How Does the Most Probable Energy Minimize Helmholtz Free Energy?

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Homework Help Overview

The discussion revolves around demonstrating that the most probable energy minimizes the Helmholtz free energy, with a focus on the relationship between probability, entropy, and free energy in statistical mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the definition of probability in relation to energy states and question the reasoning behind taking the logarithm of the probability. There are attempts to connect this to the Helmholtz free energy equation.

Discussion Status

Some participants have offered insights into the mathematical relationships involved, while others express confusion regarding the steps and connections being made. Multiple interpretations of the logarithmic relationship and its implications for free energy are being explored.

Contextual Notes

There is a noted lack of clarity on the assumptions regarding the probability distribution and its relation to the Helmholtz free energy, as well as the specific definitions of terms like Omega and delta E.

captainjack2000
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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 probability 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)?
 
captainjack2000 said:
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.
 

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