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According to wikipedia "... the average density of a supermassive black hole (defined as the mass of the black hole divided by the volume within its Schwarzschild radius) can be less than the density of water..."

(source :https://en.wikipedia.org/wiki/Supermassive_black_hole)

So my question is: Do supermassive black holes contain a singularity or the physical laws - as we know them- are still valid, inside their event horizon?

edit to clarify some things a little bit more:

From the same article : "As with density, the tidal force on a body at the event horizon is inversely proportional to the square of the mass: a person on the surface of the Earth and one at the event horizon of a 10 million M☉ black hole experience about the same tidal force between their head and feet."

That implies -to me -that the mass is uniformly distributed inside the volume of its Schwarzschild radius. If this statement is accurate what a free falling observer into the black hole would experience?

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  • $\begingroup$ We don't know what is really happening inside the horizon, or better, we don't have a theory for what may be happening close to where general relativity predicts the singularity. Whether general relativity even holds near the horizon is not known. There is no evidence that it doesn't, and I don't believe there is any theoretical reason to assume that it doesn't, which makes black holes rather boring objects on the outside. The inside is probably shielded from measurements, anyway, so in essence we don't even have to care. $\endgroup$ – CuriousOne May 28 '16 at 12:29
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    $\begingroup$ Regarding the edit: remember that if you're outside the mass, the distribution of the mass doesn't affect the behavior of the gravitational characteristics as long as the mass is spherically symmetric. A point mass like a singularity and an extended, uniform-density mass like the Earth (or close enough, anyway) have the same gravitational field outside. As would a thin spherical shell, concentric spherical shells, and uniform collections of smaller point masses would also approximately match. The density and the tidal force are unrelated. $\endgroup$ – Asher May 28 '16 at 13:24
  • $\begingroup$ Your understanding of tidal forces is incorrect. The tidal force would be the same if the black hole was uniform or a singularity at the center. (which Asher beat me to pointing out) $\endgroup$ – userLTK May 28 '16 at 13:25
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The important quantity associated with a black hole is the event horizon area. The volume contained inside is not what one would think of as $V = 4\pi r^3/3$. More on the volume later. The important quantitiy is the area of the event horizon. The reason is that from the perspective of an exterior observer this is the limit of observation. Everything that falls into the black hole is observed to have its observed time intervals on a clock dilated or slowed as radiation it emits is red shifted arbitrarily far.

The Schwarzschild metric for a nonrotating black hole of mass $M$ gives the line element $$ ds^2 = \left(1 - \frac{2m}{r}\right)dt^2 - \left(1 - \frac{2m}{r}\right)^{-1}dr^2 – r^2d\Omega^2~m = GM/c^2 . $$ For null rays we have the interval is zero $ds = 0$ and we proceed to compute the clock time $t$ on a standard coordinate frame of a very distant observer for the time it takes a photon to radially escape from some distance $R$ form a black hole $$ \int^Tdt = \int_R^\infty \left(1 - \frac{2m}{r}\right)dr = R - 2m ln(R - 2m) $$ clearly this becomes infinite as $R \rightarrow 2m$. Physically this means that everything that made the black hole, including the original star that imploded into it are “pasted” right above the event horizon. It is as if the black hole has a geological layer cake structure of everything that went into it. Notice there is no reference at all to anything in the interior. The observer on the outside, which is the wise place to remain if you wish to keep living, witnesses everything about the black hole as pinned on the event horizon, and this is one basis for the holographic principle.

All of the stuff that composes the black hole forms the entropy of the black hole. The Bekenstein entropy for the area of a black hole event horizon $$ S = k~\frac{A}{4\ell_{pl}^2} $$ where the horizon area is $A = 4\pi R^2$ $= 16\pi m^2$ and $\ell_{pl} = \sqrt{G\hbar/c^3}$ is the Planck unit of length. We can see that the area of a black hole can be written according to $N$ Planck areas of a black hole, for the Planck area $A_{pl} = 4\pi\ell_{pl}^2$ and the entropy of the black hole is then given by $S = \pi k N$. From here there are all types of interesting connections with quantum information theory, but I will defer on that for now.

