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With the free-surface elevation denoted by $\zeta$, the rotating shallow water equations over a fixed bottom areThe momentum gradient is that of the surface elevation, not depth alone; bottom slope enters through the depth and continuity equation. Taking ${\partial }_x$ of meridional momentum minus ${\partial }_y$ of zonal momentum gives the vorticity equationThe terms quadratic in velocity gradients combine into ${\omega }_r{\nabla }_H\cdot {\mathbf{u}}_H$; the constant Coriolis parameter contributes $f{\nabla }_H\cdot {\mathbf{u}}_H$. Continuity gives $D_{H}h/Dt=-h{\nabla }_H\cdot {\mathbf{u}}_H$. Dividing the two equations proves the exact shallow-water potential vorticity invariant

For the quasi-geostrophic approximation, let $U_s$ and $\ell$ be representative velocity and horizontal length scales. Require the Rossby number $\mathrm{Ro}=U_s/(|f|\ell )\ll 1$, a shallow layer, and slow evolution on time $\ell /U_s$. Put $L_R=\sqrt{g{h}_{00}}/|f|$ and assume $\mathrm{Bu}=(L_R/\ell {)}^2$ is order one. Leading momentum then gives geostrophic balance, and a quasi-geostrophic streamfunction can be chosen asIts surface-height scale satisfies $|\zeta |/h_{00}=O(\mathrm{Ro}/\mathrm{Bu})\ll 1$. Take the bottom-height ratio of the same small order, with gentle slopes, so $\eta =\zeta -b$ obeys $|\eta |/h_{00}\ll 1$. Relative vorticity is order $\mathrm{Ro}f$, and expanding the exact potential vorticity givesBecause the leading geostrophic velocity is horizontally nondivergent, ageostrophic advection of the anomaly is one order smaller. At the first evolving order the shallow-water quasi-geostrophic potential vorticity equation isThe constant reference $f/h_{00}$ has zero gradient and drops out. These assumptions also require all topographically induced velocities to retain a small Rossby number; the small bottom-height condition alone is not a full justification of quasi-geostrophic balance.

The uniform eastward current has ${\psi }_0=-Uy$ up to an additive constant, so ${\zeta }_0=-(fU/g)y$. With $b_0=-(fU/g)y$, surface and bottom slopes are equal and $h=h_{00}$. Its relative vorticity vanishes and $Q_0=0$. This sloping background is used locally where its surface and bottom departures remain within the small-height regime.

Add the compact ridge and set $\psi =-Uy+\varphi (x)$. Then $u=U$, $v={\varphi }^{\prime }(x)$, and $Q$ depends only on $x$. The steady equation becomes $U{Q}_x=0$. Uniform incoming potential vorticity fixes that constant to zero, giving the cosine-ridge shallow-water geostrophic responseLet $a=\pi L/2$, $\lambda =L_R^{-1}$ andThe forcing is even, so the decaying response is even. A cosine particular solution inside the ridge leaves an even homogeneous term; outside, retain only the decaying exponential. Since the forcing is bounded and contains no delta function, both $\varphi$ and ${\varphi }^{\prime }$ are continuous at the edges. Write $\varphi =A\cos (x/L)+D\cosh (x/L_R)$ inside and $\varphi =B{e}^{-(|x|-a)/L_R}$ outside. Matching at $x=a$ givesSolving yields $D=A(L_R/L)e^{-a/L_R}$, henceA globally decaying homogeneous solution that is continuously differentiable is zero, so this matched solution is unique.

A steady streamline is a constant-$\psi$ curve. If its upstream ordinate is $y_{\infty }$, it therefore hasFor $f>0$, $\varphi$ is positive: the flow first turns toward increasing $y$, passes the crest at its largest deflection, then returns to its upstream ordinate. Reversing the sign of $f$ reverses the deflection. For $L=3L_R$, the center deflection is $A[1+e^{-3\pi /2}/3]/U$ and each tail decays on scale $L_R$. The figure shows several streamlines and the ridge support; the deflection extends beyond the ridge.

One compatible realization of the initial condition has $h(x)=h_{00}-\hat{b}(x)$ and $\zeta =b_0(y)$. Its stated zonal velocity then satisfies $u(x)h(x)=U{h}_{00}$, so the initial mass flux is constant and $v=0$. The data given only far upstream do not uniquely determine the initial height everywhere; this is the natural mass-flux-compatible choice. It is not fully balanced over the ridge: $fu(x)$ differs from $-g{\zeta }_y=fU$. The resulting transverse acceleration initiates geostrophic adjustment and launches fast inertia-gravity waves.

