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Potential-vorticity inversion is the diagnostic recovery of the flow from its advected vortical scalar, using the balance relation and specified boundary conditions. Its usefulness is a separation of tasks: transport the scalar by the current velocity, invert to find the new velocity, and repeat. The inverse is spatially nonlocal even though the transport law is local. The four models differ chiefly in the operator being inverted and the information needed at boundaries.

In two-dimensional vortex dynamics, planar incompressibility permits $\mathbf{u}=(-{\psi }_y,{\psi }_x)$, and vertical vorticity is $\zeta ={\nabla }^2\psi$. The curl of inviscid momentum givesThere is no vortex stretching in strictly planar motion, so vorticity is materially conserved. The two-dimensional potential-vorticity inversion is the Poisson equation ${\nabla }^2\psi =\zeta$. On the plane, with suitable integrability and far-field behavior, its Green's function is $(2\pi {)}^{-1}\log |\mathbf{x}-{\mathbf{x}}^{\prime }|$, givingDifferentiation gives the planar Biot-Savart kernel. Here ${\psi }_h$ is a harmonic function fixed by boundary, circulation and far-field data. A rigid impermeable wall is a streamline, so $\psi$ is constant on it; constants or circulations around distinct holes require specification. The vorticity alone cannot distinguish two flows differing by a permitted harmonic contribution, such as a uniform current on an unbounded plane. A constant added to $\psi$ has no physical effect.

In nonrotating layerwise-two-dimensional vortex dynamics, strong stable density stratification constrains motion to almost horizontal layers. In the ideal independent-layer limit, vertical velocity vanishes at leading order, $(u,v)=(-{\psi }_y,{\psi }_x)$ andhold separately at each fixed $z$. Height is a parameter: the horizontal inverse Laplacian contains no vertical derivative. With background buoyancy $N^{2}z$, the leading Ertel potential vorticity is $Q=\mathbf{\omega }\cdot \nabla b\simeq N^2\zeta$, so the materially advected normalized scalar is $Q/N^2\simeq \zeta$. The same horizontal Poisson equation is inverted layer by layer, using each layer's boundary data. This is an asymptotic constrained model, not arbitrary three-dimensional Euler motion with vertical velocity set to zero: its validity requires small Froude number and controlled vertical scales. At finite buoyancy frequency, the weak vertical motion and tilted isopycnals restore coupling. The inversion may be independent by layer while the reconstructed fields vary with height. The horizontal Euler equations for an inviscid fluid determine the pressure gradient, while hydrostatic balance relates its vertical derivative to buoyancy.

In shallow-water quasi-geostrophic inversion, begin with the exact shallow-water potential vorticity $(f+\zeta )/h$. Linearizing its anomaly about depth $H$ gives, after multiplying by $H$, $\zeta -f_0\eta /H$. For small Rossby number, geostrophic balance gives $\psi =g\eta /f_0$, while $\zeta ={\nabla }_h^2\psi$. Keeping the beta plane background gradient givesThus $q-\beta y$ is inverted with the modified Helmholtz equation $({\nabla }_h^2-L_D^{-2})\psi =q-\beta y$, under the specified boundary and far-field conditions. The velocity follows from $\psi$ and surface elevation from $\eta =f_0\psi /g$. A horizontal Fourier mode of the anomaly satisfiesThe Rossby deformation radius therefore limits the reach of the balanced response. Layer stretching, represented by $-\psi /L_D^2$, is absent from ordinary planar vorticity inversion. Unlike the pure Poisson equation, the finite-deformation operator also responds to the constant part of $\psi$, which specifies mean elevation rather than an arbitrary streamfunction offset. Specified bottom topography adds a known source to $q$; it is subtracted along with the background before inversion.

In stratified quasi-geostrophic inversion, continuously varying vertical displacement replaces the shallow layer's single stretching term. For positive $N^2(z)$ and constant nonzero $f_0$, the three-dimensional quasi-geostrophic potential vorticity isGeostrophic balance and hydrostatic balance give velocity $(-{\psi }_y,{\psi }_x)$ and buoyancy perturbation $b=f_0{\psi }_z$. Invert the anisotropic elliptic boundary value problem using interior $q$, prescribed top and bottom buoyancy as Neumann boundary conditions, and the lateral, far-field and global-flow data. Boundary buoyancy evolves by its own material conservation when the boundary is impermeable. It cannot be discarded: the Eady edge wave in the preceding solution has zero interior potential vorticity anomaly but nonzero boundary buoyancy and a nonzero induced flow.

