How precisely does a star collapse into a black hole?

Question

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).

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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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

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.