Consider a paramagnet, which is a material with \(n\) spins per unit volume each of which may each be either “up” or “down”. The spins have energy \(\pm mB\) where \(m\) is the magnetic dipole moment of a single spin, and there is no interaction between spins. The magnetization \(M\) is defined as the total magnetic moment divided by the total volume. Hint: each individual spin may be treated as a two-state system, which you have already worked with above.

1. Find the Helmholtz free energy of a paramagnetic system (assume \(N\) total spins) and show that \(\frac{F}{NkT}\) is a function of only the ratio \(x\equiv \frac{mB}{kT}\).

2. Use the canonical ensemble (i.e. partition function and probabilities) to find an exact expression for the total magentization \(M\) (which is the total dipole moment per unit volume) and the susceptibility \begin{align} \chi\equiv\left(\frac{\partial M}{\partial B}\right)_T \end{align} as a function of temperature and magnetic field for the model system of magnetic moments in a magnetic field. The result for the magnetization is \begin{align} M=nm\tanh\left(\frac{mB}{kT}\right) \end{align} where \(n\) is the number of spins per unit volume. The figure shows what this magnetization looks like.

3. Show that the susceptibility is \(\chi=\frac{nm^2}{kT}\) in the limit \(mB\ll kT\).

Find the equilibrium value at temperature \(T\) of the fractional magnetization \begin{equation} \frac{\mu_{tot}}{Nm} \equiv \frac{2\langle s\rangle}{N} \end{equation} of a system of \(N\) spins each of magnetic moment \(m\) in a magnetic field \(B\). The spin excess is \(2s\). The energy of this system is given by \begin{align} U &= -\mu_{tot}B \end{align} where \(\mu_{tot}\) is the total magnetization. Take the entropy as the logarithm of the multiplicity \(g(N,s)\) as given in (1.35 in the text): \begin{equation} S(s) \approx k_B\log g(N,0) - k_B\frac{2s^2}{N} \end{equation} for \(|s|\ll N\), where \(s\) is the spin excess, which is related to the magnetization by \(\mu_{tot} = 2sm\). Hint: Show that in this approximation \begin{equation} S(U) = S_0 - k_B\frac{U^2}{2m^2B^2N}, \end{equation} with \(S_0=k_B\log g(N,0)\). Further, show that \(\frac1{kT} = -\frac{U}{m^2B^2N}\), where \(U\) denotes \(\langle U\rangle\), the thermal average energy.

We have the following equations of state for the total magnetization \(M\), and the entropy \(S\) of a paramagnetic system: \begin{align} M&=N\mu\, \frac{e^{\frac{\mu B}{k_B T}} - e^{-\frac{\mu B}{k_B T}}} {e^{\frac{\mu B}{k_B T}} + e^{-\frac{\mu B}{k_B T}}}\\ S&=Nk_B\left\{\ln 2 + \ln \left(e^{\frac{\mu B}{k_B T}}+e^{-\frac{\mu B}{k_B T}}\right) +\frac{\mu B}{k_B T} \frac{e^{\frac{\mu B}{k_B T}} - e^{-\frac{\mu B}{k_B T}}} {e^{\frac{\mu B}{k_B T}} + e^{-\frac{\mu B}{k_B T}}} \right\} \end{align}

1. List variables in their proper positions in the middle columns of the charts below.

2. Solve for the magnetic susceptibility, which is defined as: \[\chi_B=\left(\frac{\partial M}{\partial B}\right)_T \]

3. Using both the differentials (zapping with d) and chain rule diagram methods, find a chain rule for:

   \[\left(\frac{\partial M}{\partial B}\right)_S \]

4. Evaluate your chain rule. Sense-making: Why does this come out to zero?