Forming a Cube from a Sphere with Tetratic Order

Composed of square particles, the tetratic phase is characterized by a fourfold symmetry with quasi-long-range orientational order but no translational order. We construct the elastic free energy for tetratics and find a closed form solution for $\pm 1/4$ disclinations in planar geometry. Applying the same covariant formalism to a sphere, we show analytically that within the one elastic constant approximation eight $+1/4$ disclinations favor positions defining the vertices of a cube. The interplay between defect–defect interactions and bending energy results in a flattening of the sphere towards superspheroids with the symmetry of a cube.

Figure 1

The change of the angle ${\alpha }^{(+)}$ (2) upon encircling a $+1/4$ disclination (solid line, top configuration) and of the angle ${\alpha }^{(-)}=1/4\mathrm{Im}\log (r{e}^{i\phi })$ (dashed line, bottom configuration). The tetratic ${\text{model}}^{(+)}$ gives the energetically favorable solution with ${\mathcal{F}}_{\text{tetra}}^{(+)}/{\mathcal{F}}_{\text{tetra}}^{(-)}=4/{\pi }^2$ (3).

Figure 2

Tetratic order: on a projected plane $z=x+iy$ with $8(+1/4)$ disclinations (5) at the vertices of two concentric squares (a) and twisted by $\pi /4$ (b); stereographic projection on a sphere $(\mathrm{a})\to (\mathrm{c})$ and $(\mathrm{b})\to (\mathrm{d})$, similar to [3]. (e) The free energy difference between (d) and (c) configurations within the ${\text{model}}^{(-)}$ (nematiclike), the integrand $\pm 8\pi {\tan }^7(v/2)/((1+\cos v)[1-{\tan }^{16}(v/2)])$ [14].

Figure 3

(a) Inside: superspheroid parametrized by (6) with $p=6$. Tetratic order with 8 ($+1/4$) disclinations at $v=\pm \pi /4$ and $u=\pi /4$, $3\pi /4$ projected on the superspheroid [14]. (b) The normalized tetratic (squares) and bending (circles) energies, plotted as a function of the power $p$ responsible for the flattening of the superspheroid ($p=2$ is a sphere, $p\to \infty$ is a cube). We choose the angle cutoff to be $\epsilon =(\delta /a)2^{1/p-1/2}$ with $\delta =0.01$ and $a$ determined by assuming a constant area of the surface $\mathcal{A}=4\pi R^2$.