sinusoid. The USNO might have tried to adjust for some small unexplained error, by zeroing the discrepant RA, with a cubic curve, at 1930 (approx. date of the catalog), 1850 (approx. beginning of Neptune data) & 1890 (midpoint)(the three roots of the sinusoid). This attempt would begin to fail markedly a quarter cycle away, before 1830 and after 1950, as Fig. 8a shows.
...

Today's comment:

The Vol. XII USNO Neptune ephemeris, as plotted by Standish, has a nonsinusoidal error. Real orbits are sums of sinusoidal functions (in a way, the epicycles of Ptolemy have evolved into Fourier terms). The Vol. XII error, on the other hand, resembles that due to use of a cubic polynomial term, and therefore must be the work of man. Polynomials often are used to enhance the accuracy of slowly varying constants such as precession, so it's likely that the (basically) cubic polynomial error in USNO Vol. XII, is due to the limited range of accuracy of a polynomial used to describe empirically, 1.5 cycles of an unexplained sinusoidal residual. This is appropriate despite its limitations, because the USNO's primary mission is accuracy by any means.

Newcomb's 1866 ephemeris, applied to 1866-1889 USNO Neptune data, apparently uses an extra second order harmonic to describe Barbarossa's tidal effect. The small residual arises from the difference between the period, 0.5*164.8=82.4 yr, of Newcomb's second order harmonic term, and the actual period of Barbarossa's tidal effect, 0.5/(1/164.8-1/2780)=87.59 yr. Yesterday and today I composed and ran the following program:


REM program name: USNONEWC.NEP
REM lang. QBasic
REM This program finds the (linear & quadratic) time & opposition dependence
REM of USNO 1866-1889 Neptune residuals vs. Newcomb's 1866 ephemeris.

PRINT : PRINT
pi = 4 * ATN(1): pi180 = pi / 180: n = 0: nn = 0
DIM sunra(12): PRINT "Read sun RA"
REM sunra is RA of sun in radians, on 15th of mo. in 1880, per Nautical Alm.
FOR i = 1 TO 12
READ a, b
REM datacheck
IF a < 0 OR a > 23 OR b < 0 OR b > 59 THEN GOSUB 3000
sunra(i) = pi180 * (a * 15 + b / 4)
NEXT
DIM nep(100, 4): DIM x(100): DIM y(100): PRINT "Read Neptune data"
REM nep's components are: time from 1866.0 in yrs; no. observations for mo.;
REM residual vs. Newcomb's ephemeris in " (ave. for mo.);
REM RA of Nept. in radians, at 1st observation of mo.;
REM RA of sun in rad, 15th of mo., per 1880 Naut. Alm.
FOR i = 1 TO 100
READ a, b, c, d, e
IF a = -1 GOTO 100
REM datacheck
IF a < 0 OR a > 23 OR b < 1 OR b > 12 OR d < -200 OR d > 400 THEN GOSUB 3000
IF c < 1 OR c > 15 OR e < 0 OR e > 240 THEN GOSUB 3000
n = n + 1: nn = nn + c
nep(i, 1) = a + (b - .5) / 12
nep(i, 2) = c
nep(i, 3) = d / 1000 * 15
nep(i, 4) = e / 4 * pi180 + pi - sunra(b)

REM adjusts for independent effect of distance from opposition
REM nep(i, 3) = nep(i, 3) + .285 * nep(i, 4) ^ 2

NEXT

REM find linear & quad effect, of distance from opposition
100 PRINT "n = "; n; "; nn = "; nn
FOR i = 1 TO n
x(i) = nep(i, 4): y(i) = nep(i, 3)
NEXT
PRINT : PRINT "Effect of distance from opposition"
GOSUB 1000

REM find lin & quad effect of time
200 FOR i = 1 TO n
x(i) = nep(i, 1): y(i) = nep(i, 3)
NEXT
PRINT : PRINT "Effect of time"
GOSUB 1000

REM find dependence of distance from opposition, on year of observation
300 FOR i = 1 TO n
x(i) = nep(i, 1)
y(i) = nep(i, 4)
REM for dependence of sq. of dist. fr. opp., on yr, use line below:
REM y(i) = y(i)^ 2
NEXT
PRINT : PRINT "Dependence of dist. from opposition, on time"
GOSUB 1000

END

REM finds corr. coeff. & best fit slope, vs. x & vs. x^2
1000 PRINT "Corr. coeff. & slope vs. x: ";
GOSUB 1100
FOR i = 1 TO n
y(i) = y(i) - slope * (x(i) - mx): x(i) = (x(i) - mx) ^ 2
NEXT
PRINT "Detrended corr. coeff. & slope vs. (x-mx)^2: ";
GOSUB 1100
RETURN

REM calc corr coeff & slope
1100 sx = 0: sy = 0: u = 0: v = 0: w = 0: m = 0
FOR i = 1 TO n
wt = nep(i, 2): sx = sx + x(i) * wt: sy = sy + y(i) * wt: m = m + wt
NEXT
mx = sx / m: my = sy / m
FOR i = 1 TO n
wt = nep(i, 2): w = w + (y(i) - my) * (x(i) - mx) * wt
u = u + (x(i) - mx) ^ 2 * wt: v = v + (y(i) - my) ^ 2 * wt
NEXT
cc = w / SQR(u * v): slope = cc * SQR(v / u)
PRINT "corr. coeff "; cc; " slope "; slope;
PRINT " mean x "; mx; " mean y "; my;
PRINT " Fisher's z = "; SQR(n - 3) * LOG((1 + cc) / (1 - cc))
RETURN

END

REM sun RA at Greenwich mean noon on 15th of month per 1880 Nautical Alm.
REM hr & min; nearest minute; zero signifies month not used
2000 DATA 19,47,0,0,0,0,0,0,0,0,0,0,7,41,9,41,11,35,13,24,15,25,17,34

REM from House of Reps. Misc. Documents v. 101, pp. B151-157;
REM USNO Neptune positions 1866-1889;
REM see main program for more explanation
2101 DATA 0,7,2,85,50
DATA 0,8,12,116,49
DATA 0,9,10,91,47
DATA 0,10,13,88,45
DATA 0,11,10,56,42
DATA 0,12,10,109,40
DATA 1,1,1,90,40
DATA 1,7,2,60,58
DATA 1,8,6,16.5,58
2110 DATA 1,9,10,22,56
DATA 1,10,14,36,53
DATA 1,11,4,67.5,49
DATA 1,12,7,19,48
DATA 2,8,4,11,66
DATA 2,9,7,70,64
DATA 2,10,9,64,62
DATA 2,11,12,47.5,59
DATA 2,12,6,65,57
DATA 3,1,2,80,56
2120 DATA 5,9,3,40,91
DATA 5,10,2,25,87
DATA 5,11,1,-30,83
DATA 5,12,2,-15,82
DATA 6,9,1,80,98
DATA 6,11,8,-101,93
DATA 7,10,1,20,105
DATA 7,11,8,54,101
DATA 7,12,1,-10,98
DATA 8,10,6,118,113
2130 DATA 8,11,7,84,111
DATA 8,12,9,37,108
DATA 9,1,2,115,107
DATA 9,9,5,158,125
DATA 9,10,5,90,123
DATA 9,11,11,125,120
DATA 9,12,8,182.5,117
DATA 10,9,2,95,134
DATA 10,10,8,124,131
DATA 10,11,4,172.5,128.5
2140 DATA 10,12,7,134,125
DATA 11,10,4,115,139
DATA 11,11,7,149,137
DATA 11,12,4,137.5,134
DATA 12,9,1,80,151
DATA 12,11,4,152.5,146
DATA 13,9,2,165,160
DATA 13,10,2,190,157
DATA 13,11,5,174,154
DATA 13,12,1,200,151
2150 DATA 14,1,2,170,150
DATA 14,11,1,260,162
DATA 14,12,8,287.5,161
DATA 15,10,3,177,175
DATA 15,11,4,267.5,172
DATA 15,12,8,190,170
DATA 16,1,1,200,167
DATA 16,11,1,240,181
DATA 17,9,1,260,195
DATA 17,11,5,260,192
2160 DATA 17,12,7,181,189
DATA 18,11,6,165,200
DATA 18,12,4,172.5,198
DATA 19,1,2,230,195
DATA 19,11,3,190,209
DATA 19,12,7,221,207
DATA 20,1,1,300,204
DATA 20,11,3,117,218
DATA 20,12,6,122,216
DATA 21,1,1,230,214
2170 DATA 21,11,7,221,229
DATA 21,12,6,248,226
DATA 22,1,5,248,222
DATA 22,12,4,280,235
DATA 23,1,3,280,232

DATA -1,0,0,0,0

END

REM datacheck
3000 PRINT "! incongruous data value at "; i
RETURN

END


Trying to read Newcomb's mind, I think he would have fitted the data with, essentially, an empirical series of sinusoids, all harmonics of the presumed Neptune period. Surely Newcomb would have admitted an ad hoc second order harmonic term, because of the possibility of yet another remaining distant outer planet, as LeVerrier already in 1846 publicly had suspected and as Todd in 1877 would predict and seek. Because of resonances, small undiscovered planets nearer than Neptune likely would give higher harmonics too. Newcomb's series wouldn't exactly equal the theoretical Newtonian perturbed elliptical orbit. I think Newcomb would have published the orbital elements of the ellipse which most resembled his series of sinusoids, while using the empirical series itself to calculate positions.

Above, I've estimated that Barbarossa's effect on Neptune's longitude would have period 87.59 yr (so, it isn't exactly a second harmonic in period), would cross the time axis downward in 1890, and would have (semi)amplitude 0.2144"/87.5/(2*p


Reply author: Joe Keller
Replied on: 06/15/2008 19:11:13
Message:

Barbarossa's tidal effect on Neptune's longitude, spans 5/4 cycle of a sine wave from 1846 to 1956 (zeros at 1846, 1890, 1934; peaks 1868 & 1956; trough 1912). Eckert et al published in 1951, but for a simple approximation, let's call that 1956. Eckert's purpose wasn't accurate prediction; his purpose was to find discrepancy, from that Neptune orbit which would be implied by known solar system bodies. So, Eckert mainly was adjusting position, period, eccentricity and perihelion, for best fit to observed longitude. With Neptune's small eccentricity, this amounts to adding some small constant, plus a small first order harmonic, to the function theta(time).

Such a fit would use the bottom half of a first order sine wave, intersecting the second order curve of Barbarossa's perturbation, in 1868, 1912, & 1956, and tangent to it in 1912. The maximum discrepancy would be only half the (semi)amplitude of Barbarossa's perturbation of Neptune. A small constant term could halve this again, to only 5.6"/4 = 1.4" max. The fit would worsen < 1868, but these data are sparser and less accurate anyway; in Parks Library at ISU, I found no USNO Neptune observations earlier than 1866, in any of the House or Senate documents.

