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Homework Help: Average Energy of Boltzmann Distribution

  1. Jul 29, 2010 #1
    The question is:

    Write an expression for the average energy of a set of particles obeying Boltz-
    mann statistics and each having energy E = bz2, where b is a constant and
    z is a variable. Hence, show that the average energy per degree of freedom
    for each particle is 1
    2kBT; where kB is Boltzmann's constant. You should use
    the standard integrals shown at the end of the question.

    You are given:

    [integral from -infinity to infinity]exp(-ax2)dx = SQRT(PI/a)

    and

    [integral from -infinity to infinity]x2exp(-ax2)dx = (1/2)SQRT(PI/a3)


    I've taken the average energy as [integral from -infinity to infinity]E.Aexp(-E/KBT)dE which gives:

    [integral from -infinity to infinity]2b2z3exp(-bz2/KBT)dz by changing variable to z (dE/dz=2bz)

    But this is not the standard result so I can't proceed!
     
  2. jcsd
  3. Jul 29, 2010 #2
    Try integration by parts:

    [tex]\int u\,dv=uv-\int v\,du[/tex]

    where [itex]u=E[/itex] and [itex]v=-k_BT\exp\left[-E/k_BT\right] [/itex]
     
  4. Jul 30, 2010 #3
    I just tried it and ended up with -2Ab7/2
     
  5. Jul 30, 2010 #4
    oops!


    I mean -2Ab^7/2.b^7/2.KB^-3/2.T^-3/2
     
  6. Jul 30, 2010 #5
    I think your problem is two-fold:
    1. You're not using the correct formula for the average value
    2. You're putting in [itex]E=bz^2[/itex] when you don't really need to
    Both points are explained a little more below

    [tex]\langle E\rangle=\frac{1}{k_BT}\int_0^\infty E\exp\left[-\frac{E}{k_BT}\right]dE[/tex]

    then using integration by parts from my previous post:

    [tex]\int_0^\infty E\exp\left[-\frac{E}{k_BT}\right]dE=\left.-E\exp\left[-\frac{E}{k_BT}\right]\right|_0^\infty +\int_0^\infty k_BT\exp\left[-\frac{E}{k_BT}\right]dE[/tex]

    The first term on the right is zero at the limits, so then

    [tex]\langle E\rangle=\frac{1}{k_BT}\int_0^\infty k_BT\exp\left[-\frac{E}{k_BT}\right]dE=k_BT[/tex]
     
  7. Jul 30, 2010 #6
    Where does the additional term 1/KBT come from?

    That appears to work but I need to show the average energy is (1/2)KBT.
     
  8. Jul 30, 2010 #7
    [tex]
    \int_{0}^{\infty}{E^{n} \, \Exp\left[-\frac{E}{k \, T}\right] \, dE} = (k \, T)^{n + 1} \, \int_{0}^{\infty}{x^{n} \, e^{-x} \, dx} = \Gamma(n + 1) \, (k \, T)^{n + 1}
    [/tex]

    [tex]
    \Gamma(n + 1) = n!
    [/tex]
     
  9. Jul 30, 2010 #8
    The [itex]1/k_BT[/itex] comes from the normalization constant A that you have in your first post.

    If you are requiring that to be the answer then I suppose that [itex]z[/itex] is a speed? If that's the case, then

    [tex]f(E)=A\exp\left[-\frac{E}{k_BT}\right]=\sqrt{\frac{b}{\pi k_BT}}\exp\left[-\frac{bz^2}{k_BT}\right][/tex]

    Then the average speed-squared is

    [tex]\langle z^2\rangle=\sqrt{\frac{b}{\pi k_BT}}\int_{-\infty}^\infty z^2\exp\left[-\frac{bz^2}{k_BT}\right][/tex]

    You can then use your second given equation and you should be able to get the answer:

    [tex]\langle z^2\rangle=\frac{k_BT}{2b}\rightarrow\langle E\rangle=b\langle z^2\rangle=\frac{1}{2}k_BT[/tex]
     
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