Einstein Solid and Sterling's Approximation

[Ai52487963]

Homework Statement

Show that the multiplicity of an Einstein solid with large N and q is

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

Homework Equations

N! \approx N^N e^{-N} \sqrt{2 \pi N}

The Attempt at a Solution

Well, I've done thus so far:

<br /> \Omega(N,q) = \frac{(q+N-1)!}{q!(N-1)!} \approx \frac{(q+N)!}{q!N!}<br /> <br /> ln(\Omega) = ln(q+N)! - lnq! - lnN <br /> \par<br /> \approx (q+N)ln(q+N) - (q+N) - qlnq+q - NlnN + N = (q+N)ln(q+N) - qlnq - NlnN<br /> <br />

I feel like I'm close, but I've no idea where to go from here.

 

[Ai52487963]

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?

 

[LmidgitD]

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:

ln(q+N)^{q+N)}-lnq^{q}-Nln^{N}

Then simplifying:

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

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

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

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