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Next: Homework Problems Up: Ensembles and Thermodynamic Potentials Previous: General Case: How to

Subsections


Equivalence of Ensembles

Canonical and Grand Canonical Ensembles Revisited

Compare the partition function and grand partition function:

\begin{displaymath}
\begin{gathered}
 Q = \sum_i \exp(-E_i/kT)\  \Xi = \sum_{i,N} \exp(-E_i/kT+\mu N/kT) 
 \end{gathered}\end{displaymath}

we see that  
 \begin{displaymath}
 \Xi = \sum_{N} Q_N\exp(\mu N/kT)\end{displaymath} (2)
We can express one partition function through other!

Two connections between canonical and grand canonical ensembles:

1.
A system in canonical ensemble can be divided into parts. Each part can be considered in grand canonical ensemble with $\mu=\mathit{const}$!
\begin{figure*}
 \InputIfFileExists{can-grandcan.pstex_t}{}{}
 \end{figure*}
2.
A system in grand canonical ensemble can be considered as a set of systems, each of them in canonical ensemble with $N=\mathit{const}$
\begin{figure*}
 $\underbrace{\InputIfFileExists{grandcan-can.pstex_t}{}{}}_{\displaystyle\text{Grand
 Canonical Ensemble, $\mu=\mathit{const}$}}$ 
 \end{figure*}

These two ways reflect a deep connection between the function and its Legendre Transform.

Through this method we can connect any two ensembles!

Method of Maximal Term
Or
Did Your Teachers Cheat You?

In our calculations we often substituted $\left\langle N\right\rangle$ for N. But in canonical ensembles $N=\mathit{const}$, in grand canonical ensemble $N\ne\mathit{const}$. Did we cheat?

Theorem:
suppose we have a sum  
 \begin{displaymath}
 s = \sum_{i=1}^N t_i,\quad t_i\gt
 \end{displaymath} (3)
and

\begin{displaymath}
N\to\infty, \quad t_{\max}\sim e^N
 \end{displaymath}

Then at $N\to\infty$

\begin{displaymath}
\ln s \approx \ln t_{\max}
 \end{displaymath}

Proof:
For any N

\begin{displaymath}
t_{\max}<s<Nt_{\max}
 \end{displaymath}

Take log:

\begin{displaymath}
\ln t_{\max} < \ln s < \ln t_{\max} + \ln N
 \end{displaymath}

At $N\to\infty$ we have $\ln t_{\max}\sim N \gg \ln N$, and the theorem is proven.
Consequence:
In the sum (2) all terms are $\sim
 e^N$. In the thermodynamic limit

\begin{displaymath}
\Xi\approx Q_{\overline{N}}\exp(\overline{N}\mu/kT)
 \end{displaymath}

with such N that

\begin{displaymath}
\overline{N}\mu + kT\ln Q_{\overline{N}} \mapsto\max
 \end{displaymath}

or

\begin{displaymath}
\mu = \left(\frac{\partial A}{\partial N}\right)_{T,V}
 \end{displaymath}

Your teachers did not cheat you--at least in the thermodynamic limit!

next up previous
Next: Homework Problems Up: Ensembles and Thermodynamic Potentials Previous: General Case: How to

© 1997 Boris Veytsman and Michael Kotelyanskii
Tue Sep 9 22:39:08 EDT 1997