Put $q=\delta\theta$, $p=\delta\phi$, so $\theta=\pi/2-q$ and $\phi=p$. The <quadratic Lagrangian> is obtained by taking
$$
\mathcal L_2=sq\dot p-
\frac{\rho_s}{2}\left[(\nabla q)^2+(\nabla p)^2\right].
$$
The two real fluctuations form a coordinate and its <canonical momentum> rather than two independent modes. Their <Euler-Lagrange equations> combine to give
$$
\ddot q+\left(\frac{\rho_s}{s}\right)^2\nabla^4q=0,
$$
and similarly for $p$. The <ferromagnetic magnon> therefore has the quadratic <dispersion relation>
$$
\omega_k=\frac{\rho_s}{s}k^2.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 337 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-337/2/d/solution