# Canonical ensemble, energy, heat bath

I am studying through the book Thermodynamics and Statistical Mechanics by Walter Greiner and I’ve got a couple of doubts when I was reading about the classical ensembles, specially the Canonical ensemble (chapter 7, pages 159-160, Springer, 2004).

In the case of the canonical ensemble it is considered that the system has a fixed temperature due to the contact with the termal reservoir and it can assume a large range of values of energy (I think, but I am not sure, that these different values of energy are attained by the system – in termal contact with the reservoir – until the final equilibrium state is attained.)

In the page 160, about the probability $p_{i}$ of finding the system $S$ in a microstate $i$ with an energy $E_{i}$, it is told:

“If $S$ is a closed system, $p_{i}$ will be proportional to the number of microstates ${\Omega}_{S}(E_{i})$. Analogously, $p_{i}$ is proprotional to the number of microstates in the total closed system for which $S$ lies in the microstate with the energy $E_{i}$. Obviously this is just equal to the number of microstates of the heat bath for the energy$E–E_{i}$, since $S$ assumes one microstate $i$ [...]”

What I don’t understand is: How “obviously” is that? Because to me it’s not obvious.

After that, I don’t understand too how to get the Equation (7.4), since we the expansion is “with respect to $E_{i}$”.

$$k{\ln}{\Omega}_{R}(E-E_{i}){\approx}k{\ln}{\Omega}_{R}(E)-{\frac{\partial}{{\partial}E}}(k{\ln}{\Omega}_{R}(E))E_{i}+...,$$ where $E$ is the total energy, $E_{R}$ is the energy of the reservoir, $E_{S}=E_{i}$ is the energy of the system in a given microstate and $E=E_{R}+E_{S}$.

If anybody could help, I really would appreciate.

• Could you add the equation and what is expanded? Not everyone has your book lying next to them. – ACuriousMind Jul 16 '14 at 18:28
• @ACuriousMind, and what about the paranthesis "(I think, but I am not sure, that these different values of energy are attained by the system – in termal contact with the reservoir – until the final equilibrium state is attained.)"... Is it true? – Poli Tolstov Jul 16 '14 at 18:35
• Your equation is simply a Taylor serie, $f(E-E_i) = f(E) - f'(E)\,E_i+...$ , where you keep only the first derivative term because $E_i << E$ – Trimok Jul 17 '14 at 10:21

If you think it again, it is not that difficult to understand. First, assume that $S$ and the reservoir are an isolated system. In any such system, it is an intuitive assumption (axiom, due by symmetry) that the system will spend the same amount of time on each of these states (the total energy is a constant, $E$). So if you know that $S$ is in a specific microstate $i$, with energy $E_i$, you also know that the likelihood of the system being on this state is also proportional to the number of microstates compatible with the fact that $S$ is on $E_i$ (a single state), which is the the same as the number of states in the reservoir consistent with it, that is all those states in which the reservoir is in a state of energy $(E-E_i)$ (that is, the total energy less the energy of $S$).