The constant that travels

What decay never forgets builds its whole argument on one line of closure — the dissipation is $A u^3/l$, with $A$ a constant of order one — and then, in its own list of limits, says plainly what that line is worth: “$A$ 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.”

This is the test, and the drift does more than change the exponent. It produces a second family of decay laws which overlaps the first, so that two of the four available answers sit five per cent apart while disagreeing about everything that produced them. The consequence is that the measurement the whole subject has been making for eighty years — fit a power law to a decaying energy, read off the exponent, argue about which invariant it implies — cannot answer the question it is asked, and a different measurement can.

The coefficient is not a model, and it is not a constant either #

Start by writing down what $A$ actually is, because the usual presentation hides it inside a closure and makes it look like an assumption when half of it is an identity.

Define the dissipation coefficient in the obvious way, $C_\varepsilon = \varepsilon\, l/u^3$, with $u$ the root-mean-square of one velocity component and $l$ the integral scale. For homogeneous isotropic turbulence the dissipation is exactly $15\nu u^2/\lambda^2$, where $\lambda$ is the Taylor microscale — that is a definition of $\lambda$ rather than a result. Substituting and tidying gives

$$C_\varepsilon = \frac{15\,(l/\lambda)}{Re_\lambda}, \qquad Re_\lambda = \frac{u\lambda}{\nu}.$$

There is no physics in that. It is two definitions rearranged, and every turbulence that has ever existed satisfies it. Computed along four different decays under two different closures, the two sides agree to $5\times10^{-16}$, which is the arithmetic and not an achievement.

What the identity does is make the choice visible. To fix $C_\varepsilon$ is to fix $l/\lambda$ in proportion to $Re_\lambda$, and to fix $l/\lambda$ is to make $C_\varepsilon$ fall as the reciprocal of $Re_\lambda$. There is no third possibility and there is no neutral position: a decay model must say something about the scale separation, and calling the coefficient constant is one of the things it can say rather than the absence of saying anything.

The classical choice is $C_\varepsilon$ constant, and the cascade argument is what recommends it: the large scales hand energy down at a rate $u^3/l$ set by their own turnover, the small scales take whatever arrives, and nothing in that sentence has a viscosity in it. It is the same reasoning that produces the dissipative anomaly, which is the best-established strange fact in the subject, and it is the step the closure problem cannot take on its own. It follows from it that the scale separation must grow with the Reynolds number, and it does, in every equilibrium flow it has been measured in.

The other choice is what is measured in the near field of a grid — within the first forty or fifty mesh lengths, and out to several hundred behind a fractal grid — where $C_\varepsilon$ is observed to fall as $Re_\lambda$ falls, keeping $l/\lambda$ fixed at a value set by the inlet rather than by the local state. That is an empirical scaling and is treated here as an input whose consequences are computed. Nothing on this page derives it, and the argument does not need it to be true: it needs only that it is available, because an available alternative that changes the answer is enough to make the answer not a measurement of what it was thought to measure.

What each closure does to the decay #

The rest follows from one equation. Energy is lost only to dissipation, and the large scales hold an invariant $K l^p$ — $p = 3$ for Saffman’s $k^2$ spectrum, $p = 5$ for Batchelor’s $k^4$, exactly as What decay never forgets sets it up. That fixes $l$ in terms of $K$, and the closure fixes $\varepsilon$ in terms of both.

With a constant coefficient, $\varepsilon = A K^{3/2}/l$ and the answer is the one that essay gives: $K \propto t^{-2p/(p+2)}$, which is 1.2000 and 1.4286.

With $l/\lambda$ held at $c$, the dissipation is $15c^2\nu K/l^2$ instead, and the same separation of variables gives

$$K \propto t^{-p/2},$$

which is 1.5000 and 2.5000. Integrating the two-equation model by Runge–Kutta, with the invariant recomputed rather than advanced so that it holds along the run to $2\times10^{-15}$, returns 1.1995, 1.4278, 1.4998 and 2.4996 against those four.

