Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 71 / 4

Let $a={\eta }_0$ and $\chi =kx-\omega t$. For a deep-water gravity wave, ${\omega }^2=gk$ and the first-order velocity potential is $(a\omega /k)e^{kz}\sin \chi$. ThereforeRelative to mean parcel labels, integrate at fixed zeroth-order position to obtain the oscillatory fluid displacementThe Stokes drift is the period mean of the displacement correction to the first-order velocity field:Thus $\boxed{u_s={\eta }_0^{2}k\omega e^{2kz}}$, in the direction of propagation. A parcel samples faster forward motion near the top of its orbit and weaker backward motion lower down. This is the leading second-order drift of small-steepness waves, $k{\eta }_0\ll 1$, relative to the Eulerian mean velocity.

Choose the positive-frequency branch with $k>0$, let $K=|\mathbf{k}|$, and write $\mathbf{k}=K(\cos \theta ,\sin \theta )$. The internal gravity wave dispersion relation is $\omega =Nk/K=N\cos \theta$. The phase velocity and group velocity areThey are perpendicular: the phase advances normal to the crests while energy travels along them. For a general signed $k$, the positive frequency is $N|k|/K$ and the horizontal component of group velocity changes sign with $k$.

If $\omega >N$, there is no real wave vector satisfying this relation. For prescribed real horizontal wavenumber, $m^2=k^2(N^2/{\omega }^2-1)<0$, so a bounded boundary response is an evanescent wave, rather than a freely propagating plane wave.

For a single incompressible plane wave, internal-wave polarization gives ${\mathbf{u}}_1\cdot \mathbf{k}=0$. Its oscillatory displacement ${\mathbf{\xi }}_1$ is parallel to ${\mathbf{u}}_1$ and therefore also perpendicular to $\mathbf{k}$. Every spatial derivative of the plane-wave field is proportional to a component of $\mathbf{k}$, soThere is no leading Stokes drift for that single wave. The fluid parcels oscillate along straight lines within phase planes and never sample a spatial change of the wave along their displacement. This conclusion does not prohibit cross terms between different wave vectors.

Assume $\kappa ,\omega >0$ and impose upward group velocity in $z>0$. The two positive-frequency components then have wave vectors ${\mathbf{k}}_1=(\kappa ,m)$ and ${\mathbf{k}}_2=(-\kappa ,m)$ with $m<0$. Equal lengths are automatic, and orthogonality gives $m^2={\kappa }^2$. HenceThe given orthogonal-wave condition is therefore possible only at this forcing frequency. The upward radiation condition excludes the $m=+\kappa$ branch.

Use superposition with vertical amplitude $W$ in each component and internal-wave polarization:The two sets of crests have slopes $+1$ and $-1$; their energy propagates upward and outward at angles of $45^{\circ }$, while their phases travel downward. Their interference makes fixed horizontal bands of vertical-velocity nodes, at $\cos \kappa x=0$, and horizontal-velocity nodes, at $\sin \kappa x=0$.

Put $A=\kappa x$, $B=\kappa z+\omega t$. The first-order displacement relative to mean parcel labels isThe horizontal displacement correction vanishes identically, whileConsequently the formal cross-wave Stokes drift isIt has alternating downward and upward columns, no depth decay in this ideal infinite plane-wave field, and zero horizontal-period-averaged vertical transport. Each wave alone has zero drift; the interference terms create this result.

This calculation is a kinematic second-order consequence of the linear field, not a complete prediction of mean transport. In particular, an impermeable oscillating lower boundary cannot have that nonzero mean normal parcel velocity. Its first-order displacement is ${\zeta }_b=(2W/\omega )\cos (\kappa x)\sin \omega t$. Expanding the exact kinematic boundary condition on $z={\zeta }_b$ givesThis Eulerian mean velocity cancels the computed normal Stokes drift at the material boundary. Away from it, the second-order mean-flow and buoyancy equations must also be satisfied. Thus the alternating drift is a valid formal cross-wave effect, but treating it alone as persistent transport with zero Eulerian mean would be physically incomplete. A prescribed permeable boundary would require its own second-order flux specification instead.