Initially the relative vorticity is zero, but exact shallow-water potential vorticity is $f/[h_{00}-\hat{b}(x)]$, not the uniform upstream value. That localized anomaly is materially transported downstream, approximately at $U$ in the leading quasi-geostrophic approximation; it is not annihilated by adjustment. Fresh incoming parcels of uniform PV subsequently supply the ridge region, whose balanced potential-vorticity inversion approaches the steady solution above. The fast waves can carry unbalanced energy away on the inertia-gravity time scale, short compared with the advective time at small Rossby number. In an open domain this can leave the local balanced response; an ideal closed or reflecting domain need not settle monotonically, because undamped waves can return. Radiation, advection of the initial PV anomaly, and any weak damping must be distinguished from irreversible loss of PV.

Choose a fixed vertical datum so the physical free surface is at $H+\zeta$, with $H$ constant; its depth above the bottom is $h=H+\zeta -b$. The gradient of the constant datum does not enter the pressure force. With $D_H={\partial }_t+u{\partial }_x+v{\partial }_y$, the rotating shallow water equations areThe fluid is hydrostatic with depth-independent horizontal velocity at this order. The pressure-gradient force uses the free-surface slope, not the layer-depth slope: $\nabla \zeta =\nabla h+\nabla b$.

Let $q=v_x-u_y$ be relative vertical vorticity, $q_a=f+q$ the absolute vorticity, and $\delta =u_x+v_y$. Taking the horizontal curl of the two momentum equations and expanding the advective derivatives givesMass conservation is $D_{H}h=-h\delta$. The two stretching factors cancel, proving the exact shallow-water potential vorticity conservation lawThe denominator is $h$, the actual column thickness that changes under horizontal convergence; an arbitrary surface elevation $\zeta$ does not obey that column mass equation. There is no $w{\partial }_z$ term because the reduced layer quantities have no vertical dependence. The vertical motion has already supplied the layer-thickness equation, rather than remaining an independent coordinate of this model.

Now take $H=h_{00}$ and assume $|q|\ll |f|$, $|\zeta |,|b|\ll H$. To first order,For a small Rossby number, leading geostrophic flow satisfies $fv=g{\zeta }_x$ and $fu=-g{\zeta }_y$. ThusFor $f\ne 0$, the barotropic deformation radius is $L_R=\sqrt{gH}/|f|$. Multiplication of the first-order PV by $H$ givesThis is the required shallow-water quasi-geostrophic potential vorticity, including the bottom contribution.

Write $S=fϵ/H$ and $L=L_R$. The initial state has $\psi =0$, so ${\mathcal{Q}}_i=f+(f/H)b$. The specified homogenization sets ${\mathcal{Q}}_f=f$ in $|y|<a$ and leaves it unchanged outside. Hence the requested change isFor $f>0$ this is a descending straight segment from $+Sa$ to $-Sa$, with zero exterior values and jumps at the two edges. Endpoint values do not affect the inversion. It is the anomaly, not the full final PV profile, that appears as the source below.

For the zonal mean, $v=0$, $u=-{\psi }^{\prime }(y)$, and potential-vorticity inversion requireswith decaying disturbance at both infinities. The odd source makes the unique decaying solution odd. A particular interior solution is $S{L}^{2}y$, so writeBoth $\psi$ and ${\psi }^{\prime }$ must be continuous at $y=\pm a$: a jump would introduce a delta function or its derivative into a source that has only finite jumps. Matching at $a$ givesand thereforeDifferentiate the interior streamfunction to obtainThe exterior profiles needed for the sketches are $u=(D/L)e^{-(|y|-a)/L}$ and $\zeta =(f/g)\mathrm{sgn}(y)D{e}^{-(|y|-a)/L}$.

For $L\gg a$, put $A=a/L$, $Y=y/L$, so $|Y|\le A$. Uniform Taylor expansion givesConsequentlyFor $f>0$ the leading interior current is westward and approximately parabolic.

For $L\ll a$ and away from the edges, $u\simeq -S{L}^2$ and $\zeta =(f/g)\psi \simeq ϵy$: the surface follows the bottom slope centrally. Near $|y|=a$, the exponentially thin corrections give positive edge jets of width $L$ and peak scale $SLa/2$ when $f>0$. Those jets decay in the exterior. The elevation is odd, has opposite-sign extrema just inside the edges, and decays to zero outside; at the positive edge $\zeta (a)\simeq ϵ(a-L)/2$. The following sketch uses the full matched solution with $L/a=0.08$. For negative $f$, the current and PV-change signs reverse, while this elevation profile does not.

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.

