# Einstein solids: multiplicity of large systems

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 2.17-2.19.

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

$\displaystyle \Omega=\binom{q+N-1}{q} \ \ \ \ \ (1)$

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 derivation of this approximation for the high temperature case ${q\gg N}$ (lots more energy quanta than oscillators) is given in Schroeder’s book, so I’ll deal here with the low temperature case ${q\ll N}$.

Writing out the binomial coefficient

 $\displaystyle \binom{q+N-1}{q}$ $\displaystyle =$ $\displaystyle \frac{\left(q+N-1\right)!}{q!\left(N-1\right)!}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \frac{\left(q+N\right)!}{q!N!} \ \ \ \ \ (3)$

where we’ve the fact that if we multiply a very large number like ${\left(q+N-1\right)!}$ by a merely large number like ${N}$, the original very large number is essentially unchanged.

We can now take logs and use Stirling’s approximation for the log of a factorial

$\displaystyle \ln n!\approx n\ln n-n \ \ \ \ \ (4)$

We get

 $\displaystyle \ln\Omega$ $\displaystyle \approx$ $\displaystyle \left(q+N\right)\ln\left(q+N\right)-q-N-q\ln q+q-N\ln N+N\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(q+N\right)\ln\left(q+N\right)-q\ln q-N\ln N \ \ \ \ \ (6)$

If we now make the assumption that ${q\ll N}$, we get

 $\displaystyle \ln\Omega$ $\displaystyle \approx$ $\displaystyle \left(q+N\right)\ln\left[N\left(1+\frac{q}{N}\right)\right]-q\ln q-N\ln N\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(q+N\right)\left[\ln N+\ln\left(1+\frac{q}{N}\right)\right]-q\ln q-N\ln N\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle q\ln N+\left(q+N\right)\frac{q}{N}-q\ln q\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle q\ln\frac{N}{q}+q+\frac{q^{2}}{N}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle q\ln\frac{N}{q}+q \ \ \ \ \ (11)$

where to get the third line we’ve used the approximation ${\ln\left(1+x\right)\approx x}$ for ${\left|x\right|\ll1}$, and in the last line we’ve negelected the ${q^{2}/N}$ term in the ${q\ll N}$ limit. Exponentiating this result gives the approximate value for ${\Omega}$:

$\displaystyle \Omega\approx\left(\frac{Ne}{q}\right)^{q} \ \ \ \ \ (12)$

[The corresponding result in the high temperature case is ${\Omega\approx\left(qe/N\right)^{N}}$ which could have been predicted easily, since ${q}$ and ${N}$ appear symmetrically in the approximation 3.]

This result applies also to the two-state paramagnet with ${N}$ magnetic dipoles and ${N_{\downarrow}}$ energy quanta, since the system is formally equivalent to an Einstein solid (we’re distributing the energy quanta among dipoles rather than oscillators). The multiplicity of the paramagnet is then

$\displaystyle \Omega\approx\left(\frac{Ne}{N_{\downarrow}}\right)^{N_{\downarrow}} \ \ \ \ \ (13)$

Finally, we can use Stirling’s approximation on 2 directly to get an approximation for the case where ${N}$ and ${q}$ are any large numbers, without one necessarily being much larger than the other. We have

 $\displaystyle \Omega=\binom{q+N-1}{q}$ $\displaystyle =$ $\displaystyle \frac{\left(q+N-1\right)!}{q!\left(N-1\right)!}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{q!}\frac{N}{N!}\frac{\left(q+N\right)!}{\left(q+N\right)}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{N}{q+N}\frac{\left(q+N\right)!}{q!N!} \ \ \ \ \ (16)$

Stirling’s approximation for a large factorial is

$\displaystyle n!\approx\sqrt{2\pi n}n^{n}e^{-n} \ \ \ \ \ (17)$

so we get

 $\displaystyle \Omega$ $\displaystyle \approx$ $\displaystyle \frac{N}{q+N}\frac{\sqrt{2\pi\left(q+N\right)}\left(q+N\right)^{q+N}e^{-\left(q+N\right)}}{2\pi\sqrt{qN}q^{q}N^{N}e^{-\left(q+N\right)}}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{N}{2\pi q\left(q+N\right)}}\frac{\left(q+N\right)^{q+N}}{q^{q}N^{N}}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{N}{2\pi q\left(q+N\right)}}\left(\frac{q+N}{q}\right)^{q}\left(\frac{q+N}{N}\right)^{N} \ \ \ \ \ (20)$

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