Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 344 / 3 / a / Solution

A nematic director is an unoriented axis: $\mathbf{n}$ and $-\mathbf{n}$ describe the same molecular alignment. A vector average cannot represent this head-tail symmetry. The orientational second moment $⟨n_i{n}_j⟩$ is a symmetric second-rank tensor, and removing its isotropic part gives the traceless second-rank tensor $Q_{ij}\propto ⟨n_i{n}_j-{\delta }_{ij}/2⟩$. It vanishes for an isotropic angular distribution and transforms correctly under rotation matrices.

Every nonzero real symmetric traceless $2\times 2$ matrix has eigenvalues $\lambda /2,-\lambda /2$ and can be written $Q=\lambda (\mathbf{n}{\mathbf{n}}^{\mathsf{T}}-I/2)$ by choosing the eigenvector of the positive eigenvalue as director and $\lambda \ge 0$. Consequently $\mathrm{Tr}{Q}^2={\lambda }^2/2$. The bulk Landau-de Gennes free energy becomesFor $a<0$, its minimum occurs at ${\lambda }^2=-a/b$. The elastic density $K|\nabla \cdot Q{|}^2/2$ is nonnegative and vanishes for a uniform field. Thus, with no boundary condition imposing a texture, the pointwise lower bound is attained byThe negative-amplitude representation is equivalent in two dimensions to rotating the director by $\pi /2$, so the positive choice loses no tensor minima. This is minimization of the specified coarse-grained functional, not a claim of true thermodynamic long-range nematic order at every temperature in an infinite two-dimensional system. The PDF explicitly uses the divergence elastic term; the TeX's $|\nabla Q{|}^2$ loses the dot and would give different elastic coefficients.