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A The problematic 1/N! in canonical partition function

  1. Jan 4, 2017 #1

    ShayanJ

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    The canonical partition function in classical statistical mechanics is calculated by ## Q_N(V,T)=\frac 1 {N! h^{3N}}\int e^{-\beta H(\mathbf q,\mathbf p)}d^{3N}q \ d^{3N}p ##. The ## \frac 1 {N!} ## is there to prevent the Gibbs paradox. But now consider a system of N particles that have no interaction with each other and only interact with external potentials. This means that their canonical partition function can be written as ## Q_N(V,T)=\frac{[Q_1(V,T)]^N}{N! h^{3N}} ## where ## Q_1(V,T)=\int e^{-\beta H(q,p)}d^3 q \ d^3p ##. So for ## N=N_1+N_2 ## particles of the same type, I'll have ## Q_{N_1+N_2}(V,T)=\frac{[Q_1(V,T)]^{N_1+N_2}}{(N_1+N_2)! h^{3(N_1+N_2)}} ##.
    Now suppose there are two types of particles differing in a parameter which appears in their Hamiltonians. So for a system consisting of ## N_1 ## particles of the first kind and ## N_2 ## particles of the second kind, the partition function is ## Q^{(1)}_{N_1}(V,T)Q^{(2)}_{N_2}(V,T)=\frac{[Q^{(1)}_1(V,T)]^{N_1}[Q^{(2)}_1(V,T)]^{N_2}}{N_1!N_2! h^{3(N_1+N_2)}} ##. Now if I take the limit that the parameter is the same for the two kinds of particles, I expect to get ##Q_{N_1+N_2}(V,T)##, but instead, I get ##Q^{(1)}_{N_1}(V,T)Q^{(2\to 1)}_{N_2}(V,T)=\frac{[Q^{(1)}_1(V,T)]^{N_1}[Q^{(2\to 1)}_1(V,T)]^{N_2}}{N_1!N_2! h^{3(N_1+N_2)}} ## which is different from what I expect because instead of ## (N_1+N_2)! ## in the denominator, it has ## N_1!N_2! ##. It seems that the factor introduced to solve the Gibbs paradox, is itself causing a paradox. What is wrong here? What is it that I don't see?
    Thanks
     
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  3. Jan 4, 2017 #2

    TeethWhitener

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    I'm pretty sure this is your problem right here. The particles are only indistinguishable when the parameters are identical. You can't continuously tune into or out of indistinguishability. So if the first kind of particle is indexed by a parameter ##a## like so: ##Q^{(a)}##, and the second kind of particle is indexed like so: ##Q^{(b)}##, you can't continuously walk ##b## into ##a##. In other words, the condition for indistinguishability (where the additivity condition ##(N_1+N_2)!## would apply) is ##\delta_{ab}##.
     
  4. Jan 4, 2017 #3

    ShayanJ

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    Interesting point. But now consider a system whose behavior can be described using a model with free electrons and some free quasi-particles with electric charge equal to -e and effective mass that can be tuned using e.g. an applied electric field. You're saying that if I tune the mass of the quasi-particles to be exactly equal to the mass of the electrons, these quasi-particles can still be distinguished from the electrons? Is there such a model?
     
  5. Jan 4, 2017 #4

    TeethWhitener

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    In this hypothetical scenario, if the only observables were mass and charge, and a quasiparticle had the same mass and charge as an electron, then I don't think they'd be distinguishable.
     
  6. Jan 4, 2017 #5

    ShayanJ

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    So just imagine my first post is about such a model and I'm tuning the mass of the quasi-particles to be equal to the mass of the electrons and there is no other difference between them. So now you also seem to agree that I should get ## Q_{N_1+N_2}(V,T) ## in the limit. And we're back to the problem!
     
  7. Jan 4, 2017 #6

    TeethWhitener

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    Yeah that's a pickle. I'll have to think a little harder about it. Maybe they are distinguishable. If you're able to tune the mass, then wouldn't that qualify as a measurement?
    Edit: I'm not sure how much of this is applicable to classical statistical mechanics.
     
  8. Jan 5, 2017 #7

    ShayanJ

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    Well, people do such things in labs all the time and don't have to deal with measurements. So I guess its not. But I also somehow understand why its not a measurement.
    Good point. Maybe classical statistical mechanics is simply not able to take into account indistinguishability satisfactorily.
     
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