Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 344 / 3

The antisymmetric velocity-gradient tensor $\mathbf{\Omega }$ is the local angular velocity of a rigid-body rotation. Under such a rotation a material vector must rotate at exactly the same angular velocity, independently of its molecular constitution. The coefficient of $\mathbf{\Omega }\cdot \mathbf{p}$ is therefore fixed to one by rotational covariance, also called material objectivity. In contrast, the symmetric strain-rate tensor $\mathbf{D}$ deforms rather than merely rotates the material, so its flow-alignment coefficient $\xi$ is material dependent.

During a small incompressible displacement $u_i=v_i\Delta t$, pure advection changes the polar field byThe free-energy change is $\delta F=\int h_i\delta p_i,d\mathbf{r}$, where $h_i=\delta F/\delta p_i$. Integrating the translational term by parts and separating the antisymmetric and symmetric parts of ${\nabla }_i{u}_j$ identifies the order-parameter stress. Up to an arbitrary isotropic pressure, its three contributions obeyThe first term is the distortion or Ericksen force density, the second transfers antisymmetric rotational torque, and the third is the symmetric flow alignment of a polar order parameter contribution.

For a uniform field, $\mathbf{v}\cdot \nabla \mathbf{p}=0$, and the stated flow has $\mathbf{\Omega }=0$ andThe equation $D\mathbf{p}/Dt=-\Gamma \mathbf{h}$ becomesThus

The shifted molecular field is the gradient ofThus $A=\mathrm{diag}(a+\alpha ,a-\alpha /2,a-\alpha /2)$.

If $\alpha <0$, the $x$ mode softens first atThe extensional flow already selects the $x$ axis, and ordering chooses one of its two polar directions. The transition is continuous and spontaneously breaks the remaining inversion symmetry $p_x\mapsto -p_x$.

If $\alpha >0$, the degenerate transverse modes soften first atThis transition is also continuous. The flow preserves rotations about the $x$ axis, while the ordered vector chooses an azimuthal direction in the $yz$ plane and spontaneously breaks that $SO(2)$ symmetry.

For $\alpha <0$, choose one ordered state $\mathbf{p}=(p,0,0)$ with $p^2=(-a-\alpha )/b$. Stationarity gives $\tilde{\mathbf{h}}=0$, henceUniformity makes ${\nabla }_i{\Sigma }_{ij}^{(1)}=0$, so its spatially constant part can be absorbed into the pressure. Since $\mathbf{h}$ is parallel to $\mathbf{p}$, the antisymmetric term ${\Sigma }^{(2)}$ vanishes. The symmetric term isup to isotropic pressure. Using $\alpha =-\xi \gamma /\Gamma$, its only deviatoric component is ${\Sigma }_{xx}^p=({\xi }^2\gamma /\Gamma )p^2$.