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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.

Luis
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  • Can you explain why the calculation of "density" using the Schwarzchild Radius "doesn't work in this case"? – mcFreid Feb 17 '14 at 02:19
  • @mcFreid Please see similar posts, for example, http://physics.stackexchange.com/q/2268/. – Luis Feb 17 '14 at 02:38
  • 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. – mcFreid Feb 17 '14 at 02:54
  • @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. – Luis Feb 17 '14 at 03:01
  • 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... – mcFreid Feb 17 '14 at 03:16
  • @mcFreid I would think that before the entire universe collapses, you would see the galaxies and clusters collapsing onto themselves first. Since the majority of galaxies contain a super massive blackhole, this being the dominant density of the galaxy, you would end up with stand alone black holes and possibly other galaxies merging with each other. Effectively you would have numerous super massive black holes collapsing on each other, each having the density of water. Ultimately they would all combine into yet the final water density black hole. – Luis Feb 17 '14 at 03:41
  • First, you are assuming the universe is going to collapse - which is an unknown area of physics. Secondly, you are assuming a process of HOW it is going to collapse, which is yet another unknown area of physics. And, you are assuming that all black holes will have the density of water, which is not necessarily true. Like I said, the density is proportional to the inverse mass squared. Lastly, you are assuming that if two black holes merge, each with the density of water, that the resultant black hole will have the density of water. This is also not true as $d_f \neq d_1 + d_2$ – mcFreid Feb 17 '14 at 03:47
  • There are a number of models for the universe, some showing expansion and others collapsing, the most re-known is the so called Big Crunch. Since, as you said, we don't know, all these models are as good as the others. One of the things I've noted about the collapsing models is that the ultimate result will be a black hole. I am not assuming, I am asking, if the final black hole would end up with the density of the water. I understand that $d_f not equal d_1+d_2$, but why not in this case? If I mix water with water, I still get the density of water. And back to the Shcharzschild radius.. – Luis Feb 17 '14 at 03:57
  • there is still a problem using these formulas. – Luis Feb 17 '14 at 03:58
  • mixing water with water is NOTHING like two black holes merging. As I mentioned, the density is proportional to the inverse mass squared. Thus if you have two black holes, one with massses $m_1$ and $m_2$, then the density of the merged black hole will be proportional to $\frac{1}{(m_1+m_2)^2} \neq \frac{1}{m_1^2} + \frac{1}{m_2^2} \neq \frac{1}{m_1^2}$ so we see that if one of the black holes has the density of water, the merged black hole will not. (note: I included the last $\neq$ to show that even if you could sum densities, the merged black hole would still not be the density of water. – mcFreid Feb 17 '14 at 04:11
  • Ok, you're right. That makes sense. In any event, how would you calculate the density of a super massive black hole for the universe having completely collapsed? – Luis Feb 17 '14 at 04:25
  • The density is the total mass $m$ divided by the volume $v$ where the volume is just the volume of a sphere with radius equal to the Schwarzchild radius. From the formulas you quoted, you should be able to get the density as a function of mass and then plug in the mass of the universe (which is not really a well-known number either, but current estimates put it on the order of $10^{53}$ kilograms. – mcFreid Feb 17 '14 at 04:31
  • The above formulas yield silly results. Note that $r_s$ turns out larger than the current radius of the obvervable universe. Then the density turns out to be less than the current matter density, which is a contradiction from what we know about black holes. As the user in the above mentioned post states "..you can’t use the Schwarzschild metric to identify a horizon inside the sphere of uniform density" and ".. the Schwarzschild geometry describes a vacuum spacetime. So you can't use it for a spacetime filled with matter". The user suggests to use the FRW, but I don't think these equations – Luis Feb 17 '14 at 05:04
  • show the formation of a black hole. – Luis Feb 17 '14 at 05:05

1 Answers1

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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.

John Rennie
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  • @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. – Luis Feb 17 '14 at 19:42
  • @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? – Luis Feb 17 '14 at 19:48
  • 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? – Luis Feb 17 '14 at 21:31