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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.

For a zonal background $U(y)$, its relative vorticity is $\overline{\xi }=-U_y$. The meridional absolute vorticity gradient is thereforeAt $30^{\circ }\mathrm{N}$,A localized jet stream of velocity scale $U=134\mathrm{m}{\mathrm{s}}^{-1}$ and width $L$ hasFor $L=1000$--$2000\mathrm{km}$ this is about $1.7$--$6.8$ times the planetary contribution, so wind curvature cannot generally be neglected. The coefficient in this scaling depends on the chosen jet profile and width convention.

Linearizing the vorticity equation now gives the Rossby-wave equation for a sheared zonal currentFor a qualitative estimate at the jet core, freeze the local coefficients. At a smooth maximum $U_{yy}<0$, so $Q_y>\beta$: the restoring absolute vorticity gradient is stronger. The local Rossby-wave dispersion relation in a zonal jet gives, for the same local $U$ and zonal wavenumber $k$ with $l=0$,soThe envelope in the diagram would advance eastward faster, while individual phase contours would advance more slowly eastward, become stationary if $Q_y=U{k}^2$, or move westward if $Q_y>U{k}^2$. The separation between envelope and phase speeds increases. On a plot with time downward, compare horizontal displacement at fixed time: the envelope moves farther right, while a given phase moves less far right, or left after reversal. If the frequency changes sign for a fixed signed $k$, conjugate the mode to restore the convention $\omega \ge 0$; the physical propagation is unchanged.

This conclusion concerns the concave-down jet core. On a flank with $U_{yy}>0$ the correction has the opposite sign; the value at the edge of the Hadley cell alone does not specify the curvature everywhere. Indeed the angular-momentum-conserving profile in 1(a) is a rising inner branch, not a complete smooth jet maximum. A global sheared-current normal mode requires its meridional structure and boundary conditions; the local calculation is the qualitative estimate requested.

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.