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For constant-density incompressible inviscid flow with conservative body force, Euler equations for an inviscid fluid give ${\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla (p/\rho +\Phi )$. Using $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$, take the curl to obtain ${\partial }_t\mathbf{\omega }=\nabla \times (\mathbf{u}\times \mathbf{\omega })$. Since both divergences vanish, the vector identity expands this toWithout incompressibility or the constant-density/barotropic hypothesis, extra compression or baroclinic terms are present; the printed equation uses these flow assumptions.

Let $\kappa (t)=\dot{h}/h$, with $h>0$. Axisymmetric mass conservation is $r^{-1}{\partial }_r(r{u}_r)+{\partial }_z{u}_z=0$. If $u_r$ is independent of $z$, then $u_z$ is affine in $z$. The impermeable moving-plate conditions $u_z(r,\pm h,t)=\pm \dot{h}$ force $u_z=\kappa z$, and integrating the radial equation gives $u_r=-\kappa r/2+C(t)/r$. Regularity at the axis excludes $C(t)/r$. The meridional part of the axisymmetric inviscid flow between moving parallel plates is thereforeFor no swirl, $u_{\theta }=0$ completes the field. Continuity alone does not determine azimuthal motion; this must be specified separately. More generally, prescribe a smooth regular initial profile $U(r)$ independent of $z$. The azimuthal part of Euler equations for an inviscid fluid isso conservation of $r{u}_{\theta }$ along radial particle paths gives $u_{\theta }(r,t)=\lambda U(\lambda r)$, where $\lambda =\sqrt{h/h_0}$. Together with the meridional components above, this gives a family of complete fields, with a suitable pressure determined by the radial and vertical equations. The field is therefore not unique without rotational data.

With the stated solid-body rotation, $u_{\theta }=\Omega (t)r$ and $\mathbf{\omega }=2\Omega {\mathbf{e}}_z$. The vorticity equation gives $2\dot{\Omega }=2\Omega {\partial }_z{u}_z=2\Omega \kappa$, henceFor a fluid particle, $\dot{r}=-\kappa r/2$, so $r^{2}h$ is constant and $r^2\Omega$ is constant. This is conservation of angular momentum about the axis: radial spreading when the plates approach decreases the angular velocity. The meridional and rotating fields satisfy the inviscid equations with an appropriate pressure; no tangential no-slip condition is imposed on inviscid plates.

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.

Choose an origin near the bounded loop and write ${\mathbf{r}}^{\prime }$ for the source position. The magnetic vector potential and its far-field multipole expansion arewith the loop held fixed as $r=|\mathbf{r}|\to \infty$. The first term integrates to zero since $C$ is closed. Define its oriented vector area byThe second equality is Stokes theorem, applied componentwise to any oriented spanning surface. In particular the vector area is independent of the choice of spanning surface. From ${\oint }_{C}d(r_i^{\prime }{r}_j^{\prime })=0$, the matrix ${\oint }_C{r}_i^{\prime }d{r}_j^{\prime }$ is antisymmetric, which gives ${\oint }_C(\mathbf{r}\cdot {\mathbf{r}}^{\prime })d{\mathbf{r}}^{\prime }=\mathbf{S}\times \mathbf{r}$. HenceThe orientation follows the current by the right-hand rule. For $r\ne 0$, taking the curl of this leading magnetic vector potential gives the magnetic dipole fieldFor a planar loop, $\mathbf{S}$ is the signed area times the unit normal vector. For a nonplanar loop it is the oriented vector area, rather than the scalar area of an arbitrarily chosen surface.

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 $K={\rho }_0|\mathbf{u}{|}^2/2$ be the kinetic energy per unit volume. Dot the momentum equation with ${\rho }_0\mathbf{u}$. The Coriolis acceleration does no work because $\mathbf{u}\cdot ({\mathbf{e}}_z\times \mathbf{u})=0$. Incompressibility turns the advection and pressure terms into divergences, yieldingThus $\mathbf{F}=(K+P)\mathbf{u}$ is the energy flux and $S={\rho }_0{\Omega }^2(3x-2qz)u_x$ is the source in the requested kinetic-energy form.

