Here I report the reasoning from which my question comes. According to:
O.V. Misochko, Muneaki Hase, K. Ishioka, and M. Kitajima. Transient bose–einstein condensation of phonons. Physics Letters A, 321(5–6):381 – 387, 2004.
one cannot have the phonon Bose-Einstein condensation at equilibrium due to zero
chemical potential.
Physically I expect that as $T\to0$ the vibrational energy of a Crystal tends to its zero point energy (considering the nuclei of the crystal as quantum particles and according to the uncertainty principle).
So at $T=0$ the number of phonons should be the smallest possible and all should be in the ground state. As far as I know the number of phonons in a crystal depends on the Temperature through the Bose-Einstein distribution:
$$
n_{({\bf q},s)}=\frac{1}{e^\frac{\hbar \omega({\bf q})}{k_B T}-1}
\tag{1}
$$
The reason for which the Phonons have zero chemical potential, $\mu=0$, is that they do not have to be fixed in number, so when finding the phonon distribution function we have only one Lagrange multiplier for the energy which turns out to be related to $k_B T$.
The phrase remarked above made me think that $\mu=0$ should imply that it is more convenient to destroy phonons at $T=0$ instead of having them condensed in the ground state.
I have also noticed that an acoustic phonon at $\Gamma$ has zero energy and so the occupation number from Eq. (1) should diverge…
concerning this, I have read the last answer to this question:
Could the chemical potential of a Bose gas be zero?
and since the dispersion relation of an acoustic phonon in gamma has zero second derivative, its mass is zero and it seems me that phonons in $\Gamma$ can be seen as analogous to blackbody photon mentioned in that answer. Anyway that answer is still too qualitative to be satisfactory to me, for this reason I am still asking this question.
I studied the Bose-Einstein condensation a gas of a fixed number, $N$, of Bosons. In that case I had $\mu <0$ at finite $T$ and $\mu \to 0$ in presence of the condensate. This is also confirmed in the article
G. Cook and R. H. Dickerson. Understanding the chemical potential. American Journal of Physics, 63(8):737–742, 1995.
which treats also the case of photons but unfortunately it does not explain why for photons the BEC does not occur.
Summing up,
I have some ideas why BEC does not happen for phonons with $\mu=0$ but I am still not satisfied. I would appreciate any help and some good references for this question.
Best Answer
I think you pretty much have all the right ideas in your question.
In a conventional BEC, particle number is conserved. Therefore, as T goes down towards 0, $\mu$ goes up towards the lowest state to conserve particle number. So the number of particles in the lowest state goes up ... eventually becoming macroscopically large, and that's the BEC (in the non-interacting situation).
For phonons, particle number is not conserved, and $\mu$ is always 0 (in equilibrium). Therefore, as T gets lower but $\mu$ stays the same, the occupation of every single mode goes down. You can see that directly from the formula.
PS: You referred to the zero-frequency acoustic phonon. Well, it's only zero-frequency in the textbook calculation for an infinitely-large crystal. Real objects have finite size, and the lowest phonon frequency corresponds to the fundamental acoustic mode of the object. In everyday terms, try whacking your crystal with a stick, and listen to the ringing sound it makes. What frequency is it? Maybe 300Hz for a small object? Well then, 300Hz is the lowest-frequency phonon. The occupation of this lowest-frequency 300Hz phonon mode decreases with temperature, just like all the other phonon modes.