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? Jun 3, 2021 at 3:09
• 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).

• 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. Jun 27, 2021 at 19:49