To decide about conservation of energy, absorb the ordinary radial shearing-sheet tidal potential ${\Phi }_t=-3{\Omega }^2{x}^2/2$ into the density. Its material derivative cancels the $3x$ work term:For $q=0$ this is the usual conservative Jacobi energy in a shearing sheet, subject to vanishing boundary energy flux. For $q\ne 0$, the extra body force has nonzero curl, $\nabla \times [-2{\Omega }^{2}qz{\mathbf{e}}_x]=-2{\Omega }^{2}q{\mathbf{e}}_y$, so no scalar potential can absorb it. A potential $2{\Omega }^{2}qxz$ would necessarily introduce a vertical force $-2{\Omega }^{2}qx{\mathbf{e}}_z$, absent from the model. Therefore this vertical-shear model does not generically conserve the standard total mechanical energy: its maintained shear/irradiation background can supply or remove energy. This is the energy balance of the incompressible vertical-shear model.

Let $\mathbf{e}=\mathbf{X}/R$, $s=\mathbf{e}\cdot \mathbf{Be}$, and $A_{\parallel }=\mathbf{A}\cdot \mathbf{e}$. Applied forces and couples below are forces and torques on the solid; by force balance their values are also the strengths exerted by that solid on the fluid. This fixes the signs in Faxén's first law and Faxén's rotational law. To hold the passive sphere fixed, set its translation and angular velocity to zero in those laws. The incident two-mode tensorial squirmer flow givesOnly the stresslet part has vorticity:Consequently, for fixed $\mathbf{A},\mathbf{B}$ as $a/R\to 0$,The leading force is the radial stresslet force. Its next correction, $O(\mu a^4|\mathbf{A}|/R^3)$, comes from the incident potential dipole. The next $\mathbf{B}$ correction, $O(\mu a^5\Vert \mathbf{B}\Vert /R^4)$, contains both the swimmer's $r^{-4}$ field and the Laplacian term in Faxén's first law. If the relevant coefficient vanishes, these next terms must be retained instead of calling the vanished term a nonzero leading approximation. Further exchanges in the method of reflections for Stokes flow are smaller still.

The passive sphere reflects primarily a Stokeslet with strength ${\mathbf{F}}_h$. At the swimmer, its velocity is $(\mathbf{I}+\mathbf{ee}){\mathbf{F}}_h/(8\pi \mu R)$. The unchanged surface slip velocity adds the free swimming velocity to the incident-flow contribution in Faxén's first law. ThusThe leading change is $O(a^3\Vert \mathbf{B}\Vert /R^3)$; its next correction is the displayed $O(a^4|\mathbf{A}|/R^4)$ term. Finite-radius Faxén's first law corrections, the passive sphere's reflected stresslet, and its rotlet contribute at order $a^5\Vert \mathbf{B}\Vert /R^5$ to translation.

For a Stokeslet at the origin,The curl of the first term is $\mathbf{F}\times \mathbf{x}/r^3$; that of the second is also $\mathbf{F}\times \mathbf{x}/r^3$, before multiplying by $1/(8\pi \mu )$. Hence ${\mathbf{\omega }}_F=\mathbf{F}\times \mathbf{x}/(4\pi \mu r^3)$. At the swimmer use displacement $-\mathbf{X}$. The leading $\mathbf{B}$ force is parallel to $\mathbf{X}$, so its Stokeslet has zero vorticity there. This eliminates the putative $O(a^3\Vert \mathbf{B}\Vert /R^4)$ rotation. The potential dipole itself has no vorticity, but the holding force it induces is not radial. Applying Faxén's rotational law to its reflected Stokeslet yieldsThe passive sphere's rotlet and reflected stresslet can affect that last order, so they cannot restore the larger missing rotation. When $\mathbf{A}\times \mathbf{X}=0$, the boxed coefficient is zero and the higher-order terms determine any rotation.

For Darcy-Bénard convection, the dimensionless Darcy law is $\mathbf{u}=-\nabla p+R\theta {\mathbf{e}}_z$, with incompressibility and a heat equation about $T_0=1-z$. Here $R$ is a Darcy thermal-time Rayleigh number, with permeability replacing the squared-depth dependence in the fluid-layer Rayleigh number. The stated stream function convention gives $\mathbf{u}=(-{\psi }_z,0,{\psi }_x)$, so $w={\psi }_x$. The vertical-plane curl of Darcy law gives ${\nabla }^2\psi =R{\theta }_x$; omitting the quadratic perturbation advection from the heat equation gives ${\theta }_t={\psi }_x+{\nabla }^2\theta$.

On the impermeable connected square boundary, zero normal velocity makes $\psi$ constant along the boundary; choose that constant as zero. The prescribed conductive boundary temperature fixes $\theta =0$ on all four sides. Thus the perfect-conductor wording is interpreted through the supplied homogeneous perturbation data: the side temperatures retain the background vertical profile, rather than imposing a different isothermal side state.

