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A moving boundary excites an internal gravity wave through its kinematic boundary condition. For wavenumber $k$ and angular frequency $\omega$, the vertical velocity field satisfies $W^{\prime \prime }+k^2(N^2/{\omega }^2-1)W=0$. The radiation condition selects outward group velocity; when $\omega >N$, the bounded response is an evanescent wave.

For constant buoyancy frequency, density forcing $r_0{e}^{-\mu z}\cos kx\cos {\omega }_{0}t$ gives two frequencies $\omega =\pm {\omega }_0$. The quasi-geostrophic streamfunction satisfies ${\lambda }_{\omega }^2=(N^2/f_0^2)(k^2+\beta k/\omega )$ and $C_{\omega }=ig\mu r_0/(2{\rho }_0{f}_0\omega )$. With zero streamfunction at the bottom, a positive ${\lambda }_{\omega }^2$ selects the decaying solution $C_{\omega }(e^{-\mu z}-e^{-{\lambda }_{\omega }z})/({\mu }^2-{\lambda }_{\omega }^2)$. For $\beta ,k>0$, the westward branch propagates vertically when ${\omega }_0<\beta /k$; a radiation condition selects positive vertical group velocity, not merely bounded amplitude. Coincident exponential exponents are handled by the finite limit $-C_{\omega }z{e}^{-\mu z}/(2\mu )$.

For a slowly switched-on plane internal gravity wave in the half-space $z>0$, the method of multiple scales gives transport of its amplitude at the vertical group velocity. For ${\omega }^2=N^2{k}^2/(k^2+m^2)$, $c_{gz}=-\omega m/(k^2+m^2)$. Initially undisturbed fluid and forcing at the bottom require $c_{gz}>0$. The outgoing radiation condition therefore selects $m<0$ when $\omega >0$, although the vertical phase velocity is downward.

For the proposed plane wave, define the positive squared effective wavenumberThe horizontal Laplacian contributes $-(k^2+l^2){\psi }^{\prime }$ and the vertical stretching operator contributes $-(f_0^2/N^2)m^2{\psi }^{\prime }$. Substitution in the Linear Rossby-wave equation therefore givesThe intrinsic phase velocity is westward for positive $\beta$. A stationary disturbance with $\omega =0$ and $k\ne 0$ must have its westward intrinsic phase velocity exactly cancelled by the eastward background current. Solving for the squared vertical wavenumber yieldsFor Earth, with northward $y$, $\beta =2\Omega \cos \varphi /a>0$ in both hemispheres away from the poles. Thus an oscillatory vertical structure requiresThe background flow must be eastward but not too strong for the chosen horizontal scale. Long horizontal wavelengths have the largest allowed speed. An easterly current, $U<0$, gives $m^2<0$ and an evanescent wave; so does an eastward current exceeding the upper bound. At equality $m=0$ and the vertical group velocity vanishes. A zonally uniform disturbance, $k=0$, is a degenerate stationary pattern with no vertical Rossby wave energy propagation, so it cannot be obtained by cancelling $k$ in the preceding calculation.

A real $m$ alone does not select upward propagation. Differentiate the dispersion relation at fixed $k,l$ to obtain the vertical group velocity:The outgoing radiation condition for a surface source requiresFor the convention $k>0$, take $m>0$. The stationary phase lines then obey $dx/dz=-m/k<0$: they tilt westward with height while wave energy travels upward. The real conjugate component $(-k,-l,-m)$ describes the same physical upward-propagating wave.

The figure suggests a winter extratropical path from the lower terrestrial atmosphere into the stratosphere, where the mean zonal flow is eastward. The summer stratosphere has westward flow and blocks such stationary Rossby waves in this idealization. A zero-wind surface is a stationary-wave critical layer in a shear flow; it also bounds the winter westerlies aloft near the upper part of the diagram. For a slowly varying real wind profile, local propagation needs $m^2(z)>0$ along the path, sufficient surface forcing, and no intervening absorbing critical layer in a shear flow. A layer where the eastward wind becomes too strong is a turning barrier for the given horizontal wavenumbers. The uniform-current result is a local guide, not a complete prediction through a sheared spherical terrestrial atmosphere.

