= Solution
Apply <mass conservation> to a fixed horizontal region $D$. The layer mass is $\rho\int_Dh\,dx\,dy$ and the outward horizontal flux is $\rho\int_{\partial D}h(u,v)\cdot\mathbf n\,ds$. The <divergence theorem> and arbitrary choice of $D$ give
$$
\boxed{h_t+(hu)_x+(hv)_y=0.}
$$
This depth-integrated law retains free-surface motion; neglecting the vertical velocity in horizontal momentum is not the assumption that the depth cannot change.
Take the free-surface pressure to be constant. <hydrostatic pressure> gives $p=p_{\rm atm}+\rho g(h-z)$, so the horizontal pressure forces are $-g\eta_x,-g\eta_y$. Linearization about a resting layer gives the <linearized shallow water equations>
$$
u_t-fv=-g\eta_x,\qquad v_t+fu=-g\eta_y,\qquad
\eta_t+h_0(u_x+v_y)=0.
$$
Taking the curl of the momentum equations yields $\zeta_t=-f(u_x+v_y)$. Combining with continuity gives the <linearized shallow-water potential-vorticity anomaly>
$$
\boxed{\partial_t\left(\zeta-\frac f{h_0}\eta\right)=0,\qquad Q=Q_0(x,y).}
$$
For $Q_0=0$, $\zeta=f\eta/h_0$. Taking the divergence of momentum gives $\partial_t(u_x+v_y)-f\zeta=-g\nabla_h^2\eta$. Eliminate the divergence with continuity to obtain
$$
\boxed{\eta_{tt}-gh_0\nabla_h^2\eta+f^2\eta=0.}
$$
For $\theta=k(x-ct)$, substitution of $\eta=\eta_0\cos\theta$ gives the <inertia-gravity wave> <dispersion relation>
$$
\boxed{\omega^2=gh_0k^2+f^2,\qquad\omega=kc,\qquad c^2=gh_0+f^2/k^2.}
$$
The associated purely oscillatory <velocity field> is
$$
u=U\cos\theta,\qquad v=V\sin\theta,\qquad
U=\frac{c\eta_0}{h_0},\quad V=\frac{f\eta_0}{kh_0}=\frac f\omega U.
$$
These formulas satisfy all three linearized equations and the zero-anomaly condition. A separately added spatially uniform inertial oscillation is not part of this monochromatic wave.
To first order in amplitude evaluate the wave at a particle's equilibrium coordinate $x_*$, so $\theta_* = kx_*-\omega t$. Integrating its <velocity> gives displacements relative to the orbit centre
$$
\xi=-\frac U\omega\sin\theta_*,\qquad
\upsilon=\frac V\omega\cos\theta_*,\qquad
\boxed{\frac{\xi^2}{(U/\omega)^2}+\frac{\upsilon^2}{(V/\omega)^2}=1.}
$$
These are <particle ellipses for rotating shallow-water waves>, with axis ratio $V/U=f/\omega<1$. For the stipulated positive signs, the velocity at phases $0,\pi/2,\pi,3\pi/2$ points respectively \b[right, up, left, down], with magnitudes $U,V,U,V$. The particle orbits are clockwise as time advances, since the phase decreases.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/ib/paper-4-wave-velocities.png]
{title=Horizontal velocity directions and particle ellipse for a rotating shallow-water wave}
Back to article page
