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

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.