For scale, at $60^{\circ }$ latitude $\beta \simeq 1.14\times 10^{-11}{\mathrm{m}}^{-1}{\mathrm{s}}^{-1}$. A planetary zonal wavenumber $s$ has local $k=s/(a\cos \varphi )$. Taking $l=0$ gives the most permissive limitThe resulting limits are approximately $116$, $29$ and $13\mathrm{m}{\mathrm{s}}^{-1}$ for $s=1,2,3$. Therefore the long planetary waves are favored; the strong core of the winter vortex can exclude shorter ones even though it is westerly. Nonzero meridional wavenumber lowers these limits further. A compressible log-pressure coordinate model would also contain a density-scale-height correction; it is absent from the equation supplied here and should not be silently inserted.

Interchanging the Northern and Southern Hemispheres changes the sign of $f_0$, but only $f_0^2$ enters the dispersion relation and vertical group velocity. With the same eastward $x$ and northward $y$, $\beta$ stays positive. Thus the propagation criterion is the same in both hemispheres: it selects the winter westerlies, irrespective of which hemisphere is winter. Actual amplitudes can differ because mountains, land-sea contrasts, forcing spectra and mean winds differ between hemispheres; those source differences do not reverse this propagation criterion.

Let $\Omega$ be the solid exterior to the cavity, and choose $\mathbf{n}$ outward from the solid, hence into the cavity on its boundary $S$. Subtraction of the incident solution from the total solution gives the scattered displacement field equationsThe traction-free cavity condition for the total field gives ${\sigma }_{ij}^s{n}_j=-{\sigma }_{ij}^0{n}_j$ on $S$. The scattered field has zero initial displacement and initial velocity before the pulse reaches the cavity and satisfies an outgoing radiation condition, excluding an additional incoming elastic wave. Equivalently one seeks the causal response to this prescribed boundary traction. Although the incident plane pulse extends across an infinite plane, the causal scattered field from the bounded cavity has finite propagation distance at each finite time.

To prove energy uniqueness for traction-driven elasticity, take the difference $\mathbf{w}$ of two solutions with the same initial conditions, body force, boundary traction and outgoing condition. It satisfiesFor a large enclosing ball $B_L$, multiply by ${\dot{w}}_i$ and integrate over $\Omega \cap B_L$. Integration by parts givesSymmetry of stress replaces the last derivative by ${\dot{e}}_{ij}(w)$, and major symmetry of the elastic stiffness tensor makes that integral the derivative of the strain energy. ThereforeThe cavity contribution is zero. On any fixed finite time interval choose $L$ beyond the causal support of the scattered difference; the outer contribution is then zero as well. This gives conservation of its nonnegative elastic energy. Positive mass density and positive strain energy imply that zero initial energy forces $\dot{\mathbf{w}}=0$ everywhere; the zero initial displacement field then gives $\mathbf{w}=0$. The same argument works for finite-energy outgoing solutions by taking the limit of enclosing surfaces and excluding incoming energy. Hence the causal scattered field ${\mathbf{u}}^s$ is unique. Specifying the incident field and cavity traction alone without the initial and outgoing conditions would not establish uniqueness.

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.

Let $S(z)=d{\rho }_s/dz$, $N^2=-gS/{\rho }_0>0$, and define the quasi-geostrophic streamfunction by $\psi =\tilde{p}/({\rho }_0{f}_0)$. Use horizontal scale $L$, vertical scale $D$, velocity scale $U$ and advective time $L/U$. The required quasi-geostrophic approximation has small Rossby number $\epsilon =U/(|f_0|L)$, $\beta L/|f_0|=O(\epsilon )$, and Burger number $N^2{D}^2/(f_0^2{L}^2)=O(1)$. Hydrostatic, thin-layer scaling has $D/L\ll 1$; density variations are small compared with ${\rho }_0$, as required by the Boussinesq approximation. The leading horizontal flow is nondivergent, with vertical velocity $w=O(\epsilon UD/L)$. Retain forcing of the same slow order as density advection, $r=O(U\tilde{\rho }/L)$; stronger forcing need not preserve this balanced scaling.

