The decay inside the four-fifths law

The constant that travels ended on a proposal. The decay exponent of grid turbulence cannot separate the two closures for its dissipation, because the four answers they produce overlap. What separates them is the dissipation coefficient itself, $C_\varepsilon = \varepsilon l/u^3$: held constant by one closure, falling as one over the Taylor-scale Reynolds number in the other, so that over a decay in which $Re_\lambda$ halves the two differ by a factor of two. And there is a way to measure $\varepsilon$ that goes through no closure at all — the one exact result, $-D_{LLL} = \tfrac45\varepsilon r$, which turns a measured third moment of velocity differences into a dissipation rate with nothing fitted. The essay’s own last line named the catch: whether the exact law is usable at the Reynolds numbers a decaying grid flow reaches.

This essay settles that. The law is exact in a limit, and a decaying flow at a grid’s Reynolds number is not in it; the terms that the limit throws away are computable, and one of them is made by the decay itself. They bias the measurement heavily. They also bias it in a known direction and by a known amount, and the proposal survives, with a correction attached.

The law that is actually exact #

The four-fifths law is not a free-standing result. It is what remains of the Kármán–Howarth equation — the exact evolution equation for the second-order velocity correlation in homogeneous isotropic turbulence — when two of its terms are dropped. Kept, the equation reads

$$-D_{LLL}(r) = \tfrac45\,\varepsilon r \;-\; 6\nu\,\frac{\partial D_{LL}}{\partial r} \;+\; \frac{3}{r^4}\int_0^r s^4\,\frac{\partial D_{LL}}{\partial t}\,ds.$$

The second term is viscous. The one exact result computed it, and showed it takes the law back below a few Kolmogorov lengths. The third term is the decay term, and it is there because the flow is losing energy: the second-order structure function at every separation is falling in time, $\partial D_{LL}/\partial t < 0$, so the term is negative and subtracts from the third moment. In stationary, forced turbulence something else stands in its place; in freely decaying turbulence it is exactly this.

The four-fifths law is the statement that at separations far above the viscous scale and far below the integral scale both terms are negligible. Whether such separations exist at a given Reynolds number is the question, and it is the question a real Reynolds number usually answers badly.

Computing both terms #

The structure function needs a spectrum, and the spectrum here is the one a relation with no turbulence in it used: Pope’s model, with its two shape constants solved at each Reynolds number so that it carries exactly the energy and the dissipation it is given. $D_{LL}(r)$ is then an integral over the spectrum with the isotropic kernel, and its derivative in $r$ is a difference.

Its derivative in time needs the decay. A power-law decay at fixed viscosity, $K \propto t^{-n}$, fixes everything: over a short time $dt$ the energy falls by $\varepsilon\,dt$, the dissipation falls as $t^{-n-1}$, and the Taylor-scale Reynolds number follows from the new velocity and dissipation. The spectrum is refitted at the later instant and the structure function differenced. The model spectrum is assumed to keep its shape as it decays, which is a statement about the model and not a solution of the dynamics — but it is the same assumption every scaling argument for decay makes, and it has two checks.

The first is energy. Far beyond the integral scale $D_{LL}$ tends to twice the variance of one component, so its time derivative must tend to $-\tfrac43\varepsilon$ exactly, since the energy is lost at the rate $\varepsilon$. At thirty integral scales the computed derivative is 1.000001 times that. The second is the time step: the result does not move by more than five parts in ten thousand between steps of a three-hundredth and three ten-thousandths of the decay time.

What the decay term is, physically #

It helps to see why a decaying flow has this term at all, because it is not a correction in the usual sense. It is where part of the energy goes.

The four-fifths law is a statement about a flux. In the inertial range the energy passing each scale on its way down to the viscous scales is $\varepsilon$, and the third moment measures that flux. In a stationary flow every scale holds its energy steady, so the flux through each is the same, and it is the dissipation. In a decaying flow every scale is losing energy of its own as well. The energy passing a separation $r$ on its way down is then the dissipation less the rate at which the scales smaller than $r$ are emptying — because those scales are supplying part of the dissipation below them out of their own store, and it does not have to arrive from above.

So the third moment at a separation measures a flux that is smaller than the dissipation by the rate of loss of everything below that separation, and the decay term is exactly that rate. The larger the separation, the more of the flow’s energy lies below it and the larger the term, until beyond the integral scale the whole store is below and nothing needs to pass at all. What decay never forgets followed the same energy at the other end of the spectrum, where the largest scales keep an invariant while the rest runs down; here it is the small and middle scales running down, and the cascade is carrying less than the dissipation because they are.

