# How is the distribution probability in the canonical ensemble derived?

I'm confused by the derivation of the canonical ensemble, namely the origin of the probability density, that is the Boltzmann factor. Here's what I have:

We have a system of particles with $(N_{tot},V_{tot},E_{tot})$ in thermodynamical equilibrium that we divide into two subsystems, $A$ and $B$. Let $(N,V,E)$ describe $A$, and let $(N',V',E')$ describe $B$. We have the relations $$N+N' = N_{tot}$$ $$V+V' = V_{tot}$$ $$E+E' \approx E_{tot}$$

where in the last one we've neglected any potential interaction between the particles, arguing that the number of degrees of freedom in any interaction is orders of magnitude smaller than the total amount of particles, because interactions are short ranged.

Now, consider the probability that subsystem $A$ has energy $E$. This is the ratio of the number of microstates of the entire system wherein $A$ has energy $E$ to the number of microstates such that the entire system has $E_{tot}$:

$$P(A \mbox{ has }E)= \frac{\Omega_{tot}(A \mbox{ has }E)}{\Omega_{tot}(E_{tot})}$$

but since we've neglected the interactions, we can factor the numerator:

$$P(A \mbox{ has }E)= \frac{\Omega(E)\Omega ' (E')}{\Omega_{tot}(E_{tot})}$$

where the omegas follow the same convention with primes and subscripts as the other variables. But now I don't understand how to get the Boltzmann factor... this relies on having

$$P(A \mbox{ has }E) = C \cdot \Omega ' (E')$$

But I don't see why that's true... $\Omega(E)$ isn't constant, is it?

-
–  joshphysics Jul 7 '13 at 20:44
Thanks, I remembered this answer of yours but unfortunately I'm not yet at a level where I can understand it. So I guess for the moment I'm stuck with trying to understand my subpar derivation :) –  DepeHb Jul 7 '13 at 20:54

Courtesy of Kittel's Elementary Statistical Physics:

Like Kittel, I'm going to call the two subsystems S (for subsystem) and R (for reservoir), instead of A and B.

The first key: instead of counting the states of S with exactly energy E_s (a set of volume 0 in the full phase space), consider the states within a small amount $\delta E_s$ of $E_s$. Then the probability $dp$ that S's energy falls within that range is:

$$dp = C d \Gamma_{tot} = C d \Gamma_s \Delta \Gamma_r$$

where $d \Gamma_{tot}$ is the volume of system phase space for which A is in the range $[E_s-\delta E_s, E_s]$, that volume factors into subsystem and reservoir phase space volumes, $d \Gamma_s$ is the volume of the subsystem phase space with the desired energy, and $\Delta \Gamma_r$ is the volume of reservoir phase space that keeps the total energy at $E_{tot}$.

Now the (dimensionless) entropy of the reservoir $\sigma_r$ is:

$$\sigma_r(E_r) = \log \Delta \Gamma_r$$

We want to Taylor-series expand this entropy about the total energy to get a term proportional to $E_s$. The reservoir energy $E_r$ is:

$$E_r = E_{tot} - E_s$$

Now the second key: So that we can truncate the Taylor series after the first term, we assume that the reservoir is much, much bigger than S, so that $E_s << E_{tot}$. Then:

$$\sigma_r(E_r) \approx \sigma_r(E_t) - \frac{\partial \sigma_r (E_{tot})}{\partial E_{tot}} E_s = \sigma_r(E_t) - \frac{E_s}{\tau}$$

where $\tau$ is the (normalized) system temperature, by definition (where again we're employing the assumption that the reservoir is much larger than the subsystem, so that the same temperature characterizes both reservoir and system). Plugging in:

$$dp = C \Delta \Gamma_r d \Gamma_s = C e^{\sigma_r(E_r)} d \Gamma_s = A e^{-E_s/\tau} d \Gamma_s$$

where $A = C e^{\sigma_r(E_{tot})}$ is a constant (since it's evaluated at $E_{tot}$.)

And there you are.

-