Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. May 7, 2018 #1
    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. jcsd
  3. May 7, 2018 #2

    Dale

    Staff: Mentor

    Don’t forget to include the detector in your setup. Is it free falling or co-accelerating?
     
  4. May 7, 2018 #3
    Sorry, the co-accelerating detector. The observer sits on the surface also.
     
  5. May 8, 2018 #4

    PeterDonis

    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.
     
  6. May 8, 2018 #5
    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?
     
  7. May 8, 2018 #6

    haushofer

    User Avatar
    Science Advisor

    An accelerating observer observes an event horizon, like an observer at rest near a BH. An observer on earth doesn't.
     
  8. May 8, 2018 #7

    Khashishi

    User Avatar
    Science Advisor

    There's some discussion on this here.
    https://physics.stackexchange.com/questions/2696/do-all-massive-bodies-emit-hawking-radiation
    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.
     
  9. May 8, 2018 #8

    PeterDonis

    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
  10. May 8, 2018 #9

    PeterDonis

    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.
     
  11. May 8, 2018 #10
    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.
     
  12. May 8, 2018 #11

    Dale

    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.
     
  13. May 9, 2018 #12
    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.
     
  14. May 9, 2018 #13

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    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:
    3mCaL.png

    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).
     
  15. May 9, 2018 #14

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  16. May 9, 2018 #15

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2017 Award

    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?"
     
  17. May 9, 2018 #16
    We need A level answers. It is better to have some equations rather than something else. Thanks in advance.
     
  18. May 9, 2018 #17

    PeterDonis

    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.
     
  19. May 10, 2018 #18

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2017 Award

    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.
     
  20. May 10, 2018 #19
    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.

    [Moderator's note: Question about EP deleted; it was already asked in a previous post that has been moved to a new thread.]
     
  21. May 10, 2018 #20

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2017 Award

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Loading...