The second family’s exponents will look familiar to anyone who has read the essay directly below this one. $p/2$ is exactly what the final period gives, and the coincidence is structural rather than numerical: in the final period the dissipation is $c\,\nu K/l^2$ because the equations have gone linear, and here it is $15c^2\nu K/l^2$ because the scale separation has been frozen. The same functional form integrates the same way whatever put it there.

That is worth stating as a warning rather than as an elegance. A measured decay exponent of 2.5 is consistent with the very beginning of a decay, out of equilibrium behind a grid, and with the very end of one, after the turbulence has stopped — two states separated by ten decades of time and by the entire existence of the cascade. The exponent does not distinguish them; the Reynolds number obviously does, and that is the point: the exponent needs a second measurement beside it before it says anything.

What a frozen scale separation means about the flow #

A constant $l/\lambda$ is easy to write down and worth translating, because it is a strong statement about what the turbulence is doing rather than a tuning choice.

The scale separation is how many steps there are between the eddies that hold the energy and the eddies that lose it. Where the energy goes is the argument that the number of steps grows with the Reynolds number: raise $Re$ and the small scales retreat, the range between them widens, and the rate at which energy is handed down stays put because it is decided at the top. That is why $C_\varepsilon$ is constant in equilibrium and why $l/\lambda$ must grow.

Freezing the ratio says that the range between the two ends is being held by something other than the local state — in the near field of a grid, by the grid. The turbulence has not had time to arrange its own scale separation, so it is still wearing the one the apparatus gave it, and the dissipation is consequently set by the wrong end: $15c^2\nu K/l^2$ has the viscosity in it explicitly, because with the small scales pinned there is nothing to make it cancel.

The Reynolds-number histories say the same thing in one line each. Under the equilibrium closure with Saffman’s invariant, $Re_\lambda$ falls as $t^{-1/10}$ — slowly, because the growing scale separation partly offsets the falling velocity. Under the non-equilibrium closure it falls as $t^{-1/4}$, two and a half times faster, because nothing offsets anything. A flow in the second state is spending its Reynolds number much more quickly than the classical picture allows, which is a second observable consequence and one that compounds with the first: the drift in $l/\lambda$ that discriminates the two hypotheses is a drift against a Reynolds number which is itself moving at different rates in the two cases.

The five per cent that is not a disagreement about a number #

Now put the four exponents on one axis with the published measurements beside them.

Grid-turbulence exponents cluster between 1.15 and 1.45 — that is the decay essay’s summary of eighty years of measurement, and it is a borrowed number rather than one computed here. Batchelor’s invariant with a constant coefficient gives 1.4286. Saffman’s invariant without one gives 1.5000. They are 5.0 per cent apart, they sit at and just above the top of the measured band, and they agree about nothing else: one has a growing scale separation and a conserved Loitsyansky integral, the other a frozen scale separation and a conserved Saffman integral.

An experiment that measures 1.45 has therefore measured something compatible with both, and the practice of reading an exponent as evidence for an invariant requires the closure to be known in advance. It is not known in advance. It is the thing the near-field measurements put in doubt.

Nor does the degeneracy come from the asymptotic idealisation. Fitting each integration over one decade — the window a wind tunnel has — at a hundred initial eddy times returns 1.406 and 1.496 for the colliding pair, which is the same five per cent reached by the route an experiment takes. Early windows read low for all four, because the self-similar state has not been reached, and that is a separate and well-known source of scatter which makes the discrimination worse rather than better.

The plot everybody makes, which decides nothing #

If the exponent cannot separate the two hypotheses, the obvious next question is what can, and the obvious first candidate is the Taylor microscale — because the two closures disagree about it explicitly, one holding $l/\lambda$ fixed and the other letting it grow.

It turns out that the standard microscale measurement is precisely the one that cannot help.

