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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.

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Your metric is:

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

and Einstein's Equation:

$$G_{μν} = -\frac{3}{L^{2}}g_{μν}$$

(Assuming that $L$ is the dS radius then the equation represents an AdS spacetime not dS. For a dS spacetime $G_{μν} = \frac{3}{L^{2}}g_{μν}$). Clearly there are two unknown functions and one expects two independent equations(the $θθ$ equation is related to $rr$ and $tt$). The $tt,rr,θθ$ equations are the following:

$$\frac{e^{-2 (\Lambda (r)+\Phi (r))} \left(-\left(L^2+3 r^2\right) e^{2 \Lambda (r)}-2 L^2 r \Lambda '(r)+L^2\right)}{L^2 r^2} =0$$

$$\frac{e^{-4 \Lambda (r)} \left(\left(L^2+3 r^2\right) e^{2 \Lambda (r)}-L^2 \left(2 r \Phi '(r)+1\right)\right)}{L^2 r^2}=0$$

$$\frac{\frac{3 r}{L^2}+e^{-2 \Lambda (r)} \left(\left(r \Phi '(r)+1\right) \left(\Lambda '(r)-\Phi '(r)\right)-r \Phi ''(r)\right)}{r^3} =0$$

The first equation is a differnetial equation for $Λ$. Integrating we get:

$$\Lambda (r)=-\frac{1}{2} ln \left(\frac{-c_1+L^2 r+r^3}{L^2 r}\right)$$

Using the above the second equation is solved:

$$\Phi (r)=\frac{1}{2} \left(ln \left(-c_1+L^2 r+r^3\right)-ln (r)\right)+c_2$$

,where $c_1,c_2$ are constants of integration and plugging the results in the third equation we can see that $Λ,Φ$ satisfy the equation. The metric will read:

$$d s^{2}=-(\frac{e^{2 c_2} \left(-c_1+L^2 r+r^3\right)}{r}) d t^{2}+(\frac{L^2 r}{-c_1+L^2 r+r^3})d r^{2}+r^{2} d \Omega^{2}$$

For the appropriate asymptotic behavior we shall identify:

$$c_2 = - ln(L^2)/2$$

and comparing with the vaccum solution: $c_1 = ML^2$.

Now the line element takes the form:

$$ds^{2} =- (1-M/r +\frac{r^{2}}{L^2})dt^2 + (1-M/r +\frac{r^{2}}{L^2})^{-1}dr^2 + r^2d\Omega^2$$

If you never tried to obtain (doing all calculations by hand) at least one of the famous solutions to Einstein's equations i strongly recommend you to do it.

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