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Away from forcing, set $F=0$. For a linear sinusoidal disturbance $A=\mathrm{Re}[\hat{A}{e}^{i(kx-\omega t)}]$, the reduced equation givesSince $\sigma ,\beta ,{\alpha }_1>0$, the phase velocity $c_p=\omega /k=\sigma k^2/(\beta {\alpha }_1)$ is positive for every nonzero wavenumber. The group velocity isThus both the linear phase and wave packet travel downstream; the sign of $k$ does not introduce an upstream branch.

Now seek a permanent nonlinear wave with $\xi =x+ct$, for which the propagation velocity is $-c$. Substitution gives $\sigma A^{(3)}-\beta {\alpha }_{1}c{A}^{\prime }-{\alpha }_{2}A{A}^{\prime }=0$. Integration with vanishing far-field data givesMultiplying by $A^{\prime }$ and integrating again, with zero integration constant, givesHere ${\alpha }_2>0$, because ${\gamma }_0>{\gamma }_1>0$ and $R<R+1$. A smooth nonzero wave approaching zero requires $c>0$: for $c<0$ the right-hand side is negative at sufficiently small nonzero $A$. At $c=0$ the allowed positive branch has no nonzero turning point, so it cannot form a smooth localized pulse.

When $c>0$, the nonzero turning point is $A_m=-3\beta {\alpha }_{1}c/{\alpha }_2$. The energy curve is positive between $A_m$ and zero and negative below $A_m$, giving a homoclinic negative pulse. The positive branch has no turning point and cannot return to zero. ThereforeAn explicit solution, also confirming smooth exponential tails, isThis establishes the opposite propagation directions in the annular displacement equation: linear waves move downstream while its localized nonlinear depression waves move upstream.

Let the layer occupy $0<z<h$, with a free surface at $z=0$ and rigid base at $z=h$. Write an SH-wave as $u_y=Y(z)e^{i(kx-\omega t)}$. The equation and boundary conditions giveThus $Y=A\cos (qz)$ and the guided SH modes between rigid and free planes haveThe propagating branch has $\omega >{\omega }_n$ and $k=\sqrt{{\omega }^2-{\omega }_n^2}/{\beta }^{\prime }$. Hence the rigid-base approximation to Love waves predictsAt each threshold $c_n\to \infty$ and $U_n\to 0$. As frequency rises, the phase velocity decreases towards ${\beta }^{\prime }$ and the group velocity increases towards ${\beta }^{\prime }$. Frequencies below ${\omega }_n$ have imaginary horizontal wavenumber and do not give propagating guided modes. The infinite threshold phase speed does not violate causality: the limiting group speed is zero. These quarter-wave thresholds differ from the actual Love-wave cutoff frequencies of an elastic substrate.

Write $c=\sqrt{gH}$ and $R_d=c/|f|$, with $f\ne 0$. For the profiles below take $f>0$, so $\alpha =L/R_d$ agrees with the stated parameter; changing the sign of $f$ reverses the balanced meridional velocity but not the height profile. The linearized shallow water equations areTaking the horizontal curl gives $(v_x-u_y{)}_t=-f(u_x+v_y)=f{\eta }_t/H$. Hence the signed linear shallow-water potential vorticity anomaly is conserved:The initial data are independent of $y$, so the solution remains independent of $y$ and $v_x=f(\eta -{\eta }_0)/H$. Differentiating continuity in time and using the zonal momentum equation gives

For a horizontal Fourier mode with wavenumber $k$, put ${\Omega }_k^2=f^2+c^2{k}^2$. The solution splits into a balanced component and inertia-gravity waves:These waves have frequencies $|\omega |\ge |f|$, phase velocity $\omega /k$, and group velocity $c^{2}k/\omega$, whose magnitude is less than $c$. They have zero perturbation potential vorticity and carry the unbalanced part away. On the unbounded line, the localized difference between the initial and balanced profiles disperses, leaving a local steady limit. No damping is needed for this local geostrophic adjustment; in a closed periodic domain, undamped waves would persist and a literal pointwise steady limit would generally not exist.

