Why doesn't an electron sitting on the Earth radiate?

[stevendaryl]

Because Hawking radiation and Unruh radiation aren't the same thing.

I guess I had the idea that they were---that Unruh radiation + the equivalence principle implies Hawking radiation.

 

[PeterDonis]

I meant the radiation in both cases is a manifestation of doing QFT in a noninertial frame.

I don't think this is correct. Unruh radiation is a manifestation of the fact that which state of a quantum field is the vacuum state is different for accelerated observers vs. inertial observers. This is because accelerated observers and inertial observers have worldlines that are integral curves of different timelike Killing vector fields, and therefore they define "energy" (as the conserved Noether current associated with a timelike KVF) differently. And the definition of "energy" in turn defines which quantum field state is the state of lowest energy, i.e., the vacuum or ground state. All of this can be analyzed in an inertial frame; the key fact is the proper acceleration of the accelerated observer's worldline, not the frame used for the calculation.

Hawking radiation is a manifestation of the fact that, in a particular curved spacetime (Schwarzschild spacetime in Hawking's original derivation; Schwarzschild-anti de Sitter in some more recent versions), a quantum field state that looks like a vacuum to observers at infinity in the far past, looks like a thermal state with a radiation temperature equal to the Hawking temperature of the black hole to observers at infinity in the far future. Obviously this analysis cannot be done in an inertial frame, since there are no global inertial frames in a curved spacetime; but it also doesn't have to be done in Schwarzschild coordinates, which are the closest analogue to Rindler coordinates in Schwarzschild spacetime. Again, the key fact is how the spacetime curvature interacts with the quantum field, not the frame in which the calculation is done.

 

[stevendaryl]

I don't think this is correct. Unruh radiation is a manifestation of the fact that which state of a quantum field is the vacuum state is different for accelerated observers vs. inertial observers.

That sure sounds to me like a manifestation of doing quantum field theory in a noninertial frame. I don't understand the distinction you're making.

Hawking radiation is a manifestation of the fact that, in a particular curved spacetime (Schwarzschild spacetime in Hawking's original derivation; Schwarzschild-anti de Sitter in some more recent versions), a quantum field state that looks like a vacuum to observers at infinity in the far past, looks like a thermal state with a radiation temperature equal to the Hawking temperature of the black hole to observers at infinity in the far future.

Well, the derivation of the Hawking temperature in Wikipedia (https://en.wikipedia.org/wiki/Hawking_radiation#Emission_process) actually does use the Unruh effect plus the equivalence principle.[/QUOTE]

 

[PeterDonis]

That sure sounds to me like a manifestation of doing quantum field theory in a noninertial frame.

No, it's a manifestation of doing QFT for an accelerated observer.

I don't understand the distinction you're making.

You can analyze an accelerated observer using an inertial frame. And your choice of frame can't affect any actual physics. When you say "a manifestation of doing QFT in a noninertial frame", to me you are implying that your choice of frame can affect the physics--that if you did the same QFT for the same observer in an inertial frame, somehow that would make the Unruh effect vanish. It wouldn't.

the derivation of the Hawking temperature in Wikipedia

You should know better than to give Wikipedia as a source, particularly in an "A" level discussion.

One obvious problem with the Wikipedia presentation: Hawking radiation is observed at infinity, but the analogy between a black hole horizon and a Rindler horizon only applies to accelerated observers very close to the horizon. So no such analogy can explain why Hawking radiation is observed at infinity--or, to put it another way, it can't explain why the hole loses mass due to Hawking radiation, which requires radiation escaping to infinity. That requires a global analysis, in which any analogy between the black hole horizon and a Rindler horizon, and therefore between Hawking radiation and Unruh radiation, breaks down.

 

[PeterDonis]

It's probably worth emphasizing at this point that the question of whether black holes emit Hawking radiation, and if so, what exactly is the process that causes it, is still an open question. Various physicists have opinions, but they're not all consistent, and nobody has an experimentally confirmed theory that covers this regime (QFT in strongly curved spacetime) because we can't do experiments in it, so there's no way of resolving the issue by asking Nature to vote.