Barbarossa's perturbation of Neptune has period ~ 88 yr, but really, Neptune's period is only 164.8, not 88*2. For an approximation, let's fix the 1912 point of the first order fitting sinusoid. The fitting sinusoid has not only about twice the period but also about twice the amplitude of the actual Barbarossa perturbation. Thus for the hypothetical case of no Barbarossa at all, the predicted 1795 (Lalande) residual is cos(360*(1912-1795)/164.8) * 2 * 5.62" - 1.405"(from the small constant term giving least max residual 1868-1956) - 5.62" (because the midline of the first order sinusoid is at the peak of the Barbarossa sinusoid) = -9.82". For the actual Barbarossa case, 1795's longitude is perturbed sin(360*(1890-1795)/87.59) * 5.62" = +2.85".

So in this approximation Eckert should actually find a longitude residual of

-9.82"+2.85" = -6.97"

and if the main approximation error is corrected, by raising the first order sinusoid 0.2144"/yr * (1956-1951) * 0.5 (because I've already split the difference to minimize the max error), we get

-6.97" - 0.54" = -7.51"

for 1795. Eckert did find (Rawlins, Astronomical Journal 75:856+, 1970, Table II; citing Eckert, Brouwer & Clemence, Astron. Papers Am. Ephemeris 12, 1951) -6.4", assuming a negligible mass (e.g., the currently accepted 0.002 Earth mass) for Pluto. Rawlins thought Eckert's figure should be adjusted to -8.6". So the effect of Barbarossa, after partial removal by best-fit orbital adjustment, equals the Lalande residual, as estimated by Eckert et al and by Rawlins.


Reply author: Joe Keller
Replied on: 06/15/2008 20:21:53
Message:

sample letter

(This is the first email letter I've sent about this subject, that got a response from a professional astronomer besides Dr. Van Flandern. The responder is a department chair, too.)


June 15, 2008

Dear Prof. *******,

I'm a Harvard math major who has compiled much evidence regarding "Planet X", which I've discovered and named Barbarossa. I've written a computer program which analyzes Neptune positions for tidal influence; the result is that the direction and tide due to Barbarossa are roughly what Lowell thought they were. I've analyzed residuals from Eckert's (1951), the USNO Vol. XII (1929) and Newcomb's (1866) ephemerides, and found that all of them seem to be due to one and the same Barbarossa, minus whatever ad hoc fitting corrections were used in the ephemeris.

The direction and tidal strength of Barbarossa are about the same as estimated by Todd, Lowell, or Harrington. I also calculate that Standish's correction to Neptune's mass is far too small to make such residuals go away, so, I don't know why Standish doesn't find them. Maybe Standish made the graduate student do it over and over until it came out zero.

Maran's numerical experiment on the stability of Trans-Neptunian Objects, also supports a Barbarossa of this tidal strength, as do simple outer solar system orbital precession resonances I discovered myself. Articles purporting to disprove "Planet X" (Barbarossa), have big holes in their arguments, and often explicitly acknowledge that they do.

The unexplained correlation of the multipoles of the "cosmic" microwave background, with the ecliptic, caused me, motivated by a novel physical theory of the CMB, to search the USNO-B catalog online, for excess inconsistent magnitudes near the (+) CMB dipole. These might be due to automated misidentification of Barbarossa and its moons, and stars, with one another. Later I realized that there is much evidence of a moving (i.e., intra-solar system) nebula in this direction. Also IRAS might have detected Barbarossa's bow shock.

I found disappearing dots on sky survey plates yielding an accurate and believable orbit. Amateur astronomers Joan Genebriera (Spain; 16", Tenerife), Steve Riley (U.S.A.; 8", S. California) and Robert Turner (England; automated 14", Tenerife) aiming at my coordinates, prospectively captured electronic images, presumably of Barbarossa or its moons, consistent with this ephemeris. Most of these images aren't starlike, but there might be novel reasons for that.

Working backwards from my own, through Tombaugh's, now through Adams or LeVerrier's methods, I'd be ready for Herschel's method of simply looking through a telescope, except that I can't afford the time or money (many say it's easy but few do it). The Barbarossa system, Barbarossa together with its moons (which include one identified and two inferred giant planets in their own right), seems to have 11 Jupiter masses. Barbarossa itself, with ~ 8 Jupiter masses, is near the theoretical boundary between cold brown dwarf and giant planet. At equilibrium temperature (at est. 197.7 AU) it could escape IRAS detection. The apparent magnitude is only +18 or +19; this is consistent with somewhat smaller than quantum-theoretical size, albedo in the low theoretical range (like the alkali metal "black smoker" brown dwarf) and/or an obscuring nebula.

Maybe your observatory would like to look for Barbarossa. Coordinates and other information have been posted by me on the messageboard of Dr. Tom Van Flandern, at www.metaresearch.org.

For more than a year, I've contacted hundreds of persons about this. The response of messageboards, except for Dr. Van Flandern's, has been either total silence, or, usually, one form or another of expulsion. The response of professional astronomers (except for Dr. Van Flandern, who does not agree with all my ideas, but who has suggested books and articles, and has formulated some hypothetical questions) has been silence. I've written (mail, not email) government officials and Congressmen; the only response there, has been a postcard saying, basically, I'd like to help you son but you don't vote in my district.

Sincerely,
Joseph C. Keller, M. D.
B. A., Harvard, cumlaude, 1977


Reply author: Joe Keller
Replied on: 06/18/2008 13:07:28
Message:

Above, I've calculated the orbit of Barbarossa assuming its mass is negligible compared to the sun's. The actual 0.0103::1 mass ratio alters the calculated Barbarossa-sun distance from 197.7 to 198.4 A.U. However, this alteration affects Barbarossa's coordinates by only ~ 2" or less.


Reply author: Joe Keller
Replied on: 06/21/2008 19:54:48
Message:

Mass-Distance Scaling

When an outer planet perturbs an inner one (both circular orbits) the amplitude of the 2nd harmonic term of tangential tidal force, is approximately proportional to the inner radius, r, for fixed R and M. So the 2nd harmonic amplitude determines r*M/R^3, and M is known if the outer radius, R, is, or vice versa. (Based on 2nd harmonic amplitude alone, Barbarossa could be 198.4/5=39.7 AU from the sun with mass 3430/5^3=27 Earth masses, but such a planet would be only a magnitude dimmer than Neptune.)

For r/R = 0.1, the 1st harmonic amplitude (of tidal force) is only 0.012x the 2nd; for r/R = 0.5, it's 0.066x. The halved frequency of the first harmonic helps the amplitude of displacement (force twice integrated) fourfold, so for r/R = 0.5, the 1st harmonic displacement is 4*0.066=0.264x the 2nd.

Suppose Lowell, Harrington, and Maran, each in their own way, essentially determined the first harmonic of perturbing force. For the first harmonic, the formula r*M/R^3=const., is drastically inaccurate. Above, I used instead M/R^2*f(r/R), where alpha=arcsin(r/R), and f is either

(sec^2-cos)(alpha) or (sec-cos^2)(alpha).

The former expression gives the tangential tidal force at quadrature; the latter expression multiplies this by the sine of the angle at the sun (at quadrature as seen from the perturbed planet). Either of these expressions will maintain approximate proportionality to the actual 1st harmonic amplitude as r varies. For the range r/R = 0.1 to 0.5, the error is never more than 10%. The Maclaurin series in r/R of these expressions are, resp., (3/2)*(r/R)^2+(9/8)*(r/R)^4+..., and (3/2)*(r/R)^2+(3/8)*(r/R)^4+... .

The Maclaurin series in r/R, for the actual convolution for the 1st harmonic amplitude, using the series of 1st derivatives of Legendre polynomials, is found to be proportional to (3/2)*(r/R)^2+0.94*(r/R)^4+... . As expected from these series, exact computation shows that the former trigonometric expression above, becomes (for r/R increasing from 0.1 to 0.5) 4% too big, and the latter 10% too small. Above, I've shown that either expression implies that the 1st harmonic amplitudes given to Uranus by Lowell's or Harrington's, and to TNO's by Maran's estimates, are about the same as those given by Barbarossa.

The longitudes of Lowell's & Harrington's estimates, and also of my variational estimate, agree with each other, but are greater than that of Barbarossa. This might be because these methods (certainly my variational method) include all unknown gravitational and possibly pseudogravitational forces, not only Barbarossa.

On the other hand, Eckert et al (with or without Rawlins' refinement) effectively restricted attention to the difference between the 2nd harmonic due to Barbarossa, and a best-fitting 1st harmonic approximating it. Eckert's and Rawlins' values constitute a +/- 1" range which, considering the slopes of both the sinusoids in 1795, confirms my position for Barbarossa, +/- 2.5deg longitude.


Reply author: Joe Keller
Replied on: 06/23/2008 19:35:09
Message:

The electron in a hydrogen atom has kinetic energy equal to 0.5*alpha^2 times its rest mass (i.e., rest energy), where alpha = 1/137.036, is the fine-structure constant. This constant is regarded as a "fundamental" physical constant. That is, its value is caused by some unknown, likely statistical, mechanism, which might recur for gravitational as for electrical fields.

Barbarossa's mean orbital potential energy happens to be -0.5*alpha^2, and its mean orbital kinetic energy 0.25*alpha^2, times the difference, between the potential energy at the surface of Barbarossa and the potential energy at the surface of the sun. Barbarossa itself apparently has somewhat less than 1/100 solar mass and theoretically about 1/10 solar diameter (or maybe less, which could account for its dimness). (The factor of 2 between this gravitational case, and the hydrogen atom case, might be from the gyromagnetic ratio g=2, or some other reason.) If the -0.5*alpha^2 (mean potential energy) formula is exact, then Barbarossa's semimajor axis should be

2*136.036^2*1392000km/2/149598000km /(1 - 0.00845*7/9/0.1) = 184.3 AU

using my best estimates of Barbarossa's individual mass (0.00845*7/9 solar masses) and diam (0.1 solar diameters).

Known distant (arbitrarily, > 100 AU, thus beyond the, somewhat disputed, "brown dwarf desert") brown dwarf companions roughly confirm the foregoing. Gleaned from the internet:

HD3651B, 480AU, 20-40 Jup mass, primary star is type K0V
AB Pictoris, 275 AU, 13.5 Jmass, primary K2V
GQ Lupi, 103 AU, 1-42 Jmass, primary K7eV
CD-33 7795, 100 AU, 20 Jmass, primary M1

These detected brown dwarf companions are those that are unusually massive and hot. My theory above predicts that if Barbarossa's mass were 0.04 solar masses (42 Jmass) its distance would be 287 AU.

