# Two-state paramagnet as an Einstein solid

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 2.11.

As a more-or-less real world example of an Einstein solid, we can look at a two-state paramagnet which is a material composed of magnetic dipoles which can align parallel or antiparallel to an external magnetic field. The restriction to two states comes from quantum mechanics, since the magnetic moment can take on only certain discrete values. The two-state paramagnet is the simplest case; in other magnets, the dipole can take on more than two states relative to the applied field.

We can model the two-state paramagnet as an Einstein solid where the role of the oscillators is played by the dipoles, and an energy quantum is the difference in energy between a parallel and antiparallel dipole. The antiparallel dipole has higher energy since a torque must be applied to twist the dipole against the field.

Using this model, we can picture two interacting paramagnets as two interacting Einstein solids. As before, we have two solids, ${A}$ and ${B}$, containing ${N_{A}}$ and ${N_{B}}$ dipoles 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

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

Consider a system with ${N_{A}=N_{B}=100}$ and ${q=80}$. Using Maple to calculate the binomial coefficients (Maple has a ‘binomial’ function that does this automatically) and produce the plot, we have

 $\displaystyle \Omega_{overall}$ $\displaystyle =$ $\displaystyle \binom{80+100+100-1}{80}=2.12\times10^{71}\ \ \ \ \ (3)$ $\displaystyle \Omega_{A}$ $\displaystyle =$ $\displaystyle \binom{q_{A}+99}{q_{A}}\ \ \ \ \ (4)$ $\displaystyle \Omega_{B}$ $\displaystyle =$ $\displaystyle \binom{q-q_{A}+99}{q-q_{A}}=\binom{179-q_{A}}{80-q_{A}}\ \ \ \ \ (5)$ $\displaystyle Prob\left(q_{A}\right)$ $\displaystyle =$ $\displaystyle \frac{\Omega_{A}\Omega_{B}}{\Omega_{overall}} \ \ \ \ \ (6)$

A plot of the probabilities looks like this:

As we’d expect, the most probable state is ${q_{A}=q_{B}=40}$, with a probability of 0.075. The least probable state is either ${q_{A}=0}$ or ${q_{A}=80}$, both of which have probability ${7.87\times10^{-20}}$.

## 3 thoughts on “Two-state paramagnet as an Einstein solid”

1. Will

But in a two-state paramagnet, a single particle can hold a maximum of one energy unit, so the combinatorics is different from that of an Einstein solid. I used the regular “n choose m” function for the multiplicities and got a noticeably different probability, 11.471%, for the maximum case where q_a = 40.