A pair's memory shapes the cloud it spreads into

The further apart, the faster they part solved Richardson’s equation for the distance between two marked parcels in turbulence. Richardson proposed in 1926 that separations diffuse, with a diffusivity that grows as the four-thirds power of the separation itself, because larger eddies push larger pairs apart; Kolmogorov’s dimensional argument later made the four-thirds law inevitable. The equation has an exact self-similar solution. Its mean-square separation grows as $t^3$ with a constant of $1144/81$ in units of the diffusivity’s prefactor, and its cloud has a definite shape: a density falling as $\exp(-b\,r^{2/3})$, sharply peaked at small separations with a long tail of pairs that separated fast, and a kurtosis — the fourth moment over the square of the second — of 3.76, against a Gaussian cloud’s 5/3.

The essay ended by naming what the diffusion equation assumes and turbulence does not do. A diffusion has no memory: the push a pair receives at each moment is independent of the push the moment before. Real pairs keep their relative velocity for about the turnover time of eddies of their own size, because the eddy that is separating them persists that long — the same eddies whose energy passes down the cascade at the rate the four-fifths law fixes. This essay gives the pairs that memory and asks what it changes — whether the cube law survives, what happens to its constant, and whether the cloud’s shape moves towards or away from what is measured.

A pair with a velocity to remember #

The model follows each pair’s separation vector $\mathbf{R}$ and its relative velocity $\mathbf{V}$ in three dimensions. Each component of the velocity is an Ornstein–Uhlenbeck process: it relaxes towards its mean over a correlation time $\tau$ and is kicked at random, so that its variance is $\sigma^2$. Both depend on the separation. The correlation time is a fixed multiple $\beta$ of the turnover time of eddies of size $r$, $t_r = r^{2/3}$ in units where the dissipation and the prefactor are one; the variance is chosen so that $\sigma^2\tau$ equals Richardson’s diffusivity $K = r^(4/3)$. The product is what a random walk with steps of that size and that duration diffuses at, so as $\beta$ goes to zero the model’s pairs become Richardson’s diffusion, and at finite $\beta$ they have a memory of exactly $\beta$ turnover times at the same diffusivity. The one parameter is the memory.

One piece of the model is subtle enough to have taken two attempts. Because the velocity’s variance grows with separation, a pair that has drifted outwards is kicked harder than one that has drifted in, and without a correction the model’s memoryless limit is not Richardson’s equation but one with an extra inward drift: the overdamped limit of such a velocity carries a flux $-\tau\nabla\sigma^2$ that a diffusion does not have. Relaxing the velocity towards $+\tau\nabla\sigma^2$ instead of zero removes it. The first guess, relaxing towards the gradient of the diffusivity itself, is twice as large, and it was the check against Richardson’s exact constant that refused it.

The constant, recovered and then lost #

The check is the model’s own limit. At a memory of a hundredth of a turnover time the model’s constant $g$ is 12.5, and at half that 13.1; the approach is linear in $\beta$, and extrapolated to no memory the constant is 14.07 against Richardson’s exact 14.12. At the smallest memory the kurtosis has risen to 3.6 and is still climbing towards 3.76. The model contains Richardson’s diffusion as its limit, which is what makes the rest of its answers comparable with his.

The rest are not small. At a memory of a tenth of a turnover time the constant is 5.2; at three-tenths, 2.2; at one turnover time, 0.58.

The first figure shows the fall. At the same diffusivity — the same product of velocity variance and correlation time — a pair with memory spreads more slowly than one without, and by a large factor. The reason is the separation dependence of everything. A memoryless pair that is kicked outwards is immediately kicked by the stronger eddies of its new separation, and keeps accelerating away; a pair with memory carries the velocity appropriate to its old, smaller separation for a while, and the positive feedback that gives Richardson’s diffusion its explosive $t^3$ growth is delayed by that while. The delay is a fixed fraction of each turnover time, so it costs a fixed fraction of the growth, and the $t^3$ law’s constant falls while its exponent does not.

The cube law does not care #

The second figure confirms that the exponent survives. At every memory, once the release separation has been forgotten, the mean-square separation grows as the cube of time. That is inevitable: with a memory that is a fixed multiple of the turnover time, the model has no scale of its own, and the only combination of the dissipation and the time with the dimensions of an area is $\varepsilon t^3$. Kolmogorov’s argument fixes the law; it says nothing about the number in front of it, and the number is what memory changes.

