Solving differential equation for the Schwarzschild metric with cosmological constant

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

Attempt:

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.

• – Qmechanic Mar 11 '19 at 15:43

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 $$e^{2\beta} = \frac{1}{-\frac{\Lambda r^2}{3}+1+\frac{c_1}{r}}$$