What decay never forgets

Grid turbulence is the simplest experiment in the subject. Push air through a wire mesh, and downstream of it there is a nearly homogeneous, nearly isotropic turbulence with nothing driving it, decaying as it convects.

Ask how fast it decays and the answer is a power law, $K \propto t^{-n}$, and the value of $n$ has been argued about for eighty years. This essay is about why it is arguable at all, which is more interesting than the argument.

The rate cannot come from the fluid #

Start with what the viscosity does, because the answer is nothing.

Where the energy goes establishes the cascade: energy enters at the large scales, passes down through the inertial range without loss, and is dissipated at the small ones. The rate at which it is dissipated is therefore set at the top, by the large eddies handing it over, and not at the bottom by the viscosity.

The limit that is not the value is the sharp version: the dissipation is $\nu$ times the square of a velocity gradient, the gradient rises by exactly the factor the viscosity falls by, and the product stands still across six decades of $\nu$.

So the decay rate is $\varepsilon = A u^3/l$ with $A$ of order one and no viscosity anywhere. Every question about how fast a turbulence decays is a question about $u$ and $l$.

Three lines of algebra, and the whole answer #

Write $K = u^2$ for the energy and $l$ for the integral scale. Energy is lost only to dissipation:

$$\frac{dK}{dt} = -\varepsilon = -A\frac{u^3}{l}.$$

That is one equation and two unknowns. Something has to relate $u$ and $l$, and the something is a conserved quantity of the largest scales — a number the cascade cannot change, because the cascade moves energy downward and the largest scales have nothing above them.

Suppose that quantity is $u^2 l^p$. Then $l = (C/u^2)^{1/p}$, the equation separates, and the solution is a power law with

$$n = \frac{2p}{p+2}, \qquad K \propto t^{-n}, \qquad l \propto t^{2/(p+2)}.$$

Three lines, and the exponent is determined by one number.

The two candidates, which differ by nineteen per cent #

Which $p$? That is decided by the shape of the energy spectrum at the very smallest wavenumbers — the scales larger than any eddy, where the spectrum is a property of how the turbulence was made rather than of anything the turbulence does.

Saffman’s case has $E(k) \to Lk^2$ as $k \to 0$. The conserved quantity is the integral of the two-point correlation, $u^2l^3$, so $p = 3$ and

$$n = \tfrac65 = 1.200.$$

Batchelor’s case has $E(k) \to Ik^4$. The conserved quantity is Loitsyansky’s integral, $u^2l^5$, so $p = 5$ and

$$n = \tfrac{10}{7} = 1.4286.$$

Nineteen per cent apart in the exponent. Both are computed here from the closed form to $10^{-14}$ and recovered by integrating the model equation numerically at 1.19973 and 1.42819, with the invariant held along each run to better than $10^{-9}$ — which is the check that the integration is solving the equation it was given.

Reading the exponent formula #

The formula $n = 2p/(p+2)$ repays a minute’s attention, because its shape says something about what a conserved large-scale quantity can and cannot do.

At $p = 0$ — which would be conserving the energy itself, with no length in it — the exponent is zero and nothing decays, which is right: if the energy is conserved it does not decay.

As $p \to \infty$ the exponent tends to 2, so no invariant of this form can produce a decay faster than $t^{-2}$. That is a real constraint and it is worth having: a measured exponent above two cannot be explained by any large-scale invariant at all, and a measurement that produced one would be evidence for something else entirely — anisotropy, a boundary, or the final period.

And the exponent is strictly increasing in $p$, so a turbulence with longer-range correlations — smaller $p$, a shallower spectrum at small $k$ — decays more slowly. That is the physically sensible direction: long-range correlations mean the large scales hold more of the energy and the cascade has further to carry it.

What the difference costs #

Nineteen per cent in an exponent does not sound like much, and over the times a turbulence is watched for it is a great deal.

Two boxes, identical energy, identical integral scale, identical spectra over every decade a probe could measure — differing only in the shape of $E(k)$ below the smallest wavenumber anything resolves. After one decade of time they are 1.34 apart in energy. After three they are 3.63.

And the gap grows: it goes as $t^{10/7 - 6/5}$, so it is not a prefactor to be absorbed. Two further decades multiply it by 2.9.

That is the refutation this essay carries. The turbulence has forgotten the mesh, the geometry, the phase of every eddy and the whole of its own history — and it has not forgotten one number, and the number sets the rate at which everything else happens.

Which is a strange thing for a cascade to allow #

It is worth pausing on why the largest scales can hold anything at all.

