Black hole and singularity

[Ranku]

Is there a theorem which proves that for a black hole to form, matter must gravitationally collapse to a singularity?

 

[Vanadium 50]

A singularity is not a physical object, so "no". There may be a way to rephrase your question so the answer is different. but that would be up to you.

 

[Nugatory]

The standard Schwarzschild solution of the Einstein Field Equations (the one that people are talking about when they just say “black hole” without qualification) describes a eternal black hole which has existed forever, not formed by gravitational collapse in the past. So we aren’t going to find a theorem that says we can’t have a black hole without collapse - we already have a counterexample.

In some EFE solutions (the Schwarzschild black hole is one, as is the black hole formed by Oppenheimer-Snyder collapse) a singularity is necessarily implied by the math. So in that sense we have a theorem of the form “If the math of GR applies everywhere, then there must be a singularity”. However, that’s just the math of GR - it is possible, even likely, that GR is not a complete description of the spacetime near where the singularity would be so this theorem is of limited practical significance.

 

[martinbn]

Is there a theorem which proves that for a black hole to form, matter must gravitationally collapse to a singularity?

The defining property of a black hole is the event horizon and for the formation of an event horizon you do not need "matter to collapse to a singularity" (whatever you mean by this). For example imagine many stars arranged on a sphere with very large radius and centre Earth. They will fall towards us and an event horizon will form here and expand until it the shell of stars crosses it. And you will not notice anything until much later when the stars come here.

 

[Ibix]

Is there a theorem which proves that for a black hole to form, matter must gravitationally collapse to a singularity?

You can even make a black hole out of EM radiation, at least in theory - the word to look up is kugelblitz.

A question that might be what the OP means (and I think is interesting anyway) is: do you have to have a singularity somewhere for an event horizon to exist? I think you do because you have to have a place other than $\mathcal{I}^+$ where light rays can terminate. I'm not sure how that's possible without a singularity somewhere in spacetime (noting @martinbn's comment that you can have an event horizon before the singularity exists). Or is there some topological oddity I'm not aware of?

 

[martinbn]

You can even make a black hole out of EM radiation, at least in theory - the word to look up is kugelblitz.

A question that might be what the OP means (and I think is interesting anyway) is: do you have to have a singularity somewhere for an event horizon to exist? I think you do because you have to have a place other than $\mathcal{I}^+$ where light rays can terminate. I'm not sure how that's possible without a singularity somewhere in spacetime (noting @martinbn's comment that you can have an event horizon before the singularity exists). Or is there some topological oddity I'm not aware of?

The singularity theorems (here the original one of Penrose) imply that a singularity will form if you have a trapped surface and the usual energy conditions are satisfied.

 

[PeterDonis]

do you have to have a singularity somewhere for an event horizon to exist? I think you do

One has to be careful here since the term "event horizon" is overloaded; the cosmological horizon in de Sitter spacetime, which is nonsingular everywhere, is also called an "event horizon". (The term is also used to describe the maximal future development of a comoving observer's past light cone in models like the best fit model of our current universe.)

A better formulation might be: do you have to have a singularity in any spacetime in which a black hole region, defined as a region not in the causal past of future null infinity, exists? (The event horizon in such spacetimes is simply the boundary of the black hole region.) This rules out de Sitter since de Sitter has no future null infinity at all; its conformal structure is different. (That is also true of models like the best fit model of our current universe.)

Formulated that way, I don't know of any counterexamples, but I also don't know of a proof.

 

[Vanadium 50]

imply that a singularity will form

Please don't say that. A singularity is a mathematical concept, akin to a division by zero, not a physical one. Equations can become singular. "Where is this equation undefined" is a valid question, even if one cannot put the undefinedness into a box.

 

[martinbn]

Please don't say that. A singularity is a mathematical concept, akin to a division by zero, not a physical one. Equations can become singular. "Where is this equation undefined" is a valid question, even if one cannot put the undefinedness into a box.

My impression is that this "formation of singularity" is the standard terminology.

 

[Nugatory]

My impression is that this "formation of singularity" is the standard terminology.

Certainly no one will misunderstand what you’re saying.
V50 does have a point about that terminology being inaccurate (and unfortunately likely to mislead laypeople) but life is too short to say “the solution of the EFE that we are considering has a singularity” every single time.

 

[PeterDonis]

My impression is that this "formation of singularity" is the standard terminology.

The actual "singularity theorems", and more generally the global methods for analyzing spacetimes for which the canonical text is Hawking & Ellis, define "singularity" as geodesic incompleteness: there exist geodesics in the spacetime that cannot be extended past some finite value of their affine parameter. The concept of a singularity "forming" makes no sense in this formulation. The spacetime geometry doesn't "form"; it just "is".

life is too short to say “the solution of the EFE that we are considering has a singularity” every single time.

And even that statement doesn't necessarily capture the real issue of geodesic incompleteness, since it is not as well understood as it should be that that is what "singularity" means in the actual mathematical framework in question.

 

[Nugatory]

Everyone following the thread, please please try to focus on the question. (Some off topic posts have been deleted.)

(And when you see a miscategorized thread start, it's best to just report it. We can correct with a few mouse clicks, and the reports themselves provide us with useful information about how people are understanding the levels. And further discussion belongs in the feedback forum)

 

[Bosko]

For example imagine many stars arranged on a sphere with very large radius and centre Earth. They will fall towards us and an event horizon will form here and expand until it the shell of stars crosses it. And you will not notice anything until much later when the stars come here.

