Requiem for Relativity

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16 years 5 months ago #20199 by Joe Keller
Replied by Joe Keller on topic Reply from
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*pi)=5.623". Conveniently, 1795 (Lalande's unwitting but usefully accurate Neptune position; see Rawlins, Astronomical Journal 75:856+, 1970, "The Great Unexplained Residual in the Orbit of Neptune") and 1853 (near the midpoint of the 1846 to, at latest, 1866, post-discovery data available for Newcomb's 1866 ephemeris) lie on the same level of this actual sinusoid. A constant term, and the amplitude and phase of the exact second harmonic sinusoidal correction term (implicitly) used, are three parameters which can be fitted to Lalande's point, to the 1853 point (i.e., approximate midpoint of the post-discovery data) and to the 1853 time derivative of the actual (not exactly second harmonic) sinusoid.

I did so. This is about the best fit Newcomb could achieve with the harmonics he likely used. Third and higher harmonics help little because Newcomb's base and prediction epochs are 1/8 cycle apart; for the third harmonic this is 3/8 cycle, so the values of its derivatives at the base epoch bear little relation to their values at the prediction epoch.

Using the above program, I found the actual residuals. The best-fit (using correlation coefficients) quadratic polynomial describing the residuals, is

1.755" + 0.1302" * (time - 1875.187)
+ 0.00294 * ((time - 1875.187)^2 - 50.8985)

Season-minus-opposition (i.e., signed angular distance of the observation, from opposition) also correlated with the residuals, but only through its correlation with time. On the other hand, the squared distance from opposition, did correlate, at -1.9 sigma, with the residuals, independently of time. The best-fit quadratic polynomial including this (season-minus-opposition)^2 effect, is

1.889" + 0.1266" * (time - 1875.187)
+ 0.00320 * ((time - 1875.187)^2 - 50.8985)
- 0.285" * (distance in radians, from opposition)^2

The r.m.s. distance from opposition was 0.619 radian. Using this r.m.s. value to get corrected, smoothed observations, I find for residuals from Newcomb's presumed best-fit exact second harmonic (discussed above)

1866.0 +0.724"
1877.5 +1.927"
1889.0 +3.976"

(The residuals according to the fit which ignored the distance of the observations from opposition, hardly differed from these.)

For comparison, the predicted residuals (assuming, as above, Barbarossa's tidal effect, and Newcomb's harmonic term best fitting it) are

1866.0 +0.552"
1877.5 +1.627"
1889.0 +2.400"

Integrating by Simpson's rule, we find that this theory, of a difference between Newcomb's exact second harmonic, and the actual slightly longer second harmonic due to Barbarossa, explains 69% of the magnitude of the area under the residual curve, which is to say, 90% of the residual variance (excluding short-term variations).

Similarly, Eckert, Brouwer & Clemence fit Neptune's orbit. Their method was to find the closest fitting Newtonian orbit. They avoided empirical correction terms. Their purpose was to determine the discrepancy implied by Lalande's 1795 postion. They were adjusting the Newtonian orbit, in effect adding a first-order harmonic only (the higher order harmonics would be small due to Neptune's small eccentricity). I found that their procedure, given Barbarossa's tidal force, should imply exactly the Lalande residual that it did. I'll write about that tomorrow.

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16 years 5 months ago #20821 by Joe Keller
Replied by Joe Keller on topic Reply from
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.

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16 years 5 months ago #20822 by Joe Keller
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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

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16 years 5 months ago #20201 by Joe Keller
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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.

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16 years 5 months ago #20314 by Joe Keller
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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.

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16 years 4 months ago #20826 by Joe Keller
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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 1st, and of the 3rd spherical harmonic -43% of the 2nd).

However, today I found yet another prop for my theory that there is something very important about the distance, 52.6 AU from the sun. The gravitational acceleration due to the sun at 52.6 AU, is the geometric mean, of the acceleration due to the sun at Earth's orbit, and Hubble's constant times the speed of light (that is, 80.0 km/s/Mpc * c, or 7.7*10^(-8) cm/s^2). This is a few percent greater than most estimates of Hubble's constant, and a few percent less than most estimates of the Pioneer anomalous acceleration (which I and, in their own way, mainstream relativists before me have argued, is essentially also a manifestation of Hubble's constant).

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