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4.2 The Partition Function

Take-home message: Far from being an uninteresting normalisation constant, $Z$ is the key to calculating all macroscopic properties of the system!

The normalisation constant in the Boltzmann distribution is also called the partition function:

$\mbox{\LARGE\colorbox{yellow}{\rule[-3mm]{0mm}{10mm} \
$\displaystyle Z=\sum_j e^{-\varepsilon_j/k_{\scriptscriptstyle B}T}$  }}$
where the sum is over all the microstates of the system.

How can a constant be a function? Well for a given system and reservoir, that is fixed temperature, particle number, volume or magnetic field (as appropriate), $Z$ is a constant. But if the temperature etc are allowed to vary, then $Z$ is a function of them: $Z=Z(T,N,V)$ or $Z=Z(T,N,B)$. (The dependence on $V$ or $B$ comes through the energies of the microstates $\varepsilon_i$)

Why are we emphasising this? Because if we know $Z$, we can calculate all macroscopic properties of the system - energy, pressure, magnetisation, entropy...

For instance the average energy $\left\langle E \right\rangle $ (actually an ensemble average) is

\begin{displaymath}
\left\langle E \right\rangle = \sum_i \varepsilon_i  p_i = ...
...}T}\over
\sum_j e^{-\varepsilon_j/k_{\scriptscriptstyle B}T}}
\end{displaymath}

The top line is like the bottom line (the partition function) except that each term is multiplied by $\varepsilon_i$. We can get the top line from the bottom by differentiating by `` $1/(k_{\scriptscriptstyle B}T)$''. This is a bit awkward, so we introduce a new symbol

\begin{displaymath}
\beta\equiv\frac 1 {k_{\scriptscriptstyle B}T}
\end{displaymath}

giving

\begin{displaymath}
\left\langle E \right\rangle = -{1\over Z}{\partial Z \over \partial \beta}_{\scriptscriptstyle N,V}
\end{displaymath}

or
$\mbox{\large\colorbox{yellow}{\rule[-3mm]{0mm}{10mm} \
$\displaystyle E=-{\partial \ln Z\over \partial \beta}$  }}$
(where--contrary to the strict instructions given earlier--we will take it for granted that it is particle number and volume or magnetic field constant that we are holding constant.)

From the energy we can find the heat capacity:

\begin{displaymath}
C_V=\left({\partial \left\langle E \right\rangle \over \partial T}\right)_{\!\scriptstyle V,N}.
\end{displaymath}

We have found the average energy, but there will be fluctuations as heat is randomly exchanged between the system and the heat bath. These are given by

\begin{displaymath}
(\Delta E)^2=\left\langle E^2 \right\rangle -{\left\langle E \right\rangle }^2
\end{displaymath}

It can be shown that $(\Delta E)^2$ is related to the heat capacity,

\begin{displaymath}
(\Delta E)^2=(k_{\scriptscriptstyle B}T)^2{C_V\over k_{\scriptscriptstyle B}}
\end{displaymath}

For a normal macroscopic system the average energy is of the order of $N k_{\scriptscriptstyle B}T$ and the heat capacity is of the order of $N k_{\scriptscriptstyle B}$. Thus

\begin{displaymath}
{\Delta E\over E}\approx {1\over \sqrt N}
\end{displaymath}

For a system of $10^{24}$ atoms, $\Delta E/ E\approx 10^{-12}$ and so fluctuations are unobservable. There is no practical difference between and isolated system of energy $E$ and one in contact with a heat bath at the same temperature.

References



Subsections
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Next: 4.3 Entropy, Helmholtz Free Energy and Previous: 4.1 The Boltzmann Distribution
Judith McGovern 2004-03-17