# Interacting Einstein solids: rectangular peak in multiplicity graph

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 2.22.

In a system composed of two interacting Einstein solids, the multiplicity function, which gives the number of microstates as a function of the number of energy quanta ${q_{A}}$ in solid ${A}$, is very sharply peaked about the point where the quanta are evenly distributed between the two solids. The width of the peak is approximately ${q/\sqrt{N}}$ for a system containing a total of ${q}$ energy quanta and ${N}$ oscillators.

We can get another, somewhat rougher, estimate of this width by first calculating the total number of microstates ${\Omega_{total}}$ accessible to the system (that is, the total number of microstates summed over all possible macrostates), and then finding the number of microstates ${\Omega_{mp}}$ in the most probable macrostate (even distribution of energy quanta). If we then assume that the peak in the graph is rectangular rather than Gaussian, then the width ${w}$ of the peak is found from the area of the rectangular peak, according to

$\displaystyle w=\frac{\Omega_{total}}{\Omega_{mp}} \ \ \ \ \ (1)$

To apply this, let’s consider a simple system where the two solids each have ${N}$ oscillators and the total number of quanta is ${q=2N}$. For large ${N}$ and ${q}$, we can use the approximation derived earlier for the number of microstates in a solid with ${q}$ quanta and ${n}$ oscillators (I’m using a lowercase ${n}$ here to distinguish it from the ${N}$ in the problem):

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

To find ${\Omega_{total}}$, we can combine the two solids into one composite solid since we’re interested in all microstates, no matter how the quanta are divided up between the two solids. In this case ${q=2N}$ and ${n=2N}$, so we get

 $\displaystyle \Omega_{total}$ $\displaystyle \approx$ $\displaystyle \sqrt{\frac{2N}{2\pi\left(2N\right)\left(4N\right)}}\left(\frac{4N}{2N}\right)^{2N}\left(\frac{4N}{2N}\right)^{2N}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2^{4N}}{\sqrt{8\pi N}} \ \ \ \ \ (4)$

To find ${\Omega_{mp}}$, we need to separate the solid into its constituent parts ${A}$ and ${B}$, and assign quanta so that ${q_{A}=q_{B}=\frac{q}{2}=N}$. The total number of microstates for this particular macrostate is then

 $\displaystyle \Omega_{mp}$ $\displaystyle =$ $\displaystyle \left[\sqrt{\frac{N}{2\pi\left(N\right)\left(N\right)}}\left(\frac{2N}{N}\right)^{N}\left(\frac{2N}{N}\right)^{N}\right]^{2}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2^{4N}}{4\pi N} \ \ \ \ \ (6)$

The term in square brackets is ${\Omega}$ for one of the solids, since it contains ${n=N}$ oscillators and ${q_{A}=N}$ energy quanta. Since the other solid is identical, we just square the result to get ${\Omega_{mp}}$.

The width of the peak is then

$\displaystyle w=\frac{4\pi N}{\sqrt{8\pi N}}=\sqrt{2\pi N} \ \ \ \ \ (7)$

[The Gaussian width is ${q/\sqrt{N}=2\sqrt{N}}$ and since ${\sqrt{2\pi N}\approx2.5\sqrt{N}}$ the rectangular approximation isn’t actually all that bad.]

The total number of macrostates in the two-solid system is ${q+1=2N+1\approx2N}$, so the fraction of macrostates that have reasonably large probabilities is

$\displaystyle \frac{\sqrt{2\pi N}}{2N}=\sqrt{\frac{\pi}{2N}} \ \ \ \ \ (8)$

For a macroscopic solid with ${N=10^{23}}$, this fraction comes out to around ${4\times10^{-12}}$. This shows just how unlikely it is that a macroscopic system will ever be found with an energy distribution significantly different from the most probable case.

## 3 thoughts on “Interacting Einstein solids: rectangular peak in multiplicity graph”

1. Pingback: Entropy | Physics pages

2. blipp

Should change “microstates” to “macrostates” in the second last paragraph there. Had me confused for a little while.