# Does a non-spherical black hole have distribution of mass like an empty body, solid body or pointed object?

Suppose a non-spherical (say, rotating or under distortion of another gravity source) black hole.

Does it have its mass distributed as if all the mass was on its surface, or as if the mass were distributed over its volume as some density or it would behave like a body with all its mass in the center?

For a spherical BH it would be all indistinguishable, but what about a non-spherical case?

Caveat reader: the current community voting on this answer suggests it doesn't have enough disclaimers about how mixing classical and relativistic calculations is a recipe for saying things that don't make sense in an erudite-sounding way. There should be one such disclaimer after about every sentence. See also the comments. Having said that:

You can probe the mass distribution of a spherically symmetrical object by spinning it and measuring its moment of inertia, $$I = J/\Omega$$. A solid classical sphere with mass $$M$$ and radius $$R$$ has $$I_\text{solid}=\frac25MR^2$$, while a thin spherical shell with the same mass and radius has the larger $$I_\text{shell}=\frac23MR^2$$ because more of the mass is further from the rotation axis.

A rotating black hole has a nonlinear relationship between $$J$$ and $$\Omega$$. Using this notation (see also), including the gravitational radius $$R_G = GM/c^2$$, there's a parameter $$-\pi/2 \leq \Phi \leq \pi/2$$ which characterizes the rotation by

$$a = \frac{J}{Mc} = R_G\cos\Phi$$

In this case $$\Phi=0$$ (or $$a=R_G$$) corresponds to a maximally-rotating Kerr black hole and $$|\Phi|=\pi/2$$ collapses to the non-rotating case. The rotating black hole has "outer" and "inner" event horizons, with radii

$$r_\pm = R_G\cdot(1\pm\sin\Phi) = R_G \pm \sqrt{R_G^2 - a^2}$$

The outer radius, $$r_+\to2R_G$$, is the Schwartzchild radius, the size of the event horizon in the non-rotating limit. There are also angular frequencies associated with these horizons,

\begin{align} \Omega_\pm &= \frac{c\cos\Phi}{2r_\pm} = \frac{ca}{2R_G^2 \pm 2R_G\sqrt{R_G^2 - a^2}}, \end{align}

although interpreting $$\Omega$$ as the angular frequency of a rigid classical object raises some thorny questions. The definition can be solved for the specific angular momentum $$a$$:

\begin{align} a = \frac{J}{Mc} &= \frac\Omega{c} \frac{4R^2}{1 + (2R\Omega/c)^2} \quad(\text{both of }\Omega_\pm) \\ J &= \frac{4}{1 + (2R\Omega/c)^2}\times MR^2\Omega \end{align}

This suggests you might consider a "moment of inertia" $$I=J/\Omega$$ of

\begin{align} I_\text{slow} &\approx 4MR_G^2 \approx Mr_+^2 \\ I_\text{max} &= 2MR_G^2 = 2Mr_+^2 \end{align}

That's interesting. When finding moments of inertia in classical physics one always finds $$I=fMR^2$$ by dimensional analysis, and if the radius $$R$$ is the maximum size of the rotating object one always finds $$f\leq1$$. For both limits of the moment of inertia here, the coefficient is $$f\gt1$$, which (combined with the default assumption of spherical-ish symmetry) suggests that $$R_G$$ is an underestimate of the classical size of the rotating mass distribution. If you wanted a sphere of mass $$M$$ to have the same classically-computed moment of inertia as a non- or slowly-rotating black hole with that mass, you'd make a spherical shell or a uniform-density sphere with radius larger than the Schwartzchild radius of the event horizon. (A thin hollow shell would go at $$\sqrt6 R_G = 1.23 r_+$$.) The maximally-spinning black hole, which has an outer event horizon of size $$r_+=R_G$$, likewise has "too large" of a moment of inertia.

Conclusion: Using classical moment-of-inertia considerations to analyze data on $$J/\Omega$$ for rotating black holes would lead you to require that some or all of their mass distribution were outside of their event horizons.

I personally find that kind of satisfying in a hand-waving, non-mathy kind of a way. After all, we no longer interact with matter that has crossed inside of the event horizon. If the mass-energy distribution of a black hole were actually found inside of its event horizon, wouldn't we be unable to interact with it? In electromagnetism, energy is stored in the fields, and the gravitational field of a black hole certainly extends outside of its horizon, so perhaps it's not nutty to locate some of the energy density near but outside of the event horizon. But this hand-wavy interpretation probably would not survive contact with a careful relativist.

• The more basic question that the OP could have started with is "Does a non-spherical black hole have distribution of mass?" The answer is no. There is no sensible way to define such a thing. The fact that the calculation in this answer doesn't give a result that makes sense is a sign of this. If we really believed that it was meaningful to talk about where the mass is located, then we would be able to tell whether, in the case of a black hole formed by gravitational collapse, the infalling matter had reached the horizon "now," or had reached the singularity "now." We can't. – user4552 Dec 9 '18 at 23:05
• Also, I think this answer could benefit from some kind of disclaimer to the effect that there is really no $\Omega$ in the Newtonian sense; the linked answer by gj255 has a note at the end about this. – user4552 Dec 9 '18 at 23:06
• What if the sphere has imaginable radius? – Anixx Dec 10 '18 at 2:56
• @BenCrowell Your first comment would make a nice complementary answer. I've tried to address your second. – rob Dec 10 '18 at 4:18
• @Anixx Do you mean an "imaginary" radius, like if there were enough angular momentum that $a>R$ and $r_+$ became complex? I think that the usual interpretation of the Kerr metric is that $a\to R$ is an unreachable limit, in the same way that $v\to c$ is an unreachable limit in special relativity. – rob Dec 10 '18 at 4:22