Understanding ensembles, micro/macrostates, equilibrium, reversibility and more

1. Jul 5, 2012

shockingpants

Hi,

I just completed a semester of undergrad course in thermodynamics/stat mech and while I find it somewhat easy to simply dump the right equations in different situation to get the right answer, its the fundamentals that has continued to baffle me. I shall be as concise as I can and hopefully, someone will be able to shed light on my queries.

Please correct anything I state if its unclear or just plain wrong!

This is my understanding of a microcanonical ensemble. It is an ensemble where every accessible state of an isolated system has a total energy (between E and E+ΔE). I shall define these states to be micro states as well. By the ergodic hypothesis, given enough time and/or systems, each of these states will have an equal chance of being sampled. In fact, this only applies at equilibrium since different microstates share the same probability only at equilibrium. (AKA, the equal a priori probability postulate only holds at equilibrium) If I assume that the number of states do not depend on any other parameter, then the total number of states Ω$_{tot}$ does not change, i.e. entropy of the system is constant.

If this is right thus far, I shall move on to talk about the canonical system, where my questions really lie.

Last edited: Jul 5, 2012
2. Jul 5, 2012

shockingpants

Suppose we have 2 systems of similar degrees of freedom/number of molecules that can interact with one another through exchange of energy only (the energy exchanged is really heat Q). They are isolated from the rest of the world and thus E$_{tot}$ = E$_{R}$+E$_{S}$ is constant. The total number of states is equal to Ω$_{tot}$=Ω$_{R}$Ω$_{S}$, which is dependent on E$_{S}$. Suppose we maximize Ω$_{tot}$ such that dΩ$_{tot}$/dE = 0, the resulting equality will lead us to the definition of temperature. In a special case where system R is a heat reservoir or water bath, its dlnΩ$_{R}$/dE$_{R}$ aka $\frac{1}{k_{B}T}$ is constant.

My first question is really on Ω$_{tot}$. I have defined Ω$_{tot}$ as the number of stats when E$_{S}$ is equal to a specific value i.e. Ω$_{tot}$ = Ω$_{tot}$(E$_{S}$), which changes as E$_{S}$ changes. However, if we sum (assume quantum treatment, hence discrete energy packs) Ω$_{tot}$(E$_{S}$) over all possible E$_{S}$, we get Ω$_{sum}$ = $\sumΩ_{tot}(E_{S})$.

So we have 4 types of no. of states
1. Ω$_{S}$(E$_{S}$)
2. Ω$_{R}$(E$_{tot}$-E$_{S}$)
3. Ω$_{tot}$(E$_{S}$)=Ω$_{R}$Ω$_{S}$
4. Ω$_{sum}$ = $\sumΩ_{tot}(E_{S})$

Is the canonical ensemble essentially an ensemble consisting of every single state in Ω$_{sum}$?

If I wanted to calculate a probability of a system S in a macrostate E$_{S}$, is it simply just using $\frac{Ω_{tot}(E_S)}{Ω_{sum}}$?

If I rewrote (and taylor expanded) the numerator of $\frac{Ω_{R}(E_{tot}-E_S)\cdotΩ_S(E_S)}{Ω_{sum}}$ to get $\left(Ω_{R}(E_{tot})-\frac{dΩ_R}{dE_R}E_S\right)$g$_{s}(E_S)$ and performed some manipulations, I would end up with the boltzman distribution right?

In this case, what is equilibrium? Is it a state when every single microstate in Ω$_{sum}$ has an equal probability of being sampled, or is it a state when Ω$_{tot}$(E$_{S}$) is at a maximum?

Last edited: Jul 5, 2012