Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 65 / 1 / a

Differentiating the velocity potential gives $u=a\omega e^{kz}\cos (kx-\omega t)$ and $w=a\omega e^{kz}\sin (kx-\omega t)$. Let $A=a{e}^{kz}$ and $\vartheta =kx-\omega t$. To first order in the wave steepness $ka$, evaluate these velocities at the initial parcel position and integrate from time zero:These expressions satisfy both initial conditions and distinguish initial and mean parcel labels in a surface wave. Their actual short-time limits areThe printed expressions omit the integration constants and therefore cannot be the small-time displacement from the prescribed initial point. For example, at $x=0$, they give $Z(0)-z=a{e}^{kz}\ne 0$. This is a genuine inconsistency, rather than a missing step in the calculation.

The intended oscillatory orbit is recovered by using mean parcel coordinates instead. Set $x_c=x+A\sin (kx)$ and $z_c=z-A\cos (kx)$ to this order. ThenThus a deep-water gravity wave produces approximately circular parcel orbits with radius $a{e}^{k{z}_c}$, decaying exponentially with depth. The approximation is one of small amplitude over a wave cycle, not an assertion that these oscillatory coordinates vanish at time zero.

The leading Stokes drift comes from evaluating the oscillatory velocity at the displaced particle position. Taylor expansion givesHere $u_x=-ak\omega e^{kz}\sin \vartheta$ and $u_z=ak\omega e^{kz}\cos \vartheta$. Inserting the initial-position displacements from the preceding solution yieldsIn particular the instantaneous difference is zero at $t=0$, as it must be. The unaveraged constant equality in the PDF is incompatible with its initial labels. Averaging over a period $\mathcal{T}=2\pi /\omega$ removes the oscillatory term:Equivalently, using mean parcel labels and the purely oscillatory displacements gives $a^{2}k\omega e^{2kz}({\sin }^2\vartheta +{\cos }^2\vartheta )$ at this order. This recovers the intended constant result. The period-mean drift is in the wave-propagation direction and decreases as $e^{2kz}$. The corresponding second-order vertical difference is $a^{2}k\omega e^{2kz}\sin (\omega t)$ for initial labels and has zero mean. The resulting horizontal displacement per cycle is $u_s\mathcal{T}$; closed first-order circles do not imply zero second-order transport.