Every grid experiment plots $\lambda^2$ against distance downstream, because the plot is a straight line whose intercept locates the virtual origin, and a virtual origin is needed before any decay can be fitted at all. Under the non-equilibrium closure $\lambda \propto l \propto \sqrt{\nu t}$, so $\lambda^2$ is linear in time. Under the equilibrium closure $\lambda \propto \sqrt{l/u}$, and the invariant makes $l/u$ exactly proportional to $t$ whatever $p$ is, so $\lambda^2$ is linear in time there too. The measured exponents of $\lambda$ across the four cases are 0.49996, 0.49993, 0.49980 and 0.49976, and the largest gap between any two of them is $1.7\times10^{-4}$.

So the one plot made in every study of decaying turbulence is the same straight line under every hypothesis in the family, and the fact that it comes out straight has been read for decades as a confirmation of a picture it does not test. It confirms self-similarity and it confirms that the microscale is diffusive, both of which are true in all four cases. It says nothing about which.

The systematic that is larger than the signal #

There is a worse consequence of the $\lambda^2$ plot being uninformative, and it is not that a useful measurement was missed. It is that the plot is not idle: its output goes into the exponent.

A decay is measured against distance downstream, and the power law is fitted in $(x - x_0)$ rather than in $x$, because the turbulence does not begin at the grid. The virtual origin $x_0$ has to be found before any exponent can be fitted, and it is found from the intercept of the $\lambda^2$ line. So the quantity that cannot distinguish the hypotheses is used to set a parameter that the fitted exponent depends on.

Refitting the same integrations over the same one-decade window with the origin displaced gives the size of it. Batchelor-equilibrium returns 1.4061 with the origin right, 1.5592 with it ten per cent early and 1.2401 with it ten per cent late. Saffman-non-equilibrium returns 1.4958, 1.6585 and 1.3195. A ten per cent error in the origin moves the answer by about 0.16, and the whole gap between the two hypotheses is 0.0714.

Ten per cent of the window’s centre is not a large error in a virtual origin. It is a few mesh lengths in a tunnel, comparable to the scatter with which the $\lambda^2$ intercept is usually determined, and it is systematic rather than random: a single fit inherits a single displacement, so averaging more data does not remove it. The published scatter of 1.15 to 1.45 is entirely consistent with one decay measured by several groups whose origins differ by a few per cent, which is a duller explanation of eighty years of argument than either invariant and cannot be excluded by anything on this page.

What it does establish is an ordering. Before a decay exponent can be read as evidence about a closure — let alone about an invariant — the origin has to be pinned to better than a couple of per cent by something that is not the $\lambda^2$ intercept. Until it is, the largest term in the answer is an artefact of the procedure, and the two quantities the argument is about are both underneath it.

What does separate them, and how slowly it moves #

The quantity that does discriminate is the ratio $l/\lambda$ itself, and it is a harder measurement than any of the above.

Under the non-equilibrium closure the ratio is fixed, exactly, because that is what the closure declares — the flat line is a statement of the input rather than a result. Under the equilibrium closure it moves as $Re_\lambda$, which for Saffman’s invariant falls as $t^{-1/10}$: a factor of 2.47 across the seven decades of the run, and 26 per cent across the one decade an experiment has.

Twenty-six per cent in a ratio of two lengths, each measured from a correlation function, is a demanding measurement but not an impossible one, and it is the measurement worth making. It also has the merit of being a shape rather than a value: no absolute calibration is needed, only whether the ratio holds still or drifts, and the two hypotheses say opposite things about that.

The size of the thing being argued about is worth seeing directly. Along the non-equilibrium run the coefficient rises by a factor of 14.1 — the number that the model below calls $A$, describes as “of order one”, and holds fixed while integrating over seven decades of time. Along the equilibrium run it is flat to $9\times10^{-16}$, which it must be, since it is the input. That is the honest picture of what a closure is: not a small correction that was neglected, but a quantity of unknown behaviour that was assigned a value — the same shape as the mixing length, where one line of assumption produces the whole structure of a wall profile and the agreement establishes less than it appears to.

What a settling chamber is now owed #

The decay essay ends on the one place a decay exponent is used rather than argued about: a wind tunnel’s settling chamber, a duct inlet, an engine intake — all of them exercises in letting turbulence decay for a distance, and all of them sized by an exponent. Between $n = 1.20$ and $n = 1.43$ the distance needed to reduce the intensity tenfold differs by a factor of about 1.5, and that essay’s honest conclusion was that the only reliable practice is to measure the decay behind a grid of a given geometry and use it for grids of that geometry.

