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I have read this question:

It turns out that black holes aren't really objects, but rather regions in spacetime. In fact, this is so true that black holes are what we call vacuum solutions: there isn't matter anywhere in the spacetime. All of the mass of the black hole is there due to effects of gravity itself. Another way of thinking it is that a black hole is so collapsed that its mass is entirely due to gravitational energy. It is a bit harder to grasp this concept, but once you get it, the rest is simpler. The black hole stays there because it isn't "made" of anything. There isn't a star just below the event horizon waiting to come out. There is nothing there, but gravity.

What exactly makes a black hole STAY a black hole?

While it is widely accepted that certain black hole solutions are vacuum solutions, known black holes are hypothesized to be collapsed (neutron) stars, so the collapsed remnant must be there somewhere (or maybe it transforms into the energy of the gravitational field).

Question:

  1. What do we mean when we say that black holes aren't made of anything?
Qmechanic
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Árpád Szendrei
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    The Schwarzschild solution is static, eternal, and the only thing in its universe. That also applies to the Kerr solution. So they can't apply exactly to real astrophysical objects, but they make excellent approximations. But it can be confusing when people talk about black holes and blur the distinction between those ideal theoretical black holes and the real astrophysical things we apply those ideal models to. – PM 2Ring Nov 14 '22 at 23:23
  • The Schwarzschild and Kerr solutions have $T^{\mu\nu}=0$ everywhere (except the singularity, which by definition of a manifold isn’t part of the spacetime manifold). So… they have no energy density, no momentum density, and no stress. – Ghoster Nov 15 '22 at 05:09
  • Related: https://physics.stackexchange.com/q/735441/276316 and links therein. – William Martens Nov 15 '22 at 10:06
  • I think it might be a really, really, really badly worded attempt to say this. – J.G. Nov 15 '22 at 10:12
  • @J.G. sorry to clarify, what exactly is badly worded? the question ? or some of the comments? – William Martens Nov 15 '22 at 10:13
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    The paragraph the OP quoted. – J.G. Nov 15 '22 at 10:14
  • It depends on the theory of quantum gravity. If you like loop quantum gravity, there is this: https://en.wikipedia.org/wiki/Planck_star – lvella Nov 15 '22 at 14:37
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    @J.G. Or may be you think wrong and it’s not an attempt to say what it doesn’t and it’s worded just fine. – safesphere Nov 16 '22 at 14:00

5 Answers5

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The mass is still there, inside the event horizon. The thing that you observe, the black hole, is a region of space which is so curved not even light can escape. But that is why your quote says a black hole isn't an object: the actual region that is black isn't made of anything.

So a small note on it being a vacuum solution. Using general relativity we can write down an exact solution of gravity for a perfectly spherically symmetric body (with the rest of the universe a vacuum). With spherically symmetric I mean that the density of the body only depends on the radius from the center. Because the equations in general relativity are so complicated this is one of the few exact solutions that are known. And actually, it is not a full solution. If $R$ is the radius of the body then we can divide the space into two regions: $r<R$ being inside the body and $r>R$ being the vacuum. For $r>R$ the solution is known at is called the Schwarzschild metric. For $r<R$ the solution is generally complicated and I don't know if an exact solution is generally possible.

For the Schwarzschild metric it doesn't matter how the matter is distributed, as long as it is spherically symmetric. So we could squeeze the matter matter together and the gravity outside would still look the same. That is basically what a black hole is. A bunch of matter which gets squeezed together until the escape velocity at the surface exceeds the speed of light, which results in the matter being squeezed together indefinitely. But from the outside we don't notice this squeezing anymore because the gravitational field doesn't change outside the black hole. In fact, it is not known what happens exactly to the matter inside a black hole. From simply following GR one would expect the matter to get denser indefinitely until a singularity is reached. I don't know what the current consensus about this is. If you know let me know in the comments.

