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Homework Help: Statistical mechanics - microstates & entropy

  1. Feb 18, 2010 #1
    1. The problem statement, all variables and given/known data
    a) Derive an asymptotic expression for the number of ways in which a given energy E can be distributed among a set of N, one-dimensional harmonic oscillators, the energy eigenvalues of the oscillators being [itex](N+\frac{1}{2})\hbar\omega, n=0, 1, 2, ...[/itex].
    b)Find the corresponding expression for the "volume" of the relevant region of the phase space of the system. Establish the correspondence between the two results, showing that the conversion factor [itex]\omega_0[/itex] is precisely [itex]h^N[/itex].
    c) In addition, derive the exact number of ways in which the energy can be divided among the N oscillators (before taking the asymptotic limit). Calculate the entropy for this system in the thermodynamic limit. Express the internal energy E in terms of the temperature T and oscillator number N, and plot this function versus T. Also, plot the heat capacity versus temperature. Find the asymptotic limits of the energy for large and small temperature.

    2. Relevant equations
    [itex]S=k_Bln\Omega[/itex] (Entropy)
    [itex]lnN!\approx NlnN-N[/itex] (Stirling's Approximation)

    3. The attempt at a solution
    So for a given energy E, [itex]\frac{E}{\hbar\omega}=k[/itex], where k is a positive integer. Let [itex]m_i[/itex] denote the number of oscillators whose energy (per unit [itex]\hbar\omega[/itex]) is equal to i. Obviously then we know
    [tex]\sum_{i=1}^k m_i=N[/tex]

    Now the number of possible arrangements is the total number of ways we can divide N oscillators into k groups of sizes [itex]m_1, m_2, ..., m_k[/itex]. Then the number of ways to get a group of size [itex]m_1[/itex] is just
    [tex]\binom{N}{m_1}[/tex]

    Similarly, we want [itex]m_2[/itex] oscillators of the remaining [itex]N-m_1[/itex] oscillators, which is just
    [tex]\binom{N-m_1}{m_2}[/tex]

    Continuing this line of thought, the total number of arrangements should be
    [tex]\Omega=\binom{N}{m_1}\binom{N-m_1}{m_2}...\binom{N-m_1-m_2-...-m_{k-1}}{m_K}=\frac{N!}{m_1!m_2!...m_k!}[/tex]

    This would then be the exact expression that is wanted in part c, right? I can get the entropy from this by using Stirling's approximation, so
    [tex]S=k_B(NlnN-N-\sum_{i=1}^k(m_ilnm_i-m_i))=k_B(NlnN-\sum_{i=1}^k(m_ilnm_i))[/tex]

    Intuitively, I think the m's should go like
    [tex]m_i=e^{-\frac{E_i}{k_BT}}[/tex]

    So when we sum over all i we get
    [tex]N=\sum_{i=1}^ke^{-\frac{E_i}{k_BT}}[/tex]

    That's where I get stuck, I have a relation between N, T and [itex]E_i[/itex], but I can't figure out how to turn it into E=f(N,T). Also, I did the problem backwards finding the exact expression for the number of arrangements first (assuming it's right). Can I simply turn the exact expression into an asymptotic form, or is there a way to derive it without knowing the exact form?
     
  2. jcsd
  3. Feb 18, 2010 #2
    This does not seem right to me. How many oscillators can have energy k?
     
  4. Feb 19, 2010 #3
    The oscillators can have energy in amounts of [itex]\hbar\omega, 2\hbar\omega, ... k\hbar\omega[/itex], the only constraint being that the total energy of all oscillators is E. [itex]m_1[/itex] is counting the number or oscillators with energy [itex]\hbar\omega[/itex], similarly for the other m's.

    For example say we had 6 oscillators and [itex]E=11\hbar\omega[/itex], then one distribution of energy might be {1, 1, 2, 5, 1, 1} and in this case we have [itex]m_1=4, m_2=1, m_5=1, m_i=0 (i=3, 4, 6, 7, 8, 9, 10, 11)[/itex]. Summing all the m's gives back 6, the total number of oscillators. I believe this is true in general as each of the N oscillators can only be in one of the [itex]m_i[/itex] subsets, so summing over the number of oscillators in each subset should return the total number or oscillators. The only problem is this might be complicated by the fact I realized E can have half integer values for odd N.

    The only other thought I had was we know
    [tex]E=\sum_{i=1}^N\hbar\omega(n_i+\frac{1}{2})=\hbar\omega\left(\frac{N}{2}+\sum_{i=1}^Nn_i\right)[/tex]

    So the number of ways to distribute the energy between the oscillators is equal to the number of non-negative integer solutions for the expression
    [tex]\sum_{i=1}^Nn_i=k, k=\frac{E}{\hbar\omega}-\frac{N}{2}[/tex]

    After some searching on google, I found a formula for the number of solutions that can be proven by induction (k will always be an integer as E will be half integer for odd N and integer for even N),
    [tex]\Omega=\binom{k+N-1}{k}=\binom{\frac{E}{\hbar\omega}+\frac{N}{2}-1}{\frac{E}{\hbar\omega}-\frac{N}{2}}=\frac{\left(\frac{E}{\hbar\omega}+\frac{N}{2}-1\right)!}{\left(\frac{E}{\hbar\omega}-\frac{N}{2}\right)!\left(N-1\right)!}[/tex]

    If this is correct, the entropy can be found for large N by using Stirling's approximation, but how would you get an asymptotic form for just [itex]\Omega[/itex] from the above expression?
     
