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In a charge-free and current-free vacuum, Maxwell equations giveTake the curl of Faraday's law and use the curl of the curl identity $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$. The divergence term vanishes by Gauss's law, while the time derivative of the magnetic curl supplies $-{\mu }_0{ϵ}_0{\partial }_t^2\mathbf{E}$. ThusThis is the transverse electromagnetic wave equation.

Use the fully developed flow branch of Hartmann flow. Translation invariance along the walls makes the velocity and induced magnetic field functions of $z$ alone. Incompressible flow and zero normal velocity at the walls give $u_z^{\prime }=0$ and hence $u_z=0$. The zero divergence of the magnetic field makes $B_z$ constant, equal to the imposed $B_0$. There is no forcing in $y$; the homogeneous $y$-components obey the same coupled viscous-resistive equations as the $x$-components, with zero boundary data. Multiplying them by $u_y$ and $b_y/({\mu }_0\rho )$, integrating by parts and adding givesThus $u_y=b_y=0$, establishing the asserted forms on this fully developed flow branch. This is a symmetry reduction of the steady channel model, rather than a claim that every possible flow in a channel is translation invariant.

The current density and Lorentz force density areThe advective acceleration vanishes because $\mathbf{u}\cdot \nabla =u(z){\partial }_x$ acts on fields independent of $x$. The $x$-component of the magnetohydrodynamic momentum equation therefore givesThe $z$-component must also balance: it requires ${\partial }_z(p+b^2/(2{\mu }_0))=0$. Hence a compatible pressure is $p(x,z)=p_{\ast }-\rho Gx-b(z{)}^2/(2{\mu }_0)$; the specified streamwise pressure gradient does not require the ordinary pressure to be uniform in $z$. This accounts for the magnetic pressure of the induced field.

The resistive induction equation with constant magnetic diffusivity gives ${\partial }_t\mathbf{B}=\nabla \times (\mathbf{u}\times \mathbf{B})+\eta {\nabla }^2\mathbf{B}$. Since $\mathbf{u}\times \mathbf{B}=(0,-B_{0}u,0)$, its curl has $x$-component $B_0{u}^{\prime }$. Steadiness consequently givesFinally the no-slip boundary condition gives $u(\pm L)=0$, and the normal magnetic field boundary condition gives $b(\pm L)=0$. The coupled equations and all wall conditions follow directly from momentum balance and magnetic induction.

Use the usual setting of a continuously differentiable solenoidal vector field defined on all radial segments from the origin, including the origin. More generally an open star-shaped set containing zero suffices. Differentiation under the integral sign givesApply divergence and curl of a cross product with the second field $x$, using $\nabla \cdot x=3$ and $(D\cdot \nabla )x=D$:The scaling identity in part (a) now converts the last expression to a one-variable derivative:Continuity at the origin ensures the lower endpoint is zero. Hence the radial vector potential of a solenoidal vector field isThe order of the cross product matters: $x\times D$ would produce $-B$. This construction is not valid on every punctured or multiply connected domain. For example $B=x/|x{|}^3$ is solenoidal away from zero, but its radial integral diverges at zero and it has nonzero flux through a surrounding sphere. The stated formula therefore implicitly needs the regular radial-domain hypothesis, rather than just local zero divergence.

Here both the divergence and vorticity vanish:The flow is incompressible and irrotational. With $\mathbf{u}=\nabla \varphi$ and $(u_x,u_y)=({\psi }_y,-{\psi }_x)$,The streamlines are the level sets $x^2-y^2=\text{constant}$. To determine the arrows, put $r=x+y$ and $s=x-y$. A particle satisfies $\dot{r}=-2r$, $\dot{s}=2s$. Thus the separatrix $y=x$ approaches the stagnation point at the origin, whereas $y=-x$ leaves it; the other streamlines are hyperbolas following those asymptotes. The origin itself is a stationary particle.

The divergence and scalar vorticity areThus the flow is irrotational but not incompressible on the plane. A velocity potential satisfying $\mathbf{u}=\nabla \varphi$ isIndeed its two partial derivatives are the two velocity components. A usual two-dimensional stream function would satisfy $u_x={\psi }_y$, $u_y=-{\psi }_x$ and hence have identically zero divergence. No such stream function exists on an open region where this divergence is not identically zero; the line $y=0$ is not a two-dimensional flow domain.

