= Solution
Since the particle starts at the origin, its <Lagrangian trajectory> satisfies $\boldsymbol X(t)=\int_0^t\boldsymbol v(s)ds$. Differentiating the mean-square displacement gives
$$
\frac{d}{dt}\langle|\boldsymbol X(t)|^2\rangle
=2\left\langle\boldsymbol v(t)\cdot\int_0^t\boldsymbol v(s)ds\right\rangle
=2\int_0^t\langle\boldsymbol v(t)\cdot\boldsymbol v(t-\tau)\rangle d\tau.
$$
Here 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 $R_L(0)=\langle|\boldsymbol u|^2\rangle$. A second integration yields the exact <Taylor turbulent dispersion> formula
$$
\boxed{\langle|\boldsymbol X(t)|^2\rangle=2\int_0^t(t-\tau)R_L(\tau)d\tau.}
$$
For times short compared with <velocity> decorrelation, continuity at zero gives $R_L(\tau)\simeq R_L(0)$ throughout the integral. Thus
$$
\boxed{\langle|\boldsymbol X(t)|^2\rangle\simeq\langle|\boldsymbol u|^2\rangle t^2.}
$$
The 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>. Therefore
$$
\boxed{\langle|\boldsymbol X(t)|^2\rangle\sim2\langle|\boldsymbol u|^2\rangle t_Lt.}
$$
This is diffusive transport: successive long-time displacements lose memory of one another. In an isotropic three-dimensional <diffusion> equation, $\langle X^2\rangle=6D_Tt$, so $D_T=\langle|\boldsymbol u|^2\rangle t_L/3$. 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.
Back to article page
