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Minimising helmholtz free energy

  1. May 3, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that the most probable energy minimises the Helmholtz free energy.


    2. Relevant equations
    F=E-TS(E) where S(E) is the entropy of te system of given energy E.

    3. The attempt at a solution
    Not sure how you would 'show' is ????

    P(E) = 1/Z *weight funciton*exp(-beta E)
     
  2. jcsd
  3. May 3, 2009 #2
    The probablity that a system at temperature T is in a state r with energy E_r is given by exp[-E_r/(k T)]/Z

    The probability that the system has an energy between
    E and E + dE is thus the probability that the system is in any particular state with eneergy E times the number of states inside the energy range from E to E + dE. The probability density P(E) as a function of energy is thus:

    P(E) = Omega(E)/(delta E) exp[-E/(k T)]/Z

    where delta E is the energy resolution used to define Omega(E). If you take the logarithm, use that S = k Log(Omega), then you find the desired result.
     
  4. May 3, 2009 #3
    I'm sorry but I am really not following what you said. Why would you take the logarithm of the probability? and how does this relate to the free energy F=E-TS(E)?
     
  5. May 3, 2009 #4
    If you take the log then E - TS pops out. And that's the free energy.
     
  6. May 3, 2009 #5
    The probability is maximal if the logarithm of the probability is maximal and vice versa. If you take the logarithm then you see that:

    Log[P(E)]= -F(E) + constant

    where the constant does not depend on E.
     
  7. May 4, 2009 #6
    Last edited: May 4, 2009
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