The bounded steady height satisfies the modified Helmholtz equationOddness removes the even homogeneous solution in the central region. The exterior solution is bounded and tends to the appropriate plateau. Matching both height and its first derivative at $x=\pm L$ gives the geostrophic adjustment of a finite-width height ramp:For example, writing the interior solution as $hx/L+B\sinh (x/R_d)$, matching its slope to the decaying exterior at $L$ gives $B=-h{e}^{-\alpha }/\alpha$.

The steady meridional momentum equation gives $u_s=0$, and the zonal equation gives the geostrophic balance $f{v}_s=g{\eta }_s^{\prime }$. ThusThe height is odd and monotone, and the meridional current is even, with maximum $v_s(0)=gh(1-e^{-\alpha })/(fL)$ for $h>0$.

For $\alpha \ll 1$, the initial ramp is narrow compared with the Rossby deformation radius. To leading order,Near the origin the height rises with slope $h/R_d$, rather than $h/L$; the current has width $R_d=L/\alpha$ and peak $gh\alpha /(fL)$. For $\alpha \gg 1$, the central height remains nearly $hx/L$ and the central velocity nearly $gh/(fL)$. Around each end of the ramp there is a matching layer of width $R_d=L/\alpha$, with height correction of order $h/\alpha$; the exterior current decays over the same width. At $x=\pm L$ the current is approximately half its central value. In both limits the zonal current is zero. These are the broad rotational adjustment of a narrow disturbance and the weak edge adjustment of a broad disturbance, respectively.

Use the energy normalization specified for the height energy: the linear total energy density is $(u^2+v^2)/2+g{\eta }^2/(2H)$. The linear equations implyFor $\alpha \ll 1$, let the potential energy loss mean the positive difference between the initial and final values, per unit transverse length. Although both absolute potential energies diverge on the infinite line, their difference is finite. The leading step approximation givesThe finite ramp changes this leading expression by a relative $O(\alpha )$ correction. The initial kinetic energy is zero and the final gain isThe remainder $\Delta V-\Delta T\simeq g{h}^2{R}_d/H$ is energy carried away by the outgoing inertia-gravity waves. This explains why the ratio is less than one without invoking dissipation: conservation of energy includes the waves, not only the final balanced flow.

Let $R=\overline{u^{\prime }{v}^{\prime }}$ be the eddy momentum flux, $F=\overline{{\rho }^{\prime }{v}^{\prime }}$ the eddy density flux, and $S=d{\rho }_s/dz<0$. The PDF's hydrostatic equation is $g\overline{\rho }=-{\overline{p}}_z$; the converted TeX mistakenly replaces the pressure derivative by a density derivative. With $N_0^2=-gS/{\rho }_0$, defineThe residual mean circulation is nondivergent because $v_y^{\ast }+w_z^{\ast }={\overline{v}}_{ay}+{\overline{w}}_{az}$. Substituting in the original mean momentum and density equations gives the transformed Eulerian mean equationswith the unchanged geostrophic and hydrostatic balances. Here the Eliassen–Palm flux components areThe eddy buoyancy flux has been absorbed into the residual transport, while both types of wave forcing enter the momentum equation through the Eliassen–Palm flux divergence.

For geostrophic disturbances, the Taylor identity for quasi-geostrophic flux givesFor example, zonal averaging removes $\overline{{\psi }_x^{\prime }{\psi }_{xx}^{\prime }}$ and converts the remaining horizontal and vertical terms in $\overline{{\psi }_x^{\prime }{q}^{\prime }}$ into derivatives of $\overline{{\psi }_x^{\prime }{\psi }_y^{\prime }}$ and $(f_0^2/N_0^2)\overline{{\psi }_x^{\prime }{\psi }_z^{\prime }}$, giving precisely the displayed fluxes.

Locally, for a basic flow with nonzero meridional PV gradient $Q_y$, the quasi-geostrophic wave pseudomomentum is $\mathcal{A}=\overline{q^{\prime 2}}/(2Q_y)$. Multiplying the linear perturbation PV equation by $q^{\prime }/Q_y$ giveswhere $\mathcal{D}$ is a PV source or sink. In a locally uniform conservative Rossby wave packet, $\mathsf{F}=\mathcal{A}{\mathbf{c}}_g$. Thus the flux tracks group velocity rather than the generally different phase velocity, with a signed wave activity if $Q_y<0$.

