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With velocity gradient $A_{ij}={\partial }_i{v}_j$, set $\Omega =(A-A^T)/2$. A vector rigidly rotating with the material has material derivative $D_t\mathbf{p}=-\Omega \mathbf{p}$, so its corotational derivative vanishes. Its unit rotational coefficient is fixed by an objective time derivative; adding $-\xi D\mathbf{p}$ describes the material's flow alignment of a polar order parameter.

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.

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.