Universal, and one of five

What decay never forgets ends with a paragraph it did not earn. Having spent the whole essay showing that a decaying turbulence remembers exactly one number — an integral over the largest scales, fixed before the decay begins, deciding every exponent measured afterwards — it turns to the very end of the decay, where the Reynolds number has fallen far enough that the nonlinear term no longer matters, and says that there the exponent is 5/2 and universal, “because a linear equation has no room for an invariant to choose between”. It then records, under its own list of limits, that the 5/2 was quoted rather than computed.

Computing it takes an afternoon and gives a different answer. The linear equation has all the room in the world for an invariant, because the invariant was never in the equation: it was in the initial condition, and a linear equation carries an initial condition better than a nonlinear one does. The final period decays as $t^{-(m+1)/2}$, where $k^m$ is the shape of the energy spectrum at wavenumbers smaller than any eddy — the same $m$, from the same unmeasurable corner of the same spectrum, that decided the turbulent stage. Batchelor’s $k^4$ gives 5/2. Saffman’s $k^2$ gives 3/2.

What the final period is, stated exactly rather than approximately #

The final period is usually introduced as an approximation — the nonlinear term is small — and it is cleaner to define it as the case in which that term is absent. Then there is no error to estimate and nothing to be careful about, and the question of whether a real flow is in it is separated from the question of what happens when it is.

With the nonlinear term gone, every Fourier mode of the velocity field decays on its own, at a rate the viscosity and its own wavenumber set between them:

$$\frac{\partial E(k,t)}{\partial t} = -2\nu k^2 E(k,t), \qquad E(k,t) = E(k,0)\,e^{-2\nu k^2 t}.$$

That is the entire dynamics, and it is worth pausing on how little is left. Modes do not exchange energy, so there is no cascade; the spectrum’s shape is not maintained by anything, so there is no similarity hypothesis to make; and the rate at which a mode dies depends on the mode and not on the flow. The turbulence has become a collection of independently rotting Fourier components, which is what “the equations are linear” means when it is written out.

The energy is the integral of that over all wavenumbers, and the integral is where the exponent comes from:

$$K(t) = \int_0^\infty E(k,0)\,e^{-2\nu k^2 t}\,\mathrm{d}k.$$

The exponential is a window that closes. At time $t$ it has already killed everything above $k \sim (2\nu t)^{-1/2}$ and has barely touched anything below it, so as the decay proceeds the window retreats towards the origin, and what is left under the curve is decided by whatever the spectrum does there. Substituting $s = k\sqrt{2\nu t}$ turns the integral into a constant:

$$K(t) \to C\,(2\nu t)^{-(m+1)/2} \int_0^\infty s^m e^{-s^2}\,\mathrm{d}s = \tfrac12\,C\,\Gamma\!\left(\tfrac{m+1}{2}\right)(2\nu t)^{-(m+1)/2},$$

where $E(k,0) \to Ck^m$ as $k \to 0$. The exponent is $(m+1)/2$ and the prefactor has a gamma function in it — 0.886227 for $m = 2$ and 1.329340 for $m = 4$, both computed here by a Lanczos approximation and checked against the exact half-integer values rather than looked up.

The relation to the turbulent stage’s arithmetic is immediate once both are written in the same variable. That stage conserved $u^2 l^p$ and gave $n = 2p/(p+2)$, and the two conventions line up as $p = m+1$ — Saffman’s $k^2$ with $u^2l^3$, Batchelor’s $k^4$ with $u^2l^5$. So the whole of a decay is

$$n_{\text{turbulent}} = \frac{2p}{p+2}, \qquad n_{\text{final}} = \frac{p}{2},$$

two functions of one number that nobody can measure.

The state being described is worth naming plainly. The cascade has stopped: there is no flux between scales, so the one exact result — which fixes the third moment of the velocity differences at minus four fifths of that flux times the separation — reads nought equals nought, and every skewness in the field decays to zero. A turbulence in its final period is turbulent in vocabulary only.

The amplitude a linear decay is not able to move #

The claim that the invariant survives into the final period is stronger here than it was in the turbulent stage, and it is worth being precise about the difference, because the two are usually run together.

In the turbulent stage the invariance of $u^2 l^p$ is an assumption about the nonlinear term: the argument is that the nonlinear term conserves a particular moment of the two-point correlation provided correlations fall off fast enough with distance, and What decay never forgets records that one of the two candidate moments — Loitsyansky’s — turned out not to be conserved after all, because the long-range pressure correlations it assumed to decay do not.