AccidentalTaylorExpansion
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    It might be worth noting that the term "singularity" for what's at the centre of a black hole comes from maths, and it means a point where a mathematical model "doesn't work". i.e. if you draw a graph y = 1/x, the point x = 0 is a singularity because there is simply no valid value of y you can use in your graph. As far as I know it is a pop science misconception that there is an infinitely dense physical object at the centre of a black hole called a singularity. It is supposed be read as "our maths breaks down here, so we will need different maths to tell us anything about it". – Ben Nov 16 '22 at 03:21
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    @Ben, an objection to your sentence "It is supposed be read as our maths breaks down here". In math any function has its domain of definition, i.e. argument's range where the function is defined (finite). In case of $f(x)=1/x$ function it is the open set $(0,\infty)$. – JanG Nov 16 '22 at 15:11
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    This answer contradicts my understanding of and a plain reading of the claim presented in the question. You seem to be saying that there really is matter in there somewhere, whereas I think the claim the OP is asking about is that there really isn't any matter in there, only gravitational energy. Is this discrepancy intentional? – John Bollinger Nov 16 '22 at 17:39
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    @JohnBollinger Yes that is correct. To be honest I have only done an introductory course in general relativity so I could be wrong but that is indeed how I view black holes. Take a look at the Penrose diagram in this answer https://physics.stackexchange.com/a/474525/93729. Why would the matter suddenly disappear once a black hole forms? – AccidentalTaylorExpansion Nov 16 '22 at 19:37
  • @AccidentalTaylorExpansion which course? I am curious; (sorry this is probably a bit off topic but) still* I am actually right now trying to find a course I can enroll in; – William Martens Nov 16 '22 at 19:44
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    @WilliamMartens It was just called General Relativity. It is a master's course for theoretical physics. I don't know if you're doing a bacholor's currently but if so I don't think GR is ever taught at that level. In case you are interested in that kinda stuff you might look into cosmology courses or something related. Or something like manifolds/differential geometry if you're more into math. – AccidentalTaylorExpansion Nov 16 '22 at 19:57
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    Yes; thanks! I do not do any physics study (uni level) but; yes I am really the mathematically inclined one I guess; thanks again. – William Martens Nov 16 '22 at 20:04
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    Ok. I do not, myself, make any claims about the nature of black holes. However, I'm not prepared to accept that this answer's explanation of black holes is a reasonable characterization of what the original claim in question ("black holes aren't made of anything") means in its context. And that, after all, is the question that was actually posed. It would be nice if this answer devoted at least a few words to that. – John Bollinger Nov 16 '22 at 20:19
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    @JohnBollinger My bad, I misread your comment slightly. I agree with the last part of your comment. I asked the person that originally posted the answer for clarification and based on that I might modify my answer in the future. – AccidentalTaylorExpansion Nov 16 '22 at 22:58
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    @JanGogolin My way of thinking about it was that you can define a function f(x) = 1/x over the domain (0, ∞), and that function is well-defined and continuous everywhere in its domain; it has no singularities. But you can also define a different function with the same equation over a larger domain, such as (-∞, ∞) and that has a singularity at x = 0. (You could exclude the singularity and keep the negative half of the range by arbitrarily excluding the single value 0 from the domain, but then the domain itself is discontinuous, which seems just as awkward as having a singularity?) – Ben Nov 17 '22 at 02:06
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    @JanGogolin If your objection is that you define a singularity to be excluded from the domain of a function (because the maths breaks down and doesn't produce a valid value for the function), then sure, I'm happy to accept that as a possible definition. On that view, the singularity at the centre of the black hole would be a point excluded from the domain of the functions that are used to model the black hole: then those functions officially do not say anything about what happens there. So my point remains exactly the same. – Ben Nov 17 '22 at 02:11
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    @Ben my objection have only concerned the wording "... math breaks down here". I do not think physics break down, too. I would prefer to say that we do not quite understand what happens at the star center by begin of forming black hole. In the spacetime described by interior Schwarzschild metric for $r_{S}/R=8/9$ the central pressure diverges like $p=4/\kappa\cdot 1/r^2$ whereas the central energy density remains constant and the initial event horizon arises. I see no singularity there. – JanG Nov 17 '22 at 19:08
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    @WilliamMartens Why would we say we don't know what happens there if the mathematical model produced sensible results? We'd just say "this is what our model says, though we don't have experimental confirmation". My understanding is that we think we need new physics to deal with conditions extremely close to (or even at) the centre of a black hole because the mathematics has a singularity; either we need a physical understanding of how singularities in our model actually behave in the universe, or we need a new more accurate theory that doesn't have singularities there. – Ben Nov 17 '22 at 23:36
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    And @JanGogolin, that exact equation you quoted $p=4/κ⋅1/r^2$ has a mathematical singularity at $r=0$, surely? I fully admit to having no expertise in general relativity (I just know some maths), but I think you may be reading more into my informal "maths breaks down here" language than I intended. I was just trying to illustrate the concept of a mathematical singularity; a point where a particular equation has no solution. I do not mean that there is something in black holes that makes maths or physics stop working, or anything like that. – Ben Nov 17 '22 at 23:44
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    @Ben yes, I just don't like words "singularity" and "break down" in that context. In my view, metric represents the whole universe therefore it should not be singulair (mathematically). If you look at Einstein field equations and their solutions for static spherically symmetric perfect fluid sphere you will see that it is not metric that is singular at $r=0$ but the derived from metric components pressure. By the way, $r=0$, means a sphere in the limit of vanishing surface area $A$ ($r=\sqrt{A/4/pi}$) or the inverse of its Gaussian curvature $K$ ($K=1/r$). It is not radius. – JanG Nov 18 '22 at 11:33
  • @Ben +1 for ur comment (the reply) thanks; I was probably misunderstand the previous reply. – William Martens Nov 18 '22 at 12:09
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That's one of the things I was thinking about when I wrote the disclaimer at the bottom hahaha. I'm glad you asked, so I can comment on it!

