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Take a positive initial separation $0<l_0\ll \eta$. In the dissipation range, the velocity field is smooth, so the difference of the two Lagrangian trajectories obeys the linearized separation equation $\dot{\mathbf{l}}\simeq (\nabla \mathbf{u})\mathbf{l}$. Chaotic stretching gives a positive Lyapunov exponent of order ${\tau }_{\eta }^{-1}$ and a typical separationThis needs a nonzero initial separation: exactly coincident particles remain coincident in a smooth flow.

Once $l$ lies in the inertial range, the Kolmogorov 1941 theory gives a relative speed of order $(ϵl{)}^{1/3}$. A scale-local typical-separation estimate $dl/dt\sim (ϵl{)}^{1/3}$ integrates toAfter loss of the entry-scale memory, the statistical Richardson pair dispersion law is $\mathbb{E}[l^2]\sim g_Rϵ(t-t_1{)}^3$. Thus the characteristic separation grows as $(t-t_1{)}^{3/2}$ while $\eta \ll l\ll L$. This is a statistical scaling closure, rather than a deterministic equation for every pair.

For $l\gg L$, the particle velocities become approximately independent. If their Lagrangian velocity autocorrelations have a finite Lagrangian integral time $T_L\sim L/U$, the diffusive large-scale pair dispersion has effective relative diffusivity of order $UL$. After a further velocity-decorrelation transient,This is the long-time mechanism of Taylor turbulent dispersion applied to the difference of two decorrelated trajectories. The root mean square separation eventually loses memory of the $L^2$ term.

Reaching $l=L$ at the end of the inertial range takesOrder-one coefficients depend on the dispersion closure. The initial exponential stage can take a long time if $l_0$ is extremely small, but the inertial-range stage is of order one outer eddy turnover time.

The three-regime turbulent pair-separation model in the sketch matches the regimes continuously for illustration. Its coefficients are schematic; the universal claims here are the scaling powers under the stated stretching, locality and decorrelation assumptions.