For an interesting take on the subject, George Ellis's presentation from 2014 is worth reading:

http://www.iap.fr/vie_scientifique/seminaires/Seminaire_GReCO/2014/presentations/Ellis.pdf

It's a summary of two papers (which are referenced) that develop the hypothesis that Hawking radiation is emitted by trapped surfaces, and compares and contrasts this with other views.

 

[PeterDonis]

George Ellis's presentation

Btw, Ellis uses the term "apparent horizon" to mean "trapped surface"; he also refers to at least two kinds of trapped surface, OMOTS and IMOTS. "MOTS" means "Marginal Outer Trapped Surface", which in ordinary language means "a surface at which radially outgoing light just fails to move outward--it stays at the same radius". The "O" and "I" in front refer to "outer" and "inner" surfaces that both have this property in a model in which an object like a star collapses to a black hole (this is the kind of model that Ellis's diagrams refer to).

 

[stevendaryl]

No, it's a manifestation of doing QFT for an accelerated observer. You can analyze an accelerated observer using an inertial frame.

I don't understand the distinction you're making. To me, to talk about observers means describing things in the observer's (accelerated) frame. If you are describing things from the point of view of an inertial frame, then the fact that the "vacuum" is different for accelerated observers does not seem relevant. You stick to the inertial frame, and use that notion of vacuum.

And your choice of frame can't affect any actual physics. When you say "a manifestation of doing QFT in a noninertial frame", to me you are implying that your choice of frame can affect the physics--that if you did the same QFT for the same observer in an inertial frame, somehow that would make the Unruh effect vanish. It wouldn't.

Of course not. But the derivation of Unruh effect comes from applying QFT to an accelerated frame. Maybe it could be derived in any frame, but that's not how it was derived.

You should know better than to give Wikipedia as a source, particularly in an "A" level discussion.

I was trying to diplomatically suggest that maybe you're wrong. I'm very unconvinced by this discussion. You seem to be saying that Unruh and Hawking radiation are two very different effects, but everything else I have read have suggested that Hawking radiation and Unruh radiation are the same phenomena, connected by the equivalence principle. For example:

http://www.physics.princeton.edu/~mcdonald/accel/unruhrad.pdf

As remarked by Unruh, this phenomenon [Hawking radiation] can be demonstrated in the laboratory according to the principle of equivalence: an accelerated observer in a gravity-free environment experiences the same physics (locally) as an observer at rest in a gravitational field. Therefore, an accelerated observer (in zero gravity) should find him(her)self in a thermal bath of radiation characterized by temperature $T = \frac{\hbar a}{2πck}$ , (2) where a is the acceleration as measured in the observer’s instantaneous rest frame.

 

[stevendaryl]

Okay, here's a slide show that definitely states that Hawking and Unruh radiation are not the same thing, and that both come into play near a black hole:

https://www.univie.ac.at/lunch-seminar/talks2016/Barbado_SS2016.pdf

 

[PeterDonis]

I don't understand the distinction you're making. To me, to talk about observers means describing things in the observer's (accelerated) frame.

This is obviously wrong. You can talk about accelerated observers using an inertial frame. You can talk about inertial observers using an accelerated frame. You can talk about any observers using any frame you like that contains their worldlines.

If you are describing things from the point of view of an inertial frame, then the fact that the "vacuum" is different for accelerated observers does not seem relevant. You stick to the inertial frame, and use that notion of vacuum.

I think you are misunderstanding the term "vacuum". Let me try to restate things in a completely frame-independent manner.

Consider two particle detectors, each of which is coupled to the same quantum field in flat Minkowski spacetime. One detector is following an inertial worldline; the other is following a worldline with constant proper acceleration.

