The range a real Reynolds number does not have

Where the energy goes derives the minus five thirds law in three lines and then measures a slope of $-1.658$ over four and a half decades. It is careful to say what it is doing: the slope is measured off a drawing of the dimensional argument, which is a check on the drawing.

This essay asks what a real Reynolds number would have given instead, and the answer is not a small correction.

What the argument is a statement about #

The dimensional argument has two hypotheses and both are about separation.

A band of scales small enough to have forgotten the forcing — so nothing about how the turbulence was made can appear. And large enough not to feel the viscosity — so $\nu$ cannot appear either. In such a band the only quantity available is $\varepsilon$, and dimensions then force

$$E(k) = C_K\,\varepsilon^{2/3}k^{-5/3}.$$

Both hypotheses hold in a limit. The first needs $k \gg 1/L$ and the second $k \ll 1/\eta$, so what is being asserted is that $L/\eta$ is large — and $L/\eta = Re^{3/4}$, so it is large only if the Reynolds number is enormous.

The model spectrum, and why one is needed #

Measuring how wide the band actually is needs a spectrum that is right at both ends, and the standard one is Pope’s:

$$E(k) = C_K \varepsilon^{2/3} k^{-5/3} f_L(kL)\,f_\eta(k\eta),$$

with $f_L$ suppressing the large scales and $f_\eta$ cutting off the small ones, and two constants fixed by requiring that the spectrum integrate to the energy and that $2\nu k^2E$ integrate to the dissipation.

Nothing in it is fitted to this essay’s argument. It is the shape everybody uses, its constants are solved for here — the energy integral comes out at $1.000000$ of $K$ and the dissipation integral at $0.999998$ of $\varepsilon$ — and what is measured on it is what a measurement measures.

The fixed point that determines the two constants is worth a note, because getting it backwards is easy: a larger cut-off constant $c_\eta$ delays the exponential and therefore raises the dissipation integral. Driving it the other way converged perfectly well on a spectrum whose dissipation was a third of $\varepsilon$.

How wide the band is #

Measured on that spectrum, the band within 0.05 of $-5/3$ is

Ordinary laboratory grid turbulence runs at $Re_\lambda$ between 30 and 200. Over that whole range the inertial band is less than a decade wide, and at the bottom of it there is no band at all.

That is not an argument that Kolmogorov was wrong. It is an argument about what a spectrum measured in a wind tunnel is showing, which is a curve that passes through $-5/3$ rather than a range that sits at it.

What those numbers mean for a particular flow #

It is worth converting once, because $Re_\lambda$ is not a number most readers carry.

A grid in a small wind tunnel gives $Re_\lambda$ around 40 — the same flow what decay never forgets is about, measured a few metres downstream. A large research tunnel reaches 200 or 300. A jet at laboratory scale reaches 500. The atmospheric surface layer on a windy day is in the thousands, and the highest values ever measured — in the wake of a ship, or in a cryogenic helium facility — are around ten thousand.

So the table above says that everything up to and including a good laboratory jet has under a decade of inertial range, and that two decades requires the atmosphere.

The corresponding scale separation is the more intuitive number. The grid nobody can build computes it in the direction that matters for a simulation: $L/\eta = Re^{3/4}$ and the number of grid points is $Re^{9/4}$. Here the same ratio is read the other way — $L/\eta$ is 22 at $Re_\lambda = 30$, 1,467 at 500 and 24,000 at 3,200 — and an inertial range needs a comfortable margin at both ends of it, so a separation of a thousand buys about one clean decade in the middle.

The local slope, which is never flat #

Plotting $d\ln E/d\ln k$ makes the point directly. At every Reynolds number the curve rises from the large-scale end, crosses $-5/3$, and falls away into the dissipation range. At none of them does it stop.

What grows with the Reynolds number is the length of the crossing, not the existence of a flat region — the curve becomes locally flatter near its crossing, and the tolerance within which it counts as flat is a choice.

The compensated spectrum is a summit, not a plateau #

The same statement in the form everybody plots: multiply by $k^{5/3}$ and divide by $\varepsilon^{2/3}$, and a true inertial range is a horizontal line at $C_K$.

What a finite Reynolds number gives is a single maximum. At $Re_\lambda = 200$ it reaches 1.4938 and the band over which it is within one per cent of its own peak is 0.30 decades; at $Re_\lambda = 3{,}200$ it reaches 1.5000 and the band is 2.06 decades.

So the Kolmogorov constant is recovered — to four figures at the high end — and the plateau is not. A summit that happens to be at the right height is a different object from a range, and only one of them is what the theory predicts.

What a fit reports, and the sign change in its error #

The practical question is what happens when somebody does the thing everybody does: fit a straight line to a decade of the spectrum and quote the slope.

