Requiem for Relativity

Hi,
On the issue of a binary companion, and precession.

This is lengthy, and relatively difficult to follw, son cling with me for a moment. And please correct me if I'm wrong.

Let's assume an orbital period for a binary companion of 3600 years, for simplicity.

Both the binary(let's call it Barbarossa) and the Sun will orbit their barycenter once in 3600 years. That would be equivalent to a rate of precession for the Earth of 360 degrees/3600 years = 6 arc min/year. The actual rate of Earth's precession is ~ 50 arc sec/year. Even assuming that all of that is due to the Sun's curved path around its barycenter, and assuming a retrograde(similar to Earth's precession) orbit, 6 times that amount is missing from actual Earth's precession.

So, either or:
1) Barbarossa's period must be much bigger (around 10 times the actual
estimation.)
2) Some yet unknown mechanism must account for that difference in
precession rates. By example: Earth must precess by their own
gravitational interactions with Sun and Moon(so called lunisolar
precession), an amount 6 times bigger than it's actually stated, in the opposite direction of Sun's orbit, for the observed value to be 50
arc secs/year.
3) Something is very wrong/unaccounted for in actual precession
calculations. And indeed, with star positions, astronomical coordinates, etc.

Have you heard about so called Bessel's corrections/reductions? I've
read that they were introduced in the past because things (like noon) didn't fall where they were expected.

More recently astronomers have introduced so called "leap seconds",
probably to account for the same.
See www.siriusresearchgroup.com/articles/second.shtml
for a good critic of leap seconds, and food for thought.

Well, I can be wrong, but I think that I have comprehended the issue:

This is by no means easy to visualize. The first thing is to visualize the solar system movement through space, and particularly, the Sun movement.
If the Sun is in a binary orbit, it will describe an elliptical movement around the barycenter it conforms with its binary.
Let's assume that Barbarossa's data is correct. So we have, approximately:
Barbarossa-Sun mean distance: 298 AU
Barbarossa mass: 4 Jupiter masses
Barbarossa location: near the ecliptic plane (~ -9 degrees DEC.)
So:
Barbarossa's period: ~3600 years (for simplicity.)
Sun-Barycenter distance: 1.13 AU ~= 1 AU

Now, for all practical issues, we can "forget" about Barbarossa, and
consider only that the Sun orbit is circumscribed and concentric with
Barbarossa's orbit, with a radius of around 1 AU, and that the Sun will complete a revolution around its barycenter in roughly 3600 years.

All of the solar system is supposedly also moving toward Hercules, so in reality the Sun movement is more like a corkscrew, an helicoid or lemniscata.
But we can forget about that movement for the moment, and concentrate
in the "elliptical" Sun movement.

Have you done that? did you include the Earth, and its axial tilt? what about the other planets, and their markedly different axial tilts?

Done?

So, my conclusions:
On the issue of parallax:
We'll have two parallaxes, so to speak. One parallax, the classic one,
is due to the orbit of the Earth around the Sun, and amounts to 2 AU/year.
The other one, the "new" one, is due to the orbit of the Sun around its barycenter, and will amount to 2.26 (~= 2) AU/1800 years. That is: 0.001255 AU/year, or 187828.437 km/year, which must correspond to an arc of 6 arc min/year of Sun's displacement in the sky, related to the fixed stars.
Side note: Both parallaxes must yet be corrected for the presupposed solar system displacement toward Hercules.

On the issue of precession:
We'll have two precessions, too.
The classic(lunisolar) precession, amounting to ~ 50 arc secs/year, which is effectively due to the gravitational pull of Sun and Moon acting on an oblate Earth. You can clearly observe this precession observing the movement or the Earth axis related to Polaris, by example.
A "new", apparent precession, due to the Sun movement around its
barycenter, amounting to roughly 6 arc min/year. You wouldn't be able to observe this "precession" looking at the relation between Earth axis and Polaris!(because the Earth axis wouldn't precede at all due to this.) Except over long periods of time, and at first it will look more like a "displacement" than like a rotation; it will look as a complete “rotation”(of the Earth itself in space around a "center", not of its axis of rotation, and neither of the Earth around its own axis) only after 3600 years. A rotation with a radius of roughly 1 AU, due to the Earth following the Sun's movement in its binary orbit.

