# Are black holes very dense matter or empty?

The popular description of black holes, especially outside the academia, is that they are highly dense objects; so dense that even light (as particle or as waves) cannot escape it once it falls inside the event horizon.

But then we hear things like black holes are really empty, as the matter is no longer there. It was formed due to highly compact matter but now energy of that matter that formed it and whatever fell into it thereafter is converted into the energy of warped space-time. Hence, we cannot speak of extreme matter-density but only of extreme energy density. Black holes are then empty, given that emptiness is absence of matter. Aren't these descriptions contradictory that they are highly dense matter as well as empty?

Also, if this explanation is true, it implies that if enough matter is gathered, matter ceases to exist.

(Sorry! Scientifically and Mathematically immature but curious amateur here)

• I don't know where you heard that black holes are empty... of course they are not empty, if they were, then they would behave exactly like "empty space" (which is also not empty, but that's a different post). The problem is that we don't know what state of matter is in there and how spacetime behaves under those circumstances. There is no such thing as "pure energy". Energy is the ability to perform work and black holes can perform work, but that doesn't mean that they are somehow made of energy. A black hole is simply yet another state of matter. – CuriousOne Mar 29 '16 at 6:20
• Actually, a black hole of a given mass has about the same properties (gravity etc) as any other object of the same mass..unless of course you are near the horizon, when it starts to go bizarre. – GRrocks Mar 29 '16 at 6:23
• Sometimes even competent physicists seem to be saying questionable things when they are being asked to over-simply on tv. Having said that, it is indeed a misconception that black holes are "made of highly condensed matter", but the logical alternative to that is not that they are "empty". It's a lot more complicated than either. – CuriousOne Mar 29 '16 at 6:57
• Possible duplicates: physics.stackexchange.com/q/18981/2451 and links therein. – Qmechanic Mar 29 '16 at 10:05
• I think the primary question is whether it's possible to detect which idea (or neither) is correct. That is, if there's no observable difference in the way the black hole behaves or experiment that could prove one explanation to be true, it's not Physics. – JimmyJames Mar 29 '16 at 18:14

The phrase black hole tends to be used without specifying exactly what it means, and defining exactly what you mean is important to answer your question.

The archetypal black hole is a mathematical object discovered by Karl Schwarzschild in 1915 - the Schwarzschild metric. The curious thing about this object is that it contains no matter. Techically it is a vacuum solution to Einstein's equations. There is a parameter in the Schwarzschild metric that looks like a mass, but this is actually the ADM mass i.e. it is a mass associated with the overall geometry. I suspect this is what you are referring to in your second paragraph.

The other important fact you need to know about the Schwarzschild metric is that it is time independent i.e. it describes an object that doesn't change with time and therefore must have existed for an infinite time in the past and continue to exist for an infinite time into the future. Given all this you would be forgiven for wondering why we bother with such an obviously unrealistic object. The answer is that we expect the Schwarzschild metric to be a good approximation to a real black hole, that is a collapsing star will rapidly form something that is in practice indistinguishable from a Schwarzschild black hole - actually it would form a Kerr black hole since all stars (probably) rotate.

To describe a real star collapsing you need a different metric. This turns out to be fiendishly complicated, though there is a simplified model called the Oppenheimer-Snyder metric. Although the OS metric is unrealistically simplified we expect that it describes the main features of black hole formation, and for our purposes the two key points are:

1. the singularity takes an infinite coordinate time to form

2. the OS metric can't describe what happens at the singularity

Regarding point (1): time is a complicated thing in relativity. Someone watching the collapse from a safe distance experiences a different time from someone on the surface of the collapsing star and falling with it. For the outside observer the collapse slows as it approaches the formation of a black hole and the black hole never forms. That is, it takes an infinite time to form the black hole.

