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I The partition function

  1. Jun 10, 2016 #1
    I was reading about partition function. I noticed that there are two approaches toward partition function.

    The first approach:
    Suppose we are dealing with a closed system where the system is composed of heat bath (R) and inside it there is a very small system (E), the two systems are in thermal equilibrium with temperature T.
    This is called the canonical partition function. I will not write the derivation but the canonical partition function obtained is [itex]Z=\sum_{l=0}^{Ω_{s}} e^{-E_{l}/k_{B}T}[/itex] Where [itex]E_{l}[/itex] is the energy of the system (E) and [itex]Ω_{s}[/itex] is the total number of microstates for system (E).
    [itex]p(E_l)=e^{-E_l/k_BT}/Z[/itex]
    The system (E) +(R) is isolated.
    To emphasis here the system is composed of a heat reservoir (R), which is a huge system and a very small system (E).
    The second approach:
    we have here a big isolated system (G), the number of microstates at thermal equilibrium is approximately equal to t{nj*}, where t{nj*} is the number of microstates corresponding to the distribution{nj*} which is the average distribution. The average distribution is the thermal distribution.
    So [itex]Ω\approx t{n_j*}[/itex] ( when N the number of particles is very large). This means that the thermal distribution is the most probable distribution.

    We have that t{nj}=N!/nj! then ln(t)=lnN!-∑ln(nj!)⇒lnt=(NlnN-N)-∑(njln(nj)-nj) Note : nj is the number of particles in the state j.

    The most probable distribution correspond to the maximum t, which satisfies the equation d(lnt )=-∑dnjln(nj)=0.
    As N and U are fixed then d(N)=∑d(nj)=0 and d(U)=∑εjdnj=0 where εj is the energy corresponding to the state j.

    Using Lagrange multiplier method, we will get ∑(-ln(nj*)+α+βεj)dnj=0 for any values of α and β.
    then -ln(nj*)+α+βεj=0 then"" nj*=exp(α+βεj)"" which is the boltzmann distribution.
    the partition function is defined here by Z=∑exp(βεj) where β=-1/kT where k is the boltzmann constant.


    Now lets divide the isolated system (G) into two systems "p" and "q".
    where the system "p" has energy Up and number of particles Np,and "q" has energy Uq and number of particles Nq.
    At thermal equilibrium, we have for system "p" nj=exp(αp+βεj) and for system "q" nj'=exp(αq+βεj') Where nj and nj' are the number of particles that have energies εj and εj' respectively at thermal equilibrium.

    For system"p" Zp=Σexp(βεj) and for "q" Zq=∑exp(βεj').

    My question: why the partition function is defined differently in the second case( where I divided the isolated system into two systems p and q),as εj here does not represent the energy of the system but the energy of specific particles in the system ? Is it because it is not considered an canonical ensemble as there is no heat bath?
    How the partition function can be different from case to case and why?Thanks!
     
  2. jcsd
  3. Jun 11, 2016 #2
    I think the second approach you mention is the boltzmann-distribution function for the energies of individual classical particles. (as in the Maxwell-Boltzmann distribution for velocities)

    Because particles are indistinguishable according to QM, you need an ensemble description for statistical physics, which means you are looking at the distribution of the whole system. This is the first approach, commonly called the Gibbs distribution.
     
  4. Jun 11, 2016 #3
    and it is the partition function I mentioned in the next approach. Can we name it the molecular partition function?
    So for the system "p" in the next approach, the partition function Z is the partition function of a single particle. As "p" contains Np particles, can we obtain the partition function of the whole system by multiplying the partition functions, such that Ztotal=Z^N?If No, is it because the system here is not canonical?
     
  5. Jun 11, 2016 #4
    You can for particles that are considered distinguishable, for indistinguishable particles Ztotal=Z^N/(N!)
     
  6. Jun 11, 2016 #5
    Ok.
    Suppose that the particles in system "p" are distinguishable, does Ztotal here is the same of the Z that is related to the Gibbs approach?
     
  7. Jun 11, 2016 #6
    Yes, the Gibbs approach works everytime if you use the proper set of states. If there are two particles that can either be in state A or B, the ##\Omega_s## contains four elements in the distinguishable case (AA, AB, BA, BB) and three in the indistinguishable case (AA,AB=BA,BB). Other constraints on the allowed states such as pauli exclusion, can be added as well.
     
  8. Jun 12, 2016 #7
    But the system I mentioned in the original post("p" and"q") can't be treated in canonical ensemble, since there is no heat bath, the systems "p" and "q" have the same dimensions, so I think the partition function related to "p" is not canonical.
    So ( I think) that's why we go to the molecular partition function, where we treated every particle by its own(correct me if I'm wrong). So how comes in this case that Ztotal^N =Z(canonical)?
     
  9. Jun 12, 2016 #8
    Might be, thermodynamics of small systems is always a bit tricky. In the limit of big systems, canonical, microcanonical and grandcanonical become equivalent.
     
  10. Jun 12, 2016 #9
    The problem in statistical physics is that the concepts are missed up together.
    There is no a clear method where we can use canonical ensemble and where we can't.
    Some texts books emphasize that a heat bath must exist to consider canonical ensemble, while others deals with systems that do not exchange energy with a heat bath,but with normal systems, and although treat it as an canonical ensemble!
    I'm really confused!
     
  11. Jun 12, 2016 #10
    I would say that the proper choice of the ensemble is rather a technical detail, it doesn't really matter for viewing the big picture, so you shouldn't focus too much on it.
     
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