1
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.