Average Energy of Boltzmann Distribution

  • Thread starter mamela
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  • #1
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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!
 

Answers and Replies

  • #2
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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]
 
  • #3
6
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I just tried it and ended up with -2Ab7/2
 
  • #4
6
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oops!


I mean -2Ab^7/2.b^7/2.KB^-3/2.T^-3/2
 
  • #5
383
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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]
 
  • #6
6
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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.
 
  • #7
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5
[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]
 
  • #8
383
0
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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