Unruh Effect

PeterDonis

In Wald's book Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics, he discusses a similar question; he analyzes the case of a particle detector on an accelerating worldline, where the quantum field is in a state that, according to an inertial observer, is vacuum. There is a nonzero probability of a state transition, which is described as follows (I think I've got this right--I don't have my copy of the book handy to check):

* From the viewpoint of an observer who is accelerated with the detector, the detector absorbs a particle and registers a corresponding increase in energy, and changes its motion slightly as a result of the absorbed particle's momentum.

* From the viewpoint of an inertial observer, the detector *emits* a particle, and registers a corresponding *decrease* in energy and a change in momentum due to "radiation reaction".

The underlying quantum field viewpoint is that there is a state transition from the "inertial" vacuum state to the "accelerated" vacuum state (which the inertial observer does *not* see as a vacuum state), and a corresponding state transition in the detector, as required by the appropriate conservation laws. It all hangs together, but I admit it is certainly not intuitive.

yuiop

OK, this sort of makes sense. It suggests to me that the Unruh particles only interact with accelerating objects and the accelerating observer only observes the Unruh particle from the effects on his accelerating detectors and does not see them in transition.

Next question. As I understand it, the Unruh particles are only observed coming from behind the accelerating observer from the location of the Rindler horizon. Any intervening cold wall between the Rindler horizon and the accelerating observer casts a shadow that seems to block the Unruh radiation. Now let's say that we have one accelerating object that has an intervening wall between it and its Rindler horizon and another accelerating object that does not have an intervening wall between it and its Rindler horizon. Would the inertial observer now see a radiation reaction and particles being emitted from the accelerating observer without an intervening wall and no radiation reaction or particles emitted from the accelerating object with no intervening wall behind it? Is not radiation reaction an intrinsic part of accelerating particles and shouldn't it be independent of any wall behind the accelerating object?

stevendaryl

That's not exactly correct. A particle's charge doesn't change with energy, so a negative energy electron is still an electron, and not a positron. In the old-fashioned "Dirac Sea" interpretation of quantum field theory, it was postulated that all the negative-energy electron states were filled. An energetic photon could boost a negative energy electron to make it positive energy. This would produce a new, positive-energy negatively-charged particle (an ordinary electron) and also would leave a "hole" in the negative-energy states, which means a net positive charge among negative energy states. This hole could be reinterpreted as a positron.

Anyway, according to this (as I said, old-fashioned) way of looking at things, a positron is not a negative energy electron, it is the absence of a negative energy electron.

By the way, this way of viewing a hole in an otherwise filled shell of electron states as a positively-charged particle is no longer used in quantum field theory, but is still commonly used in solid-state physics, especially semiconductors.

Demystifier

But in the case of real (hermitian) field, such as electromagnetic field, antiparticles do not exist. And yet, Unruh effect exists for such fields as well. Thus, you are right that negative-energy particles are not virtual, but they are not antiparticles either. Negative-energy particles simple do not exist in the physical Hilbert space, essentially because such states would have negative norm.

In fact, the Rindler vacuum is a superposition of various Minkowski-particle states all of which have a POSITIVE energy. Technically, this superposition of positive energy states is known as a squeezed state.

Minkowski particles and Rindler particles are states in the same Hilbert space. Two definitions of particles correspond to two definitions of particle-number operator which do not commute. The fact that the same state |0> may be either vacuum or not-vacuum with two different definitions of the particle-number operator is analogous to the more familiar fact that the same spin-1/2 state |down> may be either down or not-down state with two different definitions of the z-axis. More physically, just like rotation of the Stern-Gerlach apparatus (which measures spin) may turn down-state into a not-down state, acceleration of a particle detector may turn vacuum state into a not-vacuum state.

PeterDonis

This is true of any kind of particle; we only observe them from their effects on detectors, we don't observe them "in transition". The "Unruh particles" are bona fide particles just like any others. As Demystifier said, the fundamental physics here is that the "inertial" and "accelerated" particle number operators (physically, the particle number operator is basically what a "particle detector" realizes) do not commute; they have different sets of eigenstates, so a state that is an eigenstate of the inertial operator (such as the vacuum state, with eigenvalue zero, so there is zero probability of detecting a particle) is *not* an eigenstate of the accelerated operator (so there is a nonzero probability for it to detect a particle).

I'm not sure this is true; can you give a reference that leads you to this conclusion? The "cold wall" scenario you're describing doesn't match up with what I know about how the Unruh effect works (which isn't a lot).

yuiop

I read about this a long time ago but I cannot recall where. A quick Google turned up this mention in a Wikipedia article http://en.wikipedia.org/wiki/Unruh_effect#Calculations

PeterDonis

Ah, I see; basically the idea is that the Unruh effect only occurs if the Rindler horizon is "visible", as it would be for linear acceleration in free space, but is not for the two counterexamples they give (but note the "citation needed"):

* For acceleration in a circular path, there is no Rindler horizon (I think that's right) and no Unruh effect (I think that's also right but I'm not sure).

* For the "acceleration" of objects at rest on the Earth's surface, the Rindler horizon is not "visible" because the spacetime is curved--the Earth is only 4000 miles in radius but the Rindler horizon would be a light-year away for a 1 g acceleration. So there is no Unruh effect here either (again, I think that's right for the Earth--but note that for a black hole, there is Hawking radiation, the mathematics of which is very similar to that of the Unruh effect).

