Possible correlation between sun's spin and precession of orbit of planets

[grav]

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.

[Tensor]

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.

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.

[grav]

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.

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...

[Tensor]

I am only using the sun's spin. I know the rest of the precession is caused by the other planets as they orbit, 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...

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.

[grav]

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.

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.

[Tensor]

I'm looking forward to the day that I can slip in a one word post.

Why?

[publius]

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

[publius]

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

[grav]

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?

[grav]

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.

[grav]

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

[publius]

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

[Squashed]

Why?

Excellent!!!

[hhEb09'1]

keep it up, grav!

[grav]

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

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.

[Tensor]

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.

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.

[grav]

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.

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.

[hhEb09'1]

Oh, so he needed help with it, too, huh? Just kidding.

No

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

[publius]

Grav,

It's interesting and certainly made me raise an eyebrow that your
vR/c worked out so well, but it just looks like an amazing coincidence.

Your formula predicts the precession is a function of angular velocity, and so a non-spinning source mass would not have this effect. GR says any mass, rotating or not, has this effect.

So as the sun's rotational speed varied over the life of the solar system, the precession of Mercury should have varied as well. I suspect this would have long term consequences, resulting in different orbits over time.

I don't know if there are any other systems where this effect can be measured with the precision we can measure the solar system (maybe some binary star system?) Now, GR would predict the precession depended only on the mass and distance, and you would say the stars' rotational speed and radius was the main factor. Your formula would fail when 3r_s is much different than vR/c.

-Richard

[Nereid]

Maybe the (first) double pulsar?

[hhEb09'1]

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).

Why is the effect calculated as the difference between the two limbs, and not integrated over the surface?

And why is the effect per orbit, shouldn't there be some sort of integration over time?

[grav]

Why is the effect calculated as the difference between the two limbs, and not integrated over the surface?

It is "integrated" over the "equator". I figured it for half of the difference from the center to each side. Otherwise the effect would be twice as great. I am now trying to determine whether there should be some effect according to the tilt of planets (or the sun) as well. I am unsure, however, whether the effect is related to the Doppler effect or "frame dragging". For the Doppler effect, it would seem that the precession should progress in the opposite direction of the orbit of a planet, considering that the spin of the sun and the orbits are in the same direction. If it is a combination of the two effects, and frame dragging is twice the value, then the precession might be found with two times the amount for frame dragging minus one for the Doppler effect.

And why is the effect per orbit, shouldn't there be some sort of integration over time?

It is found over time, the time of orbit. That is the original formula. I then changed the units to find it for the precession per century. Now, that would be the precession per century according to the present rate of precession. If the precession, or any other values, steadily change the average distance of an orbit, or the sun's spin changes over time, then the actual precession might itself change very slightly after a century has passed.

[grav]

I don't know if there are any other systems where this effect can be measured with the precision we can measure the solar system (maybe some binary star system?) Now, GR would predict the precession depended only on the mass and distance, and you would say the stars' rotational speed and radius was the main factor. Your formula would fail when 3r_s is much different than vR/c.

I would like to see how it affects stars orbitting the galaxy. But I seem to have trouble finding the numbers for this type of calculation. Does anybody know of any specific numbers for the characteristics of the galaxy (and stars) I can use for this?

The extra precession would act like extra velocity, and make stars appear to be travelling much faster than they should. One can picture this by considering a star orbitting the galaxy in a circular orbit. One could not tell that the orbit was being "turned" with the extra precession. It would appear the the star itself was simply travelling much faster than it should for that orbit. Also, for a star that is orbitting at the edge of the galaxy, the R/d ratio would be large (approximately 1), so the precession should be large as well. That is to say, the apparent orbit of the star might include the additional velocity of the galaxy as a whole, so the further one looks from the center to the edge, the faster the stars would be travelling above what they should otherwise be according to ordinary Newtonian gravity (according to thinking of the galaxy as a point mass instead of considering the gradient and spin), and will then fall off with distance past the edge.

I will also look into how it affects binary systems as well.

[hhEb09'1]

It is "integrated" over the "equator". I figured it for half of the difference from the center to each side. Otherwise the effect would be twice as great.

"integration" doesn't appear to be the same as integration, though.

Seriously, I don't see any justification, in quantity or quality, for the sum. Especially since the rotational speed of the surface of the Sun varies with latitude.

It is found over time, the time of orbit. That is the original formula. I then changed the units to find it for the precession per century.

I understand that. But again, why is it per time of orbit? I don't see the justification for that either. Why isn't it per unit of time, rather than per orbit? What is the justification for that analysis?

[trinitree88]

grav. As your first post in this thread is Aug 15, and you seem to be claiming first thoughts as you go...eight days earlier Aug 7, the claim that the precession of the perihelion of Mercury should proceed in the direction of the spinning of the central body was already posted. It was also included in the neutrino sea gravitational theory written in April 1982. Somehow your first claims are going to have to accomodate people who have already done that, decades earlier. Cheers. Pete

[publius]

Grav,

It seems to me you formula is indeed in (fractions of an) orbit per orbit, or revolutions per revolution (what fraction of 360 degrees the perihelion or other reference point on the ellipse shifts per orbit). You write:

vR/c * 1/T_mercury. That converts to fractional revolution shift per time. You then multiply by the time per century. This is equivalent to multiplying by orbits per century -- ie time per century over time per orbit is orbits per century. You then multiple by the number of degrees per orbit/revolution (=360 of course), then by 3600 to convert to arc seconds.

So, the dimensions of your vR/c is (fractions of orbit per orbit)*length.

This is a remarkable coincidence that vR/c for the sun is very close to GR's 3*r_s, or 6pi*r_s in radians shift per orbit. But so far, you haven't shown how this actually makes an orbit precess.

