Temperature vs Phonons: Exploring the Relationship

[Tipi]

Hi!

I was looking at the high- and low-temperature limits of the specific heat in the quantum theory of cristals (Ashcrof&Mermin, Chap. 23).

To get the behavior under these limits, one consider first the case where T is large compared with all the phonon frequencies and second, when T is low compared to these frequencies.

But, the temperature shouldn't be (in some way) proportionnal to the phonon frequency? If this was right, then the low limit $$\omega\gg T$$ would be a non-sense.

So I realize that I don't really understand the relation between temperature and phonons. Sure, I know that the number of phonon of each type will come to play, but I can't make a whole picture of all that in my head.

Can someone try to explain, or give some refs where this is clearly explained?

Thanks a lot,

TP

 

[Manchot]

The phonon dispersion relation (the ω-k relationship) is determined only by the lattice properties of the solid, and is not a strong function of temperature. Each point on the ω-k diagram corresponds to some vibrational mode of the system, and since phonons are bosons, the probability that a phonon exists in any given mode is given by Bose-Einstein statistics. Put another way, at every frequency/phonon energy, you have some density-of-states determined by the dispersion relation. But only a fraction of those states are filled, and Bose-Einstein statistics tell you how many are filled at a certain temperature.

 

[Tipi]

The phonon dispersion relation (the ω-k relationship) is determined only by the lattice properties of the solid, and is not a strong function of temperature. Each point on the ω-k diagram corresponds to some vibrational mode of the system, and since phonons are bosons, the probability that a phonon exists in any given mode is given by Bose-Einstein statistics. Put another way, at every frequency/phonon energy, you have some density-of-states determined by the dispersion relation. But only a fraction of those states are filled, and Bose-Einstein statistics tell you how many are filled at a certain temperature.

Hi Manchot,
Thanks for your answer!

To make my question more precise, thit is the answer that satisfied my curiosity :

There is two quantities that link phonon and temperature : the frequency of the phonon and its probability in the overall distribution. If you isolate T in the distribution, you get :
$$T=\frac{\hbar\omega_s(\mathb{k})}{k_B\ln(\frac{1+n_s(\mathb{k})}{n_s(\mathb{k})})}$$

where $$\omega$$ is the frequency and $$n$$ its associated probability. So for a given probability, the temperature is proportional to the frequency but for a given frequency the more the probability is small, lower is the temperature.

My question was something like : How can you obtain small temperature from hign phonon frequencies. The answer is simply that these frequencies must have low probability.

Your comments are welcome,

TP

 

[Manchot]

^ Yes, that is correct. At phonon energies considerably higher than the temperature, the occupation fraction is small and so there simply aren't many phonons present.