For constant $N,f_0$, the stretched coordinates for quasi-geostrophic inversion $Z=Nz/|f_0|$ turn the inversion operator into the three-dimensional Laplacian. Its infinite-space Green's function is $-1/(4\pi |\mathbf{X}-{\mathbf{X}}^{\prime }|)$ in $(x,y,Z)$ coordinates. Hence a horizontal disturbance of scale $L$ communicates vertically over scale $|f_0|L/N$, explicitly coupling layers. In a bounded domain with homogeneous difference data, multiplying the difference of two inversions by its streamfunction and applying integration by parts givesThis proves uniqueness up to a constant when the relevant boundary terms vanish; a mean or pressure reference fixes the remaining constant. Compatible boundary fluxes are needed for a purely Neumann boundary-value problem.

Across all four models, inversion provides the induced flow, while scalar transport supplies time evolution. The independent horizontal inverse in the first two models becomes a finite-deformation inverse in shallow water and a vertically coupled elliptic inverse in continuous stratification. The principle applies within the stated balance or constrained model. In the unrestricted Boussinesq equations or shallow water equations, gravity waves and divergent motion can share the same potential vorticity, so the vortical scalar alone does not reconstruct every possible flow.

Let $\theta =T-T_0>0$ and ${\theta }_1=T_1-T_0$. The aperture spans the enclosure's depth $H$, so the incoming volume flux, equal to the outgoing volume flux, isOnly one of these equal exchange fluxes multiplies the temperature difference in the heat budget: incoming fluid brings $q{T}_0$, outgoing fluid removes $qT$. With constant mass density and specific heat capacity, those factors cancel and the well-mixed ventilation temperature balance isIntegrating from ${\theta }_1$ givesThe cooling is algebraic rather than exponential because the buoyancy-driven single-opening exchange flow weakens as its driving reduced gravity decreases. The formula assumes negligible wall heat capacity, no heat input, the specified aperture geometry, and a uniform interior temperature maintained by mixing.

With a low-level opening, the dense incoming cold fluid tends to spread along the floor instead of falling through the warm interior. A cold lower layer and warm upper layer develop, giving stable density stratification. The incoming fluid can short-circuit back out through the low opening after that layer occupies it, leaving warm fluid above poorly ventilated. Thus the uniform-temperature assumption is generally inappropriate: cooling becomes vertically nonuniform and substantially slower for the upper warm fluid. If mixing were artificially maintained, the same lumped budget would still apply; it is the loss of that assumption, rather than a reversal of buoyancy, that changes the physical outcome.

In the top-hat plume model, vertical velocity $w(z)$ and reduced gravity $G(z)$ are uniform over $-b<x<b$. The ambient has zero vertical velocity, negligible ambient turbulence and a prescribed $T_0(z)$. Thenand the equations becomeEquivalently, the ordinary differential equations for the three unknowns areThe last equation displays dilution by fluid entrainment as well as loss of relative buoyancy while rising through a stably stratified ambient. The model assumes a steady slender two-dimensional turbulent plume, large Reynolds number, the Boussinesq approximation, constant $\beta$ and $\alpha$, negligible heat loss, negligible imposed crossflow, and negligible pressure departure from ambient hydrostatic pressure. The factors of two count both entraining edges; $b$ is a half-width. A finite source also requires its volume, momentum and buoyancy fluxes as boundary data.

Let the closed box have ceiling height $H$ above the source. The printed width $2R$ alone does not specify this height, so it must appear as an additional geometrical parameter. Take $t=0$ here when the plume has first reached the ceiling and its thin outflow has spread across the box. The leading filling box model assumes a negligible source volume, insulated boundaries, fast horizontal spreading, slow evolution compared with the plume transit time, negligible diffusion, and $b\ll R$. Below the interface the ambient still has its original temperature, so the local plume flux remains $q(z)=2\alpha w_{\ast }z$ from part (c).

The ceiling outflow creates a warm upper region. Fluid entrainment draws the original lower fluid into the plume, making that upper region deepen. The filling-box first front is the lower material boundary of fluid already modified by plume discharge, separating it from the initially unmodified ambient; it need not bound a uniformly mixed upper layer. With negligible occupied plume area, volume conservation at its height $h(t)$ givesThereforeIf source switch-on is the time origin instead, replace $t$ by elapsed time after the initial transit and spreading stage; the quasi-steady model does not resolve that stage.

