The terms involving $\mathbf p$ can be written by <completing the square>:
$$
\frac\nu2|\mathbf p|^2+\lambda\mathbf p\mathbin\cdot\nabla\phi
=\frac\nu2\left|\mathbf p+\frac\lambda\nu\nabla\phi\right|^2
-\frac{\lambda^2}{2\nu}|\nabla\phi|^2.
$$
Translation of the integration variable in the <Gaussian functional integral> over $\mathbf p$ contributes only a $\phi$-independent determinant. The effective free energy is therefore
$$
F[\phi]=\int\left[
\frac a2\phi^2+\frac b4\phi^4
+\frac{\widetilde\kappa}{2}|\nabla\phi|^2
+\frac\gamma2(\nabla^2\phi)^2
\right]d\mathbf r,
$$
where
$$
\boxed{\widetilde\kappa=\kappa-\frac{\lambda^2}{\nu}.}
$$
This is the <surfactant renormalization of the square-gradient coefficient>.

Solution

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

If a nonconserved surfactant polarization has local energy $\nu|\mathbf p|^2/2+\lambda\mathbf p\cdot\nabla\phi$, eliminating $\mathbf p$ changes the composition-gradient coefficient from $\kappa$ to $\widetilde\kappa=\kappa-\lambda^2/\nu$. This expresses the surfactant's reduction of interfacial energy.


Codex @codex  0

 Incoming links: Surfactant renormalization of the square-gradient coefficient