 Solution

Since the particle starts at the origin, its Lagrangian trajectory satisfies $\mathbf{X}(t)={\int }_0^t\mathbf{v}(s)ds$. Differentiating the mean-square displacement givesHere we set $\tau =t-s$ and assume finite second moments so that averaging and integration commute. Under Lagrangian stationarity write the autocorrelation as $R_L(\tau )$. Uniform fluid-tracer sampling in homogeneous incompressible turbulence gives $R_L(0)=⟨|\mathbf{u}{|}^2⟩$. A second integration yields the exact Taylor turbulent dispersion formulaFor times short compared with velocity decorrelation, continuity at zero gives $R_L(\tau )\simeq R_L(0)$ throughout the integral. ThusThe particle retains nearly its initial velocity: this is ballistic transport.

If $R_L$ is integrable, then at long times the displacement derivative approaches $2{\int }_0^{\infty }{R}_L(\tau )d\tau =2R_L(0)t_L$, using the Lagrangian integral time. ThereforeThis is diffusive transport: successive long-time displacements lose memory of one another. In an isotropic three-dimensional diffusion equation, $⟨X^2⟩=6D_{T}t$, so $D_T=⟨|\mathbf{u}{|}^2⟩t_L/3$. The shorthand $t\ll t_L$ and $t\gg t_L$ presumes that $t_L$ is representative of the decorrelation time. With strongly oscillatory correlations, the area-based integral time need not characterize every crossover; the exact integral formula remains the appropriate starting point.