# Minimising helmholtz free energy

## Homework Statement

Show that the most probable energy minimises the Helmholtz free energy.

## Homework Equations

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

## The Attempt at a Solution

Not sure how you would 'show' is ????

P(E) = 1/Z *weight funciton*exp(-beta E)

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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.

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)?

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

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.