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-   -   Einstein Solid and Sterling's Approximation (http://www.physicsforums.com/showthread.php?t=300500)

 Ai52487963 Mar17-09 05:24 PM

Einstein Solid and Sterling's Approximation

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

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

2. Relevant equations
$$N! \approx N^N e^{-N} \sqrt{2 \pi N}$$

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

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

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

 Ai52487963 Mar18-09 01:49 AM

Re: Einstein Solid and Sterling's Approximation

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 Sep13-09 01:12 PM

Re: Einstein Solid and Sterling's Approximation

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?

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