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Joe Keller

USA
944 Posts

Posted - 23 Feb 2007 :  13:38:19  Show Profile  Reply with Quote
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.
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Joe Keller

USA
944 Posts

Posted - 23 Feb 2007 :  15:16:43  Show Profile  Reply with Quote
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.
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Joe Keller

USA
944 Posts

Posted - 23 Feb 2007 :  18:55:45  Show Profile  Reply with Quote
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.
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Joe Keller

USA
944 Posts

Posted - 23 Feb 2007 :  22:57:22  Show Profile  Reply with Quote
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.
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Stoat

United Kingdom
964 Posts

Posted - 24 Feb 2007 :  05:11:09  Show Profile  Reply with Quote
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 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

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
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Larry Burford

USA
2064 Posts

Posted - 24 Feb 2007 :  09:14:39  Show Profile  Reply with Quote
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.
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Joe Keller

USA
944 Posts

Posted - 24 Feb 2007 :  12:05:00  Show Profile  Reply with Quote
[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.
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Joe Keller

USA
944 Posts

Posted - 24 Feb 2007 :  12:44:28  Show Profile  Reply with Quote
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.
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nemesis

84 Posts

Posted - 24 Feb 2007 :  12:50:19  Show Profile  Reply with Quote
Shouldn't Barbarossa show a large annual parallelax, much larger than it's proper motion?
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Joe Keller

USA
944 Posts

Posted - 24 Feb 2007 :  12:55:47  Show Profile  Reply with Quote
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.
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Joe Keller

USA
944 Posts

Posted - 26 Feb 2007 :  12:51:22  Show Profile  Reply with Quote
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.

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Joe Keller

USA
944 Posts

Posted - 26 Feb 2007 :  16:15:14  Show Profile  Reply with Quote
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.
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Joe Keller

USA
944 Posts

Posted - 27 Feb 2007 :  20:56:52  Show Profile  Reply with Quote
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.
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Joe Keller

USA
944 Posts

Posted - 28 Feb 2007 :  20:24:49  Show Profile  Reply with Quote
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).
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Joe Keller

USA
944 Posts

Posted - 28 Feb 2007 :  20:44:05  Show Profile  Reply with Quote
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.
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Joe Keller

USA
944 Posts

Posted - 04 Mar 2007 :  21:37:43  Show Profile  Reply with Quote
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.
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Joe Keller

USA
944 Posts

Posted - 05 Mar 2007 :  00:59:02  Show Profile  Reply with Quote
Object #2 isn't on the Aladin plate.
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Joe Keller

USA
944 Posts

Posted - 05 Mar 2007 :  11:01:52  Show Profile  Reply with Quote
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.
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Joe Keller

USA
944 Posts

Posted - 05 Mar 2007 :  11:24:08  Show Profile  Reply with Quote
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.
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Joe Keller

USA
944 Posts

Posted - 05 Mar 2007 :  11:31:28  Show Profile  Reply with Quote
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.
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