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At separations well above the integral scale of turbulence, two fluid velocities can decorrelate spatially. If the relative-velocity autocorrelation $R_{\mathrm{rel}}$ is stationary and integrable, the displacement identity used in Taylor turbulent dispersion gives the displayed long-time linear mean-square growth. With speed $U$ and Lagrangian integral time of order $L/U$, its relative diffusivity is of order $UL$. The characteristic separation grows as $t^{1/2}$, unlike inertial-range Richardson pair dispersion.

For stationary statistics along a Lagrangian trajectory $\mathbf{X}$, the velocity autocorrelation depends on the lag $s$. Integrating velocity twice gives the mean-square displacement $2{\int }_0^t(t-s)R_L(s)ds$. If $R_L$ is integrable, the long-time displacement is diffusive. Its normalized one-sided area is the Lagrangian integral time.

Let ${\sigma }_v^2=⟨v(0{)}^2⟩$, assumed finite. The displacement is $\Delta X(t)={\int }_0^{t}v(s)ds$. By stationarity, its two-time covariance is $⟨v(s)v(r)⟩={\sigma }_v^2\rho (s-r)$, with $\rho$ even for a real scalar process. Integrating over the two triangles of the square gives the Taylor turbulent dispersion identityA sufficient condition for ordinary long-time variance growth is ${\int }_0^{\infty }|\rho (s)|ds<\infty$ with ${\int }_0^{\infty }\rho (s)ds>0$. The dominated convergence theorem then givesMore generally the necessary asymptotic condition for a finite positive variance-based diffusion coefficient is convergence of ${\sigma }_v^2{\int }_0^t(1-s/t)\rho (s)ds$ to that coefficient. Absolute integrability is sufficient, not necessary. A zero integral does not give nondegenerate ordinary diffusion. Moreover a second-moment calculation alone does not establish a normal distribution or an entire diffusive scaling limit; those require additional probabilistic assumptions.

For the specified positive autocorrelation, direct integration gives, when $\alpha \ne 1,2$,Thus the power-law velocity-correlation dispersion has the following regimes. If $\alpha >1$, the Lagrangian integral time is $1/(\alpha -1)$ and ordinary variance growth holds, with $D={\sigma }_v^2/(\alpha -1)$. At $\alpha =2$ the exact expression is $2{\sigma }_v^2[t-\log (1+t)]$. If $\alpha =1$,For $0<\alpha <1$,The last two regimes spread faster than ordinary diffusion and have no finite long-time diffusion coefficient. Their slowly decaying velocity memory is responsible. At short times all these correlations instead give the ballistic law $⟨\Delta X^2⟩\sim {\sigma }_v^2{t}^2$.

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.

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.