Write $\zeta =(\nabla \times \mathbf{u})\cdot {\mathbf{e}}_z$ and $q=\zeta -f\eta /h_0$, so the PDF's vector potential vorticity is $\mathbf{Q}=q{\mathbf{e}}_z$. The local TeX loses some of this boldface. Taking the vertical curl of the momentum equation gives ${\zeta }_t=-f\nabla \cdot \mathbf{u}$, while mass conservation gives ${\eta }_t=-h_0\nabla \cdot \mathbf{u}$. ThusThis is the linear perturbation of shallow-water potential vorticity, scaled by $h_0$; it should not be confused with the full nonlinear conserved ratio $(f+\zeta )/h$.

Since $\nabla \cdot ({\mathbf{e}}_z\times \mathbf{u})=-\zeta$, taking the divergence of momentum gives $(\nabla \cdot \mathbf{u}{)}_t=f\zeta -g{\nabla }^2\eta$. Differentiate mass conservation and use $\zeta =q+f\eta /h_0$ to obtainFor $\mathbf{Q}=0$, substitution of the stated plane wave gives the linear rotating shallow-water dispersion relation, with $c=\sqrt{g{h}_0}$:The two even branches approach $\pm c|k|$ at large $|k|$ and have values $\pm |f|$ at $k=0$. The original dispersion diagram below also compares the speed magnitudes.

For $k\ne 0$, the phase velocity and group velocity magnitudes areThe printed convention $e^{i(kx+\omega t)}$ means the signed phase velocity is $-\omega /k$ and the signed group velocity is $-d\omega /dk$. The speed magnitudes above are independent of that sign convention.

For $f\ne 0$, longer wavelengths have greater phase velocity but smaller group velocity: individual crests and wave packets therefore answer “faster or slower” differently. At long wavelength, $|k|\ll |f|/c$, the frequency is nearly the inertial frequency $|f|$, $|c_p|\sim |f|/|k|$, and $|c_g|\sim c^2|k|/|f|$. Rotation has its largest relative effect here, and the waves are strongly dispersive. At short wavelength, $|k|\gg |f|/c$, both speeds tend to $c$ and rotation gives only a small, weakly dispersive correction. If $f=0$, the gravity waves are nondispersive, with speed $c$; at $k=0$ the phase velocity formula is undefined.

The Jacobian determinant of ${\varphi }_t$ is $e^t$, while every coordinate and every coordinate differential is multiplied by $e^t$ under ${\psi }_t$. Hence the pullback of a differential form operation givesTo compare directly with Cartan's magic formula, note $d\mu =0$ and computeThe first two differentiated terms vanish because their coordinate differentials repeat; the last contributes $dz\wedge dx\wedge dy=\mu$. AlsoTheir sum is $2\alpha$, as required. Equivalently the divergence of $X$ relative to the standard volume form is one.

Let the constant mass density be $\rho >0$ and put ${\mathbf{v}}_a=\mathbf{B}/\sqrt{{\mu }_0\rho }$, the Alfvén velocity. Since both the velocity and the magnetic field have zero divergence, the magnetic tension and magnetic pressure decomposition of the Lorentz force givesThe Newtonian gravitational potential $\Phi$ remains in $\psi$; uniform mass density does not justify dropping a prescribed gravitational acceleration. Adding and subtracting the two equations proves the Elsässer variable equationsHere ${\mathbf{z}}^{\pm }=\mathbf{u}\pm {\mathbf{v}}_a$; each Elsässer variable is transported by the other.

Define $I_{\pm }={\int }_V|{\mathbf{z}}^{\pm }{|}^{2}dV$. Taking the dot product with $2{\mathbf{z}}^{\pm }$ gives the Elsässer energy invariant balanceThe divergence theorem proves $d{I}_{\pm }/dt=0$ whenever the net boundary flux vanishes. In particular, $\mathbf{u}\cdot \mathbf{n}=\mathbf{B}\cdot \mathbf{n}=0$ makes ${\mathbf{z}}^{\pm }\cdot \mathbf{n}=0$, so both fluxes vanish pointwise. Periodic boundary conditions or decay at infinity are also sufficient. These boundary conditions matter: fixed volume alone gives no conservation.

