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Let $H=h_{00}+\zeta$ and write the horizontal shallow water equations on the constant f-plane asKeeping only terms linear in the disturbance about rest gives the linearized shallow water equationsDefine the horizontal divergence $\delta =u_x+v_y$ and relative vorticity $q=v_x-u_y$. Taking respectively the divergence and vertical curl of the momentum equation yieldsTogether with ${\zeta }_t=-h_{00}\delta$, this provesThe nonlinear shallow-water potential vorticity is $Q=(f+q)/H$. The nonlinear vertical vorticity equation gives $D(f+q)/Dt=-(f+q)\delta$, while mass conservation gives $DH/Dt=-H\delta$; their ratio is therefore materially conserved. Expanding about rest,Advection of its constant background contributes nothing at first order, explaining why the displayed linear anomaly is conserved at each fixed position.

A further time derivative eliminates $q$ and $\zeta$ from the divergence equation:For a plane wave with horizontal wavevector $(k,l)$, write $K=\sqrt{k^2+l^2}$. More directly, substituting the plane wave into the three linear equations gives a coefficient matrix with determinant proportional to $\omega [{\omega }^2-f^2-c_0^2{K}^2]$. Thus the full linear rotating shallow-water dispersion relation has one balanced branch and two wave branches:The latter are inertia-gravity waves. The positive branch starts at $f$ with zero slope and tends to the straight line $c_{0}K$; the negative branch is its reflection. The requested sketch, including the stationary branch, is

For the positive-frequency branch, the magnitudes of the phase velocity and group velocity areThe group velocity vector is $c_0^2(k,l)/\omega$; the negative branch reverses its direction. Near $\omega =f$, namely $K{R}_D\ll 1$ with the Rossby deformation radius $R_D=c_0/f$, $c_p\sim f/K$ is large while $c_g\sim c_0^{2}K/f$ is small. The $K=0$ limit is an inertial oscillation, for which a spatial phase speed is not defined. Far above $f$, both speeds approach $c_0$, the shallow water gravity-wave speed, with $c_p>c_0$ and $c_g<c_0$ at finite $K$.

On the zero-frequency branch, geostrophic balance givesFor a nonconstant plane wave, $\hat{q}=-(g{K}^2/f)\hat{\zeta }$: relative vorticity and elevation have opposite signs. They can both be nonzero and spatially structured, whereas the horizontal divergence vanishes identically. A constant elevation is also a zero-frequency solution, with zero velocity and zero relative vorticity.

For the initially resting ridge, set ${\zeta }_0(x)=ϵh_{00}{e}^{-x^2/L^2}$ and assume an unbounded domain, uniform in $y$. Initial potential vorticity fixesIn the balanced remainder, insert $q_s=(g/f){\zeta }_s^{\prime \prime }$ to obtainThe decaying Green's function of this modified Helmholtz equation gives the Geostrophic adjustment of a Gaussian height ridge explicitly:The initial outward pressure force creates a cross-ridge current; the Coriolis acceleration turns that current and builds the along-ridge geostrophic flow. The remainder is radiated inertia-gravity waves, not a dissipative disappearance of energy.

For precision, use the Fourier transform $\hat{\zeta }(k)=\int \zeta (x)e^{-ikx}dx$. Initial rest implies ${\zeta }_t(x,0)=0$, and the conserved anomaly givesConsequentlyThese formulas satisfy both initial conditions. They also show that the waves carry no linear potential vorticity anomaly: $q=(f/h_{00})(\zeta -{\zeta }_0)$ and $\delta =-{\zeta }_t/h_{00}$ reconstruct the remaining fields, with $u$ and $v$ obtained from their derivatives. On the infinite line, dispersive wave packets propagate away and decay locally, leaving the balanced ridge and its oppositely directed flanking currents. Conservation of energy still holds globally. In a reflecting finite basin, standing waves may persist, so a permanent pointwise relaxation is not guaranteed without dissipation.

If $L\gg R_D$, most of the initial elevation remains balanced: ${\zeta }_s\simeq {\zeta }_0$ with small smoothing corrections. If $L\ll R_D$, the balanced ridge spreads to width $R_D$, and away from the original narrow coreThus the final central height is much smaller than the initial one and the wave component is substantial. The integral of the balanced elevation equals the initial integral, since the Green's function has unit integral; the oscillatory component has zero net added volume.

Write $\zeta =(\nabla \times \mathbf{u})\cdot {\mathbf{e}}_z$ and $q=\zeta -f\eta /h_0$, so the PDF's vector potential vorticity is $\mathbf{Q}=q{\mathbf{e}}_z$. The local TeX loses some of this boldface. Taking the vertical curl of the momentum equation gives ${\zeta }_t=-f\nabla \cdot \mathbf{u}$, while mass conservation gives ${\eta }_t=-h_0\nabla \cdot \mathbf{u}$. ThusThis is the linear perturbation of shallow-water potential vorticity, scaled by $h_0$; it should not be confused with the full nonlinear conserved ratio $(f+\zeta )/h$.

Since $\nabla \cdot ({\mathbf{e}}_z\times \mathbf{u})=-\zeta$, taking the divergence of momentum gives $(\nabla \cdot \mathbf{u}{)}_t=f\zeta -g{\nabla }^2\eta$. Differentiate mass conservation and use $\zeta =q+f\eta /h_0$ to obtainFor $\mathbf{Q}=0$, substitution of the stated plane wave gives the linear rotating shallow-water dispersion relation, with $c=\sqrt{g{h}_0}$:The two even branches approach $\pm c|k|$ at large $|k|$ and have values $\pm |f|$ at $k=0$. The original dispersion diagram below also compares the speed magnitudes.

For $k\ne 0$, the phase velocity and group velocity magnitudes areThe printed convention $e^{i(kx+\omega t)}$ means the signed phase velocity is $-\omega /k$ and the signed group velocity is $-d\omega /dk$. The speed magnitudes above are independent of that sign convention.

For $f\ne 0$, longer wavelengths have greater phase velocity but smaller group velocity: individual crests and wave packets therefore answer “faster or slower” differently. At long wavelength, $|k|\ll |f|/c$, the frequency is nearly the inertial frequency $|f|$, $|c_p|\sim |f|/|k|$, and $|c_g|\sim c^2|k|/|f|$. Rotation has its largest relative effect here, and the waves are strongly dispersive. At short wavelength, $|k|\gg |f|/c$, both speeds tend to $c$ and rotation gives only a small, weakly dispersive correction. If $f=0$, the gravity waves are nondispersive, with speed $c$; at $k=0$ the phase velocity formula is undefined.

Wave dispersion means that phase velocity depends on wavenumber or, equivalently, wavelength. Wave packets generally travel with group velocity, which need not equal phase velocity. The linear rotating shallow-water dispersion relation ${\omega }^2=f^2+c^2{k}^2$ has faster long-wave phase motion but slower long-wave packet motion when $f\ne 0$.