Small deviations from equilibrium and Lagrange multipliers

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SUMMARY

The discussion centers on the necessary and sufficient condition for small deviations from equilibrium as outlined in "Principles of Statistical Mechanics" by Amnon Katz. Specifically, it addresses the requirement that ##\exp(-\alpha N)## can be expanded in powers of ##\alpha##, retaining only the first-order term. Participants confirm the notation and express a need for clarification on the statistical functions ##f## and the derivation of the Lagrange multiplier ##\alpha##. The conversation emphasizes the importance of these concepts in understanding equilibrium in statistical mechanics.

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Ted Ali
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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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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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