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Let $\zeta$ denote free-surface height and $h=\zeta -b$ the water depth. The inviscid rotating shallow water equations with constant Coriolis parameter $f$ areIt is $\zeta$, not $h$, that supplies the horizontal pressure gradient. For a fixed bottom, mass conservation equivalently gives $D_{H}h=-h(u_x+v_y)$.

Write $\xi =v_x-u_y$ for relative vorticity. Differentiate the second momentum equation in $x$ and the first in $y$, then subtract. The pressure terms cancel. Expanding the differentiated advection terms givesConsequently the absolute vorticity $q_a=\xi +f$ satisfies $D_H{q}_a=-q_a{\nabla }_H\cdot {\mathbf{u}}_H$. Combining this with mass conservation yields exact shallow-water potential vorticity conservation:A sloping bottom changes the depth and hence the stretching, but contributes no curl to the gravitational pressure force.

For the quasi-geostrophic approximation, choose constant reference depth $H_0$, speed $U$, and horizontal length $L$. Require Rossby number $\mathrm{Ro}=U/(|f|L)\ll 1$, slow evolution on the advective time $L/U$, smooth horizontal scales, and $L/L_D=O(1)$ or smaller, where $L_D=\sqrt{g{H}_0}/|f|$ is the Rossby deformation radius. Write $\zeta ={\zeta }_0+\eta$, $b=b_0+\delta b$, and $h=H_0+\eta -\delta b$. The leading momentum balance is geostrophic balance:It gives $\eta /H_0=O(\mathrm{Ro}(L/L_D{)}^2)$, while we require $\delta b/H_0=O(\mathrm{Ro})$ and slopes on the chosen horizontal scale. Relative vorticity divided by $f$ is also $O(\mathrm{Ro})$. ThusThe derivative of the constant first term vanishes. In advecting the first-order anomaly, the ageostrophic flow contributes only at the next order. Therefore the shallow-water quasi-geostrophic potential vorticity equation isThis retains the balanced vortical branch, including Rossby waves driven here by the topographic potential-vorticity gradient. It filters the freely oscillating inertia–gravity branches of the full shallow water equations, and does not retain independent boundary Kelvin wave dynamics. Small Rossby number alone is insufficient for the reduction: fast waves must be absent at leading order, as in well-prepared balanced initial data.

In the channel take $H_0=h_{00}$ and measure the bottom anomaly from mid-channel: $\delta b=\epsilon (y-L/2)$. Define $\beta =\epsilon f/H_0$. The resting potential vorticity is $\beta (y-L/2)$, and the linear Rossby wave equation isFor nonzero zonal wavenumber $k$, wall conditions $\psi =0$ selectThe $k=0$ modes are stationary. Write $K^2=k^2+(n\pi /L{)}^2+L_D^{-2}$ and take real $A$ for the following sketch. The total potential vorticity is $q=\beta (y-L/2)-K^2\psi$. For a contour whose undisturbed position is $y_0$, its small meridional displacement isThus potential-vorticity conservation says a northward displacement in a positive background gradient carries a negative anomaly, $q^{\prime }=-\beta \delta y$. Potential-vorticity inversion turns this into positive $\psi$ and anticyclonic motion around the bulge. At a positive cosine crest, $v={\psi }_x$ is northward on its western flank and southward on its eastern flank. Advection raises the contour on the west and lowers it on the east, so the crest moves west. Algebraically, $\delta y_t=v$ and $\delta y=(K^2/\beta )\psi$ give $\omega /k=-\beta /K^2$, the same phase velocity as above. For $\beta <0$ all propagation directions reverse.

For a square basin use a nonzero-frequency normal mode $\psi =\mathrm{Re}\{\Psi (x,y)e^{-i\omega t}\}$, with $\Psi =0$ on all walls. Its equation isSet $\alpha =\beta /(2\omega )$ and $\hat{\Psi }=e^{i\alpha x}\Psi$. Substitution cancels the first derivative and givesThe square's Dirichlet Laplacian eigenfunctions are $\hat{\Psi }=A\sin (m\pi x/L)\sin (n\pi y/L)$, with $m,n$ positive integers. Consequently the square-basin topographic Rossby mode hasFor $\beta >0$, the positive-frequency branch therefore tends to $\beta L/[2\pi \sqrt{m^2+n^2}]$ as $L/L_D\to 0$, as required. The negative-frequency solution is its complex-conjugate representation for a real disturbance; the physical frequency magnitude is $|\beta |/[2\sqrt{{\pi }^2(m^2+n^2)/L^2+L_D^{-2}}]$.