In a exterior region without matter to a stationary black hole, spherical symmetric, where the cosmological constant is not zero. From the Cartan's structural equations for space without torsion, we get the following non zero Ricci's tensor:

$$ R_{r0} = 0 \\ R_{00} =e^{-(U + V)}[\partial_r ( e^{(-V + U)} \partial_rU)] + \frac{2}{r} e^{-2V} \partial_rU \\ R_{rr} = -e^{-(U+V)}[\partial_r(e^{(-V + U)} \partial_rU] + \frac{2}{r} e^{-2V} \partial_rV \\ R_{\theta \theta} = R_{\phi \phi} =\frac{e^{-2V}}{r}(\partial_rV - \partial_rU) + \frac{1}{r^2} (1 - e^{-2V}) $$

Then I am asked to find components of the metric resolving Einstein's equations and right here I'm kinda stuck, I can write the general Einstein's equation:

$$ G_{\mu \nu} + \Lambda g_{\mu\nu} = R_{\mu \nu} - \frac{R}{2}g_{\mu \nu} + \Lambda g_{\mu\nu} = \kappa T_{\mu \nu}$$

and then how can find the components of the metric using this?


Corrected Einstein's Equation


Rereading the exercise, I noticed I had more information that I didn't wrote on this post.

The metric components are given by:

$$ds^2 = g_{\mu \nu} dx^{\mu} dx^{\nu} = \eta_{\mu \nu} \omega^{\mu} \otimes \omega^{\nu} $$

and $$ \begin{align} \omega^0 = e^{U(r)dt} \\ \omega^1 = e^{V(r)dt} \\ \omega^{\theta} = rd\theta \\ \omega^{\phi} = r\sin\theta d\phi \end{align} $$

From this I can compute the metric and write:

$$ g_{\mu \nu} = diag(-e^{2U(r)}, e^{2V(r)}, r^2 , r^2 \sin^2 \theta) $$

So for the component $00$, we write:

$$R_{00} - \frac{R}{2}g_{00} + \Lambda g_{00} = 0 \\ \Leftrightarrow R_{00} - \frac{g^{\alpha \beta} R_{\alpha \beta}}{2}g_{00} + \Lambda g_{00} = 0 $$

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    $\begingroup$ First, what you wrote as the Einstein equations is not correct. By definition, $G_{\mu\nu}\equiv R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}$. The Einstein equations are then $G_{\mu\nu}+\Lambda g_{\mu\nu}=\kappa T_{\mu\nu}$, where $\kappa=8\pi G/c^4$, $\Lambda$ is the cosmological constant, and $T_{\mu\nu}$ is the stress-energy tensor. Since you are working in vacuum, $T_{\mu\nu}=0$, so the Einstein equations are $G_{\mu\nu}+\Lambda g_{\mu\nu}=0$. (...) $\endgroup$ – Andrew Jan 12 at 15:58
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    $\begingroup$ (...) Second, in general the Einstein equations are a set of non-linear second order PDEs for the metric. However, since you are dealing with a spherically symmetric, static spacetime, you will really end up with an ODE where everything is a function of $r$ only. So you want to express the equations in a form where this structure is manifest. As a starting point, try to write out the $00$ component of the Einstein equations explicitly in terms of $U$ and $V$. Can you see how it gives you an ODE? If that works, try doing the $0i$ and $ij$ components. $\endgroup$ – Andrew Jan 12 at 16:01
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    $\begingroup$ @RFeynman So let's take your first line, $R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R+\Lambda g_{\mu\nu}=0$. Then let's look at the $00$ component, $R_{00}-\frac{1}{2} g_{00} R + \Lambda g_{00}=0$. We want to extract an equation for $U$ and $V$. So you need to plug in the $00$ component of the metric, $g_{00}$, the Ricci scalar $R$, and the $00$ component of the Ricci tensor, $R_{00}$, in terms of $U$ and $V$. In your question you actually have $R_{00}$ already. You should also be able to write out the metric $ds^2=g_{\mu\nu} dx^\mu dx^\nu$ to get $g_{00}$. The Ricci Scalar is $g^{\mu\nu}R_{\mu\nu}$. $\endgroup$ – Andrew Jan 12 at 16:24
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    $\begingroup$ Also please be careful. In your edited question you write down $g^{\mu\nu}R_{\mu\nu} g_{\mu\nu}$. This notation "doesn't parse" -- in the Einstein summation convention, repeated indices are summed over. A given index should either appear twice, once upstairs and once downstairs , in which case it is summed over and is a dummy index. OR, the index should appear once, either upstairs or downstairs, in which case it is a "free" index and is not summed over. A correct way to write this term is $g^{\alpha \beta} R_{\alpha \beta} g_{\mu\nu}$ -- there is an implicit sum over $\alpha,\beta$. $\endgroup$ – Andrew Jan 12 at 16:32
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    $\begingroup$ The $00$ component is $R_{00}-\frac{1}{2} g_{00} R + \Lambda g_{00}=0$. To compute $R$, you need to sum $g^{\alpha \beta} R_{\alpha \beta}$ over $\alpha$ and $\beta$. Here, note that $g^{\alpha\beta}$ is the inverse metric, which is different from the metric $g_{\alpha \beta}$. Meanwhile, you already have written an expression for $R_{00}$ in your question. You should also have written down an expression for $g_{\mu\nu}$ somewhere, in order to have computed the Ricci tensor -- from this expression, you can get $g_{00}$ as well as the inverse metric. $\endgroup$ – Andrew Jan 12 at 16:53

After some tips from @Andrew, I could solve the exercise:

We start by writing Einstein's equation in vaccum ($T_{\mu \nu} = 0$):

$$G_{\mu \nu} + \Lambda g_{\mu\nu} = R_{\mu \nu} - \frac{R}{2}g_{\mu \nu} + \Lambda g_{\mu\nu} = 0 \\ \Leftrightarrow R_{\mu \nu} - \frac{g^{\alpha \beta}R_{\alpha \beta}}{2}g_{\mu \nu} + \Lambda g_{\mu\nu} = 0 $$

Then, we plug the component we want in $\mu $ and $\nu$ , then we solve a ODE for U and V and get the components the metric.


To find the inverse metric we use, due to the fact that the metric is diagonal: $$ g_{\mu \nu} = \frac{1}{g^{\mu \nu}}$$


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