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A fluid displacement measured from an initial position must vanish at the initial time. Integrating a harmonic velocity therefore introduces phase-dependent constants. Mean-orbit labels absorb those constants and leave purely oscillatory displacements. The instantaneous displacement correction to velocity can differ under the two reference conventions, but its period mean gives the same leading Stokes drift.

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.

The dyed particle satisfies the Lagrangian trajectory equationsFor the proposed approximation, $x=O({\epsilon }^2)$ andAlsoThe initial conditions hold, verifyingthrough order ${\epsilon }^2$.

Over one period, the periodic terms return to their initial values while the secular term changes $x$ by ${\epsilon }^2\pi$. Hence the dyed particle has Stokes driftto this order, despite the zero Eulerian mean velocity.