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Statistical Physics

  1. Apr 3, 2010 #1
    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

    [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]

    2. Relevant equations

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

    3. 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.
    Last edited: Apr 3, 2010
  2. jcsd
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