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

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

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

$$\beta\equiv\frac 1 {k_{\scriptscriptstyle B}T}$$

giving

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

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:

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

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

$$(\Delta E)^2=\left\langle E^2 \right\rangle -{\left\langle E \right\rangle }^2$$

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

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

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

$${\Delta E\over E}\approx {1\over \sqrt N}$$

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

• Mandl 2.5
• Bowley and Sánchez 5.2
• Kittel and Kroemer 3