# Two-state paramagnet: numerical solution

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problems 3.17 – 3.18.

We can apply the formulas for entropy, temperature and heat capacity to a real-life system by looking at a two-state paramagnet. This is a system of ${N}$ magnetic dipoles which, when placed in a magnetic field ${B}$, align themselves so that their magnetic moment ${\mu}$ points either parallel or antiparallel to the field. The energy of a dipole that is aligned with the field is lower, and we’ll call it ${-\mu B}$, so that the antiparallel dipole has energy ${+\mu B}$, and the total energy of the system is

$\displaystyle U=\mu B\left(N_{\downarrow}-N_{\uparrow}\right)=\mu B\left(N-2N_{\uparrow}\right) \ \ \ \ \ (1)$

where an up arrow indicates parallel alignment and a down arrow antiparallel.

The net magnetization is then

$\displaystyle M=\mu\left(N_{\uparrow}-N_{\downarrow}\right)=-\frac{U}{B} \ \ \ \ \ (2)$

The multiplicity of states is the same as a set of ${N}$ coins with ${N_{\uparrow}}$ heads, so

$\displaystyle \Omega=\binom{N}{N_{\uparrow}}=\frac{N!}{N_{\uparrow}!\left(N-N_{\uparrow}\right)!} \ \ \ \ \ (3)$

For small systems, we can find the entropy directly as

$\displaystyle \frac{S}{k}=\ln\Omega \ \ \ \ \ (4)$

For ${N_{\uparrow}=98}$, we get ${U/\mu B=-96}$, ${M/N\mu=0.96}$, ${\Omega=4950}$ and ${S/k=8.507}$.

For each value of ${N_{\uparrow}}$ from 0 up to ${N}$, we can evaluate ${U}$ and ${S}$ from the formulas above and then plot ${S}$ versus ${U}$ (I used Maple for the plot):

For ${-100\le U/\mu B<0}$, the curve is a ‘normal’ entropy curve in that the entropy increases with increasing energy, and the curve is concave down.

From this we can get the temperature

$\displaystyle \frac{1}{T}=\frac{\partial S}{\partial U} \ \ \ \ \ (5)$

Thus the temperature increases with energy in the region ${-100\le U/\mu B<0}$. At ${U=0}$, however, the derivative is zero implying an infinite temperature, and for ${U>0}$, the slope is negative, indicating a negative temperature. Since, in this case, negative temperatures occur at higher energies than positive temperatures, we have to interpret a negative temperature as actually being higher than a positive one, in fact, higher than an ‘infinite’ positive temperature.

In this case, it’s probably better to use the entropy of the system as a physical measure of what’s going on, since the second law implies that the system will tend to the energy with the greatest entropy, which is ${U=0}$. At this energy, there are equal numbers of up and down dipoles, so the system is maximally randomized.

For a system with ${N=100}$, we can approximate 5 by taking finite differences. Thus for a given value of ${N_{\uparrow}}$ we can estimate the temperature as

 $\displaystyle T\left(N_{\uparrow}\right)$ $\displaystyle =$ $\displaystyle \frac{\Delta U}{\Delta S}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{U\left(N_{\uparrow}+1\right)-U\left(N_{\uparrow}-1\right)}{S\left(N_{\uparrow}+1\right)-S\left(N_{\uparrow}-1\right)} \ \ \ \ \ (7)$

For ${N_{\uparrow}=98}$, we get

$\displaystyle T\left(98\right)=\frac{U\left(99\right)-U\left(97\right)}{S\left(99\right)-S\left(97\right)}=0.541\frac{\mu B}{k} \ \ \ \ \ (8)$

This allows us to plot temperature versus energy:

We can see the flip over from ${+\infty}$ to ${-\infty}$ as the energy increases through zero.

The heat capacity can be obtained similarly as

 $\displaystyle C$ $\displaystyle =$ $\displaystyle \frac{\Delta U}{\Delta T}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{U\left(N_{\uparrow}+1\right)-U\left(N_{\uparrow}-1\right)}{T\left(N_{\uparrow}+1\right)-T\left(N_{\uparrow}-1\right)} \ \ \ \ \ (10)$

where we use 7 to calculate the temperatures. For ${N_{\uparrow}=98}$ we have

$\displaystyle C\left(98\right)=0.310Nk \ \ \ \ \ (11)$

The plot of ${C}$ versus temperature is

[The plot does actually extend down to 0 at ${T=0}$ if we use the analytic solution, but because we’re dealing with discrete values of ${N_{\uparrow}}$, it cuts out early.] This curve is similar in shape to that of an Einstein solid at low temperatures.

Finally, we can plot the magnetization as a function of temperature:

If we start off with ${T>0}$ and lower the temperature towards zero, the magnetization gradually increases until at ${T=0}$, the system is frozen into the state where all dipoles are parallel to the field. As we increase ${T}$ to ${+\infty}$, we approach maximum randomness with equal numbers of dipoles pointing up and down so ${M\rightarrow0}$. Increasing the temperature beyond ${+\infty}$ so it becomes negative (starting at ${-\infty}$) the magnetization again increases, but now the dipoles are aligned antiparallel to the field, eventually saturating as ${T\rightarrow0}$ from the negative side.