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

[center o bass]

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?

 

[Cthugha]

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.

 

[center o bass]

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.

 

[Cthugha]

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.

 

[center o bass]

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 $\nu$. "

 

[Demystifier]

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.