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

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1. May 7, 2018

craigthone

I have some puzzles when I combine the Equivalence principle and Unruh's effect.
Equivalence principle says that acceleration is equivalent to a uniformly gravitational field, or locally equivalent to gravitational field. Then the electron sitting on earth should have Unruh's effect, i.e. radiate. What is wrong with the argument?

2. May 7, 2018

Staff: Mentor

Don’t forget to include the detector in your setup. Is it free falling or co-accelerating?

3. May 7, 2018

craigthone

Sorry, the co-accelerating detector. The observer sits on the surface also.

4. May 8, 2018

Staff: Mentor

Not quite. It says that accelerating in flat spacetime is locally equivalent to being at rest in a uniform gravitational field.

No, that's not what the Unruh effect says. The Unruh effect says that an accelerating detector should detect radiation from the quantum vacuum. However, the intensity of the radiation for a 1 g acceleration is way, way too small to actually detect.

5. May 8, 2018

craigthone

Forget the electron, do observers (detector) sitting near earth see Unruh's radiation? I know the answer is no, but I do not have good explainations. Maybe I should ask in flat space for accelerating observer what is the quantum state corresponding to vacuum for static observer in the earth?

6. May 8, 2018

haushofer

An accelerating observer observes an event horizon, like an observer at rest near a BH. An observer on earth doesn't.

7. May 8, 2018

Khashishi

There's some discussion on this here.
I think the consensus is no, you don't get Unruh radiation because the Unruh radiation is perceived as coming from the Rindler horizon. But Earth has no horizon. The equivalence principle only applies locally - an accelerating frame looks like gravity. But an accelerating frame is equivalent to a constant gravitational field which extends to infinity. There is a Rindler horizon, which is some distance behind the accelerating frame where not even light can catch up with the center of the frame. But Earth's gravity isn't a constant gravitational field, and there isn't a horizon.

8. May 8, 2018

Staff: Mentor

Yes.

I already answered this. We don't see Unruh radiation on Earth because it's way too faint for us to detect.

The Rindler horizon is not an event horizon; it's not the boundary of a region that can't send light signals to future null infinity. There are some local similarities between the two, but they're not the same.

[Edit: removed final statement, not correct.]

Last edited: May 8, 2018
9. May 8, 2018

Staff: Mentor

I know this is a common heuristic, but I'm not sure it's actually true. The actual derivation of the Unruh effect does not rely on the presence of the Rindler horizon. It relies on the fact that, in quantum field theory, which state is the "vacuum" depends on your state of motion. In particular, if the quantum field in flat spacetime is in a vacuum state with respect to inertial observers, that same state will not be a vacuum state with respect to an accelerated observer; it will be a thermal state at the Unruh temperature, and therefore the accelerated observer will observe radiation at that temperature. That is true right at the location of the accelerated observer; there's no need to reach out to the Rindler horizon.

10. May 8, 2018

sweet springs

OP 's concern is radiation from the event horizon, but as a layman I have a more primitive question of radiation from the charge itself.
Dose a charge sitting on the floor under gravitation of the Earth or accelerating rocket emit light by radiation? I do not think so. I would like to know the condition of radiation more in detail.

11. May 8, 2018

Staff: Mentor

So am accelerating electron (in an elevator) will not produce any radiation that can be detected by a co-accelerating detector. By the equivalence principle the same is what we expect for the charge and detector at rest in gravity.

12. May 9, 2018

craigthone

I do not think so.
From QFT in Rinler space, the argument of Unruh's effect takes a special state $\rho=\exp(-2\pi H)$ where $H$ is the Rindler Hamiltonian and the Minkowski boost, this state $\rho=\exp(-2\pi H)$ is Minkowski vacuum. While for quantum fields on the earth, expanding around the surface, we do not have such state, and no compactification of Euclidean time and no temperature. The picture is totally different from near black hole horizon case. So there is no Unruh radiation rather than to weak to detect.

13. May 9, 2018

stevendaryl

Staff Emeritus
This picture tries to give an intuitive reason that an accelerated charge radiates. Imagine that you have a charged particle moving along the x-axis at a constant speed. The electric field associated with that situation points toward the moving particle. Now, at some point, the particle abruptly halts. Nearby the particle, the electric field points toward the particle. But the information that the particle's motion has changed hasn't reached far-away places, so the electric field continues to point toward where the particle would have been, if it kept traveling at the same speed. Because electric fields have to be continuous in vacuum, there must be a transition region, spreading out spherically at the speed of light, where the electric field rapidly changes from one configuration to the other. This region of rapidly changing electric (and magnetic, since that changes, too) fields is a spherical burst of electromagnetic radiation. This is shown in the following picture:

To the extent that this intuitive picture is accurate, I think we can conclude that there is no radiation as viewed from a reference frame in which the situation is static. If the electric field far away from the charge isn't changing with time, then there shouldn't be a need for the radiation, which sort of acts like a "correction" to the field that propagates at the speed of light.

