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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
    you need to remember the identities

    [tex] 1+x+x^{2}+.................= (1-x)^{-1} [/tex]

    [tex] x+2x^{2}+3x^{3}+................= -x(1-x)^{-2} [/tex]

    setting [tex] x=e^{\frac{-\hbar \omega}{k_{b}.T}} [/tex] you can sum both series and get finite answers for U

    Oh man , i wish physics were sooo easy nowadays too
  4. Jun 15, 2010 #3
    I don't really see 'how' the step is made either - it doesn't look like a natural one-liner, although with sufficient ingenuity you can show why that it is correct (zetafunction has shown the level of ingenuity which is sufficient). The 'easier way' is to do the sum earlier on - by first writing down the partition function for a single oscillator, and working from there.

    Have you met partition functions in the context of statistical physics before?
  5. Jun 15, 2010 #4
    Thanks for the replies :D

    zetafunction, thank you for those formulas. I tried using them and I got the correct formula out, but with a negative sign in front from the negative x in the second formula above. Where do those identities come from?

    peteratcam, no I'm not familiar with partition functions. How would that work in this case?
  6. Jun 15, 2010 #5
    Well the partition function is the sum over all energy states:
    [tex]Z = \sum_i e^{-\beta E_i}[/tex]
    All thermodynamic information can be derived from the partition function, for example the Helmholtz Free Energy is [tex]-kT \ln Z[/tex] and the internal energy is [tex]-\partial \ln Z/\partial \beta[/tex].

    For a harmonic oscillator, calculating the partition function is as simple as knowing how to sum a geometric series. And then you can take derivatives to obtain the energy.

    Look in any university level statistical physics book for more details.
  7. Jun 15, 2010 #6
    The identity for 1+x+x^2+x^3+... = (1-x)^(-1) is just the binomial expansion. The other one is obtained by noticing that it is related to the derivative of the first.
  8. Jun 16, 2010 #7
    ok thanks genneth, that makes sense.
    So if you take the derivative of the first and multiply by x, you get x(1-x)^-2 not -x(1-x)^-2, which gives the required result.
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