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Why does photons of a given frequency satisfy the Boltzmann distribution?

  1. Oct 8, 2014 #1
    A mode of frequency ##\nu## has energy ##E_n = h \nu##. In terms of photons, the interpretation that I have read several places, is that this correspond to ##n## photons of energy ##h \nu##. Furthermore, it is stated that the probabilty of finding ##n## photons at frequency ##\nu## is given by
    $$p(n) = e^{-nh\nu}/Z,$$
    where Z is the partition function (for example in: http://disciplinas.stoa.usp.br/pluginfile.php/48089/course/section/16461/qsp_chapter10-plank.pdf)

    This correspond to Boltzmann statistics, and I'm a bit confused by this since photons are supposedly bosons. Should'nt this instead be Bose-Einstein statistics?
  2. jcsd
  3. Oct 8, 2014 #2


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    Yes, Bose-Einstein statistics are the correct approach. However, in the classical limit the differences between the two distributions vanish. For example for bosonic atoms, there is not really a difference between the distributions for high temperatures and diluted gases.

    For photons, using Boltzmann statistics is fine if you are dealing with large energies because the additional "-1" becomes negligible and Maxwell-Boltzmann statistics become a good approximation to Bose-Einstein statistics.
  4. Oct 8, 2014 #3
    I agree that this is correct in the classical limit. However in http://disciplinas.stoa.usp.br/pluginfile.php/48089/course/section/16461/qsp_chapter10-plank.pdf the complete planck law is derived by assuming that the probability that a single mode is in a state of energy E=nhν (a state of n photons) is given by a Boltzmann distribution. Hence, the derivation does not consider any limit.
  5. Oct 8, 2014 #4


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    Sorry, I do not have access to that PDF file on my phone right now, but are you completely sure that the partition function Z they use is really the classical Maxwell-Boltzmann one?

    If they just consider the number of photons in each energy state, they apply BE statistics. If they also take multiplicities into account and you see a lot of factorials, it is most likely the classical partition function.
  6. Oct 8, 2014 #5
    Might the confusion lie in the difference between modes vs photons or photons at frequency ##\nu## vs photons at any frequency?

    To quote directly what is stated:

    "The probability that a single mode has energy ##E_n = n h\nu## is given by
    $$p(n) = \frac{e^{-E_n/kT}}{\sum_{n=0}^\infty e^{-E_n/kT}}$$
    where the denominator ensures that the sum of probabilities is unity, the standard normalization procedure. In the language of photons this is the probability that the state contains ##n## photons of frequency ##\nu##. "
  7. Oct 10, 2014 #6


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    Bose-Einstein distribution involves an average over all n, giving the probability of a given frequency. It is the summation over all n that gives the characteristic Bose-Einstein form of the distribution.

    When n is fixed, there is no any difference between classical (Maxwell-Boltzmann) and quantum (Bose-Einstein) statistics.
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