At marginal stability put $q=\sqrt{R}$ and $P=\psi +iq\theta$. The two coupled equations implyThe gauge transformation for conducting-square Darcy onset $P=e^{-iq(x-1/2)/2}F$ removes the first derivative:The square's Dirichlet Laplacian eigenfunctions are $\sin (m\pi x)\sin (n\pi z)$, with eigenvalues of $-{\nabla }^2$ equal to ${\pi }^2(m^2+n^2)$, $m,n\ge 1$. The first marginal value is thereforeIt is genuinely the first instability threshold, not just an isolated neutral value. Eliminating $\psi$ gives the temperature linear operator $L_R={\Delta }_D+R{\partial }_x{\Delta }_D^{-1}{\partial }_x$, where ${\Delta }_D$ is the Dirichlet realization of an elliptic operator for the Laplacian. Integration by parts shows the second term is self-adjoint and nonnegative, since its quadratic form is $-⟨{\theta }_x,{\Delta }_D^{-1}{\theta }_x⟩\ge 0$. Thus the spectrum is real, starts negative at $R=0$, and its leading eigenvalue cannot cross into growth without a zero eigenvalue. The smallest zero value is the one found above.

Let $S=\sin \pi x\sin \pi z$, $\xi =x-1/2$ and $a=q_c/2=\sqrt{2}\pi$. Multiplying the lowest real $F=S$ by independent real and imaginary constants gives two real physical eigenfunctions:The subscripts indicate even and odd parity about $x=1/2$. Because $S$ is even in $\xi$, the first has even $\psi$ and odd $\theta$, while the second reverses those parities. They are linearly independent and satisfy all four Dirichlet boundary conditions. The lowest complex $F$ eigenfunction is simple over the complex numbers, but its arbitrary complex multiplier supplies two independent real pairs; this explains the physical degeneracy.

Consider a horizontal fluid layer under the Boussinesq approximation heated from below, rotating uniformly about its vertical axis. To make the linear formulas definite, take depth one, stress-free impermeable fixed-temperature plates, a horizontally infinite or sufficiently large periodic domain, and negligible centrifugal-buoyancy modifications. Rigid plates change the vertical eigenfunctions and thresholds and require their own boundary value problem; the explicit formulas below use the free-slip model. Let $P=\nu /{\kappa }_T$ be the Prandtl number, $\mathrm{Ta}=(2\Omega d^2/\nu {)}^2$ the Taylor number, and $R$ the Rayleigh number. Time is again measured on the thermal diffusion time.

The linearized rotating Rayleigh-Bénard convection equations about conduction areThe Coriolis force does no work on the fluid, but it couples vertical motion to vertical vorticity and thereby changes the balance of buoyancy and viscous damping. On the plates require $w=\theta =0$ and ${\partial }_z{u}_x={\partial }_z{u}_y=0$. Write $\zeta =(\nabla \times \mathbf{u}{)}_z$. Taking a curl and a double curl removes pressure and givesA normal mode with horizontal wavenumber $k>0$, vertical index $n\ge 1$ and complex growth rate $s$ has $w=W\sin (n\pi z)e^{st+i\mathbf{k}\cdot \mathbf{x}}$, $\theta =\Theta \sin (n\pi z)e^{st+i\mathbf{k}\cdot \mathbf{x}}$ and $\zeta =Z\cos (n\pi z)e^{st+i\mathbf{k}\cdot \mathbf{x}}$. Put $a=k^2+n^2{\pi }^2$. The three equations become $a(s/P+a)W+\sqrt{\mathrm{Ta}}n\pi Z=R{k}^2\Theta$, $(s/P+a)Z=\sqrt{\mathrm{Ta}}n\pi W$, and $(s+a)\Theta =W$. Their determinant gives the rotating-convection growth-rate polynomialThis cubic includes viscous and thermal modes as well as the convective instability. It avoids excluding a root by division by $s/P+a$. Linear stability means every root has negative real part; stationary onset has $s=0$, while overstability has $s=\pm i\omega$ with nonzero angular frequency.