Leading geostrophic balance and the hydrostatic approximation giveWrite $D_g/Dt={\partial }_t-{\psi }_y{\partial }_x+{\psi }_x{\partial }_y$. The density equation becomesTaking the curl of the horizontal momentum equations at the next order and using continuity gives the stretching equationSubstitute the expression for $w$. Vertical differentiation commutes with the geostrophic material derivative acting on ${\psi }_z$ here: the extra Jacobian is $J({\psi }_z,{\psi }_z)=0$. Since $N$ depends only on height, this yields the diabatically forced quasi-geostrophic potential vorticity equationThe minus sign is appropriate to the specified density forcing; a positive upward buoyancy forcing is $R=-gr/{\rho }_0$ and gives $D_{g}q/Dt=f_0{\partial }_z(R/N^2)$.

Now take constant $N$ and linearize about rest. Put $s=f_0^2/N^2$ and $S_0=g{f}_0\mu r_0/({\rho }_0{N}^2)$. The forced equation isFor the harmonic heating response of quasi-geostrophic flow, represent the two traveling Fourier components with frequencies ${\omega }_{+}={\omega }_0$ and ${\omega }_{-}=-{\omega }_0$:Each component of the forcing has amplitude $S_0/2$, soEquivalently its vertical structure obeysThese are the time-periodic forced responses; unspecified initial data may add free Rossby waves. Selecting no incoming waves describes forcing switched on causally and then its persistent harmonic part.

For $\beta >0$, the eastward component always has ${\lambda }_{+}^2>0$. Let ${\lambda }_{+}>0$. The growing exponential would introduce an unbounded velocity and pressure disturbance at infinity, so discard it. The bottom condition ${\varphi }_{+}(0)=0$ then givesIf $\mu ={\lambda }_{+}$, this expression has the regular limit $-C_{+}z{e}^{-\mu z}/(2\mu )$, rather than a physical singularity.

For ${\omega }_0>\beta /k$, the westward component also has positive ${\lambda }_{-}^2$. Choose ${\lambda }_{-}>0$ and use the same decaying expression with minus subscripts, including the same coincident-exponent limit. Thus both components are vertically evanescent above the heating.

For $0<{\omega }_0<\beta /k$, putBoth homogeneous solutions $e^{\pm imz}$ are bounded, so boundedness is insufficient. A free Rossby wave with phase $kx+mz-\omega t$ hasThe radiation condition demands energy propagation away from the heating, upward into $z>0$. For $\beta ,k>0$ that selects positive $m$, hence $e^{imz}$ in the westward component. The resulting solution isIt satisfies ${\varphi }_{-}(0)=0$ and, away from the decaying forcing, ${\varphi }_{-}^{\prime }\sim im{\varphi }_{-}$. Its far-field amplitude does not decay in this nondissipative model: continuous forcing supplies an outgoing wave train. Thus the low-frequency westward component propagates vertically while the eastward component remains evanescent.

At the threshold ${\omega }_0=\beta /k$, the westward vertical wavenumber is zero. Excluding the linearly growing homogeneous solution and retaining the bottom condition gives ${\varphi }_{-}=C_{-}(e^{-\mu z}-1)/{\mu }^2$. This bounded limiting response tends to a constant and has zero vertical group velocity; demanding both bottom zero and decay to zero at infinity would overdetermine this threshold problem. For $\beta \le 0$, the branch labels and propagation inequality must be reconsidered; the comparison stated here assumes the usual positive beta-plane gradient.

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.

In the lower uniform half-space the causal vertical wavenumber branch has $\mathrm{Im}{k}_{\beta ,b}>0$. The downward factor $e^{i{k}_{\beta ,b}(z-z_b)}$ decays as $z\to +\infty$, whereas the upward factor grows there and represents an incoming wave from the lower infinity. The outgoing radiation condition therefore retains only the downward factor and givesFor each uniform layer $j$ of thickness $h_j$, work upwards from its known bottom load $Z_{j,b}$. With $q_j={\mu }_j{p}_{\beta ,j}$ and ${\delta }_j=\omega p_{\beta ,j}{h}_j$, part (iv) givesvelocity and traction are continuous across a bonded interface, so this top impedance becomes the bottom load of the next shallower layer. Repeat up to $z_a$ to obtain $Z_{\mathrm{in}}=Z(z_a)$. Poles can be handled by carrying $(V,T)$ or the reciprocal ratio. Since the lower load has positive real part, part (iii) actually ensures that the outgoing whole-stack impedance stays finite with positive real part at all shallower depths in the upper frequency half-plane.

