Minimising helmholtz free energy

1. May 3, 2009

captainjack2000

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. May 3, 2009

Count Iblis

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.

3. May 3, 2009

captainjack2000

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)?

4. May 3, 2009

Welshy

If you take the log then E - TS pops out. And that's the free energy.

5. May 3, 2009

Count Iblis

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.

6. May 4, 2009

gravityandlev

Last edited: May 4, 2009