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BTZ black holes are defined for the case of 1+2D gravity theory because of closed form computation. I’m wondering if there exist a $1+3D$ analogue black hole?

Edit: What I'm actually looking for is the black hole solution of Einstein-Maxwell equation with a negative cosmological constant but in $1+3D$ dimension instead of $1+2D$, as done in BTZ case.

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  • $\begingroup$ You mean black hole solutions in 1+3D with a negative cosmological constant? $\endgroup$
    – Qmechanic
    Jun 18 '20 at 3:25
  • $\begingroup$ @Qmechanic Yes! $\endgroup$
    – aitfel
    Jun 18 '20 at 3:45
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    $\begingroup$ Even with that clarification the question is too vague: which aspects of BTZ black hole do you wish to replicate in 3+1 dimensions? Being locally isometric to AdS space, or something else? $\endgroup$
    – A.V.S.
    Jun 18 '20 at 18:38
  • $\begingroup$ @A.V.S. I have to admit it was quite vague but now I have clarified what I'm looking for. $\endgroup$
    – aitfel
    Jun 19 '20 at 8:56
  • $\begingroup$ You mean Reissner Nordstrom (A)dS black holes?? $\endgroup$
    – ApolloRa
    Jun 19 '20 at 9:09
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The solution you are looking for is the RN(A)dS black hole. The line element is given by:

$$ ds^{2} = -b(r)dt^2 + \cfrac{1}{b(r)}dr^2 + r^2 (d\theta ^2 + sin^{2}\theta d\phi^{2}) $$

where: $b(r) = 1 -\cfrac{2M}{r} + \cfrac{Q^2}{r^2} -\cfrac{Λr^2}{3}$

Note that $b(r)$ is the most general solution of the equation:

$$R=4Λ \Rightarrow -\frac{r^2 b''(r)+4 r b'(r)+2 b(r)-2}{r^2} = 4\Lambda \Rightarrow$$

$$ r^2 b''(r) + 4rb'(r)+ 2b(r)- 2=-4r^2 \Lambda \Rightarrow (r^2 b'(r))' + (2rb(r))' = -4r^2\Lambda +2 $$ $$ \Rightarrow (r^2 b'(r) + 2rb(r))' = (-\cfrac{4}{3}r^3 \Lambda +2r)' \Rightarrow r^2 b'(r) + 2rb(r) = -\cfrac{4}{3}r^3 \Lambda + 2r +c_1 \Rightarrow (r^2 b(r))'$$ $$ = (-\cfrac{4}{3}\cfrac{r^4}{4}\Lambda + r^2 +c_1 r)'\Rightarrow r^2 b(r) =- \cfrac{r^4}{3}\Lambda + r^2 +c_1r +c_2\Rightarrow$$ \begin{equation} b(r) = 1 - \cfrac{r^2}{3}\Lambda + \cfrac{c_1}{r} +\cfrac{c_2}{r^2} \end{equation} ,where $R = -\frac{r^2 b''(r)+4 r b'(r)+2 b(r)-2}{r^2} $ is the Ricci Scalar for the metric ansatz i imposed. Since the electromagnetic energy tensor is traceless, we can solve form the constant Ricci scalar condition for the metric function and then substitute back to Einstein's equations to see if this configuration satisfies them. The condition also satisfies Einstein's equations obtained from: $$G_{ab} + g_{ab}Λ = T_{ab}^{EM}$$

with the constraint that $c_2=Q$. Ιnformations about the solution can be found here: https://arxiv.org/abs/1405.4931

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  • $\begingroup$ Thanks! Can angular momentum be added to it? And if so what is the name of the family of solution? $\endgroup$
    – aitfel
    Jun 19 '20 at 12:25

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