In the final period it is a theorem about the solution. The factor $e^{-2\nu k^2 t}$ tends to one as $k$ tends to zero, at every time, so the coefficient $C$ of the small-$k$ power is not conserved on average or to leading order: it is untouched. The check is to read the amplitude off the exact solution at a wavenumber far below the surviving peak, at three times six decades apart, for both candidate spectra, and see whether it has moved. It moves by $10^{-12}$, which is the arithmetic of the exponential and not the physics.

This inverts something. Every treatment of decaying turbulence introduces the invariant as the fragile part of the argument, the assumption that has to be made about the nonlinear term and that has already failed once. In the stage where the turbulence has stopped, the assumption becomes an identity — and so the memory this argument has been calling a residue is at its most secure in exactly the regime where there is nothing left to remember it.

Where the two halves of a decay agree, and it is at neither candidate #

Setting $2p/(p+2)$ equal to $p/2$ gives $p = 2$, and nowhere else. Every candidate invariant anybody has proposed has $p$ above two, so a decay always steepens when it stops being turbulent, and the size of the step is a second thing the invariant decides: from 1.200 to 1.500 for Saffman, a step of 0.300; from 1.4286 to 2.500 for Batchelor, a step of 1.0714.

The step is a measurable signature in principle and a treacherous one in practice. A decay whose exponent drifts upward as the record proceeds is a decay approaching its own end — but so is a decay whose self-similarity has not yet been reached, and so is one whose measurement window is walking off the end of the homogeneous region. The decay essay records that published exponents span 1.15 to 1.45, which is the whole distance between the two turbulent-stage candidates; the final-period pair, 1.5 and 2.5, sit outside that band and a full decade apart. A single measurement that reported 2.5 would be worth more than a hundred reporting 1.3, and this is why: it would be a measurement of the same number with ten times the leverage. It is the same argument the inertial range’s own width makes about exponents measured over too little — with the unusual feature that here the leverage is available in principle.

The length forgets what the energy remembers #

The integral scale does something the energy does not, and the contrast is the sharpest statement in the whole calculation of what a linear decay is.

In the turbulent stage the integral scale grows as $t^{2/(p+2)}$ — $t^{2/5}$ for Saffman, $t^{2/7}$ for Batchelor — because the invariant ties the length to the velocity and the velocity is falling. In the final period the surviving part of the spectrum peaks at $k \sim (2\nu t)^{-1/2}$ whatever the shape underneath it, so every length in the problem grows as the square root of the time, and the invariant has no grip on it at all.

Measured off the exact solution over three decades of time, the exponents are 0.499819 and 0.499819 — identical to six figures, which they should be, since neither has anything in it but the diffusion. That is $\sqrt{\nu t}$, the same length as the depth a shaken wall reaches into a fluid, arrived at from the opposite direction: there a clock sets a thickness, here a thickness is all that a clock can set.

So in the final period the energy carries the memory and the length does not. Two boxes of turbulence started with different spectra at the origin end up with the same integral scale at the same time and different energies in it — which is a strange enough state of affairs that it is worth saying what it means physically. The length is being set by how far momentum has diffused since the decay began, and diffusion has no memory. The energy is being set by how much was sitting in the region the diffusion has not yet reached, and that is a number fixed at $t = 0$.

The dissipative anomaly, and the regime defined by not having one #

The single strangest fact in turbulence is that the dissipation does not contain the viscosity. Squeeze the viscosity by six decades and the velocity gradients rise by exactly the factor that keeps the product standing still, which is what the limit that is not the value measures and what the whole cascade picture exists to explain: energy arrives at the small scales at a rate the large ones set, and the viscosity’s only job is to be there when it arrives.

The final period is the case where that has stopped being true, and it is worth putting a number on how completely. In the exact solution the dissipation is

$$\varepsilon = 2\nu\!\int\! k^2 E(k,t)\,\mathrm{d}k = \frac{(m+1)}{2}\frac{K}{t},$$

and rewriting the time in terms of the integral scale, which is $\sqrt{\nu t}$ times a constant, gives

$$\frac{\varepsilon L^2}{\nu K} = \text{a pure number}.$$

It is a pure number in the strongest sense: computed along a decay spanning three decades of time it is 21.205782 for the $k^2$ spectrum and 15.707963 for the $k^4$ one, drifting by $3.7\times10^{-9}$ and $3.4\times10^{-14}$ respectively. Those are $27\pi/4$ and $5\pi$, and the ratio between them is exactly 27/20.