The cascade is a downward flux: energy leaves a scale for the scales below it, and every scale is fed from above. The largest scales have nothing above them, so nothing feeds them, and what they have is what they started with.

More precisely, the conserved integral is a moment of the two-point velocity correlation taken over all separations, and the nonlinear terms of the Navier–Stokes equations turn out to leave it alone under the right conditions on how fast correlations decay with distance. Which moment is conserved depends on how fast: a turbulence with long-range correlations conserves the lower moment, one with short-range correlations the higher.

That is the whole content of “the shape of the spectrum as $k \to 0$”. It is a statement about how far apart two points can be and still know about each other, and it is fixed by whatever made the turbulence.

The Reynolds number runs itself down #

The second consequence is one a reader does not usually meet, and it changes what “decaying turbulence” means.

The integral scale grows — $l \propto t^{2/(p+2)}$, which is $t^{2/5}$ for Saffman and $t^{2/7}$ for Batchelor — because the invariant ties $u^2$ to a power of $l$ and $u$ is falling.

So the Reynolds number $ul/\nu$ goes as $t^{(2-p)/(p+2)}$: $t^{-1/5}$ for Saffman and $t^{-3/7}$ for Batchelor. It falls, in both cases, but slowly and at very different rates.

The number that is not a number is the essay about how little a critical Reynolds number determines; this is the complementary observation that in a decaying flow the Reynolds number is not even a fixed parameter. A decaying turbulence is spending its own Reynolds number, and the two candidate invariants predict wildly different lifetimes. Starting at $Re = 5{,}000$, Saffman’s turbulence reaches $Re \sim 1$ at $t \sim 3\times10^{18}$ and Batchelor’s at $4\times10^{8}$ — ten orders of magnitude apart, from a difference in a spectrum nobody can measure.

The final period, which is a different law entirely #

Eventually the Reynolds number does reach one, and at that point the nonlinear term stops mattering.

The Navier–Stokes equations then reduce to a diffusion equation for the velocity field, which is linear, and a linear problem has a decay law of its own: $K \propto t^{-5/2}$. That is the final period of decay, and it is a genuinely universal exponent because a linear equation has no room for an invariant to choose between.

The interesting statement is not the exponent but the crossover. Everything above the final period is decided by an invariant nobody can see, and everything below it is not — so the one part of the problem where universality is available is the part in which the turbulence has already stopped being turbulent.

What can actually be measured, and what cannot #

The experimental position is worth stating plainly, because it explains why eighty years have not settled it.

The exponent is measured by fitting a power law to a decaying energy against distance downstream of a grid. The measurable range is short — a wind tunnel is a few metres and the turbulence has to be homogeneous before the fit starts and still measurable when it ends — so the fit spans perhaps one decade of time. Over one decade, $t^{-1.20}$ and $t^{-1.43}$ differ by a factor of 1.7 in energy at the far end, which is measurable; but the prefactor is also free, so what is actually being distinguished is a slope on a log-log plot over one decade, with scatter.

Published exponents cluster between 1.15 and 1.45, which is the whole of the range under argument. Making the measurement sharper needs more decades, which needs a longer tunnel, and the length of tunnel required grows exponentially in the precision wanted.

That is the practical face of a limit that is approached slowly, and it is the same shape as the problem the range a real Reynolds number does not have records for the inertial range: a power law whose exponent everybody agrees about in principle and which no achievable separation of scales measures cleanly.

What the model here is, and is not #

The computation above is not a simulation of turbulence and should not be read as one.

It is a two-equation model: energy loses to dissipation, dissipation is $Au^3/l$, and $l$ follows from a conserved combination. Everything in it was put in — the cascade closure, the invariant, the constant $A$ — and what comes out is the consequence of those assumptions integrated exactly.

What that buys is a clean separation between the parts of the argument. The exponent $2p/(p+2)$ is a theorem about the model; the choice of $p$ is physics the model does not contain; and the numerical agreement to four figures between the integration and the closed form is a check on the arithmetic and nothing more.

Simulating the real thing is out of reach for the reason the grid nobody can build gives — $Re^{9/4}$ grid points — and a decaying simulation additionally has to run for many eddy turnovers with the largest scales resolved, which is the expensive direction.

The invariant that may not be one #

Honesty requires a paragraph about the status of Loitsyansky’s integral, because the argument above treats both candidates as equally respectable and they are not.