This isn't the case for an observer falling in with the star. They see the singularity form in a finite (short!) time, but ... the Oppenheimer-Snyder metric becomes singular at the singularity, and that means it cannot describe what happens there. So we cannot tell what happens to the matter at the centre of the black hole. This isn't just because the OS metric is a simplified model, we expect that even the most sophisticated description of a collapse will have the same problem. The whole point of a singularity is that our equations become singular there and cannot describe what happens.

All this means that there is no answer to your question, but hopefully I've given you a better idea of the physics involved. In particular matter doesn't mysteriously cease to exist in some magical way as a black hole forms.

• Thanks for your simplified explanation of the nuances of general relativity. I suspected that my argument (and necessarily, my understanding) is susceptible to interpretation problems due to vagueness and ambiguity resulting from relying solely on natural language with imprecise terms. – Keerthi Mar 29 '16 at 7:17
• Now I am convinced that my trouble in understanding the exotic phenomenons of the universe are largely due to naive assumptions in my understanding of "time". Perhaps a little more maturity in mathematics (lot more!) and thorough understanding of what Newtonian mechanics "takes for granted" would help me get comfortable with the mathematical & physical implications of GR. – Keerthi Mar 29 '16 at 7:17
• Could you expand a bit on what it means for an equation to "become singular"? – Dan Henderson Mar 29 '16 at 16:22
• If you consider the "No hair theorem" it doesn't really matter how the black hole was formed. Kerr-Newman solution is still a vacuum solution to Einstein-Maxwell equations, so intuitively, most of the "inside" of the black hole is empty (beside the singularity). – Alexander Mar 29 '16 at 19:11
• "For the outside observer the collapse slows as it approaches the formation of a black hole and the black hole never forms." - So the star shrinks and darkens, but the black or dark sphere it forms doesn't look like a black hole? – Cees Timmerman Mar 30 '16 at 12:52

For non-experts it is worthwhile mentioning the simpler 'Newtonian' interpretation of a black hole. Given the fact that most theorists believe that the General Theory of Relativity (GR) is actually inapplicable in the vicinity of a black hole singularity, that viewpoint is also not necessarily disfavored. (This point seems to meet some criticism; please see also my comments below to understand why I am saying this. There is two related problems there: 1) The BH is, first of all, an astrophysical object which has been conjectured to exist based on GR but is not identical to a particular singular solution of GR 2) One can debate about where a BH starts, e.g. at the event horizon, and thus how big it is. In the vicinity of a BH space-time is considerably curved this affects the meaning of distance.)

Already long before Schwarzschild (John Michell 1783) 'dark stars' were discussed. These are objects whose mass concentration is so large that their escape velocity exceeds that of light (John Michell assumed that light was composed of corpuscles as proposed by Newton - an idea which got later temporarily disfavored).

Imagine throwing such a light particle on earth. If its velocity is small it will fall back to ground. However, if it exceeds the escape velocity it will never return. The necessary velocity depends on the mass of earth, if the attractive force is given by Newtons Universal Law of Gravity, but also on the radius of earth since it defines your distance from earth's center of gravity.

This consideration leads to a 'critical radius', below which light cannot leave a given spherical mass distribution $M$ if it has the finite velocity $c$. This radius is $r = 2(GM/c^2)$ and in agreement with the critical radius of the Schwarzschild metric.

In this picture, your question for the density of a black hole is answered simply by $\rho=M/V=M/(4/3\pi r^3)$ with $r$ the critical radius given above. This is the average density of a spherical mass distribution which light cannot escape.

Besides, the term 'black hole' was introduced by John Archibald Wheeler after David Finkelstein recognized that the Schwarzschild metric had this property of dark stars.

Neither GR nor this Newtonian picture describe what happens inside of a black hole. But, based on such considerations, physicists agree that a (non-rotating) mass-distribution compressed to densities above the critical one form something that, what concerns its exterior, is described by the Schwarzschild solution of GR.

