# Finding de-Sitter Schwarzschild Metric

Given the metric in Schwarzschild coordinates

$$d s^{2}=-e^{2 \Phi(r)} d t^{2}+e^{2 \Lambda(r)} d r^{2}+r^{2} d \Omega^{2}$$

and a vacuum Einstein equation with cosmological constant given by:

$$G_{\mu \nu}=-\frac{3}{L^{2}} g_{\mu \nu}$$

where $$L$$ is a constant. I'm interested in finding the de-Sitter-Schwarzschild metric $$g_{\mu \nu}$$.

I believe the non-zero components of the Einstein tensor are:

\begin{aligned} G_{r r} &=-\frac{1}{r^{2}} \mathrm{e}^{2 \Lambda}\left(1-e^{-2 \Lambda}\right) \\ G_{\theta \theta} &=-r \mathrm{e}^{-2 \Lambda} \Lambda^{\prime} \\ G_{\phi \phi} &=\sin ^{2} \theta G_{\theta \theta} \end{aligned}

but I'm not sure where to go from here in terms of finding the metric.

• Have you consulted a textbook? I think it is in most of the GR introductory books. – Andrew Steane Nov 7 '19 at 0:09
• – Qmechanic Dec 17 '19 at 23:16

$$G_{tt} = \cfrac{e^{(-2 Λ(r) + 2 Φ(r))} (-1 + e^{(2 Λ(r))} + 2rΛ'(r))}{r^2}$$, $$G_{rr} =\cfrac{1-e^{2Λ(r)} + 2rΦ'(r)}{r^{2}}$$ $$G_{θθ} = e^{-2Λ(r)}r((-Λ'(r) + Φ'(r))(1+rΦ'(r))+rΦ''(r))$$ and as expected: $$G_{φφ}=sin^{2}θG_{θθ}$$