The non-acceleration theorem for quasi-geostrophic waves states that statistically steady conservative waves, with no wave-activity sources, sinks or independent boundary forcing, produce no balanced mean acceleration. Their flux divergence vanishes. In the present stable, impermeable mean-flow problem the homogeneous residual-circulation equation then has the zero solution, so $v^{\ast }=w^{\ast }=0$ and ${\overline{u}}_t=0$. The Eulerian mean flow can still circulate because its eddy-induced correction need not vanish. Dissipation, absorption at a critical layer, time-dependent wave activity or boundary forcing produces a nonzero flux divergence and invalidates the non-acceleration conclusion.

Use incompressibility, negligible viscosity and surface tension, and an initially irrotational flow. Take the undisturbed free surface at $z=0$ and bed at $z=-H$. The velocity potential $\phi$ satisfies Laplace equation and ${\phi }_z(-H)=0$. For a Fourier mode, writeThe linear kinematic boundary condition and linear Bernoulli equation at the free surface are ${\eta }_t={\phi }_z$ and ${\phi }_t+g\eta =0$. Eliminating $A,{\eta }_a$ gives the surface gravity wave dispersion relationThe rightward phase velocity and group velocity areThe phase velocity transports a crest; the group velocity transports a narrow-band envelope and its leading-order energy. For $kH\gg 1$, $\omega \sim \sqrt{gk}$ and $c_g\sim c_p/2\sim \sqrt{g/k}/2$: deep-water surface gravity waves are dispersive. For $kH\ll 1$, $\omega \sim k\sqrt{gH}$ and $c_g\sim c_p\sim \sqrt{gH}$, the nondispersive shallow-water dispersion relation. The boundary velocity controls the excited amplitude; it does not change the linear dispersion relation.

For a nonrotating inviscid Boussinesq fluid, start from $\nabla \cdot \mathbf{u}=0$, $D\rho /Dt=0$ andThe resting background has ${\bar{p}}_z=-g\bar{\rho }$ and $N^2=-g{\bar{\rho }}_z/{\rho }_0>0$. Write $\rho =\bar{\rho }(z)+{\rho }^{\prime }$, $p=\bar{p}+p^{\prime }$ and $\sigma =-g{\rho }^{\prime }/{\rho }_0$. Keeping first-order terms givesThe PDF has a genuine sign error: its momentum equation uses a minus sign before $\sigma \hat{\mathbf{z}}$ despite defining $\sigma =-g{\rho }^{\prime }/{\rho }_0$. A lighter parcel has $\sigma >0$ and accelerates upward, so the plus sign is necessary. Retaining both printed signs would yield negative squared wave frequency and exponential growth in a supposedly stable fluid; no oscillatory stable-wave dispersion relation follows from that inconsistent system. Alternatively one can define a downward density variable $\tilde{\sigma }=g{\rho }^{\prime }/{\rho }_0$, but then its density equation must have ${\tilde{\sigma }}_t=+N^2{w}^{\prime }$.

For the consistent system, take a plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$, with $k\ne 0$. Incompressibility gives $ku+mw=0$. Project the momentum equation perpendicular to the wavevector, eliminating pressure: $-i\omega w=k^2\sigma /(k^2+m^2)$. Combining with $-i\omega \sigma =-N^{2}w$ givesOn a smooth signed-frequency branch, the phase velocity and group velocity areTheir scalar product is zero. Equivalently, the dispersion relation is homogeneous of degree zero in $(k,m)$, so Euler homogeneous function theorem gives $(k,m)\cdot {\nabla }_{(k,m)}\omega =0$. The phase velocity moves crests normal to their phase lines; the group velocity moves wave packets and energy along those lines. Their perpendicularity explains why following individual crests does not reveal the direction of energy propagation.

