# Mean energy and preassure inolving partition function

1. Sep 20, 2010

### rayman123

1. The problem statement, all variables and given/known data
Can someone explain to me how this work out this formula

2. Relevant equations
we are supposed to find the mean energy and preasure of a gas with given partition function

3. The attempt at a solution
mean energy is given $$\overline{-}U=\sum_{r}E_{r}p_{r}$$
I know also that Boltzman's probability distribution is described by
$$p_{r}= \frac{e^{-\beta E_{r}}}{\sum_{r}e^{-\beta E_{r}}}$$

because the partition function is definied as $$z=\sum_{r}e^{-\beta E_{r}^}$$

so rewriting now the Boltzman's probablility distribution I get
$$p_{r}= \frac{e^{-\beta E_{r}}}{z}$$
1. The problem statement, all variables and given/known data

now going back to the mean energy I can write
$$\overline{-}U=\frac{1}{z}\sum_{r}E_{r}e^{-\beta E_{r}$$

These are operations I do not understand. Could someone explain them step by step ?

$$\sum_{r}E_{r}e^{-\beta E_{r}}= -\frac{\partial}{\partial \beta}\sum_{r}e^{-\beta E_{r}}= -\frac{\partial}{\partial \beta}z$$

and the final one

$$U= -\frac{1}{z}\frac{\partial z}{\partial \beta}=-\frac{\partial lnz}{\partial \beta}$$

2. Sep 20, 2010

### zhermes

The first part of the equation is how you find the (ensemble) average energy---you find the weighted average over all ensemble states. The second part is showing that you can rewrite this as the partial derivative with respect to beta; if you take the derivative of the boltzmann factor WRT beta, the energy E gets pulled down
$$\frac{\partial}{\partial \beta}e^{-\beta E_{r}}= E_{r} e^{-\beta E_{r}}$$
In the expression using the partial derivative wrt beta, the sum over boltzmann factors is the definition of the partition function. So you end up with the partial derivative of the partition function = the average energy.

Again, this is just rewriting things with derivatives. Using the chain rule, if you take the derivative of lnz, you get a 1/z term out front.

'z' is some function of beta $$z = z(\beta)$$, so using the chain rule: $$\frac{\partial ln z(\beta)}{\partial \beta} = \frac{1}{z} \frac{\partial z(\beta)}{\partial \beta}$$

Does that make sense?

Last edited: Sep 20, 2010
3. Sep 21, 2010

### rayman123

yes it does!:) thank you