Can we have a black hole without a singularity?

Question

Assuming we have a sufficiently small and massive object such that it's escape velocity is greater than the speed of light, isn't this a black hole? It has an event horizon that light cannot escape, time freezes at this event horizon, etc. However this object is not a singularity.

If a large star's mass were compressed to the size of, say, a proton, it would certainly have these properties but it would still not be a singularity as a proton does have volume.

The reason "physics breaks down" at singularities is because we cannot divide by zero, but as long as the proton-sized object has volume, physics won't "break down", yet we still have an event horizon and an object that is invisible (but not undetectable) from the outside.

I have read the answers to this related question. I'm not sure if they don't address my specific question or if I don't understand the answers.

Answer 1

Suppose you have some collection of matter that is so dense it has an event horizon where the escape velocity is greater than the speed of light. The escape velocity is obviously due to the strong gravitational field of the matter inside the event horizon, and equally obviously that matter is also pulled by it's own gravity towards it's centre of mass. Also obvious is that because the surface of your collection of matter is nearer to the centre of mass than the horizon is, the gravitational pull on it must be even stronger than the gravity at the event horizon i.e. the (hypothetical) escape velocity would be even faster than the speed of light.

The reason why this situation isn't stable is that the matter making up your object cannot resist the force of it's own gravity and is irrestably pulled inwards until it forms a singularity. At that point we have a standard black hole with an event horizon and a singularity at the centre.

To understand why the matter within the event horizon cannot avoid being pulled down into a singularity you have to do some maths. If you're interested my answer to Why is a black hole black? gives a hopefully not too scary explanation of the maths.

I think there is a semi-plausible way to explain why the matter can't avoid collapsing into a singularity, but don't take this too literally. I've mentioned above that if the escape velocity at the event horizon is the speed of light, the escape velocity inside the event horizon must be faster than light. But all forces, e.g. the electrostatic forces that hold you in shape, propagate at the speed of light. That means inside the event horizon the electrostatic force can hold matter in shape because it can't propagate outwards fast enough. This also applies to the weak and strong forces, and the end result is that no force is able to resist the inwards fall of the matter into a singularity.

Answer 2

To make this question more precise, we need to define terms a little better. The standard definition of a black hole is the following. Suppose there are points in spacetime from which it's impossible to escape to a large distance. (Technically, we want these to be points from which we can't escape to future null infinity.) If we have set of such points, then the boundary of that set is a black hole event horizon. A black hole is defined as a region of space surrounded by this particular type of event horizon. So there is nothing in the definition that directly requires a singularity.

It is certainly possible to have a horizon without a singularity. In fact, horizons are observer-dependent. In flat (Minkowski) spacetime, you can have an observer with constant proper acceleration, and for that observer, there is a horizon. Events behind the horizon can never send a signal that the observer will be able to receive. However, this horizon isn't the boundary of a black-hole event horizon, so there is no black hole.

There is a concept very similar to that of an event horizon, which is a trapped lightlike surface. This is a surface such that even if you emit light rays from it in the outward direction, the resulting surface formed by the emitted rays has a decreasing volume. If such a trapped surface exists, then the Penrose singularity theorem guarantees that the spacetime contains a singularity.

For a more in-depth discussion of this sort of thing, see Booth, http://arxiv.org/abs/gr-qc/0508107