finite wave-vector, indicating a thermodynamic instability in the anti-ferromagnetic gas. Conversely for negative $c_{2}$ the spin-dependent contact interaction leads to a gap in the quadrupolar mode spectrum.

As pointed out by Leggett [@leggett], one can also derive these equations by writing the macroscopic equations for the density and spin current: $$\begin{aligned}
\label{eq:5}
\partial_{t}\textbf{n} + \frac{\vec{\nabla}_{\textbf{R}}\cdotp\vec{\textbf{j}}}{m} =  i\Big[\textbf{V},  \textbf{n} \Big] \\\nonumber
\partial_{t}\vec{\textbf{j}} + \frac{\vec{\nabla}_{\textbf{R}}\textbf{Q}}{m} = i\Big[\textbf{V}, \vec{\textbf{j}} \Big] - \frac{1}{2}\Big\{\vec{\nabla}_{\textbf{R}}\textbf {V},\textbf{n} \Big\} - \frac{\vec{\textbf{j}}}{\tau_{D}}\end{aligned}$$ where the energy density $\textbf{Q} = \int d{\vec{p}}~(\vec{p}\cdotp\vec{p})~\textbf{F}$. In order to obtain a closed set of equations, we approximate $\textbf{Q} \approx \int d{\vec{p}}~\frac{1}{3}\langle\vec{p}\cdotp\vec{p}\rangle~\textbf{F} = \frac{p_{T}}{2}\textbf{n}$, where $p_{T} = mv_{T}$. In a collisionless gas, sufficiently close to equilibrium, these approaches are equivalent.

An advantage of the Leggett approach is that it provides access to the dispersion at large $k$ more readily than the former approach. Solving (\[eq:5\]), we get the standard relations $\omega(k) = q - \frac{k^{2}v^{2}_{T}\tau_{D}/2}{(1 + (\Omega_{int}\tau_{D})^{2})} (\Omega_{int}\tau_{D} -  i)$, and $\omega(k) = -4c_{2}n_{0} + \frac{k^{2}v^{2}_{T}\tau_{D}/2}{(1 + (\Omega_{int}\tau_{D})^{2})} (\Omega_{int}\tau_{D} -  i)$. Setting $\tau_{D}$ to infinity, we once again recover the relationships derived above.

Excitations about the $m_{F} = 0$ state
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Next we consider the case of a gas in the $m_{F} = 0$ state. As 