The kinetic energy plus magnetic energy isExpanding $|\mathbf{u}\pm {\mathbf{v}}_a{|}^2$ and using the cross-helicity definition yieldsBoth energy and cross-helicity therefore follow from the two Elsässer energy invariants. The quantity $E$ here is exactly the requested kinetic energy plus magnetic energy, without an additional gravitational term.

Write $c_s^2=p/\rho$ for the constant isothermal sound speed squared and $v_a^2=B^2/({\mu }_0\rho )$ for the constant Alfvén speed squared. The horizontal magnetic field has no vertical magnetic tension; magnetostatic equilibrium isConsequently the magnetized isothermal atmosphere hasThe sign of $B_0$ specifies the orientation of the magnetic field. The adiabatic sound speed is $v_s^2=\gamma p/\rho =\gamma c_s^2$, also constant.

For the proposed normal mode, let $A=\exp (i\mathbf{k}\cdot \mathbf{x}+z/(2H)-i\omega t)$, $a=1/(2H)$, and temporarily write $\mathbf{\xi }$ for its constant complex amplitude. The divergence amplitude is $\Delta =i{k}_x{\xi }_x+i{k}_y{\xi }_y+(i{k}_z+a){\xi }_z$. Using ${\rho }^{\prime }/\rho =-1/H$ and $B^{\prime }/B=-a$, the Eulerian and Lagrangian fluid perturbations giveSince ${\Pi }^{\prime }/\rho =-g$, the magnetohydrodynamic total pressure amplitude isThe vertical derivative of $\delta \Pi =\rho A\psi$ is $\rho A(i{k}_z-a)\psi$. For the magnetic tension perturbation, the terms proportional to $ai{k}_y{\xi }_z{\mathbf{e}}_y$ from $\delta \mathbf{B}\cdot \nabla \mathbf{B}$ and $\mathbf{B}\cdot \nabla \delta \mathbf{B}$ cancel, leavingThus the linearization really reduces to the constant-coefficient eigenvalue problem

For a static ideal magnetohydrodynamics equilibrium with conservative gravitational acceleration, the magnetohydrodynamic energy principle gives a Hermitian operator for the restoring force in the mass density weighted inner product, provided the surface terms vanish. The factor $e^{z/(2H)}$ removes the exponential mass density weight. This can be checked directly, without assuming stability: set $s=v_s^2$, $b=v_a^2$, and $q=s+b$. Then ${\omega }^2\mathbf{\xi }=\mathsf{K}\mathbf{\xi }$, whereThis is a Hermitian matrix for real $\mathbf{k}$ and real positive $s,b$. Therefore every squared frequency ${\omega }^2$ is real; negative values give exponential growth, rather than an oscillation with a complex squared frequency. In an infinite atmosphere these are generalized plane waves, not finite-total-energy global modes.

At ${\omega }^2=0$ and $k_y\ne 0$, the $y$ equation, after cancelling its magnetic terms, gives $s\Delta =g{\xi }_z$. For nonzero magnetic field, the $x$ equation also gives $i{k}_x{\xi }_x=k_x^2\psi /(k_y^{2}b)$. Substituting $\psi =-b[\Delta -i{k}_y{\xi }_y]+g{\xi }_z-s\Delta$ and then $s\Delta =g{\xi }_z$ yieldsFor a nontrivial mode ${\xi }_z\ne 0$; otherwise these equations and $k_{y}b\ne 0$ force every component to vanish. Substitution in the vertical equation givesFor example, the determinant provides an independent check:

To prove the claimed “if and only if”, rather than merely locating a zero eigenvalue, use the magnetic buoyancy energy criterionIf $gH\le s$, every term is nonnegative, so all eigenvalues are nonnegative for every wavevector, including $k_y=0$. If $gH>s$, choose $k_z=0$, any fixed $k_x\ne 0$, and ${\xi }_z=1$, ${\xi }_x=-a/(i{k}_x)$, ${\xi }_y=g/(si{k}_y)$. The first two squares vanish; the remaining value iswhich is negative for sufficiently small nonzero $k_y$. The Rayleigh quotient of this displacement is therefore negative, proving a negative eigenvalue and an unstable normal mode. This uses precisely the assumption that boundary conditions do not exclude the chosen wavelengths. Equality is a neutral threshold, not exponential growth.

