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For the shallow-water approximation, let the vertical scale be much smaller than the horizontal scale, so vertical acceleration is negligible and pressure is hydrostatic. Assume constant density, a flat impermeable bottom, a free surface under constant atmospheric pressure, and a horizontally moving column with no leading-order vertical shear. Neglect friction. Taking the bottom at $z=0$, hydrostatic pressure is $p=p_a+\rho g(h-z)$, so its horizontal gradient is independent of $z$. The horizontal momentum equations are thereforeIntegrating incompressibility from the bottom to the free surface, using zero bottom vertical velocity and the surface kinematic condition $w=h_t+u{h}_x+v{h}_y$, givesTogether these are the nonrotating shallow water equations.

Rotation with angular velocity $f\hat{\mathbf{z}}/2$ adds Coriolis force terms $-fv$ and $+fu$ to the left sides of the $u$ and $v$ equations, respectively. The centrifugal potential is absorbed into the background pressure and effective gravity. ThusWrite $\zeta =v_x-u_y$ and $d=u_x+v_y$. Taking the horizontal curl, including the derivatives of the advecting velocity, yieldsFor example the curl of the Coriolis force contribution is $fd$, and the curl of nonlinear advection supplies $\zeta d$. Since $q_a=f+\zeta$ and $D_{H}h=-hd$, division givesThis is exact shallow-water potential vorticity conservation within the rotating shallow water equations.

For the quasi-geostrophic approximation, take a horizontal scale $L$, speed $U$, mean depth $H$, and slow time $L/U$. Require small Rossby number $\mathrm{Ro}=U/(|f|L)\ll 1$ and small surface departure $\eta =h-H$. Introduce the Rossby deformation radius $L_R=\sqrt{gH}/|f|$ and Burger number $\mathrm{Bu}=L_R^2/L^2=O(1)$. Geostrophic balance gives $g\eta /L\sim |f|U$, hence $\eta /H\sim \mathrm{Ro}/\mathrm{Bu}\ll 1$. The Froude number is $U/\sqrt{gH}=\mathrm{Ro}/\sqrt{\mathrm{Bu}}\ll 1$. These assumptions select slow balanced motion, with fast gravity-wave transients absent at leading order.

Define the quasi-geostrophic streamfunction by $\psi =g\eta /f$. The leading geostrophic flow is $(u_g,v_g)=(-{\psi }_y,{\psi }_x)$, which is horizontally nondivergent. Expanding the exact shallow-water potential vorticity givesThe correction is smaller than $f/H$ by $O(\mathrm{Ro})$. Advecting it by the leading geostrophic flow, while the constant leading term has zero gradient, givesHere potential-vorticity conservation means that $Q$ is carried unchanged by the leading horizontal trajectories. Potential-vorticity inversion means solving the elliptic equation $({\nabla }_H^2-L_R^{-2})\psi =Q-f$, with appropriate boundary or decay conditions, to recover $\psi$ and thus the velocity. Conservation evolves the scalar field; inversion obtains its velocity field.

For a mean current $U(y)=-{\bar{\psi }}_y$, substitute $Q=\bar{Q}+Q^{\prime }$ and $\psi =\bar{\psi }+{\psi }^{\prime }$ into potential-vorticity conservation. Retaining first-order terms givesLet $\Delta Q=\bar{Q}(0^{+})-\bar{Q}(0^{-})>0$. Differentiating the mean inversion relation gives ${\bar{Q}}_y=-U^{\prime \prime }+L_R^{-2}U$. Away from $y=0$ this is zero, so decay at both infinities gives $U=U_0{e}^{-|y|/L_R}$. A velocity jump would introduce a derivative of a delta function, absent from ${\bar{Q}}_y$, so $U$ is continuous. Integrating through zero now gives $-[U^{\prime }]=\Delta Q$, and therefore

For the regular interfacial normal mode ${\psi }^{\prime }=\varphi (y)e^{ik(x-ct)}$, the linear equation readsIn each region, choose zero perturbation potential vorticity, giving ${\varphi }^{\prime \prime }-(k^2+L_R^{-2})\varphi =0$. This supplies a regular solution even if $U=c$ at isolated locations away from the interface. Continuity and decay require $\varphi =A{e}^{-\gamma |y|}$, where $\gamma =\sqrt{k^2+L_R^{-2}}$. Integrating the equation across zero givesSince $[{\varphi }^{\prime }]=-2\gamma A$, the nonzero normal mode hasThis potential-vorticity step edge wave is an interfacial Rossby wave. Displacing the potential vorticity jump creates alternating anomalies; their inverted velocity displaces the interface further with an intrinsic propagation opposite to the positive mean current. The mechanism requires a nonzero jump, or more generally a gradient, of the mean potential vorticity. It does not require planetary variation of $f$. Reversing the jump reverses the intrinsic propagation; setting the jump to zero removes this interfacial restoring mechanism.

