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Unruh Effect

  1. Feb 5, 2012 #1

    pervect

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    Would it be fair to describe the Unruh effect by saying that from the perspective of an accelerated observer, some virtual particles in an inertial frame become real particles in the accelerated frame?

    Wald talks about how the appropriate transformations between inertial and accelerated frames (bogoliubov transformation) map positive frequencies to negative frequencies (page 414).

    I think it is correct to say that the negative frequency particles would have negative energy and hence be virtual particles, leading to my description above, but I'm not terribly confident about QFT.
     
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  3. Feb 5, 2012 #2

    Mentz114

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    Yes, I get that impression. The full calculation is very heavy but I found this much more accessable work some time ago. The authors quantize a scalar field in 2D Minkowski spacetime and then transform to Rindler coords. They show that the vacuum expectation is different from the unaccelerated state. Re negative frequency states they say

    I've attached the paper because I don't remember where I found it ( I think it's course notes from U. Texas).
     

    Attached Files:

  4. Feb 5, 2012 #3
    I don't think this is correct. Unruh effect is about describing same physics from different perspectives. One is inertial perspective, and other is non-inertial accelerating perspective (offten called Rindler). Main point is how you define what is particle. In inertial case you solve wave equation, and separate solutions into positive and negative frequency solution with respect to inertial time, and define vacuum and particles with respect to them (lets call them Minkowski particles and vacuum).
    Now you do the same thing in Rindler coordinates (coordinates adopted to accelerating observer), solve the wave equation in Rindler coordinates, and separate solutions into positive and negative frequencies with respect to proper time (proper time is what accelerating observer is using), and define Rindler particles and vacuum with respect to these mode functions (or solutions to wave eq.).
    Now you want to know how is Minkowski vacuum described using Rindler perspective, and you get that Minkowski vacuum is infect described as thermal bath of Rindler particles. This is telling you that concept of a particle is observer dependent, and unless you specify state of a detector being used to measure particles, not very useful.
    I found that some author's don't distinguish Rindler from Minkowski particles in a sence that they equate thermal bath of Minkowski particles with thermal bath of Rindler particles that is seen by acceleration observer. This is not apriori true because mode functions for Rindler particles are different from mode functions for Minkowski particles, or another way of saying this is that thermal radiation in inertial motion is measurably different from thermal radiation in acceleration motion.

    Bogoliubov transformation used here map positive frequency modes with respect to inertial time to positive and negative frequency modes with respect to proper time, and vice versa. If there where no mixing between the two, Minkowski and Rindler perspective would have same vacuum state, and hence would agree on particle content of a theory.
     
    Last edited: Feb 5, 2012
  5. Feb 5, 2012 #4
    I don't think negative energy particles are virtual particles, but antiparticles.
     
  6. Feb 5, 2012 #5

    Mentz114

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    True. So is the affect of acceleration to break a vacuum symmetry ? If the (effective) zero point is moved, then either particles become anti-particles or vice-versa depending on whether the zero-point is raised or lowered. I have the Dirac 'sea' in mind.

    In your first post you don't seem to be contradicting what Pervect and the paper I attached are saying.
     
  7. Feb 6, 2012 #6
    Oh.. you are right, I didn't say what exactly I was disagreeing in OP. I'm not disagreeing with paper you attached.

    I'm disagreeing on a way that OP is defining a particle. Positive and negative modes functions are not particles. So when you use Bogoliubov transformations you are not turning one particle into another.
    Historically it was not immediately realized that relativistic particles can be described only by quantized fields. At first, fields were regarded as wave functions of point particles.

    I never tough of Unruh effect in terms of Dirac sea, that is interesting observation. But I don't think it is accurate description of processes. I find Dirac see a historical curiosity, ingenious solution to problem of negative energy particles. But I don't consider this sea real and taking this sea to seriously can lead to faulty conclusions.
     
    Last edited: Feb 6, 2012
  8. Feb 6, 2012 #7

    Mentz114

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    I understand.
    Yes. I'm near novice level in QFT. Thanks for the clarifications.
     
  9. Feb 6, 2012 #8
    Yes.
    You have just [properly] described hovering outside a black hole versus free falling toward it....that's the Unruh effect....called Hawking radiation for black holes.

