0
$\begingroup$

It is possible to get the Schwartzschild metric assuming spherical symmetry, vacuum solution and Minkowski spacetime when $r \to \infty$.

Is it possible an analytic solution for a geocentric system? I mean, taking the apparent daily movement of the celestial bodies as real. So, the (apparent) trajectories of moon, sun and planets should be geodesics according to the metric.

I suppose it is necessary to assume that when $r \to \infty$ geodesics are circles, as the fixed stars does every night from an observer on earth. So it is not a Minkowski metric at infinity.

I don't know if the Godel solution is something like that.

$\endgroup$
  • $\begingroup$ By geocentric system you mean the geocentric model? $\endgroup$ – Qmechanic Oct 17 at 20:48
  • $\begingroup$ There isn’t even an exact solution for the two-body problem in GR. $\endgroup$ – G. Smith Oct 17 at 21:49
  • $\begingroup$ Yes. It is known that it is possible to choose different frames, but I've never seen an example. $\endgroup$ – Claudio Saspinski Oct 17 at 21:49
0
$\begingroup$

There isn’t even an exact solution for the two-body problem in General Relativity, much less for the $n$-body problem, even if you take the center of mass as the origin. The $n$-body problem doesn’t have an exact solution in Newtonian gravity.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

As G. Smith said, there are no exact solutions to GR describing orbiting bodies.

That aside, the center of the orbit is absolute in general relativity in the sense that, for instance, the Doppler shift and aberration of distant objects shows a yearly variation consistent with circular/elliptical motion. Of course, you can always pick a coordinate system that puts you at the center of the universe, but coordinates have no physical significance. In terms of actually measurable quantities like Doppler shift, the heliocentric (or rather center-of-mass-centric) picture is clearly more sensible than the geocentric picture.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, you are right. I think about some conditions. When $r \to \infty => d\phi/dt = cte, dr/dt = \lambda r, d\theta/dt = 0$ $\endgroup$ – Claudio Saspinski Oct 17 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.