Requiem for Relativity

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.

(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.

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.

Joe, have you posted since the 25th? The main header says there should be something from the 28th.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by nemesis</i>
<br />Joe, have you posted since the 25th? The main header says there should be something from the 28th.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

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

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.