Requiem for Relativity

Further Confirmation of Barbarossa/Frey Orbit

Using a Pentium computer, I stretched the apparent counterclockwise Barbarossa/Frey orbit (obtained using that inferred fifth point which equalized the three areal rates, thereby satisfying two equations with one adjustable parameter) in every direction and amount ("homothetic transformations"). The stretch which makes the origin (Barbarossa) a focus (tested by the sum-of-distances definition of an ellipse) gives the real orbit.

The real orbit has semimajor axis (between Frey and Barbarossa) 0.94 AU, eccentricity 0.24, and period 15.225 yr. This implies a mass Barbarossa + Frey = 0.0036 solar masses, a third of what I estimated from several effects on the solar system. So, additional moons, rings and nebular matter are likely.

This disproves the complicated orbit I estimated on this messageboard March 23, 2008. It confirms the other, simple, orbit I estimated on this messageboard Feb. 18, 2008 ( > 0.50 AU; 14.7 yr).

The normal to the Barbarossa/Frey orbit is 90 +/- 23 = 113 or 67 deg from our line of sight. Also, this normal is 48 deg from the normal to the Barbarossa/Sun orbital plane.

followup email

Dear Prof. *********,

The other three images, are from sky surveys, and amateur photos in 2007. For almost two years on Dr. Van Flandern's messageboard, I've been writing about the trajectory defined by these three. With one adjustable parameter (the Barbarossa/Frey mass ratio) the 2007 c.o.m. is within a very few arcsec of where it should be, in longitude and latitude (precisely solving two independent equations with one adjustable parameter).

Now, not only does the Dec. 2008 ******* image, assuming the same mass ratio, fall on the same trajectory (4" off) but with four images, I can define the apparent binary orbital ellipse up to one adjustable parameter. Again, I find that with proper choice of this parameter, not one, but two equations (for three equal areal speeds with Barbarossa as the origin) are satisfied (to 0.125% accuracy).

Sincerely,
Joe Keller

(to someone listed in a magazine, as an officer of an astronomy club, who asked if I'd emailed him by mistake)

"Hi *******!

I sent that notice to many officers of astronomy clubs. For almost two years, I 've been sending emails (it's not spam, because I only send the emails to those who advertise themselves as people who, likely as not, should be interested in this - often directly or indirectly tax-subsidized), driving around visiting astronomy professors, recruiting people with adequate telescopes, to image this region.

Now I have even more certain proof that there are two massive objects there in binary orbit. The math is basically freshman calculus. I majored in math, cumlaude, at Harvard, so I know how to do the math, or at least Harvard's math dept. thought I knew how to do the math. I just mention that, because the situation is, that astronomers will believe this if and only if they believe the math; but they're not going to bother to check the math unless they believe. So I've had to break a vicious cycle. It's been like making "stone soup".

Anyway, now with the binary orbit proven to about 1" accuracy, it's a certainty that it's right there, and only a matter of time (this winter? 2000 years from now?) before big telescopes take plenty of pictures.

It's like an "orphan drug" - a big advance that's ignored because no one can make a profit as the "discoverer". It already has been imaged by three amateurs, Joan Genebriera, Steve Riley, and Robert Turner, each of whom contributed huge unpaid effort to achieve their results with barely adequate equipment. It's been imaged once again by the robotic telescope at the U. of *********, which now is broken and might or might not be fixed in the next few weeks.

Sincerely,
Joseph C. Keller, M. D. (retired)

Dr Joe, Can you say for sure the math is a sure thing in that math of tha kind has caused most of the confusion in science. Observation is good though, so if the subject is seen it would prove the math works. Maybe proving the math works(in that it really does predict nature)would be more important than finding the subject. And the other way round would be a good thing too. It might be better to stop dumping on the powers that be and look at new ways around the stone in the soup. Just trying to be helpful here.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />Dr Joe, Can you say for sure the math is a sure thing in that math of tha kind has caused most of the confusion in science. Observation is good though, so if the subject is seen it would prove the math works. Maybe proving the math works(in that it really does predict nature)would be more important than finding the subject. And the other way round would be a good thing too. It might be better to stop dumping on the powers that be and look at new ways around the stone in the soup. Just trying to be helpful here.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Thanks for your input!

If Frey is at L1, the Mass Matches

Suppose Barbarossa is like the Sun, Barbarossa's most massive moon is like Jupiter, and Frey is like a Trojan asteroid or asteroids, but not at a Lagrange point leading or following; rather, at the Lagrange point L1 between Barbarossa and Barbarossa's massive moon. For convenience, let the distance between Barbarossa and Frey be 1; and the implied total mass, if Frey itself were the massive moon as in previous posts, also be 1. Let the distance from Barbarossa to the presumed center of mass be a = 0.1229 as determined from the 1954, 1986, 2007 & 2008 photos. Let Barbarossa's actual mass be m2, the massive moon's m1, and the actual distance from Barbarossa to the massive moon, r &gt; 1. There are three equations:

For the period, m1 + m2 = r^3

For the center of mass, m2*a = m1*(r - a)

For Frey to be at the Lagrange point L1, the extra centrifugal and centripetal forces balance, m1/(r-1)^2 = m2 - (1-a)

Eliminating m1 & m2 gives a quintic equation with a root r = 1.39, m2::m1 = 10.3::1, and m2 + m1 = 0.0098 solar masses, close to the mass estimated indirectly in several other ways above.