Forcing at the integral scale leaves the cascade alone

The decay inside the four-fifths law found that grid turbulence cannot show Kolmogorov’s exact law cleanly at any Reynolds number a tunnel reaches. The third moment of the velocity differences, which the law fixes at $-\tfrac45\varepsilon r$, rises through the viscous range and turns over before it gets there: at a Taylor-scale Reynolds number of 200 it peaks at three-quarters of four-fifths, and the shortfall closes only as the two-thirds power of the Reynolds number. Part of that shortfall was the decay itself — a term in the exact balance that a decaying flow has and a stationary one does not.

That essay ended on the obvious comparison. Put a stationary flow, forced at its largest scales, into the same balance at the same Reynolds number and see how much of the shortfall goes away. If the answer were “a little”, the shortfall would be what finite Reynolds number does to everything. The answer is “nearly all of it”: a quarter becomes three and a half per cent. And the calculation that gives it gives something more general — a rule for how fast the shortfall closes in any flow, read off the spectrum of whatever is putting energy in or taking it out.

The same balance, with a different source #

The four-fifths law is what is left of the Kármán–Howarth equation when two terms are negligible. In a decaying flow those terms are viscous and decay:

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

A stationary flow has no time derivative, but it has to be driven, and the driving enters the same equation in the same place. The cleanest way to see the two side by side is in wavenumber. The spectrum’s own budget is $\partial E/\partial t = T - 2\nu k^2E + F$: transfer, dissipation and a forcing spectrum $F(k)$ whose total is the dissipation $\varepsilon$. A decaying flow has $F = 0$ and a spectrum that falls in time; a stationary one has $\partial E/\partial t = 0$ and a forcing that holds it up. Either way there is a source term — $-\partial E/\partial t$ for the decay, $F$ for the forcing — and carried through the isotropic kernel that turns a spectrum into a structure function, its share of the balance, as a fraction of $\tfrac45\varepsilon r$, comes out as

$$\int_0^\infty \frac{S(k)}{\varepsilon}\,\bigl[1 - 15\,c(kr)\bigr]\,dk, \qquad c(x) = \frac{3\sin x - 3x\cos x - x^2\sin x}{x^5}.$$

The bracket is the whole of the physics. It is zero at $kr = 0$ and rises to one when $kr$ is large, as $(kr)^2/14$ at first. So the source term at a separation $r$ counts the part of the source at wavenumbers above about $1/r$ — the energy that enters, or is released, at scales smaller than the separation. That energy does not have to be carried down through the separation by the cascade, and the third moment, which measures what the cascade carries, falls short by exactly that much. Where the energy goes put it in words: the four-fifths law is the statement that everything dissipated at the small scales was delivered from above. The source term is the part that was not.

Two flows on one spectrum #

To make the comparison fair, both flows get the same spectrum: Pope’s model, fitted at each Reynolds number to carry exactly the energy and the dissipation it is given — the spectrum a relation with no turbulence in it was built on. The viscous term then comes out identical for the two, and every difference in the result is the source. The forcing is a narrow band in wavenumber, log-normal in shape, centred at the spectrum’s peak and half a unit wide in $\ln k$ — a factor of about 1.6 either side, which is roughly what a simulation that forces only its first two or three wavenumber shells does.

The first figure is the result at a Taylor-scale Reynolds number of 200. The viscous term is the one the one exact result computed, nearly the whole balance at a Kolmogorov length and falling steeply. The forcing term is invisible until the separation approaches the integral scale. Between them the third moment of the forced flow climbs to 0.965 of four-fifths and holds there for a while before the forcing takes it back. The decaying flow, whose curve is drawn faintly behind it, turns over at 0.747 and much sooner.

The peak also moves. The forced flow’s summit sits at about ninety Kolmogorov lengths, three times further out than the decaying flow’s, because the term that ends it does not begin to bite until the separation is a sizeable fraction of the integral scale. At 200, the integral scale is a few hundred Kolmogorov lengths. The forced flow uses almost all of the room between the two ends; the decaying flow uses the bottom third of it.

Where each source puts its energy #

Why the difference is so large is clearest in the sources themselves. The second figure plots each against wavenumber, per unit of $\ln k$ so that area on a logarithmic axis is energy.

The forcing band is where it was put. It peaks at the spectrum’s peak and has fallen by more than four orders of magnitude a decade higher; nothing of it reaches the inertial range.

The decay’s source is not a choice. A decaying flow loses energy at every wavenumber in proportion to what it holds there, roughly, and in the inertial range it holds $E(k) \propto k^{-5/3}$. Per unit of $\ln k$ that is $k^{-2/3}$ — a slow fall, which is still nearly one per cent of the peak value at the wavenumber of the Kolmogorov length. So the decay is a source spread through the whole inertial range, and at any separation within it some of the energy dissipated below comes from scales below — the small eddies running down their own store, which the decaying balance described as the cascade carrying less than the dissipation because the scales beneath it are emptying.