Define the quasi-geostrophic streamfunction by $p^{\prime }/{\rho }_0=f_0\psi$. Leading horizontal geostrophic balance, $f_0\hat{\mathbf{z}}\times {\mathbf{u}}_g=-{\nabla }_{H}P$, gives $(u_g,v_g)=(-{\psi }_y,{\psi }_x)$. The hydrostatic approximation gives $P_z=\sigma$, henceThus $D_g$ is the material derivative along the leading geostrophic flow, and the vertical derivative of pressure determines the buoyancy perturbation. This also gives thermal wind: $u_{gz}=-{\sigma }_y/f_0$ and $v_{gz}={\sigma }_x/f_0$.

The balanced regime is favored by small Rossby number, a shallow aspect ratio, and a Burger number $N^2{H}^2/(f_0^2{L}^2)$ of order one. Equivalently the stratified Froude number $U/(NH)=\mathrm{Ro}/\sqrt{\mathrm{Bu}}$ is small. On a beta plane, require $\beta L/|f_0|=O(\mathrm{Ro})$, and take slow, initially balanced motion away from the equator. Under the traditional approximation, the horizontal Coriolis force generated by vertical velocity is small because vertical velocity is small, while the vertical rotational force is small compared with the leading hydrostatic pressure and gravity terms. Strong stratification and small aspect ratio justify retaining only the vertical component of Earth's rotation in this regime; this argument is not uniform near the equator.

Put $A(z)=f_0^2/N^2(z)$. The buoyancy equation gives $w=-f_0{D}_g{\psi }_z/N^2$. Substitution into the vertical vorticity equation yieldsThere is an important commutator cancellation. With $J_{xy}(a,b)=a_x{b}_y-a_y{b}_x$,Also $D_{g}f=\beta {\psi }_x$ for $f=f_0+\beta y$. Adding terms therefore proves potential-vorticity conservation:This is the three-dimensional quasi-geostrophic potential vorticity.

At an impermeable stationary lower boundary $z=b(x,y)$, the kinematic condition is $w={\mathbf{u}}_H\cdot {\nabla }_{H}b=D_{g}b$. For small elevation and slope, evaluate this at the reference level $z=0$ to leading order. Combining it with the buoyancy equation gives the rigid-boundary buoyancy condition for quasi-geostrophic wavesThe quantity in parentheses is the boundary displacement plus its buoyancy-equivalent displacement.

Now take constant $N,f$ and the mean quasi-geostrophic streamfunction $\bar{\psi }=-(U_0+\Lambda z)y$. Its mean buoyancy perturbation is $-f\Lambda y$, consistent with thermal wind, and its interior potential vorticity is constant. In the boundary condition, the perturbation velocity $v^{\prime }={\psi }_x^{\prime }$ advects the mean boundary buoyancy gradient. Keeping this contribution givesIn particular the $-\Lambda {\psi }_x^{\prime }$ term must not be omitted.

The linear interior potential-vorticity equation is $({\partial }_t+U(z){\partial }_x)Q^{\prime }=0$. Starting with zero interior perturbation potential vorticity, and with forcing only through the boundary, it remains zero. For a steady solution this is also the regular zero-interior-PV choice, continued through any isolated zero of $U$. The inversion equation for a horizontal mode isThe printed convention $K>0$ takes $f>0$; for either hemisphere the decay rate is defined using $|f|$. Boundedness aloft selects the decaying mode. Writewith real part understood. The linear boundary condition becomesFor $K{U}_0+\Lambda \ne 0$, a steady quasi-geostrophic mountain wave is thereforeOn $y=0$, its real meridional velocity isFor the positive-parameter convention $k,ϵ,f,U_0>0$ and $K{U}_0+\Lambda >0$, the velocity minimum lies one quarter wavelength downstream of each topographic maximum; its maximum lies one quarter wavelength downstream of each topographic minimum. If the velocity prefactor changes sign, maxima and minima interchange. The condition $K{U}_0+\Lambda >0$ alone does not fix that prefactor if $U_0$ is allowed to be negative. The upper panel shows the positive-current convention.

