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From the perspective of an observer inside a black hole's horizon, where does the energy for Hawking radiation come from?

+ 6 like - 0 dislike
179 views

Would energy be seen to "flow" to the outside of the black hole? Through what mechanism?

This post imported from StackExchange Physics at 2015-04-11 10:34 (UTC), posted by SE-user user1247
asked Mar 6, 2012 in Theoretical Physics by user1247 (530 points) [ no revision ]

3 Answers

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The observer inside the event horizon does not observe Hawking radiation--- this is a pure external effect.

This post imported from StackExchange Physics at 2015-04-11 10:34 (UTC), posted by SE-user Ron Maimon
answered Apr 15, 2012 by Ron Maimon (7,535 points) [ no revision ]
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Not Hawking Radiation per se, but surely some mechanism by which energy is disappearing must be observed from the inside of the black hole.

This post imported from StackExchange Physics at 2015-04-11 10:34 (UTC), posted by SE-user user1247
Why should it be? There is no local mechanism--- the inside is a construct of the quantum states. You don't see stuff flying out in the future region. There is what I consider the likely possibility that for charged and rotating black holes you end up going through the middle, missing the singularity and coming out the same black hole. It is in the outgoing leg of your journey, on your way to exiting the event horizon that you might see the Hawking radiation. I didn't work it out, and I don't think this past region was the intention of your question.

This post imported from StackExchange Physics at 2015-04-11 10:34 (UTC), posted by SE-user Ron Maimon
Maybe I'm not understanding your response. Are you saying that for an inside observer, the black hole would not appear to be losing mass?

This post imported from StackExchange Physics at 2015-04-11 10:34 (UTC), posted by SE-user user1247
@user1247: Yes, I am saying that. The inside observer can't measure the mass of the black hole.

This post imported from StackExchange Physics at 2015-04-11 10:34 (UTC), posted by SE-user Ron Maimon
I'll accept your answer if you edit it to make it explicitly clear that you saying that the inside of a black hole does not lose mass/energy in its own reference frame. As it stands, it looks like you are just saying that Hawking radiation is an external effect -- something that I already understood. I was asking about the mechanism by which mass/energy is conserved; it sounds like the answer is something along the lines of "infinite time dilation" or "causal disconnection" or something rather than a mechanism by which energy is transferred across the horizon.

This post imported from StackExchange Physics at 2015-04-11 10:34 (UTC), posted by SE-user user1247
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@user1247: The inside does not have a reference frame, so I can't make it clear--- the inside time is different, it points towards the middle. I also don't know the answer in the extremal case, where you can trace the stuff, so I don't want to say anything too precise, just that the time inside goes inwards to increasing r, and the time outside is "sideways" to the black hole. This coordinate tipping over of time is important, but it makes causal notions inside completely disconnected from outside, different notion of time.

This post imported from StackExchange Physics at 2015-04-11 10:34 (UTC), posted by SE-user Ron Maimon
@RonMaimon I think I see now; it's like why if you could get to the center of the Earth you'd be weightless. That the mass loss would have to 'fall' from the EH to the singularity isn't something I'd've expected; but seems reasonable in retrospect.

This post imported from StackExchange Physics at 2015-04-11 10:34 (UTC), posted by SE-user Dan Neely
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This is an incomplete answer, but I would bring two important points here:

1) When observer moves freely in GR, he/she observes locally flat spacetime up to quadratic corrections in 3-distance from his/her worldline. Hence, if you are falling into a black hole, you shall see the same as what you would see, if you were in Minkowski spacetime throughout all your infall, unless you start measuring the tidal forces. Hence, you shall see no Hawking radiation generated around you at all, until you come so close to the singularity, that the tidal forces shall become very strong and you shall eventually observe local generation of particle-antiparticle pairs around.

However, this doesn't exclude the possibility of observing some flux falling with you from the infinity into the black hole.

2) The notion of particles in GR depends on your reference frame. If you imagine an observer located above the event horizon, accelerated in such a way that he/she has stationary spatial coordinates, he/she will not be moving freely due to acceleration and shall see particles/antiparticles generated around, some of them falling under the horizon, and some of them flying away from the black hole. If you consider some static asymptotically flat coordinate system, you shall see that some particles in this system shall escape to infinity and shall be visible to the remote observers, moving freely with respect to that system.

To conclude, you shall not observe Hawking radiation generated around you, though you might possibly see already existing particle-antiparticle background (possibly, exactly Hawking radiation).

This post imported from StackExchange Physics at 2015-04-11 10:34 (UTC), posted by SE-user Alexey Bobrick
answered Jun 22, 2012 by Alexey Bobrick (0 points) [ no revision ]
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An observer inside a black hole is impossible because a black hole has no internal structure (otherwise it would be possible to transfer information from inside by moving massive bodies.

Also no information can be under the horizon because this would mean information loss which is impossible thermodynamically.

So any black hole is an uniform body without internal structure.

This post imported from StackExchange Physics at 2015-04-11 10:34 (UTC), posted by SE-user Anixx
answered Jul 22, 2012 by Anixx (30 points) [ no revision ]

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