If the Barbarossa system contained only Barbarossa and Frey (not the Freya and Lowell I've hypothesized, to explain Frey's precession and Barbarossa+Frey's ecliptic longitude discrepancy, resp.) then my best estimate of Barbarossa's individual mass in solar units would be 0.00845*0.8771=0.0074115=1/134.9 ~ 1/137. Maybe such is the usual mass for a star's main distant hyperjovian companion.

Recently I estimated the mass of the Barbarossa system as 0.0103 solar masses, based on hypothetical resonances of orbital precessions due to Barbarossa in the outer solar system. Earlier (see above letter to Lowell family, Aug. 30, 2007) I got practically the same estimate, 0.0104 solar masses, based on hypothetical period equality, of the Edgeworth Belt's typical orbital precession due to Barbarossa and the Edgeworth Belt's typical precession due to the known solar system (these estimates assumed Barbarossa's orbit is circular with radius 197.7 AU). My theory of the CMB temperature and dipole (also mentioned in my letter to the Lowell family) gives (recalculated June 25, 2008) 0.008763 solar masses for the Barbarossa system (carried to precision O(m^3), neglecting the known planets, and assuming circular orbit with radius 198.4 AU because of the reduced-mass correction).

The precession effects are functions of mass*mean(orbital radius^(-3)). On the other hand, because Barbarossa's position seems to be within a degree of the (+) CMB dipole, the dipole is a function of instantaneous orbital radius. The orbital energy is a function of the semimajor axis. (All my radii neglect the mass of the known planets, which alters Barbarossa's radius by less than 1 part in 2000, for a given period.)

Last year, I emailed Lowell Observatory, telling them that Barbarossa's position coincided, in heliocentric longitude, with the Jupiter/Saturn heliocentric conjunction of 1981 (a "triple conjunction" as seen from Earth), if Jupiter & Saturn were corrected to mean circular orbits. I had taken Barbarossa's orbit to be circular. Alternatively, Barbarossa might have been near its apse, so that no mean circular correction to its longitude was needed; if so, then Barbarossa's apse is near 172.6 ecliptic longitude.

Now let's solve a 5x5 system of eqns.:

(1) r=a/(1-e)

(2) r=a*(198.4/a)^0.75

(3) 1 + 1.5*e^2 + 1.875*e^4 = a^(-3) / mean(r^(-3))

(4) 197.7^(-3) / mean(r^(-3)) = m / 0.01035

(5) r^2 / 198.4^2 = m / 0.008763


Explanation: r is 1981 radius, a is semimajor axis, e is eccentricity, m is mass.
(1) says that Barbarossa's aphelion happened to occur at the time of the heliocentric Jupiter/Saturn conjunction (~1981.3), nearly the midpoint (~1980.8) of the 1954-2007 range of my Barbarossa identifications in online sky surveys and prospective amateur photos. This is justified above, as the reason no eccentricity correction (mean circular correction) needed to be applied to Barbarossa's orbital longitude, to equate it with the Jupiter/Saturn conjunction point. If perihelion is assumed instead of aphelion, the equations become inconsistent.
(2) says that Barbarossa's average angular speed for 1954-2007 is that of a circular orbit at 198.4 AU (using the standard "reduced mass" equation).
(3) comes from the polar equation of an ellipse, and Kepler's equal-areas-in-equal-times law. It's accurate to O(e^6).
(4) adjusts the mass determined from precession resonances, to the true effective distance.
(5) adjusts the mass determined from my theory of the CMB dipole, to the true effective distance.

This determines the semimajor axis as 184.5 AU, in perfect agreement with the quasi-atomic theory above. The instantaneous (1981 aphelion) radius is 194.8 AU. The mass (of the Barbarossa system) becomes 0.00845 solar masses. The eccentricity is 0.05295. By comparison, the eccentricities of Jupiter & Saturn are 0.048775 & 0.055723, resp. The arithmetic mean of the eccentricities of Jupiter & Saturn is 0.052249; the mean weighted by mass*(major axis)^2 is 0.052279; the mean weighted by mass*aphelion^2 is 0.052303.

To first order, Jupiter's and Saturn's directed eccentricities differ (subtracting vectorially using the law of cosines, then multiplying by 2.4833/(2.4833-1))(J:S value from Franklin & Soper, AJ 125:2678+, 2003) so as to amount to an equivalent eccentricity of 0.11009 with perihelion 138.9, for the cyclical advancement and retardation of the J:S resonance point.

The "Great Inequality" of Jupiter & Saturn, is known only to the range 900+/-60 yr (Vladimir Ladma, www.sweb.cz). This corresponds to a mean period for the advancing Jupiter/Saturn resonance, of 2700+/-180 yr, i.e., a major axis for Barbarossa (using standard "reduced mass" correction) of 185.8 to 203.1 AU. The usually given Jupiter & Saturn periods, 11.86 & 29.46 yr, give a period of 2758 yr and a major axis (using "reduced mass") of 197.3 AU.

So, Barbarossa's major axis in my model (184.5 AU), corresponds to a period at the lower range of accepted values for the "Great Inequality" (the reason for the uncertainty, is that the nearness of the J/S frequency ratio to 2.5, causes mathematical divergence and chaos). My model gives an eccentricity for Barbarossa, which is suspiciously near the mass*radius^2 weighted mean of the eccentricities of Jupiter & Saturn. The eccentricity and apse of Barbarossa differ grossly from those required to follow the J/S conjunction point, but "in the mean" (i.e., replacing all ellipses with approximating circular orbits) Barbarossa does follow (one of) the J/S conjunction points, about as nearly as possible considering Barbarossa's inclination.

The asymmetric (apparent) radial velocities on outwardly progressing shells of galaxies, approach equivalence to the CMB dipole, when the shells reach radius ~ 100 Mpc. This suggests (but doesn't prove) that the CMB dipole isn't a solar system phenomenon. Another objection to my solar system theory of the dipole, is that for higher spherical harmonics of the CMB anisotropy, prediction greatly exceeds observation (the coefficient of the 2nd spherical harmonic term is -36.81% = -1/e of the


Reply author: Stoat
Replied on: 06/24/2008 05:29:20
Message:

Hi Joe, over in cosmicsurfers thread I give a link to the papers by Consoli, where he re examines all of the aether drift experiments. There's definitely something there, worthy of investigation. I do think though that its unfortunate that this is seen as some sort of slight breeze. It looks as though there's a slight anisotropy in light speed for the vacuum, about 3.45E-9 metres but a larger figure for air, about 72 metres. An entrained aether bubble round any mass object, drops the speed of the aether wind to very low values, depending on the refractive index of the material through which the light of the interferometer passes.

So rather than thinking of it as a breeze, perhaps we should think of it as a very peculiar windshield glass. Using my speed of gravity, where c^2 / b^2 = h (where b is the speed of gravity.) I had to find a scaling factor to fit the measured aether drift, it turned out that the fine structure constant was the best fit.

I'm still trying to make sense of it. The implication is that a diamagnetic material lets say, will have a different "awareness" of its motion than a paramagnetic material. Hmm, pretty bizarre.


Reply author: cosmicsurfer
Replied on: 06/24/2008 14:14:34
Message:

Hi Joe and Stoat, In measuring h with a "moving-coil watt balance," electric current generates egual balanced upward motion of magnetic pressures against the downward motion of gravity. What if bulk charge at 4D scale levels much like a stretched rubber band is the graviton negative charged bec from bare electrons that emit virtual photon/positron/electron pairs, while the return wave of antigraviton bulk 4d scale charge is paired with forward motion of positive space. This looped current is conducted across 4D space instantaneously between the positive and negative regions forming the energy carrier wave between bare electrons and bare positrons. It is a chicken or egg scenario which came first? The fact that electric current generates opposing gravito-magnetic forces in the "moving-coil watt balance," in measuring h it is obvious that the two forces are entwined. John


Reply author: Joe Keller
Replied on: 06/24/2008 16:54:19
Message:

quote:
Originally posted by Stoat

Hi Joe, over in cosmicsurfers thread I give a link to the papers by Consoli, where he re examines all of the aether drift experiments. There's definitely something there, worthy of investigation. I do think though that its unfortunate that this is seen as some sort of slight breeze. It looks as though there's a slight anisotropy in light speed for the vacuum, about 3.45E-9 metres but a larger figure for air, about 72 metres. An entrained aether bubble round any mass object, drops the speed of the aether wind to very low values, depending on the refractive index of the material through which the light of the interferometer passes.

So rather than thinking of it as a breeze, perhaps we should think of it as a very peculiar windshield glass. Using my speed of gravity, where c^2 / b^2 = h (where b is the speed of gravity.) I had to find a scaling factor to fit the measured aether drift, it turned out that the fine structure constant was the best fit.

I'm still trying to make sense of it. The implication is that a diamagnetic material lets say, will have a different "awareness" of its motion than a paramagnetic material. Hmm, pretty bizarre.



Excellent post! Thanks for putting it on this thread!


Reply author: Joe Keller
Replied on: 06/24/2008 17:09:24
Message:

quote:
Originally posted by cosmicsurfer

Hi Joe and Stoat, In measuring h with a "moving-coil watt balance," electric current generates egual balanced upward motion of magnetic pressures against the downward motion of gravity. What if bulk charge at 4D scale levels much like a stretched rubber band is the graviton negative charged bec from bare electrons that emit virtual photon/positron/electron pairs, while the return wave of antigraviton bulk 4d scale charge is paired with forward motion of positive space. This looped current is conducted across 4D space instantaneously between the positive and negative regions forming the energy carrier wave between bare electrons and bare positrons. It is a chicken or egg scenario which came first? The fact that electric current generates opposing gravito-magnetic forces in the "moving-coil watt balance," in measuring h it is obvious that the two forces are entwined. John



Thanks for posting this. If Barbarossa exists, it is to a first approximation the sun's lone "planet". Most main-sequence stars show a rough proportionality of mass and radius, so, the potential gravitational energy at a main-sequence star's surface is roughly a constant of nature, and of course magnetic phenomena occur on stellar surfaces. Maybe these phenomena could arise from your theory.


Reply author: Stoat
Replied on: 06/25/2008 03:03:07
Message:

Hi Joe, I worked out this speed of gravity from the electromagnetic/gravitational couple. I reasoned that there had to be something pretty important about the ratio of the speed of light to the speed of gravity. It took me weeks to discover that the ratio of the squares was equal to h.

Try that speed of gravity in place of c in e = mc^2 The gravitational energy is the same number as the electromagnetic frequency of the mass!

The electromagnetic Lorentzian 1- v^2 / c^2 and the gravitational Lorentzian
1 - c^2 / b^2 are proportionate. The number one in the first equation is c^2 / c^2 and in the second b^2 / b^2 but I think its safe to assume that we are looking for a one to one correspondence here.