As seem from the outside, the collapse of a star never really finishes. The object just gets extremely redshifted to the point it is no longer visible. One can then argue whether that can or not be called a black hole as seem from the outside. What is sure is: from the outside, you won't ever see the matter stop falling, and hence there surely is still matter there at least on the outside. I'm afraid I don't know the details of star collapse well enough to give a much better discussion. In that answer, I overlooked these details because OP didn't have a Physics background. Considering these nuances of stellar collapse makes the overall picture way more complicated and I considered it wouldn't be suitable for that particular answer.

Furthermore, as other people mentioned in the comments and in other answers, the Kerr and Schwarzschild solutions are vacuum solutions which approximate the exterior part of a black hole extremely well. We can actually get a grasp of black hole evaporation with a Schwarzschild black hole that was not formed out of stellar collapse (Kerr black holes are more complicated), and in this sense black holes are made of gravity alone.

I should also comment on

maybe it transforms into the energy of the gravitational field

This is a more technical way of understanding what I meant by "There is nothing there, but gravity". You have gravity being sourced by gravity itself. However, there is a nuance: gravitational energy is not well-defined in General Relativity. One can't define a covariant stress-energy-tensor for gravity because it is always possible to choose coordinates in which the metric is Minkowski and the Christoffel symbols vanish, and hence the stress tensor would need to vanish as well. Some sense can be obtained by using pseudotensors. Of course, all of this is just a way of interpreting the fact that the Einstein equations are nonlinear.

You might also be interested in this answer of mine: Addition of gravitational fields in general relativity.

