How do we solve for Einstein's equation in the vacuum with a cosmological constant, in the static spherically symmetric situation?


Following Sean Carroll's Spacetime and Geometry (p.195), I write the equations

\begin{align}R_{tt} - \frac 12 R g_{tt} + \Lambda g_{tt} &= 0 \tag{1}\\ R_{rr} - \frac 12 R g_{rr} + \Lambda g_{rr} &= 0\tag{2}\end{align}

And contract $R_{\mu\nu}-\frac 12 R g_{\mu\nu} + \Lambda g_{\mu\nu} = 0$ with the inverse metric to obtain $R = 4\Lambda$, which I plug back in $(1)$ and $(2)$.

Then, as in Carroll, I multiply $(1)$ by $\exp[-2(\alpha-\beta)]$, where $\alpha$ and $\beta$ are functions of $r$ and add it to $(2)$ to obtain:

$$\frac 2r \partial_r(\alpha +\beta) = \Lambda e^{-2\beta}\left(1-e^{-4(\alpha+\beta)}\right)\tag{3}$$

Can we solve this differential equation?

Addendum: I simplified $(3)$ to

$$\frac 2r \partial_r(\alpha +\beta) = \Lambda \left(1-e^{-4(\alpha+\beta)}\right)\tag{3*}$$

and obtained

$$\Lambda r^2 + 4C = \ln\left(e^{4(\alpha+\beta)}+1\right)$$

where $C$ is a constant of integration.


1 Answer 1


First, you should try to get a relation between $\alpha$ and $\beta$, it will simplify a lot the calculations and expressions.

For example, using the $rr$ and $tt$ components you can arrive to the expression $$\frac{2}{r}\partial_r\alpha + \frac{2}{r}\partial_r\beta=0 $$ which means that $\alpha = -\beta$.

With this result in one hand, you can plug it in the $tt$ equation and get a solution in terms of the exponential $$e^{2\beta} = \frac{1}{-\frac{\Lambda r^2}{3}+1+\frac{c_1}{r}}$$

If you won't find the way to get these answers, tell me and I will post a full resolutions with all the steps.


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