    This from the paper in post #2 is consistent with

    John Baez's description of Hawking Radiation:

    "An observer at rest has his own definition of a vacuum: it is the state in which he sees no particles. An accelerated observer also has his own vacuum, using the same definition.
    We will show that these two vacuums are not the same, so that an accelerated observer actually sees particles in the inertial observer's vacuum."

    http://www.obscure.org/physics-faq/Relativity/BlackHoles/hawking.html



     
    Last edited: Feb 6, 2012
  10. Feb 6, 2012 #9

    Mentz114

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    Quoted by Naty1 above ( can't work out where it's from :redface: )

    Hmmm... can the accelerated observer ever 'see' the inertial observers vacuum ? Or am I misunderstanding ?
     
  11. Feb 6, 2012 #10
    I'll like to point out that this view is not completely accurate. What autor, Naty1 is quoting, is describing is Unruh effect in extended Schwarzschild spacetime. The analog of Rindler vacuum state in extended Schwarzschild spacetime is known as Boulware vacuum and analog of Minkowski vacuum is known as Hartle-Hawking vacuum.

    I'll paraphrase R.M.Wald from his book Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics - page 129.

    "This thermal property of Hartle-Hawking vacuum state often is presented as though it were a derivation of effect of particle creation by a Schwartzschild black hole (Hawking radiation). This viewpoint is not correct."

    Than he explains why is Hartle-Hawking state highly unphysical, which is somewhat technical, that's why I'm not reproducing it now.

    "The difference in nature between the "Unruh effect" and the "Hawking effect" of particle creation by black holes is dramatically illustrated by considering the case of the Kerr metric. As already noted above, it has been proven that no analog of Hartle-Hawking vacuum state exists in Kerr spacetime. Thus, there is no analog of "Unruh effect" in Kerr spacetime. However, there is no corresponding difficulty with the derivation of particle creation in the case where gravitation collapse produces a Kerr black hole."
     
    Last edited: Feb 6, 2012
  12. Feb 6, 2012 #11
    Let me correct my self.. Actually this link gives pretty good description of Hawking radiation.

    Naty1 by saying
    gave what author in the link calls watered-down version, and this version is usually given because it simplifies math considerably. That "watered-down" version is what I was opposing in my previous post.
     
    Last edited: Feb 6, 2012
  13. Feb 6, 2012 #12
    Hi MiljenkoM, I don't really understand all of your post #10.....

    Do you agree this is or is not accurate:

    I think its fine...its one explanation; but there could well be other intepretations.


    MY limited understanding is that the Rindler horizon associated with the Unruh effect and a Schwarzchild horizon, for example, associated with a black hole are pretty much equivalent descriptions regarding the appearance of 'thermal radiation' ...
    but I do not know all the math so am happy to learn more.


    Not as I understand things. In each case (Unruh, black hole) acceleration introduces a horizon and with with it a different view of the energy of the vacuum state. So as in other aspects of relativity, different observers make different observations. In the Black Holes case for example, a free falling observer sees no horizon and passes that theoretical boundary without incident...and is unable to make any detection of it. An accelerating observer, however, in this case meaning a stationary observer near the horizon will be fried
    by thermal radiation in an instant.

    This description SUGGESTS something different between the Unruh and Black Hole situation, but other than the fact that in the Unruh effect the horizon is way distant [maybe infinity, I'm not positive] and the Black Holes horizon is right nearby, I'm unsure what else may be different.

    ok, found this.....Here is what Wikipedia offers:

    http://en.wikipedia.org/wiki/Unruh_effect#Vacuum_interpretation

    # I think what they mean is this:
    Elsewhere I have read that just as an acelerating [stationary]black hole observer sees the positive energy photons appear on the observer's side of the horizon, so an accelerating Unuh observer sees a similar radiation on the observer side of the Unruh horizon.
    On the other hand, as has been noted in other discussions in these forums, nothing in the math reportedly discusses the 'appearance' of virtual particles...this description was an 'intuitive' description which Hawking ,for one, used and said was not specifically related to his own mathematics.
     
    Last edited: Feb 6, 2012
  14. Feb 7, 2012 #13
    Mentz:
    Quoted by Naty1 above ( can't work out where it's from )


    This quote is from the introduction to the paper you reference in Post #2.

    That statement is perfectly consistent with other sources I have read.
     