The buoyancy jump requires distinguishing ambient fluid at the front from the plume crossing it. Just after the first ceiling outflow, the ambient buoyancy immediately above the nascent front isOutside the narrow plume, ambient temperature is transported without diffusion. If $G_a$ is its buoyancy relative to the original fluid, its equation isThe first front follows precisely these material trajectories, so the temperature of the fluid immediately above it is conserved. The initially unmodified fluid below has $G_a=0$. Thus the ambient jump isBy contrast, the contemporaneous plume immediately below the front has $G_p(h,t)=w_{\ast }^2/h(t)$, which grows as the front descends. That plume value is not the ambient jump across the material first front. Equating the two would inadvertently impose a uniformly mixed upper layer and discard the stable density stratification of the standard filling box model. Diffusion or extensive interfacial mixing can smooth and alter the ideal jump.

The filling box model requires the first outflow to spread across the enclosure and remain above unmodified fluid, while a narrow nearly vertical plume rises below it. Several mechanisms can defeat those assumptions. A tall narrow box makes the plume a substantial fraction of the width, so its compensating return velocity is no longer small and modifies both fluid entrainment and momentum flux. A fast ceiling current can descend the sidewalls and overturn the initially stable upper region. In both cases a horizontally uniform slowly descending filling-box first front is inappropriate. Strong plume displacement from the centre, imposed crossflow, large source volume, vents or substantial wall cooling likewise alter the volume and heat budgets.

A useful diagnostic is whether the ceiling outflow's inertial acceleration, of order $U_c^2/{\ell }_c$, overwhelms the restoring buoyancy of the nascent layer. If $U_c^2/(G_H{d}_c)$ is not small for its depth $d_c$, the outflow can penetrate or overturn the stratification rather than remain trapped above it. This is a diagnostic Froude number, not a universal threshold without specifying the wall-current geometry and mixing. Low Reynolds number invalidates a constant turbulent entrainment coefficient, and thermal diffusion or wall heat loss can erase the sharp first-front jump. Thus large aspect ratio, strong return circulation, overturning, ventilation or diffusion require a model beyond the standard filling-box approximation.

Use the Boussinesq approximation and a top-hat plume model. Let $v(z)<0$ be the uniform ambient return velocity outside the plume. Negligible source volume flux and an impermeable cylindrical wall require zero net transport through every horizontal section:The entrainment speed must be specified relative to that moving ambient. Taking it to be $\alpha (w-v)$ givesThe surrounding unmodified fluid has zero buoyancy relative to the initial fluid. Neglecting thermal or compositional loss therefore givesLet $B_s$ denote the conserved specific source buoyancy flux: $B_s=B_o/\pi$ if $B_o$ is the actual integrated source flux, or $B_s=B_o$ if the source strength has already been normalized by $\pi$.

To obtain the usual integral-model singularity one must also state a pressure closure. Here take the whole-section mean pressure to have only its reference ambient hydrostatic pressure, and neglect wall force and extra vertical turbulent-stress fluxes in the integral budget. Internal momentum transfers between plume and ambient cancel in that whole-section balance. The remaining upward force is the plume buoyancy, givingThe two momentum fluxes add, even though $v$ is negative: vertical momentum transported downward also contributes $v^2$ to the flux through an upward-oriented horizontal section. These three flux balances, with the algebraic return-velocity constraint, govern $b,w,g^{\prime }$ in the confined plume with compensating return flow.

The pressure-neglecting closure is an integral approximation, not a claim that the return fluid has zero acceleration. If a mean pressure departure $p_{\ast }(z)$ is retained, the whole-section momentum balance instead contains $-(R^2/{\rho }_0)p_{\ast }^{\prime }(z)$, along with any wall-force or stress contribution. The entrainment hypothesis alone does not specify that function. The singularity calculated below belongs to the stated pressure-neglecting model; it is not forced by continuity alone or asserted to be an infinite velocity in the real fluid.

Use $p$ for disturbance pressure divided by the constant inertial density. The ideal, nonrotating Linearized Boussinesq equations areHere $N^2>0$ is a prescribed time-independent buoyancy frequency squared. Taking the scalar product of momentum with velocity gives ${\partial }_t(|\mathbf{u}{|}^2/2)=-\nabla \cdot (p\mathbf{u})+\sigma w$, using incompressibility. Multiplying the buoyancy equation by $\sigma /N^2(z)$ gives ${\partial }_t[{\sigma }^2/(2N^2)]=-\sigma w$. No spatial derivative of $N$ is needed in this cancellation. ThusThis is positive disturbance energy per unit inertial density; its flux is pressure work.