I've been trying to think about the aether drift problem in a way thats comprehensible. I thought of it in Star Trek terms. Scotty has to give his annual lecture at the fleet academy. A cadet asks why the inertial dampers aren't perfect. So Scotty calls up an image of his great granddad, who happened to be a chief engineer on a Royal Navy warship.

The image shows a bunch of officers all wearing ear protectors. Scotty then gets the replicator to magic up a pair. He pulls them apart to show a series of foam layers of differing densities. He points out that these layers preferentially knock out certain frequencies but allow the officers to talk to each other.

He then sets the cadets some home work. He asks them to wear a pair of these ear protectors while they look at the famous null result of the MM experiment. Its actually an unexpected result, rather than a null.

The layers in ear protectors are colloids, and this suggests to me that any mass has an atmosphere of its own "space," with an inverse fourth power fall off. A multi layered colloid in keeping with Planck's idea of pairs of resonating particles, that make up the entrained aether. It so happens, that I dont think we need pairs but we have to think of any particle having a dual character.

(Edited) a slight error with that previous post. For air I've given a figure of about, the speed of light plus or minus 72 metres per second. Then I've given the figure for a lab vacuum in terms of a metre run, rather than a run of 2.99792458E 08 metres. So change that 72 to 4.33264313473E-09 for a metre run.

Now I think that for the spatial vacuum pi metres per second might well be important. Putting that in I get a figure of 8.24874107849E-12 metres per metre run.

I did e mail that italian guy to ask if anyone had tried to construct a slow light interferometer. The crystal might only be a centimetre long but to slowed light the interferometer arms would be any size you like.


Reply author: Stoat
Replied on: 06/27/2008 04:35:29
Message:

Thinking a bit more about this. We seem to have two sorts of refractive index. An electromagnetic one and a gravitational one. The electromagnetic one.
v^2/ c^2 = 1 /eta^2
In the case of the vacuum,
c^2 / c^2 = 1
The gravitational refractive index is going to be
c^2 / b^2 = 1 / eta^2 = 3.88482813146E 16
(This would differ a good deal if we say that the speed of gravity is 20 billion times c. This is based on my estimate of a speed of gravity of 1.16464217444E 25 metres per second)

To work out an aether drift we take the electromagnetic refractive index of something and multiply it by the fine structure constant, then half that. We get a result in kilometres per second.

Lets take an electron of Compton wavelength radius and stop it stone dead. That will give us an aether drift of minus infinity. But thats an electron with absolutely no angular momentum. I would argue that an electron, at its edge has an angular momentum of h and an angular velocity of c. So, lets stop our electron but allow it to spin. Then we get the speed of light, plus or minus, 1.18E-169 metres per second. Thats down to the leading edge and the trailing edge.

Whats changing in terms of the fine structure constant? My best guess at the moment, is that its the permitivity constant (its not constant) Water and ice for instance will have differing permitivity values, because of their very structure.

As we are looking at refractive index changes, we are looking at changes in wavelength, frequency doesnt change after all. That suggests that frequency and permeability stay the same.


Reply author: Stoat
Replied on: 07/03/2008 04:18:48
Message:

Hi Joe, lets assume that aether drift is real. Our solar system is moving in the direction of Vega, at about 208 km/s. The other much studied object in that region is the ring nebula. For much of the solar system's journey, we have been moving in the direction of a gas cloud that formed Vega. Its less than half a billion years old.

Is there anything odd about it, that could be explained by an assumed aether drift? I think we have to write off some things as being simply coincidences. Vega was once our pole star, and its pole points almost directly at us.

It's spinning very fast. 93% of its allowed velocity, much more and it would rip apart. So there's gravitational darkening, its hotter at the poles than at the equator. That might be worth taking a look at in terms of aether drift.

It has a very low metalicity. That means it's short of all elements above helium, and not just metals. That could mean that those elements are there but arent showing up, or that Vega formed from a cloud of gas that was low in everything but hydrogen. That might be worth looking at. This has to be the same cloud of dust that formed our sun. Why did the late comer Vega, get such a low metal ratio?

On that refractive index times the fine structure constant thing from my previous post, that has to be wrong, I'm missing something there. I'll have to kick the equations round to make a table of converted refracted indices. My first thoughts are that the r.i. of the vacuum is not quite one.


Reply author: Joe Keller
Replied on: 07/03/2008 22:57:20
Message:

quote:
Originally posted by Stoat

Hi Joe, lets assume that aether drift is real. Our solar system is moving in the direction of Vega, at about 208 km/s. ...and its pole points almost directly at us.



Hold that thought! Dayton Miller remarked that the angular momentum vector of our solar system (approximately, that of any of the planets' orbits, or of the sun's rotation) differs only 6deg from Miller's best (i.e., Mt. Wilson) ether drift determination. A few years ago I found that, contrary to some published claims, the orbital axes of binary stars are significantly nonrandom when considered as a function of position on the celestial sphere.


Reply author: Stoat
Replied on: 07/04/2008 02:26:55
Message:

Hi Joe, interesting that, Vega's pole points almost directly at us, as I said, five degrees of tilt. So do you think what I remarked on as merely coincidence is something more? Our moving makes the dust cloud that became Vega have its axis pointing at us?

(Edited) Perhaps our solar system dumped its angular momentum into the cloud that was to become Vega. That could explain why Vega is close to actually tearing itself apart. Is there another star in the southern hemisphere? Dayton Miller did think we were moving in that direction and did name a star but I can't for the life of me remember what it was called.


Reply author: Stoat
Replied on: 07/05/2008 04:37:21
Message:

I thought I'd put up an image of our solar system heading toward Vega. On the right can be seen the rough position of Joe's planet Barbarossa but I'd stuck it into "Stars in your Backyard" as Nemesis. The image is still good to give people an idea of just how far out this planet is.

I suppose we have to think of how angular momentum is dumped from a "failed" binary system. It cannot be a gradual process, Sudden periodic energy impulses. Transferred to the cloud that would become Vega, It spins up and throws off its heavier content. Then it takes a long time to slow down enough to let star formation kick in.

Now that seems to work with electromagnetic angular momentum transfer but but what about gravitational transfer? Aether particles in Vega's gas cloud are going to be affected by a near instantaneous force. They have to wait years for the electromagnetic angular momentum to arrive.

[img]http://farm4.static.flickr.com/3172/2637885033_1c45aaefa8_o.jpg[/img]


Reply author: Stoat
Replied on: 07/07/2008 07:13:19
Message:

I get 1.000000007717000698 for the refractive index of the vacuum.

(Edited) I think what's interesting about this, is that if we pushed the speed of gavity upward, then refractive indexes would bunch up together. WE wouldn't see any aether drift. Conversely, lower the speed of gravity towards that of light and it would be obvious that we are moving through an aether.


Reply author: Joe Keller
Replied on: 07/08/2008 22:28:56
Message:

quote:
Originally posted by Stoat

I thought I'd put up an image of our solar system heading toward Vega. On the right can be seen the rough position of Joe's planet Barbarossa...


Thanks for these posts, and for this beautiful chart!


Reply author: Joe Keller
Replied on: 07/08/2008 22:54:35
Message:

As Barbarossa shepherds J:S, so the outer planets once shepherded E:M & V:E?


Above, I've attributed the J:S=5:2 orbital resonance discrepancy, to shepherding by Barbarossa. Two other planetary orbital small-integer near-resonances, originally could have been shepherded by one and the same planet near Jupiter's orbit.

The Venus:Earth=5:3 orbital near-resonance, has two resonance points which progress around the ecliptic in 16.40 yr (my data are from an astronomy text published c. 1980). The Earth:Mars=2:1 near-resonance, progresses in 15.81 yr. In the early solar system, these might have been exactly synchronous and shepherded by a planet with period ~16 yr.

Jupiter's period is ~12 yr, but if Saturn once was Jupiter's moon, their period, assuming no gain or loss of energy, would have been ~14 yr. Including Uranus & Neptune in the menage would give ~15 yr. If other members, adding to a few Earth masses, eventually became Edgeworth-Kuiper objects or left the solar system, the result could be exact.


Reply author: Stoat
Replied on: 07/09/2008 05:26:20
Message:

Hi Joe, I suppose that this will depend to some extent on how we see the planets forming in the first place. I still like the ideas of Lytleton, that Mercury spins out of a proto Venus, Mars from a proto Earth and Saturn from a proto Jupiter. The most popular view now seems to be that of cataclysmic collisions between proto planets.

On the question of Aether drift, may I ask you a question, as you are much better at maths than I am? Weve got a speed of gravity that is much much greater than the speed of light. We can write it as 1 - h it struck me that it would be a rather interesting cubic equation (1 - h)^3 but I dont want to do it, as cubic equations are horrible animals. [:(][xx(]Can I get away with saying that this is the distance per second but look at it as the distance over two seconds? That way I can say Ive got very nearly a cube of eight units, minus a very very small amount. So roots of 2, and
(-1+ sqrt (-3)) and (-1- sqrt (-3)) or tidied up a bit (-1 + 1.7321i) and (-1 - 1.7321i)


Reply author: Joe Keller
Replied on: 07/09/2008 10:50:48
Message:

quote:
Originally posted by Stoat

...cubic equation (1 - h)^3 ...


When I took undergraduate statistical mechanics from Dr. Wendell Furry at Harvard, my classmates told me that Furry had testified in front of the House Un-American Activities Committee. I went to Widener Library and looked up his testimony down in the dungeonlike "stacks". As I read Furry's testimony, Furry seemed to have no fear of them at all, had refused compliance or contradicted them with utter impunity. Of course anyone else is invited to read Furry's public-record testimony and disagree with my assessment if they wish.

That was when I realized that maybe McCarthyism was only some kind of show or sham, that McCarthy might have been a mere dupe. A lot of Hollywood actors moaned and groaned about McCarthyism for decades, when really, being asked to appear in front of the Committee, made them heroes of the Left and was the best thing that ever happened to some of those mediocrities. "Oh, his career was ruined, he 'had to' go back to France and make movies there instead for a few years."

I say, "asked to appear", because, contrary to what some lawyers say, Congress isn't a court and has no power to subpoena anybody. That's called "separation of powers". Bills of attainder, i.e., laws saying, "Enacted by Congress: Joe Keller is a communist/ failed to tell us what he should have told us/ whatever/ and shall be punished" are explicitly prohibited by the Constitution. There's no such thing as "contempt of Congress" unless one is an officer of the Executive Branch obligated to enforce their laws. The only way to punish anybody lawfully in the U.S., is to take them to court. If Congress wants me to testify about Communism or anything else, they can sue me.

