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I understand that the typical density of the super massive black hole is close that of the water. It is also my understanding that this density is not true matter density because the volume used to calculate the density is measured from event horizon (so I suppose this is rather potential gravitational energy density?).

If it was the case that the universe were to collapse into a super massive black hole, can we still assume that it would be close to that of the water? If so, why, and if not, how would you calculate the density? Note that the equations using the Shwarzschild Radius don't work in this case.

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  • $\begingroup$ Can you explain why the calculation of "density" using the Schwarzchild Radius "doesn't work in this case"? $\endgroup$ – mcFreid Feb 17 '14 at 2:19
  • $\begingroup$ @mcFreid Please see similar posts, for example, physics.stackexchange.com/q/2268. $\endgroup$ – Luis Feb 17 '14 at 2:38
  • $\begingroup$ That question doesn't seem much related to yours. The issue with using the Schwarzchild metric in that question is because the question was asking about us being inside the event-horizon of a black hole (in which case the Schwarzchild metric does not describe the geometry of space-time). But your question doesn't involve that situation. $\endgroup$ – mcFreid Feb 17 '14 at 2:54
  • $\begingroup$ @mcFreid Ok then let's take for example the density formula $d_c = \frac {3m}{4 \pi r_s^3}$ That requires to have $r_s$ calculated, which is, $r_s = \frac {GM} {c^2}$. I believe this is going to yield a radius near as large as the entire universe. $\endgroup$ – Luis Feb 17 '14 at 3:01
  • $\begingroup$ right, so the density is proportional to the inverse mass squared. Because of this, you would get a density much less than that of water if the mass is equal to the mass of the universe (though, as you pointed out, the density is not uniform). I guess I'm confused by your question... When you say "we can still assume that it would be close to that of water": what assumption are you talking about? It's just a calculation that for certain masses, the density of a black hole is close to water. For other masses, it won't be... $\endgroup$ – mcFreid Feb 17 '14 at 3:16
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The density of a black hole is defined simply as the mass within the event horizon divided by the volume within the event horizon. This gives an average density, but doesn't imply that the density is uniform within the event horizon. So when you hear statements like the density of a supermassive black hole is the same as water don't take this too literally.

I suppose it has some meaning in that the average density can be related to the curvature at the event horizon, and therefore to the tidal forces you would experience if you jumped in. Even so I doubt anyone outside the popular science media would attach much significance to the density of a black hole.

If you're interested, the Schwarzschild radius is given by:

$$ r_s = \frac{2GM}{c^2} $$

and if you take the volume to be $4/3 \pi r_s^3$ that gives an average density of:

$$ \rho = \frac{3c^2}{8\pi Gr_s^2} $$

Set this equal to 1000 kg per cubic metre (density of water) and you get a radius of about $4 \times 10^{11}$ and this corresponds to a black hole mass of about $10^8$ solar masses, which I guess counts as supermassive. Note however that the average density is proportional to $r_s^{-2}$, which is proportional to $m^{-2}$, so if you took a larger mass like the whole observable universe the average density of the corresponding black hole would be lower than water - vastly lower in fact.

Note that I referred to the observable universe in my comments above. Your question just refers to the universe but the universe as a whole can't collapse into a black hole. In a black hole the event horizon divides the interior of the black hole from the rest of the universe. But the whole universe can't form a black hole because there is nothing outside it for the event horizon to divide it from.

The universe as a whole does have an average density, and you can calculate this using the FLRW metric. If you wind time back towards the Big Bang the average density of the universe increases and at some point it would be the same density as water. The universe seems unlike to recollapse in the future, so in the future its average density is just going to keep decreasing.

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  • $\begingroup$ @John_Rennie Thank you for the explanation. I this answer my question, but perhaps not in the way you intended. First, on your comment "the whole universe can't form a black hole because there is nothing outside it for the event horizon to divide it from", confirms to me that you cannot use the Schwartzchild formulas, but just because you cannot use these formulas does not mean that the whole universe cannot collapse into a black hole. There are a number of models that support the idea of a black hole collapse, which is what I am trying to understand at the moment. $\endgroup$ – Luis Feb 17 '14 at 19:42
  • $\begingroup$ @John_Rennie you are also giving me a very good tip when you state "If you wind time back towards the Big Bang the average density of the universe increase..", because I could in fact do that and see where that takes me. However, I will be still stuck trying to find at what point the density begins to yield a black hole. I thought that perhaps I can use the Schwartzschild Newtonian field equations where $g$ is used. Once $g --> c$ and $R=r_s$ then I may conclude that I have a black hole (if that is even possible). What are your thoughts on this? $\endgroup$ – Luis Feb 17 '14 at 19:48
  • $\begingroup$ One more comment I would like to make. If it was in fact the case that the Schwartzchild equations applied in this case, then I would imagine a black hole but with the $R_s$ that of the entire universe: so a very dark universe where all matter and energy has collapsed into a ball of high density, except that space-time has not really collapsed. Effectively the size of the universe spacetime is still very large. But then in this case, the radius and density of the ball would be more useful than $R_s$ and $\rho$ based on $R_s$. Don't you think? $\endgroup$ – Luis Feb 17 '14 at 21:31

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