Now a very important question arises:
Why we didn't detect this apparent precession(and the associated
parallax)?
My answer is that we did, but decided to ignore it! For all practical
purposes, astronomers consider the Sun fixed, and the planets in orbit
around a fixed point. So the only "correction" astronomers supposed they must introduce, was to account for the very evident(because you can see it in a relatively short time, and also in the change of the Earth axis related to Polaris) lunisolar precession effect. The other
"precession" (in reality, it's not a precession of the Earth; physically speaking, it is the movement of the Sun) "got lost" into the corrections.
That is, it is actually "hidden" in what we consider to be the duration of the sidereal year!
For that very reason, from time to time astronomers must introduce "leap" seconds, to account for the acceleration of the Sun around its barycenter (to correct the correction, so to speak.)
We relate these leap seconds to the slowing down of the Earth, but, I dare to say: that is a mistake.
The main reason they are there is because the Sun is not only moving, but it is also moving in an accelerated way around its Barycenter.

A rough estimate (based on the previous values) is that the "real" sidereal year would be ~140 minutes shorter, and that those extra 140 minutes are there today to "align" everything up, disregarding that way the Sun's own movement, in order to be able to treat it as stationary.

The most important point is: Apparent precession will not be directly observed, because it is not reflected/based in a real movement of the Earth axis of rotation(better said: not reflected as a precessionary kind of movement; it will only slowly "displace" the Earth, without changing its obliquity). So., it will easily be confused (and hidden) into the duration of the sidereal year.
The only ways to detect it are to find Barbarossa, or to measure the length of the sidereal year against the backdrop of fixed stars.
(See siriusresearchgroup.com for details on the later.)

Of course, all this stands(if it's correct, as I think it is) for any binary orbit, and will be gradually more difficult to detect/observe, the longer the orbital period.

Best regards!
Mauro

(email sent 15 min. ago)

Hi Prof. *********!

I haven't opened your more recent email yet. I wanted to finish my best prediction and give that to you first.

My best prediction, reworking the data from the beginning, for Barbarossa's geocentric celestial coordinates at the Jan. 20, 2009 epoch, is

RA 11h 28m 10.6s
Decl -9deg 16' 49"

This includes the small correction for binary motion.

The prediction I sent two days ago (then, I simply gave the change from my Dec. 22, 2008 position) amounted to

RA 11h 28m 9.6s
Decl -9deg 17' 0"

If I add a correction to that, for binary motion, it's

RA 11h 28m 9.4s
Decl -9deg 16' 58"

Sincerely,
Joe Keller

Hi Prof. *********!

I noticed that the star to the east of my "best" coordinates, labeled magnitude 19.5 on the jpg version that I have of the Jan. 20, 2009, U. of Iowa photo, is listed in the USNO-B catalog, as Red1 magnitude 18.03 & Red2 magnitude 18.22. Likewise, the star (or galaxy) to the south, labeled magnitude 20.7, is listed as R1 19.18 & R2 19.27. Glancing at only a few other stars, I see that USNO-B 0806-0230182, at 11:28:23.9, -9:18:58, listed as R1 19.81 & R2 19.33, is far less impressive than the star labeled 20.7, and well might be overlooked. Likewise USNO-B 0807-0228782, at 11:28:14.1, -9:17:17, listed as R1 [blank] & R2 19.34.

I've never found Barbarossa or Frey on a Blue sky survey. My magnitude estimates come almost entirely from comparison to the R1 & R2 magnitudes, according to USNO-B, on Red sky surveys. So, I have been looking at objects roughly a magnitude dimmer than what is labeled "19.5" on this photo, roughly equal to what is labeled "20.7", and only a fraction of a magnitude brighter than many objects hardly imaged at all.

If the limiting magnitude is ~ +21, as measured by this instrument, that would translate to ~ +21-(20.7-19.2) = ~ +19.5, as I've been estimating against Red sky surveys. This hardly seems sufficient, since my best estimate of the magnitudes, by comparison with Red sky surveys, has been ~ +19.

The two position predictions I gave, are based on different assumptions. The first prediction (11:28:9.4, -9:16:58) though less precise, might be more accurate. There is a density at about 11:28:10.0, -9:17:09 (determined with a drafting right angle, with first order correction for differing horizontal and vertical scales on my screen). This is only 14" from my rough estimate of the ~ 194" change in position (mainly due to Earth parallax) between Dec. 22 and Jan. 20.

Sincerely,
Joe Keller

---

Date: Fri, 23 Jan 2009 18:42:14 -0600
Subject: Re: Jan. 20 U. of Iowa photo
From: *********
To: josephkeller ---
CC: *********

Joe,

Sorry, nothing within ~ 60 arcsec at that position to a limiting magnitude ~21 on the Jan 20 image (see attached.)