Finally the plasma beta is $\beta =p/[B^2/(2{\mu }_0)]=2c_s^2/v_a^2$. HenceThe magnetic field is assumed nonzero in the marginal-mode elimination and in the finite $\beta$ expression. The positive-sound-speed, nonmagnetic limit remains covered by the energy identity; for the usual $\gamma >1$ it is stable. Parker instability occurs because magnetic pressure supports part of the atmosphere's weight while the gas can drain along gently bent magnetic field lines. The resulting buoyant displacement can release Newtonian gravitational potential energy; choosing long wavelengths along the magnetic field makes the stabilizing magnetic tension small. The completed-square calculation quantifies this competition, including the adiabatic process restoring force.

Let the free surface be at $r=R_1$ and take the Newtonian gravitational potential to vanish at infinity. Mass conservation fixes $R_1=(3M_1/(4\pi {\rho }_1){)}^{1/3}$. The enclosed mass is $m(r)=4\pi {\rho }_1{r}^3/3$, and hydrostatic equilibrium requires $d{p}_0/dr=-{\rho }_{1}Gm(r)/r^2$. With vacuum outside, $p_0(R_1)=0$, givingThe pressure is zero outside. A prescribed constant external pressure would simply add that constant to $p_0$.

The bulk Eulerian and Lagrangian fluid perturbations obey $\delta \rho =0$ because the mass density is uniform and the Lagrangian displacement has zero divergence. The linearized Euler momentum equation in the interior isTaking the curl gives ${\omega }^2\nabla \times \mathbf{\xi }=0$. Thus every nonzero-frequency normal mode has an irrotational flow displacement; the simply connected space formed by the interior admits $\mathbf{\xi }=\nabla U$ and incompressibility gives the Laplace equation ${\nabla }^{2}U=0$. For a regular center, one spherical harmonic component is $U=A_l{r}^l{Y}_l^m$, with $|m|\le l$.

It is essential to retain the surface gravity perturbation of a uniform-density sphere. Although the bulk mass density perturbation is zero, the displaced density discontinuity givesThe Poisson equation implies that $\delta \Phi$ is continuous and its outward radial derivative has the jumpUsing regularity at the center and decay at infinity, writeThe derivative jump is $-(2l+1)C_l{R}_1^{l-1}{Y}_l^m$, so $C_l=-4\pi G{\rho }_{1}l{A}_l/(2l+1)$.

Integrating the bulk Euler momentum equation gives $\delta p/{\rho }_1={\omega }^{2}U-\delta \Phi$ for $l\ge 1$. At the free surface the Lagrangian pressure perturbation vanishes:Combining these relations yields the Kelvin stellar mode frequencyThe $2l+1$ choices of $m$ are degenerate because of spherical symmetry. For $l=1$, $r{Y}_1^m$ is a linear combination of coordinates in a Cartesian coordinate system, so $\nabla U$ is a constant displacement: the zero frequency is rigid translation of the whole isolated body. Such a translation has no restoring force. The constant $l=0$ potential generates no displacement; a radial breathing motion is excluded by incompressibility and a regular center.

The printed irrotational assertion needs the nonzero-frequency qualification, or restriction to the potential incompressible stellar surface modes. It is false for all neutral displacements: $\mathbf{\xi }=\mathbf{\Omega }\times \mathbf{r}$ has zero divergence, zero normal displacement at the surface, and $\delta p=\delta \Phi =0$, so it is a zero-frequency displacement, but $\nabla \times \mathbf{\xi }=2\mathbf{\Omega }\ne 0$. This does not change the Kelvin stellar mode spectrum just derived.

