Small deviations from equilibrium and Lagrange multipliers

In summary, according to "Principles of Statistical Mechanics" by Amnon Katz, for small deviations from equilibrium, the expression ##\exp ( -\alpha N )## must be able to be expanded in powers of ##\alpha## with only the first order term kept. The necessary condition for this is that the statistical function ##f## takes the form ##\exp(\Omega + \alpha N)##, and the Lagrange multiplier ##\alpha## is obtained through the correct derivation.
  • #1
Ted Ali
12
1
Homework Statement
Consider the grand canonical ensemble, for a system A with N identical particles. This system can exchange particles with a reservoir A'. We consider only small deviations from equilibrium. What is the necessary and sufficient condition that the Lagrange multiplier ##\alpha## of N, must satisfy?
Relevant Equations
The Lagrange multiplier ##\alpha## is: $$\alpha =\left( \frac{\partial \ln \Omega}{\partial N'} \right)$$.
According to the book "Principles of Statistical Mechanics" by Amnon Katz, page 123, ##\alpha## must be such that ##\exp ( -\alpha N ) ## can be expanded in powers of ##\alpha## with only the first order term kept. Is this the necessary and sufficient condition for small deviations from equilibrium?

Thank you in advance,
Ted.
 
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  • #2
Hello Ted.

Looking through your expressions, I guess that you are using the same notation of the book ("Principles of Statistical Mechanics" by Amnon Katz).

Two points:

What is the expression of your statistical functions ##f##'s?
I assume that it should be something like this for system ##A##,

$$f=\exp(\Omega\ +\ \alpha N )\ ,$$
right? (please correct me if I am wrong).

May you elaborate on how you obtained the Lagrange multiplayer ##\alpha##? If your derivation is correct, then you have the necessary condition.
 

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