Okay. So I've been trying to fit the precession of the planets with the time gravity would take to travel the distance to the planets and the distance the sun might move during that time. Didn't work. So last night, I'm sitting here thinking about how the orbit of Mercury would effect the sun's motion to begin with. Even if it did to any significant degree, one must also figure in how the other planets effect it as well. So Mercury's precession should be effected by the sun's motion due to all planets, probably a little herky-jerky, but not enough to do anything major. Besides, all of the motions of the planets might about even out in some way, so there should be very little motion of the sun anyway.

So I was back at square one. I thought, what is it about the sun that is different than the Newtonian values by considering it a stationary point? Well, it is not a point, of course. There is a gradient across its width. So, what is it about its width that is different from one side to the other? Well, it spins, of course. And then I had it.

Okay. So the sun spins. The Doppler effect directly toward the sun would then be 1+(sin 0)v/c on one side and 1-(sin 0)v/c on the other, where v is the velocity at the equator of the sun. This would average out across its gradient as ([1+(sin 0)v/c]+[1-(sin [i]0[i])v/c])/2=1. Well, no luck there. Gravity simply evens out across the gradient directly toward the sun. But what about tangent to it? In this case, we would take the difference between the gravity acting on one side of an object to the other, averaged over the gradient from the center, and the result is ([1+(cos 0)v/c]-[1-(cos 0)v/c])/2=(cos 0)v/c. The cosine of the gradient is R_sun/(R_sun²+d²)^1/2. But even for a planet as close as Mercury, the radius of the sun is not much compared to the distance of Mercury, so the entire formula reduces to simply (v/c)(R_sun/d).

Well, that's it. That's the whole formula. No way it could be that simple, you say? Let's find out. By the way, another way to think about what is going on here is to imagine the velocity of rotation of the sun at the surface, (v/c), and then project it to some other distance, as if the gravitational field rotates with the sun itself. It is the same as a gravitational field projecting itself in a straight line with the original velocity of a body, even after the body has changed directions. And the same thing with EM waves, I suppose.

The rotation of the sun will act as if it would turn an entire elliptical object as large as the solar system with it, so also an orbit little by little. The most common form of units for precession seems to be that of arc-seconds per century. This formula is a ratio for the amount of rotation of orbit per orbit. So we will convert it. To do so, we simply multiply it times the ratio of the time per century to the time per orbit, and then by degrees of orbit. The formula is now (v/c)(R_sun/d)(t_century/T_orbit)(360 degrees * 3600). Now we will apply it to Mercury and see what happens. First, though, we will replace v for the sun with 2piR_sun/T_sun, which amounts to the same thing. Our entire formula is now precession_Mercury=2piR_sunt_century(360 degrees * 3600)/T_suncd_MercuryT_Mercury. Next we must find the average distance of Mercury from the sun. Its closest distance is D_c=4.6*10¹⁰ meters and its farthest is D_f=7*10¹⁰ meters. We could find the average with (D_c+D_f)/2=5.8*10¹⁰ m, or (D_cD_f)^1/2=5.675*10¹⁰ m, or we could take the distance Mercury would be if it were to orbit in a perfect circle in the same time for orbit, or d_Mercury=(GM_sunT_Mercury²/4pi²)^1/3=5.791*10¹⁰ m. I'm actually not sure which one should be used here, but the last one seems most likely, and it doesn't differ much from the others anyway.

So finally, here we go. For R_sun=6.96*10⁸ m, T_sun=2.193*10⁶ sec, d_Mercury=5.791*10¹⁰ m, T_Mercury=7.6005*10⁶ sec, and t_century=3.1536*10⁹ sec, we get a precession of Mercury of 42.992 arc-seconds per century. Not bad, huh? Of couse, this can also be applied to the other planets as well. It could also be applied to stars orbitting the galaxy, since the galaxy also rotates. For stars orbitting on the outer edge, R_galaxy/d=1, so they would have a large precession. But since this precession is within their orbit, it would simply add to their velocity, and they will orbit much faster than it would seem they should. This might explain that whole dark matter thing as well.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

It seems you are calculating the entire precession for Mercury, if so, it's wrong. The total precession of Mercury's perihelion is approx 5600 arcseconds/century. General precession (cause by the orbit not being fixed) is approx 5025 Arcseconds/century. The tugs of the other planets add approximately 531 of those arcseconds. The main deal about the 43 arcseconds percentury, is that Newtonian predictions (which include the oblateness of the sun (less than 1 tenth of an arcsecond/century) predicted approx 5557, which was short of the observed value by 43 arcseconds/century. BTW, there are sites all over the internet using the sun's spin, the sun's oblateness and other thing in trying to explain that 43 arcseconds.

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Some try to tell me, thoughts they cannot defend,... - Moody Blues.

I am only using the sun's spin. I know the rest of the precession is caused by the other planets as they orbit and so forth, but I only meant to find the small difference for that of the sun. I thought at first that the motion of the sun was causing this, as effected by the planets, but that led to no end. I figured out the additional precession caused by the sun's spin, however, within an hour of working on it . It was that simple. Of course, I guess that could go either way, since I have not yet had a chance to fully explore the consequences, which is why I named this thread Possible correlation...

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Ahhhh, ok. Sorry about that. I didn't see anything in your post about the total precession and your calculation lead me to believe that you were finding all of the precession. My fault for misunderstanding.

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Some try to tell me, thoughts they cannot defend,... - Moody Blues.