Also recall the equivalence of mass and energy. GR treats both forms in the same way. They both curve space-time and get influenced by this curvature. When matter, e.g. some gas, falls in a black hole it gets disrupted and it is probably better thought of as energy (maybe elementary particles but maybe also something that cannot be called a particle).

One of the strongest principles in physics is energy conservation. The matter that falls into a black hole is presumably not destroyed but converted into some form of energy. You can then ask what the energy density of a black hole is. The answer, as above, depends on where the black hole starts. In the case of a non-rotating black hole, a preferred definition is that it begins at the Schwarzschild or critical radius.

The energy of such theoretical solutions of GR that represents a BH is stored in the field configuration, i.e. in the curvature of space time. There is no reference to the state of the matter that formed it. The situation is however that astrophyiscal BHs gain their mass/energy by feeding on matter, e.g. that of a collapsing star which may have initially formed it and later possibly also matter in its environment. How this matter behaves and what happens to it while it is digested (how it is disrupted and what forms it takes on intermediate stages) is not described by GR which is a theory only for the gravitational force. It does already not account for what happens to matter that approaches the event horizon from outside (how it lights up and radiates). Also GR alone does not answer if and how BHs radiate of and loose energy in the form of Hawking radiation.

The question what happens to the matter that falls into black holes can therefore not be answered by reference to GR alone. They can however be addressed within Quantum Field Theory combined with classical (non-quantum) GR. For its ultimate fate, somewhere inside the event horizon, we would also need something like a quantum version of General Relativity.

• Given the fact that most theorists believe that the General Theory of Relativity is actually inapplicable in the vicinity of a black hole. Um, what? GR is how we understand black holes at all. – HDE 226868 Mar 29 '16 at 14:03
• A BH in GR is a mathematical singularity in a field theory. Most physicists agree that this theory is the classical limit of a quantum theory yet to be identified (because of such problems with GR). The straight-forward approach to a quantum gravity fails (because of a singularity) if the curvatures are as strong as those of BH's. Therefore, GR is in this case a-priori not the appropriate theory. This does however not mean that the particular prediction of the existence of BH's is wrong. Similarly, the failure of Newtons theory did not mean that its prediction of dark stars was wrong. – highsciguy Mar 29 '16 at 14:12
• @highsciguy: the horizon is not in the vicinity of the singularity. – Jerry Schirmer Mar 29 '16 at 15:00
• For the ad-hoc quantisation of gravity to break down you don't need to go all the way to the singularity. It breaks down due to a different kind of singularity -> renormalizability. The OP asked about the matter inside the BH. Since it is a matter of definition where a BH starts, we need to contemplate that somewhere on the way to the BH singularity a quantum version of GR applies, supposedly. We thus don't know the appropriate theory and can only hope that its limit is close enough to GR. – highsciguy Mar 29 '16 at 15:10
• To rephrase it: The Schwarzschild solution of GR is one which holds in an idealized settings which ignores known details but also the fact that we don't really know what happens in these extreme situation. It is furthermore singular which most see as indication that it is wrong in an environment of the singularity. GR alone does not tell us what happens to the matter inside the BH. – highsciguy Mar 29 '16 at 15:24

At points where the escape velocity exceeds the speed of light, you would expect all matter to be falling inward faster than the speed of light. However, it would take infinite energy to make any matter reach the speed of light. I would instead GR to make time greatly slow down for any matter approaching the speed of light, and therefore give an outer shell with time stopped, and something much harder to understand within that shell.

(fellow amateur here) I believe that the notion that black holes are empty comes from that fact that all the matter that gets sucked past the event horizon is packed into an infinitival small ball of infinite density, thus it would be the same as if you had a hollow sphere the size of earth, and one particle in the center, one would say that the sphere is empty, and to my knowledge that is why most people would say that a black hole is empty.

• You should check the equations too see that BH appear connected to singularity – Mikey Mike Mar 29 '16 at 15:23

## protected by Qmechanic♦Mar 29 '16 at 14:31

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