 Solution

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.