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Internal Energy of Degenerate Fermi ideal gas to the 4th power

  1. Dec 3, 2016 #1
    1. The problem statement, all variables and given/known data
    We are asked to derive the expression for the internal energy of an ideal Fermi degenerate gas using Sommerfeld expansions, writing out terms up to the fourth order in ##(\frac{T}{T_F} )## , that is, we must determine ## \alpha ## in the following expression: $$ U= \frac{3}{5}N\varepsilon_F \left\{ 1 + \frac{5\pi ^2}{12} {\left( \frac{T}{T_F}\right)}^2 + \alpha {\left( \frac{T}{T_F}\right)}^4 \right\}$$
    The method is basically the one used in the text of Kubo on Statistical Mechanics.

    2. Relevant equations
    We're treating the 3-D "particle in a box" model with periodic boundary conditions, so that the energy spectrum is given by: $$\varepsilon_k = \left( \frac{\hbar^2 k^2}{2m} \right) , k=\left( 2n\pi /L \right)$$
    We are also supposing that the number of particles is sufficiently large for we to approximate the summations characteristic of a discrete spectrum by integrals. We then have:
    $$
    N= \gamma V \int_0^\infty D(\varepsilon)f(\varepsilon) d\varepsilon, \\
    U= \gamma V \int_0^\infty \varepsilon D(\varepsilon)f(\varepsilon) d\varepsilon \\
    $$
    where ## \gamma ## takes the spin degeneracy into account and:
    $$
    D( \varepsilon ) = \frac{1}{4 \pi^2} {\left( \frac{2m}{\hbar ^2} \right)}^{3/2} \varepsilon ^{1/2} \equiv C \varepsilon ^{1/2} ,\ \ f(\varepsilon) = \frac{1}{exp[\frac{\varepsilon - \mu}{k_bT}] + 1}.
    $$
    The Sommerfeld expansion is also given, whereby we can solve integrals of the form ## I = \int_0^\infty f(\varepsilon) \psi(\varepsilon) d{\varepsilon}## :
    $$
    I = \int_0^{\mu}\psi(\varepsilon) d\varepsilon + \frac{1}{2!}I_2 {\left( \frac{d \psi}{ d \varepsilon} \right)}_{\varepsilon= \mu} + \frac{1}{4!}I_4 {\left( \frac{d^3 \psi}{ d\varepsilon^3} \right)}_{\varepsilon= \mu} + ...
    $$
    Within approximations, the integral ##I_2## and ##I_4## are taken to equal, respectively, ## \frac{\pi^2 (k_bT)^2}{3} ## and ## \frac{7 \pi^4}{15}(k_bT)^4##
    3. The attempt at a solution

    Since U must be given in powers of ##T/T_F ##, we must first use the equation for N in order to write #/mu# in terms of T and #T_F#. Therefore, for N:

    $$
    \psi(\varepsilon) = \varepsilon^{1/2} \implies N = \gamma VC \left \{ \frac{2}{3}\mu^{3/2} + \frac{\pi ^2}{12} (k_b T)^2 \mu^{1/2} - \frac{3}{8} \frac{1}{4!} \mu^{\frac{-5}{2}}I_4 + ... \right \} \\
    $$
    $$
    \frac{N}{\gamma VC}= \frac{2}{3} \mu^{3/2} \left[ 1 + \frac{\pi^2}{8}\left (\frac{k_b T}{\mu} \right)^2 + \left( \frac{7 \pi^4}{640} \right) \left (\frac{k_bT}{\mu} \right)^4 \right]
    $$
    Now, after this expression Kubo (and other books I've checked) make some kind of expansion around ## \mu_o ## and arrive at the following expression:
    $$
    \mu = \mu_o \left[1- \frac{\pi^2}{12} \left( \frac{k_b T}{\mu_o}\right)^2 - \frac{\pi^4}{80} \left( \frac{k_b T}{\mu_o}\right)^4 \right]
    $$
    and , no matter how I try, I can't get anything like that last expression (I managed to get the first term by truncating the series and using a binomial expansion to the first order, but when I include the next power I can't make it work)
     
  2. jcsd
  3. Dec 3, 2016 #2

    TSny

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    I have a copy of Kubo. Are you OK with his derivation up through equation (6) on page 241?
    upload_2016-12-3_21-45-17.png
     
  4. Dec 4, 2016 #3
    Hi! Well, no, I don't get the approximation that leads to equation (6), though I'm ok with the one above it. Don't know how to solve that equation.
     
  5. Dec 4, 2016 #4

    TSny

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    To get the second equation of (6) from the first equation, write the bracket part of the top equation as ##f(x) = \left(1 + a x + b x^2 \right )^{-2/3} ## where ##x = \left( \frac{kT}{\mu} \right)^2## and do a Taylor expansion about ##x = 0## to second order in x.

    Or, are you having difficulty getting to the first equation of (6)?
     
  6. Dec 4, 2016 #5
    Wow, thanks a lot! I was carrying the expansion around the ## \mu_o##. So, let me see if I got it right: I make a Taylor expansion around x=0, but that implies T=0, and, for that reason and the definition of the fermi energy, I wrtite ##\mu_o ## insead of ## \mu##?
     
  7. Dec 4, 2016 #6

    TSny

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    You are just treating ##x = \left(\frac{kT}{\mu} \right)^2## as small compared to 1. In the Taylor expansion, you will still write ##x## with ##\mu## rather than ##\mu_0##. Note how the bracketed expression in the second equation in (6) is still in terms of ##\mu##.

    You will still have some work to do to get from (6) to (7), where the right-hand side is in terms of ##\mu_0##.
     
  8. Dec 4, 2016 #7
    From (6) to (7) must I do another expansion (around ##\mu_o##)? There is no equation relating ## \mu_o \text{and } \mu## that I remember, but neither my expansions around ##\mu_o## nor T=0 lead to the desired expression :(.
     
  9. Dec 4, 2016 #8

    TSny

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    To fist order in ##x =\left( \frac{kT}{\mu} \right)^2##, (6) tells you that ##\mu = \mu_0 \left\{1-\frac{\pi^2}{12} \left( \frac{kT}{\mu} \right)^2 \right\}##.

    Convince yourself that this is still accurate to first order if you replace ##\mu## by ##\mu_0## in the last term on the right. So, you get the first-order approximation given in Kubo below equation (6).

    To get the second-order approximation in ##\left( \frac{kT}{\mu} \right)^2##, you need to go back to (6). In the second term inside the { }, you need to substitute for ##\mu## the first-order expression given just below (6). This term can then be approximated to order ##\left( \frac{kT}{\mu} \right)^4##. The third term in the { } in (6) is easy to approximate to order ##\left( \frac{kT}{\mu_0} \right)^4## by just letting ##\mu = \mu_0##.
     
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