# Why does angular momentum shorten the Schwarzschild Radius of a black hole?

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.

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.