Níckolas Alves
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  • If that is true (and I do believe it is true) why the black hole information paradox is so important? – lvella Nov 15 '22 at 14:41
  • @ivella, these (few) physics gurus at the top of the field who are discussing these things (on the level of Hawkings etc.) are very nitpicky. :) At this level it's all about principles (i.e. whether the Universe as a whole is allowed to loose information). The thought that things can pass the Schwarzschild radius and then be "gone" is simply unbearable... – AnoE Nov 15 '22 at 16:05
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    @AnoE that is not the issue at all. It is okay for things to go beyond the Schwarzschild radius and I think most researchers are okay with that. The discussion concerns what happens once the black hole completely evaporates (if it completely evaporates) – Níckolas Alves Nov 15 '22 at 20:23
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    @Ivella Hawking radiation occurs in this stellar collapse scenario, even though an external observer never sees the black hole form. The thing with information loss is that the external observer eventually will see (according to Hawking's calculations) the spacetime where the black hole evaporated completely. Hence, stuff that should have fallen inside the black hole seems to have vanished from the Universe completely. Some researchers consider this to be in conflict with Quantum Mechanics, some consider this to be a prediction of Quantum Mechanics. It is still a matter of debate – Níckolas Alves Nov 15 '22 at 20:25
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    “You have gravity being sourced by gravity itself” - To clarify, the gravity of a BH is not sourced by the gravitational energy of the BH. The BH equation is not “curvature equals energy”, but “the Ricci curvature equals zero”. There is no energy in the equation. Then the Weyl curvature (gravity) is defined by the parameter $M$. We interpret $M$ as mass by comparing to the Newtonian gravity, but this mass does not contribute to the stress-energy tensor, which remains zero everywhere. Gravity propagates through vacuum. When it is spherically symmetric and non-zero, it is a BH with no source. – safesphere Nov 16 '22 at 14:35
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    @NíckolasAlves, "As seem from the outside, the collapse of a star never really finishes." That is I wanted always to ask. Gravitational waves have been discovered due to merging of two black holes. In this process one black hole "collapsed" into the second one, or other round speaking matter had been swallowed in a finite time of the outside observers. – JanG Nov 16 '22 at 15:30
  • @safesphere I fully agree. Do you think the current version of the answer is misleading (and hence I should edit it) or are you mentioning this in addition to the answer? – Níckolas Alves Nov 16 '22 at 16:14
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    @JanGogolin Strictly speaking, none of those black holes have fully formed from our point of reference. Both the individual black holes and the merged black hole are actually still "shells" of infalling matter. It takes infinitely long for an external observer to see the black hole form. However, those objects are ridiculously well described by black hole metrics, and hence it is extremely common to simply call them black holes and treat them as such. This is a really interesting question, and I think it deserves a post of its own if you think this explanation was too short – Níckolas Alves Nov 16 '22 at 16:17
  • @NíckolasAlves agreed, I would want a question like this I am very fascinated. – William Martens Nov 16 '22 at 16:20
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    @NíckolasAlves I think the answer is fine. As in illustration, you must apply a force to bend a flexible rod, but once you join the opposite ends, it stays as a circle on its own. Similarly, once gravity is strong enough to bend spacetime to a critical point of creating a black hole, then no source is needed for the black hole to remain. "Nothing, but gravity" is the best description of a classical black hole. – safesphere Nov 17 '22 at 17:30
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    @NíckolasAlves "the collapse of a star never really finishes" - Similarly, a pendulum never stops bouncing. Its amplitude drops exponentially due to the air resistance, but per the idealized classical equations never reaches zero. In reality though any exponential process ends fast. The gravitational collapse is exponential and physically ends (reaches the Planck scale as the physically meaningful limit) in about one second by a remote clock. Assuming no charge or rotation, the result is externally a Schwarzschild black hole. – safesphere Nov 17 '22 at 18:07
  • @safesphere I would like to say the same but I have one objection: due to Einstein (and me -:) ) without matter there is no spacetime. Is it the spherical symmetry of space without matter some kind of "gravitational" matter? – JanG Nov 17 '22 at 19:24
  • @JanGogolin "Matter" is energy, which can take different forms. Some of these forms contribute to the stress-energy tensor, others don't. For example, the kinetic energy does not curve spacetime. As matter collapses, all its energy is converted to kinetic, so in the end the stress-energy tensor becomes zero (as in a vacuum solution), but the total energy is conserved and represented in the geometry of spacetime. You could call it "matter" in the sense that a black hole does have mass and inertia, but this "matter" is not at any specific location and the stress-energy tensor is zero everywhere. – safesphere Nov 17 '22 at 21:53
  • @safesphere totally d'accorr, however, I would argue that spacetime and matter already existed before gravitational collapse had started. The static spherically symmetric vacuum spacetime is simply the final stage of it. Parameter $M$ in Schwarzschild vacuum solution is a remnant of the forgoing forming process. – JanG Nov 18 '22 at 11:21
  • @JanGogolin Correct. During the collapse, matter in the form of the stress/energy tensor is converted to energy in the form of the spacetime curvature with a zero stress/energy tensor. The parameter $M$ is the conserved total energy that remains the same before and after the collapse. – safesphere Nov 19 '22 at 06:33
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A black hole twists space and time around so that time (the future) points inwards towards the singularity. All the possible worldlines of infalling particles go in to the central point and then have nowhere else to go. We don't know what happens next - Einstein's theory doesn't say.

Some think they stop there, forming an object of infinite density. Some think an unknown theory of quantum gravity takes over and some new unknown-to-current-physics effect stops the collapse. The singularity is replaced by some string-theory 'fuzzball' object, or something similar. Some think that they 'fall off the edge of space' and disappear. (Like, if you hold a length of rope at one end and wiggle it, you can get a wave to pass down it to the other end. What 'happens to' the wave when it gets to the end of the rope? Where does it go?) Some exclude the singularity not because they think it necessarily actually disappears but just to keep the maths simple - General Relativity assumes for a lot of its working that spacetime is a 'manifold', which cannot have edges or corners. Some think that they squeeze through the point and emerge into some other domain. (e.g. 'Journey Beyond the Schwarzschild Black Hole Singularity' Arraya et al.) Common sense beliefs derived from our ordinary experience of conservation of matter don't necessarily extend to such extreme circumstances. We don't know.

Nor does it really matter, for the purposes of understanding the 'source' of the black hole's gravity, because the singularity is in the future of every particle that has yet to hit it, and causality requires that the future has no effect on the present. The gravity you experience outside a black hole is entirely due to matter falling into the black hole in the distant past, before it reaches the event horizon.

We can draw a picture of the spacetime around a black hole stretched and twisted so that at every point the lightcones point up the page, like in flat Minkowski space. These are Kruskal-Szekeres coordinates. In this diagram, light always travels at $45^\circ$ angles to the vertical, and local time around any point flows up the page.

