Are empty black holes possible

When just considering GR without evaporation nor QM, is an empty (containing no matter or anything) black hole possible ?

Let's say that there is only GR and nothing else (no matter or boson fields), and that at time t in some coordinate the metric is a black hole, how will it "evolve" ?

• No mass-energy = no gravitational attraction = no event horizon = no black hole. – HDE 226868 Feb 27 '15 at 20:32
• I believe the answer to your first question is "no", but am not certain. I think if you begin with no singularities and no matter that the metric will stay regular, but I do not have the machinery to prove it. – Geoff Ryan Feb 27 '15 at 20:44
• Your second question is somewhat different: I have no idea how the metric would relax from a (for instance) Schwarzschild solution. It will not release gravitational radiation because its spherically symmetric... but apart from that I have no idea. Anyone know what happens in Electromagnetism? – Geoff Ryan Feb 27 '15 at 20:53
• A static (Schwarzschild) black hole 'contains' no matter; indeed, there is no matter anywhere in the spacetime. See, for example: scienceworld.wolfram.com/physics/EternalBlackHole.html – Alfred Centauri Feb 28 '15 at 4:10
• Ok Alfred I think this answer the second part. – agemO Feb 28 '15 at 10:28

Strictly speaking the two well known uncharged black hole metrics (Schwarzschild and Kerr) are vacuum solutions. This means the stress-energy tensor is zero everywhere except at the singularity where it is undefined. This is what Alfred means when he says "A static (Schwarzschild) black hole 'contains' no matter". However this strikes me as a bit of a cheat since there is a mass associated with the black holes. The geometry becomes the flat spacetime Minkowski metric if we take the associated mass to zero, so it seems to me that a reasonable answer to your question is that we cannot have a black hole that doesn't contain anything.

The nearest I know of is a hypothetical object called a geon. The geometry of a geon is maintained by the gravitational self energy associated with it, so it can exist even though there is no matter/energy within it. However, as far as I know no-one has managed to find a stable geon though I don't think it has been proved they can't exist.

I am going to answer the question 'When just considering GR without evaporation nor QM, is an empty (containing no matter) black hole possible?' and I will omit the 'anything' part, because is an ambiguous term.

And the answer is yes, they are possible. As stated by another person here, Schwarzschild black holes and the rotating and charged versions are vacuum solutions of the Einstein equations. We can associate a mass to these black holes, i.e. an energy scale that gives us its size. But, where is this mass if there is no matter?

Well, the Schwarzschild metric,

$$ds^2=-\left(1-\frac{2GM}{r}\right)dt^2 + \left(1-\frac{2GM}{r}\right)^{-1}dr^2 + r^2d\Omega_2^2,$$

has a geometrical singularity in $r=0$. The mass is encrypted in this singularity, in the topology of the spacetime.

So here we have a black hole with some kind of mass that we can associate to it that comes from the topology of spacetime, but there is no matter anywhere.

Answering the second question, as you can see this metric doesn't depend on time, it's an static solution, so it will not evolve into anything.

Yes, it is totally possible. You need to use classical GR with continuous matter distributions and it requires energetic continuous matter with zero rest mass, and it is an unstable setup, so you have to do it 100% perfectly with absolutely no margin for any error. Let's see how.

You can take a sphere of radius $R$ in Minkowski space, and take a Schwarzschild solution where the event horizon is a spherical shell of surface area $4\pi R^2$. Cut out the interior of the Schwarzschild solution and glue the exterior and the sphere together (that's possible since they have the same surface area). The discontinuity implies there is a source term at that surface, so you need to place energetic matter there, and it has to fit on the shell perfectly. It could be almost any matter on the shell if you only want the black hole to be empty just for a little bit, but in that case it's only a shell that hasn't yet formed the singularity.

If you want the black hole to stay empty, then since the matter must stay on that shell it has to be moving outwards radially and move at the speed of light, so it must be energetic but have zero rest mass. Then it just sits on the event horizon forever. It tries to move away at the speed of light, but the event horizon is the exact place where when you do that you stay on the same surface of surface area $4\pi R^2$.

This situation is unstable because any slight perturbation will cause it to collapse into a singularity. It is a black hole because there is a mathematical event horizon, but it is a completely empty (and flat) spacetime on the inside. It is only a mathematical event horizon since if anything actually tried to cross the event horizon that would trigger the collapse into a singularity, so if you want an event horizon with emptiness on the inside, you can't let anything cross it (well, that's fair because then it wouldn't be empty on the inside).

My explanation is done in the style of just writing down a manifold, then computing what the Einstein tensor is, then finding a stress-energy tensor that can equal it, the manifold-first style. In this case the stress-energy tensor doesn't violate any energy conditions, is singular in the sense that a finite amount of energy is confined to a vanishingly thin spherical shell, and has no rest mass since the source moves outwards at the speed of light. Known gravitational sources without rest mass (just energy, momentum, stress etc.), such as classical electromagnetic waves, can't be confined to a vanishingly thin shell, because they need space in which to undulate in order to propagate. So I see no known way to build this shell (and since it is unstable, it couldn't be done in practice anyway).

Consider a thin spherical dust shell. By the usual arguments, the gravitational field inside the shell is exactly zero -- this is valid both for GR and Newtonian gravity. Over time, the radius of the shell will decrease under the action of gravity until eventually the radius of the shell becomes smaller than its Schwarzschild radius. At that point, the shell will continue to collapse until it becomes a black hole.

In other words, an event horizon will form where there previously was none. But since any observer inside the shell cannot exceed the speed of light, an event horizon will be retroactively formed in the region inside the shell corresponding to light rays that wouldn't have enough time to escape the region enclosed by a sphere with radius equal to the dust shell's Schwarzschild radius.

The end result is that in this situation an event horizon is formed, albeit for a finite amount of time, in a region with exactly zero curvature and no matter fields whatsoever.

I would go further and say it's not only possible, it is most likely mandatory. As John Rennie implies, the issue of a black hole's mass is not very clear and he alludes to the issue of gravitational self-energy. Some time back, I was trying to find out what role is played by the self-energy in the external gravitational field. How does the mass of the outer field add into the accounting of the black hole's total mass?

I was surprised to run across a very interesting paper proposing that the mass of a black hole resides spread out through its external gravitational field and nowhere else. The paper claims that the mass of a black hole resides, wholly and solely, in the gravitational field outside the hole.

This calculation of the total gravitational field energy of a black hole (or any spherical object) was made in 1985 by the Cambridge astrophysicist Donald Lynden-Bell (founding director of the Royal Astronomical Society) and Professor Emeritus J. Katz of the Racah Institute Of Physics.