For $z=(e^{-i\phi}\cos(\theta/2),\sin(\theta/2))^T$,
$$
z^\dagger\dot z=-i\cos^2\left(\frac\theta2\right)\dot\phi.
$$
Thus $i\kappa z^\dagger\dot z=(\kappa/2)(1+\cos\theta)\dot\phi$. Dropping the <total derivative> $(\kappa/2)\dot\phi$ and writing $s=\kappa/2$ gives
$$
\mathcal L_{
m WZ}=s\cos\theta\,\dot\phi.
$$
The leading rotationally invariant <gradient energy> is $(\rho_s/2)(\nabla\mathbf n)^2$, so an effective Lagrangian is
$$
\mathcal L=s\cos\theta\,\dot\phi-
\frac{\rho_s}{2}\left[(\nabla\theta)^2+sin^2\theta(\nabla\phi)^2\right].
$$

Solution

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

 Solution

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