1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Einstein Solid and Sterling's Approximation

  1. Mar 17, 2009 #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. Mar 18, 2009 #2
    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. Sep 13, 2009 #3
    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?