# Stress-energy tensor given Einstein tensor [duplicate]

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?

• Yes, people do this all the time. Dec 22, 2021 at 19:12
• 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.
– J.G.
Dec 22, 2021 at 19:47
• 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 $$T_{\mu\nu}=\frac{c^4}{8 \pi G}({R_{\mu\nu}-\frac12Rg_{\mu\nu}})$$?
– aygx
Dec 22, 2021 at 19:54
• 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.
– J.G.
Dec 22, 2021 at 19:57