Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Temperature vs phonon

  1. Dec 14, 2008 #1
    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 shouldnt be (in some way) proportionnal to the phonon frequency? If this was right, then the low limit [tex]\omega\gg T[/tex] would be a non-sense.

    So I realize that I dont 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 cant 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
     
    Last edited: Dec 14, 2008
  2. jcsd
  3. Dec 16, 2008 #2
    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.
     
  4. Dec 17, 2008 #3
    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 :
    [tex]T=\frac{\hbar\omega_s(\mathb{k})}{k_B\ln(\frac{1+n_s(\mathb{k})}{n_s(\mathb{k})})}[/tex]

    where [tex]\omega[/tex] is the frequency and [tex]n[/tex] 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
     
  5. Dec 17, 2008 #4
    ^ 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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Temperature vs phonon
  1. Phonon Dispersion (Replies: 2)

  2. Phonon waves (Replies: 19)

  3. What is a phonon? (Replies: 4)

  4. About phonons (Replies: 2)

Loading...