Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 344 / 2

A real symmetric traceless $2$ by $2$ matrix has eigenvalues $\lambda$ and $-\lambda$ and can be writtenfor a unit nematic director $\mathbf{n}\sim -\mathbf{n}$. Since $Q_{ij}{Q}_{ji}=2{\lambda }^2$, the uniform Landau free energy isFor $a<0<b$, its nonzero minima satisfyand henceThe signs $\pm {\lambda }_0$ amount to exchanging the two orthogonal eigenvectors, so the chosen convention takes ${\lambda }_0\ge 0$.

The bulk potential gives fluctuations in $\lambda$ a nonzero restoring force, whereas slowly rotating the nematic director costs only gradients. At lengths much larger than the amplitude correlation length, it is therefore the leading gradient expansion approximation to set $\lambda ={\lambda }_0$ while retaining $\mathbf{n}(\mathbf{r})$.

DifferentiatinggivesSubstitution into the elastic term yields

Let $\mathbf{n}=(\cos \theta ,\sin \theta )=(c,s)$ with $\theta =\theta (x)$. ThenThe squared norm is $({\theta }^{\prime }{)}^2$ becauseThus the one-elastic-constant nematic free energy becomeswith

The anchoring conditions are $\theta (0)=0$ modulo $\pi$ and $\theta (L)=\pi /2$ modulo $\pi$, since the nematic director identifies angles differing by $\pi$. Their difference can therefore be $m\pi /2$ for any odd integer $m$. The Euler-Lagrange equation ofis ${\theta }^{\prime \prime }=0$, so every stationary solution has the formIts free energy per unit length in the $y$ direction isThe smallest possible $m^2$ is one, giving exactly the two degenerate global minima $m=1$ and $m=-1$.

Traverse the large rectangle counterclockwise, taking its lower edge at $y\to -\infty$ and upper edge at $y\to +\infty$. Along the lower edge the director angle changes by $-\pi /2$; along the upper edge, traversed from $x=L$ to $x=0$, it changes by another $-\pi /2$. The anchored director is constant along the two vertical edges. The net continuous angle change is thereforeThe topological charge of a two-dimensional nematic disclination enclosed by a circuit is $q=\Delta \theta /(2\pi )$, soA nonsingular director field on the enclosed disk would have zero winding. At least one nematic disclination must therefore lie inside.

The least costly configuration contains one charge-$-1/2$ nematic disclination, placed near $(L/2,0)$ by the reflection symmetries of the boundary data. Away from its small core, the director interpolates as smoothly as possible between the wall anchoring and the two far-field textures. Locally around the defect one may sketchThe elastic energy of an isolated defect scales as $q^2\log (R/a_{\mathrm{core}})$. Splitting the required total $-1/2$ charge into additional allowed half-charge defects is impossible without also adding compensating defects, which raises both the logarithmic elastic energy and the positive core energy. The single centered $-1/2$ defect is therefore the lowest-energy topology, up to smooth distortions and symmetry-related core placement.

For general odd $m_{+}$ and $m_{-}$, the same rectangular circuit givesEvery elementary two-dimensional nematic defect has $|q|=1/2$, so the minimum number is

In three dimensions the director takes values in the real projective plane, whose fundamental group is $\mathbb{Z}/2\mathbb{Z}$. The signs of half-charge line defects are no longer distinct topological classes, and two such lines can annihilate by escape into the third dimension. Hence only the parity of $N_{2D}$ remains:When one is required, it is a disclination line extending through the unbounded $z$ direction.