# Does geodesics from solving full field equations are same as path from energy-momentum tensor?

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?

• It seems that you are asking something along the lines of 'is general relativity self-consistent'. The (obvious) answer is yes.
– Danu
Jun 16, 2014 at 18:13
• @Danu is it obvious? i can't understand how it is so. Jun 16, 2014 at 18:23
• That GR is consistent, yes. How this manifests itself in your particular scenario, I'm not sure
– Danu
Jun 16, 2014 at 18:26
• Jun 16, 2014 at 20:59
• @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" Jun 16, 2014 at 22:19

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.
• 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. Jun 16, 2014 at 22:17