At $K{U}_0+\Lambda =0$, the free boundary Eady edge wave has zero frequency in the ground frame. If $kϵU_0\ne 0$, a steady amplitude cannot satisfy the forced amplitude equation. With $A(0)=0$, integration givesConsequentlyThe symbol multiplying $i$ here is the horizontal wavenumber $k$, as in the PDF. Taking the real part and differentiating in $x$ givesFor $U_0=-\Lambda /K>0$, positive $ϵ,f$, and $t>0$, the meridional-velocity minima coincide with topographic maxima, and its maxima with topographic minima. This is the lower panel. The change from quadrature to opposition is the phase of a resonantly forced topographic edge wave.

The stationary relief continuously excites a boundary Rossby wave whose intrinsic propagation exactly cancels mean advection. The forcing therefore remains coherent with that wave; with no damping or radiation loss for this trapped mode, its linear amplitude grows secularly. This is a resonant topographic quasi-geostrophic wave. Unlimited growth is a conclusion of the linear idealization: eventually the small-disturbance approximation fails.

The assertion that resonance must be unsteady needs nonzero forcing. If $k=0$, then $b_x=0$ and the boundary current does not cross the relief; if $U_0=0$, the forcing also vanishes. In either case the zero disturbance is a steady solution even if $K{U}_0+\Lambda =0$. For example, choose $k=0$, $\ell \ne 0$, and $\Lambda =-K{U}_0$; this is a concrete exception to an unconditional resonance claim. When both horizontal wavenumbers vanish, $K=0$ and the stated decaying-wave ansatz itself does not apply.

A precise stratification–rotation analogy uses constant $N$, ideal fluid, and independence of one horizontal coordinate. For the stratified system take ${\partial }_y=0$, $v=0$, and allow prescribed forces $(F_x,0,F_z)$. The possibly nonlinear Boussinesq equations areFor the unstratified system choose right-handed coordinates $(X,Y,Z)=(z,y,-x)$, rotation $2\mathbf{\Omega }=N{\mathbf{e}}_Z$, and ${\partial }_Y=0$. Its rotating Euler equations, with the static gravitational and centrifugal contributions absorbed into the background pressure, arewhere $D_R={\partial }_t+U{\partial }_X+W{\partial }_Z$. The identification isThen $D_R=D$, $P_X=p_z$, and $P_Z=-p_x$, proving the component-by-component equivalence. In particular, buoyancy perturbation maps to transverse rotating velocity, and stratified horizontal velocity maps to rotation-axis velocity. The mapping permits arbitrary prescribed forces in the two-dimensional plane, but no transverse force: a transverse force in the rotating problem would create a source in the stratified buoyancy equation. A transverse stratified force would also invalidate $v=0$. Constant $N$ is essential to this constant-rotation equivalence.

A physical free surface introduces extra structure beyond the interior equations. After subtracting hydrostatic pressure, its zero total pressure condition gives $p=g\eta$ to first order, together with the kinematic condition ${\eta }_t=w$ for the stratified vertical displacement. The mapping takes this normal velocity into $U$, whereas an actual rotating free surface normal to its physical vertical $Z$ has normal velocity $W$ and a displacement along $Z$. The image of the stratified horizontal surface is a vertical surface in the rotating coordinates. Consequently the physical free-surface problems are not analogous, although deliberately mapped rigid boundaries or mathematical boundary data can be.

For the first prescribed force, the steady linear stratified equations are satisfied byReal parts are understood. The horizontal pressure gradient balances the force; the vertical pressure gradient is balanced by buoyancy; and $w=0$ satisfies buoyancy conservation and incompressible flow. Thus neither $u(z)$ nor $b(z)$ is fixed by the steady equations alone.

