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Einstein solid, Sterling approximation

  1. Jan 24, 2008 #1

    nicksauce

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    1. The problem statement, all variables and given/known data
    Use Sterling's approximation to show that the multiplicity of an Einstein solid, for any large values of N and q is approximately
    [tex]\Omega(N,q) = \frac{(\frac{q+N}{q})^q(\frac{q+N}{N})^N}{\sqrt{2\pi q(q+N)/N}}[/tex]


    2. Relevant equations
    [tex]\Omega(N,q) = \frac{(N+q-1)!}{q!(N-1)!}[/tex]
    [tex]\ln(x!) \simeq x\ln(x) - x[/tex]


    3. The attempt at a solution
    I see where the terms in the numerator come from, but I cannot see where the terms in the denominator come from. Specifically, the squareroot, and the factor of 2 pi*q. When I grind out the math, I get that the denominator should be (q+N)/N. Help anyone?
     
    Last edited: Jan 24, 2008
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  3. Jan 24, 2008 #2

    nicksauce

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    Solved it.... Turns out I had to use the other Sterling's approximation

    [tex]N! = N^Ne^{-N}\sqrt{2\pi N}[/tex]
     
  4. Jan 24, 2008 #3
    Sterling' s approximation is

    [tex]n! \approx \sqrt{2\,\pi\,n}\left(\frac{n}{e}\right)^n[/tex]

    To get your formula fist get rid of 1 at [tex](N+q-1)!,\,(N-1)![/tex]

    I think that in the formula

    [tex]\Omega(N,q) = \frac{(\frac{q+N}{q})^q(\frac{q+N}{N})^N}{\sqrt{2\pi q(q+N)/N}}[/tex]

    the denominator must be [tex]\sqrt{2\,\pi\,q\,N/(q+N)}[/tex]
     
  5. Jan 24, 2008 #4
    :smile: You solved it while I was typing!
     
  6. Jan 24, 2008 #5

    nicksauce

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    Thanks... it's the thought that counts anyway!
     
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