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To my knowledge if you calculate the coordinate time (time experienced by an observer at spatial infinity) it takes an infinite amount of time for an object to fall past the horizon of a Schwarzschild black hole. Doesn't this imply that it takes an infinite amount of coordinate time for a Schwarzschild Black Hole to form since the last bit of in-falling matter won't ever ross the horizon as observed by someone at spatial infinity? If so, is it possible for other types of black holes (Kerr etc.) to form in finite coordinate time?

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Related: physics.stackexchange.com/q/5031/2451 , physics.stackexchange.com/q/21319/2451 and links therein. –  Qmechanic Aug 23 '12 at 20:59

1 Answer 1

Pick some maximum visible wavelength of light (say, the radius of the solar system). And let's consider only initial source frequencies below the rate at which, say, one photon per year is emitted. In a finite time, all of the light leaving the matter distribution below this frequency, as observed by a distant observer, will be redshifted beyond your maximum wavelength. It will appear practicably indistinguishable from a black hole.

The actual plunge phase of this process will happen very, very quickly (think days, not centuries), so the object will go from emitting in the visible to essentially dark in a very short period of time.

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thanks for pointing this out. I agree that all visible light will be redshifted away of the visible spectrum, but will it form a black hole in the formal sense of the word? –  DJBunk Aug 24 '12 at 2:50
    
@DJBunk: yes, that's what a formal sense of the word is--it will be observationally indistinguishable from a black hole and inconsistent with any other object with a well-defined theoretical backing. If it quacks like a duck, then I betchya that it is a duck. –  Jerry Schirmer Sep 23 '12 at 0:14
    
I betcha it isn't –  Nathaniel Nov 22 '12 at 3:39
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...by which I mean, in the case of black holes, there's an awfully big difference between being observationally indistinguishable from a black hole and actually being one. That difference is the no hair theorem. If an object actually is a black hole then you run into problems with what happened to the information associated with the matter that it was originally made of, whereas if it merely quacks like a black hole then all that matter is really still there, just outside the event horizon, but just not practically observable, and in this case there's no deep issue about the information. –  Nathaniel Nov 22 '12 at 3:45
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So given my limited knowledge of GR I think this answer is probably correct (+1), but I think it's important to note that it's essentially a negative answer (collapsing stars very rapidly become good approximations to black holes, but never actually become "true" black hole solutions to Einstein's equations), and it's also interesting to note that if it is correct it solves the black hole information paradox completely, without needing to invoke string theory or black hole complementarity or anything else beyond general relativity itself. –  Nathaniel Nov 22 '12 at 3:50

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