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I am trying to derive the Kerr-Newman-Ads black hole in Eddington-Finkelstein coordinates, but I got stuck. The Kerr-Newman-Ads black hole in Boyer coordinate has the following form

$$d s^{2}=-\frac{\Delta_{r}}{\rho^{2}}\left[d t-\frac{a \sin ^{2} \theta}{\Xi} d \phi\right]^{2}+\frac{\rho^{2}}{\Delta_{r}} d r^{2}+\frac{\rho^{2}}{\Delta_{\theta}} d \theta^{2}+\frac{\Delta_{\theta} \sin ^{2} \theta}{\rho^{2}}\left[a d t-\frac{r^{2}+a^{2}}{\Xi} d \phi\right]^{2},$$

where

$$ \begin{array}{c} \rho^{2}=r^{2}+a^{2} \cos ^{2} \theta, \quad \Xi=1-\frac{a^{2}}{l^{2}}, \quad \Delta_{r}=\left(r^{2}+a^{2}\right)\left(1+\frac{r^{2}}{l^{2}}\right)-2 m r+z^{2}, \quad \Delta_{\theta}=1-\frac{a^{2}}{l^{2}} \cos ^{2} \theta . \end{array} $$ To change to the Eddington-Finkelstein coordinates, one needs to eliminate the $g_{rr}$ term.

Usually, the following coordinate transformation is applied:

$$ du= dt+ g(r)dr , \quad d\varphi= d \phi +h(r)dr. $$

However, the problem is we only have one equation for two unknown variables.

My question is for Kerr-Newman-Ads black hole in Boyer coordinate, which coordinate transformation is needed to switch to Eddington-Finkelstein coordinates. Does the Eddington-Finkelstein form exist for Kerr-Newman-Ads black hole? I would also appreciate it if anyone could directly show me the Kerr-Newman-Ads black hole in the Eddington-Finkelstein form with a reference.

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    $\begingroup$ Maybe can be useful arxiv.org/pdf/2007.04354.pdf $\endgroup$
    – Pipe
    Commented Jul 15, 2023 at 14:40
  • $\begingroup$ You posted this same question previously. Did you delete it and repost? If so, please don’t do that. $\endgroup$
    – Ghoster
    Commented Jul 15, 2023 at 17:07
  • $\begingroup$ @Ghoster Oh, sorry, I am testing it to make sure I didn't encounter a bug because I remember if I add new content to the question description then I could see it on the home page, but I didn't see my question. So I just test whether I can see my question on the home page by reposting this question. This will not happen again $\endgroup$
    – David Shaw
    Commented Jul 15, 2023 at 20:02
  • $\begingroup$ @Pipe Thank you, it solves my problem. $\endgroup$
    – David Shaw
    Commented Jul 15, 2023 at 20:06
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    $\begingroup$ @Yukterez Thanks. Do you know if there is a Kerr-Schild form for Kerr-Newman-AdS? $\endgroup$
    – Ghoster
    Commented Jul 16, 2023 at 17:47

1 Answer 1

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DavidXiao wrote: one needs to eliminate the $g_{\rm rr}$ term

If we take the Kerr Newman De Sitter metric in $\rm \{t, \ r, \ \theta, \ \varphi\}$ Boyer Lindquist coordinates in natural units of $\rm G=c=k_B=1$ and apply the transformation

$\rm dt=du-dr \ \left(a^2+r^2\right) \chi/\Delta \ , \ \ d\varphi =d\phi -dr \ a \ \chi /\Delta $

with $\rm a$ for the spin, $\rm Q $ the charge, $\rm \Lambda=3 \ H^2$ the cosmological constant and

$\rm \Delta =\left(a^2+r^2\right) \left(1-\Lambda \ r^2/3\right)-2 \ M \ r+Q ^2\ , \ \ \chi =a^2 \ \Lambda /3+1$

we get the $\rm \{u, \ r, \ \theta, \ \phi\}$ Null coordinate form with the covariant metric components

$g_{\rm uu}=\rm-\frac{3 \ \left(a^2 \sin^2 \theta \ \left(a^2 \ \Lambda \cos^2 \theta +3\right)+a^2 \left(\Lambda \ r^2-3\right)+6 \ M \ r+\Lambda \ r^4-3 \ \left(r^2+Q ^2\right)\right)}{\left(a^2 \ \Lambda +3\right)^2 \left(a^2 \cos^2 \theta +r^2\right)}$

$g_{\rm rr}=0 $

$g_{\rm \theta\theta}=\rm -\frac{3 \ \left(a^2 \cos^2 \theta +r^2\right)}{a^2 \ \Lambda \cos^2 \theta +3} $

$g_{\rm \phi\phi}=\rm-\frac{9 \ \left(\frac{1}{3} \left(a^2+r^2\right)^2 \sin^2 \theta \ \left(a^2 \ \Lambda \cos^2 \theta +3\right)-a^2 \sin^4 \theta \ \left(\left(a^2+r^2\right) \left(1-\Lambda \ r^2/3\right)-2 \ M \ r+Q ^2\right)\right) }{\left(a^2 \ \Lambda +3\right)^2 \left(a^2 \cos^2 \theta +r^2\right)}$

$g_{\rm ur}=\rm -\frac{3}{a^2 \ \Lambda +3}$

$g_{\rm u\phi}=\rm \frac{3 \ a \sin^2 \theta \ \left(a^2 \ \Lambda \ \left(a^2+r^2\right) \cos^2 \theta +\Lambda \ r^2 \ \left(a^2+r^2\right)+6 \ M \ r-3 \ Q ^2\right)}{\left(a^2 \ \Lambda +3\right)^2 \left(a^2 \cos^2 \theta +r^2\right)}$

$g_{\rm r\phi}=\rm \frac{3 \ a \sin^2 \theta }{a^2 \ \Lambda +3}$

$g_{\rm r\theta}=0 $

$g_{\rm u\theta}=0 $

$g_{\rm \theta\phi}=0 $

and the electromagnetic vectorpotential

$\rm A_{\mu}=\left\{\frac{3 \ r \ Q }{\left(a^2 \ \Lambda +3\right) \left(a^2 \cos^2 \theta +r^2\right)},\frac{3 \ r \ Q }{a^2 \ \left(\Lambda \ r^2-3\right)+6 \ M \ r+\Lambda \ r^4-3 \ \left(r^2+\mho ^2\right)}, \ 0, \ -\frac{3 \ a \ r \ Q \sin ^2 \theta }{\left(a^2 \ \Lambda +3\right) \left(a^2 \cos^2 \theta +r^2\right)}\right\}$

If you replace $\rm \chi \to \chi^2$ in the transformation rule you get $g_{\rm ur}=-1$, but the $g_{\rm rr}$ is then no longer $0$. For more details see here and here and the links therein.

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