    Last edited: Feb 7, 2012
  15. Feb 7, 2012 #14

    Mentz114

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    Naty1, I wasn't doubting it. I just don't know what it means. How can one see particles in another observers vacuum ?

    As I understand Beckenstein, the horizon changes the entropy so the accelerated observer detects particles in their vacuum.
     
  16. Feb 7, 2012 #15
    Hi, Naty1... I'll try to make my points more clear in this post, and if you have any questions after, please ask.

    Well, there is parse nothing particularly wrong with putting it like that, but I don't think it is very accurate description of effect. It raises a lot of questions if you put it like that, for example: why only some virtual particles, how do you decide which? Now, the main problem with this is that you are using classical concept of particles that involves poin-like objects moving along trajectories. Now we know that this concept does not actually apply on subatomic scales. As you know, one needs QFT in which basic object are not particles but quantum fields. Quantum states of the fields are interpreted in terms of corresponding particles. Experiments are than described as computing probabilities for specific field configuration and so on....
    As I sad in my previous post, I got a feeling that OP was thinking that mode functions are wave functions of particles, and that Bogoliubov transformations are turning one particle into another...
    Since Unruh is quantum field effect, I don't see any point in making it "particle" effect.


    Agreed, horizon is crucial in derivation of Hawking radiation. Consider for example neutron star, whose radious may be close to Schwarzschild radius but now there is no horizont, and neutron stars as far as I know do not emit Hawking radiation. But with Unruh effect horizon is only crucial for thermal spectrum of detected particles. You can set for example mind experiment with model detector that is accelerating only for finite time, than there is no horizon, all the photons will eventually catch up with detector. And you'll find out that transition probability for detector ≠0, meaning, it is detecting particles but with different spectrum than thermal spectrum [arXiv: gr-qc/0306022v2]. BTW how do you set a link in this forum?

    Horizon for accelerating observer is 1/a away from him, where a is acceleration, but I don't think this is important.
    Main difference is that Unruh effect is flat space effect, and Hawking is curved space.
    Now you can draw analogy between this two affects. First you need to find analog to Minkowski vacuum state, state that is regular throughout Schwarzschild spacetime, and this state is called Hartle-Hawking vacuum. State that is analog to Rindler vacuum (vacuum defined by accelerating observer) in case of black hole is called Boulware vacuum. And now by analogy with Unruh effect you find that Hartle-Hawking vacuum is thermal state, same as you will find that Minkowski vacuum is thermal state in Unruh effect.

    Now, my point in post #10 was that Unruh and Hawking effect are not infect analog, where one in flat spacetime, other in curved. And way to illustrate this is to look at Kerr black hole, where there is no Hartle-Hawking vacuum, and you can't draw analogy to Unruh effect, but nevertheless there is emission of particles from black hole (Hawking effect).
     
    Last edited: Feb 7, 2012
  17. Feb 7, 2012 #16
    You can't if you are by yourself :) . But let say that you have a friend named Peter, and you fly by him in accelerating rocket.

    You: Dude, what do you see?
    Peter: Dude, I don't see anything, this must be vacuum.
    You: Strange, my thermometer is showing temperature.
     
    Last edited: Feb 7, 2012
  18. Feb 7, 2012 #17

    Mentz114

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    But that means the particles are in my frame, not in his. Direct contradiction to
     
  19. Feb 7, 2012 #18
    Hmm... I think you are interchanging vacuum with frame. There is no contradiction, or am I missing something :/ .
     
  20. Feb 7, 2012 #19

    Mentz114

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    No, I'm not. In your scenario it is I, the accelerated observer who sees particles, thus they are in my vacuum. That contradicts this
    (my bold)

    I may not know QFT but I know English and that is a contradiction. Either the particles are in both observers vacua, or one of those statements is wrong. According to Beckenstein, the one I quote above is wrong. I don't think it is even as good as wrong - it is devoid of meaning.
     
  21. Feb 7, 2012 #20
    I know nothing at all about this but I find it fascinating. :smile:
    Also - and despite my complete lack of knowledge of this topic - I dare to expand on that line of argument: if indeed this is about field theory, then I guess that acceleration could imply radiation. And some people may refer to the prediction of possible photon detections as "particles", even if nothing material is implied. Am I on the right track?
     
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