The field $\mathbf p=-p(r)\widehat{\mathbf r}$ has angle $\theta=\varphi+\pi$. Its defect is at $r=0$ and has $q=+1$. Since
$$
\partial_r\mathbf p=-p'\widehat{\mathbf r},
\qquad
\frac1r\partial_\varphi\mathbf p
=-\frac p r\widehat{\boldsymbol\varphi},
$$
one has
$$
(\partial_i p_j)(\partial_i p_j)
=(p')^2+\frac{p^2}{r^2}.
$$
The local free-energy density is therefore
$$
\boxed{
\mathcal F
=\frac a2p^2+\frac b4p^4
+\frac\kappa2\left[
\left(\frac{dp}{dr}\right)^2+\left(\frac p r\right)^2
\right]
}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 344 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-344/2/d/solution