How Does the Most Probable Energy Minimize Helmholtz Free Energy?

captainjack2000
Messages
96
Reaction score
0

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)
 
Physics news on Phys.org
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)?
 
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.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top