Let's put an observer on one of the stars.
During the fall a shell of homogeneously distributed stars will reach the photon sphere.

What will the observer see?
According to my understanding: An infinite plane homogeneously filled with stars

If at that moment we let the observer calculate the force of gravitational pull of other stars on his star.
What will he get as a result?
Zero, in my humble opinion

What I mean is, that if the above mentioned stars are arranged so densely that they form a photon sphere, they will not fall towards the center due to gravity.

Neither an event horizon nor a central singularity will form.

 

[PeterDonis]

During the fall a shell of homogeneously distributed stars will reach the photon sphere.

What will the observer see?
According to my understanding: An infinite plane homogeneously filled with stars

Your understanding is incorrect.

If at that moment we let the observer calculate the force of gravitational pull of other stars on his star.
What will he get as a result?
Zero, in my humble opinion

Your humble opinion is wrong.

if the above mentioned stars are arranged so densely that they form a photon sphere, they will not fall towards the center due to gravity.

This is incorrect.

Neither an event horizon nor a central singularity will form.

Both of these claims are incorrect. Your scenario is just a version of the model first investigated by Oppenheimer and Snyder in their classic 1939 paper. Both an event horizon and a singularity do form in that model.

 

[Vanadium 50]

Boslo. Boslo. Bosko. Its OK not to know something. In that case, it may be more helpful to ask them in your own thtead than to chime in on someone else's. Try it! It works!

 

[Bosko]

Your understanding is incorrect.

Ok. What a star-filled photon sphere looks like to an observer from one of the stars?

The meaning of the question is whether that sphere of stars will continue to fall and form an event horizon.
Will it continue to fall after that and end up in the central singularity.

Boslo. Boslo. Bosko. Its OK not to know something. In that case, it may be more helpful to ask them in your own thtead than to chime in on someone else's. Try it! It works!

In Schwarzschild's solution there are two singularities $r=0$ and $r=r_s$

Your position is that: The central singularity is an imperfection of the mathematical model.

Do you think that the singularity $r=r_s$ is not a real singularity but a "wrong interpretation" of the solution?

 

[martinbn]

Ok. What a star-filled photon sphere looks like to an observer from one of the stars?

The meaning of the question is whether that sphere of stars will continue to fall and form an event horizon.
Will it continue to fall after that and end up in the central singularity.

Your questions show that you think in non relativistic terms. This is not just wrong here it is meaningless. Before trying to understand black holes, horizons, singularities and so on you need to learn enough relativity.

In Schwarzschild's solution there are two singularities $r=0$ and $r=r_s$

Your position is that: The central singularity is an imperfection of the mathematical model.

Do you think that the singularity $r=r_s$ is not a real singularity but a "wrong interpretation" of the solution?

There is no singularity at $r=r_s$.

 

[Ibix]

Ok. What a star-filled photon sphere looks like to an observer from one of the stars?

If the stars were eternally hovering there, you would see an infinite number of them, yes, in a hall-of-mirrors kind of way. But they are not hovering - they are infalling at considerable speed. Because there's only flat spacetime inside the shell, you'd just see a Dopplered and aberrated sphere of stars everywhere below you. You might even see the sphere occupying some of the sky above you due to lightspeed delay, although I'm not 100% certain of that.

The meaning of the question is whether that sphere of stars will continue to fall and form an event horizon.

Of course they'll continue to fall. What's going to stop them?

Will it continue to fall after that and end up in the central singularity.

Classically, the singularity isn't in the center, but it is where all the matter would end up. In reality, we expect GR to stop being accurate somewhere inside the horizon, so we don't know.

Your position is that: The central singularity is an imperfection of the mathematical model.

It's almost certainly a symptom of our model not being correct, true.

Do you think that the singularity $r=r_s$ is not a real singularity but a "wrong interpretation" of the solution?

It's a singularity in Schwarzschild's coordinate system, closely analogous to the issues with latitude and longitude at the poles - what's your longitude if you stand at the north pole? Just like the poles, there is no actual physical singularity there, and you can pick a different set of coordinates (Eddington-Finkelstein, Gullstrand-Painleve or Kruskal-Szekeres are well known ones) that are well behaved through the horizon.

 

[PeterDonis]

What a star-filled photon sphere looks like to an observer from one of the stars?

They will see stars if they look in the tangential direction (along the photon sphere), including multiple images of stars (including their own) from light that has traveled around the photon sphere multiple times. But they will see empty space if they look radially inward or radially outward.

The meaning of the question is whether that sphere of stars will continue to fall and form an event horizon.
Will it continue to fall after that and end up in the central singularity.

Yes to both. But the answer to those questions has nothing whatever to do with how the stars look when the shell of stars has fallen inward to the photon sphere.

In Schwarzschild's solution there are two singularities $r=0$ and $r=r_s$

The one at $r = r_s$ is only a coordinate singularity. It has no physical meaning and it poses no issues with the model (just choose coordinates that don't have it).

Your position is that: The central singularity is an imperfection of the mathematical model.

No, it's not. The entire mathematical model is perfectly self-consistent. The locus $r = 0$ is not part of the mathematical model.

Most physicists believe that, because of the curvature singularity at $r = 0$, the model will become less and less accurate as that locus is approached, and some new physics, such as some kind of quantum gravity, will take over. But that is a physical issue that might or might not turn out to be valid. It is not a mathematical issue at all.

Do you think that the singularity $r=r_s$ is not a real singularity but a "wrong interpretation" of the solution?

It's a coordinate singularity. See above.