Compare the two regimes at the same state — the same energy, the same integral scale, different viscosity. The turbulent stage dissipates at $AK^{3/2}/L$, with no $\nu$ in it. The final period dissipates at $c\,\nu K/L^2$, which is proportional to it. The ratio of the two is one over a Reynolds number, which is the sharpest available statement of what the crossover is: the final period is not the stage where the turbulence has become weak, it is the stage where the dissipation has started to depend on the fluid again. A decay is turbulent for exactly as long as the anomaly holds, and the kink in the composite curve is the anomaly switching off.

That also explains, without any further argument, why the length stops carrying the memory. In the turbulent stage the length is set by where the invariant says it must be relative to the energy; in the final period it is set by how far momentum has diffused, and diffusion is a property of the fluid. The two halves of the decay are memory-governed and fluid-governed in exactly the same places.

The length that emerges is a familiar one. $\sqrt{\nu t}$ is the depth a shaken wall reaches into a fluid and the thickness a sucked layer settles at, and its arrival here says the same thing those two say: where nothing but diffusion is happening, there is exactly one length and a clock to set it.

Twenty-two decades before the kink #

The obvious question is where the change of exponent happens, and the honest answer is that this calculation does not know. The two power laws are each exact in their own regime and the crossing between them is the one place where the nonlinear term is neither dominant nor negligible, which is precisely the case neither stage solves.

What can be done is to place the crossing by the condition that makes physical sense — the Reynolds number reaching order one — and then to be explicit that the order-one number is a choice. The turbulent stage supplies the Reynolds number’s own decay, $Re \propto t^{(2-p)/(p+2)}$, which is $t^{-1/5}$ for Saffman and $t^{-3/7}$ for Batchelor, and solving that for the time gives the crossing.

From $Re = 5{,}000$, Saffman’s turbulence reaches $Re = 1$ after $3.1\times10^{18}$ eddy times with $6.4\times10^{-23}$ of its energy left, and Batchelor’s after $4.3\times10^{8}$ with $4.7\times10^{-13}$. Moving the criterion from $Re = 1$ to $Re = 5$ moves the first to $10^{15}$ and the second to $10^{7}$ — three and one and a half decades of time, which sounds like a large sensitivity and is the same three and one and a half decades on a plot spanning twenty-two. The placement of the kink is uncertain by much less than the distance to it.

Twenty-two decades of energy is the number worth carrying. The final period of a laboratory-scale turbulence is not late in its decay; it is after the decay has destroyed everything that could be called a signal, and the fact that the exponent there is exactly computable is of no use to anybody trying to measure it in that flow.

Why the exponent that gets measured is the one that can be reached #

Which makes it puzzling that the final period has been measured, repeatedly, since the nineteen-forties, and that what those measurements report is 5/2.

The resolution is in the same figure read the other way. The wait is short only if the decay starts at a low Reynolds number, and it is far shorter for Batchelor’s invariant than for Saffman’s, because Batchelor’s Reynolds number falls more than twice as fast.

From a Reynolds number of ten — which is what a grid tunnel run slowly and measured far downstream gives — Batchelor’s turbulence reaches $Re = 1$ in 215 eddy times and Saffman’s in $10^5$. From thirty, 2,800 against $2.4\times10^7$. The first of each pair is a length of tunnel; the second is not.

So the final period is reachable in a $k^4$ turbulence and unreachable in a $k^2$ one, and the exponent that has been measured is the exponent belonging to the case that could be measured. That is not a criticism of the measurements, which are careful and say what they measured. It is an observation about what “universal” has been doing in the sentence: an exponent observed in every experiment that reached the regime, where reaching the regime selects the class of turbulence, is not evidence that the class does not matter.

The modern position that grid turbulence is Saffman’s makes this sharper rather than softer. If the decay in a grid tunnel is a $k^2$ decay, then either the classical final-period measurements were made on turbulence that was not in the Saffman class — plausible, since they were made behind fine grids at low speeds, in apparatus and at Reynolds numbers quite unlike the ones the Saffman classification was argued from — or the final period was not reached and what was measured was the approach to it. Distinguishing those needs the small-$k$ spectrum of the particular flow, which is exactly the quantity a box of finite size cannot supply.

