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.

3 thoughts on “Two-state paramagnet: analytic solution

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