# Is the mass of a black hole somewhere at all?

The Schwarzschild vacuum solution describes the gravitational field and thus the spacetime curvature of a black hole. The mass of a black hole is sometimes associated with its singularity as pointed out in an answer here: Black holes: where is its mass? In a singularity or on the horizon?

An answer to the question If black holes are just empty vacuum of space inside, then what causes the curvature? explains

"So realistically, the cause of the curvature would be stress-energy that is outside of the vacuum solution, in the part of the universe not described by the Schwarzschild metric."

The cause of the stress-energy is the mass $$M$$ of the black hole. If $$M$$ is associated with the singularity it is not a part of the manifold (corresponding to "outside of the vacuum solution") which seems to mean that the mass is not at a locality describable by coordinates.

Is the mass $$M$$ nowhere? Is it merely represented by spacetime curvature?

• Related: math.ucr.edu/home/baez/physics/Relativity/BlackHoles/… "In this sense the black hole is a kind of "frozen star": the gravitational field is a fossil field". But that doesn't exactly answer your question of "where's the mass?" Jun 30, 2021 at 18:21
• Note that the pure Schwarzschild BH is an ideal vacuum solution: it's eternal, not formed by the gravitational collapse of matter, and there's nothing else in its universe. Jun 30, 2021 at 18:27
• Why don't you like Dale's answer in the question you linked? I think it clears up any misunderstanding Jun 30, 2021 at 19:52
• @anna do you equate stress-energy with mass and say the mass is "in the part of the universe not described by the Schwarzschild metric."? Note the Schwarzschild solution describes a universe. There is no outside.
– timm
Jul 1, 2021 at 7:31
• Yes to your question. The solution applies as far as geometry goes, in analog with euclidean, but how the stress energy tensor, which is the mass/energy part relates to the local approximation cannot be seen , as it is a local solution. Jul 1, 2021 at 7:41

In general relativity, in general mass and energy are not well-defined globally since they are not conserved. So the question is in a sense meaningless in general, but black holes may be an exception... except that the location of the mass still remains iffy.

Locally everything is fine: the local flow of mass and energy is described by the stress-energy tensor, and there is energy conservation locally. Globally there is no good way of defining mass globally so it is conserved, in general. The fundamental problem is that energy gets its conservation from time translation invariance, and general spacetimes does not have that symmetry.

But the Schwarzschild spacetimes does! That means one can define the Komar mass. Indeed, it can be calculated by integrating a volume integral, and it does produce the "right" answer. But the volume stretches out to spatial infinity. Making "quasi-local" measures appears tricky.

Note that a collapsing material object turning into a Schwarzschild black hole seems to put all the mass into $$r=0$$. But this is a dynamic spacetime so the Komar mass is not defined until after the end of the collapse, not during it.

So where does this leave us? If you say "amount of mass-energy" is a measure of something conserved related to the stress-energy tensor, then it is clearly just zero in the Schwarzschild metric. The relativistic mass measures are all global, and do not correspond to any "where" of the mass. I have no idea of all the quasi-local measures.

In the end, maybe one should turn the question around: does it matter?

• it seems very "Machian" that Komar mass cannot be consistently defined until you encompass all the available spacetime in its volume integral. Those quasi-local measures should encode interesting things about the uncertainty of the rest of spacetime in its boundary Jun 30, 2021 at 17:22
• Please note we talk about the Schwarzschild vacuum solution, not about the gravitational collapse. - If you say the stress-energy "is clearly just zero in the Schwarzschild metric" you seem to mean "zero" in the universe described by the metric. Do you mean the stress-energy is outside the Schwarzschild universe?
– timm
Jul 1, 2021 at 8:15
• @timm What does that "outside the Schwarzschild universe" even mean? Note that collapse solutions (having nonzero stress-energy in some region) that approach Schwarzschild (no stress-energy anywhere) would seem to give a link between everyday localizable mass and the nonlocal Komar mass... except that selecting the right spacelike slice to do the integration becomes fraught. Jul 1, 2021 at 10:02
• @AndersSandberg Exactly! That's why I asked this question. So stress-energy is not inside the metric and outside makes no sense. The cause of stress-energy is mass. Again, is it nowhere?
– timm
Jul 1, 2021 at 15:37

I think it is best to understand this by thinking about the way the curvature is established before the black hole forms. The Einstein equation is very much a local cause-and-effect type of equation. The curvature of spacetime is established at each event by the matter at that event and by the continuous nature of spacetime; the way it connects to neighbouring regions. In particular, the curvature of spacetime in the vicinity of some mass is caused in this way. If the mass collapses within an event horizon then the curvature outside the horizon is maintained because spacetime has the kind of continuity which the field equation describes. Each region of spacetime is, if you like, holding the neighbouring regions in a kind of 'tension' (speaking loosely). The role of the singularity in all this is to announce that 'here the description you have been using runs out of validity'. So what do we do? We note that the Schwarzschild-Droste solution is indeed a solution to the field equation so we conjecture that it does describe spacetime in the region where it holds (i.e. everywhere except at the singularity).