Assume $U>0$, $k=\alpha >0$, constant $N$, negligible rotation, dissipation and reflection from infinity, small $k{\eta }_0$ and $m{\eta }_0$, and no wave overturning. The terrain is stationary, so laboratory frequency is zero and intrinsic frequency is $\hat{\omega }=-kU$. Propagation requires $kU<N$. PutThe kinematic boundary condition at the linearized bed is $w(x,0)=U{\eta }_x=-kU{\eta }_0\sin (kx)$. The upward radiation condition selects the positive $m$ branch because $c_{gz}=-\hat{\omega }m/(k^2+m^2)>0$. ThusThe internal-wave polarization and every other perturbation follow from this vertical velocity and the linear equations.

In the frame moving with the fluid, $x^{\prime }=x-Ut$, the phase is $k{x}^{\prime }+mz+kUt$, andThe intrinsic energy ray slopes upward and upstream, $dz/d{x}^{\prime }=-k/m$, along a constant-phase line of an internal gravity wave. At fixed intrinsic frequency the spatial wave equation is hyperbolic, with the two characteristic curve line slopes $dz/d{x}^{\prime }=\pm k/m$; the radiation condition selects the upward ray of the appropriate horizontal propagation branch.

In the terrain frame, phase lines remain fixed and the pattern phase velocity is zero. A transported packet hasIts energy path slopes upward and downstream, $dz/dx=m/k$, and need not coincide with a stationary phase line. One should not add $(U,0)$ to the chosen normal intrinsic phase velocity and call that the normal ground-frame phase velocity: tangential motion along a phase line is arbitrary; zero laboratory frequency makes the entire crest pattern stationary. The sketch distinguishes stationary crests, intrinsic characteristics and transported energy.

If $kU>N$, $m$ is imaginary and the bounded field is an evanescent wave; at $kU=N$ the vertical group velocity vanishes and an ordinary upward radiation interpretation degenerates.

Use translating coordinates $Z=z-vt$ and a profile $\psi (\mathbf{x},t)=\Phi (x,y,Z)$; the background condensate remains at rest in the laboratory. Then ${\psi }_t=-v{\Phi }_Z$, and the Gross–Pitaevskii solitary wave equation isThis is a coordinate translation of a traveling solution, rather than an additional phase boost to change the background flow.

Set $\Phi =1+f+ig$ with small real $f,g$. Since $|\Phi {|}^2=1+2f+O(f^2+g^2)$, the real and imaginary linearized equations areFor a longitudinal sinusoid, choose $f=A\cos (kZ)$ and $g=B\sin (kZ)$. Their coefficients obeyA nonzero solution therefore requires $4v^2=k^2+2$, giving the positive-direction phase velocityEquivalently the time-dependent linear Bogoliubov spectrum is ${\omega }^2=k^2(k^2+2)/4$ and $v=\omega /k$. Its long-wavelength sound speed is $1/\sqrt{2}$, which becomes $\sqrt{\mu /m}$ in physical units using ${\ell }_0/t_0=\sqrt{2\mu /m}$. For a general wavevector $\mathbf{k}$, the stationary resonance is $4v^2{k}_z^2=k^2(k^2+2)$; specifying only its magnitude does not determine the velocity unless its direction is also fixed. This sinusoidal dispersion describes linear radiation about the background, not an amplitude-independent dispersion law for every nonlinear solitary wave; localized subsonic waves do not require a real oscillatory far-field wavenumber.

Substitution of $e^{i(kx-\omega t)}$ gives the dispersion relation $\boxed{\omega (k)=Uk+k^3}$. The phase velocity is $U+k^2$, whereas the group velocity is $U+3k^2$. Thus nonzero-wavenumber crests move more slowly than their wave packet; the velocities coincide at zero wavenumber.