Black hole in Kruskal-Szekeres coordinates

(To connect this picture to the point of view of a distant observer seeing 'a spherical black hole floating in space', tilt your head $45^\circ$ to the right. The singularity (thick black line) is the timeline of the point at the centre of the black hole, at radius zero, and the event horizon (dashed lined) is at a fixed radius away from it, which diagrammatic scale distortions have expanded into a 'trumpet' shape centred on the horizon.)

You can see the collapsing star that forms the black hole shaded orange on the left of the diagram. Light from it travels up the page at $45^\circ$ to the vertical. Because of the twisting of spacetime, it moves parallel to the event horizon, stuck on the boundary. It is like water is flowing inwards down a drain, and ripples spread outwards at exactly the speed the water is flowing inwards, hanging on the boundary forever.

Observers hovering outside the black hole, or falling into it, can only see events happening on their past light cone - the two lines pointing downwards at $45^\circ$ to the vertical. When you look towards a black hole, you can see its entire history right back to the beginning, apparently frozen in time - it only looks black because the last few seconds of light emitted are spread out thinly over millions of years.

Likewise, they can only see the same region gravitationally. And like the light, the gravity of the collapsing star is also trapped on/near the event horizon. Even though the matter has long since fallen into the hole and met its fate, the gravitational information 'emitted' from its descent into the hole is still in the process of escaping.

The Schwarzchild black hole is a 'vacuum solution', in that it is the shape of empty space surrounding an infinitely dense point mass. It approximates a real black hole, which is the shape of the empty space surrounding a collapsing star. The gravity of the collapsing matter remains behind, trapped forever on the event horizon by the matter's past distortion of spacetime, even long after the star itself has collapsed and fallen into the singularity.

Nullius in Verba
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    +1 for saying "we don't know" while at the same time making the answer very interesting! – AnoE Nov 15 '22 at 16:14
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    How can the gravity of collapsing matter remain behind "even long after the star itself has collapsed and fallen into the singularity"? How can one remain behind as an artifact of extreme time dilation but not the other? You say "it only looks black because the last few seconds of light emitted are spread out thinly over millions of years" which seems to imply the same thing is happening to gravity, yet why is it not "diluted" over millions of years like light is? – Michael Nov 15 '22 at 23:45
  • The end of the paragraph under the diagram is a bit confusing, especially "the event horizon (dashed lined) is [...] centred on the horizon". – PM 2Ring Nov 16 '22 at 14:29
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What do we mean when we say that black holes aren't made of anything?

I risk the answer.

A black hole in general relativity is a spacetime that have gravity (converging geodesics) but no matter (zero stress-energy tensor). The simplest black hole describes the Schwarzschild vacuum metric that represents static spherically symmetric spacetime free of matter.

Now, the point is that spherical symmetry in general relativity is a feature of spacetime and not matter. The matter merely inherits it through Einstein Field equations (EFE). Imagine a flat Lorentzian spacetime. It has no gravity because its geodesics do not converge. To cause them converging one needs a piece of matter that when compressed to some critical compactness ratio ($r_{S}/R$) gives rise to the initial event horizon at the center. Further compression shifts the event horizon to the surface of the matter ball what is a very sketchy description of the black formation process. The end result of that process is a spacetime without matter disconnected physically by the event horizon from the hypothetical "interior" transient spacetime.

JanG
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It is confusing because he is still using the word"gravity". One must stop thinking of gravity as a force at this point, and remember that gravity is an artifact of time- space. so as time stops, "gravity-space" becomes infinite, and where space stops, "gravity- time" becomes infinite. But all of this is only describing what is observed with light! One must, at this point understand that the universe exists beyond our ability to see it with light, As the merging black holes prove with their measurable gravity waves. We measure gravity waves with changes in space/ time, not with light! So we"see"the wave as it is across time and space, not light waves. Like the difference between amplitude modulation and frequency modulation. In this case F.M. is the detection of gravity waves using wave interference across space, and A.M. is the detecting light from the star across time. They both tell the story, but from different perspectives. When you combine the two, you get an extra dimension of data that correlates between them. These data DO NOT EXIST ALONG THE SAME DIMENSION! One must stop assigning all data to the same dimension! Light is not the end- all, be- all of the universe. The universe exists beyond your ability to see it with light. As we have clearly defined, light has limits, remember that when contemplating the universe. Close your eyes, and see the universe in your mind, without the light that enters your eyes. The limits of light are the deceptions of the nature of the universe. You must "see" past them.

Joseph Wimsatt
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