Batchelor’s $u^2l^5$ was believed conserved for two decades and then shown not to be: the long-range pressure correlations that were assumed to decay fast enough do not, and the integral drifts slowly. Saffman’s $u^2l^3$ is conserved under weaker assumptions and is on firmer ground.

The modern position is that both regimes exist, that which one a given turbulence is in depends on how it was generated, and that a turbulence generated in a way that produces a $k^2$ spectrum at small $k$ stays in Saffman’s class. Grid turbulence appears to be Saffman’s; some numerical initial conditions are deliberately Batchelor’s.

None of that changes the structure of the argument, which is the point of this essay. It changes which number is right, and it leaves intact the statement that a number fixed at $t = 0$ decides a rate measured arbitrarily far in the future.

Where else a memory survives a cascade #

The pattern is worth recognising because it is not confined to decay.

Where the inverse cascade stops is the two-dimensional version: energy going the other way, arriving at the largest scale the domain has, and piling up there because there is nothing above it to take it away. The largest scale is again the place where a cascade cannot help, and again what happens there is decided by something outside the cascade — in that case a friction, here an initial condition.

And the moment a spectrum cannot hold is the general warning: a spectrum records the second-order statistics and nothing else, so a field with the right spectrum can have no cascade in it at all. Here the reverse point is being made — a corner of the spectrum nobody measures decides a rate everybody measures.

What a designer would want from this, and cannot have #

There is a practical use for decay laws and it is worth saying why the ambiguity matters outside the argument.

Turbulence generated by a grid, a screen or an obstruction upstream of something sensitive decays as it convects — the wake of the obstruction being the source, in the sense what a jet keeps makes precise for a free shear flow — and the question “how far downstream is it quiet enough” is answered by an exponent. A wind tunnel’s contraction, a settling chamber, the inlet of a duct, the intake of an engine — all of them are exercises in letting turbulence decay for a distance.

Between $n = 1.20$ and $n = 1.43$, the distance needed to reduce the intensity by a factor of ten differs by a factor of about 1.5. That is a real difference in the length of a settling chamber, and there is no way to compute which exponent applies to a given grid short of measuring it.

So the honest engineering position is a correlation: measure the decay behind a grid of a given geometry, and use it for grids of that geometry. Counting what matters is the essay about how much dimensional analysis can do before a measurement is needed, and here it does a great deal and stops one step short.

What a spectrum at $k \to 0$ actually is #

A closing note on the quantity that decides everything here, because “the spectrum as $k$ tends to zero” is a phrase that sounds like an idealisation and is not.

$E(k)$ at small $k$ describes correlations between points far apart — further apart than any eddy. In a box of size $L$ the smallest available wavenumber is $2\pi/L$, so the shape below that is not measured and cannot be: it is a statement about a hypothetical infinite domain.

What can be measured is the integral the shape implies, which is a moment of the two-point correlation over all separations, and the two candidate invariants are the two lowest such moments that converge. Whether one or the other converges depends on how fast the correlation falls off with distance, and that is set by how the turbulence was made — by whether the stirring left long-range correlations behind it or not.

So the number that decides the decay exponent is not unmeasurable in principle. It is a moment of a correlation function, and measuring it means measuring correlations at separations much larger than the integral scale, where the signal is small and the required record is long.

Limits recorded rather than smoothed over #

The closure is $\varepsilon = Au^3/l$ with $A$ constant. That is the standard assumption and it is known to be imperfect: $A$ drifts with Reynolds number during a decay, and there is a body of work arguing that the drift changes the exponent. Nothing here tests it.

The invariants are asserted, not derived. Deriving $u^2l^3$ or $u^2l^5$ from the Navier–Stokes equations means examining how the two-point correlation behaves at large separations, which is a different and much harder calculation than anything on this page.

The self-similarity is assumed. Writing $K \propto t^{-n}$ presumes that the decay is self-similar, which is a hypothesis about the spectrum keeping its shape. Real decays approach self-similarity slowly and from a state that is not.

And the final-period exponent of 5/2 is quoted. It follows from a linear diffusion problem with an assumed initial spectrum, and neither the problem nor the spectrum is solved here.

The residue #

The limit is $t \to \infty$, and everything about the turbulence goes with it: the energy, the velocity, the Reynolds number, the memory of the grid.

What survives is a single number — the value of an integral over the largest scales, fixed before the decay began — and that number sets the exponent of every power law measured afterwards, for ever.

It is the most extreme case in the collection of a limit leaving a residue. A limit that erases everything, and a residue that decides the rate of the erasure.