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?


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


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:

  1. Kerr black hole horizons and infinite redshift surfaces

  2. What's the inner ergosphere in a Kerr black hole?

  3. Escape velocity from a rotating black hole

  4. Exceeding the Kerr black hole spin limit

  5. Free-falling from rest into a Kerr black hole

  6. For a given mass, how big can a Kerr black hole get?

  7. Particle falling into a Kerr black hole

  8. Particle crossing the outer event horizon of a Kerr black hole

  9. What is exactly rotating in a rotating black hole?

  10. What kind of volume does the event horizon of a Kerr black hole enclose?

  • $\begingroup$ 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. $\endgroup$ – Prahar Jun 27 at 19:49
  • $\begingroup$ @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. $\endgroup$ – SG8 Jun 28 at 21:06

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