Unravelling Superdiffusion in Turbulent Flows — Epoche C2
Which displacement is superdiffusive, and which is not Whether the mean-squared displacement of a particle carried by turbulence grows faster than linearly in time depends on which displacement is meant, and the earlier version of this essay gave the wrong answer for one of the two cases. It asserted that weakly inertial particles, those whose response time is comparable with the smallest eddy time, exhibit a long-time mean-squared displacement scaling as $t^{\beta}$ with $1 Taylor's theorem, and what it forbids Take one Cartesian component $u$ of the velocity along a Lagrangian trajectory in stationary homogeneous turbulence, with variance $\overline{u^2}$ and normalised autocorrelation $R(s) = \overline{u(t)\,u(t+s)}\,/\,\overline{u^2}$. The displacement is $X(t) = \int_0^t u(s)\,\mathrm{d}s$, so $$\langle X^2(t)\rangle = \int_0^t\!\!\int_0^t \langle u(s)u(s')\rangle\,\mathrm{d}s\,\mathrm{d}s' = 2\,\overline{u^2}\int_0^t (t-s)\,R(s)\,\mathrm{d}s.$$ The second equality is the whole content of G. I. Taylor's 1921 paper: stationarity makes the integrand depend only on $|s-s'|$, and integrating over the band of the square at fixed lag $s$ contributes a length $t-s$, which is where the weight comes from. Two limits follow immediately, and each carries its reason. Short times. For $t \ll$ the correlation time, $R \to 1$ and the integral gives $t^2/2$, so $\langle X^2\rangle \to \overline{u^2}\,t^2$. The exponent is $2$ because the particle has not yet forgotten its initial velocity: displacement is velocity multiplied by time. Long times. If the Lagrangian integral timescale $T_L = \int_0^\infty R(s)\,\mathrm{d}s$ converges, then for $t \gg T_L$ the factor $t-s$ is dominated by $t$ and $\langle X^2\rangle \to 2\,\overline{u^2}\,T_L\,t$, Fickian, with diffusivity $\overline{u^2}T_L$. The exponent is $1$ because the displacement is a sum of many effectively independent increments. The consequence is sharp. The only route to an asymptotically superdiffusive single-particle mean-squared displacement is the divergence of $\int_0^\infty R(s)\,\mathrm{d}s$ — an infinite Lagrangian integral timescale, which is what a heavy-tailed velocity correlation or a Lévy-flight velocity statistic would supply. Stationary turbulence at finite Reynolds number supplies no such thing: $R$ decays on the timescale of the energy-containing eddies, and the integral converges. Taylor's hypotheses — stationarity of the Lagrangian velocity and an integrable autocorrelation — are exactly the ones that hold here, which is why the result is a constraint on the answer rather than one model among several. Does particle inertia open an escape? It does not, and the reason is worth stating because it settles the case the earlier version got wrong. For a small heavy sphere the Maxey and Riley equation of motion for a particle in a non-uniform flow reduces, when the particle density greatly exceeds the fluid's and the particle is smaller than the smallest flow scale, to Stokes drag alone, with $\tau_p$ the viscous response time over which the particle relaxes to the surrounding fluid velocity: $$\dot{\mathbf v} = \frac{\mathbf u(\mathbf x_p(t),t) - \mathbf v}{\tau_p},$$ which is a first-order low-pass filter acting on the fluid velocity sampled along the particle path. A linear filter of that kind cannot manufacture a non-integrable correlation from an integrable one; it lengthens the correlation time — for an exponential input autocorrelation the particle's integral timescale becomes roughly $T_L + \tau_p$ — and reduces the velocity variance, but leaves both finite. Taylor's conclusion therefore carries over to inertial particles with modified constants. Their long-time transport is Fickian. What the earlier version was probably describing is the crossover, and it deserves to be shown explicitly because it is the commonest source of spurious anomalous exponents in the literature. Take the simplest integrable case, $R(s) = e^{-s/T}$. Then $\langle X^2\rangle = 2\overline{u^2}T^2\left(x - 1 + e^{-x}\right)$ with $x = t/T$, and the local logarithmic slope $\beta(t) = \mathrm{d}\ln\langle X^2\rangle/\mathrm{d}\ln t$ is $$\beta = \frac{x\left(1 - e^{-x}\right)}{x - 1 + e^{-x}},$$ which runs monotonically from $2$ to $1$: $t/T$ 0.1 0.5 1 2 5 10 50 $\beta$ 1.97 1.85 1.72 1.52 1.24 1.11 1.02 Any fit over a window that does not span the crossover returns a number between $1$ and $2$, and the number depends on the window rather than on the physics. The operational test is whether the exponent is stable when the fitting range is moved and widened; an exponent that is not is a crossover, not a scaling law. The same table disposes of the earlier version's suggestion that heavy particles might show subdiffusion in intermediate regimes: since $\mathrm{d}^2\langle X^2\rangle/\mathrm{d}t^2 = 2\overline{u^2}R(t)$, the mean-squared displacement is convex wherever $R$ is positive, and $\beta$ cannot fall below $1$. Subdiffusion requires $R$ to change sign, which happens in flows with trapping structures, with a mean rotation, or with confining boundaries — not in homogeneous isotropic turbulence. The Stokes number, made concrete The classification of regimes in the earlier version is sound and worth keeping, but its two parameters were left as definitions. Both can be given numbers, and doing so shows where the interesting regime actually sits. The particle response time follows from balancing the inertia of a rigid sphere of radius $a$ and density $\rho_p$ against Stokes drag in a fluid of density $\rho_f$ and kinematic viscosity $\nu$: setting $m_p\dot{\mathbf v} = 6\pi\rho_f\nu a(\mathbf u - \mathbf v)$ with $m_p = \tfrac43\pi a^3\rho_p$ gives $$\tau_p = \frac{\tfrac43\pi a^3 \rho_p}{6\pi\rho_f\nu a} = \frac{2}{9}\frac{\rho_p}{\rho_f}\frac{a^2}{\nu},$$ the factor $2/9$ being simply $(4/3)/6$. The Kolmogorov time and length, built from $\nu$ and the mean dissipation rate per unit mass $\varepsilon$, are $\tau_\eta = (\nu/\varepsilon)^{1/2}$ and $\