0
$\begingroup$

I have checked several references for the derivation of the probability function of the canonical ensemble. I have seen two (essentially similar) approaches. Both assume a system is placed in a large reservoir:

  1. Study the probability that the system is in a given microstate. Assert that this probability is proportional to $\Omega_S$ where $\Omega_S$ is the multiplicity of the system. For example, https://www2.oberlin.edu/physics/dstyer/StatMech/CanonicalEnsemble.pdf.
  2. Study the probability that the reservoir is in a given microstate. Assert that this probability is proportonal to $\Omega_R$ where $\Omega_R$ is the multiplicity of the reservoir.

Both of these seem unsatisfactory to me. In both cases it seems like the probability should be proportional to $\Omega_R \cdot \Omega_S.$ Why is it okay to neglect this multiplication?

Improve this question
$\endgroup$
6
  • $\begingroup$ The following lecture notes may be of interest to you on this topic. While this is framed in the context of thermodynamics and not statistical mechanics, the derivation ultimately gives the Boltzmann factor, whose normalization constant is the canonical partition function. physics.byu.edu/faculty/colton/docs/phy123-winter11/… $\endgroup$
    – Matt Hanson
    Aug 9 at 5:24
  • $\begingroup$ @MattHanson Thank you. The lecture notes you linked to make the same assumption that I am questioning. Why should the probability that the system is in a certain state be proportional to the entropy of that state, rather than the entropy of the entire universe? $\endgroup$
    – Jbag1212
    Aug 9 at 14:32
  • $\begingroup$ I think I see your problem. Basically, if all of the states are the same energy (isolated system, so that is by definition true), then the only thing that can govern the probability of occupying a given state is how many of them there are: the multiplicity. I believe that justifies the assumption. $\endgroup$
    – Matt Hanson
    Aug 9 at 14:47
  • $\begingroup$ @MattHanson Hello, true - but in the canonical ensemble the system is not isolated - it is in thermal equilibrium with a reservoir and free to exchange energy. $\endgroup$
    – Jbag1212
    Aug 9 at 14:56
  • 1
    $\begingroup$ @MattHanson I realize now I was being silly and misreading what had been asserted. The arguments had been asserting that $P_s \propto \Omega_R$ where $P_s$ is the probability of observing the system in a particular microstate and $\Omega_R$ is the entropy of the reservoir. I was misreading it as $P_s \propto \Omega_s$ or $P_R \propto \Omega_R$ $\endgroup$
    – Jbag1212
    Aug 9 at 21:14

1 Answer 1

Reset to default
1
$\begingroup$

You're probably misunderstanding the oberlin document. They consider the class of states where system is in definite microstate, so $\Omega_S=1$. Then $\Omega_S\Omega_R=\Omega_R$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Hmm interesting. But why should I care about that class of systems? If I want to know the probability of the reservoir having energy $E$, shouldn’t I consider the macrostates rather than microstates of the system? $\endgroup$
    – Jbag1212
    Aug 9 at 12:32
  • $\begingroup$ Or to put it another way, I agree that the conditional probability $P_r (\text{system has energy E}| \text{reservoir is in fixed microstate})$ is proportional to the entropy, but we are not interested in this conditional probability. $\endgroup$
    – Jbag1212
    Aug 9 at 14:33
  • $\begingroup$ Nevermind, I was just misreading. I thought they were asserting $P_S \propto \Omega_S$ or $P_R \propto \Omega_R$, but they were really asserting $P_S \propto \Omega_R$ for a particular microstate. $\endgroup$
    – Jbag1212
    Aug 9 at 19:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.