The interesting case is a charged particle undergoing constant proper acceleration. From the point of view of a Rindler observer, the situation is static, so there should be no radiation. But from the point of view of an inertial observer, the situation is not static, and it's also not the constant-velocity case. I'm not sure what the electric field looks like to the inertial observer.

A charge at rest on a planet shouldn't radiate, according to this heuristic, because the distant electric field is unchanging (in the frame in which the situation is static, which is someone at rest in the Schwarzschild coordinate system).

14. May 9, 2018

stevendaryl

Staff Emeritus
I have to caution readers that I'm not at all claiming to be an expert in this. I've just given an intuitive heuristic that may reflect a misunderstanding on my part.

15. May 9, 2018

Staff Emeritus
This thread - as is the case for this topic, as well as it's evil twin, radiation of a charge in free fall - is a bit of a mess. But before I respond to the mention of my name, is this really at the A-level? It seems not to be. I am not going to spend a lot of time on writing an A-level answer only to get a reply like "What is a multipole?" or "What 16 components?"

16. May 9, 2018

craigthone

We need A level answers. It is better to have some equations rather than something else. Thanks in advance.

17. May 9, 2018

Staff: Mentor

Technically, you're correct, an observer standing on Earth's surface is not in vacuum, there's air around him and ground below him. But imagine, for example, someone standing on an airless planet (or on the Moon). There is vacuum around that person, and they are accelerated, so they should in principle observe Unruh radiation. That's all I was trying to say.

18. May 10, 2018

Staff Emeritus
It's confusing two things. Are you saying this electron should radiate because it is charged? Or are you saying it should radiate because of the Unruh effect, which has nothing to do with charge? Until we get the question straight, it's unlikely we will be able to provide a good answer.

19. May 10, 2018

craigthone

In the process of discussing, I think the first question (this electron should radiate because it is charged) is answered by Dale using equivalence principle correctly.

The second question is Forget the electron, do observers (detector) sitting near earth see Unruh's radiation? I know the answer is no, but I do not have clear argument.

20. May 10, 2018

Staff Emeritus
The heuristic is that Unruh radition is a consequence of applying the Equivalence Principle to Hawking radiation. If you go the other way, you get Hawking radiation back - i.e. the geometry of a black hole, not the geometry of a planet.

If you are wondering why BH's emit Hawking radiation and planets don't, the easiest way to see this is conservation of energy. A BH can get energy by shrinking its horizon. Since a planet doesn't have a horizon, there's nowhere for the energy to come from.

Last edited: May 10, 2018
21. May 10, 2018

Staff: Mentor

Very well explained!

22. May 10, 2018

Staff: Mentor

I would put this the other way around: Hawking radiation is (heuristically) the result of applying the EP to Unruh radiation. Unruh radiation can be derived rigorously using QFT in flat spacetime. Hawking radiation, if you want to try to derive if from first principles, has to be derived using a semi-classical approximation--QFT in a fixed curved background spacetime--which is not rigorous, and nobody AFAIK has figured out how to make it rigorous.

First, I think you mean "A BH can lose energy by shrinking its horizon".

Second, a planet has lots of other ways of losing energy, so I don't think "not being able to lose energy" is a good heuristic for why planets don't emit Hawking radiation. I think a much better heuristic (which is based, not on Hawking's original argument, but on how our understanding has developed in the decades since he first published his result) is that Hawking radiation is associated with trapped surfaces, not event horizons. For a black hole, the two coincide (or almost coincide, since a real hole will be accreting matter), but the difference is that the trapped surface is a local phenomenon--radially outgoing light does not move outward--so it's much easier to see how an object can "tell" whether it's "allowed" to emit Hawking radiation (a planet has no trapped surface, so no Hawking radiation, even though a planet can radiate in lots of other ways).

23. May 10, 2018

Staff Emeritus
I mean it can transfer energy from the horizon to the radiation field. A planet cannot do this.

And the PF quibbling begins!

You know what I meant. Yes, you can make this more complicated. No, that doesn't help with the understanding.

I knew that getting involved in this morass of a thread was a mistake. Lesson learned.

24. May 10, 2018

Staff: Mentor

I'm not sure the answer is no. See my post #10. In practice, Unruh radiation for an object accelerating at 1 g is many, many orders of magnitude too weak for us to detect, so there's no way of testing the question experimentally.

25. May 10, 2018