How to check if a curved metric satisfies Einstein's equations?

Question

So I have come across threads on how to check the vacuum solution of Einstein's field equations. But say if someone give me an AdS-like metric, or the AdS C metric for example. If I by inspection don't recognize the exact form of the metric, how would I be able to run a consistency check to see if the metric solves Einstsins equation?

Basically I can compute the Einstein tensor, $G_{\mu \nu}$. But then what?

I actually get something very complicated in my particular case. I don't know how to eye-ball it and say anything about it.

If the metric were Minkowski I know that the Einstein metric should vanish. But again, what if I want to check if it is AdS and check that it explicitly solves the Einstein equations?

Answer 1

The snarky answer to this question is that every metric solves Einstein's equation for some stress-energy tensor. To find that stress-energy tensor, simply calculate the Einstein tensor, multiply it by $c^4/8\pi G$ (in whatever units you like), and that's the corresponding stress-energy tensor. This method of generating solutions to Einstein's equation is sometimes given the tongue-in-cheek name Synge's method.

This method is akin to generating solutions to Maxwell's equations by writing down some set of scalar and vector potentials depending on space and time and then figuring out what charges and currents produce them by taking the appropriate divergences and curls. Of course, the charges and currents that are the "sources" of the fields may not be terribly physically interesting (spread out over all of space, diverging as $t \to \infty$, etc.) Similarly, if I write down an arbitrary metric then the corresponding stress-energy tensor may not be physically meaningful.

One particularly nice result that you might find, though, is that when you calculate $G_{ab}$ it works out to be some scalar multiple of the metric $g_{ab}$. In that case, the spacetime is well-described by a "cosmological constant" stress-energy tensor, meaning that it could be anti-de Sitter, an AdS C-metric, or something else entirely. Conversely, if $G_{ab}$ is not directly proportional to $g_{ab}$, then the metric does not correspond to an AdS-like spacetime.

Answer 2

In general you need to plug the metric in Einstein's equation

$$G_{\mu\nu} +\Lambda g_{\mu\nu} = T_{\mu\nu}$$

that arises from the variation of the action

$$ S = \int d^4x \sqrt{-g} \left(\frac{R-2\Lambda}{2} + \mathcal{L}_{M}\right)~,$$

where $T_{\mu\nu}$ comes from the matter lagrangian $\mathcal{L}_{M}$.

If your spacetime satisfies all components of Einstein's equation then it is a solution of the equation.

In the case of AdS spacetimes, the $g_{tt}$ component of the metric function asymptotes to

$$g_{tt}(r\to \infty) \sim Cr^2 $$

where $C$ will be related to the cosmological constant.

I don't think there's a way out of this procedure, expect from remembering that Minkowski and Schwarzchild spacetimes are solutions of the vaccum field equations ($T_{\mu\nu} =\Lambda=0$), (A)dS spacetime is a solution of the Einstein-$\Lambda$ field equations ($T_{\mu\nu}=0$) etc.

Answer 3

There is actually a simpler way that is more intuitive in answering the question (cf Wald 1984, pg 36-40,72). I usually don’t respond to posts so my etiquette and/or format may not be as clean but here we go.

The quick answer is that the Einstein tensor is divergence free and is created so that it is always true for Einstein’s formulation. Use the metric to build the Einstein tensor which is constructed with Riemann tensor contracted to the Ricci tensor and Ricci scalar.

Where the Christoffel symbols are defined in the following way:

Noting the divergence of the Einstein Tensor:

Then take the divergence and you are done. If it goes to zero then it is divergence free and are solutions.

Like all of GR it is mostly physical intuition and a whole lot of bookkeeping. I hope this helps.