Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 344 / 2

A polar liquid crystal distinguishes the two ends of each constituent. Its orientational order is described by a vector $\mathbf{p}$, and $\mathbf{p}$ and $-\mathbf{p}$ represent different states. A nematic liquid crystal has head-tail symmetry, so its director obeys $\mathbf{n}\sim -\mathbf{n}$. Its lowest-rank faithful nematic order parameter is the symmetric traceless tensorwhich is unchanged by $\mathbf{n}\mapsto -\mathbf{n}$.

The Fourier transform sends each spatial derivative to $i{q}_i$, so at Gaussian levelFor $\kappa <0<\gamma$, minimizing over $s=q^2\ge 0$ givesThe minimum kernel is $G(q_0)=a-{\kappa }^2/(4\gamma )$. Gaussian fluctuations first diverge when this vanishes, so the nonzero-wavevector soft-mode sphere becomes unstable at

For candidate (i), $\mathbf{p}=p_0\hat{\mathbf{p}}\cos q_{0}x$. Since $G(q_0)=\hat{a}:=a-a_c$, $⟨{\cos }^2{q}_{0}x⟩=1/2$, and $⟨{\cos }^4{q}_{0}x⟩=3/8$, its mean-field free-energy density isFor $\hat{a}<0$, stationarity givesand substitution yieldsFor $\hat{a}\ge 0$, the minimum is $p_0=0$. The amplitude therefore vanishes continuously as $p_0\propto (a_c-a{)}^{1/2}$ on approaching $a_c$ from below, which is a continuous mean-field transition.

For candidate (ii), $|\mathbf{p}|=p_0$ everywhere and each component has wavevector magnitude $q_0$. The quadratic density is therefore $\hat{a}{p}_0^2/2$, while the quartic density is $b{p}_0^4/4$ without trigonometric averaging:For $\hat{a}<0$,Since $(-1/4)/(-1/6)=3/2$, the helical structure's free energy is $50\%$ more negative than that of candidate (i).

Every term in the free energy contracts the vector components with the Euclidean inner product: it depends only on $|\mathbf{p}{|}^2$, $|\mathbf{p}{|}^4$, $({\nabla }_i{p}_j)({\nabla }_i{p}_j)$, and $|{\nabla }^2\mathbf{p}{|}^2$. A constant $\mathcal{R}\in SO(3)$ preserves all these contractions and commutes with spatial differentiation. Thereforeso every constant rotation of candidate (ii) has the same free energy. The orientation of the rotation plane of the polar helical smectic is thus continuously degenerate.

Let ${\mathbf{e}}_1,{\mathbf{e}}_2$ span the rotating plane and let $\mathbf{n}={\mathbf{e}}_1\times {\mathbf{e}}_2$ be its normal. For modulation along $\hat{\mathbf{x}}$,and period averaging givesFor $\lambda >0$, this is minimized by $\mathbf{n}\parallel \hat{\mathbf{x}}$, so the rotation plane is perpendicular to the modulation direction and the helix is transverse. For $\lambda <0$, it is minimized energetically by maximizing the bracket: $\mathbf{n}\perp \hat{\mathbf{x}}$, so the modulation direction lies in the rotation plane. Thus either sign lifts the full rotational degeneracy, leaving only the rotations consistent with its selected relative orientation. A sufficiently small negative $\lambda$ does not overcome the stabilizing higher-gradient terms.