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The Enstein field equations are $R_{\mu\nu} - \frac{R} {2} g_{\mu\nu} = - \frac{8\pi G} {c^4} T_{\mu\nu} $

In vacuum, it's assumed that $R_{\mu\nu} = 0$.

The energy momentum tensor has to be zero in vacuum. Now, I'm wondering, why it's derived

$R_{\mu\nu} =0 $ in the vacuum and not

$R_{\mu\nu} - \frac{R} {2} g_{\mu\nu} = 0$?

BarrierRemoval
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1 Answers1

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These are actually equivalent.

If $R_{\mu\nu}=0$ then $R=R_\mu^\mu=0$.

For the other direction we assume that $R_{\mu\nu}=\frac{R}{2} g_{\mu\nu}$, so then taking the trace on both sides gives $R_\mu^\mu=\frac{R}{2}d$ since the trace of the metric tensor is the number of dimensions. So in $d=4$ it gives $R=2R\implies R=0$ which we can substitute in to get $R_{\mu\nu}=0$.

Jean Du Plessis
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  • Thank you! Could you please quickly show that the trace of the Schwarzschild metric tensor is 4? – BarrierRemoval Mar 15 '23 at 17:32
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    Sure, the trace will be $g_{\mu\nu}g^{\mu\nu}$, which by the definition of the inverse metric tensor (as well as the symmetry of the metric) is equal to $\delta_\nu^\nu$ which is clearly just $1+1+1+1$ in $d=4$. For the Swartzschild metric specifically, you would use this definition of the inverse metric tensor in order to calculate its components, so anything involving the actual components would be superfluous. – Jean Du Plessis Mar 15 '23 at 22:28