Setting $s=0$ gives the stationary neutral curve of rotating convectionFor oscillatory convection, set $s=i\omega$ and equate real and imaginary parts. Eliminating $R$ yieldsThe oscillatory neutral curve of rotating convection is physically admissible only when ${\omega }^2>0$. Thus $P<1$ and sufficiently rapid rotation are necessary; a formal minimum with ${\omega }^2\le 0$ is not a Hopf bifurcation. Convection begins at the smaller of the stationary minimum and the admissible oscillatory minimum. The angular frequency approaches zero where a fixed-wave-number stationary and oscillatory threshold meet; that is a double-zero limit of the cubic, requiring a different slow-time reduction.

For wavenumber selection in rotating convection, the lowest vertical index is $n=1$. To compare indices, set $k^2=n^2{\pi }^{2}y$ at fixed $y>0$. Both neutral thresholds have a positive term proportional to $n^4$ and a rotation term independent of $n$, while the admissible angular frequency squared decreases with $n$. Their minima therefore cannot improve on $n=1$. Set $t=k^2$, $m={\pi }^2$. Minimizing each neutral curve givesFor the latter equation also check ${\omega }^2>0$, or minimize over the admissible set instead. Without rotation, $k_c=\pi /\sqrt{2}$ and $R_c=27{\pi }^4/4$. At rapid rotation, $k_s\sim (\mathrm{Ta}{\pi }^2/2{)}^{1/6}$ and $R_{s,c}\sim 3(\mathrm{Ta}{\pi }^2/2{)}^{2/3}$: rotation narrows the horizontal convection rolls and raises their threshold. In a finite box only discrete wavenumbers are permitted, so the minimum must be taken over those allowed modes.

The rapid-rotation comparison has $R_{o,c}/R_{s,c}\sim 2P^{4/3}/(1+P{)}^{1/3}$. Its equality gives $8P^4-P-1=0$, with positive root about $0.67660$. Below this value, sufficiently rapid rotation can make oscillatory convection the primary instability; low Prandtl number allows inertial motions to interact with the thermal field before viscosity damps them. This estimate describes the selected neutral-curve comparison in the stated free-slip asymptotic model, rather than claiming every $P<1$ has oscillatory primary onset at every rotation rate.

Linear selection does not determine the nonlinear planform or its stability. A weakly nonlinear expansion with a solvability condition gives amplitude equations. A stationary convection roll amplitude has the Landau amplitude equation $A_T=rA-g|A{|}^{2}A$; a supercritical branch requires $g>0$ and has $|A{|}^2=r/g$, whereas $g<0$ requires higher-order saturation and permits subcritical behaviour. The coefficient depends on the physical parameters and boundary conditions, and should not be assumed positive merely because the linear threshold is known.

A nonzero-frequency Hopf bifurcation supplies oppositely travelling convection roll amplitudes $Z_{+},Z_{-}$. At a fixed orientation, a cubic normal form iswith complex $g,h$ and the understood symmetry exchanging propagation directions. For $r>0$ and $g_r=\mathrm{Re}g>0$, a travelling-roll branch has one nonzero amplitude, of squared modulus $r/g_r$; it is stable to the opposite travelling amplitude when $h_r=\mathrm{Re}h>g_r$. A standing-roll branch has equal squared moduli $r/(g_r+h_r)$, and is stable within this pair when $g_r>h_r$ and $g_r+h_r>0$. The imaginary parts shift angular frequencies. These conditions concern amplitude perturbations in that reduced subspace; orientation and modulation modes still have to be tested. This is the competition of travelling and standing convection rolls.

Near simultaneous stationary and oscillatory thresholds, a codimension-two bifurcation requires retaining both types of mode. After selecting an orientation and fixing a steady spatial phase, a schematic nonresonant steady–Hopf mode interaction isHere $a$ is real, $z$ is the chosen complex oscillatory amplitude, $g_s,h_s$ are real, and $g_o,h_o$ may be complex. Assume $r_s>0$, $g_s>0$ for a radially stable pure steady branch, and $r_o>0$, $g_{o,r}=\mathrm{Re}(g_o)>0$ for a radially stable pure oscillatory branch. Steady convection rolls suppress the oscillatory mode if $r_o-h_{o,r}{r}_s/g_s<0$, where $h_{o,r}=\mathrm{Re}(h_o)$; oscillatory convection rolls suppress the steady mode if $r_s-h_s{r}_o/g_{o,r}<0$. These are transverse tests for branches whose existence and radial stability have already been checked. Mixed states have positive intensities $u=a^2$, $v=|z{|}^2$ solving $g_{s}u+h_{s}v=r_s$, $h_{o,r}u+g_{o,r}v=r_o$. If $D=g_s{g}_{o,r}-h_s{h}_{o,r}\ne 0$, then $u=(r_s{g}_{o,r}-h_s{r}_o)/D$, $v=(g_s{r}_o-h_{o,r}{r}_s)/D$. For positive $u,v$ and the positive self-saturation coefficients above, the intensity Jacobian matrix has negative trace and determinant $4uvD$, so the mixed state is attracting in intensities when $D>0$ and is a saddle equilibrium when $D<0$; phase and other-mode perturbations remain separate tests. Depending on the cross-couplings, one obtains coexistence or competition and bistability; degeneracies or resonances require extra terms. A full travelling/standing-wave competition must retain both $Z_{\pm }$, not just this illustrative one-mode $z$. If the Hopf angular frequency tends to zero, averaging over fast oscillations fails and a double-zero normal form must keep the corresponding two slow variables. A crossing of global minima at distinct wavenumbers is instead an interaction of distinct modes, not automatically the same double-zero problem.