Initial rest and switching on in the unbounded fluid remove this ambiguity by causal pressure adjustment by internal waves. Work with the $e^{imz}$ coefficient and let $\hat{F}(q)$ be the horizontal Fourier transform of $f^{\prime }(x)$. For an abrupt switch at time zero, elimination in the linear equations givesAt $q=0$ the bracket cancels the apparent singularity at every finite time. The persistent pressure term is the symmetric antiderivative of the force: the inverse transform of $-i\hat{F}/q$, interpreted as a Cauchy principal value, is $f(x)-\epsilon /2$. The oscillatory part propagates away. To see why its zero-frequency part does not leave an arbitrary offset, note that $\Omega (q)\sim cq$ with $c=N/m$; its long-wave contribution consists of oppositely moving fronts. At fixed $x$, the two far-field values of $f-\epsilon /2$ are $-\epsilon /2$ and $+\epsilon /2$, whose average is zero. Nonzero horizontal wavenumbers give dispersive transients that decay locally. Smooth finite-duration switching changes the transient, not this selected steady limit. ThereforeA spatially constant pressure gauge is immaterial. The long-wave group velocity $c=N/m$ gives the adjustment scale: the elapsed time since switching must satisfy $Nt\gg 1$ and $(N/m)t\gg |x|+x_1$, and exceed the switching duration. These are asymptotic conditions, not an assertion of exact equilibration after a finite time; the full dispersion relation allows tails. The same long-wave limit gives $u_{\infty }=\epsilon m{e}^{imz}/(2N)$ if desired. Equal-amplitude outward fronts leave opposite pressure offsets and the same velocity on their two sides.

For the force depending instead on $y$, any steady linear horizontal momentum balance requires $p_x=f^{\prime }(x)e^{i\ell y}$ and $p_y=0$. Their mixed derivatives disagree wherever $f^{\prime }\ne 0$, since ${\partial }_y{p}_x=i\ell f^{\prime }(x)e^{i\ell y}$ but ${\partial }_x{p}_y=0$. Hence no steady linear velocity field can satisfy the equations, irrespective of whether it is $x$-independent. This statement concerns the linearized problem; nonlinear inertial balances were excluded by the assumptions.

For an unsteady solution put $\mathbf{u}=(-t{\varphi }_y,t{\varphi }_x,0)$ and take $\sigma =0$. Its horizontal divergence is zero. The pressure derivatives needed in momentum balance areTheir compatibility is exactlyThus the Poisson equation supplies an actual pressure and the proposed velocity solves both momentum equations; the vertical momentum and buoyancy equations hold because $w=\sigma =0$. This flow grows linearly in time, so it remains a valid small-disturbance prediction only while its velocities are small.

As $x_1\to 0$ with total force $\epsilon$ fixed, $f^{\prime }\to \epsilon \delta (x)$. Set $\varphi =\Phi (x)e^{i\ell y}$. Away from the source ${\Phi }^{\prime \prime }-{\ell }^2\Phi =0$, and decay at both ends requires $\Phi =A{e}^{-\ell |x|}$. Integrating across the source gives the jump ${\Phi }^{\prime }(0^{+})-{\Phi }^{\prime }(0^{-})=-i\ell \epsilon$. Since this jump is $-2\ell A$, $A=i\epsilon /2$, soThis verifies the distributional source as well as the decay condition.

The corresponding rotating problem is not expected to reproduce this evanescent accelerating flow. Its forcing varies in $Y$, precisely the coordinate whose independence was required in the stratification–rotation analogy. Moreover, $x$ maps to the rotation-axis coordinate $-Z$. At small Rossby number, the Taylor–Proudman theorem gives ${\partial }_Z{\mathbf{u}}_R=0$ in the unforced exterior regions. Such nonzero columns cannot decay with $|Z|$, hence cannot reproduce an $e^{-\ell |x|}$ tail. There is also a direct unsteady obstruction to copying the proposed flow: in a homogeneous fluid rotating about the mapped $x$ axis, a mode with nonzero axial wavenumber has inertial wave frequency of magnitude $N|q|/\sqrt{q^2+{\ell }^2}$, rather than the zero-frequency horizontal vortical response of the stratified problem. The steady or slowly varying exterior rotating response must be columnar or accompanied by outgoing inertial waves; the decaying stratified solution is outside the partial analogy.

