Lagrangian trajectory

A Lagrangian trajectory follows one marked fluid particle and solves

A fluid element is an infinitesimal material parcel followed along a Lagrangian trajectory.

A material volume is a region containing the same fluid particles at every time: its boundary moves with the fluid velocity field. Its mass is fixed by mass conservation, while its volume can change in compressible motion. Applying the Reynolds transport theorem to its volume gives $d|V(t)|/dt={\int }_{V(t)}\nabla \cdot \mathbf{u}dV$. A material volume element is the infinitesimal version.

A material volume element contains the same fluid particles as it moves and deforms. If its edges transform with the local velocity gradient, its volume changes at a rate equal to the trace of that gradient, giving $D(\delta V)/Dt=(\nabla \cdot \mathbf{u})\delta V$. Mass conservation states $D(\rho \delta V)/Dt=0$, not that its volume is always fixed. These two identities give the material form of the continuity equation, $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$.

Stokes drift is the difference between the mean velocity of fluid particles and the Eulerian mean velocity at fixed positions. It appears at second order because particles sample the oscillatory velocity at displaced positions.

For prescribed one-dimensional velocity $u=u_0\sin (kx-\omega t)$, let $\alpha =k{X}_0$, $\theta =\alpha -\omega t$ and expand the Lagrangian trajectory as $x=X_0+u_{0}a+u_0^{2}b$. The Taylor expansion of $\dot{x}=u(x,t)$ gives $\dot{a}=\sin \theta$ and $\dot{b}=ka\cos \theta$. Initial position $X_0$ requires $a=(\cos \theta -\cos \alpha )/\omega$ andThe secular velocity $k{u}_0^2/(2\omega )$ is Stokes drift, even though the Eulerian time average vanishes. This fixed-time expansion requires $|k{u}_0/\omega |\ll 1$ and is not uniform once the secular displacement changes the phase by order unity.

Two plane internal gravity waves can each have zero Stokes drift but give a nonzero drift through interference. The displacement of one wave need not be perpendicular to the other wave's wave vector. For upward waves with wave vectors $(\kappa ,-\kappa )$ and $(-\kappa ,-\kappa )$ and vertical amplitudes $W$, the cross-wave drift is $(0,-2\kappa W^2\cos (2\kappa x)/\omega )$ with $\omega =N/\sqrt{2}$. It has zero spatially averaged vertical flux. It cannot on its own satisfy an impermeable oscillating boundary: the second-order Eulerian mean velocity supplies the necessary normal correction there.

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.