Thanks to Furry, I understand the concept of "units" as well as "tensors". "Units" and "tensors" are basically the same thing. In both cases, some numerical equations must hold true no matter what the "units of measurement" or "transformation of coordinates". For example, if I say, "beta = 1/sqrt(1-v^2)", that's not a tensor equation; it holds true only in units of measurement for which c=1. On the other hand, if I say, "beta=1/sqrt(1-(v/c)^2)", that's a tensor equation, or put another way, "the units cancel". It holds true no matter what my units of measurement are.

"Units" can be a guide to which equations might or might not be universally true. Einstein used this concept a lot when he sought tensor equations describing gravity. It seems to me that this concept might find application in Mr. Turner's work as it progresses.


Reply author: Joe Keller
Replied on: 07/09/2008 18:48:59
Message:

(duplicate posted to "Planetary Science...Origin of Solar System...Is the Sun a Binary?" thread, because the search feature currently misses most of my posts, and I want to ensure that I and others can find this one)


quote:
["Nemesis", previous page] This leaves an object that is massive but emits very little electromagnetic radiation.


Dr. Van Flandern response: That is rather contradictory. Massive implies hot, and hot implies radiating abundantly.

quote:
["Nemesis"] Possible candidates could be a brown dwarf, or a collapsed object like a neutron star, a supernova remnant. I favor the latter possibility myself.


Dr. TVF: Doesn't that require that the Sun had a companion that went supernova? That would have wiped out the planets and changed the Sun drastically from a normal G-type star.

quote:
["Nemesis"] The best way to detect it may be through occultation of background stars. This latter method may be the only way to pick up a collapsed object, or maybe gravitational lensing of background objects.


Dr. TVF: The gravitational lensing should indeed be strong and evident. But the whole sky has been surveyed many times, which is how we get proper motions of stars. And region where gravitational lensing was going on would distort all the proper motions in that vicinity. No such region has been seen. So no dark companion with significant gravity exists. (par.) There is now yet another direct test for this, one that I also mentioned to Cruttenden. Pulsar timings allow us to detect any unknown accelerations of the Sun's motion through space, because they would displace the Earth a bit closer to pulsars in some direction, but farther away from pulsars in the opposite direction. So the arrival times of pulsar signals would be changed by a certain predictable pattern across the sky. This has long been known as a test for the possible existence of undiscovered planets of significant mass. No such displacement signal is seen, meaning the Sun is not undergoing any significant acceleration from an unknown cause. -|Tom|-


Joe Keller's comment:

A more detailed version of the following information, was posted piecemeal, subsequent to the above 2006 post by Dr. Van Flandern, by me on the "Requiem for Relativity" thread.

The last statement, about pulsar timing, especially refers to the Oct. 2005 article of Zakamska & Tremaine of Princeton, which to my knowledge has not been superseded in accuracy. They claimed only the ability to detect a Jupiter at 200 AU (or a solar mass at 200*sqrt(1000)=6000AU=0.1 light year). These authors mixed nonparametric and parametric statistics, which might have caused them to overestimate the sensitivity of their test. Also, my own investigation has revealed that millisecond pulsars (commonly a kiloparsec distant, and who knows what is really happening in the vast intervening space?) show a median Pdot/P which happens to equal the Hubble constant. This peculiar and unexplained coincidence of Pdot/P with the so-called "Hubble expansion", casts doubt on the simple model of acceleration which underlies Zakamska and Tremaine's conclusion.

Regarding gravitational lensing, Gaudi & Bloom, Astrophysical Journal 635:711+, Dec. 2005, state in their abstract:

"...Gaia (launch date 2011). A Jupiter-mass object at 2000 AU is detectable by Gaia over the whole sky above 5 sigma, with even stronger constraints if it lies near the ecliptic plane. ..."

Until now I've omitted microlensing from my Barbarossa discussion, because I saw this article. Here is a claim that sometime after 2011, a planned satellite better than Hipparcos *will be* able to detect a Jupiter at 2000AU. Surely Gaia will exceed Hipparcos not only in accuracy but in the number of stars observed; and Hipparcos exceeded ground-based astrometry in accuracy. I haven't yet found any article claiming that Hipparcos has ruled out a companion of any mass at any distance in any part of the sky (if anyone knows of one, please tell me!).

Good point about the supernova remnant, Dr. Van Flandern!

Regarding mass and temperature, the state of the art (1990s) calculation in the mainstream literature, is that a 4.6 billion yr old brown dwarf is roughly the same temperature whether it is slightly above or slightly below the critical mass for (brief) nuclear reaction ignition. The present temperature of the brown dwarf is very sensitive to the (very small) theoretical thermal conductivity of the degenerate matter. However, in 4.6 billion years, there's a lot that can go wrong with a theoretical model of bulk physical properties, untestable in the lab. Unexpected mechanisms of convection might occur; or, gravitational energy might be deposited mainly on the surface, not the interior, to begin with, depending on the mechanism of accretion.

Dr. Van Flandern, I'd like to work for Mr. Cruttenden!


Reply author: nemesis
Replied on: 07/09/2008 22:19:19
Message:

Joe, in that original thread I remember commenting I wasn't so certain a supernova explosion hundreds or thousands of AUs from the sun would have had such a drastic effect. This would have been very early in the solar system's history, with no life yet (it would certainly have an adverse effect now!) Far from "wiping out" the planets it my have stabilized the young solar system by sweeping it clear of dust and gas. I don't know enough about stellar evolution to comment on it affecting the sun's evolution from a normal G-star.


Reply author: Joe Keller
Replied on: 07/10/2008 20:48:25
Message:

More about gravitational lensing

An Astronomical Almanac says a star in quadrature with the sun, has its position altered gravitationally 0.004". So, starlight grazing Barbarossa (~0.01 solar masses) at 1AU distance (equivalent to ~15' at 200AU) has its position altered 2*0.004"*0.01 = 80 microarcsec. Grazing Barbarossa at 1 solar radius (equivalent to ~5" at 200AU) would give the famous 1.8" * 0.01 = 18 milliarcsec. The former is equivalent to stellar parallax at 12kpc, but the latter is equivalent to parallax at 50pc, and could be measured from the ground, though it would be more difficult than a parallax, because only one night's observation would be obtainable.

Alas, only 2*5/200000*360/40000 = 0.5/10^6 of the sky is within 5" of Barbarossa's entire track, which would be only 0.5 star in the "million star" 11th mag printed catalog, or 0.05 Hipparcos star. Only ~ 0.5/10^6/2780 = 0.2/10^9 of the sky would pass within 5" of Barbarossa *each year*, which would be only 0.2 star in the USNO-B 21st mag billion star catalog. Only rarely would stars as bright as Barbarossa itself, pass near enough Barbarossa to measure gravitational refraction with ground telescopes; and, then, why not look at Barbarossa itself instead?


Reply author: Joe Keller
Replied on: 07/10/2008 22:27:19
Message:

In Defense of Cruttenden

Howard Carter's "Tomb of Tut" (vols. 1-2, 1923; vol. 3, 1933) was co-authored with "A. C. Mace", which according to ISU's library catalog was the pseudonym of Arthur Cruttenden. Apparently Walter Cruttenden comes from a line of Egyptologists. Mommsen championed the use of monument and inscription data in Roman history. Newcomb, Eckert, Brouwer, Clemence, and Rawlins all carefully studied Lalande's 18th century Neptune positions; Cruttenden, like Mommsen, takes this beyond old papers, to even older stones.

According to Heiskanen & Moritz' "Physical Geodesy" (Freeman, 1967) Sec. 9.6, pp. 349-350, Helmert in 1884 already had determined Earth's quadrupole ("J2") to 0.4% accuracy by considering its torque on the moon's orbit. Essentially repeating Helmert's work with artificial satellites, improved this accuracy to 0.02% or better, already by 1966. Various orbiters now have done similarly for Mars.

Earth's orbit around the sun precesses only 0.1"/yr (Mars' orbit, only a little more)(vs. 20deg/yr for the moon's orbit around the Earth) and almost all of that 0.1", is due to the planets. So, it's impossible to measure directly how much Earth's rotation is torquing Earth's orbit; i.e., how much Earth's orbit really is torquing Earth's rotation, assuming only Newton's third law of motion. However, the measured precessions of Earth and Mars (for Mars, see Edvardsson & Karlsson, AJ 135:1151+, 2008; Bouquillon & Souchay, A&A 345:282+, 1999) conform accurately to Newton's law of gravity, given the quadrupole values measured with satellites.

Cruttenden's first, astronomical, thesis is basically that the sun has a massive, distant, undetected companion. Cruttenden's second, physical, thesis is basically that the phenomenon of planetary precession is yet improperly explained. All I've written above on this messageboard, about a planet Barbarossa near 200 AU, could be used to defend Cruttenden's first thesis. In the remainder of this message, I make four lines of defense for Cruttenden's second thesis:

1. Bouquillon & Souchay, 1999, Table 5, p. 294, say Mars' J2 = 1964/10^6 (IAU value). This is 81% more than Earth's J2 = 1082.63/10^6, as published, inter alia, in Bomford's Geodesy, 4th ed., p. 418. Earth is only 40% denser than Mars; their days are practically equal. In the homogeneous case, Mars' J2 should be only 40% greater than Earth's. The case of a small dense core, gives the same J2 as the homogeneous case. Mars' J2 seems to me to be too big to be consistent with Newton's laws. Even Earth's J2 is suspiciously near the "hydrodynamic" (ball of water, or ball of sand) upper limit, and Mars seems to me to exceed that limit.

Maybe an extra precession, w2, equal for all planets, is imposed externally. If Mars' J2 were, as I calculate it should be, only 1.4x Earth's, then solar (for Mars) and lunisolar (for Earth) precession would impose, I calculate, 15x more precession on Earth than on Mars; call these precessions 15*w1 & w1 for Earth & Mars, resp. Solve the system

15*w1 + w2 = 50288mas/yr
w1 + w2 = 7576mas/yr
(Mars precession from Folkner et al, Science 278:1749+, 1997)

to find w2=4525mas/yr. Perhaps reversing the usual cause and effect, J2 then alters, to that consistent, via Newton's laws, with the total precession. This extra precession corresponds to a period, for the entire solar system, of 286,400 yr, a circular orbit of 4345AU = 0.07 lightyear (somewhat closer than the Type M7 Proxima Centauri is, to its companions), with solar gravitation at that distance, 3.18/10^8 cm/s/s. Mars' orbit is about a degree from the principal plane of the solar system, so suppose w1 & w2 differ in direction by a degree. Then w2 changes Mars' obliquity by as much as 4525*sin(25)*sin(1)=33mas/yr. Folkner (1997) thought the true confidence interval for Mars' obliquity change, is [-15,+17] mas/yr, i.e., 5 sigma, vs. the statistical confidence interval 1 +/- 3 mas/yr. The relative positions of the solar system's principal plane, Mars' orbit, and Mars' axis, multiply 33mas/yr by a sine not quite small enough to bring it into Folkner's confidence interval.

