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Einstein's formula for specific heat

  1. Jun 15, 2010 #1
    I'm working through a derivation of Einstein's formula for specific heat and I'm stuck.

    So far I've been working off Planck's assumption of quantised energy [tex] E=n\hbar\omega [/tex] and the energy probability [tex] P(E)= e^{\frac{-E}{k_b T}} [/tex], using the fact that the mean expectation energy is [tex]\langle E \rangle= \frac{\sum_n E P(E)}{\sum_n P(E)}[/tex] to get total energy [tex]U=3N\langle E \rangle=\frac{3N\sum_n n\hbar\omega e^{-n\hbar\omega/k_b T}}{\sum_n e^{-n\hbar\omega/k_b T}}[/tex]

    The next step is where my problem is. The derivation I am studying says the above expression is equal to [tex]3Nk_b T\left[\frac{\hbar\omega/k_b T}{e^{\hbar\omega/k_bT}-1}\right][/tex], which when differentiated wrt T gives the Einstein formula, but I don't see how that step is made.

    Any ideas?
  2. jcsd
  3. Jun 15, 2010 #2
    I haven't found how to derive the result from your way yet, but here is an idea:
    The numerator is somewhat like [tex]A = e^{-1} + 2e^{-2} + ...[/tex] and the denominator is somewhat like [tex]B = 1 + e^{-1} + e^{-2} + ...[/tex]
    Notice that: [tex]A = (B-1) + (B-1)e^{-1} + (B-1)e^{-2} + ...[/tex] and that B is a geometric series and easily computed.
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