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

Phonon Energy and Density of States

  1. Mar 10, 2015 #1
    Hi all,

    In Charles Kittel (Introduction to Solid State Physics) He writes :

    U (Total Phonon Energy ) = Σkp((ħ*ωk,p)/((exp(ħ*ωk,p/τ))-1))

    I understand this, but then he integrate over k and multiply by density of states :

    U (Total Phonon Energy ) = ∑p∫dω*Dp(ω)*((ħ*ωk,p)/((exp(ħ*ωk,p/τ))-1))

    I understand the Integration, but why multiply by density of states, if he wants to change the variable dk to dω why not just use the dispersion relation i.e k=g(ω) so dk=(the first dervative g(ω))*dω , dk=dω/Vg

    so it be :

    U (Total Phonon Energy ) = ∑p∫dω*(1/Vg)*((ħ*ωk,p)/((exp(ħ*ωk,p/τ))-1))

    Thanks in Advance.
    Last edited: Mar 10, 2015
  2. jcsd
  3. Mar 11, 2015 #2


    User Avatar

    Staff: Mentor

    As soon as you go from a sum to an integral, you need to introduce the density of states. The number of states in the range ##(k,k+dk)## or ##(\omega,\omega+d\omega)## depends on ##k## and ##\omega##, respectively.
  4. Mar 11, 2015 #3
    Ok, that makes senses, but whats the physical meaning that at ω=0, we have a the Density of States g(ω) = (N/π)*√(M/K).

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook