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Thanks to time dilation, a distant observer watching a man fall in to a black hole will only see him asymptotically approach the event horizon. So how do black holes ever get bigger?

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Just because the distant observer sees it, doesn't make it true. – chharvey Nov 10 '12 at 1:41

In the frame of reference of the black hole (if there is such a thing), the man is indeed absorbed by the black hole. But the information will never reach the observer since any photon emitted past the event horizon (which is not the black hole surface) will never reach the observer.

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I don't think this answers the question. By the way, the frame of reference "of the black hole" is not well-defined, and the even[t] horizon is not a bad definition for a black hole "surface"... – hwlin Nov 10 '12 at 4:15
I thought your question implied that it would take an infinite amount of time for the man to fall in the black hole. Which is not true. Also, I always thought the size of the black hole was smaller or equal to the Schwarzschild radius, which is the event horizon. – FrenchKheldar Nov 10 '12 at 4:39

I don't think there's a paradox here: The infalling observer will observe himself passing the horizon. The observer at infinity does not see this happen, but she will notice, with time, that the region from which she never receives signals has expanded.

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This question crops up a lot, and I think it illustrates an interesting aspect of GR. Specifically what exactly does the question "How do black holes accrete mass?" mean?

In GR most questions don't make sense unless you specify what observer you're asking about or whether you want an observer independant reply. In the case of objects falling into a black hole the observer independant reply is easy because the geodesic of an object falling into a black hole is easy to calculate. See any introductory GR textbook, or if you want to ask a new question about this I'll be happy to post the details.

But most people asking the question have an observer in mind, and typically it's the Schwarzschild observer i.e. the observer sitting stationary at infinity (or far enough away to be effectively infinitely far away). In this case you're quite correct that this observer will never see an object reach the event horizon, but then the Schwarzschild observer will never even see the black hole form.

What the Schwarzschild observer can do is measure the curvature at some non-zero distance from the event horizon, and from this infer that the black hole must exist. If the observer does this, then throws some mass into the black hole and remeasures the curvature they will find it has increased. This allows them to infer that the mass of the black hole has increased i.e. the black hole has accreted mass.

It's tempting to get excessively philosophical and ask what we mean by "exist", but I think this temptation should be resisted as long as the observer independant calculation gives a clear result. If I do measurements of some object and those measurements are consistent with the Schwarzschild metric of a black hole then I think I'm justified in concluding that the object is a black hole.

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Thanks so much, great answer. And yes I meant an observer at infinity. What do you mean by "the observer will never see the BH form?" The Schwarzschild solution describes an eternal BH, so no one will see it form, right? – hwlin Nov 17 '12 at 19:35
Yes, the Schwarzschild metric strictly describes an eternal black hole, but it's a good approximation for black holes of a finite age. Well, arguably black holes of a finite age don't exist, but you know what I mean :-) – John Rennie Nov 18 '12 at 10:31
I don't understand. We have observed black holes in binary systems and other various places, but we are observers near infinity. How can we observe these black holes if we can never have seen them form? – PyRulez Jul 13 '15 at 1:58
@PyRulez: we have never observed a black hole. However we have observed objects so compact and massive that we now they would form a black hole given infinite time. Note that although a black hole takes an infinite time, a collapsing object collapses to an object very similar to a black hole very quickly. For example if you were to look at Sagittarius A$^*$ it would be indistinguishable from a black hole even though technically it isn't (yet). – John Rennie Jul 13 '15 at 5:37
@JohnRennie That's what I thought. There aren't any singularties either, aren't there? – PyRulez Jul 13 '15 at 6:51

Firstly, just considering the behaviour of infalling matter in a Schwarzschild metric doesn't directly answer the "how can they possibly grow?" question, since this metric describes an unchanging "eternal" black hole with constant mass. The other problem is that the traditional event horizon, which is often described as the defining feature of the black hole, only makes sense in the context of the entire picture of the spacetime, i.e. including its full time evolution. This is because the EH is defined as the boundary of the region in which photons are unable to escape to future null infinity, and we don't know for a fact that a photon can make such an escape until we have a picture of the entire spacetime laid out before us.

For this reason, people have considered other, more local (i.e. not requiring the whole of space-time), black hole definitions based on trapping horizons, and dynamical horizons. Here is a reference describing a dynamical horizon. In this ref fig. 4 shows a Penrose diagram of the Vaidya spacetime, in which infalling radiation causes the dynamical horizon to grow. Surprisingly, the traditional event horizon, however, reaches back even into the flat (blue) region. In the right hand picture, infall of radiation stops at $\nu=\nu_0$ and from then on the black hole looks like a Schwarzschild black hole.

Additional information on the Vaidya-Schwarzschild scenario is available here and a general treatment of various options for horizon definition here.

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