The entropy of a Schwarzschild black hole is located near the horizon, and the moment of inertia of a Schwarzschild black hole is $MR^2$. Both aspects imply that the mass of a Schwarzschild black hole is distributed around the horizon, or at least near it.

But a Nobel prize was just given, in 2020, for a black hole singularity theorem.

So where is the mass of a Schwarzschild black hole located: near the horizon or in a singularity?

This question is important because singularities cannot exist in nature: nothing is infinite in nature. Neither infinitely small, nor infinitely dense, nor infinitely hot. So the simple answer is clear: it is not in a singularity. Therefore, the question needs to be rephrased:

Is the black hole mass located near the center, or it is located near the horizon?

  • $\begingroup$ Related answer. $\endgroup$
    – rob
    Oct 21 '20 at 3:14
  • $\begingroup$ How does the moment of inertia of a Schwarzschild black hole being $MR^2$ imply that its mass is located "around the horizon, or at least near it"? If the mass of a Schwarzschild black hole were to be located around/near the horizon, one would expect the moment of inertia to be $\frac{2}{3}MR^2$ as one would for a spherical shell of mass $M$ and radius $R$. [...] $\endgroup$
    – Dvij D.C.
    Oct 21 '20 at 3:15
  • $\begingroup$ [...] If we actually go by your reasoning, then since $MR^2$ is the moment of inertia of a ring, we would be forced to believe that the mass of a Schwarzschild black hole is situated in the form of a ring at its horizon. It goes without saying that this would be patently absurd due to the spherical symmetry of the Schwarzschild solution. $\endgroup$
    – Dvij D.C.
    Oct 21 '20 at 3:16
  • $\begingroup$ Dvij, you are right about the prefactor. Still, the appearance of R^2 shows that the mass is not located at the central singularity: the moment of inertia would be zero or nearly zero in that case. $\endgroup$
    – Christian
    Oct 21 '20 at 4:42
  • $\begingroup$ ADM Mass $\endgroup$
    – safesphere
    Oct 21 '20 at 6:15

In classical general relativity, the mass of a Schwarschild black hole is associated with its singularity. The simplest argument for this, is that the Schwarzschild metric (and its full analytic extension) is a solution to the vacuum Einstein equations. Hence there is no mass, anywhere in the manifold, "hence" the mass must be associated with the only part that is not on the manifold, the singularity.

However, this is not a very satisfying argument. It does not take into account the many subtleties surrounding the definition of mass in general relativity. It is therefore worth looking at a more rigorous argument. The right tool for the job is the Komar mass. (The ADM mass is only defined at spatial infinity, and therefore does not allow us to ask where the mass is located)

The Komar mass can be defined for any stationary spacetime.

The Komar surface integral is given by

$$ M = \frac{c^2}{4\pi G}\oint_S \nabla^\mu K^\nu dS_{\mu\nu} ,$$ where $S$ is a 2-dimensional closed spacelike surface, $S_{\mu\nu}$ is the surface element of $S$, and $K^\mu$ is a time-like Killing vector field normalized such that $K^\mu K^\mu = -1$ at spatial infinity. It measures the mass contained within the boundary $S$.

If we look at the ingoing Eddington-Finkelstein extension, and calculate the Komar surface integral for a surface with fixed radius $r$ and advanced time $v$ (left as an exercise for the curious reader), we find that is always equals $M$ the total mass of the Schwarzschild black hole. This tells us that the mass of this manifold is located around the singularity at $r=0$, and in particular shows that there is no mass associated with region around the horizon.

Of course, this is the answer according to classical general relativity. In a theory of quantum gravity the answer might end up being substantially different. For example, in the string theory inspired fuzzball picture, the mass would be associated with a quantum state that has the approximate size of the horizon.

  • $\begingroup$ And how does this explain the moment of inertia of a Schwarzschild black hole? $\endgroup$
    – Christian
    Oct 22 '20 at 19:06
  • $\begingroup$ @Christian It doesn't. However, the identification of a moment of inertia of a black hole is somewhat unclear physically to begin with, because a black hole doesn't have a clear angular velocity to begin with. $\endgroup$
    – mmeent
    Oct 23 '20 at 6:30
  • $\begingroup$ mmeent, if the moment of inertia is not explained, the classical explanation is not the correct description of a real black hole. Or, ore precisely: the Schwarzschild solution is not the correct description of a real black hole. So one cannot conclude that the mass is in the center. And there are no singularities anyway - that is an unphysical concept. $\endgroup$
    – Christian
    Oct 24 '20 at 12:25

The mass of a black hole is associated with the horizon (and nearby external environment), otherwise the merger of two black holes could not release gravitational radiation. The mass of merged black holes is less than the sum of the two black hole masses (see LIGO data) giving rise to the gravitational energy that is released in the merger.

For more details see the comments to this question and the answers to its duplicate.


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