For constant $N$, write a plane internal gravity wave with wave vector $\mathbf{K}=({\mathbf{k}}_h,m)$ and phase $\theta ={\mathbf{k}}_h\cdot {\mathbf{x}}_h+mz-\omega t$. Let $k_h=|{\mathbf{k}}_h|>0$ and $K^2=k_h^2+m^2$. In complex amplitudes, incompressibility and horizontal momentum giveConsequently $\hat{p}=-\omega m\hat{w}/k_h^2$. Substitution into vertical momentum, $-i\omega \hat{w}=-im\hat{p}+\hat{\sigma }$, yieldsThe same dispersion relation follows directly from the displacement equation: its two terms become $K^2{\omega }^2\hat{\zeta }$ and $-N^2{k}_h^2\hat{\zeta }$. The positive-frequency branch is $\omega =N{k}_h/K$; the other branch is its sign reversal.

Choose real vertical-displacement amplitude $A$. The internal-wave polarization can be writtenVelocity and displacement lie along the phase planes, perpendicular to $\mathbf{K}$, and are a quarter cycle out of phase. Buoyancy is opposite in sign to vertical displacement. In the requested two-dimensional sketch take $k<0$, $m<0$ and $\omega >0$: the phase velocity $\omega \mathbf{K}/K^2$ points down and left. Then $\mathbf{e}=(-m/k,1)$ points up and left, and positive $p$ accompanies positive $w$ and negative $u$. Negative-pressure bands have the opposite velocity. Multiplication by pressure therefore transports energy up and left in both signs of band.

Differentiating the positive-frequency dispersion relation gives the group velocityIn particular $\mathbf{K}\cdot {\mathbf{c}}_g=0$. Since $|\mathbf{e}{|}^2=K^2/k_h^2$, the wavelength averages areUsing $N^2={\omega }^2{K}^2/k_h^2$ in the last expression provesFor $k,m<0$ the group velocity points up and left, perpendicular to the down-left phase velocity. The pressure sign pattern, rather than the direction of moving crests, determines the energy-flux direction shown below.

For the moving-boundary problem, first take the fluid to occupy $z>0$ with the boundary near $z=0$. An outgoing boundary-forced internal gravity wave must have $c_{gz}>0$. With the specified $k>0$ and $\omega >0$, the radiation condition therefore selectsIf instead the fluid occupies $z<0$, radiation requires $c_{gz}<0$ and the positive root for $m$. The fluid half-space is needed to select a unique sign; the following formulas use the upper half-space.

For slowly modulated internal-wave boundary forcing, put $\zeta =A(T,Z)e^{i\theta }$, $T=\mu t$, $Z=\mu z$. Acting on $A$ by ${\partial }_t\mapsto -i\omega +\mu {\partial }_T$ and ${\partial }_z\mapsto im+\mu {\partial }_Z$, the displacement equation has leading term $(K^2{\omega }^2-N^2{k}^2)A$. Its order-$\mu$ term isThus characteristics carry the boundary amplitude into the fluid:Real parts are understood. To the same leading order,Slow-envelope corrections to these polarization formulas are order $\mu$. At $T=2$, the amplitude is $ϵ$ for $0\le Z\le c_{gz}$, falls smoothly across $c_{gz}<Z<2c_{gz}$, and is zero beyond $2c_{gz}$. This translating ramp is also shown in the figure.

The mean boundary work per unit horizontal area is $\overline{W}={\rho }_{00}\overline{p(0)w(0)}$ at quadratic disturbance order. The small boundary displacement permits evaluation at the unperturbed plane. From the flux calculation,The finite causal wave train has energy integralDifferentiation, or integration of the local energy law with zero flux beyond the front, proves

Finally, distinguish the oscillatory velocity from the second-order Eulerian mean $\overline{u}$. The displacement polarizations giveThe supplied mean-flow relation therefore yields ${\overline{u}}_t=(k/\omega ){\overline{E}}_t+O({\mu }^2)$ at quadratic wave order. Equivalently, the horizontal momentum flux is $\overline{uw}=(k/\omega )c_{gz}\overline{E}$, and envelope-energy transport gives the same acceleration through ${\overline{u}}_t=-{\partial }_z\overline{uw}$.

In the frame moving horizontally at $c=\omega /k$, the boundary pattern is stationary and the initial mean current is $-c$. Subtract that background's kinetic energy before integrating over the infinite half-space. The internal-wave work in a phase-speed frame is thenIts rate of change isThe mean-current cross term supplies the quadratic energy loss in this moving frame; the lab-frame mean kinetic energy alone is fourth order in wave amplitude. The equality is at the leading slow and quadratic orders used throughout the forcing calculation.