Let ${\Sigma }_{d0}$ be the uniform dust background and put ${\delta }_d={\Sigma }_d^{\prime }/{\Sigma }_{d0}$. The linearized continuity equation is ${\partial }_t{\delta }_d=-{\partial }_x{v}_x^{\prime }$ because background shear has no divergence or axisymmetric advection. Using the late entrained velocity $v_x^{\prime }=u\cos \theta$ givesThus the coherent travelling wave has maximum fractional excess $|ku|/\Omega$, and its crest density is ${\Sigma }_{d0}[1+|ku|/\Omega ]$ when the stationary part is absent. This is the density contrast of an entrained dust wave, and requires $|ku|/\Omega \ll 1$ in the present linear approximation. Its fractional oscillating density amplitude is the same as the gas's long-wavelength amplitude, so the coherent wave produces no additional dust-to-gas concentration ratio at this order.

Velocity damping alone does not erase the time-independent $C(x)$; initial density and the history of the forcing are needed to fix it. The printed data therefore determine the travelling-wave contrast, rather than a unique total late-time concentration for arbitrary initial histories. As an explicit example, switch on the prescribed gas epicycle at $t=0$ with initially uniform dust and ${\mathbf{v}}^{\prime }=0$. The corrected velocity pair gives $v_x^{\prime }=u(1-e^{-ϵ\Omega t})\cos \theta$. Integrating continuity then leavesat late times. It is smaller by order $ϵ$ in the stipulated weak-coupling regime, and confirms the stated leading wave-crest estimate while showing why arbitrary stationary density structure is not removed by the velocity transient.

Take $z$ upwards and write $\zeta =z+Wt$, so that the unperturbed interface is $\zeta =0$, with carbon dioxide above and water below. Use the sharp-interface, incompressible flow version of Darcy's law, with constant porosity and permeability of a porous medium, no capillary pressure, and effectively unbounded layers. The distinction between Darcy flux and interface speed matters: the base flux is $q_z=-\varphi W$, since the printed $W$ is the speed of the interface. Thus the base pressure gradients satisfyThe mass density ordering is stabilizing, whereas injection of the less viscous fluid into the more viscous fluid produces the Saffman–Taylor instability.

For linear stability analysis, displace the interface by $\eta =\hat{\eta }{e}^{\sigma t+i\alpha x}$. Taking the divergence of Darcy's law shows that the perturbation pressure satisfies the Laplace equation. Decay away from the interface gives, for $a_{\alpha }=|\alpha |>0$,The kinematic boundary condition and continuity of normal Darcy flux giveMeanwhile pressure continuity, evaluated on the displaced interface, givesEliminating $A,B$ yields the buoyancy-modified Darcy fingering dispersion relation:Every nonzero wavenumber decays when $W<W_c$, grows when $W>W_c$, and is neutral at equality, whereThe zero wavenumber is a uniform displacement and is neutral. Without a short-scale regularization such as surface tension, this model has no fastest-growing finite wavenumber: on the unstable side its growth rate increases without bound as $|\alpha |$ increases. Finite layer depths would change the dispersion relation; none are specified here.

Taking $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and a year of $365.25$ days, the data giveThe numerical paragraph supplies no porosity, so it cannot determine a unique numerical interface speed. If $W$ were instead intended to mean Darcy velocity, the numerical critical value would be $103.2\mathrm{m}\mathrm{y}{\mathrm{r}}^{-1}$, and the interface would move at $W/\varphi$. These two velocity conventions must not be conflated.

Take $h=2k_c$ and $D=d/dz$. Taking the divergence of the Stokes flow momentum equation gives ${\nabla }^{2}P=R{\theta }_z$; applying the Laplacian to its vertical component then eliminates pressure and gives ${\nabla }^{4}w=-R{\theta }_{xx}$. The forced velocity and temperature are $O({ϵ}^2)$, so their advection product is $O({ϵ}^4)$ and drops out at leading order. The steady heat equation reduces to $w_0+{\nabla }^2{\theta }_0=0$.

Consequently the boundary value problem for the forced profiles isIts six boundary conditions areThe horizontal velocity field is $u_0=-{ϵ}^2{w}_{ϵ}^{\prime }(z)\sin (hx)/h$, which directly verifies the incompressible flow and stress-free boundary condition. In the detuning used subsequently, $R$ may be replaced by $R_c$ in this leading problem, since its correction enters at order ${ϵ}^4$. The vertical profiles depend linearly on $\gamma$.

