Question

Use Stirling's approximation to show that the multiplicity of an Einstein solid, for any large values of$N$andlocalid="1650383388983" $q$,is approximately

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

The square root in the denominator is merely large, and can often be neglected. However, it is needed in Problem$2.22$. (Hint: First show that$\Omega =\frac{N}{q+N}\frac{(q+N)!}{q!N!}$. Do not neglect the$\sqrt{2\pi N}$in Stirling's approximation.)

Short answer

The square root and the large values are neglect

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

Step-by-step solution

Macroscopic solid

For any macroscopic solid, both $q$and $N$are large numbers (on the order of Avogadro's number, or${10}^{23}$) so the factorials in $\Omega$are very large numbers, not calculable on most computers. To get estimates of $\Omega$we can use Stirling's approximation for the factorials. The multiplicity is given by:

$\Omega (N,q)=\left(\begin{array}{c}q+N-1\\ q\end{array}\right)$

Writing out the binomial coefficient:

$\left(\begin{array}{c}q+N-1\\ q\end{array}\right)=\frac{(q+N-1)!}{q!(N-1)!}$

we can write the factorial $(q+N-1)$! in the following form:

$(q+N-1)!=(q+N-1)!\times \frac{q+N}{q+N}$

we can write the factorial $(q+N-1)$! in the following form:

$(q+N-1)!=(q+N-1)!\times \frac{q+N}{q+N}$

but,$(q+N-1)!(q+N)=(q+N)$so

$(q+N-1)!=\frac{(q+N)!}{(q+N)}$

and also we can write the factorial $(N-1)$! in the following form:

$(N-1)!=(N-1)!\times \frac{N}{N}=\frac{N!}{N}$

Assume that $q$and $N$are large numbers, so:

Substitution 

substitute from (2) and (3) into (1), so:

$\Omega (N,q)=\left(\begin{array}{c}q+N-1\\ q\end{array}\right)=\frac{N}{(q+N)}\frac{(q+N)!}{N!q!}$

Assume that $q$and $N$are large numbers, so we can use Stirling's approximation for the factorials:

$n!\approx \sqrt{2\pi n}{n}^n{e}^{-n}$

so we get:

$$\begin{gathered} N!\approx \sqrt{2\pi N}{N}^N{e}^{-N} \\ q!\approx \sqrt{2\pi q}{q}^q{e}^{-q} \\ (q+N)!\approx \sqrt{2\pi (q+N)}(q+N{)}^{(q+N)}{e}^{-(q+N)} \end{gathered}$$

substitute from $\left(5\right),\left(6\right)$and $\left(7\right)$into $\left(4\right)$, so:

$\Omega (N,q)\approx \frac{N}{(q+N)}\frac{\sqrt{2\pi (q+N)}(q+N{)}^{(q+N)}{e}^{-(q+N)}}{\left(\sqrt{2\pi N}{N}^N{e}^{-N}\right)\left(\sqrt{2\pi q}{q}^q{e}^{-q}\right)}$

cancel $e^\{-(q+N\left)\right\}withe^\{-q\}e^\{-N\}$so:

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

$\to \Omega (N,q)\approx \frac{N\sqrt{2\pi (q+N)}}{(q+N)\left(\sqrt{2\pi q}\right)\left(\sqrt{2\pi N}\right)}\frac{(q+N{)}^{(q+N)}}{N^N{q}^q}$

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

Substitution 

but, $(q+N)^\left\{\right(q+N\left)\right\}$=$(q+N)^\left\{q\right\}(q+N)^\left\{N\right\}$, so:

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

$\to \Omega (N,q)\approx \sqrt{\frac{N}{2\pi q(q+N)}}{\left(\frac{q+N}{q}\right)}^q{\left(\frac{q+N}{N}\right)}^N$

$\to \Omega (N,q)\approx \frac{{\left(\frac{q+N}{q}\right)}^q{\left(\frac{q+N}{N}\right)}^N}{\sqrt{\frac{2\pi q(q+N)}{N}}}$