Calculation of the grand canonical ensemble

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2.5 Calculation of the grand canonical ensemble

Maximize

$S^{'} = - k \sum_{i} p_{i} ln (p_{i})$

(2.25)

with the restrictions

$0 = \sum_{i} p_{i} U_{i} - U, and 0 = \sum_{i} p_{i} - 1, and 0 = \sum_{i} p_{i} N_{i} - N .$

(2.26)

Introducing the Lagrange parameters $α$ , $β$ , and $γ$ the variation of the function

$δ [S^{'} - k α (\sum_{i} p_{i} - 1) - k β (\sum_{i} p_{i} U_{i} - U) - k γ (\sum_{i} p_{i} N_{i} - N)] = 0$

(2.27)

without restrictions leads to

$- ln (p_{i}) - 1 - α - β U_{i} - γ N_{i} = 0 .$

(2.28)

Defining again

$\frac{1}{Z} = exp (- 1 - α)$

(2.29)

we find

$p_{i} = \frac{1}{Z} exp (- β U_{i} - γ N_{i}) and Z (β, V, γ) = \sum_{i} exp (- β U_{i} - γ N_{i}) .$

(2.30)

We get

$U = \sum_{i} p_{i} U_{i} = \frac{\sum_{i} exp (- β U_{i} - γ N_{i}) U_{i}}{\sum_{i} exp (- β U_{i} - γ N_{i})} = - (\frac{\partial ln (Z)}{\partial β}) : = U (β, V, γ)$

(2.31)

and

$N = \sum_{i} p_{i} N_{i} = \frac{\sum_{i} exp (- β U_{i} - γ N_{i}) N_{i}}{\sum_{i} exp (- β U_{i} - γ N_{i})} = - (\frac{\partial ln (Z)}{\partial γ}) : = N (β, V, γ)$

(2.32)

i.e.

$S = k ln (Z) + β k U + γ k N .$

(2.33)

The total derivative is:

$\begin{matrix} \begin{aligned} \frac{d S}{k} & = (\frac{\partial ln (Z)}{\partial β}) d β + (\frac{\partial ln (Z)}{\partial γ}) d γ + (\frac{\partial ln (Z)}{\partial V}) d V + U d β + β d U + N d γ + γ d N \\ = (\frac{\partial ln (Z)}{\partial V}) d V + β d U + γ d N \end{aligned} \end{matrix}$

(2.34)

$S = S (V, N, U)$

(2.35)

and $S$ is the Legendre transformed of $k ln (Z)$ .
Let

$(\frac{\partial S}{\partial U}) : = \frac{1}{T} and (\frac{\partial S}{\partial N}) : = - \frac{μ}{T} .$

(2.36)

$β = \frac{1}{k T}, and γ = - \frac{μ}{k T} .$

(2.37)

Following again the procedure for the calculation of the free energy we find

$Ω = U - μ N - T S$

(2.38)

and

$Ω (T, V, μ) = - k T ln (Z (T, V, μ)) .$

(2.39)

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