  5. Feb 19, 2010 #4
    Can't some of the oscillators have 0 energy? Maybe I'm wrong about this, but I would assume that when E=0 all the oscillators have [tex]1/2\hbar\omega[/tex] energy since they can never have zero energy. In that case, you wouldn't have to worry about the half integer multiples since energy units come in integer multiples.

    But now that I think of it your summation would work with that change as long as you summed starting with i=0 instead of i=1. Not sure if that helps...
     
  6. Feb 19, 2010 #5

    gabbagabbahey

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    Homework Helper
    Gold Member

    No, the ground state energy of a 1D harmonic oscillator is [itex]\frac{\hbar\omega}{2}[/itex], the oscillators can't have an energy less than their ground state energy.

    You are missing all the half-integer multiples here. This, combined with the fact that [itex]E[/itex] must a half-integer multiple of [itex]\hbar\omega[/itex] for odd [itex]N[/itex], is responsible for the failure of your original method.

    This looks correct to me. To find an asymptotic expression for [itex]\Omega[/itex], try taking the natural logarithm of both sides of your equation and calculating [tex]\lim_{N\to\infty}\ln(\Omega)[/tex] first.
     
  7. Feb 19, 2010 #6
    Ah right, what I meant was to treat the lowest possible energy of the system as having zero energy, in which case
    would be correct, right? Either way, the OP can ignore me since it sounds like you have a better grasp of the problem than I do.
     
  8. Feb 19, 2010 #7
    So if we apply Stirling's approximation,
    [tex]ln\Omega=\left(\frac{E}{\hbar\omega}+\frac{N}{2}-1\right)ln\left(\frac{E}{\hbar\omega}+\frac{N}{2}-1\right)-\left(\frac{E}{\hbar\omega}+\frac{N}{2}-1\right)-\left(\frac{E}{\hbar\omega}-\frac{N}{2}\right)ln\left(\frac{E}{\hbar\omega}-\frac{N}{2}\right)+\left(\frac{E}{\hbar\omega}-\frac{N}{2}\right)-\left(N-1\right)ln\left(N-1\right)+\left(N-1\right)[/tex]
    [tex]ln\Omega=\left(\frac{E}{\hbar\omega}+\frac{N}{2}-1\right)ln\left(\frac{E}{\hbar\omega}+\frac{N}{2}-1\right)-\left(\frac{E}{\hbar\omega}-\frac{N}{2}\right)ln\left(\frac{E}{\hbar\omega}-\frac{N}{2}\right)-\left(N-1\right)ln\left(N-1\right)[/tex]
    [tex]ln\Omega=ln\left(\frac{(k+N-1)^{(k+N-1)}}{k^k(N-1)^{(N-1)}}\right)[/tex]
    [tex]\Omega=\frac{(k+N-1)^{(k+N-1)}}{k^k(N-1)^{(N-1)}}=\frac{\left(\frac{E}{\hbar\omega}+\frac{N}{2}-1\right)^{\left(\frac{E}{\hbar\omega}+\frac{N}{2}-1\right)}}{\left(\frac{E}{\hbar\omega}-\frac{N}{2}\right)^{\left(\frac{E}{\hbar\omega}-\frac{N}{2}\right)}(N-1)^{(N-1)}}[/tex]

    These expressions are pretty unweildly, but I think some simplification can be done,
    [tex]\Omega=\left(\frac{\frac{E}{\hbar\omega}+\frac{N}{2}-1}{\frac{E}{\hbar\omega}-\frac{N}{2}}\right)^{\frac{E}{\hbar\omega}-\frac{N}{2}}\left(\frac{\frac{E}{\hbar\omega}+\frac{N}{2}-1}{N-1}\right)^{N-1}=\left(1-\frac{N-1}{\frac{E}{\hbar\omega}-\frac{N}{2}}\right)^{\frac{E}{\hbar\omega}-\frac{N}{2}}\left(\frac{\frac{E}{\hbar\omega}-\frac{N}{2}}{N-1}-1\right)^{N-1}[/tex]

    I want to try and simplify further, as taking derivatives of S to find T looks to be quite tedious in this form, but it appears to be in it's most simple form. This will serve as the asymptotic form for part a then, right? So,
    [tex]\Omega=\frac{\left(\frac{E}{\hbar\omega}+\frac{N}{2}-1\right)!}{\left(\frac{E}{\hbar\omega}-\frac{N}{2}\right)!\left(N-1\right)!}[/tex]

    should be the exact number of ways to distribute energy as requested in part c?
     
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