 Solution

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