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Einstein field equations seem very nonlinear of second order derivative. Expressing LHS of Einstein field equations in purely metric tensor and derivatives, it consists somewhat of second-order derivatives of metric tensor. RHS of course contains stress-energy tensor.

This seems to mean that when calculating metric tensor at some spacetime, it depends on future metric tensor (at that spacetime point and surrounding points) as well.

But geodesic equation needs metric tensor to derive the least action path for matter fields/particles.

  1. The first question: So when calculating metric tensor, are Einstein field equations and geodesic equations coupled together for calculation?

  2. The second question: Suppose we wish to compute metric tensor at some given spacetime. Only the form of stress-energy tensor is given. Do we need entire form of stress-energy tensor at every spacetime to compute metric tensor at some given spacetime, or can we localize stress-energy tensor and use only that information?

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EFE & Geodesic Equations

To answer the first question, the only information needed to pose and solve the geodesic equations is the metric tensor, $g_{\mu\nu}$. Thus, if $g_{\mu\nu}$ is unknown, one must solve,

$$R_{\mu\nu}-\frac12 g_{\mu\nu} R = 8\pi T_{\mu\nu}$$

and then solve the geodesic equations. I don't see how you could think they are coupled in some way and need to be solved together, since the solutions to the geodesic equation are geodesics which do not enter into the Einstein field equations.


Localised Stress-Energy Tensors

As for the second question, it is sufficient to know $T_{\mu\nu}$ say in some region $\Sigma \subset M$ of the space-time and one can solve the Einstein field equations to know the form of the metric in this region.

This is done all the time; one may consider some shell with a surface stress-energy, and consider the form of space-time within and outside the shell, two separate regions.

If we know the metric only in some region of space-time, due to the fact we only know the stress-energy for a particular region, then we cannot make any inferences about how it behaves outside. However, there are strong constraints on how the metric behaves at the boundary of two regions described by different stress-energy tensors.

Among those is the Israel junction condition, which relates the jump in extrinsic curvature on both sides to a stress-energy at the boundary between both regions.


Further Resources

For more information on the junction condition and metric tensors localised to a particular region of space-time, see Singular hypersurfaces and thin shells in general relativity.

For a pedagogical resource, see section 32 of Misner, Thorne and Wheeler's Gravitation on the collapse of stars as well as Brane-Localised Gravity by Mannheim.

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  • $\begingroup$ Do the geodesics enter the EFE through the evolution of $T_{\mu \nu}$? How else would we know the value of $T_{\mu \nu}$ at later points in time? If it is governed by the geodesics, then the geodesic equations would be coupled to the EFE. $\endgroup$ – jnez71 Oct 9 '19 at 21:05
  • $\begingroup$ @jnez71 So, you have a system with some $T_{\mu\nu}$, and then the matter follows geodesics. But then, because it's moved, you don't have the same $T_{\mu\nu}$ which means a different $g_{\mu\nu}$ and hence different geodesics, and so forth. In practice, we are thankful if we can even solve for the metric analytically or numerically. If we want to fully model the evolution of a system in general relativity, we resort to numerical simulation, and that is a whole field. $\endgroup$ – JamalS Oct 9 '19 at 21:09
  • $\begingroup$ Thanks for your quick reply. That makes sense to me! But it feels in conflict with your statement that the geodesic equations are not coupled to the EFE. It sounds like quite the opposite, since as you explain, the geodesics govern the evolution of $T_{\mu \nu}$, i.e. the RHS of the EFE. Maybe I am misunderstanding something about your answer to the OP. $\endgroup$ – jnez71 Oct 9 '19 at 21:10
  • $\begingroup$ @jnez71 They are kind of coupled, but not in a way that allows you in principle to solve them like a coupled set of differential equations. Yes, $g_{\mu\nu}$ and its derivatives appear in the EFE and the geodesic equation (through the Christoffel symbols). However, there is no term that takes into account the change to $T_{\mu\nu}$ given the geodesic motion, and there is nothing in the geodesic equation that takes into account $T_{\mu\nu}$, it's purely geometric. $\endgroup$ – JamalS Oct 9 '19 at 21:23
  • $\begingroup$ @jnez71 If you want to learn numerical relativity, read Shapiro's book, but be warned you have to go through a lot of gymnastics with the equations of GR before you can get anywhere near even trying to solve things numerically. A large part of the book is dedicated simply to putting everything in a form where you can start to think how to discretise. $\endgroup$ – JamalS Oct 9 '19 at 21:25

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