 Solution

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.