To do that, you need to write your force equation. Standard way is to do it in polar coordinates, r and theta, then write the equations of motion for the r and theta coordinates. You'll need to show how your vR/c changes the r motion, so that it takes a little more than 2pi radians (or 360 degrees) for r to get back where it started.

That is, for a regular ellipse, r(theta) is periodic with period 2pi. r repeats itself every revolution, making a simple closed curve in one revolution of theta. With precession, r is still periodic, but the period is slightly more (or slightly less) than 2pi, so the ellipse shifts slightly each revolution.

Now, if you can get vR/c to do that, I'll be even more impressed.


-Richard

[publius]

Grav,

The way you've formulated this, it can't be frame dragging/gravitomagnestism. Gravitomagnetism works just like regular magnestism, with the gravitomagnetic acceleration being a Lorentz-like
v x B_g, where B_g is the gravitomagnetic flux density. The v here will be the velocity of the orbiting mass. So for Mercury, you'd have to calculate B_g then add a v x B_g to the equations of motion.

Now, as I was pointing out in the Pioneer anomaly thread, Oliver Heaviside (one of the founders of EM theory) tried this on a hunch that a gravitomagnetic force due to the sun's rotation might account for the precession of mercury.

It didn't work out. The effect was way too small and in the wrong direction. Heaviside assumed that gravity would propagate at c, just like EM. Rather than using a "mu", since gravity uses a G rather than a 1/4pi*epsilon, Heaviside used an 'H' for the gravitomagnetic force constant, with H = mu/4pi.
So c^2 = G/H --> H = G/c^2. That makes the gravitomagnetic force very small.

So whatever your vR/c is, is it not gravitomagnetism, at least as we would traditionally define it. To do it properly, you need to integrate over a rotating spherical mass distribution.

Start out with a sphere of constant density that is rotating with some constant angular speed. Now, use your "doppler effect thingy" for each mass element and integrate that over the sphere to get the total force at field point.

Then you write the equation of motion (in polar coordinates, provided it still says orbits would be confined to a plane ).

Now, if after all that, you can show a precession that agrees with GR as I mentinoned in the previous post, I'll really be impressed.

-Richard

[grav]

This one was kind of a surprise, so I really don't know what to make of it either yet. All I did was to figure that as the sun turns, the left-right motion doesn't matter much because the same angles would almost cancel out from front to back. And there is no top-bottom motion, so all that was left was the front-back motion, or v/c for each point along the equator. This would be positive v/c for one side and -v/c for the other. But it is integrated linearly over the distance to the center, or v/2c and -v/2c, for a difference between the sides of v/c acting on a planet in orbit. The gravity is (R/d)² at a distance d than at the surface, but since this is linear along the equator and the path of orbit, I considered as only (R/d) (over the gradient). So the whole equation became (v/c)(R/d).

I actually didn't think at first that this was all there was to it. That is why it took me a whole hour to figure it out. I actually figured out this first part within five minutes, but then was stumped to figure out where I should go from there. I kept trying to add more to it, and it didn't take.

One thing that I thought might do me in after I posted, though, was that I considered this for a solid rotating object. As far as I know, the sun is a plasma, and so the velocities might not fall linearly to the center, or in proportion to the distance from the center. But then I considered it for free particles, particles that aren't physically attached, but move freely according to the sun's gravity. The gravity felt at any point within a mass is simply proportional to the distance from its center. So at half the radius, the acceleration of gravity is half as great as at the surface. Since a*d=v², and a and d are each 1/2 of the value at the surface, then v is also 1/2. For d=1/3, a also equals 1/3, so then so does v. It is linear after all, just like in a solid body.

As far as the integration itself is concerned, I had originally thought that I would have to go back and correct for this. My original values were intended to be just an approximation, to get an idea of what I was dealing with, and they were to be made more precise later. But to my surprise, it worked out perfectly the first time. So why mess with perfection?

This does not mean that I am not trying to work out the integration, however. I have made a few attempts to duplicate my results by going through the points of a rotating sphere one by one. It gets rather complex, however, and may take some time. It is also difficult to contemplate the formulas for each angle to the angle of orbit at different velocities for each point. The main problem, however, is that I still don't know specifically how gravity should behave at each point in this case, especially since we are getting away from ordinary Newtonian physics. If I am capable of eventually reproducing the results by going through a mass point by point like this, then I should automatically know the precise formula for the gravity of point masses in motion, and it would be a great day indeed.

[grav]

It would appear that the precession of the planets due to the sun's spin might have some similarities with this effect, as provided by Tensor in a somewhat related thread.

[north]

This one was kind of a surprise, so I really don't know what to make of it either yet. All I did was to figure that as the sun turns, the left-right motion doesn't matter much because the same angles would almost cancel out from front to back. And there is no top-bottom motion, so all that was left was the front-back motion, or v/c for each point along the equator. This would be positive v/c for one side and -v/c for the other. But it is integrated linearly over the distance to the center, or v/2c and -v/2c, for a difference between the sides of v/c acting on a planet in orbit. The gravity is (R/d)² at a distance d than at the surface, but since this is linear along the equator and the path of orbit, I considered as only (R/d) (over the gradient). So the whole equation became (v/c)(R/d).

I actually didn't think at first that this was all there was to it. That is why it took me a whole hour to figure it out. I actually figured out this first part within five minutes, but then was stumped to figure out where I should go from there. I kept trying to add more to it, and it didn't take.

One thing that I thought might do me in after I posted, though, was that I considered this for a solid rotating object. As far as I know, the sun is a plasma, and so the velocities might not fall linearly to the center, or in proportion to the distance from the center.

if you look into fluild dynamics and viscosities this should help.