# Derivation of the Boltzmann Distribution

1. Feb 23, 2010

### scorpion990

I'm using McQuarrie's "Statistical Mechanics" for a class, and I'm not quite understanding the the derivation of the Boltzmann Distribution. I'm going to go through it, and then ask a few questions along the way.

All right. You start with a canonical ensemble with N, V, and T fixed. Heat can be exchanged between microsystems, but matter cannot. Each microsystem has a spectrum of energies E1, E2, E3, ... that are repeated for degeneracy as needed. The occupation numbers of each energy state are denoted by a1, a2, a3, etc, and represent the number of microstates that are in energy stated E1, E2, E3, etc.

Imagine that each energy level is a "box" that contains as many particles as its occupation number. A specific distribution of occupation numbers can be achieved in:
$$W(a) = \sum^{n}_{i=1}\frac{A!}{\prod^{n}_{i=1}a_{i}!}$$
So that:
$$ln(W(a)) = Aln(A) - A - \sum^{n}_{i=1}ln(a_{i}!) =Aln(A) - A - \sum^{n}_{i=1}(a_{i}ln(a_{i})-a_{i})$$
Subject to the constraints:
$$\sum^{n}_{i=1}a_{i}=A$$
and
$$\sum^{n}_{i=1}E_{i}a_{i}=\zeta$$

Now.. Here's where many books and I disagree... They say that the set of equations derived by maximizing the above system is:
$$-ln(a_{i})-\alpha-1 - \beta E_{i} = 0$$

However, when I do it out, I get:
$$-ln(a_{i})-\alpha - \beta E_{i} = 0$$

I have no idea where their extra -1 term come from.

In addition, I don't understand most books' method of finding alpha. can somebody guide me through this process more clearly? Thanks!

2. Feb 24, 2010

### torquil

3. Feb 24, 2010

### sweet springs

Hi.

-1 comes from Stirling's series, the asymptotic expansion of the logarithm.

Regards.