Surfactants accumulate at an interface and orient their two chemically distinct ends toward their preferred fluid components. The relaxation found in part c lowers the coefficient of the <gradient energy>, and hence generally lowers the <surface tension>. If the reduction makes $\widetilde\kappa<0$, the positive fourth-gradient term proportional to $\gamma$ can stabilize structure at a nonzero <wavevector>, producing the modulated correlations characteristic of a <microemulsion>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 344 / 1 / d / Solution

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
$$
\mathcal L=s\cos\theta\,\dot\phi-
\frac{\rho_s}{2}\left[(\nabla\theta)^2+sin^2\theta(\nabla\phi)^2\right].
$$

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

A gradient energy penalizes spatial variation of an <order parameter>. Its leading isotropic form is usually proportional to the squared spatial gradient.

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 Incoming links: Gradient energy