2. Jupiter, Venus, and Mercury have axes nearly perpendicular to their orbits (Mercury & Venus have slow spin-orbit locked rotation). Earth, Mars, Saturn, Neptune, and Uranus have greatly tilted axes. As seen from the sun, Mercury's mean angular diameter is slightly less than Earth's; but it is Mercury, Venus, and Jupiter, which have the largest angular diameters at perihelion. Here might be a form of gravity quantization: perhaps the sun "sees" the quadrupole of planets differently, when they subtend small angles. This could affect precession.

3. The coordinates of the planets' poles are found in, inter alia, the 1984 American Ephemeris and Nautical Almanac, p. E87. Uranus, Neptune, & Mars cluster between RA 295 & 318, while Saturn is only 6deg from Earth's pole, and these five lie approximately on a smooth curve. Not only are four planetary tilts curiously similar in magnitude; the spin angular momentum vectors of these planets plus Uranus cluster on a curve (while the other three large planets cluster at the ecliptic pole).

The five planets' spins make what NMR spectroscopists call the "magic angle" ( arccos(1/sqrt(3))=54.7deg ) with WW Campbell's approximate compromise solar apex motion vector (still called the "standard apex" as late as 2003; Drobitko & Vityazev, Astrophysics 46:224+, p. 229, 2003). At the bottom of the curve, Uranus' spin is 51deg from Campbell's apex (RA 270 Decl +30). At the top of the curve, Earth & Saturn are 60 & 63deg, resp. Mars is ~ 43deg. Most solar apex determinations find RA < 270 and Decl > 30, which equalizes the distances even more. Intermittent torque around the solar apex axis might maintain this configuration.

4. The ~ 300,000 yr presumed common precession period, might be the rotation period of an ether island or sphere, like a clear ball of jello centered on the sun, orbiting some body so that its same face always is toward it; or it might rotate in the absence of any such body. A stress in the ether, causing an extra force proportional to 1/sqrt(r), could compensate so that the orbital periods of the planets suffer no net effect.

*********

Mars has the same day as Earth, perhaps because Mars was, until geologically recently, Earth's distant moon. Maybe Luna was a moon of Mars and got cratered by asteroids on the side away from Mars. When Earth lost Mars, Earth took Luna but flipped it sunny side down, so we see the maria. Until the recent Earth/Mars separation, Luna wasn't close enough to Mars or Earth, to change their days much.

The Earth/Mars separation is geologically recent enough that Luna hasn't had time to slow Earth much. The date of the separation can be determined from when Mars froze. Maybe synchronized orbital pumping by Jupiter did it. Maybe benevolent ETs foresaw that primates would wreck Earth, so they put the spare (Mars) in the deep freeze, with our name on it in the form of monuments, so we couldn't reach it until we matured. Cydonia looks like a Planet of the Apes face because that was ET's best guess a few million years ago, or maybe ET was betting on the Planet of the Apes scenario after Homo "sapiens" fails, due to tampering with the nucleus of the atom, or tampering with the nucleus of the cell.

The Jacobi limits of Earth and Mars are just big enough for the foregoing model. It helps a lot, that tidal forces drop off as the cube of distance.

"[Q] values in the range from 10 to 500 are found for the terrestrial planets and satellites of the major planets. On the other hand, Q for the major planets is always larger than 6*10^4."

- Goldreich & Soter, Icarus 5:375, 1966

http://xoxviagra.co


Reply author: Jim
Replied on: 07/11/2008 13:03:23
Message:

JK, It might be helpfuf to slow down and look at some of the details in more depth. One of many details that jump out is the precession of the moon being 20 degrees per year. The moon orbits the earth 12.5 times during that period of time and the precession is slower or faster for each of the 12.5 orbits by a lot. How would you explain that observation? In the current dominate belief system it is said the moon is moving away from the Earth at an acceleration rate equal to the mystery motion observed at Pioneer and else where. The sun should show a red shift from two forces, acceleration and gravity. Lots of stuff makes no sense at all.


Reply author: Joe Keller
Replied on: 07/13/2008 16:47:57
Message:

Matese et al, IAU Symposium #229, 2006, calculate that the orbit of Sedna is anomalous according to present knowledge, but could result from an undiscovered Neptune-mass perturbing planet at 2000 AU or nearer. Thanks to the person who reminded me of this!


Reply author: Jim
Replied on: 07/14/2008 17:39:50
Message:

One new idea that might muck up your detailed calculations is the stuff from supernova events. Its not known how much stuff is ejected from a SN event or where it goes but some of that stuff must come our way-don't you think? For every SN event since the get go of the galaxy a very tiny part must have come toward the sun and or coming our way. That stuff would arrive millions or billion of years after the SN event and would better explain comets, meteors and other messy bits that cause problems for modelers. You can only gloss over so much detail by careful calculation and invention. But, why do that when SN events can be part of the cause?


Reply author: Joe Keller
Replied on: 07/15/2008 18:18:19
Message:

quote:
Originally posted by nemesis

Joe, in that original thread I remember commenting I wasn't so certain a supernova explosion hundreds or thousands of AUs from the sun would have had such a drastic effect. ...



Thanks for your input! The articles I've seen suggest that supernovas so rarely have surviving planets, that in one rare instance of an apparent planet of a pulsar, they thought it was likely a stellar fragment instead. Let's make a calculation:

Utrobin, Astronomy Letters, abstract, Dec. 2005:

"...accordingly, the explosion energy of SN 1987A is (1.50 0.12) 10^51 erg..."

Let a planet have radius r, density 1 in cgs units, and distance R from a supernova like SN1987A. The planet's gravitational self-energy is

16*pi^2/9*6.673/10^8 * r^5

while with albedo, say, 50%, the absorbed energy is

50% * 1/4*(r/R)^2 * 1.5*10^51

so planetary survival barely occurs when

r^3*R^2 = 1.6*10^56 in cgs units.

For R=200AU, a low-density (Jovian-like) planet must have r=2615km. (A higher density helps because self-energy = density^2 for given r; on the other hand, once the planet starts to blow up, it's bigger and receives more energy.) For R=5AU, r=30580km, so by this estimate the giant planets, even Uranus, would survive, even if the sun went supernova. All the planets except Pluto would survive a supernova at 200AU.


Reply author: Joe Keller
Replied on: 07/15/2008 18:21:11
Message:

quote:
Originally posted by Jim

...SN events can be part of the cause...



Good point! Thanks for posting!


Reply author: Jim
Replied on: 07/15/2008 18:40:08
Message:

A planet in a SN event would have an altered orbit at least because some of the mass of the star going SN is ejected from the system. How can the ejected mass from a SN event be estimated? It seems to me all SN events originate in a mass range of 5 to 20 solar mass stars but maybe all the mass is ejected at some of the events and not other events. The total energy of the event can be reasonably estimated but how would the ejected mass be estimated?


Reply author: Joe Keller
Replied on: 07/15/2008 22:52:21
Message:

quote:
Originally posted by Jim

...The total energy of the [supernova] event can be reasonably estimated but how would the ejected mass be estimated?



I don't know offhand, but it seems to me that maybe there would be articles about that in the mainstream literature. If you have a college library (2 or 4 yr) in your town, the reference librarian might be willing to set you up on "Web of Science", basically an online paperless Science Citation Index. Even if you're not a student, they might do this as a community service; I've had things like that done for me before. Often, these articles are quite technical, but the abstract or, at the end, conclusion or discussion often are surprisingly non-technical. Let me know what you find out, perhaps post it to this thread.


Reply author: Joe Keller
Replied on: 07/15/2008 23:50:57
Message:

In Defense of Cruttenden, Part II

Suppose that the "physical, dynamical, or Machian" frame isn't the same as the "extragalactic, stellar" frame. That is, our physical laws hold in a local frame ("ether island" or sphere; "big jello ball") that's rotating vis a vis the frame established by distant galaxies (or, for practical purposes, stars in our own galaxy).

Then, in the "stellar" frame, there's Coriolis acceleration. Suppose the principal angular momentum vector of the solar system is near the axis of rotation of the big jello ball, which counter-rotates. I've mentioned several reasons to think 52.6 AU (a function of the sun's mass and fundamental physical constants) is special; for one thing, the Edgeworth-Kuiper belt seems to end basically about there. Maybe it's the radius of the big jello ball ("ether island").

Let's suppose there's a centripetal "extra sunward acceleration" proportional to 1/sqrt(r), and taking the value 0.5*H*c, at 52.6 AU, where H is Hubble's parameter & c the speed of light. For now, I'll use H = 74.43 km/s/Mpc, because it will be shown below to correlate with 1/3 of Mercury's General Relativistic apse precession, neglecting the eccentricity correction. This H value is near the result of a meta-analysis of H determinations, which was published in the 1990s.

(The NASA website lambda.gsfc.nasa.gov, as of July 18, 2008, lists determinations of H, which they call "H0", via 15 different theories, all based on 5-yr WMAP data. Determination of H from CMB data is now thought by many to be the most accurate. Sometimes upper & lower error bars were slightly unequal; then, I averaged them. I made two meta-analyses of these 15 determinations of H, using elementary parametric statistics. For the first meta-analysis, I found the arithmetic mean of the 15, weighted by 1/sigma^2; the result is 71.42 km/s/Mpc. For the second meta-analysis, I used the CRC normal curve table to find the log of the product of the 15 one-tailed p values, at 66, again at 68 & again at 70 km/s/Mpc. Quadratic extrapolation to the max log gave 71.25 km/s/Mpc.)

The median "millisecond pulsar" deceleration Pdot/P, is about H per sec, according to my count of the pulsar catalog. If everything in our part of the universe (except millisecond pulsars) is accelerating that fast, then Hubble's redshift law is explained simply: things were slower in the past. We might wonder why millisecond pulsars here in our own galaxy are decelerating by H relative to us, i.e., not really changing; but, let's proceed.

If the "extra sunward acceleration" just cancels the Coriolis acceleration on a prograde circular planetary orbit (so we think the "stellar" frame is alright), then the big jello ball's retrograde rotation is 28.65"/century (exactly 2/3 Mercury's General Relativistic apse advancement!) corresponding to period 4.52 million yrs (thus maybe in 1:1 spin orbit resonance with something less massive than the sun, in circular orbit 27,350 AU = 0.432 light yr distant).

The ratio of the jello ball radius to Mercury's major axis, is 52.6/0.387 = 135.9, and the ratio of the major axis of this presumed very distant companion, to my best estimate of Barbarossa's major axis, is 27,350/184.52 = 148.2 (if I use my best estimate of Barbarossa's present distance assuming circular orbit, 198.4 AU, as a possibly better indication of the true major axis, the ratio becomes 137.85). So, both these distance ratios approximate the fine structure constant, 1/137.036.

