# Homework Help: Statistical Physics

1. Apr 3, 2010

### ian2012

1. The problem statement, all variables and given/known data

Consider a thermal system at temperature T where the probability of finding the system
in a microstate r with energy Er is given by an arbitrary probability distribution pr that is
normalised so that Sum(pr) = 1.

Let kB denote Boltzmann’s constant and consider the Boltzmann distribution

$$p^{B}_{r}= \frac{e^{-\beta E_{r}}}{Z}$$

where Z is the partition function and beta = 1/kB.T

Now we want to compare an arbitrary probability distribution

$$p_{r}$$ with the Boltzmann distribution $$p^{B}_{r}$$. Let

$$S = - k_{B} \sum p_{r} ln p_{r}$$ denote the entropy of the system characterised by an arbitrary probability distribution and

$$S^{B} = - k_{B} \sum p^{B}_{r} ln p^{B}_{r}$$

denote the entropy of the system characterized by the Boltzmann distribution. By adding and subtracting

$$k_{B} \sum p_{r} ln p^{B}_{r}$$

we can write:

$$S - S^{B} = k_{B}\sum (-p_{r} ln p_{r} + p_{r} ln p^{B}_{r} - p_{r} ln p^{B}_{r} + p^{B}_{r} ln p^{B}_{r} )$$

Assuming that the two probability distributions pr and pBr yield the same mean energy <E>, show that

$$S - S^{B} = k_{B} \sum p_{r} ln \frac{p^{B}_{r}}{p_{r}}$$

2. Relevant equations

$$<E> = \sum p_{r} E_{r}$$

3. The attempt at a solution

$$<E> = \sum p_{r} E_{r} = \frac{1}{Z} \sum E_{r} e^{-\beta E_{r}}$$

I was thinking you can cancel the summation and conclude the arbitrary distribution = the boltzmann distribution? I would then use this result and show the last two parts of the S - SB expression = 0.

Last edited: Apr 3, 2010