91Ó°ÊÓ

The number of microstates in an Einstein solid with $N$oscillators and $q$energy quanta is:

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

For any macroscopic solid, both $\mathrm{q}$and$\mathrm{N}$ are large numbers (on the order of Avogadro's number, orlocalid="1650382250464" ${10}^{23}$) so the factorials in $\mathrm{Î\copyright }$are very large numbers, not calculable on most computers. To get estimates of$\mathrm{Î\copyright }$we can use Stirling's approximation for the factorials. The derivation of this approximation for the high temperature case$q>>N$(lots more energy quanta than oscillators) is given in Schroeder's book, so at low temperature case we have$q<<N$(lots more oscillators than energy quanta). Writing out the binomial coefficient:

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

We can now take logarithms for both sides, we get,

$\ln \left(\mathrm{Î\copyright }\right)=\ln \left(\frac{(q+N)!}{q!N!}\right)$

but, $\ln \left(\frac{x}{y}\right)$=$\ln \left(x\right)-\ln \left(y\right)$and $\ln \left(x\right)-\ln \left(y\right)$=$\ln \left(x\right)+\ln \left(y\right)$, so:

$→\ln \left(\mathrm{Î\copyright }\right)$=$\backslash \ln \left[\right(q+N)!]-\backslash \ln (q!)-\backslash \ln (N!)$

use Stirling's approximation for the logarithm of a factorial:

$\ln (n!)≈nln\left(n\right)-n$

so equation (l) will become:

$$\begin{gathered} \left(\mathrm{Î\copyright }\right)=(q+N)\ln (q+N)-(q+N)-q\ln \left(q\right)+q-Nln\left(N\right)+N \\ \ln \left(\mathrm{Î\copyright }\right)=(q+N)\ln (q+N)-q\ln \left(q\right)-Nln\left(N\right) \end{gathered}$$

If we now make the assumption that$q<<N$, we get:

$$\begin{gathered} \ln \left(\mathrm{Î\copyright }\right)=(q+N)\ln \left[N\left(1+\frac{q}{N}\right)\right]-q\ln \left(q\right)-N\ln \left(N\right) \\ \ln \left(\mathrm{Î\copyright }\right)=(q+N)\left(\ln \left(N\right)+\ln \left(1+\frac{q}{N}\right)\right)-q\ln \left(q\right)-N\ln \left(N\right) \end{gathered}$$

but for$N>>q$, we have,$\ln \left(1+\frac{q}{N}\right)$ =\frac$\left\{q\right\}\left\{N\right\}$, so:

$$\begin{gathered} \ln \left(\mathrm{Î\copyright }\right)=(q+N)\left(\ln \left(N\right)+\frac{q}{N}\right)-q\ln \left(q\right)-Nln\left(N\right) \\ \ln \left(\mathrm{Î\copyright }\right)=q\ln \left(N\right)+\frac{q^2}{N}+N\ln \left(N\right)+q-q\ln \left(q\right)-Nln\left(N\right) \\ \ln \left(\mathrm{Î\copyright }\right)=\left(q\ln \right(N)-q\ln (q\left)\right)+\frac{q^2}{N}+q \end{gathered}$$

but,$q\backslash \ln \left(N\right)-q\backslash \ln \left(q\right)$$q\ln \left(\frac{N}{q}\right)$,= so:

$\ln \left(\mathrm{Î\copyright }\right)=q\ln \left(\frac{N}{q}\right)+\frac{q^2}{N}+q$

but as$q<<N$, so we can neglect the second term$\frac{q^2}{N^{\text{'}}}$, so:

$\ln \left(\mathrm{Î\copyright }\right)=q\ln \left(\frac{N}{q}\right)+q$

Exponentiating this equation gives the approximate value for$\mathrm{Î\copyright }$:

$e^{\ln \left(\mathrm{Î\copyright }\right)}=e^{\left[q\ln \left(\frac{N}{q}\right)+q\right]}$

$\mathrm{Î\copyright }={\left[e^{\ln \left(\frac{N}{q}\right)}\right]}^q×e^q$

but,$e^{lnx}=x$, so:

$\mathrm{Î\copyright }={\left(\frac{N}{q}\right)}^q×e^q$

$→\mathrm{Î\copyright }={\left(\frac{eN}{q}\right)}^q$