I want to calculate the area of event horizon for a Kerr-Newman black hole by using boyer's coordinates.

I searched a lot from web, but I could not find any information about calculating event horizon radius for kerr newman. Can anyone help me to find any information, or help me how to calculate ?

  • $\begingroup$ Is the question about how to calculate a radius, or how to calculate an area? A radius would be kind of meaningless, since it's coordinate-dependent. $\endgroup$ – user4552 May 28 '19 at 19:34
  • $\begingroup$ Yes, my main goal is area. $\endgroup$ – Ali Oz May 28 '19 at 19:37
  • $\begingroup$ I think I managed to calculate it. If there is anyone who wonders how, I can write the answer here. $\endgroup$ – Ali Oz May 28 '19 at 20:11
  • $\begingroup$ I would like to see how you did the integral. $\endgroup$ – G. Smith May 28 '19 at 20:33
  • $\begingroup$ It would be best to provide an answer if you have one so that you can close the question formally. It will stop the system having yet another open-forever question on it. $\endgroup$ – StephenG May 29 '19 at 1:47

In geometrical-Gaussian units with $G$, $c$, and $\frac{1}{4\pi\epsilon_0}$ equal to 1, the Kerr-Newman metric for a black hole of mass $M$, angular momentum $J=aM$, and charge $Q$ is

$$\begin{align} ds^2= &-\left(1-\frac{2Mr-Q^2}{r^2+a^2\cos^2{\theta}}\right)dt^2 +\frac{r^2+a^2\cos^2{\theta}}{r^2-2Mr+a^2+Q^2}dr^2\\ &+(r^2+a^2\cos^2{\theta)}\,d\theta^2 +\left(r^2+a^2+\frac{a^2(2Mr-Q^2)\sin^2{\theta}}{r^2+a^2\cos^2{\theta}}\right)\sin^2{\theta}\,d\phi^2\\ &-\frac{2a(2Mr-Q^2)\sin^2{\theta}}{r^2+a^2\cos^2{\theta}}\,dt\,d\phi \end{align}$$

in Boyer-Lindquist coordinates $(t,r,\theta,\phi)$. (When $Q$ is zero, this reduces to Wikipedia’s form for the Kerr metric. Wikipedia’s form for the Kerr-Newman metric is equivalent to the above, but seems less straightforward.)

The $g_{rr}$ component of the metric tensor is infinite when the denominator $r^2-2Mr+a^2+Q^2$ is zero. This happens at two radial coordinates,


The event horizon is at $r_+$. We want to find the area of this surface. The 2D metric on the surface $t=$ constant and $r=r_+$ is

$$ds_+^2= (r_+^2+a^2\cos^2{\theta)}\,d\theta^2 +\left(r_+^2+a^2+\frac{a^2(2Mr_+-Q^2)\sin^2{\theta}}{r_+^2+a^2\cos^2{\theta}}\right)\sin^2{\theta}\,d\phi^2$$

and the area element on this surface is

$$\begin{align} dA_+&=\sqrt{\det{g_+}}\,d\theta\,d\phi\\ &=\sqrt{(r_+^2+a^2\cos^2{\theta})\left(r_+^2+a^2+\frac{a^2(2Mr_+-Q^2)\sin^2{\theta}}{r_+^2+a^2\cos^2{\theta}}\right)}\sin{\theta}\,d\theta\,d\phi\\ &=\sqrt{(r_+^2+a^2\cos^2{\theta})(r_+^2+a^2)+a^2(2Mr_+-Q^2)(1-\cos^2{\theta})}\sin{\theta}\,d\theta\,d\phi\\ &=\sqrt{(r_+^4+a^2r_+^2+2Ma^2r_+-a^2Q^2)+a^2(r_+^2-2Mr_++a^2+Q^2)\cos^2{\theta}}\sin{\theta}\,d\theta\,d\phi. \end{align}$$

Conveniently, the coefficient of $\cos^2{\theta}$ in the square root vanishes by the definition of $r_+$,


and, using this equation to eliminate $M$ in the first term in the square root, what's left under the square root becomes the perfect square $(r_+^2+a^2)^2$. Thus the area element simplifies to the trivial-to-integrate


Integrating over $\theta$ from $0$ to $\pi$ and over $\phi$ from $0$ to $2\pi$ gives the area of the event horizon,


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  • $\begingroup$ I did exactly the same way and found same answer. Thank you so much for writing the answer! $\endgroup$ – Ali Oz May 29 '19 at 16:06
  • $\begingroup$ @AliOz Would you like to mark it as accepted? If so, thanks. $\endgroup$ – G. Smith May 29 '19 at 16:08

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