Einstein Solid and Sterling's Approximation

  1. 1. The problem statement, all variables and given/known data
    Show that the multiplicity of an Einstein solid with large N and q is

    [tex]\frac{\left(\frac{q+N}{q}\right)^q\left(\frac{q+N}{N}\right)^N}{\sqrt{2\pi q\left(q+N\right)/N}}[/tex]


    2. Relevant equations
    [tex]N! \approx N^N e^{-N} \sqrt{2 \pi N}[/tex]



    3. The attempt at a solution
    Well, I've done thus so far:

    [tex]
    \Omega(N,q) = \frac{(q+N-1)!}{q!(N-1)!} \approx \frac{(q+N)!}{q!N!}

    ln(\Omega) = ln(q+N)! - lnq! - lnN
    \par
    \approx (q+N)ln(q+N) - (q+N) - qlnq+q - NlnN + N = (q+N)ln(q+N) - qlnq - NlnN

    [/tex]

    I feel like I'm close, but I've no idea where to go from here.
     
  2. jcsd
  3. How silly of me! I just expanded out some terms and now I have the numerator, but where on Earth does the denominator come from? Should I have another -ln() term somewhere, so I can use Sterling?
     
  4. Ok, so after expanding:
    ln(q+N)!-lnq!-lnN!
    and canceling a coupel N's and q's I get:

    (q+N)ln(q+N)-qlnq-NlnN

    So I applied a few ln rules to get:

    [tex]ln(q+N)^{q+N)}[/tex]-[tex]lnq^{q}[/tex]-[tex]Nln^{N}[/tex]

    Then simplifying:

    ln([tex](q+N)^{(q+N)}/q^{q}[/tex]-[tex]lnN^{N}[/tex]

    But when I try to simplify again I come up with:

    ln([tex](q+N)^{(q+N)}N^{N}/q^{q}[/tex] - [tex]lnN^{N}[/tex]

    Which I don't believe is right, but even if it was, how do I go about recovering the 2pi n the denominator?
     
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