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This has been bugging me for some time.

As I understand it, Hawking radiation is the result of the mismatch between the vacuum state of a quantum field as seen by a free falling observer (falling directly toward the black hole) and one that is sitting at a constant radius far away from the black hole.

This is perhaps the result of my naive understanding of the subject, but it doesn't make sense to me to say that a black hole radiates, because it's observer-dependent. How do we know that the thermal bath of particles that we see when we sit at a constant radius actually causes the black hole to lose mass? Is it just a simple energy conservation argument, or is there some subtle process here that I'm missing?

Desert Coyote
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1 Answers1

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Well, here is a simple scenario:

The quantum field theoretic vacuum is seething with pair production of virtual particles everywhere. Take one such virtual pair at the event horizon of the black hole. One part of the pair, if it is going down, is grabbed by the hole and disappears while the other is with its high momentum runs away from the horizon, on shell. Where did it find the energy? Read on

A slightly more precise, but still much simplified, view of the process is that vacuum fluctuations cause a particle-antiparticle pair to appear close to the event horizon of a black hole. One of the pair falls into the black hole whilst the other escapes. In order to preserve total energy, the particle that fell into the black hole must have had a negative energy (with respect to an observer far away from the black hole). By this process, the black hole loses mass, and, to an outside observer, it would appear that the black hole has just emitted a particle. In another model, the process is a quantum tunneling effect, whereby particle-antiparticle pairs will form from the vacuum, and one will tunnel outside the event horizon.

The faraway observer is used to define the "negative energy" of the eaten-up partner.

Peter Mortensen
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anna v
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