First of all, what does "particle detector" mean? It means a quantum system with, in the simplest idealized model, two states, which we can call $|0\rangle$ and $|1\rangle$, such that the interaction Hamiltonian between the detector system and the quantum field can induce transitions from one to the other. We will suppose that, according to an observer at rest relative to the detector, the $0 \rightarrow 1$ transition is called a "particle detection" and the $1 \rightarrow 0$ transition is called a "particle emission".

Take the inertial detector first, and suppose that the quantum field is in a state $|V_\text{i}\rangle$, which we will call the "inertial vacuum" state. The key property of this state is that if the quantum field is in this state and the inertial detector is in state $|0_\text{i}\rangle$ (meaning the "0" state for the inertial detector), there is zero amplitude for a transition. That is what makes a state a vacuum state: a detector has zero probability of detecting a particle.

Now consider what the interaction of the accelerated detector with the quantum field in the state $|V_\text{i}\rangle$ looks like if the accelerated detector starts out in the state $|0_\text{a}\rangle$, meaning the "0" state for the accelerated detector. It can be shown [1] that there is a nonzero amplitude for the interaction between the quantum field and the accelerated detector to induce a $0 \rightarrow 1$ transition in the detector, i.e., a particle detection. This shows that the state $|V_\text{i}\rangle$ is not a vacuum state with respect to the accelerated detector.

In other words, saying that a given quantum field state is a vacuum with respect to one detector but not with respect to another is a perfectly objective, frame-independent statement; it's a statement about the probabilities of interactions between the quantum field and the detector causing state transitions. The state transitions themselves are also objective, frame-independent events, and all observers agree that they occur (though as we'll see in a moment, they don't necessarily agree on how to describe them in ordinary language).

It is also interesting to look at what this same process, the $0 \rightarrow 1$ transition in the accelerated detector and the corresponding change in state of the quantum field, looks like to the inertial detector. The quantum field state after the transition, which we can call $|D\rangle$, now has a nonzero amplitude to induce a $0 \rightarrow 1$ transition in the inertial detector. That is, to the inertial detector, the interaction with the accelerated detector looks like a particle emission, not a particle detection (since the state of the quantum field is now no longer a vacuum to the inertial detector, since there is now a nonzero probability of a transition). The state change in the accelerated detector, from the inertial viewpoint, is due to "radiation reaction".

[1] By the way, I rarely see the Unruh effect presented in these terms. Wald's 1993 monograph, Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics, which was the first textbook from which I learned about this subject, does it this way, and it seems to me to be a much better way to see the physics of what's actually going on than just talking about "thermal radiation".

the derivation of Unruh effect comes from applying QFT to an accelerated frame.

Let me restate this: the usual derivation of the Unruh effect comes from applying QFT using an accelerated frame. That is not the same as saying that the Unruh effect is "a manifestation of QFT in an accelerated frame". See above.

I was trying to diplomatically suggest that maybe you're wrong.

I could be as far as the actual relationship between the Unruh effect and the Hawking effect. I've already said that this is an open question; nobody knows what the right answer is, nobody knows what the right fundamental theory is, and nobody has a way of testing any of this experimentally.

I'm very unconvinced by this discussion.

If by "unconvinced" you mean you still think the Unruh effect and the Hawking effect are "the same thing", I don't think that belief is justified, for the reason I just gave (that this whole area is still an open question). If by "unconvinced" you mean that the arguments I've given don't convince you that the Unruh effect and the Hawking effect definitely aren't the same thing, you are right not to be convinced. My arguments for that are ultimately heuristic, just like everybody else's; nobody has a rigorous argument because nobody knows the right fundamental theory in this area. Probably we won't until we figure out quantum gravity, or to put it another way, until we figure out how to handle spacetime geometry itself using quantum amplitudes, the same way we handle everything else.

 

[PeterDonis]

here's a slide show that definitely states that Hawking and Unruh radiation are not the same thing, and that both come into play near a black hole

I think "definitely states" is a bit strong. This is certainly an interesting viewpoint, but I don't think we know that it's definitely right, for reasons I've already stated. (I would also like to see this viewpoint written up more formally in a paper; I don't see any reference to one in the slides.