The main thing to note, in relation to the question asked, is that the mass which caused or acted as source to the situation in spacetime at any given event is the mass in the past lightcone of that event. But the singularity is not in the past lightcone of any event. So you do not need to involve the singularity in order to understand how spacetime acquired whatever curvature it has got. Another way to put it is to say "the mass of the black hole" is that part of the total mass in a given observer's past lightcone which the observer chooses to associate with the black hole as opposed to some other object.

This kind of argument is also true of gravitation more generally. The "mass of the Sun", as far as we on Earth are concerned, is whatever mass it had 8 minutes ago.

A related question which may also help:

if I drop a mass into a black hole, will the black hole's gravity be asymmetrical before the mass reaches the singularity?

• This does not answer the question that was posed. In the "Schwarzschild vacuum solution" discussed in the question, there is not (and never was) any matter that led to the curvature. This is the answer to some other question that someone could have asked - and probably a good answer to some such question. In context here, however, it is just nonresponsive. Jun 30, 2021 at 18:05
• @Brick I see what you mean, up to a point, but I shall leave my answer (and these comments) with this final addition: this question, and many like it, bring home to me that a black hole has to be seen as a spacetime phenomenon, not just a spatial one that happens to persist. If we insist on an eternal black hole, one that somehow just is there right into past infinity, then we are talking about something unphysical so perhaps we do not have a physical answer about where the mass is, or was, or will be. We only have ways of defining its amount, such as the Komar mass. Jun 30, 2021 at 21:01
• " If we insist on an eternal black hole, one that somehow just is there right into past infinity, then we are talking about something unphysical ". In this sense yes. We talk about a model which combines Mass with curvature. The Schwarzschild universe being a vacuum solution doesn't contain the mass. Where is it, outside this universe? Or nowhere?
– timm
Jul 1, 2021 at 8:33
• @timm If indeed we are talking about a mathematical abstraction which does not model or describe anything in the physical world, then there is no answer except in the abstract 'world' of the mathematical abstraction. And that 'world' or 'universe' appears to be unable to answer the question. Jul 1, 2021 at 9:59

What you're seeing in different answers is, to some degree, differences in viewpoint about the relevance of the Schwarzschild solution rather than an answer to the specific question that you've asked. It's worth putting those in context, however, before answering your specific question. The Schwarzschild solution arises in at least two contexts that are related but different:

• It is a vacuum solution to the Einstein equations, as you noted originally, in which there's a parameter $$M$$ that defines a family of formal solutions and to which we've attached an interpretation of "mass".
• It arises as a limiting case for other processes, including but not limited to collapse processes, for long times, large distances, or both.

The interpretation of the formal parameter $$M$$ as "mass of the black hole" is probably most closely related to the second bullet rather than the first. For large distances, for example, the gravitational field of a black hole with mass $$M$$ and localized matter that you cannot resolve with total mass $$M$$ is the same. For Schwarzschild, you can work through variations of scenario, and the parameter $$M$$ will work out such that it is "as if" the black hole had that mass. This "as if" sense is not really local though, because it's about looking at the gravitational field some distance away.

Now if you look closely, you might be able to tell the difference between these scenarios on a local scale. For example, if you watched long enough or got close enough, you might see the actual matter in a collapse process, and then you could try to assign the location of the mass to the location of the corresponding matter. For a true vacuum solution, however, you can never watch long enough (the solution is static) or get close enough (there's a singularity so any finite distances is, in some loose sense, "far"). There's no real location to assign to this mass. It's "mass" only in the (non-local) "as if" sense noted above.

There is another way to view this: The reason why mass seemingly disappears for astrophysical black holes (that is, a black hole formed by collapse) is that you are using the "wrong" global time coordinate to describe the space-time. If you use "the right" global time coordinate, the matter is always there, but it evolves extremely slowly with respect to the clocks that run outside of the black hole.

Let me use the Penrose diagrams of a Schwarzschild BH to demonstrate my point: On top you have the more familiar Penrose diagram of an eternal black hole. The second diagram is the Penrose diagram of a collapsing star, where in orange is the internal solution for the internal (non-vacuum) gravitational field inside the star. The thing to notice is that the tip of the "diamond" in the region outside the black hole represents (in a compactified way) evolution of all time-like future-oriented curves as their proper-time $$\tau \to \infty$$. The tip of the diamond is the "end of time" for all external observers.

If you want a good time-type coordinate, you have to choose space-like hypersurfaces. Space-like hypersurfaces do not bend more than 45 degrees on a Penrose diagram. This means, in particular, that the $$t=const.$$ hypersurfaces are not globally space-like, they become null at the horizon. This means that the Schwarzschild $$t$$ is a bad coordinate and we should choose a new one. There is a way to choose a time-coordinate $$T$$ ($$T=const.$$ surfaces in cyan) such that

1. The coordinate singularity does not occur in your time slicing until "the end of time" for external observers, and
2. The surfaces $$T = const.$$ always contain matter!

