1
$\begingroup$

In general theory of relativity the Einstein field equations e.g. relate the geometry of space-time with the distribution of one body within it. $$R_{\mu\nu}-\dfrac{1}{2}g_{\mu\nu}R+g_{\mu\nu}\Lambda=\dfrac{8\pi G}{c^4}T_{\mu\nu}.$$

What will be the "Einstein field equations" for two or three bodies?

$\endgroup$
  • 1
    $\begingroup$ There are no bodies as such in that equation. The tensor T represents various matter and energy related "densities" in some infinitesimal element of a total field. Describing this field and solving it numerically it is brutally difficult. See einsteintoolkit.org if you want to know more about this. $\endgroup$ – m4r35n357 Jul 17 at 9:08
4
$\begingroup$

They are already included, that is a field equation. So the energy-momentum tensor on the right must include the distribution of energy (matter) and momentum in your model, all of it. Normally it is used the other way around, one takes a particular form of the energy momentum tensor to model a single "point"-like particle, by for example making the energy density into a delta function evaluated at the particles wordline.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

$T_{\mu\nu}$ represents the sources. So, if you have n bodies, you must take them into account into $T_{\mu\nu}$.

The approach is totally analogue to the definition of the sources for the Maxwell's equations: also in that case, there is no formal difference between a pointlike charge and a distribution of charge density. You just have to properly write the source term.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

The Einstein-Infeld-Hoffman equations, derived in 1938, describe the motion of $N$ gravitating point masses in General Relativity. They are an expansion in powers of $1/c$, so they are useful when the particles are not highly relativistic. The leading term (with no power of $1/c$) is simply Newtonian gravity.

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

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.