Density of supermassive black holes

Question

So I know that supermassive black holes can have "densities" less than water because black hole density scales as $1/m^2$ since $R_s = 2GM/c^2$. I am trying to reconcile this with the fact that black holes are the most compact objects for a given mass.

For instance, if I had 6 billion solar masses of material and squeezed it to the Schwarzschild radius, I would have a black hole. However, by my statement above, if I had 6 billion solar masses of water, this would occupy a volume less than the Schwarzschild radius, but would not be a black hole. How does this work? I know black holes actually have singularities (etc), but where is the flaw in my classical model of just density = M/V?

And apologies if you already know this, but at the very heavy masses of the type pictured today, observers may not even be aware they were moving inside the event horizon (and therefore doomed). — , Apr 10 '19 at 19:56
Interesting, that makes sense due to the small tidal forces, thanks! — , Apr 10 '19 at 20:32

score 1 · Answer 1 · answered Apr 10 '19 at 21:28

If I'm understanding the question correctly, then you're trying to reconcile two statements: (1) A Schwarzschild black hole consists of empty space. (2) A sufficiently compact distribution of matter is a black hole.

These statements are not in contradiction to each other, because a Schwarzschild black hole is just one model of a black hole. It's a model of an eternal black hole, which has always existed. It isn't a model of an astrophysical black hole that forms by gravitational collapse. But if you let the process of collapse run to completion, and you stop feeding the black hole, it will end up looking just like an eternal black hole.

Density of supermassive black holes

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