Let $\zeta$ denote free-surface height and $h=\zeta -b$ the water depth. The inviscid rotating shallow water equations with constant Coriolis parameter $f$ areIt is $\zeta$, not $h$, that supplies the horizontal pressure gradient. For a fixed bottom, mass conservation equivalently gives $D_{H}h=-h(u_x+v_y)$.

Write $\xi =v_x-u_y$ for relative vorticity. Differentiate the second momentum equation in $x$ and the first in $y$, then subtract. The pressure terms cancel. Expanding the differentiated advection terms givesConsequently the absolute vorticity $q_a=\xi +f$ satisfies $D_H{q}_a=-q_a{\nabla }_H\cdot {\mathbf{u}}_H$. Combining this with mass conservation yields exact shallow-water potential vorticity conservation:A sloping bottom changes the depth and hence the stretching, but contributes no curl to the gravitational pressure force.

For the quasi-geostrophic approximation, choose constant reference depth $H_0$, speed $U$, and horizontal length $L$. Require Rossby number $\mathrm{Ro}=U/(|f|L)\ll 1$, slow evolution on the advective time $L/U$, smooth horizontal scales, and $L/L_D=O(1)$ or smaller, where $L_D=\sqrt{g{H}_0}/|f|$ is the Rossby deformation radius. Write $\zeta ={\zeta }_0+\eta$, $b=b_0+\delta b$, and $h=H_0+\eta -\delta b$. The leading momentum balance is geostrophic balance:It gives $\eta /H_0=O(\mathrm{Ro}(L/L_D{)}^2)$, while we require $\delta b/H_0=O(\mathrm{Ro})$ and slopes on the chosen horizontal scale. Relative vorticity divided by $f$ is also $O(\mathrm{Ro})$. ThusThe derivative of the constant first term vanishes. In advecting the first-order anomaly, the ageostrophic flow contributes only at the next order. Therefore the shallow-water quasi-geostrophic potential vorticity equation isThis retains the balanced vortical branch, including Rossby waves driven here by the topographic potential-vorticity gradient. It filters the freely oscillating inertia–gravity branches of the full shallow water equations, and does not retain independent boundary Kelvin wave dynamics. Small Rossby number alone is insufficient for the reduction: fast waves must be absent at leading order, as in well-prepared balanced initial data.

In the channel take $H_0=h_{00}$ and measure the bottom anomaly from mid-channel: $\delta b=\epsilon (y-L/2)$. Define $\beta =\epsilon f/H_0$. The resting potential vorticity is $\beta (y-L/2)$, and the linear Rossby wave equation isFor nonzero zonal wavenumber $k$, wall conditions $\psi =0$ selectThe $k=0$ modes are stationary. Write $K^2=k^2+(n\pi /L{)}^2+L_D^{-2}$ and take real $A$ for the following sketch. The total potential vorticity is $q=\beta (y-L/2)-K^2\psi$. For a contour whose undisturbed position is $y_0$, its small meridional displacement isThus potential-vorticity conservation says a northward displacement in a positive background gradient carries a negative anomaly, $q^{\prime }=-\beta \delta y$. Potential-vorticity inversion turns this into positive $\psi$ and anticyclonic motion around the bulge. At a positive cosine crest, $v={\psi }_x$ is northward on its western flank and southward on its eastern flank. Advection raises the contour on the west and lowers it on the east, so the crest moves west. Algebraically, $\delta y_t=v$ and $\delta y=(K^2/\beta )\psi$ give $\omega /k=-\beta /K^2$, the same phase velocity as above. For $\beta <0$ all propagation directions reverse.

For a square basin use a nonzero-frequency normal mode $\psi =\mathrm{Re}\{\Psi (x,y)e^{-i\omega t}\}$, with $\Psi =0$ on all walls. Its equation isSet $\alpha =\beta /(2\omega )$ and $\hat{\Psi }=e^{i\alpha x}\Psi$. Substitution cancels the first derivative and givesThe square's Dirichlet Laplacian eigenfunctions are $\hat{\Psi }=A\sin (m\pi x/L)\sin (n\pi y/L)$, with $m,n$ positive integers. Consequently the square-basin topographic Rossby mode hasFor $\beta >0$, the positive-frequency branch therefore tends to $\beta L/[2\pi \sqrt{m^2+n^2}]$ as $L/L_D\to 0$, as required. The negative-frequency solution is its complex-conjugate representation for a real disturbance; the physical frequency magnitude is $|\beta |/[2\sqrt{{\pi }^2(m^2+n^2)/L^2+L_D^{-2}}]$.