That is also why the decay law matters so little. The term depends on how the energy below a separation is distributed and how fast it is lost relative to the dissipation, and the distribution is set by the inertial-range spectrum, and every decay law shares that. Only the overall rate depends on the exponent.

Two terms and a summit #

At a Taylor-scale Reynolds number of 200, typical of a good grid experiment some way downstream, the two neglected terms do exactly what their forms say. The viscous term is almost the whole of the balance at one Kolmogorov length and falls as a power of the separation. The decay term is negligible there and grows with the separation, crossing the viscous term at about 27 Kolmogorov lengths and taking over the whole balance beyond the integral scale, where the third moment has to vanish.

What is left for $-D_{LLL}$ is not a plateau at four-fifths. It is a summit, and a low one: the third moment peaks at 0.747 of $\tfrac45\varepsilon r$, at 32 Kolmogorov lengths, and nowhere comes closer. A measurement that finds the largest value of $-D_{LLL}/r$ and calls it $\tfrac45\varepsilon$ reads a dissipation 25 per cent too small.

The same picture at five Reynolds numbers is the one this essay is about. At 50 the peak is 0.49; at 100, 0.63; at 200, 0.75; at 500, 0.85; at 1000, 0.90. Every curve rises through its viscous range and turns over as the decay term takes hold, and none has a flat top. The separation of the peak moves out with the Reynolds number, from 17 Kolmogorov lengths at 50 to 69 at 1000, which is the viscous and energy-containing ranges drawing apart — but slowly.

Grid turbulence in a wind tunnel reaches Taylor-scale Reynolds numbers of a hundred or two, and active grids a few hundred. At every Reynolds number a decaying laboratory flow reaches, the third moment falls short of the four-fifths law by between a sixth and a half.

How slowly it closes #

The shortfall does close as the Reynolds number rises, and its rate follows from the shapes of the two terms. In the inertial range the viscous term falls as $(r/\eta)^{-4/3}$, because $D_{LL}$ grows as $r^{2/3}$, and the decay term grows as $(r/L)^{2/3}$, because the time derivative of $D_{LL}$ is proportional to $D_{LL}$ and the integral carries four more powers of the separation. The peak sits where the two are balanced, and the shortfall there goes as $(\eta/L)^{4/9}$. Since $L/\eta$ grows as $Re_\lambda^{3/2}$, the shortfall closes as $Re_\lambda^{-2/3}$.

The computed shortfall approaches that exponent from below: $-0.47$ between 50 and 100, $-0.55$ between 100 and 200, $-0.62$ between 500 and 1,000, $-0.646$ between 3,000 and 10,000. At a Taylor-scale Reynolds number of 10,000 — the atmosphere, not a wind tunnel — the peak is 0.978. Two-thirds is a slow power. Halving the shortfall from its value at 200 needs a Reynolds number near 600; getting it under five per cent needs about 3,000.

This is the same shape of result as the limit that is not the value: a law exact at infinite Reynolds number, approached so slowly that every finite measurement is a measurement of the approach.

The decay law hardly matters #

The decay term depends on how fast the structure function falls, which depends on the decay law, and the decay law is the very thing in dispute. That would make the correction circular if it mattered much. It does not. Across the four exponents the two invariants and two closures produce, from 6/5 to 5/2, the peak at $Re_\lambda = 200$ moves from 0.747 to 0.785 — four points against a shortfall of twenty-five. A faster decay, with a larger exponent, has less dissipation relative to its energy, so its structure function falls less for each unit of dissipation and the decay term is a little smaller.

So the correction can be computed from the measured Reynolds number alone, with the decay exponent entering as a detail. That is what makes the next step possible.

The bias moves, and which way #

The use the constant that travels proposed for the law is not to measure $\varepsilon$ once, but to follow $C_\varepsilon$ along a decay and see whether it moves. The bias matters to that only in so far as it changes along the decay — and it does, because the Reynolds number falls as the flow decays and the shortfall grows as it falls.

Over a stretch of decay in which $Re_\lambda$ falls from 200 to 100, the peak falls from 0.747 to 0.630. A dissipation coefficient that is truly constant would therefore appear to fall by 16 per cent, to 0.84 of its starting value. A coefficient falling as one over $Re_\lambda$ truly doubles, and would appear to rise to 1.69.

Both halves of that matter. The bias is in the opposite direction to the non-equilibrium signal, so an uncorrected measurement would understate the drift the non-equilibrium closure predicts, and would make an equilibrium decay look slightly non-equilibrium the other way — a falling coefficient, which neither closure predicts at all. And the bias is a sixth against a factor of two: even uncorrected, the two closures remain distinguishable on this stretch of decay, 0.84 against 1.69. Corrected by the peak computed at each station’s measured Reynolds number, they are 1 against 2.