Fourier evolution gives $\varphi (Vt,t)=\int A(k)e^{it\Phi (k)}dk$, where $\Phi (k)=(V-U)k-k^3$. For $V>U$ the two nondegenerate stationary points are $k=\pm \kappa$, $\kappa =\sqrt{(V-U)/3}$. At these points $\Phi (\pm \kappa )=\pm 2{\kappa }^3$ and ${\Phi }^{\prime \prime }(\pm \kappa )=\mp 6\kappa$. Expanding the phase quadratically near each point and evaluating the oscillatory Gaussian integral gives the stationary phase methodFor a real initial field, $A(-\kappa )=\overline{A(\kappa )}$, so the two terms are conjugate. The approximation assumes a sufficiently smooth decaying amplitude; if a leading coefficient vanishes, the next stationary-phase terms determine that contribution.

For $V<U$, ${\Phi }^{\prime }(k)=V-U-3k^2<0$ everywhere. There are no real stationary points, so no leading propagating packet reaches such a ray. For a Schwartz amplitude, repeated integration by parts with $e^{it\Phi }=(it{\Phi }^{\prime }{)}^{-1}{\partial }_k{e}^{it\Phi }$ makes the contribution $O(t^{-N})$ for every fixed $N$. With weaker data the decay rate depends on their regularity. This agrees with the minimum group velocity being $U$.

The free surface is a material surface. For $F(x,y,t)=y-\eta (x,t)$, the material derivative vanishes on $F=0$. Since the velocity is $\nabla \varphi$, this gives the exact kinematic boundary condition for a free-surface graphAt the surface the pressure equals the constant atmospheric pressure. Absorbing that constant into the time-dependent potential gauge in Unsteady Bernoulli equation yields the exact dynamic boundary condition for an inviscid interfaceThere is no surface-tension term in this model.

Write $\varphi =Ux+\phi$ and linearize in the small perturbation $\phi$ and the surface displacement. The base value is $F=U^2/2$, and evaluation at the displaced surface can be replaced by evaluation at zero to first order. ThusIncompressibility requires Laplace's equation ${\phi }_{xx}+{\phi }_{yy}=0$. For $k>0$, the complex perturbation $b{e}^{ky}{e}^{i(kx-\omega t)}$ satisfies this equation and decays with depth. Put $\eta =\mathrm{Re}\{a{e}^{i(kx-\omega t)}\}$ and $\Omega =\omega -kU$. The linear boundary conditions becomeThey are consistent precisely whenConsequently the two laboratory phase velocities of a deep-water gravity wave areIf propagation of a wave packet is intended, its group velocities are instead $d\omega /dk=U\pm \frac{1}{2}\sqrt{g/k}$.

Take shear-horizontal waves with displacement $\mathbf{u}=(0,0,w(x,y,t))$. This is divergence-free, so the elastic equation reduces to $\rho w_{tt}=\mu (w_{xx}+w_{yy})$. Rigid boundaries require $w=0$ at $y=0,h$, giving modes $w=A\sin (n\pi y/h)\cos (kx-\omega t)$, $n\ge 1$. With shear speed $c_s=\sqrt{\mu /\rho }$, the dispersion relation isFor $k>0$, the phase velocity is $v_p=\omega /k>c_s$ and the group velocity is $v_g=d\omega /dk=c_s^{2}k/\omega <c_s$, with $v_p{v}_g=c_s^2$.

The kinetic energy density is $T=\rho w_t^2/2$. The only strains are $e_{xz}=w_x/2$ and $e_{yz}=w_y/2$, together with their symmetric counterparts; thus the isotropic linear-elastic energy density is $W=\mu (w_x^2+w_y^2)/2$. Average over one temporal period and integrate across the layer. The sine- and cosine-squared transverse integrals are both $h/2$, givingThe guided shear-horizontal mode has average equipartition after both operations; a time average at a fixed transverse point alone need not give equality.

In the Boussinesq approximation, replace density by a constant reference density ${\rho }_0$ in inertia and incompressible flow, retaining the small density departure only in the gravitational force. Write $\rho =\bar{\rho }(z)+{\rho }^{\prime }$ and subtract the resting hydrostatic pressure. The upward buoyancy perturbation acceleration and squared buoyancy frequency areThus stable density stratification has $N^2>0$. Let $p$ henceforth denote the pressure anomaly divided by ${\rho }_0$. The ideal, nonrotating Boussinesq equations areThe buoyancy equation follows by materially conserving the total buoyancy $b_0(z)+\sigma$, where $b_0^{\prime }=N^2$; it does not assume constant $N$.

