2
$\begingroup$

As we know, if we had an energy-momentum tensor in all space-time we could obtain the metric tensor by solving field equations. Also i think if we had an energy-momentum tensor then we have distribution of matter in space-time which means we know its motion and its path. Now my question is that if we calculate the geodesics from the metric, does this geodesic is same as the path obtained from energy-momentum tensor?

$\endgroup$
  • $\begingroup$ It seems that you are asking something along the lines of 'is general relativity self-consistent'. The (obvious) answer is yes. $\endgroup$ – Danu Jun 16 '14 at 18:13
  • $\begingroup$ @Danu is it obvious? i can't understand how it is so. $\endgroup$ – Mostafa Jun 16 '14 at 18:23
  • $\begingroup$ That GR is consistent, yes. How this manifests itself in your particular scenario, I'm not sure $\endgroup$ – Danu Jun 16 '14 at 18:26
  • 1
    $\begingroup$ Related: physics.stackexchange.com/a/24382/9887 $\endgroup$ – Alfred Centauri Jun 16 '14 at 20:59
  • $\begingroup$ @Danu: particles only travel along geodesics if their mass is negligible relative to the background, and the curvature can be taken as constant over their extent. So, the answer to the OP is "in almost every case you'd ever care about, but not in general" $\endgroup$ – Jerry Schirmer Jun 16 '14 at 22:19
3
$\begingroup$

I will say that there is a subtle caveat here, that makes the answer not quite a pure "yes":

The Einstein equation factors in the back-reaction of matter -- if I have a mass distribution, then that matter will set up a gravitational field. Then, that gravitational field will tell that mass how to move. Any motion will change the gravitational field, which will then change the motion of the matter distribution. Therefore, if we solved the Einstein equation for a full matter distribution, all of these influences would be noted -- one end of the distribution would be pulled to the other, and the distribution would emit gravitational radiation as it moved, and lose matter.

Pure geodesic motion, however, only applies to particles whose mass is so small relative to the curvature around them that we can effectively ignore the gravitational field that they set up, and that are so small that the curvature of spacetime is essentially constant over the surface of the object. It is only these particles that travel along geodesics. If you relax either assumption, then the motion of the object will be predicted by the Einstein equation, but not by the geodesic equations.

$\endgroup$
1
$\begingroup$

Yes. By definition geodesics come from a metric. The way you get geodesics from the energy momentum tensor is by deriving the metric tensor from it using Einstein's equations: $$G_{\mu \nu} + \Lambda g_{\mu \nu}= {8\pi G\over c^4} T_{\mu \nu}$$ and that is the spacetime metric of general relativity. So the geodesics must be the same since they minimize the same metric.

$\endgroup$
  • 1
    $\begingroup$ I don't think this is quite what the OP is asking about. I think the question there is that, in the frame of the given metric, you have $\rho = T_{00}$ and ${\vec j}_{i} = T_{i0}$, and you therefore then know all of the motion of the test particles after you've solved Einstein's equation. But, you can also infer this from the assumption of geodesic motion. $\endgroup$ – Jerry Schirmer Jun 16 '14 at 22:17

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.