Past exam of the physics of atmospheres and oceans course of the University of Oxford / 2018 / 5 / b / Solution

In the northern hemisphere the coast is on the wave's right when looking in its direction of propagation. In the ocean-$x<0$ convention above it travels northward, with land east of the flow; for the opposite ocean side it travels southward. This follows from combining $g{\eta }_x=f_{0}v$ with a sea-level bulge that decreases offshore, rather than choosing propagation independently of trapping.

For a positive northward bulge, $v=(c/H)\eta$ is northward and strongest at its crest, and there is no cross-shore velocity in the ideal mode. Ahead of the crest the current decreases northwards, so $v_y<0$: horizontal flow convergence increases sea level, ${\eta }_t=-H{v}_y>0$. Behind the crest $v_y>0$: divergence lowers sea level. At the crest itself the alongshore derivative and instantaneous horizontal flow convergence vanish. These signs reverse appropriately for a trough or the opposite coast, while the trapping rule remains the same.

The first panel below sketches a positive bulge, alongshore velocity and the leading convergence/trailing divergence regions. Its cross-shore decay is the pressure-gradient/Coriolis balance, not a current directed into the coast.