For an $x$-periodic flow of period $P$, define $\bar{a}=P^{-1}{\int }_0^{P}adx$. An infinite-domain average with vanishing end contributions works as well. The necessary pressure condition is no mean zonal pressure gradient, $\overline{p_x}=0$: periodic pressure satisfies it, whereas a pressure containing a term proportional to $x$ does not. Use incompressible flow to write the zonal momentum equation in conservative form,Averaging its zonal derivative terms yields the exact Reynolds stress relation

For two-dimensional small disturbances about rest, the Linearized Boussinesq equations are $u_t=-p_x$, $w_t=-p_z+\sigma$, ${\sigma }_t=-N^{2}w$, and $u_x+w_z=0$. Differentiate the two momentum equations to eliminate pressure: $(u_z-w_x{)}_t=-{\sigma }_x$. Differentiating in $x$ and using $u_{xz}=-w_{zz}$ gives ${\nabla }^2{w}_t={\sigma }_{xx}$. One further time derivative givesThe equation remains valid for $N=N(z)$. No vertical derivative of $N$ was taken: ${\partial }_x^2(N^{2}w)=N^2{w}_{xx}$.

For constant $N$ and $k\ne 0$, a plane internal gravity wave hasBoth signs give the prescribed lower-boundary velocity. The radiation condition excludes energy arriving from infinity. At positive frequency, the vertical group velocity isFor a source below the fluid, energy must propagate upward, so the outgoing solution has $m<0$, even though its vertical phase velocity $\omega /m$ is downward.

To obtain this selection causally, let $T=\mu t$ and $Z=\mu z$, and replace the constant amplitude by $a(Z,T)$. The method of multiple scales treats $(z,t,Z,T)$ as independent during differentiation: the carrier derivatives hold $Z,T$ fixed, while $a_T$ holds $Z$ fixed and $a_Z$ holds $T$ fixed. Only after differentiating do we restrict to $T=\mu t$, $Z=\mu z$. Acting on the complex wave, the operators becomeWith $K^2=k^2+m^2$, the wave equation has residualThe zeroth-order term is the dispersion relation; the first-order term vanishes precisely whenIf $a_0(T)$ is the switched-on boundary amplitude, with $a_0(T)=0$ for $T<0$, the outgoing envelope is $a(Z,T)=a_0(T-Z/b)$ for $b>0$. Its characteristics carry the boundary signal into initially undisturbed fluid. If $b<0$, characteristics instead arrive from the upper half-space; zero initial disturbances and no incoming signal cannot establish the proposed boundary-forced wavetrain. This is the internal-wave envelope radiation condition.

The leading incompressible flow polarization gives $u^{\prime }=-(m/k)w^{\prime }$. Hence, to second order in amplitude and leading order in the slow modulation,There is no meridional transport here. Envelope transport gives ${\partial }_t|a{|}^2=-b{\partial }_z|a{|}^2$, so an initially vanishing mean develops asThese are the leading expressions for internal-wave momentum deposition; higher slow-modulation corrections to the stress are beyond the displayed approximation. In particular, a spatially uniform fully established wavetrain has zero stress divergence, but the front has already produced its mean flow.

The mean wave energy density per unit mass follows from ${\sigma }^{\prime }=-i{N}^2{w}^{\prime }/\omega$:Thus $\bar{u}=(k/\omega )\mathcal{E}$. In the original frame the mean kinetic energy is ${\bar{u}}^2/2=O(a^4)$. In a frame translating at $c=\omega /k$, the undisturbed velocity is $-c$, and the change in mean kinetic energy is $-c\bar{u}+O(a^4)=O(a^2)$. Indeed $\mathcal{E}-c\bar{u}=0$ at this order behind the front. Mean-flow energy is therefore essential at the same order as wave energy in the translating frame. This follows also from the frame transformation of energy, $E\mapsto E-cP$ up to the fixed background constant, and is why conservation of momentum cannot be omitted from the moving-frame conservation of energy budget.

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.