After removing the hydrostatic background pressure, the two-dimensional ideal Boussinesq equations arewith the buoyancy equation ${\sigma }_t+u{\sigma }_x+w{\sigma }_z+N^2(z)w=0$. The last equation follows by materially advecting total buoyancy $B_0(z)+\sigma$ with $B_0^{\prime }(z)=N^2(z)$. Take the streamfunction conventionso the $y$-component of vorticity is $q=u_z-w_x={\nabla }^2\psi$. Taking ${\partial }_z$ of horizontal momentum minus ${\partial }_x$ of vertical momentum cancels pressure. Incompressibility combines the remaining nonlinear terms into advection of $q$, giving $q_t+u{q}_x+w{q}_z=-{\sigma }_x$. Hence the full equations areThere is no vortex stretching in this genuinely two-dimensional velocity field.

Write $\psi ={\psi }_0(z)+\varphi$, with ${\psi }_0^{\prime }=\overline{u}=U(z)$ and background buoyancy anomaly zero. The Linearized Boussinesq equations becomeFor a stationary disturbance $\varphi =\hat{\psi }(z)e^{ikx}$ with real $k\ne 0$ and $U>0$, the second equation gives $\hat{\sigma }=(N^2/U)\hat{\psi }$. The first therefore reduces to the stationary Taylor–Goldstein equationThe restriction $k\ne 0$ is used when cancelling $ik$; a purely horizontally uniform alteration of the background is not this wave problem.

For the exponential-profile stationary stratified waves, defineSince $U^{\prime \prime }/U=M^2$ and $N/U=N_0/U_0$, all coefficients are constant. The complete vertical solution set isShifting the origin of $z$ simply changes the two constants. Vertical oscillation requires $N_0/U_0>\sqrt{M^2+k^2}$; equality gives an affine structure; below that value the structure is evanescent or growing, with admissibility determined by the physical boundaries.

The Rayleigh quotient with one Neumann endpoint must retain its upper boundary term. On a finite interval with $D=z_2-z_1>0$, choose a nonzero real vertical structure, so the denominator is strictly positive. Multiplication of the differential equation by $\hat{\psi }$ and integration by parts givesWith only ${\hat{\psi }}^{\prime }(z_1)=0$, it equals $k^2$ precisely when $\hat{\psi }(z_2){\hat{\psi }}^{\prime }(z_2)=0$. A lower Neumann boundary condition alone does not ensure this.

For $s>0$, the lower condition selects $\hat{\psi }=A\cos [\sqrt{s}(z-z_1)]$. Equality of the quotient requires $\sqrt{s}D=n\pi /2$ for a positive integer $n$: even $n$ gives an upper Neumann condition, odd $n$ an upper Dirichlet boundary condition. Thuswhich additionally restricts $N_0^2/U_0^2>M^2+n^2{\pi }^2/(4D^2)$. If $s=0$, the lower condition leaves a constant structure; it has quotient $C=k^2$, possible for a genuine wave only when $C>0$. If $s<0$, the lower condition leaves $A\cosh [\sqrt{-s}(z-z_1)]$, whose upper product is positive and whose quotient is strictly less than $k^2$. On an infinite interval, none of these nonzero lower-Neumann structures is square-integrable, so the displayed ratio of integrals is not a defined finite-integral quotient there. Ratios of truncated integrals would be a different limiting prescription.

For every mode above, including the affine and exponential cases, the disturbance satisfiesThis statement concerns the disturbance, not the sum with the background streamfunction; the latter generally has a different Laplacian eigenvalue. Its buoyancy polarization isThus no nonzero mode has a constant buoyancy-to-streamfunction ratio when $M\ne 0$.

The nonlinear exactness test for a stratified streamfunction mode now distinguishes the two equations. The disturbance's vorticity self-advection vanishes because its Laplacian is proportional to itself. But its buoyancy self-advection isFor a nonzero real sinusoidal disturbance and $M>0$, this is a nonzero second harmonic somewhere. Consequently none of the nontrivial modes with the stipulated positive $M$ is by itself an exact finite-amplitude solution of both nonlinear equations. Additional harmonics or mean corrections would be needed. In the limiting case $M=0$, $U,N,\alpha$ are constant; both self-interactions vanish, and these same single-wavenumber structures, or their superpositions with the same $C$, do solve the full ideal equations wherever their boundary conditions are admissible.

