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The textbook from which I teach physics at the end of secondary school has a question about the density of a non-rotating black hole. Because the density at the singularity is perhaps infinite or beyond the scope of GR, this can only be some "mean density" mass/volume within a sphere of a certain radius. Some authors use the Schwarzschild radius.

Why this value of the radius? How can I explain that this choice of radius makes more sense than another radius?

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    $\begingroup$ Did you read the Wikipedia page on the Schwarzschild radius? $\endgroup$ – ACuriousMind Apr 19 '15 at 19:41
  • $\begingroup$ The Schwarzschild radius is the distance fro the center at which the escape velocity of the black hole surpasses the speed of light. $\endgroup$ – Brionius Apr 19 '15 at 19:41
  • $\begingroup$ @Brionius: As far as I remember the physics lecture (I'm engineer, not physicist) this definition is based on non-relativistic calculations. You get the correct radius using this calculation however the calculation has (as far as I remember) nothing to do with the physical truth. $\endgroup$ – Martin Rosenau Apr 19 '15 at 19:50
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    $\begingroup$ Very related: Do black holes have infinite areas and volumes?. Basically, the volume of a black hole is not frame-independent, and so is not unambiguously defined. However, for a simple (non-rotating, uncharged, isolated) Schwarzschild black hole, one fairly 'obvious' choice of frame gives the same volume as the usual Euclidean formula with Schwarzschild radius. $\endgroup$ – Stan Liou Apr 19 '15 at 19:58
  • $\begingroup$ @MartinRosenau while the common Newtonian derivation of the Schwartzschild radius is a happy accident, it still is the dividing line between the point when light (or anything else) could escape. It's a convenient, accurate shorthand, but it shouldn't be taken as an excuse to apply Newtonian gravity to a non-Newtonian system. $\endgroup$ – Brionius Apr 19 '15 at 20:10
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Using the Schwarzschild radius for this purpose makes sense, because this is the radius of a sphere which becomes a black hole, if it has the given density. For example a sphere made of air at Earth density does not become a black hole if its radius is 1 meter. But if the radius is big enough, it will actually become a black hole. Even though the density is very low. The required radius is the Schwarzschild radius associated with the mass enclosed in the sphere (under given density).

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  • $\begingroup$ An implicit assumption here is that there is such a thing as 'the' radius, which is actually misleading. $\endgroup$ – Stan Liou Apr 19 '15 at 20:16
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A black hole is a region of spacetime enclosed by an event horizon. Thus, the singularity, while a fact about black holes as far as we understand them, is not an defining feature of what black hole is.

Therefore, it makes more sense to try to calculate volume (and hence indirectly, density) of a black hole according to the extent of the event horizon rather than the singularity. It so happens that for the simplest of black hole solutions, the spherically symmetric Schwarzschild black hole, the Schwarzschild radius characterizes the size of the horizon: the Schwarzschild radius is actually $\sqrt{A/4\pi}$, where $A$ is the area of the horizon. This is actually (implicitly) by definition of the Schwarzschild radial coordinate, which directly corresponds to the area of a sphere.

As I've said in the comments, a problem of talking about 'the' volume of a black hole is that it's not actually frame-independent, so calculating volume in this way is simply a conventional choice. Again, see this question for more details.

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