Boltzmann statistics

Read Chapter 13!

We should like to return now to the problem of counting all the possible ways of arranging...

  • $N$ particles
  • in a system with single-particle energy levels labelled by $j=1,2,3....$ (with energies $E_1, E_2, E_3$...)
  • where each energy level has a degeneracy (# of quantum states) of $g_j$.

The occupation number -- number of particles -- in energy level $j$ is $N_j$.

There are constraints on the total number of particles, and on the internal energy: $$\sum_i N_j=N,$$ $$\sum_i E_jN_j=U.$$

Our goal is to find the occupation numbers, N_j, compatible with a particular internal energy.

Distinguishable particles

We shall start by assuming that the particles are distinguishable.

Actually, electrons (one kind of fermion) are seriously indistinguishable from each other. (Ditto for bosons all of the same type). But we shall see that there is a limit in which these QM particles still behave like distinguishable ones.

The first energy level will have $N_1$ particles in it. The number of ways of selecting $N_1$ particles out of $N$ total particles is:

$$\frac{N!}{N_1!(N-N_1)!} = {N \choose N_1}.$$

But, now there are $g_1$ different quantum states in this level.

  • Any number of distinguishable particles can go into any of these states.
  • So, for each particle, there are $g_1$ possibilities.
  • The total number of possibilities within this energy level is
    $$g_1\times g_1...(N_1 {\rm times})=(g_1)^{N_1}$$

So, the total number of ways of arranging $N_1$ particles (out of N) in energy level 1, containing $g_1$ quantum states is: $$\frac{N!(g_1)^{N_1}}{N_1!(N-N_1)!}.$$

The second energy level: Now we pull $N_2$ particles out the remaining $N-N_1$ particles. So, the number of ways of arranging $N_2$ particles on the second level, with a degeneracy of $g_2$ will be: $$\frac{(N-N_1)!(g_2)^{N_2}}{N_2!(N-N_1-N_2)!}.$$

The total number of ways arranging the particles will be the product of these factors... $$\begin{align}w_B(N_1, N_2, N_3...N_n) & = \frac{N!(g_1)^{N_1}}{N_1!(N-N_1)!} \cdot \frac{(N-N_1)!(g_2)^{N_2}}{N_2!(N-N_1-N_2)!}\cdot...\\ & = N! \prod_{j=1}^n \frac{g_j^{N_j}}{N_j!} \end{align}.$$

Here I'm using product notation for the first time. The $\prod$ symbol is a shorthand to indicate the product of many terms: $\prod_{i=1}^n x_i=x_1\cdot x_2\cdot ... \cdot x_n$.

Convince yourself of this by writing out the third term in the product, and then looking for the pattern of what cancels...


To find the equilibrium state, all we have to do is find the maximum value of $w_B$ as we vary the occupation number $N_1, N_2,...N_n$.

There are also two constraints: $$\sum_j N_j = N,\ {\rm and}$$ $$\sum_j N_j E_j = U.$$