The band chosen here is the most favourable one available — centred on the geometric mean of the two cut-off wavenumbers, which is where a fitter who knew both would put it. The results are

Read the errors. At $Re_\lambda = 30$ the fit is too steep by 0.155, because the band is mostly dissipation range. At 100 it is too shallow by 0.045, because the band is mostly the large-scale end. Above 200 it settles into a one-sided approach and falls to 0.0005 at 3,200.

The error changes sign. So a laboratory that raises its Reynolds number from 30 to 100 and watches its fitted exponent go from $-1.82$ to $-1.62$ has watched it pass straight through the right answer, and the improvement is two errors cancelling rather than one error shrinking.

That is the refutation this essay carries, and its practical form is a warning: an exponent that agrees with theory is not evidence that the range exists.

Why the approach is logarithmic, which makes it worse #

The band’s width in decades grows as $\log_{10}(L/\eta)$ minus a constant, and $L/\eta = Re^{3/4}$. So

$$\text{decades} \approx \tfrac34\log_{10}Re - \text{constant}.$$

Each further decade of inertial range costs a factor of $10^{4/3}$ in Reynolds number — about twenty-two. Two decades of clean inertial range needs a Reynolds number five hundred times the one that gives one decade.

That is why the argument is not settled by building a bigger tunnel. It is also why atmospheric measurements — $Re_\lambda$ in the thousands, over a boundary layer a kilometre deep — are where the cleanest inertial ranges come from, and why they cost what they cost.

What happens if the band is chosen badly #

Everything above uses the most favourable band available. It is worth saying what a less careful choice costs, because the favourable one requires knowing both cut-offs in advance.

A band chosen too close to the large scales returns a slope shallower than $-5/3$, because the spectrum there is still turning over from its flat small-$k$ behaviour. A band chosen too close to the dissipation range returns a much steeper one, because the exponential cut-off is enormously steep — the local slope reaches $-7$ at $k\eta = 1$ and $-22$ at $k\eta = 4$.

The asymmetry matters. Being a factor of two too far towards the large scales costs a few hundredths of an exponent; being a factor of two too far towards the small ones costs a whole unit. So the failure mode in practice is a fit that is too steep, and the standard defence — trim the high-wavenumber end until the slope stops changing — is a procedure that stops when the two errors happen to balance.

What is missing from this model, and it is not nothing #

Pope’s spectrum is an interpolation between two asymptotes and it does not contain the bottleneck.

Real spectra are measurably shallower than $-5/3$ just above the inertial range, because the dissipation cut-off removes the small scales that would otherwise have drained the ones just above them, so those scales are left with more energy than the cascade would have given them. It shows in every careful measurement and in every high-resolution simulation, as a bump in the compensated spectrum before the fall.

This model has a hint of it — the shallowest local slope in the band is $-1.659$ rather than $-1.667$ — and no more. Nothing in this essay computes a bottleneck, and a reader should not take the summit described above as one: it is the smooth interpolation’s own maximum, and its size is a property of the interpolation.

That gap is recorded rather than papered over because the bottleneck is the one feature of a real spectrum that would make the fitting problem worse than the numbers above suggest.

The same disease elsewhere in the collection #

Recognising the shape makes it easier to spot, and it is everywhere in this subject — including in a second length at the wall, where a logarithmic overlap region is only as wide as the separation between two lengths allows.

The exponents that stop being thirds is the higher-order version of the same measurement problem: structure-function exponents are fitted over a range that barely exists, and the departures being argued about are smaller than the bias computed here.

Where the inverse cascade stops has it worse. Kraichnan’s enstrophy range is $k^{-3}$ times a logarithm to the minus one third, so the local slope is steeper than $-3$ by $1/(3\ln(k/k_f))$ and reaching $-3.01$ needs forty-eight octaves of separation. Nobody will ever measure that range either, and there the reason is not the equipment.

And a limit nothing reaches is the general statement about asymptotic regimes approached too slowly to be reached, from the dimensional-analysis side.

What the spectrum cannot tell anybody anyway #

There is a deeper limitation worth putting beside all this, and it is the subject of a neighbouring essay.

The moment a spectrum cannot hold shows that a field with exactly Kolmogorov amplitudes and independent random phases reproduces every second-order measurement — the spectrum, the correlation, the second-order structure function — and has no cascade in it at all: its third moment averages to nothing over sixty-four realisations.

So even a perfect measurement of a perfect inertial range would not establish that a cascade is happening. The spectrum is a second-order object and the cascade lives in the phases.

Between the two essays the position is: the range is narrower than anybody’s plot suggests, and even if it were not, its slope would not be evidence for the mechanism it is usually quoted as evidence for.