Physics Monkey

It seems physically reasonable to me that uniform circular motion will be accompanied by something like the Unruh effect, although the details are more complicated.

For example, Bell and Leinaas have a series of papers, 25+ years old now, where they try to interpret the known partial spin polarization of electrons orbiting in a storage ring as experimental evidence in favor the unruh effect. The original paper is I think http://www.sciencedirect.com/science...50321387900472 while an accesible discussion is at http://arxiv.org/pdf/hep-th/0101054.pdf

Another discussion of radiation in the circulating case can be found at http://arxiv.org/abs/gr-qc/9903054

I'm not sure about this "cold wall" situation, mostly because I'm not sure what that even means, but I doubt that such a thing could seriously interfere with unruh radiation.

Physics Monkey

There is a simple and neat explanation for part of this story, namely that both observers must observe an increase in the Rindler energy. The Minkowski observer sees a particle emitted which can reasonably be expected to increase the energy. On the other hand, the Rindler observer sees a particle absorbed which sounds contradictory. However, one may think of the observation of the particle at the detector by the rindler observer as a partial measurement of the state of the field. In particular, more highly excited states are more likely to lead to detection and thus, conditioned on a detection, they become more likely. Thus the average energy can actually increase as a result of a particle being absorbed. As a simple extreme example, think of a state of the radiation like [itex] \rho = \frac{N-1}{N} |0 \rangle \langle 0 | + \frac{1}{N} |N \rangle \langle N |[/itex] labeled by energy quanta. If a quanta is absorbed then the field had to be in the N state because you can't absorb from 0. The initial average field energy is 1 but the final average field energy is N-1 (because the state had to be N and you lost one unit of field energy).

This is discussed in a paper by Wald and Unruh although I'm forgetting the reference right now.

Physics Monkey

That was easy, found it: http://prd.aps.org/abstract/PRD/v29/i6/p1047_1

PeterDonis

Ah, good, I was remembering correctly:

"It is shown in detail for the simple case of a two-level detector how absorption of a Rindler particle corresponds to emission of a Minkowski particle."

yuiop

OK, that seems reasonable as in both instances, the velocity of the accelerating Rindler observer is effectively increased and momentum is presumably conserved. I also assume that while the Minkowski observer does not see the Rindler particle, that the Rindler observer does not see the emitted Minkowski particle. Now I have to ask, what if the emitted Minkowski particle (that the Rindler observer does not see) collides with a pre-existing normal particle (that both observer's see), how does the Rindler observer explain the sudden change in momentum of the normal particle after the collision?

P.S. Yet another question. By the equivalence principle, the Unruh radiation effect implies that an observer that is stationary observer outside a blackhole sees Hawking radiation and observes the black hole evaporate (possibly to nothing in a mini explosion) while a free falling observer would not be able to see the Hawking radiation and would either see no change in the mass of the black hole or would be unable to explain the loss of mass of the black hole. Which position is more accurate?

PeterDonis

All correct as far as I can see.

If there is a "pre-existing normal particle that both observers see", then the overall state of the quantum field is not a vacuum state in either sense--it's neither the Rindler vacuum nor the Minkowski vacuum. I'm not sure if there is an analysis out there of how this affects things; all the analyses I have seen of the Unruh effect assume that the field state transition is from one vacuum to the other (meaning no other "pre-existing" particles).

Edit: To speculate a bit, I would guess that in the case where a "pre-existing" particle is present, it won't look the same to the inertial observer as it does to the accelerated observer, for reasons similar to why the vacuum state for one is not the vacuum state for the other. Also, I would expect that, once the emitted Minkowski particle collides with the pre-existing particle, the Minkowski particle will no longer be in the state that the Rindler observer thinks is "vacuum"; i.e., it will now be visible to that observer. So I would guess that the Rindler observer's interpretation of events would be that his detector absorbs a particle (so the field is now in the vacuum state); then the pre-existing particle causes a particle-antiparticle pair to pop out of the vacuum (what the Rindler observer "thinks" is the vacuum), one of the pair flies off, while the other is absorbed by the pre-existing particle and changes its state. That's just a guess, though; as I said, I haven't seen any mathematical treatment of this case.

yuiop

I think perhaps you are applying a too rigid interpretation of "vacuum" here. For example Physics Monkey mentions a study of the Unruh effect in an electron storage ring. Presumably the experiment was done in an evacuated chamber approximating a vacuum, but it is well known that it is impossible to obtain a true vacuum in a laboratory because at extreme low pressure the walls of the chamber starts to evaporate and there will always be stray particles present in an experimental chamber. Another example is the Schwarzschild metric which is a vacuum solution. This does not prevent us assuming test particles and observers (with none zero rest mass) when carrying out an analysis as long as masses are insignificant compared to the mass of the gravitational body and the average vacuum density reasonably approximates a true vacuum.

PeterDonis

When I say "vacuum" I'm talking about the quantum state used in the theoretical treatment. Every theoretical treatment I have seen uses only two states, both of which are vacuum states: the Minkowski vacuum state and the Rindler vacuum state. The notion of "vacuum state" has a very precise meaning theoretically, and that's what I was talking about. I wasn't intending to say that an actual experiment would have to be done in a perfect vacuum. I was only saying that, theoretically, if there is a "pre-existing particle", then the quantum state of the field is *neither* of the states that I have seen used in theoretical treatments of the Unruh effect.

Demystifier

There are also more general theoretical notions of the "vacuum" and "particle":
http://xxx.lanl.gov/abs/gr-qc/0409054