# How precisely does a star collapse into a black hole?

I think we all heard general statements like "once big enough star burns out there is nothing to prevent the gravitational collapse ending in a black hole". But I can't remember even seeing the process described precisely.

Here's the deal: at first there is a nice object, a star. Stars can be nicely modeled by general relativity, nuclear physics and statistical physics combined and very much is known about these models and it can be observed whether they agree with things like light and neutrino fluxes, surface temperature and probably also lot of other stuff I know nothing about.

After the collapse we are left with another nice object, a black hole. We know that black holes have no hair.

The question is: what happens in-between? More precisely, between the time when all of the nuclear material has been burned out (and if possible ignore effects like reheating of the star after big enough compression) and the time where there is nothing more than just a black hole.

• Give a description of what happens during the collapse?

• How does the star "lose its hair"?

• Can the actual collapse be solved analytically?

• At what point is singularity created?

Update: I don't want to know what an outside observer will see. Instead, I'd like to find out what an individual part of the dead star will "feel" when a black hole is about to form near it. In other words, I want a complete solution (ideally analytical, but numerical would be also completely fine)

Feel free to assume anything that makes your life easier. Spherical symmetry is definitely fine. Also, if for any reason the questions don't make sense (like Cauchy problem is ill-defined in the presence of the singularity) feel free to interpret them in a way that make them sensible (e.g. assume that black hole is built from D-branes).

Also, I have a feeling that what I intended as a simple question at first ended up being pretty complex. If you think it should be split into smaller (and therefore more manageable and answerable) parts, let me know.

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I take it this is not a question that should be answered in terms of the layered burnout of the core of a large star because you are interested in what happens after the collapse of the electron degenerate core, the free-fall of the outer layers and the big rebound? I think that subsequent events are still ill-understood, especially as there are complicated angular-momentum driven goings on. – dmckee Dec 14 '10 at 21:08
@dmckee: indeed, I should have made it clear (will do) that I am not interested in the complete life of the star; just the very last part before becoming a black hole. – Marek Dec 14 '10 at 21:12
Ok, the general case can't be solved analytically. I think that there is an analytic solution only for dust. The problem of the formation of the horizon is interesting, since you have things like apparent horizons forming. I don't think you can tell much about the singularity. The general idea is that the hairs are radiated away with gravitational waves in the general case, but there are no hairs in the case of spherical collapse. There is some work of Nikos Stergioulas on the subject. I'll find the paper and I'll post it. If I'm up to it I'll try to turn that answer in to a real post. – Vagelford Dec 14 '10 at 21:26
@Vagelford: The dust solution would also be fine, if that's all that can be solved. As I said, feel free to interpret questions in any way that makes your life easier. I hope you can find that reference. – Marek Dec 14 '10 at 22:24
@Jerry, as for the Penrose-Carter: that's indeed a pity. Do you think you could get a reference to the diagram somewhere? I suppose it must've been drawn by someone already. Actually, I am going to try to find it myself. ...okay, I found some notes – Marek Dec 15 '10 at 18:43

## 8 Answers

The solution for this problem for a dust equation of state and spherical symmetry is known as the Oppenheimer-Snyder solution. You model the interior of the distribution as a FRW universe with positive spatial curvature, zero pressure, and zero cosmological constant. You model the exterior of the solution as the Schwarzschild solution cut off at a time-dependent radius. So long as the matter distribution is dust, the thing satisfies all of the junction conditions you need. See Poisson's relativity book or MTW.

A more general solution requires numerics. But one thing we can say for sure is that there is no need for the black hole to shed its 'hair' in the case of spherical symmetry--the radial dependence of the solution will just compress into the singularity eventually, or scatter out to infinity. Birchoff's theorem tells us every spherically symmetric vacuum solution must be the Schwarzschild solution (perhaps with an electrostatic charge, which is technically not vacuum). This is related to the fact that there can be no monopole radiation in relativity.

Also, the general case for this problem is very likely chaotic. Already, if the equation of state of the matter is that of a classical, spherically symmetric, Klein-Gordon field, which is a relatively simple generalization, the system exhibits a (link is a large postscript file)second-order phase transition, a result found by Matt Choptuik, and related to the settling of the Hawking naked singularity bet.