Why this is a limit essay #

The reason this sits beside its neighbours rather than in a methods appendix is the shape of the residue.

The limit is $Re \to \infty$. What it removes is the influence of both cut-offs: the forcing scale goes to infinity relative to the band and the dissipation scale to zero, and the band in between is unbounded and exactly $-5/3$.

At finite Reynolds number the residue is not a term in the spectrum. It is a bias in the measurement — the difference between what a fit returns and what the theory says — and its magnitude is set by the width of the band, which grows as the logarithm. So the residue falls, and it falls logarithmically, which is the slowest way anything falls in this subject and the reason a hundredfold increase in Reynolds number halves it.

That is the same arithmetic the flow with no solution meets in a completely different corner of the subject: an asymptotic expansion whose leading term is a logarithm of the small parameter, and which therefore never becomes accurate at any parameter value anybody can reach.

The rule of thumb #

For a reader who wants one number rather than a table: the inertial range in decades is about three quarters of the base-ten logarithm of the Reynolds number, minus two and a half.

That is the fit to the computed widths and it makes the arithmetic quick. A Reynolds number of $10^4$ gives half a decade; $10^6$ gives two; $10^8$ gives three and a half. The constant subtracted is the margin needed at each end before the spectrum is clean, and it is the expensive part: two and a half decades of scale separation buy nothing at all.

The same subtraction is why the first decade of inertial range is so much harder to get than the second. Everything before it is spent on margins.

And the measured spectrum is not the theory’s spectrum #

There is a second reason a fitted slope is not what it appears to be, and it is about what an instrument returns rather than how wide the band is.

The theory’s $E(k)$ is a three-dimensional spectrum: energy per unit wavenumber magnitude, summed over all directions. What a hot wire delivers is a one-dimensional spectrum, from a single component measured along a single line — and the two are not the same function. The one-dimensional spectrum is an integral of the three-dimensional one over all wavenumbers larger than the one being reported, so every point of it carries contributions from smaller scales. It has the same $-5/3$ exponent in a true inertial range and a different constant, smaller by $18/55$; what it does not have is the same shape near the ends of the band, because the contamination from above is exactly where the dissipation range is.

So the measured curve is smoother than the object the theory describes, and smoother in the direction that hides the very departures this essay is about.

And getting from a time series to a spatial one costs another assumption: that the turbulence is carried past the probe frozen, at the mean speed. That is good when the fluctuations are small beside the mean and degrades in proportion as they are not.

Limits recorded rather than smoothed over #

The model spectrum is a model. Every number here is a property of Pope’s interpolation with its standard constants, and a different interpolation with the same asymptotes would give slightly different widths. What is robust is the order of magnitude and the sign change, both of which follow from the two cut-offs being where they are.

The tolerance is a choice. “Within 0.05 of $-5/3$” is arbitrary; a looser tolerance gives a wider band and a tighter one a narrower. What does not depend on it is that the band grows logarithmically and that it is under a decade for laboratory Reynolds numbers.

The Taylor Reynolds number is used because it is what is quoted. $Re_\lambda$ and the large-scale Reynolds number differ by a square root and a constant, and mixing them is a standard way to be out by a factor of ten.

And nothing here is measured on a flow. This is arithmetic on a fitted curve. Its value is that the arithmetic is exact and the conclusion is about what any measurement of any flow with these two cut-offs would show.

What to do instead of fitting a slope #

There are better measurements available and they are worth naming, because the criticism above is useless without them.

Plot the compensated spectrum and look for a maximum. Its height estimates $C_K$ and its position says where the best-conditioned part of the range is. A maximum that reaches 1.5 is evidence; a fitted slope of $-1.67$ over a band chosen by eye is not.

Plot the local slope. A curve that is flat over a decade is a range; a curve that crosses $-5/3$ at a point is not, and the two are indistinguishable once a straight line has been drawn through them.

And quote the band. A slope without the wavenumbers it was fitted over is not a measurement, because the answer depends on them by more than the effect being reported.

None of the three is difficult and all of them are less flattering than a fitted exponent, which is probably why the fitted exponent survives.

The residue #

The limit is $Re \to \infty$, and in it the inertial range is infinitely wide and the exponent is exactly $-5/3$.

At any finite Reynolds number the range is a few tenths of a decade, the exponent measured over it is biased, and the bias changes sign somewhere in the middle of the range of Reynolds numbers a laboratory can reach. What survives the limit is a systematic error whose size falls as the logarithm of the Reynolds number — which is to say, an error that never quite goes away and that a bigger machine barely helps with.

The theory is right. The measurement is of something else, by an amount that is computable, and this essay computes it.