*********

Hi Maurol, I suppose that if we want to lose more than six minutes the first place to look would be in the difference between the tropical year and the sidereal year, about twenty minutes. That's explained by the couple on the earth by the sun. Moving the barycentre out so far would mean that the inner planets have to accelerate slightly for half the year , then decelerate for six months relative to the sun. That would mean that our day would vary slightly over the year. That does happen but it's explained by our orbit not being a circular one.

I don't know, I suppose we'd have to build a computer model to look for a best fit but it's an horrendous job to do.

(Edited) I just took a look at "stars in your backyard" where I had put Joe's planet into the program months ago. I've got the planet at 191 a.u. and so i changed the viewing date to 9999 A.D. Joe's planet was still close to the autumn equinox.

(Edited again) I looked at the info on Joe's planet in starry night and its year is 2640 years. About one tenth of the time it takes our earth's pole to wobble round a complete circle, 25,800 years.

Thinking a bit more about this, if we have someone here who can model it, then it might be good to talk about what we want in it.

Let's have Joe's planet having an orbital period of 2580 years. The Earth and Mercury. it might be an idea to show the sun's equator as well as the celestial equator. The reason for Mercury rather than the more obvious Jupiter is that Mercury's orbit is important in terms of the concept of curved space time.

I think we have to shove into the model a guestimatee of how fast the solar system is moving as well. From other stuff I reckon that we are moving at about 1.8% of light speed. How does the system know where it's supposed to be? If gravity takes some finite time to propagate, then Joe's planet will always be playing catch up.

I wonder if we could use Dayton Miller's data for this?

(email sent 10 min. ago)

Hi Prof. *********!

I had found, with the help of computer programs, that pairs of "disappearing dots" (Barbarossa+Frey) on sky surveys and in prospective photos, are at positions consistent with a mass ratio 0.8771::0.1229 and a circular orbit, for the center of mass, at 198.4 AU. Today I finished a new BASIC computer program, which from observed Barbarossa+Frey positions, finds the barycentric (as defined in the Astronomical Almanac, i.e., including only the familiar planets) XYZ coordinates of the Barbarossa/Frey c.o.m. under the above assumptions, using the Law of Cosines to solve the triangle. I extrapolated the Jan. 20, 2009 barycentric c.o.m., from the Dec. 22, 2008 and March 25, 2007 barycentric centers of mass. Then I converted to J2000 geocentric celestial coordinates.

To get positions for Barbarossa & Frey, from the center of mass, I assumed the same binary separation for Jan. 20, 2009, as seen Dec. 22, 2008. I estimate that the error in this assumption is negligible, because it seems that Frey is near the tip of its almost edge-on binary orbit. So, Frey's apparent distance from Barbarossa is near extremum, and its apparent circumferential binary motion is reduced, by a large factor, by perspective (the orbital path is seen almost end-on).

The crude estimates I gave a few days ago, were essentially by this method, but with rough trigonometry. The precise prediction for Jan. 20, 2009, is

Barbarossa RA 11:28:11.0, Decl -9:16:45

Frey RA 11:28:53.6, Decl -9:07:41


The other computer prediction, used to put the red circle on the Jan. 20 photo, should be adjusted very slightly to

Barbarossa RA 11:28:10.4, Decl -9:16:46

to incorporate my best current understanding of the Barbarossa/Frey binary orbit. This prediction returns to the original data, and fits a solar orbital radius varying linearly with time, to make the acceleration of the c.o.m. precisely Newtonian.

There seem to be many "hot pixels" in the Jan. 20 photo. The file name implies that it is stacked, but I would like to be sure, mainly because a stacked photo might be more sensitive.

At the first predicted position above, for Barbarossa, is a cluster of several mildly bright pixels in the Jan. 20 U. of Iowa photo, typical of what is sometimes seen for a faint galaxy. Though dimmer, these pixels resemble the pixels seen, in this same photo, at the position of the star or galaxy USNO-B 0807-0228782, at 11:28:14.1, -9:17:17, listed as R1 [blank] & R2 19.34. One might expect a planet with effective USNO-B red magnitude ~ +19.5, to look like this, in the Jan. 20 photo.

Faintness could be due to a forward-scattering nebular cloud (there is much evidence for this; see my posts on Dr. Van Flandern's website); smaller than theoretical mass, or greater than theoretical density; and/or an albedo even lower than usual for reddish outer solar system bodies. Outer solar system objects often are reddish in color, so it seems appropriate that I determined the magnitude by comparison with Red sky surveys.

Sincerely,
Joe Keller