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And I should say that, in the Oppenheimer-Snyder solution, the "big bang" joins on to the white hole singularity in the extended schwarzschild solution, while the "big crunch" joins onto the black hole singularity. – Jerry Schirmer Dec 15 '10 at 1:29
Minor comment to the answer (v4): In several of your answers Mr. Birkhoff is misspelled. This could make your answers harder to find in keyword searches. – Qmechanic Jan 12 at 18:45

The "hair" is lost via gravitational radiation. This is also known as quasi-normal ringdown, as the BH vibrates at different frequencies much like a drum (maybe a "gong" is better analogy). Any charge on the black hole will simple get shorted out by free charges in the surrounding plasma, on a very short time scale.

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Thank you. Is it also known how and when singularity and horizon form? Or is this information too much to ask for? To try to give answer myself: to tell anything about formation of the singularity one would probably need quantum gravity? And the horizon forms as soon as there is critical amount of matter, I guess. – Marek Dec 15 '10 at 18:54
@Marek: for the Oppenheimer-Snyder solution, you model the interior matter as a closed FRW model, and join on the 'big bang' sinularity to a white hole singularity, and the 'big crunch' singularity to a black hole schwarzschild singularity, and paste the spacetimes together. So the singularity forms instantaneously at exactly the point at which the FRW observers' geodesics end. The only warning those observers get is when they see an apparent horizon form around them. – Jerry Schirmer Dec 15 '10 at 19:30
@Jerry, yes I think I got the picture after our discussion above (which happened after I posted this comment). But thanks again for the clarification. – Marek Dec 15 '10 at 19:36
As somebody around here noted, it is hard to say "when" the singularity happened, because it is very much observer dependent. From an observer far from the BH (i.e., "coordinate time"), the singularity is NEVER reached. But from a free-falling observer on his way to the singularity, it is reached in roughly a light-crossing time (~10 microseconds) – Jeremy Dec 15 '10 at 21:31

I will try to do an intuitive explanation as I do not possess the mathematical knowledge for any further analysis.

I do not think, for an outside observer, singularity would form in a definite time. Rather, when the density inside is enough for an event horizon to form, i.e., the radius is smaller than the Schwarzschild radius, a black hole is said to be formed, not its singularity though. Because, for an outside observer, time comes to a complete stop at the event horizon, and only if the event horizon gets smaller, can the region inside the initial event horizon make sense for the outside observer. So if you believe that Hawking Radiation exists, then in a finite amount of time, the black hole will slowly evaporate until it's event horizon is nothing more than a singularity, then evaporate completely.

So rather than an actual forming of a singularity at a certain time for an outside observer, the mass that is "queued up" gradually condenses into a singularity, while also evaporating via hawking radiation.

As for the losing of the hair, once the event horizon forms, the hair of the materials at the event horizon or inside are lost because there are two possible fates(actually one final ultimate fate) for a particle until the forming of the singularity that I mentioned: it will be emitted as energy due to hawking radiation, or it will be the part of the last singularity which will also evaporate due to hawking radiation.

This question is actually similar to a question that I had asked and couldn't explain what I meant very clearly.

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Maybe I should have made it more clear that I am not interested in outside observers. I want to know what our equations tell us about this problem and they should describe everything, singularity included (or maybe they fail to do this; if so I want to why and where). If you need to have an observer, imagine there is one that lives inside the star. What does he experience when the black hole is about to form? – Marek Dec 14 '10 at 22:15

I'm kinda out of my depth here, but I think the balding issue can be understood in cartoon terms as an effect of the time dilation of events nearing the event horizon. From outside all the material appears to stack-up against the horizon and things appear to stop happening. If there are no dynamics there can only be static effects: i.e. electrostatics and gravitation.

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Seeing that there are two answers that look at the problem from the outside I will clarify that this is not what I am interested in at all. – Marek Dec 14 '10 at 22:15

Since you want to focus on what an individual part of the dead star will "feel" when it becomes a black hole, I think the "no hair" part is outside the scope of that; that's a condition only for someone outside of the hole. (See "externally observable" in that Wikipedia link on "no hair" you gave.) For someone falling with the collapsing body, whether or not it's a star, the body has "hair" all the way short of the singularity. (At the singularity, general relativity breaks down.) I'd say the singularity is first formed when the first chunk of matter that is below the new horizon reaches the center.

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• Give a description of what happens during the collapse?

Firstly your assumption that first the nuclear fires burn out, and then a collapse happens afterwards is just that, an assumption. An event horizon can start at a point in empty space and expand and grow into a shell that expands outwards at the speed of light until it reaches some matter far far away. And the inside could very well still have nuclear fire burning inside for a while longer before the singularity forms.

