# Average Energy of Boltzmann Distribution

1. Jul 29, 2010

### mamela

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. Jul 29, 2010

### jdwood983

Try integration by parts:

$$\int u\,dv=uv-\int v\,du$$

where $u=E$ and $v=-k_BT\exp\left[-E/k_BT\right]$

3. Jul 30, 2010

### mamela

I just tried it and ended up with -2Ab7/2

4. Jul 30, 2010

### mamela

oops!

I mean -2Ab^7/2.b^7/2.KB^-3/2.T^-3/2

5. Jul 30, 2010

### jdwood983

I think your problem is two-fold:
1. You're not using the correct formula for the average value
2. You're putting in $E=bz^2$ when you don't really need to
Both points are explained a little more below

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

then using integration by parts from my previous post:

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

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

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

6. Jul 30, 2010

### mamela

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. Jul 30, 2010

### Dickfore

$$\int_{0}^{\infty}{E^{n} \, \Exp\left[-\frac{E}{k \, T}\right] \, dE} = (k \, T)^{n + 1} \, \int_{0}^{\infty}{x^{n} \, e^{-x} \, dx} = \Gamma(n + 1) \, (k \, T)^{n + 1}$$

$$\Gamma(n + 1) = n!$$

8. Jul 30, 2010

### jdwood983

The $1/k_BT$ 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 $z$ is a speed? If that's the case, then

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

Then the average speed-squared is

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

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

$$\langle z^2\rangle=\frac{k_BT}{2b}\rightarrow\langle E\rangle=b\langle z^2\rangle=\frac{1}{2}k_BT$$