Einstein field equation in vacuum : why $G_{\mu\nu} = 0 \implies R_{\mu\nu} = 0$ Starting from the Einstein field equation (without the cosmological constant),
$$
\underbrace{R_{\mu\nu} - \frac{1}{2}g_{\mu\nu} R}_{G_{\mu\nu}} = \underbrace{\frac{8\pi G}{c^4}}_{\kappa} T_{\mu\nu}
$$
If I'm looking for the vacuum solution, why does it imply that $T_{\mu\nu} = 0$ ?
Then, if I assume that, I get
$$
G_{\mu\nu} = 0
$$
However in textbooks, it says that the Einstein field equation in vacuum reduce to
$$
R_{\mu\nu} = 0
$$
How do we obtain that ?
 A: With respect to the first question, to get the Einstein field equation, you must begin with an action of the form
\begin{equation}
S=S_{\text{EH}}+S_{\text{M}},
\end{equation}
where
\begin{equation}
S_{\text{EH}}=\frac{c^{4}}{16\pi G}\int\mathrm{d}^{4}x\sqrt{-g}\,R
\end{equation}
is the Einstein-Hilbert action, whose variation with respect to $g_{\mu\nu}$ gives you the Einstein tensor in the equation, and
\begin{equation}
S_{\text{M}}=\int\mathrm{d}^{4}x\sqrt{-g}\,\mathcal{L}_{\text{M}}
\end{equation}
is the action corresponding to matter, whose variation with respect to $g_{\mu\nu}$ gives you the energy-momentum tensor. For vacuum, there is no matter Lagrangian ($\mathcal{L}_{\text{M}}=0$), and therefore $T_{\mu\nu}=0$.
With respect to the second question, if you take the trace of the Einstein field equation, you can obtain the expression
\begin{equation}
R=-\frac{8\pi G}{c^{4}}T
\end{equation}
where $T\equiv g^{\mu\nu}T_{\mu\nu}$. Therefore you can rewrite the Einstein field equation as
\begin{equation}
R_{\mu\nu}=\frac{8\pi G}{c^{4}}\left(T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T\right)
\end{equation}
From this expression, it is clear that, if $T_{\mu\nu}=0$, then $R_{\mu\nu}=0$.
