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Canonical ensemble

  1. Aug 19, 2009 #1
    Is it true that partition function for canonical ensemble is

    [tex]Z=\sum_{\{states j\}}e^{-\beta E_j}=e^{-\beta F}[/tex]

    where [tex]F[/tex] is Helmholtz free energy?
     
  2. jcsd
  3. Aug 19, 2009 #2
    And in which case is

    [tex]Z={(\sum_{\{states of all particle\}}e^{-\beta E})}^N=[\frac{1}{h^3}\int d^3r\int d^3pe^{-\frac{\beta p^2}{2m}]^N[/tex]

    Because then it is easy to calculate


    [tex]\int d^3r=V[/tex]

    and


    [tex]\int d^3pe^{-\frac{\beta p^2}{2m}[/tex]

    is product of three Poisson integral. But when I can use this. Only for classical ideal gas?

    Can I say in every case that


    [tex]Z={(\sum_{\{states of all particle\}}e^{-\beta E})}^N=[\frac{1}{h^3}\int d^3r\int d^3pe^{-\beta E}]^N[/tex]
     
    Last edited: Aug 19, 2009
  4. Aug 22, 2009 #3
    No it's not true, the true form as I think is

    [tex]
    Z=\sum_{\{states j\}}e^{-\beta E_j}=e^{-\beta U}
    [/tex]

    where U is the internal energy of the system.

    For your second question, yes it's only for the ideal gas, because you have only energies P^2/2m, there are no other contributions like repulsion or collisions, the facts which are considered in Van der Waals equation of state. Your second question involves the microcanonical ensemble as I recall, not the canonical one, because there is not energy exchange for the system.

    For your last question, no you can't, because you have to take into consideration that the particles are indistinguishable, and so this lowers the number of configurations possible, and gives a rise to Gibbs paradox.
    The correct answer for your last question is to divide by N!. (N factorial) as well.

    I hope I answered your question, good luck
     
  5. Aug 22, 2009 #4
    Thanks! But I still have problems.
    Like if I take

    [tex]T=\sum_{ij} A_{ij}p_i p_j[/tex]

    kinetic energy like a quadratic form I will still get equation of ideal gas?

    Where [tex]\hat{A}[/tex] is symmetric matrix.
     
  6. Aug 22, 2009 #5
    I don't write [tex]N![/tex] with purpose. I have some problems with this [tex]N![/tex]. Why [tex]N![/tex] in classical statistical mechanics?
     
  7. Aug 22, 2009 #6
    In this form you're assuming there are interactions between particles i and j, therefore it's not ideal.

    And about N!, N! solves the Gibbs paradox for entropy continuity, the paradox happened when he put 2 gases in a box with a thick-less barrier. When removing the barrier, if the particles were identical, he found the entropy not changing, but if the particles are different on the sides of the barrier he found that the entropy increases, where the theory doesn't support that, and so N! solved the problem.

    The only prove of the importance of this N! comes with Quantum Mechanics. If this is not enough to convince you, imagine the following: you have N particles and N microstates, and you want to calculate the possible configurations for those particles on those states, assuming there is no degeneracy, meaning every state takes only 1 particle, we find the following:

    Classically: we assume that we can "name" the particles with numbers, the number of configurations is simply N!.

    Quantum mechanically: particles are indistinguishable, therefore exchanging particles would not change the macrostate (not microstate) and that's what we care about in Statistical mechanics, and so

    N=2 -> we have 2 configurations, meaning 2!, but exchanging won't change anything, and so we have 1 state resulting from 2!/2!.

    N=5 -> we have 5! configurations, but the same happens because all are the same, and so we need again to divide by 5! to again get the 1 state.

    And so we find that if we have N states, then we have only 1 configurations. This can be proven only with quantum mechanics, and so it happens to be a paradox before quantum mechanics, which Gibbs had solved before Quantum mechanics.

    Hope this answers :)
     
  8. Aug 22, 2009 #7
    Thanks! :D So I put [tex]\frac{1}{N!}[/tex] because entropy, Helmholtz free energy... need to be extensive quontities.
     
  9. Aug 22, 2009 #8
    Right :)
     
  10. Aug 22, 2009 #9
    Correct. TheDestroyer is wrong.
     
  11. Aug 22, 2009 #10
    Note that dividing by N! is only an approximate way of correcting overcounting. It becomes exact in the limit of a dilute gas.

    TheDestroyer's explanation is wrong, as it ignores that the classical partition function puts te paticles in all possible states, so it will count also configurations in wjhich mre than one particle is in a certain state and then you can't get a correct counting by dividing by N! The correct explanation is that if you have many more accesible states than particles, you can ignore configurations in which the particles are in the same state.
     
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