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One of the three classic tests of general relativity is the calculation of the precession of the perihelion of Mercury's orbit.

This precession rate had been precisely measured using data collected since the 1600's, and it was later found that Newton's theory of gravity predicts a value that differs from the observed value. That difference, which I am calling the anomalous precession, was estimated to be about 43 arcseconds per century in Einstein's time.

I have heard that general relativity predicts an additional correction that is almost exactly sufficient to account for that 43"/century difference, but I've never seen that calculation done, at least not correctly. Can anyone supply the details?

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I am not sure whether you've seen my answer, but I now realize that I should add this as a comment: your question is quite broad. What exactly entails the details of the calculation? And what is your background in GTR (so one knows whether they should explain basic points of GTR, or just move straight to the problem). –  Marek Nov 15 '10 at 14:49
    
@Marek: I'm actually just getting to looking at the answers now. I do have a fair amount of experience in GR (despite never having seen this particular calculation done properly before now), so I should be able to follow your answer. But what I was really hoping for was an outline of the math involved, i.e. something like a summary of one of the links sigoldberg1 posted. –  David Z Nov 15 '10 at 18:35
    
A nice paper appeared today on the arXiv about the anomalous precession of Mercury and effective theories. It also includes the derivation from GR. arxiv.org/abs/1106.1568 –  Frédéric Grosshans Jun 10 '11 at 9:32
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4 Answers

up vote 8 down vote accepted

A very detailed computation with a comparison between the classical and the relativistic solution: The Precession of Mercury’s Perihelion.

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Good find! At least based on a first impression (I only had time to skim through it of course). I do like that it includes a discussion of plugging in numerical values. –  David Z Nov 16 '10 at 4:08
    
I also skimmed (carefully, though). Looks good. By the way, see the references. Weinberg. It's obvious, right? Although I bet it's not as explicitly derived as in this work by Biesel. I don't have the book in this computer, so I'll take a look at it later. –  Robert Smith Nov 16 '10 at 4:14
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Weinberg - Gravitation and cosmology, page 194, 6. Bound Orbits: Precession of Perihelia. Very different derivation than the one developed by Biesel, with different assumptions. $\Delta \phi = 6\pi \frac{MG}{L}(\text{radians/revolution}) =43.03^{''} \text{per century}$. –  Robert Smith Nov 16 '10 at 19:13
    
Thanks, I'll have a look at that next time I can get my hands on Weinberg's book. –  David Z Nov 17 '10 at 4:11
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Try http://www.mathpages.com/rr/s6-02/6-02.htm . Caveat, I haven't looked at it carefully yet.

There is a detailed discussion at http://wapedia.mobi/en/Two-body_problem_in_general_relativity?t=3.

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Well, it goes like this: consider Schwarzschild metric and testing particle (this is Mercury) with energy $E$ and momentum $L$. Because you have enough integrals of motion, the equations basically solve themselves and you obtain an effective potential that contains basic Newton potential, the centrifugal potential and a correction term. Then you apply Binet equation and you are left with some differential equation that is not easy to solve, but essentially is an equation for conic section (as in the classical case) plus some correction terms. So you make some approximation (based on the parameters of the problem) and are left with "conic section" that precesses a little.

Now, I wonder: how much more precise do you want the above argument to be made? Do you want a complete derivation, or just to clarify some confusing point? Also, how much are you familiar with GTR? I am asking just so that I know at what level should I be explaining.

Also, see the wikipedia pages, it looks like they have some derivation there (although I did not check it and you can't always trust wikipedia).

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Is there really a need for this level of condescension? It's a perfectly reasonable question. If you don't want to take the time to type out an answer, that's fine, but you don't need to be a dick about it. –  Chad Orzel Nov 15 '10 at 14:28
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@Chad: I didn't mean to be condescending. I am genuinely interested in what level of rigour is David interesting in. I provided a sketch of the answer and wonder if I should add something more. I will rephrase the text so that it's more polite. On the other hand, I don't think someone calling me d*ck without knowing anything about me is really entitled to be giving lectures in morality ;-) –  Marek Nov 15 '10 at 14:38
    
I would guess that Marek's first language is not English, so the tone of some of his responses might not be exactly what he intends. –  j.c. Nov 16 '10 at 0:21
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@j.c.: that is certainly possible but there is also (I think more important) fact that the same miscommunication and wrong impression can happen to the native speaker even if he honestly has only the best intentions. No one is perfect. That is why I see no reason in starting ad-hominem attacks (not to mention swear words) instead of just calmly pointing out the possible problem. –  Marek Nov 16 '10 at 1:09
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Sorry this is one year late - but there is a rather detailed calculation of the precession of Mercury's orbit using General Relativity in Cornelius Lanczos book The Variational Principles of Mechanics, Dover Publications. The first edition appeared in 1949, the Dover edition in 1986.

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protected by Qmechanic Oct 1 '13 at 22:04

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