Statistical Physics Homework: Compare Arbitrary & Boltzmann Distributions

[ian2012]

Homework Statement

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

[tex]p^{B}_{r}= \frac{e^{-\beta E_{r}}}{Z}[/tex]

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

Now we want to compare an arbitrary probability distribution

[tex]p_{r}[/tex] with the Boltzmann distribution [tex]p^{B}_{r}[/tex]. Let

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

[tex]S^{B} = - k_{B} \sum p^{B}_{r} ln p^{B}_{r}[/tex]

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

[tex]k_{B} \sum p_{r} ln p^{B}_{r}[/tex]

we can write:

[tex]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} )[/tex]

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

[tex]S - S^{B} = k_{B} \sum p_{r} ln \frac{p^{B}_{r}}{p_{r}} [/tex]

Homework Equations

[tex]<E> = \sum p_{r} E_{r} [/tex]

The Attempt at a Solution

[tex] <E> = \sum p_{r} E_{r} = \frac{1}{Z} \sum E_{r} e^{-\beta E_{r}}[/tex]

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.

 

[_coyote_]

I'm not sure if this is the right way to approach the question or how I could incorporate the fact that they have the same mean energy. Any help or advice would be great :)