# Canonical ensemble derivation

1. May 16, 2012

### intervoxel

In Statistical Mechanics, the key step in the derivation of the Canonical Ensemble is that the probability of S being in the m-th state, P_m , is proportional to the corresponding number of microstates available to the reservoir when S is in the m-th state. That is

$P_m=c\Omega(E_0-E_m)$,

where E_0 is the total energy.

Where does this rule come from? It seems to be inserted ad hoc in my textbook, and even in Wikipedia.

2. May 16, 2012

### Physics Monkey

One justification for this comes from the idea of maximum entropy. Given a system about which you have limited knowledge, say only the total energy E, you can ask what probabilities you should assign to various microstates. The distribution $p(n)$ (n labels microstates at energy E) which maximizes the entropy $S = - \sum_n p(n) \log{p(n)}$ is the flat distribution $p = 1/N(E)$ which gives an entropy of $\log{N(E)}$. N(E) is the total number of states at energy E.

A similar calculations gives the boltzmann distribution with temperature T interpreted as a Lagrange multiplier that enforces the constraint that the average energy be E.

3. May 16, 2012

### Matterwave

To me the rule of maximum entropy arises from the assumption that all microstates are equally likely and not the other way around.

The idea is simply that ALL microstates are equally likely. This is a key assumption of statistical mechanics, which sounds wrong when you consider QM etc., but actually turns out to give surprisingly good answers in the correct thermodynamic limit. Thus, the probability of S in the m-th state is then the total number of microstates that correspond to S in the m-th state divided by the total number of microstates there are total (which is just a constant, if we keep the energy of the whole system constant).

The number of microstates with S in the m-th state is simply equal to what you stated above as "the corresponding number of microstates available to the reservoir when S is in the m-th state". This is obvious because there is exactly 1 state that the sub-system S is in.

4. May 17, 2012

### intervoxel

Think I've got it.

When the system is in equilibrium, T_R=T_S, S is maximized and all microstates are equally probable. Then, for instance:

Total number of microstates=100
E1'+E1=E0
E2'+E2=E0
E3'+E3=E0

Code (Text):
R   |    S
------+-------
E1' 18 |
E2' 27 |
E3' 45 |
|
.
.
|
| E3 5
| E2 3
| E1 2
--------------
90 |     10

finally, c=1/90

Is it correct?

Last edited: May 17, 2012