This has a practical edge. Measurements of $g$ from experiments and simulations scatter over a factor of several, from about a half to a few in the usual units, and part of the difficulty is that a clean $t^3$ range is hard to reach at any Reynolds number yet built. The calculation adds a second source of scatter: two flows with the same diffusivity and different memories give different constants, and a memoryless model fitted to a flow with memory reads the lower constant as a smaller diffusivity. What Richardson’s constant measures is not the diffusivity alone but the diffusivity and the memory together.

The shape is where the memory shows #

The third figure is the main result. The kurtosis of the separation — how heavy the cloud’s tails are — falls steadily as memory is added: 3.26 at a hundredth of a turnover time, 2.55 at a tenth, 2.14 at three-tenths, 1.83 at one turnover time and 1.59 at three, below a Gaussian cloud’s 5/3.

Richardson’s tails come from luck compounding. A memoryless pair that happens to separate a little faster than average meets larger eddies sooner, which separate it faster still, and a few such pairs run far ahead of the rest. With memory the luck is spread over a turnover time: a pair’s velocity is an average over a stretch of pushes rather than each push separately, and averaging suppresses the extremes. The more a pair remembers, the more alike all pairs are, and the cloud’s shape moves from Richardson’s peaked, long-tailed form towards a narrow shell.

Richardson’s peak flattens #

The fourth figure shows the whole shape rather than one number. With the smallest memory the density sits close to Richardson’s exact $\exp(-b\,r^{2/3})$: most pairs are still close together, with a long tail reaching several times the root-mean-square separation. At a memory of half a turnover time the centre has emptied and the tail shortened. At three turnover times the pairs are nearly all at about the same separation, a shell rather than a cloud, and the density at the centre has fallen far below Richardson’s.

That is the shape the essay before asked about. Its cloud, memoryless, had the kurtosis of 3.76 that particle-tracking experiments and simulations have never cleanly reproduced; measured separation clouds are usually narrower. The calculation says in which direction memory moves the shape and by how much for each amount of memory, so a measured kurtosis can be read as a measured memory: a cloud of kurtosis about 2.5 is the cloud of pairs that remember for about a tenth of their turnover time, one of about 2 for a third.

How long real pairs remember #

The model leaves the memory as a parameter, and it is fair to ask what value turbulence gives it. Two measured constants of Kolmogorov’s theory bound it. The relative velocity of a pair at separation $r$ has a variance set by the second-order structure function, $C_2(\varepsilon r)^{2/3}$ with $C_2$ about two — the relation with no turbulence in it ties its longitudinal and transverse parts together. A Lagrangian velocity’s correlation time is its variance over $C_0\varepsilon/2$, with the Lagrangian constant $C_0$ measured at about six. Putting the two together for the relative velocity gives a correlation time of about $2C_2/C_0$ turnover times — roughly 0.7.

That is an estimate, not a result: it borrows single-particle reasoning for a relative velocity, and both constants carry uncertainties of tens of per cent. But it puts real pairs well into the middle of the first figure, where the constant is a tenth or less of Richardson’s and the kurtosis near two. If the estimate is anywhere close, the memoryless diffusion is not a small approximation to turbulent pair separation but a qualitative one, right about the exponent and wrong about everything the exponent does not fix.

The same lesson as the single parcel #

Memory was never absent from turbulent dispersion; it was absent from Richardson’s equation. How far a parcel gets followed a single parcel’s wandering with Taylor’s theory of 1921, in which memory is the whole content: a parcel’s mean-square displacement grows ballistically for times shorter than its velocity’s correlation time and diffusively for times longer, and the diffusivity is the velocity variance times that time. Richardson’s pairs are the relative version, and Taylor’s two regimes are the two ends of the first figure: small memory is the diffusive end, large memory the ballistic.

A closure with no memory at all found the same thing about a different object — the eddy viscosity of the Reynolds-averaged equations, which responds to the present strain as though it had no history, and gets the stresses wrong wherever the strain changes on the eddies’ own time scale. A diffusion of separations is an eddy viscosity for pairs, and it has the same defect for the same reason.

What it does to a puff #

The practical form of the question is a puff: a pollutant released in a moment at a point, whose size and peak concentration are set by how its pairs of particles separate. A puff is not a plume, and it is the pair statistics that govern it. Memory changes it twice. The lower constant means the puff grows more slowly, so its concentration falls more slowly with time. The lighter tails mean fewer particles run far ahead, so the puff’s edge is sharper and the concentration at its centre is closer to its mean. A hazard assessment that uses Richardson’s diffusion therefore overestimates how quickly a released puff dilutes and underestimates how long its centre stays concentrated — in the direction that is not the safe one.

