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Simplifying entropy for a harmonic oscillator in the limit of large N

  1. Feb 20, 2014 #1
    1. The problem statement, all variables and given/known data
    Hey guys,

    So I have this equation for the entropy of a classical harmonic oscillator:

    [itex]\frac{S}{k}=N[\frac{Tf'(T)}{f(T)}-\log z]-\log (1-zf(T))[/itex]

    where [itex]z=e^{\frac{\mu}{kT}}[/itex] is the fugacity, and [itex]f(T)=\frac{kT}{\hbar \omega}[/itex].

    I have to show that, "in the limit of large N, this entropy becomes the following":

    [itex]\frac{S}{k}=N[1+\log(\frac{kT}{\hbar \omega})]=N[1+\log f(T)][/itex]


    2. Relevant equations

    None that I know of


    3. The attempt at a solution

    So all ive done is plugged in the expression for f(T) and f'(T) into the entropy, to get this:

    [itex]\frac{S}{k}=N(1-\log z)-\log (1-z\frac{kT}{\hbar \omega})[/itex]

    But i dont know what to do when N becomes large....
     
  2. jcsd
  3. Feb 20, 2014 #2
    Hi again...
    So first, observe that from the summation form of Z (in your previous post):
    N = z[itex]\frac{\partial}{\partial z}[/itex]logZ
    Use that with the explicit form of Z you found there, take the limit N>>1 and see what happens...
    Also, in this limit what should you do with the terms that are not multiplied by N?
     
  4. Feb 21, 2014 #3
    Hey!

    Yea you get [itex]N=\frac{zf(T)}{1-zf(T)}[/itex]. If this is to become much greater than 1, then [itex]zf(T) \rightarrow 1[/itex]. And anything thats not multiplied with N can take a hike! but i tried this, i dont get very far...but just to confirm, we only vary T, right? since we held [itex]\mu , V[/itex] constant when we found the entropy...?
     
    Last edited: Feb 21, 2014
  5. Feb 21, 2014 #4
    You have all you need to express the solution. You found that:
    S/k = N(1 – logz) – log(1 –zf)
    But then in the limit N >>1, you also found that the second term can be discarded and for the first:
    zf = 1 (or in other terms z = 1/f)
    Just put it together!
     
  6. Feb 21, 2014 #5
    Okay let me be a bit more specific with what I did.

    Start with

    [itex]\frac{S}{k}= N[\frac{Tf'(T)}{f(T)}-\log z]-\log (1-zf(T))[/itex]

    Using f(T) = kT/hbar ω, we get

    [itex]\frac{S}{k}=N[1-\log z]-\log (1-z\frac{kT}{\hbar \omega})[/itex]

    Clearly [itex]\log z = \frac{\mu}{kT}[/itex], but the problem is the expression [itex]-\log (1-zf(T))[/itex]....I tried to manipulate it as follows:

    [itex]-\log (1-zf(T))=\log (1-zf(T))^{-1}[/itex], then using a first order taylor expansion:

    [itex](1-zf(T))^{-1} \approx 1+zf(T)[/itex]

    Then the only way you can get a log(zf(T)) term is if you ignore the 1...so finally i get this

    [itex]N[1-\log z]+\log(1+zf(T))[/itex]...but im stuck!
     
  7. Feb 21, 2014 #6
    okay i'll try that...sorry i guess i was typing while you were typing!
     
  8. Feb 21, 2014 #7
    Yep i got it...thanks a bunch again!
     
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