The upper half-space has directional impedance $q_a={\mu }_a{p}_{\beta ,a}$. At $z_a$, an incident velocity $V_0$ and reflected velocity $R{V}_0$ giveTaking their ratio and solving establishes reflection from a layered SH impedance:Thus no individual deeper reflection coefficients are needed after $Z_{\mathrm{in}}$ is known: it contains all underlying interfaces and their coherent multiples. A matched load gives $R=0$; the real-frequency limits of zero traction and zero velocity give $R=+1$ and $R=-1$. For real frequency and propagating upper and lower half-space waves, the usual energy-flux balance provides a further check on the computed reflection and transmission. A stable implementation for thick layers uses the ratio update or its decaying-exponential version instead of multiplying growing transfer matrices.

Equations and conventions. Let ${\rho }_0$ be a constant reference density and define total buoyancy acceleration by $\sigma =-g(\rho -{\rho }_0)/{\rho }_0$. With reduced pressure $\Pi =p/{\rho }_0+gz$, the nonrotating ideal Boussinesq equations areDensity departures are small compared with ${\rho }_0$, so they can be neglected in inertia and mass continuity but retained in the gravitational force. The reference stratification must also change density only weakly over the depth of interest. incompressible flow is already assumed; the Boussinesq approximation additionally restricts the size of the density variations, rather than following automatically from incompressible flow. An ideal fluid here has neither viscosity nor buoyancy diffusion.

A stationary reference profile $\bar{\sigma }(z)$ has buoyancy frequencyStable stratification has $N^2>0$: an upward-displaced parcel retains its old buoyancy and has negative buoyancy relative to its new surroundings.

Vorticity equation and linearization. For motion in the $(x,z)$ plane, choose the streamfunction by $u={\psi }_z$, $w=-{\psi }_x$. The $y$-component of vorticity is then $u_z-w_x={\nabla }^2\psi$, where ${\nabla }^2={\partial }_x^2+{\partial }_z^2$. Taking the $y$-component of the curl of momentum givesThe minus sign in the buoyancy curl follows from $(\nabla \times \sigma \hat{z}{)}_y=-{\sigma }_x$.

Write $\psi =\bar{\psi }(z)+{\psi }^{\prime }$ with ${\bar{\psi }}_z=\bar{u}(z)$ and $\sigma =\bar{\sigma }(z)+{\sigma }^{\prime }$. The linear material derivative is $D_t={\partial }_t+\bar{u}{\partial }_x$. The perturbation equations areThe shear-curvature term arises because $w^{\prime }$ advects the basic vorticity ${\bar{u}}_z$. Apply $D_t$ to the first equation and use the second; $D_t$ commutes with ${\partial }_x$ and does not differentiate the height-only coefficients. Therefore

For a stationary nonzero horizontal wavenumber $k$, the buoyancy equation gives $\hat{\sigma }=N^2\hat{\psi }/\bar{u}$. Substitution into the vorticity equation givesHere ${\ell }^2$ is the Scorer parameter. Where $m^2>0$, the vertical structure is oscillatory; where $m^2<0$, it is evanescent. A rigid flat lower boundary gives $\hat{\psi }(0)=0$ for $k\ne 0$. In an upper evanescent region the growing solution is excluded by boundedness. When propagation persists to infinity, an appropriate radiation condition distinguishes outgoing from incoming energy for a forced problem.

If $\bar{u}=U_0+Sz$ with $S>0$, its curvature term vanishes. Bounded $N^2$ implies ${\ell }^2=N^2/{\bar{u}}^2\to 0$ aloft. Thus every fixed nonzero $k$ eventually lies in an evanescent region. If stratification is sufficiently strong near the bottom, a lower oscillatory region is separated from the upper evanescent region by a turning level of a stationary internal wave. Waves reflect at that level and at the rigid lower boundary, giving vertically trapped modes for the horizontal wavenumbers that satisfy both conditions. In a slowly varying example the phase condition is approximately ${\int }_0^{z_t}mdz=(n-1/4)\pi$, from a lower node and the turning-point connection; the key point is trapping by the decrease in the Scorer parameter, not a physical lid.