That answers the question the proposal was left with. The third-moment route to $C_\varepsilon$ works at grid Reynolds numbers, provided the decay term is accounted for, and it discriminates between the closures by a factor where the decay exponent discriminates by five per cent.

The checks #

The checks are three, and each could have failed. The decay loses exactly the dissipation it was given, to a part in a million, measured where the structure function is simply twice the variance. The peak is converged in the time step to five parts in ten thousand. And the shortfall’s local exponent steepens monotonically towards the $-2/3$ that the separation of the two terms predicts, reaching $-0.646$; it would not if the spectrum’s viscous or energy-containing ranges were being fitted wrongly as the Reynolds number changed. The spectrum’s own fit — energy and dissipation carried to the value given — is inherited from the model’s constructor and checked there. A scalar carried by the same flow has its own exact law with four-thirds in place of four-fifths, and a decaying scalar would have a decay term of exactly the same form; the correction computed here would carry over to it with the scalar spectrum in place of the velocity’s.

What a measurement would have to do #

The practical procedure the calculation implies has three steps, and none of them is exotic.

Measure $D_{LLL}$ and $D_{LL}$ at each station of the decay over a range of separations, which a single hot wire and Taylor’s hypothesis already give. Take the peak of $-D_{LLL}/r$ and the Reynolds number at that station. Divide the peak by the computed fraction at that Reynolds number — 0.63 at 100, 0.75 at 200 — to get $\tfrac45\varepsilon$. Better still, since $D_{LL}$ has been measured at two stations, compute the decay term directly from the measured time derivative, via Taylor’s hypothesis turning a downstream difference into a time difference, and add it back. That version does not need the model spectrum at all: it is the Kármán–Howarth equation applied to data.

Two things limit it. The dissipation that comes out is only as good as the time derivative, which is a difference of two measured structure functions and is noisy at small separations. And the dissipation lags the production in a flow that is not in equilibrium, which is the situation near a grid where the non-equilibrium closure applies; a spectrum whose shape is changing is exactly what this model assumes is not happening.

What the model assumes #

The spectrum keeps its shape. Pope’s model spectrum is fitted afresh at each instant, so the calculation assumes the decaying spectrum passes through the same family of shapes. A real decay starting from a grid does not start in that family, and in its early stages the difference is the point of the non-equilibrium argument.

Isotropy and homogeneity. Grid turbulence is close to both far enough downstream, and the Kármán–Howarth equation needs both. Near the grid neither holds.

A power-law decay. The time derivative is taken from $K \propto t^{-n}$. The result depends on $n$ only weakly, which is why that assumption is cheap — fortunately, since even the exponent of the final period, once the Reynolds number has run down, is not the universal number it is quoted as.

Nothing about intermittency. The third moment and the energy balance are fixed by the equation, and intermittency does not enter the four-fifths law or its corrections. It would enter a measurement’s noise.

Who worked it out #

Kármán and Howarth derived their equation in 1938 and Kolmogorov the four-fifths law from it in 1941. That the decay term biases the law in freely decaying turbulence was pointed out repeatedly from the 1990s, and Antonia and Burattini in 2006, and Danaila and colleagues before them, estimated it from measured and modelled spectra and found shortfalls of the size computed here, closing slowly with the Reynolds number. The non-equilibrium dissipation argument this essay tests the measurement for is the work of Vassilicos and colleagues from 2011 onwards. What is added here is the use of the correction for the specific measurement proposed — the drift of $C_\varepsilon$ along a decay — with the bias’s own drift computed against the signal it has to reveal.

Still open: the forced flow’s version #

A decaying flow has a decay term. A stationary forced flow — a jet’s far field, a turbulent boundary layer’s log region, a periodic box with forcing — has, in the same place in the equation, a forcing term or a production term, and its shape is different: forcing concentrated at large scales leaves the inertial range untouched far more completely than a decay does, and production in a shear flow brings anisotropy with it.

The next calculation puts a forcing spectrum into the same balance — a forcing band at the integral scale, with its width a parameter — and asks how much closer the third moment comes to four-fifths at the same Taylor-scale Reynolds number. If a forced flow at 200 reaches 0.9 where a decaying one reaches 0.75, the shortfall in a decaying experiment is not a Reynolds-number effect alone but a decay effect, and the Reynolds number at which a decaying flow “has an inertial range” is higher than the one at which a forced flow does. Whether it is is a matter of computing it.