Use the horizontal streamfunction convention $(u,v)=(-{\psi }_y,{\psi }_x)$, and set $a=\pi /L$, $k_d^2=f_0^2/(g{H}_0)$. Geostrophic balance gives $f_0\bar{u}=-g{\bar{\eta }}_y$, henceThe constant is fixed by the total volume or the reference surface height. To obtain shallow-water quasi-geostrophic potential vorticity, require a small Rossby number, nearly geostrophic horizontal motion and $|\eta |/H_0\ll 1$. Expanding the exact shallow-water potential vorticity givesAt leading nontrivial order $\zeta ={\nabla }_h^2\psi$, so after removing the constant background and multiplying by $H_0$,The mean potential vorticity and its gradient areWith $\psi =\bar{\psi }+{\psi }^{\prime }$ and $q=\bar{q}+q^{\prime }$, expand the Jacobian. Since both mean fields depend only on $y$, their Jacobian vanishes. The two mixed terms are $\bar{u}{q}_x^{\prime }$ and $v^{\prime }{\bar{q}}_y$, givingThe PDF has this correct mean-advection term; the TeX aid incorrectly inserts a division by $\partial x$.

Drop the quadratic perturbation Jacobian. Freeze $\bar{u}$ and ${\bar{q}}_y$ locally for a short-wave plane wave, with $K^2=k^2+l^2$ and $D=K^2+k_d^2$. Then $q^{\prime }=-D{\psi }^{\prime }$, $v^{\prime }=ik{\psi }^{\prime }$, and the local Rossby-wave dispersion relation in a zonal jet isThe intrinsic phase moves against an eastward mean flow because $\omega -k\bar{u}=-k{\bar{q}}_y/D$.

The printed wavelength bound is not a consequence of these equations. The local short-wave assumption is $K/a\gg 1$, equivalently wavelength $2\pi /K\ll 2L$; a particular cutoff $L/2$ is an optional scale-separation criterion, not a universal derived threshold. If one requires the ground-frame phase to follow the mean current, the dispersion relation only gives $K>a$, or wavelength less than $2L$. For example $K=2a,l=0$ has wavelength $L$ and a phase moving with the mean. More decisively, the stationary condition requested next fixes wavelength $2L$, contradicting a bound $L/2$ if both are imposed on the same waves.

For nonzero zonal $k$ and mean velocity, a ground-stationary wave satisfiesIt has the background length scale, so its existence should not be justified solely by the short-wave approximation. It is nevertheless an exact stationary linear solution of the Rossby-wave equation for a sheared zonal current: writing ${\psi }^{\prime }=F(y)e^{ikx}$ gives $\bar{u}[F^{\prime \prime }-(k^2+k_d^2)F]+{\bar{q}}_{y}F=0$, which reduces to $F^{\prime \prime }+(a^2-k^2)F=0$. Thus $F=e^{ily}$ with $k^2+l^2=a^2$ works globally. Indeed the total field then has $q=-(a^2+k_d^2)\psi +$ a constant, so its full Jacobian vanishes as well.

Differentiating the local dispersion relation givesOn the stationary circle this becomesFor a purely zonal wavevector, $l=0$, this is the printed expression ${\mathbf{c}}_g=2\bar{u}{K}^2\hat{\mathbf{x}}/(K^2+k_d^2)$. With $l\ne 0$ the printed expression needs a directional qualification; the general vector is the one above. Here stationarity means $\omega =0$ in the ground frame, not vanishing intrinsic frequency in the moving fluid.

A stationary obstacle can therefore anchor the phase pattern while wave energy travels away from it. The zonal group component has the mean-flow sign, so an eastward jet carries the stationary wave response downstream; tilted wavevectors also carry energy meridionally. This is the usual stationary Rossby wave wake distinction between fixed phase crests and propagating wave energy.

Use small Rossby number $U/(f_{0}l)\ll 1$, slow evolution on the advective time scale, small interface displacements relative to each layer depth, shallow hydrostatic layers, stable reduced gravity $g^{\prime }>0$, and inviscid unforced flow. On a beta plane, take $\beta l/f_0$ of the same small order as the Rossby number. The internal Burger number is retained at order unity so stratification and relative-vorticity effects can both enter the leading potential-vorticity anomaly.

