Black Hole Hawking Radiation: Frame-Dependent Virtual Particles

[MattRob]

I was reading "Black Holes and Time Warps" by Kip Thorne, and right around p.442-443 it talks about how the quantum vacuum fluctuations that give rise to Hawking radiation from an infalling frame of reference give rise to an "atmosphere" of real, non-virtual particles in an accelerated frame where the observer is near the event horizon and at rest with respect to the black hole.

Apparently, this accelerated frame of reference adds a lot of energy to these vacuum fluctuations. I'm only a Sophomore, so I really don't know the relevant GR equations, but from what I do know, energy is associated with velocity, not acceleration, so how does accelerating, even with zero velocity, add energy to these vacuum fluctuations?

 

[jartsa]

Black hole adds energy to vacuum fluctuations. Observer far away from the black hole can absorb those fluctuations. Observer near the black hole can absorb those fluctuations, if the observer is not free falling.

It's not that Hawking radiation is caused by acceleration, it's just that free falling near a black hole makes an observer unable to absorb any Hawking radiation.

 

[PeterDonis]

Apparently, this accelerated frame of reference adds a lot of energy to these vacuum fluctuations.

That's not quite what's happening. A better way of thinking of it is that, when you combine quantum field theory (which describes the fluctuations) with curved spacetime (which, in GR, is present wherever gravity is present), the concept of a "particle" being present becomes relative, not absolute.

(In fact, strictly speaking, this is even true for quantum field theory in flat spacetime; there it's called the "Unruh effect". But that's probably too much for one post.)

To the observer who is freely falling into the black hole, the quantum field state looks like a vacuum state, with no particles present (on average--there are fluctuations, but they fluctuate around an average of zero particles present). This can be confirmed by the observer's particle detector, which never registers any detection of particles.

To the observer who is accelerating in order to stay at a constant altitude above the black hole, the quantum field state looks like it contains particles. This can be confirmed by his particle detector, which registers detections of particles.

The stuff about the vacuum fluctuations giving rise to the "atmosphere" of real particles comes in when you look at the mathematics of how these two different predictions (no particle detection for the infalling observer; particle detection for the accelerating observer) are made. It turns out that the particles that are detected by the accelerating observer "correspond" in a technical sense to the vacuum fluctuations that appear when things are described from the viewpoint of the infalling observer.

The reason that the "atmosphere" of particles contains energy is simple: when the accelerating observer's particle detector detects a particle, this involves a transfer of energy. It has to: if there were no transfer of energy, the state of the particle detector would not change and no particle would be detected. As jartsa says, the only place this energy can actually come from is the black hole: each time the accelerating observer detects a particle, the mass of the black hole decreases slightly.

It's interesting to ask what this process (of the accelerating detector detecting a particle) looks like from the viewpoint of the infalling observer. From this viewpoint, there was a vacuum fluctuation that caused a particle-antiparticle pair to pop into temporary existence. Normally, such a pair would just annihilate each other; but in the presence of the accelerated detector, sometimes one particle of the pair will interact with the detector before the other one can annihilate it. When that happens, the other particle falls down the hole. The particle that was detected had to have positive energy (otherwise it wouldn't be detected), so the one that falls down the hole has to have negative energy (see below for a further comment on that), which effectively means it reduces the hole's mass a bit.

So the net effect of this process is: from the viewpoint of the accelerated detector, a particle is absorbed, transferring energy from the hole to the detector; from the viewpoint of the infalling observer, a particle is emitted by the detector and goes down the hole, and since it has negative energy, the net result is to transfer energy from the hole to the detector.

from what I do know, energy is associated with velocity, not acceleration

One particular type of energy, kinetic energy, is associated with velocity, yes. But there are other types of energy. The type of energy that is carried by the particle that goes down the hole in my description above is certainly not any ordinary kind of kinetic energy that depends on velocity.

In fact, that example illustrates the implications of the fact that energy is frame-dependent. To an observer outside the hole, the particle that goes down the hole has negative energy; but to an infalling observer inside the hole (i.e., one who has already fallen in), that same particle has positive energy. So just looking at energy, by itself, can be misleading.