# Rigorous derivation

1. Mar 3, 2006

### dextercioby

Well, i'm searching for a rigorous derivation of the famous "perihelium precession problem in General Relativity".

Did anyone do it...?

Daniel.

2. Mar 3, 2006

Staff Emeritus
What do you mean rigorous. Lillian Lieber did it in her little book "Einstein's Theory of Relativity" by deriving the Schwartzschild metric with the usual conditions of isotropy and zero fields at infinity, and then getting the approximate potential for Mercury from that and doing the math. Is that rigorous enough for you?

3. Mar 3, 2006

### Garth

Try Weinberg "Gravitation and Cosmology" pages 188 - 194.

Garth

4. Mar 3, 2006

### pervect

Staff Emeritus
Goldstein works this out in the section on time independent pertubation theory in "Classical Mechanics", though he doesn't give the details of how to get the Hamiltonian (which is presented without proof). See pg 510-511.

MTW's "Gravitation" doesn't give a complete derivation, it leaves it as an exercise (suggesting a few tricks and giving the correct answer) so you'll be better off with one of the other recommendations.

Last edited: Mar 3, 2006
5. Mar 4, 2006

### dextercioby

I don't have Weinberg's text. The question i have for you is: does this derivation use elliptic functions ? If not, it's not what i'm looking for...

Daniel.

6. Mar 4, 2006

### George Jones

Staff Emeritus
You might want to look at this.

Note: I have no idea whether this paper is of high quality.

Regards,
George

7. Mar 4, 2006

### pervect

Staff Emeritus
It's fairly easy and straightforwards to work out that the full relativistic treatment of the Schwarzschild orbit involves only replacing coordinate time with proper time, the r coordinate with the Schwarzschild coordinate by the same name (r), and adding an extra term to the Hamiltonian, proportional to 1/r^3.

I.e. letting L be angular momentum, and E be energy, and setting G=c=1 for simplicity (geometric units), we get

Newtonian theory

$$(\frac{dr}{dt} )^2 = E + \frac{2M}{r} - \frac{L^2}{r^2}$$
$$\frac{d\phi}{dt} = \frac{L}{r^2}$$

Relativistic theory
$$( \frac{dr}{d\tau} )^2 = (E^2 -1) + \frac{2M}{r} - \frac{L^2}{r^2} + \frac{2ML^2}{r^3}$$
$$\frac{d\phi}{d\tau} = \frac{L}{r^2}$$

The defintion of energy differs between the two, but in both cases E represents the constant energy energy of an orbiting body, and L represents its constant angular momentum.

Assuming that the 1/r^3 potential term is small allows one to use pertubation methods to find the perihelion shift - this is the part of the job that Goldstein does (pg 511).

There's no real need for elliptic functions with this approach.

8. Mar 6, 2006

### dextercioby

Surely there is (need for elliptic functions), as long as this problem does have an exact solution. There's no need for perturbation theory when the problem does have an exact solution. I've done some research and found it. There's no reserve upon the validity of the results presented in the attached document.

Daniel.

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9. Mar 6, 2006

### Garth

dextercioby asked for a rigorous approach, pervect I think you'll find that to justify the procedure you outlined rigorously you do need elliptic functions.

And yes dextercioby, Weinberg does use elliptic functions.

Garth

10. Mar 6, 2006

### pervect

Staff Emeritus
Does "rigorous" mean "no approximations"? If so, Goldstein's approach won't be suitable, it assumes the pertubation Hamitonian is small.

But if you're willing to accept approximations (which is certainly apropriate for the specific case of Mercury), the intergals become simpler.

11. Mar 7, 2006

### dextercioby

Sometimes, we have to accept perturbations (approximations) as the only viable solution, but in this case rigurous= exact solution.

Daniel.