# Area of Kerr-Newman event horizon

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 ?

• 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. – user4552 May 28 '19 at 19:34
• Yes, my main goal is area. – Ali Oz May 28 '19 at 19:37
• I think I managed to calculate it. If there is anyone who wonders how, I can write the answer here. – Ali Oz May 28 '19 at 20:11
• I would like to see how you did the integral. – G. Smith May 28 '19 at 20:33
• 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. – 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,

$$r_\pm=m\pm\sqrt{m^2-a^2-Q^2}.$$

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 $$r_+$$,

$$r_+^2-2Mr_++a^2+Q^2=0,$$

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

$$dA_+=(r_+^2+a^2)\sin{\theta}\,d\theta\,d\phi.$$

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

$$A_+=4\pi(r_+^2+a^2)=4\pi\left(2M^2-Q^2+2M\sqrt{M^2-a^2-Q^2}\right).$$

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