Between the two is shear. In a flow driven by a mean velocity gradient — a boundary layer, a jet, a mixing layer — energy is extracted from the mean flow by eddies of every size the shear can tilt, and Lumley’s argument of 1967 gives the production spectrum in the inertial range as $k^{-7/3}$: steeper than the decay’s by two-thirds of a power, so it reaches into the inertial range less far. The model’s third source has that tail, attached smoothly to a peak at the spectrum’s peak.

The shortfall closes at a rate the source sets #

The third figure repeats the forced balance at five Reynolds numbers. At 50 the peak is 0.863 — higher than the decaying flow reaches at 500. At 100 it is 0.929, at 500 it is 0.987, at 1,000 it is 0.994. The tops of the curves also do what a decaying flow’s never do at these Reynolds numbers: they flatten. At 200 the third moment is within a per cent of its peak over a factor of two in separation, at 500 over a factor of four, at 1,000 over a factor of seven. A decaying flow’s curve is within a per cent of its peak over a factor of two at most, at every one of these Reynolds numbers — it has a summit and nothing that could be called a plateau.

How fast the shortfall closes follows from the two terms that make it, in the same way as for the decay. The viscous term falls through the inertial range as $(r/\eta)^{-4/3}$. A source falling as $k^{-q}$ in the inertial range has a share above $1/r$ that grows as $(r/L)^{q-1}$, as long as $q$ is below three; a source that stops at the integral scale, like the band, reaches $r$ only through the bracket’s small-$kr$ end and grows as $(r/L)^2$. The peak sits where the two are balanced, and with $L/\eta \propto Re_\lambda^{3/2}$ the shortfall there closes as

$$1 - \text{peak} \;\propto\; Re_\lambda^{\,-2(q-1)/(q-1+4/3)}.$$

For the decay, $q = 5/3$, and the exponent is $-2/3$, which is where the decaying balance landed. For shear production, $q = 7/3$, and the exponent is $-1$. For a band, which behaves as $q \geq 3$, it is $-6/5$.

The fourth figure is the test. On logarithmic axes the three shortfalls fall as three nearly straight lines, and their local exponents between 1,000 and 3,000 are $-0.63$, $-0.98$ and $-1.19$, each within a few hundredths of its prediction and still steepening towards it, as the decaying one did. At a Reynolds number of 3,000 the decaying shortfall is 4.7 per cent, the sheared one 0.85 per cent and the forced one 0.16 per cent — a factor of thirty between the extremes, on one spectrum with one viscous term.

That is the finding the essay’s title states. The approach to the four-fifths law is not a property of the Reynolds number. It is a property of the Reynolds number and the source, and of the two the source matters more over the range laboratories and simulations actually occupy.

A band that reaches further behaves more like a decay #

The rule predicts something checkable at a single Reynolds number: anything that moves a forcing’s energy towards the inertial range should cost it some of its advantage. The fifth figure makes the moves at 200. Narrowing the band from half a unit in $\ln k$ to a quarter raises the peak slightly, to 0.970. Widening it to one unit lowers it to 0.940, and to one and a half, where its upper tail reaches well into the inertial range, to 0.896. Moving a narrow band up the spectrum does the same thing more directly: at twice the peak wavenumber the peak is 0.937, at four times 0.889.

Shear production, at 0.890, lands among the widened and displaced bands, well above the decay and well below a clean band. That is where a boundary layer’s log region and a jet’s far field sit, and it matters because those are the flows in which most high-Reynolds-number measurements of the third moment have actually been made. They are neither decaying nor cleanly forced, and the model says their approach to the law is intermediate between the two.

The same logic runs the other way as a check on the calculation. A source whose inertial-range tail falls as $k^{-5/3}$, exactly like the spectrum, is in every respect but its origin a decay, and the balance should not be able to tell them apart. At 200 it gives 0.741, against the decaying flow’s 0.747 — the difference being that the real decay’s loss spectrum is not a pure power law near the spectrum’s peak and in the dissipation range.

What a decaying flow needs to catch up #

The sixth figure turns the comparison into the number an experimentalist would want: for a forced flow at a given Reynolds number, how high a decaying flow’s Reynolds number must be before its third moment peaks as high. For a forced flow at 50 the answer is about 560. At 100 it is about 1,580. At 200 it is about 4,830 — a Taylor-scale Reynolds number that no grid in any wind tunnel has produced. At 500 it is beyond the calculated decays and, extended along the two-thirds slope, about 22,700.

The ratio grows, from eleven at 50 to over forty at 500, because the two shortfalls close at different rates. So “the same Reynolds number” is the wrong basis for comparing a simulation of forced turbulence with a decaying experiment, and it is increasingly wrong as the Reynolds number rises. A direct numerical simulation forced in its lowest shells at a Taylor-scale Reynolds number of a few hundred and a grid experiment at the same nominal value are not two measurements of one thing. On the third moment, the simulation is as far along as a grid flow twenty times more turbulent would be.