Weakly nonlinear convection rolls can also lose stability spatially. The real Ginzburg–Landau equation $A_T=rA+\xi A_{XX}-g|A{|}^{2}A$, with $r,\xi ,g>0$, has detuned convection rolls with $|A{|}^2=(r-\xi Q^2)/g$. Linear phase modulation gives diffusion coefficient $\xi (r-3\xi Q^2)/(r-\xi Q^2)$, so the Eckhaus instability excludes $Q^2>r/(3\xi )$ even though convection rolls exist up to $Q^2<r/\xi$. Transverse bending and mean-flow couplings impose additional restrictions in the actual rotating layer; the scalar equation is a local longitudinal example, not its complete stability theory.

Most distinctively, rotation breaks mirror symmetry between competing convection roll orientations. For two sets of convection rolls at relative angle $\vartheta$, write ${\dot{A}}_2=r{A}_2-g_0|A_2{|}^2{A}_2-g(\vartheta )|A_1{|}^2{A}_2$ and interchange indices with $\vartheta \mapsto -\vartheta$ for the other equation. The established convection roll $|A_1{|}^2=r/g_0$ is unstable to the new orientation when its linear growth rate $r[1-g(\vartheta )/g_0]$ is positive. The inequality need not be symmetric under $\vartheta \mapsto -\vartheta$ because the imposed rotation supplies handedness. This is the Küppers–Lortz instability: sufficiently rapid rotation can destabilize steady convection rolls immediately above their stationary onset to oblique convection rolls of another orientation. Successive replacements can produce time-dependent convection roll switching or heteroclinic cycles, instead of a stable single convection roll pattern. It requires three-dimensional perturbations even when the original convection roll solution is independent of its axial coordinate.

At finite Prandtl number and stress-free plates, nearly parallel convection rolls can couple resonantly to a slowly damped large-scale Eulerian mean flow. This small-angle instability of rotating convection rolls makes a regular two-roll expansion nonuniform as the angle tends to zero; one must retain the mean-flow mode. It should not be identified automatically with the finite-angle Küppers–Lortz instability or assigned a universal threshold from a one-amplitude equation. Consequently the onset type and selected scale follow from the admissible neutral curves, while persistent convection roll patterns require a separate nonlinear and sideband stability calculation.

For a time-independent current, the Maxwell equations reduce to magnetostatics:Write $\mathbf{B}=\nabla \times \mathbf{A}$ using a vector potential and choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. The vector identity for the curl of a curl gives the Poisson equationThe free-space Green function therefore givesTaking the curl with respect to $\mathbf{r}$, moving it under the integral, and using $\nabla (1/|\mathbf{r}-{\mathbf{r}}^{\prime }|)=-(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$ gives the Biot-Savart law

Use the Euler equations for an inviscid fluid with no magnetic or gravitational force. Let $w$ denote specific enthalpy. For a homentropic flow, $dw=dp/\rho$, so the pressure acceleration is $-\nabla w$. The velocity identitytherefore givesTaking the curl, using that the curl of a gradient is zero and that differentiation commutes for smooth fields, proves barotropic vorticity transport:More generally the same proof works for any barotropic fluid, replacing $w$ by a pressure potential ${\int }^{\rho }{p}^{\prime }(s)ds/s$. Here “isentropic” must supply this barotropic closure, for example through one common specific entropy throughout the fluid. Merely imposing entropy advection equation $Ds/Dt=0$ allows spatial specific entropy gradients; in that case the vorticity equation contains the additional baroclinic vorticity generation term ${\rho }^{-2}\nabla \rho \times \nabla p$.