A Rossby wave is a displacement of a materially conserved potential vorticity distribution whose induced balanced velocity propagates that displacement. In a quasi-geostrophic approximation interior, take a locally uniform zonal current $U$ and meridional background PV gradient $B={\overline{Q}}_y$. A small meridional parcel displacement $\eta$ produces $q^{\prime }=-B\eta$. The linear potential-vorticity equation isFor a shallow-water vertical mode, potential-vorticity inversion is $q^{\prime }=({\nabla }_H^2-L_D^{-2}){\psi }^{\prime }$ and $v^{\prime }={\psi }_x^{\prime }$. Substitution of $e^{i(kx+ly-\omega t)}$ givesThe same structure holds for other vertical modes, with their own deformation lengths, and for barotropic waves with $L_D^{-1}=0$. A topographic or shear-induced PV gradient can replace the planetary one: locally $B=\beta -U^{\prime \prime }$ in a barotropic zonal current, with the corresponding bottom contribution when present.

The restoring process can be seen directly by choosing $\eta ={\eta }_0\cos (kx-\omega t)$ in one horizontal dimension with no mean flow. Then $q^{\prime }=-B\eta$ and ${\psi }^{\prime }=B\eta /(k^2+L_D^{-2})$. The resulting $v^{\prime }={\psi }_x^{\prime }$ is a quarter wavelength out of phase with the displacement and requires $\omega =-Bk/(k^2+L_D^{-2})$. A PV anomaly is therefore not merely carried passively: its nonlocal balanced velocity moves the displaced contours.

For $B>0$, the intrinsic zonal phase velocity isChanging the sign of $k$ gives the conjugate representation of the same real pattern, not an eastward intrinsic phase branch. More generally the intrinsic phase-velocity vector has nonpositive zonal component $-B{k}^2/[|\mathbf{k}{|}^2(k^2+l^2+L_D^{-2})]$. This one-way phase property contrasts with acoustic and internal gravity waves, whose frequency equations have two independent propagation branches. The direction reverses when the relevant PV gradient reverses; a sufficiently strong current can make the laboratory phase move eastward. Thus one-way refers to propagation relative to the mean flow and a specified gradient, not an absolute geographic direction in every problem.

Scale dependence is supplied by potential-vorticity inversion. At short horizontal scales, a fixed PV anomaly creates ${\psi }^{\prime }\sim q^{\prime }/|\mathbf{k}{|}^2$, so the intrinsic phase speed is small, of order $B/|\mathbf{k}{|}^2$. At long scales with finite $L_D$, it approaches $-B{L}_D^2$. The group velocity need not have the phase direction:In particular, short nearly zonal waves can carry a packet eastward relative to the current while their crests move westward. A group-velocity statement requires a slowly varying, propagating packet and a smooth dispersion branch; turning regions and critical levels require further analysis.

Interior PV gradients are not the only support. In a stratified quasi-geostrophic approximation fluid the inversion operator includes ${\partial }_z[(f^2/N^2){\psi }_z^{\prime }]$. The boundary buoyancy $b^{\prime }=f{\psi }_z^{\prime }$ also supplies inversion data. A meridional boundary-buoyancy gradient is equivalent, in the inversion problem, to a boundary PV sheet. Consequently Boundary Rossby waves can exist even if the interior PV gradient vanishes. Their perturbations decay away from the boundary over a horizontal-wavenumber-dependent Rossby height; opposite boundaries can support oppositely directed intrinsic waves because their effective sheet gradients have opposite signs.

The Eady model makes this concrete. For $U=\Lambda z$, constant $N$ and horizontal wave-vector magnitude $k_h$, put $K=N{k}_h/|f|$. Zero interior PV gives ${\psi }_{zz}^{\prime }-K^2{\psi }^{\prime }=0$, and each rigid boundary has the buoyancy conditionA lower isolated boundary has ${\psi }^{\prime }\propto e^{-Kz}$, so $c_{\mathrm{lower}}=U(0)+\Lambda /K$. An upper isolated boundary has ${\psi }^{\prime }\propto e^{K(z-H)}$, so $c_{\mathrm{upper}}=U(H)-\Lambda /K$. For positive shear, the lower wave is intrinsically eastward and the upper wave intrinsically westward, entirely consistently with their opposite effective PV gradients.

In the finite-depth Eady dispersion relation, $\gamma =KH/2$. For $\gamma \gg 1$, $\tanh \gamma$ and $\coth \gamma$ both tend to one. The square root therefore tends to $\gamma -1$, givingThese are the two uncoupled Eady edge waves, not a pair of opposite branches on a single boundary. Their overlap is exponentially small in $KH$. For smaller separation, the interaction can phase-lock the counterpropagating boundary waves; the square root becomes imaginary for $0<KH<2.399\dots$, yielding Eady instability. The short-wave limit is stable and illustrates why interior-PV-only reasoning misses boundary wave propagation and baroclinic instability.

