# 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**:
$$\sum_j N_j = N,\ {\rm and}$$
$$\sum_j N_j E_j = U.$$