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

  • #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?
 

Answers and Replies

  • #2
Cthugha
Science Advisor
1,956
318
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.
 
  • #3
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.
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.
 
  • #4
Cthugha
Science Advisor
1,956
318
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.
 
  • #5
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.
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##. "
 
  • #6
Demystifier
Science Advisor
Insights Author
Gold Member
11,000
3,713
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?
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.
 

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

Replies
2
Views
505
Replies
22
Views
1K
  • Last Post
Replies
10
Views
2K
Replies
7
Views
821
  • Last Post
2
Replies
34
Views
3K
  • Last Post
Replies
1
Views
529
Replies
8
Views
1K
Top