The governing balance requires a small Rossby number, weak fractional height or density perturbations, stable stratification where relevant, and frequencies slow relative to $|f|$. The local gradient must change little across a wavelength; on a planetary beta-plane $\beta \ell /|f|\ll 1$ is the associated small parameter. A hydrostatic stratified example also requires a small vertical-to-horizontal aspect ratio and $N$ large compared with its slow frequency. For reference, midlatitude $|f|\sim 10^{-4}{\mathrm{s}}^{-1}$ and $\beta \sim 2\times 10^{-11}{\mathrm{m}}^{-1}{\mathrm{s}}^{-1}$. Atmospheric deformation lengths of order $10^6\mathrm{m}$ give long-wave speeds of order $\beta L_D^2\sim 20{\mathrm{m}\mathrm{s}}^{-1}$ and synoptic-to-planetary times of days to weeks. An oceanic baroclinic deformation length of order $3\times 10^4\mathrm{m}$ instead gives about $0.02{\mathrm{m}\mathrm{s}}^{-1}$ and much longer basin-crossing times. Structural constants and background advection matter; these are scale estimates, not universal measured speeds.

Finite-amplitude Rossby waves can overturn or strongly filament PV contours, especially near a critical layer where their phase speed approaches the local current. Smooth inviscid motion still conserves each parcel's PV. Apparent mixing comes from the resulting fine structure under coarse-graining or from weak diffusion; it can flatten a coarse-grained background PV gradient and sharpen neighboring jet-edge gradients. A simplest description of Rossby-wave PV mixing and zonal momentum is a downgradient fluxFor barotropic zonal averages with zero mean meridional transport, $u^{\prime }=-{\psi }_y^{\prime }$, $v^{\prime }={\psi }_x^{\prime }$ and $q^{\prime }={\nabla }_H^2{\psi }^{\prime }$. Zonal integration by parts gives the Taylor identity for quasi-geostrophic fluxThus mixing down a positive PV gradient produces westward acceleration in the mixing region: wave-generated Reynolds stress redistributes zonal momentum. The inversion relation ${\overline{Q}}_y=\beta -{\overline{U}}^{\prime \prime }$ then relates PV homogenization to a reshaped current. Total zonal momentum cannot simply disappear in an isolated domain; integrated stress convergence is a boundary flux, so compensating momentum changes or boundary forcing must accompany localized drag.

In small conservative waves with $B>0$, the quadratic wave activity is $\mathcal{A}=\overline{q^{\prime 2}}/(2B)$. Multiplying the linear PV equation by $q^{\prime }/B$ gives ${\mathcal{A}}_t=-\overline{v^{\prime }{q}^{\prime }}$ for a zonal average on a locally uniform current. Together with the momentum equation this yields ${\partial }_t(\overline{U}+\mathcal{A})=0$ during conservative growth or decay. In a commonly used momentum convention the Rossby-wave pseudomomentum is $-\mathcal{A}$; signs must be stated rather than inferred from the word activity. Breaking and absorption invalidate the reversible small-wave description and deposit the wave's momentum through its flux convergence. A steady undamped conservative wave cannot provide sustained local drag without a flux divergence or boundary exchange, as expressed by the non-acceleration theorem for quasi-geostrophic waves.

Rossby propagation is the slow, direction-selective motion produced by PV advection coupled to balanced inversion. Interior gradients and boundary sheets share that mechanism; inversion determines the scale dependence, while wave packets, boundary-wave interaction and breaking govern how energy and momentum reach other regions.

Write $q=bu$, $\rho ={\rho }_{\infty }$ and $r=\rho -{\rho }_s>0$. The first balance is depth-integrated momentum conservation. Its left side is the downstream increase of horizontal momentum flux. The wind contributes the applied stress $S$; the remaining term is the resultant hydrostatic force associated with the sloping base of the light layer. Because $\overline{\rho }-\rho <0$, that term opposes the wind when $b^{\prime }>0$.