So where is the matter? There are perspectives from which it never left!

• On tbe first diagram, $r=const$ lines are curved. How come $r=2M$ is a straight line? +1 Jul 11, 2021 at 7:10
• @safesphere Because the $r=2M$ hypersurface is very different from any other $r=const.$ hypersurface - it is a null surface. A way to realize this is to notice that the $g_{rr}$ component of the Schwarzschild metric in Schwarzschild coordinates changes sign at the horizon.
– Void
Jul 11, 2021 at 15:35
• What you are saying is given, but does not answer the question. A general null surface would intersect other $r=const$ lines. The only null surface that doesn’t is the null infinity. It is self evident from the diagram that the horizon is a null infinity. However, “infinity” is a mathematical term for what doesn’t exist in reality. For example, the null infinity on the right side is not a location in spacetime. It doesn’t exist, but is just an artifact of the diagram compactification. Give it a thought :) Jul 11, 2021 at 17:38
• Also notice that there is no spacelike (vertical) line for $r=0$. This means the spacelike radius of the black hole is zero, which is self evident from the metric anyway. If you set $r=0$ at the horizon, as Karl Schwarzschild did in his original solution, then you can drop non-falsifiable singular regions leaving only the right diamond with $r=0$ as the left null infinity. And no objective physical experiment can prove this wrong. Jul 11, 2021 at 18:11

The singularity is a moment in the future so the mass of the black holes exist in the future. Since space and time are connected asking where it is doesnt make any sense at all.A better question (and more correct) is:which are the space-time coordinates of this object?

• I like the point you are making but it might be appropriate to make it in a technically more careful way. For a black hole formed by gravitational collapse there is a large region of spacetime which is spacelike-separated from parts of the singularity. So, for this region at least, the singularity is not unambiguously in either the past or the future (but it is true to say that it is not in the past lightcone of any event). Jun 30, 2021 at 16:29
• @AndrewSteane The OP asked about the "Schwarzschild vacuum solution", so that definitionally excludes collapse scenarios. I don't see that your "technically more careful way" is technically applicable. Jun 30, 2021 at 17:27
• @Brick not really. Outside a spherically symmetric collapsing body the spacetime is Schwarzschild the whole time. Jun 30, 2021 at 17:32
• P.S. I did not downvote! Jun 30, 2021 at 17:33
• @AndrewSteane Agree that's a true fact but disagree that it has relevance to the question posed. Presumably a spacetime that has matter in it does not present the same conceptual issues about how you can have "mass" in a "vacuum" solution. Jun 30, 2021 at 17:38

The distribution of the mass depends on the frame of the observer.

For an external stationary observer the mass of the collapsing star is only converging to its horizon radius due to gravitational time dilation, and the density is therefore finite.

For an infalling observer the mass is confined to a radius below you and the density is still finite. All the mass will arrive at the singularity in your future, but it is not there yet since the singularity itself is in your future.

With eternal black holes there is no collapsing matter, but that topic was already covered in the other answers.

This is a long comment:

As we all learn, Euclidean geometry was "discovered" and was the way, slowly, the observations of nature by mapping the earth with measurements developed, and the theory gave the possibility to predict distances and times of travel, in addition to a standard of calculating invariant areas of earth. It has been extremely useful.

Where is the mass of the earth in this theory? We know it is a local theory and the mass of the earth is irrelevant for the particular calculations.

BUT we know by observation that it is spherical geometry that really describes the surface of the earth. The axioms that relate the measurements to the mathematics used for predictions have to change to the axioms of spherical geometry. The mass of the earth is relevant only that it can be deduced that it should exist in order to get a spherical geometry.

With Newtonian gravity the mass of the earth is in the law, used to relate measurements to the mathematical theory, of $$F=ma$$, $$F$$ observed and in the observed acceleration induced by the mass.

With General Relativity what relates the mathematically possible infinity of functions to measurements of nature is the stress energy tensor and, instead of $$F=ma$$, it is Einstein's equation, relating the metric of space to the stress-energy tensor.

$$G_{\mu\nu}+\Lambda g_{\mu\nu}=\frac{8\pi G}{c^2}T_{\mu\nu}$$

See the link for the details

Instead of the simple measurable variables of $$F$$ and $$a$$ in the law of Newtonian gravitation, there are complicated mathematical expressions on both sides. This means that there exists a great number of mathematical possibilities to both sides, depending on what one is interested to study. The answer here says that for a specific left side there exist locally consistent solutions to the right side, where the masses and energies involved do not appear, in the way that the mass of the earth is not even imagined in Euclidean geometry.