The rigid-lid pressure in two-layer flow comes from neglecting the free-surface volume displacement in the rigid-lid approximation; it does not permit setting the common horizontal pressure gradient to zero. The small surface displacement multiplied by $g$ retains a finite lid-pressure multiplier. Write $h_1=H_1+\chi$, $h_2=H_2-\chi$, and let $\Pi$ denote this common pressure potential. Leading geostrophic balance giveswith ${\mathbf{u}}_i=(-{\psi }_{iy},{\psi }_{ix})$ at leading order. Literally setting $h_1+h_2$ to a spatial constant in the momentum gradients before taking the rigid-lid limit would suppress the upper-layer pressure field and fail to produce general two-layer QG dynamics.

Taking curl of each shallow-water momentum equation and using layer continuity gives material conservation of $(f+{\zeta }_i)/h_i$. Expanding it to first order and advecting the anomaly by the leading geostrophic velocity yieldsThe two-layer quasi-geostrophic potential vorticity equations areHere the common background $f_0/H_i$ has been removed and the anomaly multiplied by $H_i$. The advection term is retained at the same slow order as the time derivative even though the leading velocity/pressure balance was linear geostrophy. On an $f$ plane simply set $\beta =0$.

Use the quasi-geostrophic streamfunction convention $u_n=-{\psi }_{n,y}$, $v_n={\psi }_{n,x}$, and write $F=k_I^2$, $f=f_0+\beta y$. In a steady, large-scale basin interior, neglect the material derivatives of relative vorticity and interfacial stretching compared with advection of planetary vorticity. With small Rossby number, weak nonlinear eddy terms, no significant interior friction or topographic forcing, and the specified forcing confined to layer 1, Sverdrup balance isFor equal depths $h$, the depth-integrated meridional transport is $h({\bar{v}}_1+{\bar{v}}_2)=hW/\beta$. Here $W$ is the normalized potential vorticity source appearing in the evolution equation. If the dimensional wind stress curl is used, its usual layer forcing is $W=({\nabla }_h\times \mathbf{\tau }{)}_z/({\rho }_{0}h)$, so $\beta h({\bar{v}}_1+{\bar{v}}_2)=({\nabla }_h\times \mathbf{\tau }{)}_z/{\rho }_0$.

For $\beta >0$ and $W<0$, the upper-layer interior transport is southward. In a closed subtropical basin, negative wind stress curl also corresponds to downwelling Ekman pumping for $f_0>0$ and an anticyclonic gyre. A northward return transport is needed to close the circulation; its narrow western boundary current requires processes outside the frictionless Sverdrup balance. The local interior equations by themselves do not specify the detailed boundary-current structure.

Expanding the layer equation shows the physical budget:The first term is the change of planetary vorticity as a fluid parcel moves north or south. The second is the change of relative vorticity. The last is vortex stretching in layered quasi-geostrophic flow: displacement of the interface changes layer thickness and thus the stretching contribution to potential vorticity. Its opposite signs in the two layer definitions express their thickness changes in opposite directions. The wind stress curl supplies or removes upper-layer potential vorticity. Each $D_n/Dt$ follows its own layer velocity, rather than a common velocity for both layers.

For uniform $W$, choose the local two-layer Sverdrup interiorUnforced background zonal currents and arbitrary additive interface offsets have been set to zero. This choice is also an exact uniform-flow solution of the stated forced equations, not just a leading balance: ${\bar{q}}_1=f_0+\beta y-FVx$, ${\bar{q}}_2=f_0+\beta y+FVx$, and ${\bar{v}}_1{\bar{q}}_{1,y}=V\beta =W$. The local interface slope can be nonzero even though its stretching contribution is constant along each basic-state trajectory. A streamfunction linear in $x$ is an interior-patch description, not a complete globally bounded basin solution.

For constant density $\rho$ and constant angular velocity $\mathbf{\Omega }$, the Euler equations for an inviscid fluid in the rotating frame isThe last two forces are gravity and centrifugal acceleration; $2\mathbf{\Omega }\times \mathbf{u}$ is the Coriolis force term on the left. Incorporate both conservative forces into the modified pressureso the right side is $-\nabla \Pi$.