For reference, the unforced free-slip convection neutral curve is $R_0(k)=(k^2+{\pi }^2{)}^3/k^2$, with $k_c=\pi /\sqrt{2}$ and $R_c=27{\pi }^4/4$. At $h=2k_c$ the neutral curve has $R_0(h)=2R_c$, so the forced second horizontal harmonic is noncritical at $R_c$, and the stated linear problem has no critical-mode resonance.

Take the layer depth as the length unit, its thermal diffusion time as the time unit, and its imposed temperature difference as the temperature unit. Then $\mathbf{u}=(u,0,w)$ is the velocity field, $P$ the pressure after the conductive hydrostatic pressure has been subtracted, $\theta$ the departure from the conductive temperature $T_0=1-z$, ${\mathbf{e}}_z$ the upward unit vector, and $R=g\alpha \Delta T{d}^3/(\nu {\kappa }_T)$ the Rayleigh number. Here $d$ is the dimensional depth, $\nu$ the kinematic viscosity, ${\kappa }_T$ the thermal diffusivity and $\alpha$ the thermal expansion coefficient. The gradient, divergence and Laplacian act on the dimensionless coordinates. The term $w$ in the heat equation is advection of the background gradient, since $-\mathbf{u}\cdot \nabla T_0=w$. Mass density variations enter only the buoyancy term under the Boussinesq approximation; the incompressible flow constraint is $\nabla \cdot \mathbf{u}=0$.

For fixed, impermeable, stress-free plates, the no-penetration boundary condition, stress-free boundary condition and perfectly conducting thermal boundary condition areFor a nonzero horizontal Fourier mode, incompressibility gives $u=i{w}_z/k$, so the velocity conditions are equivalently $w=w_{zz}=0$ on both plates. A uniform horizontal velocity is not fixed by these boundary conditions; take its mean to be zero when discussing the stationary forced pattern.

The printed small-Prandtl-number label does not generally justify these equations. In thermal-time units the full momentum equation contains ${\mathrm{Pr}}^{-1}({\partial }_t\mathbf{u}+\mathbf{u}\cdot \nabla \mathbf{u})$, where the Prandtl number is $\mathrm{Pr}=\nu /{\kappa }_T$. Dropping that term gives infinite-Prandtl-number convection or another explicitly justified Stokes flow approximation. It is not the general $\mathrm{Pr}\to 0$ limit. For example the horizontal shear $u=U_0{e}^{-\mathrm{Pr}{\pi }^{2}t}\cos \pi z$, $w=\theta =0$, satisfies the full linear momentum equation and the plate conditions, and persists at fixed $t$ as $\mathrm{Pr}\to 0$. The printed inertia-free equation would instead require this shear to vanish. The following parts solve the equations actually displayed; applying them at small Prandtl number needs an additional slaving or creeping-flow ordering.

For a scalar $\varphi$, the displayed basis identities giveTaking the divergence and differentiating the basis vectors yields the Laplacian in polar coordinatesConsequently,

The regular harmonic mode with angular dependence $\cos 2\theta$ is $C{r}^2\cos 2\theta$. Its radial derivative at $r=a$ is $2Ca\cos 2\theta$, so $C=1/(2a)$. A constant is invisible to the Neumann boundary condition, givingIf two solutions existed, their difference $u$ would satisfy ${\nabla }^{2}u=0$ and ${\partial }_{n}u=0$. Green's first identity gives ${\int }_D|\nabla u{|}^2={\int }_{\partial D}u{\partial }_{n}u-{\int }_{D}u{\nabla }^{2}u=0$, so $\nabla u=0$ and $u$ is constant. Thus these are all the solutions.

Taking the divergence of the Ampère-Maxwell equation and using the Gauss's law $\nabla \cdot E=\rho /{\epsilon }_0$ giveswhich is charge conservation from Maxwell equations ${\rho }_t+\nabla \cdot J=0$.

By Ohm's law, $J=\sigma E$, so $\nabla \cdot J=\sigma \rho /{\epsilon }_0$. Thereforewith charge-relaxation rate $\sigma /{\epsilon }_0$ and time ${\epsilon }_0/\sigma$.

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.