# If black holes are just empty vacuum of space inside, then what causes the curvature?

I have read this question:

The fundamental confusion many have about black holes is thinking that they are discrete "things" surrounded by horizons and other phenomena. But they are actually extended spacetime curvature structures (that imply the various phenomena). The singularity is not doing anything and is not responsible for the gravitational field, it is a consequence of the field.

What are tidal forces inside a black hole?

As far as I understand, as per general relativity, spacetime curvature is caused by stress-energy (not mass). This answer is using a vacuum solution to describe black holes, and you can read in the comments to that question that there is no need for any matter (or mass) to be present inside the black hole, it is just a vacuum, but spacetime itself is curved, and the gravitational field itself has the energy needed for the curvature itself. This includes the singularity itself, which in this answer is described as being "off the metric", that is not part of our spacetime, hence, it cannot cause the curvature.

Now, if the interior of the black hole, is a vacuum (the model is a vacuum solution), meaning the collapsed star's gaseous matter is not there (as far as I understand it is in the singularity), and the singularity is not part of our spacetime, then neither can cause curvature.

Again, GR describes curvature as being caused by stress energy. If there is no matter, no mass, nothing with stress-energy inside the black hole, except the singularity, but the singularity is not part of our spacetime, then what causes the curvature?

There are suggestions in the comments, that the collapsing star's gaseous matter transforms into the energy of the gravitational field itself. But I do not understand how electrons and quarks can transform into gravitons.

Still, how can the gravitational field itself cause the curvature, or how can it sustain itself? Gravity sustains itself, curvature means stress-energy in the gravitational field, and this energy causes curvature?

Question:

$$1$$. If black holes are just an empty vacuum of space inside, then what causes the curvature?

GR describes curvature as being caused by stress-energy.

This statement is slightly wrong and is the cause of your confusion here.

Technically, in GR the stress energy tensor is the source of curvature. That is not quite the same as being the cause.

An easy analogy is with Maxwell’s equations. In Maxwell’s equations charge and current density are the sources of the electromagnetic field. However, although charges are the source of the field there exist non trivial solutions to Maxwell’s equations that involve no sources. These are called vacuum solutions, and include plane waves. In other words Maxwell’s equations permit solutions where a wave simply exists and propagates forever without ever having any charges as a source.

Similarly with the Einstein field equations (EFE). The stress energy tensor is the source of curvature, but just as in Maxwell’s equations there exist non trivial vacuum solutions, including the Schwarzschild metric. In that solution there is no cause of the curvature any more than there is a cause of the plane wave in Maxwell’s equations. The curvature in the Schwarzschild metric is simply a way that vacuum is allowed to curve even without any sources.

Now, both in Maxwell’s equations and in the EFE the vacuum solutions are not particularly realistic. Charges exist as does stress energy. So the universe is not actually described by a vacuum solution in either case. So typically only a small portion of a vacuum solution is used to describe only a small portion of the universe starting at some matching boundary. A plane wave can match the vacuum region next to a sheet of current, and the Schwarzschild solution can match the vacuum region outside a collapsing star.

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. This would be in the causal past of the vacuum region including the vacuum inside the horizon. Since it is in the causal past it can be described both as the cause and the source of the curvature, with the understanding that it is strictly outside of the Schwarzschild metric which is a pure vacuum solution in which the curvature has no source.

• Hi Dale the Schwarzschildsolution describes a universe. What do you mean saying "stress-energy that is outside of the vacuum solution, in the part of the universe not described by the Schwarzschild metric."? Do you talk about something which is outside the Schwarzschild universe? Why do you think there is an outside?
– timm
Jul 1 at 7:43
• @timm as I said “typically only a small portion of a vacuum solution is used to describe only a small portion of the universe starting at some matching boundary”. The “outside” referred to stress energy outside the portion of the universe that is matched to the Schwarzschild solution.
– Dale
Jul 1 at 10:57
• The Schwarzschild metric is asymptotically flat. If r goes to infinity the Schwarzschild metric shifts to the Minkowski metric. Does "outside the portion of the universe that is matched to the Schwarzschild solution" mean flat Minkowski spacetime? Then this “outside” would be "referred to stress energy". But I am not sure if I understood you correctly.
– timm
Jul 1 at 16:03
• @timm that could be one example, but I was specifically intending that the Schwarzschild solution matches the vacuum surrounding a star, but the star itself is not vacuum so the star itself is the part of the universe not described by the Schwarzschild metric. The mass of the star is the source of curvature near the star, but it is outside of the portion of the universe that matches the Schwarzschild solution.
– Dale
Jul 1 at 19:22
• @timm and now we circle back to your first comment. I am not sure how you are missing this point when it is the key point of the answer: “typically only a small portion of a vacuum solution is used to describe only a small portion of the universe starting at some matching boundary”.
– Dale
Jul 3 at 10:52

if there is no matter, no mass, nothing with stress-energy inside the black hole, and the singularity is not part of our spacetime, then neither can cause curvature

On top of the great Dale answer - the Einstein field equations (EFEs) are local. Imagine a lone planet in the universe. Then everywhere outside the planet there is a vacuum and because EFEs are local, the solutions that will describe the outside of the planet are vacuum solutions. Non-vacuum solutions are needed only to describe interior of the planet.