Axial misalignment of 13deg (Barbarossa's inclination to the principal plane) would alter the longitude of Neptune cumulatively no more than

0.287"*(1-cos(13))*164.79/(2*pi) = +/- 0.19" (& Uranus, half that)

and would alter the change in obliquity of Mars no more than

0.287"*sin(25.2)*sin(13) = +/- 27mas/yr

which, given a moderately favorable phase, would fall in Folkner's confidence interval, [-15,17]mas/yr.

The "extra" acceleration becomes apparent for spin-stabilized (thus accurately testable because they don't need attitude thrusters) space probes not in circular solar orbit. First let's consider the Galileo probe. Approximate its orbit as elliptical, with 1AU peri- and 5AU aphelion; consider an endpoint of the minor axis as typical. I find the "extra" sunward acceleration here, project it on the orbit, then project that again (on the radius to the sun) to estimate the result of tracking from Earth. The result is 6.7/10^8 cm/s/s, vs. slightly > 8/10^8 observed by JD Anderson et al.

Also let's consider Pioneer 10 & 11. Here we must add the tidal acceleration due to Barbarossa, 0.00876 solar masses, 198.4 AU distant (circular orbit approximation). Luckily both Pioneers have been nearly in quadrature with Barbarossa (& going in nearly opposite directions). Put the Pioneers at 52.6 AU, which roughly is the midrange distance for Pioneer 10 during the relevant tracking. The total ("extra", plus Barbarossa tidal) sunward acceleration, is again 6.7/10^8 cm/s/s, vs. ~ 7.8 observed by JD Anderson et al. Furthermore, the time derivative of this theoretical total sunward acceleration, is approximately zero here, explaining the approximate constancy observed.

Now let's consider the effect of the jello ball's counter-rotation, and the compensating "extra" sunward force, on planetary apse advancement. Assuming that for small eccentricity, the rate of apse advancement is proportional to the perturbation in the first radial derivative of the acceleration (it has to be an odd derivative or the convolution is zero), the "extra" force is 3/2 as effective, as the change in Coriolis force with radius during the orbit. The latter exactly suffices to keep the apse stationary in the physical frame, so the net result, for H=74.43km/s/Mpc & infinitesimal eccentricity, is that the apse advances (3/2 - 1)* 28.65"/century = 14.33"/cent., exactly 1/3 the General Relativistic apse advancement. All the planets' apses would advance at this same rate, which of course contradicts the accuracy to which their apse advancements are thought to be explained.

Whatever the eccentricity, the gradient in Coriolis force causes apse regression just equal to the frame rotation; so, the right way to correct it for eccentricity, must be to integrate r^(-1) / r^(-2) weighted by either 1/r or 1/r^2. The former weight is more plausible: it equals the potential energy, and also causes eccentricity to increase, rather than decrease, the weighted integral of r^(-1/2) / r^(-2), by a factor 1 + e^2 * 3/16, with error = O(e^4). This increases the net apse advance by a factor 1 + e^2 * 3/16 * 3, and thus lessens the H needed (to give exactly 1/3 the General Relativistic apse advancement) to 72.70km/s/Mpc, for Mercury's e = 0.2056.

To achieve the observed lack of any net unexplained advance of the apses, it's necessary somehow to divide the slope of the 1/sqrt(r) "extra" force function, by 1.5*(1+e^2*3/16). If the "extra" force were a Bessel function, its envelope would approximate 1/sqrt(r) (see, inter alia, Jahnke & Emde, Tables of Functions, 4th ed., sec. VIII.2.a, p. 138). The Bessel function of the 2nd kind (a.k.a. Neumann a.k.a. Weber function) of order -1/3, N(-1/3)(x) (Jahnke & Emde, 4th ed., sec. VIII.1.c, p. 131) has peaks at abscissae whose ratios resemble those of the major axes of Earth, Jupiter, Saturn, Uranus and Neptune (Jahnke & Emde, 4th ed., Fig. 77, p. 141, shows the first three peaks; the peaks are thereafter almost exactly equally spaced). For Saturn, the correspondence is better with the second peak of the Bessel function of the 1st kind of order -1/3, J(-1/3)(x); Venus corresponds to the first peak of the Bessel function of the 1st kind of order +1/3, J(+1/3)(x) (Jahnke & Emde, Fig. 77 again). N(-1/3) & J(+1/3) are negatively infinite at zero; J(-1/3) positively infinite. Four of the first ten peaks of the 1/3 order Bessel functions, i.e. J or N (+/- 1/3), corre


Reply author: Jim
Replied on: 07/17/2008 13:37:25
Message:

JK, If simple explaination is found in SN events way do you go to extreme ends to explain things? A SN event would eject about 5 solar masses of stuff and there have been millions of SN events in the Milkyway since it formed so all you need to understand is some tiny part of all that mass has come into the gravational field of the sun.


Reply author: Jim
Replied on: 07/17/2008 18:59:37
Message:

JK, It could be the moons of Jupiter are captured scraps of SN events from the way back times so maybe all the planets, moons and other stuff now captured by the gravity field of the sun also came from SN events. There could have been many more SN scraps passing this way in the distant past. All that stuff would alter your calculations-don't you think?


Reply author: Stoat
Replied on: 07/18/2008 03:24:32
Message:

Lets say that our sun went type two supernova, tomorrow. As Joe pointed out, Jupiter and Saturn could survive but they would be sorry states, they would be reduced right down to their rocky cores. They would also be flung out of the solar system sling shot fashion.

Stuff like the Oort belt would stay in orbit but would be vaporised. Now it simply is never going to happen, our sun doesnt have the mass to go nova.

We can work out the probability of our solar system being within ten parsecs of a supernova, through its lifetime, as being about six events. When this probability was much higher, when we were in a stellar nursery, type two supernova would undoubtedly have had major effects upon the evolution of our system.

Type two supernova are massive young stars, they won't have planets but they will have lost angular momentum by spewing out giant jovian balls of material which later could become planets. They never get the chance as the young sun burns its fuel in a couple of million years. These gobbits, I cant even see them as proto-planets, would be torn apart by a super nova. A nova is not going to send out anything other than a hot gas of elements.

If the young sun is one of a binary, and the other star is within about 20 a.u. then when one goes nova, the other can be slung shot out at it previous orbital angular velocity. Though, even in a stellar nursery the chances of such a run away star coming anywhere near another star is miniscule.

Really, all that I can see as possible fairly Hefty bit of junk coming from a nova, would be the odd, much reduced, Jupiter or Saturn, from a type one supernova. A sun capturing one of these is so unlikely as to be ruled out completely. it would simply pass through on its way to god knows where.

This does leave the question of, what is the nature of this supernova event? Its an implosion followed by an explosive after shock. All of a sudden, all of the protons in a suns core have to change into neutrons. These are heavier than protons, so energy is needed from someplace. Gravitational energy can create mass, I would argue that electromagnetic energy is far to feeble to do the job on its own.

(Edited) Monkeying about with this, if we had a sun made of nothing but iron and hydrogen, then the iron core would be about 1.6E 25 kg. Multiply that by the proton mass divide by the neutron mass and take the result from the original mass. That give us about 2.2E 22 kg.

What I think is happening is that aether particles have to get out of the way to let the core collapse. An h amount of gravitational energy is converted to electromagnetic energy. This at the hydrogen iron boundary of the star, 2.2E 22 kg is a lot of bang for your buck over this small surface area.




Reply author: Jim
Replied on: 07/18/2008 12:37:01
Message:

Sloat, If a central mass of a system suddenly explodes in a SN event and a large fraction of the mass is no longer at the center of the system then the orbits of everything in the system are going to be altered even the Oort orbits-don't you think? How does the system survive? The mass is free of the system because of the event and its future is unknown.


Reply author: Stoat
Replied on: 07/19/2008 05:10:48
Message:

Hi Jim, lets build a scale model of our solar system. Yeah I know you hate models but this model often surprises people in pubs. The sun is a billiard ball, where is mercury? Just about everyone points a finger almost touching the billiard ball. Mercury would be two metres away from the ball. The Earth seven metres. Jupiter would be forty metres, and pluto three hundred metres. The nearest star would be two thousand kilometres distant.

So, lets have our billiard ball go supernova. If I knew that billions of tonnes of hydrogen was about to hit me, travelling at thousands of kilometres per second, i dont think i would bother getting out of bed. It would simply vaporise the inner planets. Jupiter and Saturn would gravitationally respond to the suns centre of gravity, at first. However, once they were inside of the explosion envelope, most of the mass of the sun having past them at a great lick, they would flick away like a stone from a sling shot.

Past about twenty a.u. the gravitational influence of the explosion wall cannot compete with the influence of what remains of the sun. Orbits will alter but they will still be orbits of the parent body. It would be wrong to think that the wall is now fairly tolerable, it would still destroy Neptune and pluto, though there might be a few rocks left from Neptune.

I suppose we could do the sums to see whether the dirty snowballs that make up the Oort belt would survive, that might while away a pleasant hour. Barbarossa would survive and stay in orbit.

Again though, this cannot happen to our sun. Its a moot point about stars that become type one supernova. Do they have planetary systems? Is our solar system the result of being part of a failed binary system? We have about two hundred solar systems on record at the moment. They all seem to have gas giants close in to their suns. We need a new telescope in space to try and find terrestrial type planets.

Type two supernova must play their part in Joes ideas but not so much as to throw any of his calculations out of whack. Supernova in stellar nurseries will initiate star formation through the dust clouds and deliver stores of heavier elements. They cannot deliver chunks of planets, because they havent had time to build them.


Reply author: Jim
Replied on: 07/21/2008 12:42:36
Message:

Sloat, I don't hate models-my issue is in the overuse of models and making gods of them as is the case with the BB model. The thinking process is a captive of this modeling. Anyway, if the central mass of a system gains or loses mass everything orbiting that center is effected so Neptune and whatever is in the Oort belt is going to excape if the sun happened to lose most of its mass. And back to SN events-the ejected mass from the event that came our way would have an effect on the mechanics of the solar system even if it passed through leaving none of its mass and it seem likely some of the mass would be captured by the gravity system of the sun. So, if there is no way to know how many SN events have occured and effected the solar system during the past 10 billion yeaars why not just use the probibility that SN events have had an effect.


Reply author: Joe Keller
Replied on: 07/21/2008 14:19:38
Message:

quote:
Originally posted by Jim

...The thinking process is a captive of this modeling. Anyway, if the central mass of a system gains or loses mass everything orbiting that center is affected so Neptune and whatever is in the Oort belt is going to escape if the sun happened to lose most of its mass. ...


These are interesting concepts.