On a sphere of radius $R$ rotating at angular speed $\Omega$, the Coriolis parameter is $f=2\Omega \sin \varphi$. Near latitude ${\varphi }_0$, write $y=R(\varphi -{\varphi }_0)$, use locally eastward and northward Cartesian coordinates, and expandThe beta plane retains the first northward variation of planetary vorticity while neglecting higher latitude dependence and metric curvature. A midlatitude local calculation assumes $L/R\ll 1$ and $|\beta |L/|f_0|\ll 1$; near the equator the distinct equatorial beta plane has $f_0=0$, so the midlatitude low-frequency reduction used below does not apply.

For a homogeneous shallow layer, the hydrostatic approximation gives $p=p_{\mathrm{atm}}+\rho g(\eta -z)$ and hence $p_x/\rho =g{\eta }_x$, $p_y/\rho =g{\eta }_y$, independently of depth. The approximation follows from the small aspect ratio $H/L$ and neglect of vertical acceleration. Differentiate the linear horizontal momentum equations in $z$. Their depth derivatives obeyStarting from rest gives $u_z=v_z=0$ initially, and the unique solution remains zero. This establishes depth independence of hydrostatic shallow-water flow; an arbitrary pre-existing shear would not be removed merely by taking the hydrostatic approximation. Integrating the continuity equation between the rigid bottom and the moving surface gives ${\eta }_t+H(u_x+v_y)=0$ at linear order. With the depth-integrated shallow-water transports $U=Hu$, $V=Hv$, one obtains

It is useful to make the coefficient approximation in the height reduction explicit. Put ${\mathcal{F}}_y={\partial }_t^2+f(y{)}^2$. Differentiating the two momentum equations in time and eliminating the other transport givesSince ${\partial }_y({\mathcal{F}}_{y}V)={\mathcal{F}}_y{V}_y+2f\beta V$, taking their horizontal divergence and using the continuity equation givesApplying ${\mathcal{F}}_y$ proves the exact variable-Coriolis shallow-water height equationAt the reference latitude, or after the usual local freezing of undifferentiated $f$ factors to $f_0$, this gives the height relation written with ${\mathcal{F}}_0={\partial }_t^2+f_0^2$. With $f=f_0+\beta y$ over a finite region, that constant-coefficient version is a local approximation, not an exact identity. Keeping ${\mathcal{F}}_y$ as above avoids silently commuting a variable Coriolis parameter through a spatial derivative.

For the slow Rossby wave branch, take $|\omega |/|f_0|\ll 1$, approximate ${\mathcal{F}}_0$ by $f_0^2$, and discard the two time-derivative terms on the right compared with $-f_0^2{\eta }_x$. This is the regular midlatitude long-time ordering, with nondegenerate zonal variation. If $a=f_0/c$, the result isThe printed low-frequency equation has the opposite right-hand sign. The minus sign follows directly from the preceding eliminated equation and is also the sign needed for the printed isofrequency-circle centre. An independent check uses geostrophic balance: $\psi =g\eta /f_0$, $(u,v)=(-{\psi }_y,{\psi }_x)$, and the shallow-water quasi-geostrophic potential vorticity is ${\nabla }_h^2\psi -a^2\psi +\beta y$. Linear potential-vorticity conservation gives $({\nabla }_h^2\psi -a^2\psi {)}_t+\beta {\psi }_x=0$.

For $\eta =A{e}^{i(kx+ly-\omega t)}$, the shallow-water Rossby-wave dispersion relation isFor fixed $l$ and $\beta >0$, it is odd in $k$, zero at $k=0$, negative for $k>0$, and tends to zero from below as $k\to +\infty$. Its minimum occurs at $k=\sqrt{l^2+a^2}$ with ${\omega }_{\min }=-\beta /[2\sqrt{l^2+a^2}]$. Thus both long and short waves have small frequency. The zonal phase velocity is westward, $\omega /k<0$, whereas the zonal group velocity isIt changes sign at the frequency minimum.

At fixed nonzero frequency, completing the square yields the Rossby-wave isofrequency circleThe radius is real only if $|\omega |\le |\beta |/(2|a|)$. For $\beta >0$, the branch with $k>0$ has $\omega <0$ and a circle centred on the positive $k$-axis. If instead the printed plus-sign wave equation is taken literally with this same Fourier convention, its dispersion is $\omega =+\beta k/(k^2+l^2+a^2)$ and its circle is centred at $(+\beta /(2\omega ),0)$. These two conventions cannot be mixed.