Rigid channel modes. With constant $U=\bar{u}>0$ and $N$, the vertical nodes require $m_n=n\pi /H$, $n=1,2,\dots$. Stationarity then requiresThe specified strict interval admits positive $k_n^2$ for exactly $n=1,2$. Choosing $k_1,k_2>0$, every real superposition of these nonzero-wavenumber modes isThe two independent complex constants encode the amplitudes and horizontal phases; negative wavenumbers are supplied by complex conjugation, not additional independent constants.

Exact finite amplitude. Put $\lambda =N^2/U^2$. Both modes and every superposition satisfy ${\nabla }^2{\psi }^{\prime }=-\lambda {\psi }^{\prime }$. For the total streamfunction $\Psi =Uz+{\psi }^{\prime }$, the total buoyancy is $\sigma =N^{2}z+(N^2/U){\psi }^{\prime }=(N^2/U)\Psi$. A stationary velocity $({\Psi }_z,-{\Psi }_x)$ is tangent to its streamfunction contours, so $D\sigma /Dt=0$ exactly. Moreover the nonlinear self-advection of ${\nabla }^2{\psi }^{\prime }$ vanishes because it is proportional to ${\psi }^{\prime }$ itself. The remaining vorticity equation iswhich is an identity. Both rigid boundaries remain impermeable, and the curl-free residual in momentum determines a pressure. Thus the displayed two-mode family consists of exact finite-amplitude Boussinesq solutions, with no small-amplitude restriction on the constants in these equations. Very large amplitudes can overturn the stratification or leave the physical Boussinesq regime; exact existence does not establish stability.

The two-constant statement concerns wave disturbances with $k\ne 0$. If $k=0$ is literally included, arbitrary stationary horizontal shear adjustments ${\psi }^{\prime }=F(z)$ and height-only buoyancy changes are also possible. They are changes of the basic state and are not constrained by the wave equation obtained after division by $k$. Excluding that mean-flow sector is necessary for the intended count.

Balanced variables and PV derivation. Write the disturbance reduced pressure as $p^{\prime }/{\rho }_0=f_0\psi$. Leading geostrophic balance gives $(u_g,v_g)=(-{\psi }_y,{\psi }_x)$, and leading hydrostatic balance gives buoyancy anomaly ${\sigma }^{\prime }=p_z^{\prime }/{\rho }_0=f_0{\psi }_z$. The geostrophic material derivative isIt includes horizontal advection, not an independent prognostic vertical velocity. The appropriate regime has small Rossby number, small stratified Froude number, thin aspect ratio and slow evolution; a common scaling has $[NH/(f_{0}L){]}^2$ of order one and $\beta L/|f_0|\ll 1$.

Only the vertical rotation component is retained in the traditional approximation. In the horizontal momentum balance, the horizontal part of the rotation vector multiplies the small vertical velocity; in vertical momentum, its contribution is small compared with leading hydrostatic pressure and buoyancy in the thin-layer scaling. Away from the equator these corrections are smaller by a factor of order $(H/L)\cot \lambda$. They need not be small in equatorial or strongly nonhydrostatic settings, so the approximation has a definite regime of validity.

Put $A(z)=f_0^2/N^2(z)$. Eliminating $w$ from buoyancy givesThe last step needs justification because $D_g$ and ${\partial }_z$ do not commute on arbitrary fields. Their extra terms here areSubstitution into the supplied vorticity equation givesSince $D_{g}f=\beta v_g=\beta {\psi }_x$,This is three-dimensional quasi-geostrophic potential vorticity conservation; the height-dependent coefficient is retained inside the derivative.

At a small steady lower elevation $b(x,y)$, impermeability gives $w=u_g{b}_x+v_g{b}_y=D_{g}b$, evaluated at $z=0$ to leading order. Substitution into buoyancy and division by $N^2$ gives the topographic buoyancy boundary condition for quasi-geostrophic flow,