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The Unruh effect and vacuum states

  1. Apr 27, 2012 #1
    Hi all,

    I've recently been reading about the Unruh effect, and there are a few things that I'm not very clear on:

    1) An accelerating observer will see a vacuum filled with particles in thermal equilibrium (a warm gas) where an inertial observer would see zone.

    I'd originally read that the accelerated observer would see thermal (black-body) radiation only, which I had understood as EM radiation with a specific spectrum. However, more recently I read another source which said that the vacuum see by the accelerated observer would also contain a 'soup' of charged elementary particles, e.g., leptons etc.

    So would the accelerated observer see / interact with other 'real' particles? Or are we talking about virtual particles (since these are vacuum states).

    2) If I have, let's say, a Proton, accelerating, would it's interaction with the vacuum mean interaction with other, e.g., electrons in the vacuum? Again, I'm probably getting confused here. Don't phenomena like pair production require specific energies... i.e. I would expect the proton to interact only with EM photons until it's energy (from the acceleration) was large enough to produce an e.g., electron-positron pair. That is, this should be a quantized effect.

    3) The formula for the temperature observed by the accelerating proton is straightforward, but how does this translate into an expected density of photons, leptons etc. that the proton would actually interact with?

    Hoping someone can shed some light on the above (no pun intended).
  2. jcsd
  3. Apr 27, 2012 #2
    1) Particles besides EM radiation are present in blackbody radiation but in vanishingly small quantities. All particles get emitted, with probability roughly e^(-k * energy / temperature) where k is the Boltzmann constant. Since the minimum energy of, say, an electron is its rest mass, 511 keV, we should expect that electrons are emitted by blackbodies in roughly the same quantities as 511 keV gamma rays. But unless the temperature is really huge, the e^(-kE/T) probability factor above is vanishingly tiny for E this large. Roughly speaking, to emit a decent quantity of electrons you need temperatures of order (511 keV) / k = 6 billion Kelvin. Such temperatures were achieved in the initial moments of the Big Bang and also are reached in high-energy ion collisions at particle accelerators, which do indeed emit sprays of newly created particles in addition to photons.

    In the Unruh effect, the particles perceived by the accelerating observer are quite real: they will trigger particle detectors and so on. If the accelerating observer is accelerating enough that he perceives space to be filled with blackbody radiation at, say, 350 Fahrenheit, he can cook his dinner will the blackbody photons. From the perspective of an inertial observer, the energy required to cook the food comes from whatever is powering the acceleration.

    2) Photons and massive particles with arbitrarily high energies can be emitted by a blackbody at any temperature. In a body in thermal equilibrium, each particle does not have the exact same energy. Rather energy is constantly being redistributed in a random way, and occasionally a lot of it collects in one place and a really energetic particle can be emitted. It's just that the probability for this to happen becomes very small for very high energy particles, if the temperature is not correspondingly high. So it's not like electron emission will suddenly "turn on" at some temperature. It's just that not many electrons will be emitted at temperatures well below 6 billion Kelvin.

    3) The relevant formulas are the Fermi-Dirac (for fermions) and Bose-Einstein (for bosons) distributions. The blackbody spectrum for photons is just a special case of the Bose-Einstein distribution.
  4. Apr 27, 2012 #3
    Thanks Duck! That's extremely helpful.

    Out of curiosity (to get a feeling for the magnitudes involved), if I wanted to know, e.g., the cumulative probability that a detector accelerating at 1 g in a vacuum would detect, say, an electron in the thermal bath over a 1 yr. trip (i.e., accelerating at 1 g over 1 yr. of proper time), I can plug into the Unruh temperature equation - for the next step, would I then compute the Fermi-Dirac distribution and integrate over time? (I'm basically unsure about how to compute the probability of registering a detection over a period of time, although I know the numbers here are small).

    Thanks again!
  5. Apr 28, 2012 #4


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    What about neutrinos? Are neutrinos emitted in amounts comparable to photons?
    The temperature to use for the fermion distribution is clear. What is the value to be used for the Fermi energy?
  6. Apr 28, 2012 #5
    Also, what about conservation of energy?

    If we were to accelerate, say, an electron, sufficiently quickly, it could interact with a positron in the thermal bath (from it's perspective), and the result would be a gamma ray.

    Now, let's say we accelerate the electron more slowly, and perhaps for a shorter period of time, such that acceleration process (whatever is powering the acceleration) does not impart enough energy for the above interaction to occur (i.e., we haven't put in energy equivalent to the rest mass of the positron).

    Since energy must be conserved, does the magnitude and duration of the acceleration determine what types of interactions can take place? Not sure if that's clear - I'm basically wondering what an inertial observer would see in both of the cases above?
  7. Apr 28, 2012 #6
    Hi all,

    Just to clarify the above: an accelerating observer will see a thermal bath (containing all types of particles with some probability, even though that probability may be vanishingly small).

    If I consider an electron, and accelerate the electron, then with some probability the electron should see positron in the thermal bath (i.e., in the 'vacuum' as seen by the electron) with some non-zero probability. In turn, also with a low but non-zero probability, the accelerated electron and the positron in the thermal bath should interact, producing a gamma ray.

    Here's where I become confused: if the magnitude of the acceleration is small, or the duration is short, the energy imparted to the electron -may not be sufficient to produce a positron-. So, from the point of view of an inertial observer, the accelerated electron could not possibly have interacted with the positron, producing the gamma ray, because there was not enough energy involved.

    This seems to me to be a contradiction - what am I missing?
  8. Apr 29, 2012 #7
    Interesting question:

    (although I don't have access to the references he gives here unfortunately)

    taken from this paper.
    Last edited: Apr 29, 2012
  9. Apr 29, 2012 #8
    Here's a quote from another forum on Hawking radiation vs. Unruh radiation:

    So in the Unruh case, there's no actual energy transfer? This would solve the energy conservation question I asked previously, where the particle (electron) must be accelerated sufficiently.

    I'd still like to get a better idea of what's happening physically in the Unruh case, but I'm not sure if that's possible...
  10. Apr 29, 2012 #9
    I'd be interested to know what the poster you quote means by a "real" horizon and why the Rindler horizon isn't real but the Schwarzschild one is. AFAIK the Bogoliubov transformation which turns one man's empty vacuum into another man's furnace (well, OK, a pretty cool furnace!) applies to both cases.
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