This also says something about what “having an inertial range” means. The range a real Reynolds number does not have measured it in the spectrum and found it barely there at laboratory values. Measured by the third moment, the answer depends on the flow: a forced flow at a few hundred has one by any reasonable standard, and a decaying flow at the same value does not.

What it changes for the decay argument #

The decaying balance was worked out to rescue a measurement. The constant that travels proposed following the dissipation coefficient $C_\varepsilon = \varepsilon l/u^3$ down a decay, with $\varepsilon$ taken from the four-fifths law, and the decaying balance found that the decay term biases that measurement by a quarter, drifting by a sixth. Nothing here changes that correction. What it adds is a sharper statement of where the correction comes from.

The shortfall a grid experiment sees at 200 is 25 per cent. On the same spectrum a forced flow’s is 3.5 per cent. So about six-sevenths of the grid flow’s shortfall is the decay and only one-seventh is what a finite Reynolds number would do to any flow. That is the case for computing the decay term from measured data rather than for waiting for a larger tunnel: a tunnel with twice the Reynolds number would take the shortfall from 25 per cent to about 17, and a correction for the decay takes it to the three and a half per cent a forced flow has.

It also clears up a disagreement that has sat in the literature, in which forced simulations reported the four-fifths law holding well and decaying experiments at comparable Reynolds numbers did not. On this calculation both were right, and the difference between them is not a sign of a problem in either. The way the dissipation lags its production in a flow out of equilibrium adds a third behaviour — a source that is changing in time — which neither of these steady pictures contains.

What was checked #

The seventh figure is the ledger. The decay term is computed two ways — in separation, as the decaying balance was, by integrating the time derivative of the structure function, and in wavenumber, by the source formula above applied to the decaying spectrum’s loss — and the two agree to 0.26 per cent. The discrepancy is the physical-space trapezoid, which converges on the spectral answer as the square of its grid spacing: 3.4 per cent with sixty separations, 0.9 with 120, 0.26 with 240. The decaying source removes exactly the dissipation, to a part in ten billion. The forcing term at small separations matches its series, $r^2\!\int k^2F\,dk/(14\varepsilon)$, to about one part in seventy thousand. And the forced shortfall’s local exponent steepens monotonically towards $-6/5$, reaching $-1.19$.

What the model assumes #

One spectrum for every source. The comparison holds the spectrum fixed and changes only the source, which is what makes the result clean and is also its main approximation. A real forced spectrum differs from a decaying one near the forcing band, and simulations usually show a small bump — the bottleneck — at the viscous end. Neither changes the inertial-range scaling that sets the exponents, but both would move the numbers at 50 and 100 by a few points.

Isotropy. The Kármán–Howarth equation assumes it. A shear flow is anisotropic at exactly the scales where its production is largest, and the shear source here is only the isotropic part of the energy input, placed into an isotropic balance. The exponent $-1$ belongs to that idealisation; a measured boundary layer would add anisotropic corrections that fall off as the separation shrinks.

A steady source. The forcing is constant and the decay is a power law. Neither a fan switched on nor a flow emerging from a grid has had time to settle into either.

Nothing about intermittency. The four-fifths law and its corrections are exact and do not involve it, which is why they are the right tool for this comparison. The exponents that stop being thirds belong to the other moments.

Who worked it out #

Kolmogorov derived the four-fifths law in 1941 for a flow in which the energy input sits far above the scales being measured, and Novikov gave the forced version, with a forcing correlation in place of the decay term, in 1964. Lumley’s argument for the $k^{-7/3}$ production spectrum in shear flow is from 1967. The comparison of forced and decaying approaches to the law was made with spectral closures and simulations in the 2000s — by Antonia and Burattini in 2006 among others — and found the forced approach faster, as this calculation does. What is added here is the rule that ties the rate to the source’s slope, checked on three sources at once on one spectrum, and the matching Reynolds numbers that put the difference in an experimentalist’s terms.

Still open: switching the forcing off #

A flow that is forced and then left alone passes from one of these curves to the other. The third moment starts at 0.965 at a Reynolds number of 200 and must end near 0.75 once the decay has taken over — but not at once, because the energy already in the inertial range was delivered from above and takes a cascade time of its own scale to drain. The small separations should keep the forced value longest and the large ones lose it first, so the summit would travel inwards as the flow forgets it was forced.

The calculation that answers it is a time-dependent spectral closure started from the forced spectrum with its forcing removed, carrying the source term above through the transient. How many large-eddy turnover times the forced plateau survives decides whether the start of a decaying simulation — which almost always begins from a forced state — has a clean third moment for a while or loses it at once, and whether the near-grid region of a tunnel, where production is switching off as the flow moves downstream, ought to look forced, decaying or neither.