But mathematically, we can just ignore the planet. We can declare it not being part of spacetime, and we are left with spacetime completely described by the vacuum solution, but with a hole inside it. In this spacetime with a hole, there will be no material source of the curvature, but using some common sense we can understand that our spacetime is missing something and that source is probably located at the hole, which we removed from our model for whatever reason.

If we are interested only in spacetime outside the planet, like for example NASA is, then the model with a hole is good enough and the hole and spacetime without a source bothers no one except philosophers.

A singularity is an analogical scenario, albeit a little more dramatic. A singularity is where the math breaks down. In the planet scenario, we consciously removed the planet from spacetime and can put it back if wished so, but in the case of a singularity we have no choice. Our math cannot describe it, so we are left with spacetime that has a hole inside, but we are mathematically unable to fill the hole in. The infinities that pop up forbids us to use differential geometry to extend the spacetime to get rid of the hole and bring the source back to spacetime. But still, we can use this model with a hole well enough to describe things that go on outside of it. Using some common sense though, one can understand that the matter is located at the singularity, or even better, that general relativity is probably not applicable near it and a better theory would produce more sensible solutions without any holes.

My point is, just because the source is not part of our model of spacetime, it does not mean it cannot cause curvature. Just like in the planet scenario, we can remove the planet from spacetime and create a mathematical model without sources, but physically it is understood that there is a source. We just choose not (or straight out fail) to include it in the model.

You can think of the general (Riemann) curvature as being made up of two parts: Ricci curvature and Weyl curvature. Ricci curvature is determined by the stress-energy tensor, and vice versa. Since Schwarzschild is a vacuum solution, meaning its stress-energy tensor vanishes, its Ricci curvature is also zero. Weyl curvature, on the other hand, is independent of the local matter (stress-energy). That is, at a given point, the Weyl curvature is independent of the matter at that point. Weyl curvature describes propagating effects, such as gravitational waves from a distant collision of black holes. Since Schwarzschild spacetime is curved (meaning its Riemann tensor is nonzero), and vacuum, its curvature must be Weyl curvature.

There are some mismatches between the Schwarzschild metric and its physical interpretation as a black hole formed from collapsing matter. (Don't get me wrong, Schwarzschild works well and is extremely useful.) Physically, we expect black holes to be formed from matter, such as a collapsing star. Intuitively, I would expect the matter to end up in a compressed state in the centre. Yet the Schwarzchild solution is vacuum, with no matter anywhere! The singularity ($$r\rightarrow 0$$) is not part of spacetime, so the matter is not there. Most people expect future theories to improve the situation, for example quantum gravity may remove the singularity.

So what causes the curvature? It cannot exude from inside the horizon, as I understand, because gravity cannot propagate faster than light. So imagine a collapsing spherical ball of matter. Outside it, the Schwarzschild metric holds. After the entire ball passes inside the Schwarzschild radius $$r=2M$$, it can no longer influence anything at radii outside the ball, let alone at $$r>2M$$. Hence the gravitational field is self-sustaining.

• Thank you so much! "So what causes the curvature? It cannot exude from inside the horizon, as I understand, because gravity cannot propagate faster than light." As far as I understand, the static gravitational field of the black hole extends outside the horizon. It's already there, so no need for anything to propagate faster then light from inside. But I believe you are saying that it cannot be caused by the matter that fell inside. Jul 14 at 22:24
• Yes, I meant the curvature is already there, as you write. I think it gets frozen in when the horizon forms as the matter passes $r=2M$. So it is caused by the matter, but only indirectly, at an earlier "time" before the horizon forms. Jul 16 at 1:29
• ...frozen in at the horizon, I meant, and the curvature or gravity propagates outwards and inwards from there. By using the word "frozen" I do not imply the old idea of a frozen star, which denies the formation of a black hole. Also a person who specialises in the 3+1 approach (meaning foliation into hypersurfaces), including numerical relativity or the initial value problem, may be better placed to comment Jul 16 at 1:45