The interior of a black hole is only accessible to those who enter it. This is at least the case for a classical black hole. For a quantum black hole there may be some fluctuations of the horizon which make quantum information of a black hole a superposition of states outside and inside. I will not go into that for now. For the pure Schwarzschild black hole the Penrose diagram The event horizon as seen by an observer in our universe is on the right. Once you cross the horizon it splits and the horizon separating the interior of the black hole from our universe and the other horizon separating the other universe from the interior grow apart. In this eternal black hole diagram, which is a sort of mathematical idealization, the horizons grow infinitely far apart. This means the spatial region in the interior grows, and becomes infinitely large at the singularity $r = 0$. enter image description here

I could go further in how this mathematical idealization of the eternal black hole is perturbed by the collapsing star and by Hawking radiation. The imploding surface of a star will cut this diagram in half and the region between the material surface and the horizon will grow arbitrarily. Hawking radiation in addition cuts off the size of the distance between the collapsing surface and the horizon, or between these two split horizons. The scale of this has to do with the quantum Poincare recurrence and quantum complexity of the system, which gets us into a huge area of current research.

What is happening in the interior of a black hole is then a curiosity, and we will never know what happens in the interior of a large black hole. They are too far away, which is a good thing, and the classical nature of them makes access to the interior impossible. For quantum black holes, or more likely QCD analogus of AdS/black holes, we might be able to make inferences from quantum superposition of exterior and interior states.

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There's nothing different about a supermassive black hole except that its mass is large. So, if things work for small black holes they work for large ones.

It turns out that, by the measure of density given, the density decreases with the mass of the object. This follows immediately from the formula for the Schwarzschild radius, $r=2MG/c^2$: this goes as $M$, but the volume goes as $r^3$ so density goes as $M^{-2}$.

However this notion of 'density' as 'the mass contained within the horizon divided by the volume contained (naïvely) within the horizon' is not really useful, it's just a slightly amusing number. A black hole is a vacuum solution: there is no mass within it at all, however big or small it is.

Except, of course, there is: there is mass in the singularity. But the singularity is just exactly the place where things break down, for any size black hole: large or small. And classically (so, ignoring any QM stuff at the horizon, but just using GR's picture) it's the only place where things break down, for an object of any mass: the 'normal laws of physics' are perfectly valid inside the horizon as well as outside it, with the only difference being that, inside the horizon, all future-directed timelike curves have finite length and terminate on the singularity.


Of course, simply because all future-directed timelike curves within the horizon terminate on the singularity no information can escape (according to GR). So, perhaps black holes are in fact full of faeries and unicorns: we can never know. But what GR says is that everything is ordinary except at the singularity.

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  • $\begingroup$ @CuriousOne Right of course, but I assume that the question is predicated on GR being correct at and for some way beyond the horizon. $\endgroup$ – tfb May 28 '16 at 18:40
  • $\begingroup$ But there is the crux, though.... if GR is not even correct at the horizon, then the singularity problem may never occur. We may have to replace GR with a better theory, which may or may not cause singularities. The reason why I am pointing this out is because I want people to understand the problems with making implicit assumptions about theories and their untested regimes. This is a classic case of theory being over-extended without any observational data to go by. $\endgroup$ – CuriousOne May 28 '16 at 19:59
  • $\begingroup$ @CuriousOne I'm quite sure that singularities do not occur, regardless of whether GR is good at the horizon or not (I suspect it is, but that's kind of academic from the point of view of non-infalling observers, of course). $\endgroup$ – tfb May 28 '16 at 20:08
  • $\begingroup$ Agreed, it is unlikely that anything like singularities exists. $\endgroup$ – CuriousOne May 28 '16 at 20:17
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    $\begingroup$ I understand that quantum gravity has to take over before one gets to the singularity. But why must the physics near and inside the horizon require quantum gravity? For large, i.e., supermassive, black holes the gravity at the horizon is pretty weak, about the same as on earth for a 10 million suns black hole, why would a GR description not hold up, approximately? I.e., the equivalence principle, almost no spaghettification, and most importantly, no quantum effects for a firewall to form to preserve information? $\endgroup$ – Bob Bee May 29 '16 at 4:33

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