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For a mode with streamfunction $\varphi$, deformation radius $L_D$, and constant planetary-vorticity gradient $\beta$, the Linear Rossby-wave equation isIn the long-wave limit it becomes a first-order westward transport equation.

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.

Write the velocity field as $u=U+u\prime$, $v=v\prime$. The uniform background has zero relative vorticity, and the perturbation has $\xi \prime =v{\prime }_x-u{\prime }_y$. Keeping only terms linear in the perturbations givesThe incompressibility condition permits a streamfunction $\psi$ with $u\prime =-{\psi }_y$, $v\prime ={\psi }_x$, hence $\xi \prime ={\nabla }_h^2\psi$. Thus the Linear Rossby-wave equation here isChoose a plane wave $\psi =\mathrm{Re}\{\hat{\psi }{e}^{i(kx+ly-\omega t)}\}$, with $K^2=k^2+l^2>0$. Substitution givesand therefore the local Rossby-wave dispersion relation in a zonal jetThe $Uk$ term is the background Doppler shift; the intrinsic phase velocity is westward. A real physical wave can always be represented using the conjugate wavevector if necessary to choose $\omega \ge 0$.

The relevant ground-frame zonal phase velocity and group velocity areTime increases downward in the source diagram. The individual phase contours lean towards increasing longitude as time increases, so their phase velocity is eastward. The envelope moves eastward faster: successive crests grow at the downstream edge and decay upstream. This illustrates Rossby-wave downstream development.

For the intended purely zonal wave packet, $l=0$, andConsequently, with $U,\beta >0$ and the positive-frequency choice $k>0$,There is a small interpretation qualification: if “zonally propagating” specified only the plotted direction while allowing an unknown nonzero meridional wavenumber, eastward phase velocity alone would imply $k^2+l^2>\beta /U$, not the stronger bound on $k$ alone. In this uniform model a strictly zonal group velocity with $\beta k\ne 0$ does imply $l=0$, which supplies the intended interpretation.

Linearization about rest replaces the material derivative by ${\partial }_t$, so the quasi-geostrophic potential-vorticity equation becomesInsert the separation of variables $\psi =\varphi (x,y,t)P(z)$ and imposeThe horizontal amplitude then satisfiesThe Linear Rossby-wave equation therefore governs each vertical normal mode, whose Rossby deformation radius is $c/f_0$.