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A fluid particle follows a Lagrangian trajectory, so its position satisfies $\dot{x}=u(x,t)$ rather than the velocity evaluated permanently at its initial position. Assume $\omega \ne 0$, and writeSet $\alpha =k{X}_0$ and $\theta =\alpha -\omega t$. A Taylor expansion of the velocity about $X_0$ givesThus $\dot{a}=\sin \theta$, andAt the next order,so the required second-order trajectory in a sinusoidal velocity field isBoth corrections vanish at $t=0$. All retained terms have units of length. The secular contribution gives the Stokes driftalthough the Eulerian time average of the given velocity is zero. The particle samples the wave at its displaced position, producing this second-order drift. The expansion is in the dimensionless parameter $|k{u}_0/\omega |\ll 1$, for fixed times; it ceases to be uniform when the accumulated drift changes the wave phase by order unity.

Let the Eulerian time average commute with space and time derivatives and let the statistically steady mean have no time derivative. Define $U=\overline{u}$, $V=\overline{v}$, $\hat{u}=u-U$ and $\hat{v}=v-V$, so $\overline{\hat{u}}=\overline{\hat{v}}=0$. Averaging instantaneous incompressibility givesand subtracting gives ${\hat{u}}_x+{\hat{v}}_y=0$ for the fluctuations. Start from the instantaneous streamwise Navier-Stokes equationUse incompressibility to write $u{u}_x+v{u}_y={\partial }_x(u^2)+{\partial }_y(uv)$. The averaged products are $\overline{u^2}=U^2+\overline{{\hat{u}}^2}$ and $\overline{uv}=UV+\overline{\hat{u}\hat{v}}$. The mean conservative terms reduce back to $U{U}_x+V{U}_y$ because $U_x+V_y=0$. The Reynolds-averaged momentum equation is consequentlyThe signs use the positive covariance convention for the Reynolds stress. In particular, both quadratic averages involve the fluctuations, not products of the mean $U$ with a fluctuation; the converted TeX loses the overbar/hat structure here.

In the radiation zone, both fields are transverse to $\hat{\mathbf{x}}$ and obeyThe outward Poynting vector isso its time average isAn absorbing shell receives the incident momentum rather than reversing it. Hence radiation pressure on a perfectly absorbing surface givesAt $r=R$,and thereforeIf $\vartheta$ is the angle between $\hat{\mathbf{x}}$ and ${\mathbf{p}}_0$, this isIts spherical average is ${\mu }_0{\omega }^4|{\mathbf{p}}_0{|}^2/(48{\pi }^2{R}^2{c}^2)$; the vector force integrates to zero by the inversion symmetry of the electric dipole radiation pattern.

Both $\sin (x-t)$ and $\cos (x-t)$ have zero average over one period, so $⟨u⟩=0$ at every fixed point.

The material derivative gives the particle accelerationDirect differentiation yieldsTherefore the Eulerian time average at fixed $(x,y)$ is

An Eulerian time average of an incompressible constant-density Navier-Stokes equation gives the displayed mean momentum equation. Decompose velocity into its mean and a zero-mean fluctuation. In conservative form, the averaged momentum flux is $\overline{u_i{u}_j}=U_i{U}_j+\overline{u_i^{\prime }{u}_j^{\prime }}$; mean incompressibility then turns its first term into mean advection. The fluctuation covariance is the Reynolds stress divided by density, using the positive covariance convention. Closing this extra unknown requires a turbulence model.

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.