= Solution
The shifted molecular field is the gradient of
$$
\boxed{\mathbb F_{\rm mod}=\frac12(a+\alpha)p_x^2+\frac12\left(a-\frac\alpha2\right)(p_y^2+p_z^2)+\frac b4(p_x^2+p_y^2+p_z^2)^2}.
$$
Thus $A=\operatorname{diag}(a+\alpha,a-\alpha/2,a-\alpha/2)$.
If $\alpha<0$, the $x$ mode softens first at
$$
\boxed{a_c=-\alpha},
\qquad
\boxed{\mathbf p=(\pm\sqrt{(-a-\alpha)/b},0,0)}quad(a<a_c).
$$
The 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 at
$$
\boxed{a_c=\frac\alpha2},
\qquad
\boxed{p_x=0,quad p_y^2+p_z^2=\frac{\alpha/2-a}{b}}quad(a<a_c).
$$
This 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.
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