Mass distribution of a black hole

Question

One of my references states that the moment of inertia of a black hole (as might be deduced from a safe distance outside the event horizon) is I(bh) = mr^2 where r is the radius of the event horizon. For comparison, the moment of inertia of a hollow shell is I(s) = (2/3)mr^2.

Now the moment of inertia of a thin hoop is also I(h) = mr^2, which upon initial inspection suggests that the mass of a black hole is not distributed as a thin shell at the location of the event horizon (which was my initial guess) but instead as a hoop. Thus my question is:

Does the moment of inertia of a black hole convey anything meaningful to us outside it, about how its mass is actually distributed inside or right at the event horizon? That is, is the hoop equivalence just a meaningless coincidence?

Answer 1

This is just a coincidence. Dimensional analysis immediately tells us that the moment of inertia of a black hole needs to be $I_{\rm bh} = \alpha m r_{s}^2$ for some dimensionless constant $\alpha$. The answer just happens to come out as $\alpha=1$. This does not tell us anything about the distribution of the mass. The quantities that might tell us something about the mass distribution are the gravitational multipole moments. For a Schwarzschild black hole these are all zero accept for the monopole. This is incompatible with the mass being concentrated in a ring on the horizon.