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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 $Q={\nabla }_h^2\psi +\beta y$.

Take $\beta >0$, a uniform eastward $U>0$, and the nondivergent Barotropic Rossby wave model. Its dispersion relation is $\omega =Uk-\beta k/(k^2+l^2)$. A stationary wave with $k\ne 0$ hasThe directional component wavelengths are $2\pi /|k|$ and $2\pi /|l|$ when nonzero. Differentiating the full dispersion relation before imposing stationarity gives its ground-frame group velocity:Writing $k=K\cos \theta$, $l=K\sin \theta$ shows ${\mathbf{c}}_g=U(1+\cos 2\theta ,\sin 2\theta )$. Therefore the stationary Rossby-wave ray envelope at time $\tau$, for wave packets emitted from a localized mountain throughout $0<t<\tau$, is the diskThe circle is the front of the earliest emissions; later emissions fill ray segments back to the source. The sketch represents the geometrical wave packet envelope, not compact support of the full balanced near field. All propagating wave packet energy lies downstream in this model.

If the finite-depth free-surface model of the preceding part is retained consistently, its basic surface slope adds $U/R_D^2$ to the potential-vorticity gradient. Stationarity still gives $K^2=\beta /U$ and the same wavelength, but the group velocity is reduced by $\gamma =K^2/(K^2+R_D^{-2})$; the disk center and radius become $\gamma U\tau$. This tends to the plotted nondivergent result as $R_D\to \infty$.

A westward wind has no nontrivial stationary propagating planetary Rossby-wave train in this model: $U<0$ would require $K^2=\beta /U<0$. Its advective phase motion and intrinsic westward Rossby wave phase motion cannot cancel. A local forced, evanescent response and time-dependent disturbances can still occur. The degenerate $k=0$ limit is not a propagating wave train from an isolated mountain.

The QG thermodynamic relation makes vertical velocity proportional to the material derivative of ${\psi }_z$. At a flat rigid boundary, no normal flow therefore givesor ${\hat{\psi }}^{\prime }(0)=0$ for a nonzero-frequency mode. Let $K^2=k^2+l^2$ and take the vertical structure $\hat{\psi }={\psi }_0\cos (mz)$, which satisfies this condition. Linearizing $q_t+\beta {\psi }_x=0$ gives the Baroclinic Rossby wave dispersion relationFor long horizontal waves, $K{R}_d\ll 1$, so $\omega \simeq -\beta k{R}_d^2$: the vertical-mode deformation term controls the response. For short waves, $K{R}_d\gg 1$, so $\omega \simeq -\beta k/K^2$, the Barotropic Rossby wave form. In both limits the zonal phase propagation is westward relative to the mean flow.

Linearizing conservation of three-dimensional quasi-geostrophic potential vorticity about rest givesFor $K^2=k^2+l^2$, the plane-wave ansatz yieldsThus the vertical structure equation isThe rigid-boundary eigenfunctions areSubstitution gives the Baroclinic Rossby wave dispersion relationsThe $n=0$ member is the Barotropic Rossby wave; $n\ge 1$ are baroclinic vertical modes.