How Does the Most Probable Energy Minimize Helmholtz Free Energy?

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SUMMARY

The discussion centers on demonstrating that the most probable energy minimizes the Helmholtz free energy, defined by the equation F=E-TS(E), where S(E) is the entropy at energy E. The probability density P(E) is expressed as P(E) = Omega(E)/(delta E) exp[-E/(k T)]/Z, linking the probability of a system's energy state to its entropy. Taking the logarithm of the probability reveals that maximizing the logarithm corresponds to minimizing the Helmholtz free energy, establishing a direct relationship between these concepts. A detailed explanation of this process can be found in a referenced post.

PREREQUISITES
  • Understanding of Helmholtz free energy (F=E-TS(E))
  • Familiarity with statistical mechanics concepts, particularly entropy (S) and probability density (P(E))
  • Knowledge of the Boltzmann distribution and its application (P(E) = exp[-E/(k T)]/Z)
  • Basic calculus skills, specifically logarithmic functions and their properties
NEXT STEPS
  • Study the derivation of the Boltzmann distribution in statistical mechanics
  • Learn about the relationship between entropy and probability in thermodynamic systems
  • Explore the implications of Helmholtz free energy in various physical systems
  • Read the referenced post on free energy minimization for a detailed example
USEFUL FOR

Students and researchers in physics, particularly those studying thermodynamics and statistical mechanics, as well as anyone interested in the principles of energy minimization in physical systems.

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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