Two-state paramagnet: analytic solution

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.19.

We’ve looked at a two-state paramagnet and generated some curves involving entropy, energy and temperature using numerical methods. Here we’ll derive the analytic solution.

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

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

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

The net magnetization is then

$\displaystyle M=\mu\left(N_{\uparrow}-N_{\downarrow}\right)=-\frac{U}{B} \ \ \ \ \ (3)$

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

$\displaystyle \Omega=\binom{N}{N_{\uparrow}}=\frac{N!}{N_{\uparrow}!\left(N-N_{\uparrow}\right)!} \ \ \ \ \ (4)$

We begin the analytic solution by looking at the entropy directly, and using Stirling’s approximation.

 $\displaystyle \frac{S}{k}$ $\displaystyle =$ $\displaystyle \ln\Omega\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle \left(N\ln N-N\right)-\left(N_{\uparrow}\ln N_{\uparrow}-N_{\uparrow}\right)-\left(\left(N-N_{\uparrow}\right)\ln\left(N-N_{\uparrow}\right)-\left(N-N_{\uparrow}\right)\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle N\ln N-N_{\uparrow}\ln N_{\uparrow}-\left(N-N_{\uparrow}\right)\ln\left(N-N_{\uparrow}\right) \ \ \ \ \ (7)$

We can now get the temperature from

$\displaystyle \frac{1}{T}=\frac{\partial S}{\partial U} \ \ \ \ \ (8)$

Using the chain rule, 2 and 1 we get

 $\displaystyle \frac{1}{T}$ $\displaystyle =$ $\displaystyle \frac{\partial S}{\partial N_{\uparrow}}\frac{\partial N_{\uparrow}}{\partial U}\ \ \ \ \ (9)$ $\displaystyle \frac{\partial S}{\partial N_{\uparrow}}$ $\displaystyle =$ $\displaystyle k\left[-\ln N_{\uparrow}-1+\ln\left(N-N_{\uparrow}\right)+1\right]\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle k\ln\frac{N-N_{\uparrow}}{N_{\uparrow}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle k\ln\frac{N+U/\mu B}{N-U/\mu B}\ \ \ \ \ (12)$ $\displaystyle \frac{\partial N_{\uparrow}}{\partial U}$ $\displaystyle =$ $\displaystyle -\frac{1}{2\mu B}\ \ \ \ \ (13)$ $\displaystyle \frac{1}{T}$ $\displaystyle =$ $\displaystyle -\frac{k}{2\mu B}\ln\frac{N+U/\mu B}{N-U/\mu B}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{k}{2\mu B}\ln\frac{N-U/\mu B}{N+U/\mu B} \ \ \ \ \ (15)$

Exponentiating both sides gives

$\displaystyle e^{2\mu B/kT}=\frac{N-U/\mu B}{N+U/\mu B} \ \ \ \ \ (16)$

which can be solved for ${U}$:

 $\displaystyle \frac{U}{\mu B}\left[e^{2\mu B/kT}+1\right]$ $\displaystyle =$ $\displaystyle N\left[1-e^{2\mu B/kT}\right]\ \ \ \ \ (17)$ $\displaystyle U$ $\displaystyle =$ $\displaystyle N\mu B\frac{1-e^{2\mu B/kT}}{1+e^{2\mu B/kT}}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle N\mu B\frac{e^{-\mu B/kT}-e^{\mu B/kT}}{e^{-\mu B/kT}+e^{\mu B/kT}}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -N\mu B\tanh\frac{\mu B}{kT} \ \ \ \ \ (20)$

The magnetism is, from 3

$\displaystyle M=N\mu\tanh\frac{\mu B}{kT} \ \ \ \ \ (21)$

Finally, the heat capacity at constant magnetic field is (using the derivative of the tanh function: ${d\tanh x/dx=\mbox{sech}^{2}x}$):

 $\displaystyle C_{B}$ $\displaystyle =$ $\displaystyle \frac{\partial U}{\partial T}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -N\mu B\left[\mbox{sech}^{2}\frac{\mu B}{kT}\right]\left[-\frac{\mu B}{kT^{2}}\right]\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{Nk\left(\mu B/kT\right)^{2}}{\cosh^{2}\left(\mu B/kT\right)} \ \ \ \ \ (24)$

These analytic formulas give the curves we saw earlier using the numerical solution.

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