Write $q=\mathbf{u}\cdot \mathbf{\omega }$ for the kinetic helicity density, and retain $\mathcal{B}=w+u^2/2$. Since $\mathbf{\omega }$ is a curl, $\nabla \cdot \mathbf{\omega }=0$. Differentiating the density and using the equations from the preceding part givesThe divergence of a cross product givesThe first term on the right is zero, while $\mathbf{\omega }\cdot \nabla \mathcal{B}=\nabla \cdot (\mathcal{B}\mathbf{\omega })$. Thus the kinetic helicity conservation law has fluxThe last equality for the flux uses the vector triple product identity. Integrating and applying the divergence theorem gives $d{\int }_{V}qdV/dt=-{\int }_{\partial V}{\mathbf{F}}_{H_k}\cdot \mathbf{n}dS$ for a fixed volume. Hence its total kinetic helicity is conserved when the boundary flux vanishes; for example, tangency of both velocity field and vorticity to the boundary is sufficient. The local conservation law does not by itself imply that its density follows each fluid particle unchanged.

For an axisymmetric vector field, write $\mathbf{B}=B_R{\mathbf{e}}_R+B_{\varphi }{\mathbf{e}}_{\varphi }+B_z{\mathbf{e}}_z$. Pure differential rotation givesThis vector has no azimuthal component and is independent of $\varphi$, so its curl has zero radial and vertical components. Its azimuthal component isThe final braces are $R\nabla \cdot \mathbf{B}=0$ by Gauss's law for magnetism. The ideal magnetohydrodynamic induction equation therefore yields axisymmetric magnetic winding:Differential rotation creates a toroidal magnetic field from the fixed poloidal magnetic field. A stationary field under the same purely rotational assumptions requires Ferraro's law of isorotation, ${\mathbf{B}}_p\cdot \nabla \Omega =0$.

Let $\mathbf{\zeta }=\nabla \times \mathbf{u}$ and $\mathbf{\omega }=\mathbf{\zeta }+2\Omega {\mathbf{e}}_z$ be the relative vorticity and absolute vorticity. The nonlinear acceleration identity rewrites the momentum equation asTaking the curl removes the pressure, kinetic energy gradient and shearing-sheet tidal potential. ThusBoth $\nabla \cdot \mathbf{u}$ and $\nabla \cdot \mathbf{\omega }$ vanish. The vorticity equation is consequentlyFor the absolute-vorticity flux tensor $T_{ij}=u_i{\omega }_j-{\omega }_i{u}_j$, use the flux convention $(\nabla \cdot \mathbf{T}{)}_j={\partial }_i{T}_{ij}$, with the first index denoting transport direction. This givesThe scalar $\theta$ is a buoyancy displacement variable, so dimensional consistency gives it units of length; the printed potential-temperature terminology presupposes this normalization. Its curl source directly forces only the $y,z$ components of absolute vorticity. Hence radial vorticity conservation in a shearing sheet takes the local flux formIts volume integral is constant when the net boundary flux vanishes. However $D_t{\omega }_x=\mathbf{\omega }\cdot \nabla u_x$ still permits vortex stretching; the radial component is not generally a materially conserved scalar in three dimensions. Using divergence on the second index of the stated tensor would reverse the transport sign, so the tensor convention matters.

On the simply connected domain ${\mathbb{R}}^3$, a continuously differentiable differential one-form is exact if its mixed partial derivatives agree, equivalently if its associated vector field has zero curl. Here one can verify exactness directly by observing thatsatisfiesThus $udx+vdy+wdz=dF$. By the fundamental theorem for line integrals, its integral is path independent and equalsConsequently, for every path with the stated endpoints,

Work on a beta plane with $f=f_0+\beta y$ and constant depth $H$. Assume a steady, small-Rossby number interior, negligible friction there, small sea-surface displacements, and negligible depth gradients. Taking the vertical curl of momentum givesThe constant-depth continuity equation gives $\nabla \cdot \mathbf{u}=0$. Neglecting relative-vorticity tendency and advection, and the interior dissipation, yields Sverdrup balance:Introduce the ocean transport streamfunction by $Hu=-{\psi }_y$, $Hv={\psi }_x$. Then $\beta {\psi }_x=H(\nabla \times \mathbf{W}{)}_z$. Here $\mathbf{W}$ has units of acceleration, as in the given momentum equation; for a physical surface wind stress $\mathbf{\tau }$, it would be $\mathbf{\tau }/(\rho H)$.