# Event horizon of a rotating black hole

For a non-rotating black hole, Schwarzschild radius itself forms the event horizon, but how do we find the event horizon of a rotating Kerr black hole?

• Did you check Wikipedia? Commented Jun 3, 2021 at 3:09
• Commented Jun 3, 2021 at 4:19

The event horizon' radius of black holes is an infinite-redshift surface (a one-way surface where particles can never escape to infinity). It can be computed analytically (or at least numerically) by finding the largest (real) positive root of the inverse of the component $$g_{rr}$$ of the metric tensor, i.e., by solving

$$\frac{1}{g_{rr}}=0.$$

It is common to define the emblackening factor as $$f(r) = \frac{1}{g_{rr}}$$ (Sometimes it's called the metric function). In this way, the location of event horizon is obtained by solving $$f(r)=0$$ and finding its largest positive root.

Now, consider the metric of the Kerr black hole in the Boyer–Lindquist coordinates

\begin{align*} ds^2 = &-\left(1 - \frac{2GMr}{r^2+a^2\cos^2(\theta)}\right) dt^2 + \left(\frac{r^2+a^2\cos^2(\theta)}{r^2-2GMr+a^2}\right) dr^2 + \left(r^2+a^2\cos(\theta)\right) d\theta^2\\ &+ \left(r^2+a^2+\frac{2GMra^2}{r^2+a^2\cos^2(\theta)}\right)\sin^2(\theta) d\phi^2 - \left(\frac{4GMra\sin^2(\theta)}{r^2+a^2\cos^2(\theta)}\right) d\phi\, dt, \end{align*}

where $$a=J/M$$ is the Kerr (rotation) parameter. It turns out that you have to solve the following equation

$$\frac{1}{g_{rr}}=\frac{r^2-2GMr+a^2}{r^2+a^2\cos^2(\theta)}=0 \to r^2-2GMr+a^2=0,$$

which could have two roots. The larger positive root is the location of the event horizon and it reduces to the Schwarzschild radius when $$a=0$$ as the nonrotating limit.

The above argument is valid for black holes in flat or anti-de Sitter backgrounds. But, for the case of black holes in the de Sitter (dS) space, more attention is needed. In such cases, the metric function has a negative slope at the largest root which implies a negative Hawking temperature for dS black holes (which certainly is a nonphysical property.) So, for dS black holes, the event horizon's radius is the largest positive root of the metric function with the positive slope (which ensures the corresponding Hawking temperature is positive definite in agreement with the laws of black hole thermodynamics).

Some useful links about Kerr black holes in Physics Stack Exchange:

• Your definition of the horizon does not seem to be coordinate invariant. For instance, for the Schwarzschild metric in standard coordinates $\frac{1}{g_{rr}} = 1 - \frac{2M}{r}$ so here clearly $r=2M$ is the horizon as you said. Now change coordinates $r' = r \sqrt{ 1 - 2M/r} + 2 M \tanh^{-1} \sqrt{ 1 - 2 M / r }$ and in these coordinates $g_{r'r'} = 1 \neq 0$ which would suggest that there is no horizon. Clearly this is not right. Commented Jun 27, 2021 at 19:49
• @PraharMitra - Thank you very much for you comment. Well, as you know, the answers always depend on the way the question is asked and also they depend on the level of the OP. But, I will add a note about the point you raised soon.
– SG8
Commented Jun 28, 2021 at 21:06
• Nice answer (+1) I am trying to get to zero radius event horizon (naked singularity) from your equation $r^2 - 2GMr + a^2=0$, which as you said should be the larger positive root. But from $r=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$, it seems not possible to get $r=0$. Could you show how a high enough spin $J$ can give rise to $r=0$ naked singularity in the Kerr metric, please? Commented Aug 12 at 23:01
• @James A naked singularity means there is no event horizon, but there exists an essential singularity at the origin. So, it is different from the condition that you mentioned, i.e., $r=0$, this is different and it is not relevant to naked singularities. To have a naked singularity, the condition $b^2 - 4 a c < 0$ is enough since the equation won't have any roots so there is no longer an event horizon (assuming that an essential singularity exists at the origin). And a high enough spin means $a$ can take large values which can satisfy the condition $b^2 - 4 a c < 0$ to have a naked singularity.
– SG8
Commented Aug 21 at 17:44
• Ah yes, that makes sense, thank you. Commented Aug 22 at 0:44