Paradoxes Resolved, Origins Illuminated - Requiem for Relativity
Paradoxes Resolved, Origins Illuminated
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Joe Keller

USA
956 Posts

Posted - 20 May 2007 :  15:42:02  Show Profile  Reply with Quote
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 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!
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Joe Keller

USA
956 Posts

Posted - 20 May 2007 :  16:20:06  Show Profile  Reply with Quote
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.
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Joe Keller

USA
956 Posts

Posted - 20 May 2007 :  18:55:25  Show Profile  Reply with Quote
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
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Joe Keller

USA
956 Posts

Posted - 20 May 2007 :  19:18:15  Show Profile  Reply with Quote
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.
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Joe Keller

USA
956 Posts

Posted - 22 May 2007 :  18:10:05  Show Profile  Reply with Quote
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.
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Joe Keller

USA
956 Posts

Posted - 23 May 2007 :  16:51:47  Show Profile  Reply with Quote
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.
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Stoat

United Kingdom
964 Posts

Posted - 24 May 2007 :  06:05:37  Show Profile  Reply with Quote
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.
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Joe Keller

USA
956 Posts

Posted - 24 May 2007 :  12:26:06  Show Profile  Reply with Quote
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.
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Joe Keller

USA
956 Posts

Posted - 24 May 2007 :  14:37:42  Show Profile  Reply with Quote
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.
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Joe Keller

USA
956 Posts

Posted - 24 May 2007 :  14:57:13  Show Profile  Reply with Quote
(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.
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Joe Keller

USA
956 Posts

Posted - 25 May 2007 :  15:57:59  Show Profile  Reply with Quote
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.
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nemesis

84 Posts

Posted - 29 May 2007 :  14:28:27  Show Profile  Reply with Quote
Joe, have you posted since the 25th? The main header says there should be something from the 28th.
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Joe Keller

USA
956 Posts

Posted - 29 May 2007 :  17:36:17  Show Profile  Reply with Quote
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
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Joe Keller

USA
956 Posts

Posted - 29 May 2007 :  18:29:52  Show Profile  Reply with Quote
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.
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Joe Keller

USA
956 Posts

Posted - 31 May 2007 :  16:48:51  Show Profile  Reply with Quote
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.
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Joe Keller

USA
956 Posts

Posted - 31 May 2007 :  22:30:23  Show Profile  Reply with Quote
(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'.
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Joe Keller

USA
956 Posts

Posted - 31 May 2007 :  22:54:52  Show Profile  Reply with Quote
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.
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Stoat

United Kingdom
964 Posts

Posted - 01 Jun 2007 :  08:32:39  Show Profile  Reply with Quote
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.
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Joe Keller

USA
956 Posts

Posted - 01 Jun 2007 :  15:16:04  Show Profile  Reply with Quote
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
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Joe Keller

USA
956 Posts

Posted - 01 Jun 2007 :  17:49:10  Show Profile  Reply with Quote
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.
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