Reply author: nemesis
Replied on: 07/21/2008 16:14:01
Message:

"Past about twenty a.u. the gravitational influence of the explosion wall cannot compete with the influence of what remains of the sun. Orbits will alter but they will still be orbits of the parent body. It would be wrong to think that the wall is now fairly tolerable, it would still destroy Neptune and pluto, though there might be a few rocks left from Neptune."
Is there any hard data about this "explosion wall" being able to "destroy" a large planet at Neptune's distance? Most of a supernova's energy is spent as electromagnetic radiation. I recall reading somewhere that every star in the galaxy could fit in a spherical shell the size of the solar system without touching. Space is big, really big.


Reply author: Jim
Replied on: 07/21/2008 18:34:20
Message:

nem, The SN event must be more than just light because the star before and after is a very different star. The event must eject a large fraction of the mass of the star before the event-or not--?


Reply author: Joe Keller
Replied on: 07/21/2008 22:57:29
Message:

Addendum, to Defense of Cruttenden, Part II

The value H = 72.70 km/s/Mpc, becomes 72.65 when all of the e^(2*n) terms are included.

The Bessel functions of order +/- 1/3 are related to the Airy functions, which often appear in mathematical physics, e.g. in Airy's original application, the caustics of light rays (Watson, Theory of Bessel Fns., sec. 6.4, p. 188; same for 1922 ed. or 2nd ed., 1952). Also, Bessel functions of large order can be approximated, by algebraic functions multiplied by Bessel functions of order +/- 1/3 (Watson, sec. 8.43, eq. 1, p. 249; either edition). In such approximations, the argument of the one-third order Bessel fn., is an algebraic function (of the original independent variable) which might approximate a logarithmic transformation. So, the pattern I noticed in Part II above, might occur because the actual "extra" force is a large-order Bessel function.


Reply author: Stoat
Replied on: 07/22/2008 05:12:20
Message:

Im sure Joe wants to continue with his calculations and just make a note of supernovas and their effects upon star formation. Maybe another thread someplace?

((Major aside) Thinking about balls of jelly and the cores of neutron stars, ignore this bit, as its just a note to myself to look into something. We have the light speed being absolute lorentzian of Einstein, then we have a lorentzian for the possible speed of gravity, call it b. So, 1 - c^2 / b^2 = (1 / h *2GM) / rc^2 The h is just to scale things to my proposed speed of gravity. We get, rc^2 / (1 / h *2G) - rc^4 / (1 / h *2G *b^2) = M

The first term has to equal one, at the speed of light. The second term has to equal h at the speed of light. There has to be a variable missing in that first term, or h varies from its value to the value one.)

About supernovas, the estimates are that there are about one hundred million neutron stars in our galaxy.

Could all of the stars in the galaxy fit into a ball of the radius of plutos orbit? Yes.

Is most of the energy e.m? No, the explosion is busy making new heavy elements, which release em energy but we are not seeing total mass to energy conversion at any appreciable level. The explosion envelope is mainly matter doing more than 1000 km per second. A Neptune has a heat shield atmosphere but this wall of gas and heat that hits it is truly mind blowing.

The star that remains after the event is still going to be more massive than our sun. The orbit of any planets that survive are going to change but they cannot escape. Planets within twenty a.u. can be slung shot out of the system. The chances of our capturing one are minute, the effects of one hurtling through our solar system would be minute also.

(Edited) I just took a look at the orbital velocities of our planets, they are way too low. Their orbital eccentricities would alter but they wouldn't escape. The only thing that could would be a binary companion star.


Reply author: Jim
Replied on: 07/23/2008 14:00:28
Message:

Sloat, The SN event is real and the neutron star is a model. The real event seems to indicate mass is ejected if stuff astronomers believe are SNRs really are that. There must have been billions of SN events and if half the mass of a star is ejected in the SN event there is billions of solar masses of matter flying through the universe and a very tiny fraction must come our way. It might be the SN event happens at that rare and unlikely meeting of stuff flying at just the right parameters and just the right star. If a butterfly can start a chain of events---?


Reply author: Stoat
Replied on: 07/24/2008 03:25:54
Message:

Hi Jim, I suspect that the estimate of how many neutron stars there are; and hence how many supernova, theres been, is based on the big bang theory. In another thread I argued that to get a one to one correspondence between gravitational space and electromagnetic space, the radius of the electromagnetic universe had to multiplied by eight thousand.

This was to allow light, travelling extremely slowly vis a vis gravity, to move a distance equal to the Compton wavelength.

The upshot is that our galaxy can be much older than is thought. It would have graveyards of neutron stars. These things are the mass of a sun but they have been crushed into something about the size of the Earth. They are therefore hard to see. There would be more of them than the stars we can see. I get about 8% of the galaxys mass being ordinary stars and the rest dead neutron stars.

The down side of this argument, is that neutron stars make an almighty row. They spin so fast that radio telescopes pick up their noise. Of course thats not all of them, their poles have to be roughly pointing towards us. The other points to bear in mind are, what is the mean birth rate of stars, and how many totally dead white dwarfs i.e. black cinders, are there.


Reply author: nemesis
Replied on: 07/24/2008 14:29:34
Message:

Stoat, white dwarfs are about the size of Earth. Neutron stars are much smaller, a few kilometers across, smaller than many large cities. Also, the "noisy" ones are very young, a few centuries to around a million years old. A very old one, billions of years old, would have become quiescent and cooled to maybe a dull red heat. It would be very difficult to detect, even if nearby.


Reply author: Stoat
Replied on: 07/25/2008 04:30:29
Message:

Hi Nemesis, yeah youre right, I was being a bit sloppy but I was just after rough sizes to see whether JIms model could hold up.

From the equation I gave a couple of posts up we can work out the Schwarzschild radius. For a sun of six solar masses, 1 - h = 2GM/r c^2 Might as well ignore that h and say its one. We get about 18 km radius. Now if we say that at that radius the refractive index of space becomes negative then we can alter the lorentzian to become 1 + hx up to the speed of gravity, which will give us 2 = 2GM/r c^2 Dump that two from both sides and we have another radius exactly half the Schwarzschild radius.

Instead of having a gravity well shaped like a V rotated round its vertical axis, we would have a gravity well shaped like a W. It can never become a black hole.

Now the idea that a white dwarf can radiate away to a black dwarf is fine. We wouldnt know where they are, even if they were pretty close. Neutron stars are a different ball game though. They will lose angular momentum in time, through the ginormous electrical fields they generate. I cant see them losing their energy fast. They are so close to that important radius that light has a hard job getting out, it will be red shifted enormously. As the star slows the red shift will drop over time, the problem then becomes one of, how does neutronium work?


Reply author: Jim
Replied on: 07/25/2008 13:03:38
Message:

Sloat, When did I present a model? If I did so you should know there ar no WDs or neutron stars in it-no electron either. You need to dismiss those ideas as well as several other fiction items you seem to believe are real. There is a difference between models and real events.


Reply author: Stoat
Replied on: 08/04/2008 03:23:25
Message:

Hi Jim, if you say that there have been billions of supernova events, then that is a model. The estimates are for about a hundred million supernova events. This is based on the assumption that such an event leaves us with a neutron star or a black hole. They are a rare event, about one per galaxy per century, of course we dont know what percentage are going to be black holes. The other point is that I believe the estimate to be partly based on the the big bang theory. If the universe is actually older than the big bang, then there could have been billions of super nova events. However, that would mean that there would be more of the elements heavier than iron in stellar nurseries.

Thats why we talked about dwarf stars. They are hard to detect, so we we could be way out in our estimates of how many stars there actually are in our galaxy. We need to know the mean rate of star formation in order to work out the rate of take up of heavier elements into new stars.

You then added to your model by saying that supernova events have nothing to do with neutron stars. Here I think you are simply wrong. Theres a neutron star at the centre of the Crab; the Crab has to be everyones favourite supernova, though oddly no one in Europe mentioned its occurrence!

A point that Nemesis made about the outer planets surviving. Go into a classroom with a bunch of billiard balls, to ask for volunteers for a little experiment. First billiard ball is made of ice, our sun. The next one is a bog standard billiard ball, as are the next two. A volunteer is needed to stand 300 metres away from the first billiard ball which has a tnt core. Roughly a hand grenade going off.

Who knows there might be one or two takers for that. Though I think they might be a bit surprised at the noise of a real explosion, if theyve only heard them in movies. Next up is the a bomb billiard ball, about ten tonnes of tnt. No takers at all. Then we have the h bomb billiard ball, about fifty tonnes of tnt. Again not one taker for the experiment.

Last up is the supernova billiard ball. This one will convert a sizeable percentage of its mass to pure energy. An a bomb converts 0.1% of its mass energy, an h bomb 0.5% My estimate for the supernova, based on a faster than light speed of gravity, is that we are talking in the region of an explosion with a 35% efficiency. An antimatter bomb of truly staggering proportions. Note though that the sun survives. its the very explosion that forces it to become so much denser. It would still be a star with more mass than our sun, it blows off two thirds of its mass but it was about six times the mass of our sun to start off with.

Last point to make, is that mainstream science wont accept ftl gravity. They would argue that the antimatter nature of this explosion gives us the power of it. Protons have to become neutrons and in this decay they emit heaps of positrons. They meet electrons and destroy each other. Well yes but it simply doesnt happen. When they meet we get a whole bag full of gamma rays and mesons. What we would have to argue about then is the percentage. We would still be talking about serious money here. A supernova is vastly more powerful kilo for kilo than an h bomb.





Reply author: Stoat
Replied on: 08/04/2008 07:00:44
Message:

Hi Joe, a few questions for you. We know that Newton thought the speed of gravity was effectively infinite. I have heard the ridiculous argument that Newton had no comprehension of just how fast the speed of light was, living in a time when the horse was the fastest way of getting about. That the inventor of the calculus should have made such a mistake was never explained. Newton knew full well that the speed of light was fast, just not as fast as gravity.

Right, lets take my argument that c^2 / b^2 = h is good to go, here b is the speed of gravity. We can write this as 1/ eta, which is the refractive index, and equal to one over 1.5091889611E 33 Call this n.

Now lets write the lorentzian as (1 - 1 / n)^nx

That power of nx has to equal two, so x = 1.3252151E-33 That would mean that we live on the knee of a very tight exponential curve. Also, if we flip over the two divided by the reciprocal of h we get half h. In cosmicsurfers thread I did come up with this number with regard to the theoretical mass of the higgs. I didnt pursue it as I went on to something else.

Do you happen to know Joe, whether Newton himself had looked at the integral of one over x? I assume that he must have done, he would have been looking at the lorentzian as a problem in working out true compound interest I suppose.

Other questions, what do you think of the idea of gravitational caustics? For that matter near field Fre