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According to the General Theory of Relativity, the coordinate time distance per spacetime distance traveled by a particle freely falling into a black hole gets closer and closer to $0$ as the particle approaches the event horizon

$$\mathrm{d}s^2 ~=~ -\left(1-\frac{2M}{r}\right)\mathrm{d}t^2 + \left( 1-\frac{2M}{r} \right)^{-1}\mathrm{d}r^2 + \dots.$$

Equivalently, an observer looking from the outside will never see the particle cross the event horizon, even if he/she looked for an arbitrarily long time. However, Hawking radiation implies that black holes don't last forever, but instead shrink and fade away? What happens to everything that is "currently" falling into the black hole during that time?

Edit: What this seems to imply is that as a particle gets close to the event horizon, the rate and intensity of Hawking radiation will increase and increase until most of the mass of the black hole is ejected, in the form of Hawking radiation, out towards the falling particle.

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Related: and links therein. – Qmechanic Mar 20 '14 at 0:32
I am unsure how this question was closed as a duplicate of a completely unrelated question. I am voting to re-open. – Kyle Kanos Mar 21 '14 at 0:39

2 Answers 2

If, as is canonically accepted, the black hole evaporates within a finite time, then the freely falling particle will be released from its gravitational attraction. By the premise of the question, in the observer frame the particle can not have passed the event horizon during the intervening period. This would also solve the quantum information paradox, as the information represented by the particle would never have been lost.

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From the view of the outside observer, the freely falling particle hits a Planck-temperature soup of particles and is destroyed. From the view of the freely falling particle, it passes the event horizon without incident. It is very difficult to reconcile these two views, and neither of these is the scenario indicated by your answer. – Peter Shor Mar 24 at 1:22
Thanks much for your thoughts. Are you referring to the AMPS conclusion? I really should look into that, in more detail than the recent Scientific American. Do you have a paper reference? Or, I can look for myself. – odyssoma Mar 24 at 3:38
But please note that high-temperature destruction external, or at least not internal, to the black hole would remain consistent with classical thermodynamics, and not represent a destruction of information. So that the information paradox would still be avoided. – odyssoma Mar 24 at 3:41
No, I'm referring to Hawking radiation, or Unruh radiation. One way of viewing the paradox is: if the particle never interacts with the high-temperature soup from its point of view, quantum mechanics says the information cannot be transferred to the soup. – Peter Shor Mar 24 at 10:55
Your reference to a Planck temperature soup put me in mind of some recent work by Polchinski et al. on black hole firewalls. (, Ahmed Almheiri, Donald Marolf, Joseph Polchinski, and James Sully, hence the AMPS reference.) It is about Hawking radiation, but at an extremely high level. I like the idea about defeating complimentarily through entanglement, which is then destroyed by the high energies. Lots of cites, lots of controversy. (Sorry if I’m tediously telling you things you already know.) – odyssoma Mar 25 at 0:42

Doesn't the Schwarzschild metric combined with Hawking radiation imply that nothing ever gets past the event horizon of a black hole?

No. First of all, let's set Hawking radiation aside, we don't need it. Let's look at this:

An observer looking from the outside will never see the particle cross the event horizon, even if he/she looked for an arbitrarily long time.

This is accepted as being true, because at the event horizon the "coordinate" speed of light is zero, and nothing can go faster than light. So the obvious question is how can the black hole grow? One can find various opinions on this, such as The Formation and Growth of Black Holes on Mathspages, see This favours the view that the infalling body goes to the end of time and back in finite proper time, wherein like Susskind's elephant, it's in two places at once. I'm confident that this is incorrect, and that the alternative frozen-star interpretation is correct. In consequence, the black hole grows like a hailstone. Imagine you're a water molecule. You alight upon the surface of the hailstone, but you can't pass through the surface. However other water molecules surround you and bury you. So whilst you didn't pass through the surface, the surface passed through you. In similar vein nothing ever gets past the event horizon, but the event horizon gets past it.

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Odyssoma, I'm new and can't comment above. But see by the AMPS guys and note reference 87 which is to Friedwardt Winterberg's Gamma Ray Bursters and Lorentzian Relativity. A particle like an electron doesn't actually make it to the event horizon. I'm a relativity guy, and this sounds like it goes against the grain, but I have to say I think it's correct. – John Duffield Mar 25 at 19:40

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