Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 344 / 2

Use the paper's velocity gradient convention $A_{ij}={\partial }_i{v}_j$, so $\Omega =(A-A^T)/2$ and $D=(A+A^T)/2$. In a rigid motion there is rotation but no rate-of-strain tensor: $D=0$. A polar order parameter attached to the material must rotate exactly with it. In this convention its material derivative is $-\Omega \mathbf{p}$, and its corotational derivative of a polar vector must vanish.

Thus the compensating term is $+\Omega \mathbf{p}$ with coefficient exactly one. This follows from an objective time derivative and is independent of material properties. By contrast, the response to $D\mathbf{p}$ describes deformation and flow alignment of a polar order parameter; its coefficient $\xi$ can depend on the material. The transposed velocity gradient convention is essential to the sign of $\Omega$.

Let $h_i=\delta F/\delta p_i$ be the polar molecular field, with the positive functional derivative convention used in the paper. During a pure advective displacement $u_j$, the polar order parameter changes byThe free energy change is $\delta F=\int h_i\delta p_i{d}^{d}r$. Assume periodic boundaries, or boundary conditions that eliminate the surface work, and incompressibility ${\partial }_j{u}_j=0$. Integration by parts in the advective term givesRelabelling the dummy indices in the rotational and flow alignment of a polar order parameter terms givesTherefore $\delta F=\int {\Sigma }_{ij}^p{\partial }_i{u}_j{d}^{d}r$, with the reversible stress of a polar liquid crystalFor a local free-energy density $f(\mathbf{p},\nabla \mathbf{p})$, an explicit choice isIts divergence is exactly $-p_k{\partial }_j{h}_k$, by the Euler-Lagrange equation for $h_k$. A pressure term can be reassigned in an incompressible flow; the chosen representative realizes the printed divergence without that ambiguity. The force density ${\partial }_i{\Sigma }_{ij}^p$ has mechanical power $-\int {\Sigma }_{ij}^p{\partial }_i{v}_j$, the negative of the advective free energy rate, which checks the stress sign.

The polar molecular field isFor a spatially uniform polar order parameter, $\mathbf{h}=(a+b{p}^2)\mathbf{p}$, so it is colinear with $\mathbf{p}$ and its magnitude is $|\mathbf{h}|=|a+b{p}^2|p$. The scalar factor can be negative; colinearity does not necessarily mean parallel orientation.

For the simple shear flow, the paper's velocity gradient convention givesSubstitute these matrices into $\dot{\mathbf{p}}=-\Omega \mathbf{p}+\xi D\mathbf{p}-\Gamma \mathbf{h}$. Projection along and perpendicular to the polar order parameter yieldsAssume $g\ne 0$, $\Gamma >0$, $b>0$ and $p>0$. The flow-alignment angle in planar shear must satisfyFor $\xi =-1$, use the cosine equation: $\theta =\pi /2$ modulo $\pi$, and the displayed tangent is infinite. The radial equation giveswhere the sign distinguishes the angular branches, and only positive right-hand sides are ordered solutions. Linearizing the angular equation gives $\delta \dot{\theta }=-g\xi \sin 2\theta \delta \theta$. For $|\xi |>1$, the stable angular branch therefore hasIts radial relaxation eigenvalue is $-2\Gamma b{p}^2<0$. At $|\xi |=1$ the angular linearization is marginal and $p^2=-a/b$ requires $a<0$. If $g=0$, the shear restriction disappears and any orientation is allowed for $p^2=-a/b>0$. The printed request for dependence only on $a,b,\Gamma ,g$ omits $\xi$: in general the magnitude necessarily depends on it.

At $\xi =0$, the polar order parameter obeysFor $a<0$, $b>0$, the magnitude relaxes to $\sqrt{-a/b}$ while the direction rotates continuously. Thus the likely ordered behaviour is tumbling rather than a fixed alignment.

More generally, for $|\xi |<1$ and $g\ne 0$, $\xi \cos 2\theta -1$ never vanishes, so the tumbling of a polar order parameter persists. The polar order parameter completes a full $2\pi$ rotation inobtained by integrating $dt=2|d\theta |/(|g|(1-\xi \cos 2\theta ))$ over a full rotation. An unoriented nematic director repeats after half this period, whereas a polar order parameter distinguishes opposite directions. For $\xi \ne 0$ its magnitude can also oscillate. If the polar order parameter decays to zero, its orientation is no longer a physical observable; the tumbling conclusion concerns a persistent ordered state.