Requiem for Relativity

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

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 # 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.

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.

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).

I searched within 10 deg of RA 11h18m Decl -7deg, using the PM & R1,R2 criteria above. Only 2125/13615 objects have 18.36 &lt; I &lt; 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 &lt; I &lt; 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&lt;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.

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.