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.

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    $\begingroup$ What about a spherical shell of light converging inward? A stationary external observer would observe a black hole metric to emerge with all photons getting stuck at a stretched horizon. No singularity would be part of the observer's description of the black hole. $\endgroup$
    – Johannes
    Commented Dec 18, 2012 at 14:32
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    $\begingroup$ related: physics.stackexchange.com/q/75619 $\endgroup$
    – user4552
    Commented Aug 29, 2013 at 22:33
  • $\begingroup$ @Johannes: I'm not sure exactly what you're describing, but it sounds a lot like the Vaidya metric, which does have a singularity, or else a description of something that violates the Penrose singularity theorem, which isn't possible. It doesn't really make sense to talk about whether a black hole singularity is "part of the [external] observer's description" or not. By definition what's inside the event horizon is unobservable from the outside. $\endgroup$
    – user4552
    Commented Aug 30, 2013 at 3:28

3 Answers 3


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 its own gravity towards its 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 its 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't 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.

  • $\begingroup$ Seems to me I read somewhere that if the infalling matter has angular momentum, you don't get a point singularity, but some kind of torus. Is there anything you can say about that? $\endgroup$ Commented Nov 27, 2012 at 16:55
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    $\begingroup$ If the black hole has a non-zero net angular momentum then it's described by the Kerr metric, and the singularity is a ring. It's not a torus because the ring is effectively a 1D object and has no thickness. $\endgroup$ Commented Nov 27, 2012 at 17:38
  • $\begingroup$ Interesting, thanks. And I heard time does funny things in there. $\endgroup$ Commented Nov 27, 2012 at 17:48
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    $\begingroup$ It's difficult to explain this kind of thing at a nontechnical level, and this is a laudable attempt, but I think it's fundamentally not right. These arguments are basically Newtonian, and the Newtonian expectation is that even if the infalling matter can't achieve escape velocity, that just means that it should form some kind of gravitationally bound system that can't spontaneously fly apart. We would expect something like a globular cluster or a solar system. In particular, the angular momentum barrier should make collapse to a point impossible. You need the Penrose singularity theorem. $\endgroup$
    – user4552
    Commented Aug 29, 2013 at 22:40
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    $\begingroup$ @JuanPerez Yes. For example there is a simple model of a collapsing dust cloud called the Oppenheimer-Snyder metric that we use because it's simple enough to be described exactly. In this collapse the event horizon forms before the singularity at the centre has formed. $\endgroup$ Commented Sep 6, 2021 at 14:59

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 are two different theorems in GR that address this question of whether you can have an event horizon without a singularity. They say slightly different things.

The Penrose singularity theorem

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.

This theorem is important because although we know that there is a limit on the mass of a stable neutron star (the Tolman-Oppenheimer-Volkoff limit), such limits assume static equilibrium. In a dynamical system like a globular cluster, the generic situation in Newtonian gravity is that things don't collapse in the center. They tend to swing past, the same way a comet swings past the sun, and in fact there is an angular momentum barrier that makes collapse to a point impossible. The Penrose singularity theorem tells us that general relativity behaves qualitatively differently from Newtonian gravity for strong gravitational fields, and collapse to a singularity is in some sense a generic outcome. The singularity theorem also tells us that we can't just keep on discovering more and more dense forms of stable matter; beyond a certain density, a trapped lightlike surface forms, and then it's guaranteed to form a singularity.

No-hair theorems

A different type of theorem relates more directly to event horizons. These are the black hole no-hair theorems, which apply assuming that the resulting system settles down at some point (technically the assumption is that the spacetime is stationary). Basically, the no-hair theorems say that if an object has a certain type of event horizon, and if it's settled down, it has to be a black hole, and can differ from other black holes in only three ways: its mass, angular momentum, and electric charge. These well-classified types all have singularities.

Of course these theorems are proved within general relativity. In a theory of quantum gravity, probably something else happens when the collapse reaches the Planck scale.

Observationally, we see objects such as Sagittarius A* that don't emit their own light, have big masses, and are far too compact to be any stable form of matter with that mass. This strongly supports the validity of the above calculations and theorems. Even stronger support will come if we can directly image Sagittarius A* with enough magnification to resolve its event horizon. This may happen within 10 years or so.

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

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    $\begingroup$ Wouldn't Penrose theorem simply tell you there is geodesic incompleteness, but nothing about where the mass is concentrated? $\endgroup$
    – jinawee
    Commented Dec 1, 2015 at 15:17

New development: enter image description here

Kerr's new paper


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