Inside the puff the scalar itself is being stretched into filaments and mixed by its own cascade, whose rate depends on how fast neighbouring particles separate at the smallest scales. The pair memory enters there too, at the scales where the filaments are made.

Why Richardson’s shape was hard to see #

The essay before found the memoryless cloud forgetting its release separation only after about seventy turnover times of that separation, and the range of scales over which a $t^3$ law can hold in any laboratory flow is short. Measured clouds have usually been narrower than Richardson’s, and the shortfall has been put down to those finite ranges, to the release separation not yet forgotten, and to the window every measurement averages over. The calculation adds a cause that no increase of Reynolds number removes: at any Reynolds number, a pair that remembers makes a narrower cloud. Some of the shortfall is the measurement; some of it is the physics.

Reading the memory off a cloud #

The table can be run backwards. Once the release separation is forgotten, the kurtosis depends on the memory alone — not on the dissipation rate, not on the time, not on the prefactor of the diffusivity — so a measured cloud’s shape is a reading of $\beta$ that needs none of the quantities the constant depends on. That matters because the constant is the hard number to measure: it multiplies the dissipation rate, which is itself uncertain by tens of per cent in most experiments, and it is cubed in time, so a small error in when the clock started becomes a large one in $g$. The kurtosis is a ratio of moments of one cloud at one instant. A tracking experiment that reports a kurtosis near two is, on this model, reporting a memory near one turnover time; one that reports 3.7 is reporting pairs that forget at once. The two readings, the memory from the shape and the memory from the velocity’s own correlation, are independent, and whether they agree is a test of the model that no fitted constant can pass by accident.

What was checked #

The fifth figure is the ledger. With the memory taken to zero by extrapolation from two small memories, eight thousand pairs each, the model’s constant is 14.07 against Richardson’s exact 1144/81 = 14.12, and the kurtosis is rising towards his 3.76; a second seed gives 13.88. With the velocity’s variance and correlation time held fixed at their release values, the pairs spread exactly as an Ornstein–Uhlenbeck velocity predicts, to 1.2 per cent. Every other number is a seeded ensemble of three thousand pairs, whose sampling error on the kurtosis is a few hundredths.

What the model leaves out #

One memory for every eddy. The correlation time is the same fraction of the turnover time at every separation. Real turbulence need not be so tidy, and a memory that changed with scale would bend the $t^3$ law.

Gaussian velocities. The relative velocity is Gaussian at each separation. Real relative velocities are strongly non-Gaussian — their own tails are heavy — which pushes the cloud the other way, towards heavier tails.

The dissipation range and the large eddies. The model is pure inertial range; below the Kolmogorov length pairs separate exponentially and above the integral scale they separate independently, and a real experiment sees both ends.

Direction. Pairs are more likely to separate than to approach, because turbulence stretches lines on average; the model’s velocity has no such bias beyond what its drift gives.

The convention the numbers depend on #

Units are those in which the dissipation rate and Richardson’s prefactor are one, so the diffusivity is $r^{4/3}$ and the turnover time at separation $r$ is $r^{2/3}$. The memory $\beta$ is the velocity’s correlation time over that turnover time. The kurtosis is $\langle r^4\rangle/\langle r^2\rangle^2$ of the three-dimensional separation, 5/3 for a Gaussian cloud. Pairs are released at a hundredth of a unit and sampled at three time units, when the root-mean-square separation is a few hundred times the release separation.

Who found it, and when #

Richardson proposed the four-thirds law and the diffusion equation in 1926. Taylor’s theory of single-particle dispersion, with memory, is from 1921. Thomson in 1987 gave the condition — the well-mixed criterion — that a stochastic Lagrangian model must satisfy for its particles to stay spread as the fluid is, and Thomson, Borgas and Sawford built two-particle models on it through the 1990s. Measurements of pair separation reaching towards the $t^3$ range came with particle tracking in the 2000s, by Ott and Mann, Ouellette, Bourgoin and colleagues, and the scatter in $g$ and in the cloud’s shape has been argued over since.

Still open: a velocity that is not Gaussian #

The model’s relative velocity is Gaussian, and measured relative velocities are not: at separations inside the inertial range their distribution has tails heavier than a Gaussian’s, the intermittency of the velocity differences themselves. Heavy-tailed velocities make heavy-tailed clouds, and memory makes light-tailed ones. The next calculation gives the model a velocity whose distribution has the measured flatness at each separation, and asks where the two effects balance — what memory, with realistic velocity statistics, reproduces the measured kurtosis of the cloud, and whether that memory agrees with what the velocity’s own correlation time, measured separately, says it should be.