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

For a uniform field the gradient term vanishes and minimization over $p=|\mathbf{p}|$ givesWhen $a<0$, the nonzero minimum haswhile its direction is arbitrary by rotational symmetry. A sudden quench creates many independently oriented domains. Their mismatches contain gradients and topological defects, whose motion and pair annihilation are slow, so the uniform state is not reached promptly.

Outside a defect core writeAround any closed curve on which $\mathbf{p}\ne 0$, its topological charge is the winding numberSingle-valued polar order requires $q\in \mathbb{Z}$. A nonzero value prevents the field inside the loop from being continuously deformed to a uniform nonvanishing field, so the core must contain a zero or singularity.

In polar coordinates $(r,\phi )$, a radial asterand a circulating vortexboth have $q=+1$. The continuous family $\theta =\phi +\chi$, $0\le \chi \le \pi /2$, deforms one into the other without making $\mathbf{p}$ vanish away from the core, proving their topological equivalence. The hyperbolic configuration $\theta =-\phi$ has $q=-1$.

The field $\mathbf{p}=-p(r)\hat{\mathbf{r}}$ has angle $\theta =\phi +\pi$. Its defect is at $r=0$ and has $q=+1$. Sinceone hasThe local free-energy density is therefore

Outside the core, put $p=p_0$. The angular-gradient contribution isThe omitted core contribution is finite and independent of $L$.

A defect of charge $q$ has far-field elastic energy proportional to $q^2\log (L/r_0)$. Splitting $q$ into allowed charges $q_i$ with $\sum q_i=q$ lowers $\sum q_i^2$ unless the pieces already have the smallest permitted magnitude. Polar order distinguishes $\mathbf{p}$ from $-\mathbf{p}$, so its smallest nonzero charge is $|q|=1$. A nematic liquid crystal identifies $\hat{\mathbf{n}}\sim -\hat{\mathbf{n}}$; a rotation by only $\pi$ closes a loop in its order-parameter space, permitting $q=\pm 1/2$.

Let $\chi$ be the angle between $\hat{\mathbf{p}}$ and $\hat{\mathbf{n}}$. ThenFor $\zeta ,\lambda >0$, the coupling $-\zeta \lambda p^2\cos 2\chi /2$ is minimized by $\cos 2\chi =1$, soThe uniform terms depending on $p$ reduce toTheir nonzero minimum is

An isolated nematic half-defect reverses the representative director after one circuit. Alignment would then reverse the polar vector, which is not the same physical state. Half-integer defects can therefore occur only if accompanied by a branch wall on which polar order rotates sharply or vanishes. The wall has a tension proportional to its length, so half-defects are confined in pairs in the uniformly polar phase; they can deconfine where polar order disappears.