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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

    DrClaude

    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).

    Thanks
     
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