# Interacting Einstein solids: a few examples

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 2.9 – 2.10.

Here are a few more examples of the probabilities of various macrostates in two interacting Einstein solids. As before, we have two solids, ${A}$ and ${B}$, containing ${N_{A}}$ and ${N_{B}}$ oscillators and ${q_{A}}$ and ${q_{B}}$ quanta of energy, with ${q_{A}+q_{B}=q=\mbox{constant}}$. For any particular partition of the quanta, that is, for particular values of ${q_{A}}$ and ${q_{B}}$, the total number of microstates available to the compound system is

$\displaystyle \Omega_{total}=\Omega_{A}\Omega_{B}=\binom{q_{A}+N_{A}-1}{q_{A}}\binom{q_{B}+N_{B}-1}{q_{B}} \ \ \ \ \ (1)$

The overall number of microstates is

$\displaystyle \Omega_{overall}=\binom{q+N_{A}+N_{B}-1}{q} \ \ \ \ \ (2)$

Example 1 Consider a simple system with ${N_{A}=N_{B}=3}$ and ${q=6}$ as shown in Fig 2.4 in Schroeder. Using Maple to calculate the binomial coefficients (Maple has a ‘binomial’ function that does this automatically) and produce the plot, we have

 $\displaystyle \Omega_{overall}$ $\displaystyle =$ $\displaystyle \binom{6+3+3-1}{6}=462\ \ \ \ \ (3)$ $\displaystyle \Omega_{A}$ $\displaystyle =$ $\displaystyle \binom{q_{A}+2}{q_{A}}\ \ \ \ \ (4)$ $\displaystyle \Omega_{B}$ $\displaystyle =$ $\displaystyle \binom{q-q_{A}+2}{q-q_{A}}=\binom{8-q_{A}}{6-q_{A}}\ \ \ \ \ (5)$ $\displaystyle Prob\left(q_{A}\right)$ $\displaystyle =$ $\displaystyle \frac{\Omega_{A}\Omega_{B}}{\Omega_{overall}} \ \ \ \ \ (6)$

Plugging in the numbers, we get

 ${q_{A}}$ ${\Omega_{A}}$ ${\Omega_{B}}$ ${\Omega_{total}}$ ${Prob\left(q_{A}\right)}$ 0 1 28 28 0.061 1 3 21 63 0.136 2 6 15 90 0.195 3 10 10 100 0.216 4 15 6 90 0.195 5 21 3 63 0.136 6 28 1 28 0.061

A bar chart of the probabilities looks like this:

As before, the most likely state is when the energy is equally distributed between the two solids.

Example 2 Now we’ll see what happens if one solid has more oscillators to store energy than the other one. We’ll take ${N_{A}=6}$, ${N_{B}=4}$ and ${q=6}$. We now have

 $\displaystyle \Omega_{overall}$ $\displaystyle =$ $\displaystyle \binom{6+6+4-1}{6}=5005\ \ \ \ \ (7)$ $\displaystyle \Omega_{A}$ $\displaystyle =$ $\displaystyle \binom{q_{A}+5}{q_{A}}\ \ \ \ \ (8)$ $\displaystyle \Omega_{B}$ $\displaystyle =$ $\displaystyle \binom{q-q_{A}+3}{q-q_{A}}=\binom{9-q_{A}}{6-q_{A}}\ \ \ \ \ (9)$ $\displaystyle Prob\left(q_{A}\right)$ $\displaystyle =$ $\displaystyle \frac{\Omega_{A}\Omega_{B}}{\Omega_{overall}} \ \ \ \ \ (10)$

Plugging in the numbers, we get

 ${q_{A}}$ ${\Omega_{A}}$ ${\Omega_{B}}$ ${\Omega_{total}}$ ${Prob\left(q_{A}\right)}$ 0 1 84 84 0.017 1 6 56 336 0.067 2 21 35 735 0.147 3 56 20 1120 0.224 4 126 10 1260 0.252 5 252 4 1008 0.201 6 462 1 462 0.092

A bar chart of the probabilities looks like this:

Since solid ${A}$ has more oscillators, the probabilities are skewed towards more of the quanta being stored in solid ${A}$ than in solid ${B}$.

Example 3 Now we’ll ramp things up a bit and consider two solids with ${N_{A}=200}$, ${N_{B}=100}$ and ${q=100}$. As there are 101 macrostates, we won’t list all the various states. The formulas in this case are

 $\displaystyle \Omega_{overall}$ $\displaystyle =$ $\displaystyle \binom{200+100+100-1}{100}=1.681\times10^{96}\ \ \ \ \ (11)$ $\displaystyle \Omega_{A}$ $\displaystyle =$ $\displaystyle \binom{q_{A}+199}{q_{A}}\ \ \ \ \ (12)$ $\displaystyle \Omega_{B}$ $\displaystyle =$ $\displaystyle \binom{q-q_{A}+99}{q-q_{A}}=\binom{199-q_{A}}{100-q_{A}}\ \ \ \ \ (13)$ $\displaystyle Prob\left(q_{A}\right)$ $\displaystyle =$ $\displaystyle \frac{\Omega_{A}\Omega_{B}}{\Omega_{overall}} \ \ \ \ \ (14)$

The plot is

The maximum probability of 0.073 occurs at ${q_{A}=67}$ and the minimum of ${2.69\times10^{-38}}$ at ${q_{A}=0}$. As solid ${A}$ contains ${\frac{2}{3}}$ of the oscillators, the maximum probability is when ${\frac{2}{3}}$ of the energy is stored in solid ${A}$. Notice how vanishing small is the chance of finding the system in a macrostate with anything less than about ${q_{A}=45}$. Thus even though all microstates are equally probable, it is overwhelmingly likely that the energy will be more or less evenly distributed over all the oscillators.