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Angular momentum causes the event horizon of a black hole to recede. At maximum angular momentum, $J=GM^2/c$, the Schwarzschild radius is half of what it would be if the black hole wasn't spinning.

Can someone explain why angular momentum reduces the Schwarzschild radius?

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    $\begingroup$ interesting question... does the M in that expression include its rotational energy? $\endgroup$
    – Alan Rominger
    Jun 26, 2014 at 3:15
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    $\begingroup$ @AlanSE No, M is just the mass of the black hole. It doesn't include any rotational energy. $\endgroup$
    – Drew
    Jun 26, 2014 at 3:24
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    $\begingroup$ Here is a rough, qualitative and Newtonian argument that can at least make it plausible: Put yourself in a rotating reference frame following the rotation of the BH close to $r=r_s$. The BH is seen to be at rest, but in this frame you will experience a centrifugal force. The total force on you $GMm/r^2−mv^2/r=GM_{\rm eff}m/r^2$ implies the object has a smaller effective mass: $(M_{\rm eff}−M)/M \sim -\frac{J^2}{J^2_{\rm max}}$ and therefore a smaller effective $r_s$. I think this can be done properly in GR by roughly the same approach (rotating reference frame). $\endgroup$
    – Winther
    Jun 26, 2014 at 4:15
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    $\begingroup$ I'm not sure I'd use "Schwarzschild radius" to describe the size of the inner horizon in a Kerr black hole. That seems misleading to me, especially since we generally write $r_{inner}$ as a function of $r_s$ and $J/mc$. $\endgroup$
    – John Rennie
    Jun 26, 2014 at 5:49
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    $\begingroup$ Rough, Qualitatively, Newtonian... : Consider that the total energy $E$ of the black hole is the sum of positive non gravitational energies $E_{NG}$ (mass energy, rotational energy, electrostatic energy), and a negative gravitational energy $E_G \sim -G\frac{M^2}{R}$. The black hole is a limit case : $E=0, R=R_S$; Suppose that the non gravitational energy part is made of mass energy + rotational energy, that is $E_{NG} = M + E_{ROT}$ this gives $E_{NG} +E_G=0$, that is $M + E_{ROT} -G\frac{M^2}{R_S}= 0$, so when $E_{ROT}$ is increasing (starting from zero), we see that $R_S$ is decreasing. $\endgroup$
    – Trimok
    Jun 26, 2014 at 9:14

2 Answers 2

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I will approach this question theoretically, although I feel the intuition follows nicely. If we talk about Kerr black holes - rotating black holes described by their mass and angular momentum, with no additional parameters such as charge etc. - then you can show that the radius of the event horizon is given by

$\boxed{r=M + \sqrt{M^2-a^2}}$

where $a=\frac{J}{M}$.

(This value of $r$ is found by finding where the Kerr metric blows up; hence event horizon. In fact, finding where the metric blows up involves solving a quadratic equation, so we get two values of $r$ and in Kerr black holes we therefore have two event horizons; unlike in Schwarzschild black holes.)

Regarding your first point about maximum angular momentum, if we set $G=1$ and $c=1$, the maximum angular momentum you stated is given by $a=M$ and if we plug this into our equation for $r$ above we see that we have

$r=M$.

We know that the radius of the event horizon in a Schwarzschild black hole (no rotation) is $r=2M$. So therefore we can see that at maximum angular momentum, the radius of the event horizon is half of what it would be if the black hole weren't spinning.

To this end, we can also see that at zero angular momentum, $a=0$, we have

$r=2M$

which is what we want as at zero angular momentum we of course should have the Schwarzschild radius.

Using the boxed equation for $r$ at the top, it's easy to test out different values of $a$ to see what happens to the event horizon. For example, this equation alone is sufficient to show that for $a>M$ we don't have an event horizon, in which case we have what is a called "Fast Kerr" which is just a singularity with no event horizon.

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    $\begingroup$ It's not that the complex values for the horizon dont' work in general relativity, so much as that equation has no real solution. For $a > M$, you don't have a horizon. Those solutions describe naked singularities, not black holes. $\endgroup$
    – Zo the Relativist
    Jun 28, 2014 at 12:15
  • $\begingroup$ Yes that's right, it's commonly called "Fast Kerr" $\endgroup$
    – Phibert
    Jun 28, 2014 at 12:18
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    $\begingroup$ This strikes me as begging the question: sure the Kerr metric has critical points at these values, which get smaller as the angular momentum gets larger...but why does the metric have these properties in the first place, and why is it the right metric? $\endgroup$
    – zibadawa timmy
    Sep 15, 2019 at 3:56
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Maybe a qualitative answer motivated from thermodynamics: If you let your black hole rotate, you reduce the number of symmetries of your system, this will decrease your entropy $S$ which is proportional to the surface area. The surface area however is for sure monotonic increasing with your Schwarzschild radius, therefore, if your break symmetries, $r$ will decrease.

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