For a strip of unit alongshore width and length $dx$, let $E=q^{\prime }$ be the entrained ocean-volume flux per unit area. Ocean entering the layer carries temperature $T_{\infty }$ and concentration $C_{\infty }$ but no horizontal momentum in this model. Take the reference liquid enthalpy at $T_{\infty }$ to be zero and setThe latent term is negative because ice has lower enthalpy than liquid at the same temperature. The incoming entrained water has zero relative enthalpy, and the outgoing minus incoming horizontal enthalpy flux is $[eq{]}^{\prime }dx$. Atmospheric heat loss therefore gives $[eq{]}^{\prime }=-f$. Similarly, solute conservation givesThese are the second and third balances. Their reference-flux form has already subtracted the heat and solute brought in by entrainment; those fluxes must not be subtracted again.

There is a dimensional normalization to make explicit. With the printed specific $c_p,L$, the physical heat-flux balance is $\rho [eq{]}^{\prime }=-F_{\mathrm{phys}}$. Thus $f=F_{\mathrm{phys}}/\rho$. The displayed exam equation uses its symbol $F$ for $f$, or implicitly works with density-normalized heat flux. The derivation below uses $f$ to keep these two meanings distinct.

With no incoming coastal jet or ice source, the conditions at $x=0$ areThe regular emerging layer also has $T\to T_{\infty }$, $\varphi \to 0$, $\overline{C}\to C_{\infty }$ and $b\to 0$. Prescribed nonzero inlet fluxes would replace these conditions and destroy the coast-origin power law. A zero-thickness coast does not justify independently prescribing an arbitrary nonzero velocity there.

Integration of the solute and enthalpy balances yields $\overline{C}=C_{\infty }$ wherever $q>0$, and $eq=-fx$. Local phase equilibrium now givesFor $\varphi \ll 1$, putTo leading order, $\varphi q=Kx$. The potential-energy closure implies $rg\varphi b{b}^{\prime }=S/4$, and substituting it in the momentum equation givesDividing the ice flux by this last relation gives $\varphi =(K/J)u$. If $b\propto x^p$, $u\propto x^q$, momentum gives $p+2q=1$, while the potential-energy balance gives $2p+q=1$. Thus $p=q=1/3$. Substitution fixes the coefficients:This is the dilute frazil polynya similarity solution. In particular, all three variables grow as $x^{1/3}$, $q$ grows as $x^{2/3}$, and the solution stops being dilute when $\varphi$ is no longer small. Density differences have been retained in buoyancy, but not in inertia or the normalized heat capacity.

To account for upward settling, one must distinguish retained suspension from crystals that leave it. A consistent depth-averaged extension has an upward exported ice-volume flux $R$ and an entrainment flux $E$. If all phases retain the same mean horizontal velocity within the layer, its balances becomeThe last term removes the momentum carried out by the ice; its density may be restored to ${\rho }_s$ outside the Boussinesq approximation. Consequentlyso one may no longer set $\overline{C}=C_{\infty }$. The local equilibrium relations are unchanged, but use the new bulk concentration. The frazil export enthalpy balance also shows why a simple replacement $-f\mapsto -f-LR$ has the wrong sign: exported ice carries negative relative enthalpy.

A slip speed $w$ requires a vertical particle-flux closure, for example $R=w\varphi$ for a well-mixed top boundary. A nonuniform suspension instead requires its surface concentration and buoyancy moment. The potential-energy closure must then include the gravitational energy released by rising light crystals, with scale $rgw\varphi b$, and any energy and buoyancy moment carried through the boundaries. For example, a well-mixed closure that transfers a fraction $\eta$ of this released power to entrainment and neglects additional boundary potential-energy transport would replace the original work balance byThe allocation $\eta$ and exported potential-energy flux are extra closure information, not supplied constants of the original model. The original work balance cannot be retained unchanged while treating settling as an additional loss of suspended ice. Phase-dependent horizontal speeds would also replace $q\varphi$ and $q(1-\varphi )$ by their separate phase fluxes.

It is important to compare total production, not merely the retained ice flux. LetIntegrated solute conservation gives $C(q-I)=C_{\infty }(q+R_{\ast })$, henceIntegrating the heat balance, and neglecting products of the small concentration departure with the exported flux, givesto dilute order, provided the total processed fraction $P/q$, not only the retained fraction $\varphi$, remains small. Thus upward settling reduces retained suspension, but does not increase the leading-order total production at fixed atmospheric heat loss. Beyond that order, efficient removal can reduce the subsequent sensible cooling of crystals and enhance mixing of ocean water; the usual expectation is a larger latent-heat fraction and hence greater production. Its amount, or an unconditional larger/smaller claim, needs the settling, mixing and surface-export closures just identified. At fixed $f$, energy conservation bounds the production by the available heat loss; it must not be inferred simply by observing fewer suspended crystals.