# Einstein solid, Sterling approximation

1. Jan 24, 2008

### nicksauce

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

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

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
2. Jan 24, 2008

### nicksauce

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

$$N! = N^Ne^{-N}\sqrt{2\pi N}$$

3. Jan 24, 2008

### Rainbow Child

Sterling' s approximation is

$$n! \approx \sqrt{2\,\pi\,n}\left(\frac{n}{e}\right)^n$$

To get your formula fist get rid of 1 at $$(N+q-1)!,\,(N-1)!$$

I think that in the formula

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

the denominator must be $$\sqrt{2\,\pi\,q\,N/(q+N)}$$

4. Jan 24, 2008

### Rainbow Child

You solved it while I was typing!

5. Jan 24, 2008

### nicksauce

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