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Someone answered this question by saying that black hole entropy conditions and no-hair theorems are asymptotic in nature -- the equations give an ideal solution which is approached quickly but never actually reached from the point of view of an observer outside the event horizon.

Since then I've been wondering whether singularities are ever really created, and if not, why do we worry about naked singularities?

Quick recap: to an external observer, an object falling into a black hole experiences time dilation such that it appears to take an infinite amount of time to cross the event horizon and ends up sitting frozen at the border.

So here's my reasoning: the above should also apply during the formation of the black hole in the first place. The gravitational field approaches infinite density as the constituent matter approaches a central point, but to an outside observer, it takes an infinite amount of time for the singularity to form. In other words, it never happens.

As I understand it, naked singularities are dismissed with hand-waving, "we'll fix it when we go quantum," but I don't see that as necessary. It seems to me that singularities never actually form, although event horizons clearly do.

Does this mean that we can stop worrying? What happens in naked singularity scenarios when there is no singularity yet?

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Actually, it's believed that naked singularities cannot classically form from ``ordinary'' matter except in extremely exceptional circumstances. No need to invoke quantum mechanics at all. –  Jerry Schirmer May 23 '12 at 12:53
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So here's my reasoning: the above should also apply during the formation of the black hole in the first place.

That isn't true. Black holes don't start from a point in (for example) the centre of a collapsing star and grow outwards. It's actually the opposite - the event horizon forms outside the collapsing star. That means the matter forming the black hole is already inside the event horizon and GR tells us that anything inside the event horizon falls into the singularity in a finite time. You never have to worry about whether the matter does or doesn't cross the event horizon in a finite time.

It seems odd for the event horizon to spring into existance outside the star, but it's because large black holes are easier to make than small black holes. The Schwarzschild radius depends on the mass, but if you assume a uniform star the mass depends on the star's radius cubed. This means the average density of the black hole, i.e. the mass divide by the volume within the event horizon, is lower for large balck holes than for small ones.

So I think even the most sceptical would have to concede that singularities really exist. However the question of whether naked singularities exist is another and different question. If you do the maths then GR tells you that they can be created e.g. by charging a Reissner–Nordström black hole to it's extremal value. The question is whether the maths is related to reality.

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How does this picture change if one uses dynamic and isolated horizons, rather than event horizons? The latter are somewhat aesthetically unpleasing in many ways, least of which you said, that it already exists in places which do not have high matter density and is extremely teleological in nature. –  genneth May 23 '12 at 14:43
    
I'm not sure I see the connection with singularities, naked or otherwise. The main problem here is the old chestnut that nothing can fall through an event horizon because it takes infinite Schwarzschild co-ordinate time to reach the horizon. –  John Rennie May 23 '12 at 15:12
    
I almost accepted this -- "GR tells us that anything inside the event horizon falls into the singularity in a finite time" but this is from the POV of the falling object, right? From an outside-the-horizon observer isn't it also infinite? If not, why not? –  spraff May 27 '12 at 22:24
    
@spraff: it's true that if you use Schwarzschild co-ordinates it takes infinite co-ordinate time for an object outside the event horizon to reach the even horizon. However if an object starts inside the event horizon, i.e. because the event horizon forms outside it, then even in Schwarzschild co-ordinates it will reach the singularity in a finite time (though the Schwarzschild co-ordinates don't make a lot of sense inside the event horizon so you need to be careful what you mean by "time"). This is the key point. The collapsing star is inside the horizon when the horizon forms. –  John Rennie May 28 '12 at 5:59
    
@JohnRennie, I don't think your answer really answers the question. We can separate one event - "event horizon forms" and think of what happens just before it. That would be some particle finally crossing the boundary of a Schwarzschild radius sphere. So the question is, does this event happen in finite time for an outside observer or maybe from his POV the particle will approach that boundary infinitely? –  Fixpoint Jun 22 '12 at 23:30
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no proof of a singularity can exist as any thing that enters an event horizon can not return as there accessible future only exists in the black hole other wise they have to travel faster than the speed of light which according to Einstein is always the same.

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The horizons also cannot appear, not only singularities.

Our primary result, that no event horizon forms in gravitational collapse as seen by an asymptotic observer is suggestive of the possibility of using the number of local event horizons to classify and divide Hilbert space into superselection sectors, labeled by the number of local event horizons. Our result suggests that no operator could increase the number of event horizons, but the possibility of reducing the number of pre-existing primordial event horizons is not so clear and would require that Hawking radiation not cause any primordial black hole event horizons to evaporate completely.

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See also this answer: http://physics.stackexchange.com/a/21357/1186

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