Why $R_{\mu\nu}$ is zero in the vacuum and not $R_{\mu\nu} - \frac{R} {2} g_{\mu\nu}$?

Question

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

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

Answer 1

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 $R_{\mu\nu}=0$.