1
$\begingroup$

Suppose that you want to solve the Einstein Field Equations $$R_{\mu\nu}-\frac12Rg_{\mu\nu}=\frac{8 \pi G}{c^4}T_{\mu\nu}$$ but already know all the components of the metric $g_{\mu\nu}$. If thats the case, this implies that we know the Ricci tensor $R_{\mu\nu}=g^{\rho\sigma}R_{\rho\mu\sigma\nu}$, which further implies that we know the scalar curvature $R=g^{\mu\nu}R_{\mu\nu}$ which would mean that we can find the Einstein tensor $G_{\mu\nu} \implies G_{\mu\nu}=R_{\mu\nu}-\frac12Rg_{\mu\nu}$. Since we know the Einstein tensor does that this imply we can solve for the Stress-Energy tensor by $$R_{\mu\nu}-\frac12Rg_{\mu\nu}=\frac{8 \pi G}{c^4}T_{\mu\nu} \implies T_{\mu\nu}=\frac{c^4}{8 \pi G}({R_{\mu\nu}-\frac12Rg_{\mu\nu}})?$$ Is this even feasible?

$\endgroup$
4
  • 1
    $\begingroup$ Yes, people do this all the time. $\endgroup$ Dec 22, 2021 at 19:12
  • $\begingroup$ Is that all you want to do, multiply by $c^4/(8\pi G)$? You can even rearrange to get the Ricci tensor in terms of the stress-energy, if you're interested. $\endgroup$
    – J.G.
    Dec 22, 2021 at 19:47
  • $\begingroup$ Yes, that is all I want to do for now. However, given the definiton of the SEM tensor $$T_{\mu\nu}=\frac{-2}{\sqrt{-g}}\frac{\delta S}{\delta g^{\mu\nu}}$$, can I get the same result using the definiton $$T_{\mu\nu}=\frac{c^4}{8 \pi G}({R_{\mu\nu}-\frac12Rg_{\mu\nu}})$$? $\endgroup$
    – aygx
    Dec 22, 2021 at 19:54
  • $\begingroup$ Ah, now we're getting at what you want: to derive the EFEs from an action. This article does it. Your action needs two pieces, only one of which appears in your penultimate equation; the other gives the Ricci side. $\endgroup$
    – J.G.
    Dec 22, 2021 at 19:57

1 Answer 1

2
$\begingroup$

Yes, you can obtain the stress-energy tensor from the Einstein tensor. If you do not specify the stress tensor beforehand, any Lorentzian manifold will provide a solution to the Einstein equations. However, not necessarily it will be a physically interesting solution.

As a historical example, we can consider the Alcubierre Warp Drive, which has a geometry constructed to provide a way of allowing a spaceship to travel hyperfast within General Relativity (in a super short form, the spaceship is surrounded by a bubble of spacetime which moves faster than light, but the spaceship is not locally moving faster than light since it is at rest with respect to the bubble, so nothing is violated). The original paper by Alcubierre proposes a geometry and only then computes the stress tensor, which turns out to involve negative energy densities (and we currently do not know of any sort of classical matter with these sorts of properties).

The presence of negative energy densities, or matter going on acausal motion (moving faster than light), and so on is usually undesirable on General Relativity, so it is common to impose energy conditions to forbid these sorts of issues. Computing the stress tensor from the Einstein tensor might lead to tensors that violate these energy conditions, meaning that often these stress tensors might not be realizable by any sort of matter we currently know of.

In summary, yes, you can obtain the stress tensor given the Einstein tensor. However, this does not mean the stress tensor you obtain will be physically interesting or physically realizable.

$\endgroup$
2
  • $\begingroup$ By physically interesting, this means approximation, correct? $\endgroup$
    – aygx
    Dec 22, 2021 at 23:30
  • $\begingroup$ @aygx pretty much. "Physically interesting" here means that is approximates to some degree of precision the stress-energy tensor of something that actually exists within the phenomena we are considering (well, roughly, I'm not attempting to give a mathematically precise definition hahahaha). Something with negative energy densities is not physically interesting if we only consider classical matter, but quantum effects can lead to these effects, for example. $\endgroup$ Dec 23, 2021 at 1:14

Not the answer you're looking for? Browse other questions tagged or ask your own question.