Actually, you're right. I should have included that. But it seems my posts are usually too long already, so I take effort in shortening them as much as possible. Otherwise, nobody but you and maybe a couple others will ever read them. I'm looking forward to the day that I can slip in a one word post.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Why?

__________________
Some try to tell me, thoughts they cannot defend,... - Moody Blues.

Grav,

Take a gander here:

http://www.mathpages.com/rr/s6-02/6-02.htm

Calculate values according to your formula for the other planets, cand compare to the observed anomalous (not explained by perturbations from the other planets) precession of the other planets, which is given in a table at the end.

Gravitational theory has a long history. There have been many "post Newtonian" theories devised along with other reasons to explain Mercury's precession. Einstein himself tried some other things before he arrived at what we now call GR, a tensor theory of gravity (the "potential" in general terms is a rank-2 tensor, otherwise known as the "metric" ).

You might enjoy searching and reading about the "road to GR", and the subsequent tests of it. There is also something known as PPN (parameterized post-Netwonian) formalism, which is a general framework for comparing all classical (ie non-quantum) theories of gravitation. I don't understand the details, but apparently any alternate theory of gravitation can be reduced to something that depends on 10 adjustable parameters (at least for solar system tests).

You will find your idea (which I'm not exactly clear on) would be equivalent to something in this framework (provided your idea is mathematically consistent).

-Richard

Grav,

If you've compared your formula to that table, you'll see your formula predicts the precession is inversely proportional to the radius ~1/r. So that says the Earth's precession for instance, should be about 1/2.5 that of Mercury. That is way high.

To get a precession you need a slight perturbation to the radial acceleration.

As you'll see in that above link I posted, you can write the (approximate) GR correction as term that slightly modifies the "centrifugal force" term, or you can see it as a velocity dependent effect that gets appreciable in a strong field.

-Richard

Publius,

Okay. I have done the calculations. I only did them for the first three planets, however, since that link only gave the observed precession for those three compared to GR. The ones at the end were purely GR only.

Anyway, here are the results:

*****GR**********observed*********grav
Merc ..43.0 ..............43.1+-0.5 ..............42.98805
Venus ..8.6 ...............8.4+-4.8 ................9.00649
Earth ...3.8 ...............5.0+-1.2 ................4.00776

Did I do good?

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

To simplify calculations immensely, all one really needs to know is the spin constant for the sun if we are only considering planets that orbit it. That is S_sun=1.263048411*10¹³ (the units get kind of messy). Now just plug in the orbital period (in time units of seconds) for any planet into the following equation and you have the precession. The equation is precession=S_sun/T^5/3.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Well, using the simplified formula, I went ahead and found the precession for the other planets and compared them to the link for the predicted GR values. The correlation is amazing, really.

planet**********GR**********grav

Mars................... 1.3502........... 1.39848
Jupiter.................. .0623............. .06496
Saturn.................. .0137............. .01426
Uranus.................. .0024............. .00249
Neptune................ .0008............. .00081
Pluto..................... .0004............. .00041

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Grav,

I'm sorry -- I thought you were saying the precession just went as 1/r, but you were saying the precession *per orbit* went as ~1/r.

Well, that's what GR says (approximately). If you go through that link I posted above, the precession (per orbit) formula is approximately

theta = 6pi*r_s/L,

where r_s is the Schwarzchild radius, and L is the "semilatus rectum" of the ellipse, which for our purposes for a nearly circular orbit is about the average radius. Now, the Schwarzchild radius is just 2GM/c^2, which is about 1.5km for the sun.

Now, your formula, for that is v/c*R/L, where R is the radius of the sun goes, with v = 2piR/T --> 2piR^2/cT. Using your posted values for R and T, I get 4600m for that. So 4.6km/

Now, the GR value about is in radians per revolution, so we must divide by 2pi to get the "revs per rev" figure, that is just 3r_s = 4500m.

The slight difference between L and the radius makes it even closer for Mercury.

So Grav, it's coincidence that your R*v/c comes out to be about the same as 3*r_s for the sun.

Let's investigate this coincidence further, by looking at the ratio of your constant to 3r_s.

vR/c * (c^2/6GM) = c/6G * vR/M =

c/6G *wR^2/M

Now, if any of our intrepid astrophysicists can shed some light on this ratio of wR^2/M for stars, there might be some physics in this coincidence. Otherwise, it's just a coincidence for our sun.

Note that GR predicts this for any mass, rotation has nothing to do with GR's corrections. The above comes from the straight Schwarzchild metric. Now, frame dragging for a rotating mass would add some very small additional terms.

-Richard

Excellent!!!

keep it up, grav!

Well, I don't know GR and I hadn't compared the precessions of other planets until last night, and they all worked out. I just sat down and tried to think of the most logical thing that could be happening. It took me all of an hour to figure it out this way and I got it right the first time. Einstein, however, spent three years trying to work the precession into his formulas for relativity. They didn't work out for him as they should have, but only after much manipulation did he finally hit upon a formula that worked out for the precession. So you tell me which sounds more like a coincidence.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Well, Einstein's work explains a lot more than just precession. And the reason it took so much work is Einstein was the not most adept at non-Euclidean geometry. Hilbert (to whom Einstein expained what he was trying to do)actually found the Field equations first (by about a week). Through mathematical derivation. A lot of Einstein's work was trial and error, which is why there was so much manipulation.

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Some try to tell me, thoughts they cannot defend,... - Moody Blues.

Oh, so he needed help with it, too, huh? Just kidding. Actually, a lot of my work is trial and error, too. I guess I just got lucky with this one.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

No

He helped Hilbert, one of the greatest mathematicians that has ever lived.