So what does happen for sure? The matter falls in. The local inertial frames actually had to supply an outwards pressure to cause an acceleration to keep it from infalling. You stay on the surface of the earth because in a local inertial frame, a pressure is pushing on your feet, accelerating you upwards. A collapse literally feels as simple as like feeling less pressure on your feet. Like an elevator going down. That's what it feels like.

Unless you are asking about tidal forces, and those depend on the size of the black hole. Those get stronger as you go deeper. But the strength depends on how big you are. For a gas, your size is the size of an atom, so it happens at the level below where we do the fluid approximation and all the statistical mechanics.

• How does the star "lose its hair"?

Gravitational radiation (gravitational waves) do carry some away. But the hair is from the outside. And from the outside the black hole is never finished forming.

• Can the actual collapse be solved analytically?

Examples are in "Simple analytical models of gravitational collapse" by R. J. Adler, J. D. Bjorken, P. Chen, and J. S. Liu in American Journal of Physics 73, 1148 (2005); http://dx.doi.org/10.1119/1.2117187

• At what point is singularity created?

From the outside, never. From the inside you can track each layer of the star until it reaches the singularity. For a spherically symmetric star, the creation of the singularity is not when the center layer hits the singularity, because the singularity exists like a time, not like a place or an event. For a rotating star, I'm not sure anyone has the model you are looking for.

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I would hazard a guess that aside from time dilation effects, in many case it is probably a slow process. If the mass of the core is marginal for forming a BH, you should initially get a neutron star. As the NS cools, accumulates more material, and/or sheds angular momentum it can cross a threshold for collapse to a BH.

I suspect that in many SN, you get enough angular momentum, so that you get some sort of accretion disklike thing, which again must shed angular momentum before it gets small/dense enough to collapse into the BH. And even then I suspect you might have a phase where there is a smallish BH surrounded by a large heavy accretion disk, which takes its time getting swallowed up. I'm not so sure if any of that happens with ultramassive stars (i.e > 150 times a massive as the sun), these might directly collapse. But, the more common run of the mill heavystars might not take a direct collapse trajectory, but instead spend a period of time losing various forms of energy (thermal, rotational, magnetic) before they can collapse all the way. This energy loss probably results in gamma ray bursts.

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Actually, a neutron star of sufficiently small mass is a stable end state. Neutron degeneracy pressure exists even at zero Kelvin, and that's what stabilizes them. – Jerry Schirmer Dec 15 '10 at 5:31
depends what you mean by "slow process." the actual collapse is on the order of a light-crossing time, i.e. 10-20 microseconds for a 10M_sun core. the subsequent accretion of a high-density disk in a GRB could take hundreds of milliseconds, but that still sounds pretty fast to me. – Jeremy Dec 15 '10 at 13:58
No. I'm thinking that a neutron star forms, which without spin and high internal temperature is beyond the mass limit, but is still stable while it is still hot and rapidly rotating. Once it is able to shed some of this stabilizing energy it can then collapse. – Omega Centauri Dec 15 '10 at 18:16
@Omega : that's not how it works out. If the star is over the mass limit, the NS state will have a lifetime of seconds. It might be enough for the star to lose some mass from the recoil pressure wave, but not enough to slow down the collapse much at all. – Jerry Schirmer Dec 15 '10 at 23:16
I seem to recall hearing about a supernova being observed, and then a gamma ray burst being observed from the same location a couple of days later. If we assume two things (my memory is correct), and the gammaray burst is the result of blackhole formation then we have at least one case where there has been a delay. I suspect that timescales computed from spherically symmetric simulations may not be indicative of what happens in real 3dimensional geometry. – Omega Centauri Dec 16 '10 at 5:11

I doubt that a BH is able to form. I think that before the boundary condition is reached some mechanism acts, for example conversion of mass into radiation, which expels the excess weight out of the system.

EDIT ADD 2015/11:
In this doc we can see why a BH will not form at all, as I said in my answer.
New study claims black holes are mathematically impossible
Backreaction of Hawking radiation on a gravitationally collapsing star I: Black holes? 2014/11

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after reading Omega Centauri above comment "...observed... the gammaray burst is the result of blackhole formation..." and also Valgeford comment "...the GRB has to do with the fireball of the explosion of the star..." IMO it seems that the GRB can carry enough mass/energy away of the system preventing the limit situation. – Helder Velez Jan 19 '11 at 1:11