# Prove that deSitter-Schwarzschild metric with constant Ricci scalar is Einstein

Reference: this paper page 4

Prove that the following metric is Einstein

Given deSitter-Schwarzschild metric with mass $m > 0$ and scalar Ricci $R=2$ by

\begin{align} \bigg( 1-\frac{r^2}{3}-\frac{2m}{r} \bigg)^{-1} dr^2 + r^2 g_{\Bbb S^2} \end{align}

where $g_{\Bbb S^2} = d\theta^2 + \text{sin}^2 \theta\ d\phi^2$.

Proof.

\begin{align} R_{11} =\ & \Lambda g_{11} \nonumber\\ % % \Lambda =\ & -\frac{2m}{r^3} + \frac{2}{3} \tag{1} \end{align}

\begin{align} R_{22} =\ & \Lambda g_{22} \nonumber\\ % % \Lambda =\ & \frac{m}{r^3} + \frac{2}{3} \tag{2} \end{align}

\begin{align} R_{33} =\ & \Lambda g_{33} \nonumber\\ % % \Lambda =\ & \frac{m}{r^3} + \frac{2}{3} \tag{3} \end{align}

Since $\Lambda_{(1)}$ is not the same with $\Lambda_{(2)}$ and $\Lambda_{(3)}$, then the metric is NOT Einstein. On the other hand, the Scalar Ricci is constant, thus the metric is supposed to be Einstein. And now I am confused.

• Your metric does not satisfy $R_{ab}=kg_{ab}$. Any reason to expect it does? Can you cite some reference where it is claimed that it does? Aug 2, 2018 at 23:01
• The de Sitter Schwarzschild metric has a time component too. Any reason you are ignoring that? Aug 2, 2018 at 23:31

1, First, You should check if Your metric is vacuum solution of Einstein equation with non-zero cosmological constant $\Lambda$. Vanishing $\Lambda$ will give a trivial equality (if the metric is vacuum solution).
Generally, Einstein equation with zero matter tensor can be rewritten as (Einstein metric) $$R_{\mu \nu}=\frac{2\Lambda}{d-2}g_{\mu\nu}$$
2, Your metric (as I checked) is neither the solution of $d=3$ nor $d=4$ (with a constant time-time component) Einstein equation. So I think Your problem is here.