Three small scales, and one answer #

Everything above rests on the claim that the late exponent is reading the origin of the spectrum and nothing else. That claim deserves a test that could fail, because the opposite reading is available and would be more comfortable: that the exponent is really a property of the small scales, which is where the viscosity acts.

The test is to run the same $k^4$ origin under three different small-scale cuts — a Gaussian, an exponential and a spectrum chopped off at a wavenumber — and compare the exponents. They come out at 2.49998, 2.4970 and 2.5000, a spread of 0.0030. The shape of the spectrum where a probe can see it does not reach the answer, and the shape where no probe can see it decides it entirely.

The rest of the arithmetic is checked the same way. The quadrature over wavenumber agrees with the closed form’s gamma function to $7.5\times10^{-5}$ at worst across five spectra, and the worst case is the one with a known cause: the initial spectrum’s own cut contributes $(m+1)/(8\nu t k_0^2)$ to the energy, which is $7.5\times10^{-5}$ at the time used. The exponents measured by least squares over three decades come out at 1.4999870 and 2.4999783. The small-$k$ amplitude drifts by $10^{-12}$. And the calculation refuses what it has no answer for: a spectrum going as $k^{-2}$ at the origin, which has no finite energy; a crossover on an invariant whose Reynolds number rises rather than falls; a decay run to negative time; and an initial spectrum with a cut nobody wrote.

What an exactly solvable problem still does not settle #

The crossover is not solved and is not claimed to be. Joining two exact power laws at a point is a sketch of a transition, not a calculation of one. Doing it properly means keeping the nonlinear term through the range where it is comparable to the viscous one, which is a genuinely hard problem and has no closed form; what the composite figure shows is the two ends and the distance between them.

The initial spectrum is invented. The calculation is run from $Ck^m$ times a smooth cut, and the cut is shown not to matter, which is a statement about the late answer only. How long it takes for a real spectrum to forget its own shape and reach the asymptotic form is a question about the early part of the linear decay, and nothing here measures it.

Isotropy and homogeneity are assumed throughout, and at the very end of a decay they are the assumptions most likely to have failed: the integral scale has grown as $\sqrt{\nu t}$ without limit, and in any real apparatus it eventually reaches the size of the box, after which the large scales are the walls. In a tunnel the same growth reaches the tunnel’s width. The final period as computed here is a property of an infinite domain, and every experiment in it is a race between the linear decay and the arrival of the boundary.

And $m$ is not measurable. That is the decay essay’s point, inherited whole. The quantity deciding both exponents is the spectrum at wavenumbers below the smallest the domain admits, which is a statement about a hypothetical infinite fluid. What is measurable is the integral it implies, at separations much larger than the integral scale, where the signal is small and the record must be long.

One practical note before the open question. Reaching the final period in a simulation is not the expensive problem that resolving a high Reynolds number is, because the Reynolds number is falling throughout rather than being held: the grid required shrinks as the calculation proceeds. What is expensive is the number of eddy times, and that is a different budget.

Still open: whether the steepening can be caught in the act #

The two exponents are ten decades apart in time and one full unit apart in value, and between them is a transition nobody has computed and — on the numbers above — nobody has watched. That leaves a question with a real calculation in it.

A direct simulation of decaying turbulence in a periodic box is not limited by the length of a tunnel; it is limited by how long it can be run and by the box reaching its own size — a cost the grid nobody can build prices, and one that is mild here because a decaying flow’s Reynolds number is falling rather than held. Starting from a deliberately constructed $k^4$ spectrum at a modest Reynolds number, the crossover is a few hundred eddy times away, which is affordable. The calculation is to run it through, measure the local exponent $-\mathrm{d}\ln K / \mathrm{d}\ln t$ continuously, and ask three things: whether it moves from 1.43 to 2.5 as predicted, how many eddy times the transit takes, and whether the small-$k$ amplitude — which is exact after the transition and assumed before it — is conserved through it. The last is the one worth most, because it is where the assumption and the theorem meet, and it is the only place the difference between them can be seen.

Beside it is the same construction at $k^2$, where the crossover is a hundred thousand eddy times away and the box would reach its own size long first. Whether the approach to the final period is distinguishable from the final period itself over the decade that is affordable is the question that would say whether the classical measurements could have been measuring what they thought they were.