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Heat capacity using Debye dispersion relation

  1. Mar 10, 2014 #1
    1. The problem statement, all variables and given/known data
    Using the Debye dispersion approximation, calculate the heat capacity of a harmonic, monatomic, 1D lattice. Next, find the temperature dependence in the low temperature limit. (Assume that the longitudinal mode has spring constant CL = C, and the two transverse modes both have spring constant CT = 0.2C. )


    2. Relevant equations
    [itex]C_{lattice}=k_{B}\sum[/itex]p ∫d[itex]\omega[/itex]Dp([itex]\omega[/itex]) [itex]\frac{x^{2}e^{x}}{(e^{x}-1)^{2}}[/itex] , p is over all of the modes
    [itex]d\omega[/itex] [itex]D_{p}( \omega)[/itex] = [itex]\frac{Lk_{B}T}{hv\pi}dx[/itex]
    where [itex]D_{p}( \omega)[/itex] = density of states
    [itex]c_p[/itex] = the spring constants for longitudinal and transverse
    3. The attempt at a solution
    So I should end up with 3 heat capacities to add together, two transverse and one longitudinal.

    [itex]C_{lattice}=k_{B}\sum_p \int^{\inf}_{0} \frac{Lk_{B}T}{hv\pi}\frac{x^{2}e^{x}}{(e^{x}-1)^{2}}
    \omega D_{p}(\omega) = \sum_{p}\frac{Lk_{B}T}{hv\pi}\frac{\pi^{2}}{3}K\sqrt{\frac{M}{c_{p}}}[/itex]

    [itex]C_{lattice}[/itex] = [itex]\frac{\pi L k_{B}}{3h}\sqrt{\frac{M}{c_{p}}}KT(2\sqrt{5}+1)[/itex]
    So this right here is my final solution, and I'm trying to see how far off base I am.
     
    Last edited: Mar 11, 2014
  2. jcsd
  3. Mar 11, 2014 #2

    maajdl

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    Gold Member

  4. Mar 11, 2014 #3
    That was a mistake and should've been [itex](e^{x}-1)[/itex], and the expression for [itex]C_{lattice}[/itex] is from Kittel if youre wondering where I got it.
     
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