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

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 ${\Sigma }_{xx}^p=({\xi }^2\gamma /\Gamma )p^2$.