For reflection of a Rossby wave at a meridional wall, the stationary wall preserves frequency, and its translation invariance in $y$ preserves the tangential wavenumber. Thus ${\omega }_r={\omega }_i$, $l_r=l_i$. Both $k$ values solveTheir sum and product giveAn incident wave in $x>0$ travels toward the wall in group velocity, not necessarily in phase velocity. For $\beta >0$, $0<k_i<\sqrt{l_i^2+a^2}$ has $c_{gx,i}<0$, while its partner $k_r>\sqrt{l_i^2+a^2}$ has $c_{gx,r}>0$. The double-root case has zero normal group velocity and does not describe a wave packet incident on the wall.

At leading quasi-geostrophic approximation, the impermeability condition is $u=-g{\eta }_y/f_0=0$. For $l_i\ne 0$, the boundary condition givesThe reflected height therefore has equal amplitude and a phase change of $\pi$. Equality here concerns the height or quasi-geostrophic streamfunction amplitudes in the reduced model. At the degenerate $l_i=0$, geostrophic no-normal-flow is automatically satisfied and alone does not determine their ratio; the usual homogeneous wall-streamfunction condition supplies $A_r=-A_i$ if imposed. Retaining the small ageostrophic transport changes the boundary condition towhich tends to $-1$ in the regular low-frequency ordering but is not generally of unit modulus. In particular, $l_i=0$ gives $A_r/A_i=-k_i/k_r$ in that more complete boundary relation. Thus the printed equal-amplitude assertion requires the nondegenerate leading quasi-geostrophic approximation, or an explicit homogeneous wall condition; it is not a general exact shallow-water reflection law.

For a zonal basic geostrophic flow $U(y)$, take ${\psi }_0^{\prime }(y)=-U(y)$ and write $\psi ={\psi }_0+\varphi$. The flat-bottom shallow-water quasi-geostrophic potential vorticity gradient of the basic state isLinearizing its material conservation gives the Rossby-wave equation for a sheared zonal current,For a general nonuniform jet, a global two-dimensional plane wave is not an exact normal mode: the coefficients depend on $y$. The exact zonal normal mode problem, $\varphi =\hat{\varphi }(y)e^{i(kx-\omega t)}$, iswith appropriate transverse boundary or radiation conditions. If the jet varies slowly compared with a wavelength, a local plane wave freezes these coefficients at $y_0$. Its local dispersion relation isIn the nondivergent Barotropic Rossby wave model under a rigid-lid approximation, this becomes $\omega =kU-k(\beta -U^{\prime \prime })/(k^2+l^2)$. It is exact for uniform $U$ and otherwise a local relation. Retaining the free-surface term also requires retaining the basic surface slope in $Q_0^{\prime }$; simply adding $R_D^{-2}$ to the denominator while dropping $R_D^{-2}U$ from the numerator would describe a different prescribed-background model.

Use a one-layer, hydrostatic, homogeneous fluid with reference depth $H_0$, no forcing or friction, small Rossby number $ϵ=U/(|f_0|L)\ll 1$, slow evolution, and leading geostrophic balance. For the beta plane, take $\beta L/|f_0|=O(ϵ)$; the relative free-surface displacement is also small, $\eta /H_0=O(ϵ)$, with the Rossby deformation radius retained at the chosen horizontal scale. This excludes the equatorial limit. If a bottom elevation $h_b$ is included, additionally require $h_b/H_0=O(ϵ)$ and stationary topography. Set $h_b=0$ for a flat bottom.

The quasi-geostrophic streamfunction satisfies $\psi =g\eta /f_0$, $u=-{\psi }_y$, $v={\psi }_x$, so relative vorticity is $\zeta ={\nabla }_h^2\psi$. Expanding the shallow-water potential vorticity using actual depth $H_0+\eta -h_b$ givesConsequently the depth-scaled anomaly, the shallow-water quasi-geostrophic potential vorticity, isThe leading material derivative uses the geostrophic flow. Define $J(a,b)=a_x{b}_y-a_y{b}_x$; then conservation yields $Q_t+J(\psi ,Q)=0$. The term $-\psi /R_D^2$ represents free-surface stretching, $\beta y$ the variation of planetary rotation, and $f_0{h}_b/H_0$ topographic stretching. Under a rigid lid or at scales much smaller than the Rossby deformation radius, the free-surface term is absent or negligible and the flat-bottom Barotropic Rossby wave model has $Q={\nabla }_h^2\psi +\beta y$.

Assume a small Rossby number, a beta plane, and small fractional changesAt leading order, geostrophic balance givesExpanding the reciprocal depth in the shallow-water potential vorticity givesThis is the stated shallow-water quasi-geostrophic potential vorticity $P_g$. The term ${\nabla }^2\psi$ is the parcel's relative vorticity, while $-f_0^2\psi /(g{H}_0)=-\psi /L_D^2$ is vortex stretching caused by free-surface displacement, whereis the barotropic deformation radius.