The second family of exponents makes that conclusion stronger and narrower at the same time. Stronger, because the available range is now 1.20 to 2.50 rather than 1.20 to 1.43, and the tenfold-reduction distance across that range differs by a factor of 3.5 rather than 1.5. Narrower, because the non-equilibrium closure is a near-field description, and a settling chamber is exactly the near field: the screens are forty mesh lengths upstream of where the flow is wanted, not four hundred. The regime a settling chamber operates in is the regime where the constant is least likely to be one, which is precisely backwards from the way the correlations are usually applied.

It also explains something otherwise puzzling about those correlations, which is that they work. A correlation fitted behind one geometry and used behind the same geometry carries the inlet condition with it, and the inlet condition is exactly what fixes $c$ in the non-equilibrium closure. The practice that was justified as an admission of ignorance turns out to be the right practice for a reason: in a regime where the scale separation is inherited from the apparatus, a measurement made on that apparatus is the only thing that can supply it.

What this does not say #

The non-equilibrium scaling is not derived here and is not derived anywhere. It is an observed behaviour of the near field of grid turbulence, with an inlet Reynolds number in it, and treating it as an input is the whole of its status on this page. An argument that begins by objecting to a closure put in by hand and proceeds by putting in a different one by hand has not settled anything; it has shown that the conclusion was not robust to the choice.

The inlet Reynolds number in the second closure means it cannot be a statement about the limit of small viscosity. Holding $l/\lambda$ fixed while lowering $\nu$ makes the dissipation rise without bound at a fixed state, which is not a physical limit and is a sign that the scaling belongs to a finite-Reynolds-number region with an apparatus in it. Where the near field ends and equilibrium begins is a real question and nothing here addresses it.

No spectra are computed. The model is two ordinary differential equations, and the invariant, the closure and the isotropy relation for $\lambda$ are all put in. What comes out is the consequence of those inputs integrated accurately — which is worth exactly as much as the inputs and no more, a caution the decay essay states for the same model.

And the published band is borrowed. The figure of 1.15 to 1.45 for measured exponents is quoted, and the claim that fractal-grid measurements report values above two is quoted. The arithmetic of this essay is computed; the experimental context it is placed in is not.

It is worth noting which measurements would be unaffected. Anything read off a second-order statistic — a spectrum, a correlation, a structure function of order two — carries no information about the transfer at all, which is what a spectrum cannot hold establishes. The quantity in dispute here is a rate, and a rate is a third-order object.

Still open: whether the two hypotheses can be made to disagree about something easy #

The state of the argument is that the standard measurement is degenerate, the discriminating measurement is a 26-per-cent drift in a ratio of two correlation lengths, and the case where the two hypotheses disagree loudly — Batchelor’s invariant without a constant coefficient, decaying at 2.5 — is one nobody claims to have.

The calculation that would improve this is to find a quantity the two closures separate by a factor rather than by a quarter, and the promising direction is the third moment rather than the second. The exact law for the third-order structure function fixes $-\tfrac45\varepsilon r$ with no adjustable constant in it, so a measurement of that function is a measurement of $\varepsilon$ that does not go through any closure at all. Combining it with an independently measured $u$ and $l$ gives $C_\varepsilon$ directly, once, at each station of a decay — and the two hypotheses then differ by the factor of fourteen above rather than by a slope. Whether the exact law is usable at the Reynolds numbers a decaying grid flow reaches, where the inertial range is barely there, is the question that calculation would have to settle first, and it is the reason nobody has simply done it.

Beside it is the question of which regime a simulation is in. A periodic box started from a constructed spectrum has no grid, no inlet and no near field, so the inlet Reynolds number the non-equilibrium scaling contains has nothing to refer to. Whether such a flow is therefore always in equilibrium, or whether the initial condition plays the part the grid plays, decides whether numerical decay experiments can be used to settle an argument about laboratory ones at all.