Field

Viscosity

The thin layer next to a surface that ideal flow ignores, and which supplies drag, separation and the wake.

Flow past a cylinder at Re 40. A real fluid past a circular cylinder. At low Reynolds number the flow closes up behind the body much as the ideal theory says; as it rises the flow separates and a region of reversed flow appears behind, which is where drag comes from.

Everything happens in a layer you cannot see

Air has so little viscosity that ignoring it works almost everywhere. Almost everywhere leaves out a film next to the surface, perhaps a millimetre thick, and that film decides drag, stall and whether an aircraft flies at all.

Flow past a cylinder at Re 100. A real fluid past a circular cylinder. At low Reynolds number the flow closes up behind the body much as the ideal theory says; as it rises the flow separates and a region of reversed flow appears behind, which is where drag comes from.

When the flow lets go

Every body asks the air behind it to slow down and climb back up to the pressure it started at. Sometimes the air cannot, and the moment it refuses is separation — the source of most drag, the cause of stall, and the reason a golf ball has dimples.

The two theories, side by side

The exact solution and the real flow, for the same body in the same stream. One is beautiful and predicts nothing has drag; the other is approximate and has a wake in it. Where they agree and where they part is the whole map of the subject.

How thick is thin

The boundary layer has no edge. It approaches the free stream and never arrives, so any thickness quoted for it is a convention — and the three conventions in use measure three different things, one of which is not a height at all.

A wing's two drags, and the lift at which they are equal. Friction drag and induced drag plotted against lift coefficient, with their sum above them. Friction is flat, because a surface costs the same whatever the wing is doing; induced drag rises as the square of the lift. The total is least where the two are equal, and that is also the point of best glide.

The two drags a wing pays

A wing pays for having a surface, and it pays for making lift with a finite span. One of those bills falls as it flies faster and the other rises, so there is a speed at which the total is least — and the condition for it turns out to be that the two are equal.

How much uphill a layer can take

A boundary layer running into rising pressure is climbing a hill on the last of its momentum. There is a definite steepness at which it can no longer do it, and the number is not a rule of thumb — it is where a family of solutions stops existing.

Pressure recovery along the upper surface at 6°. Surface speed and the local Falkner–Skan pressure-gradient parameter, plotted along the upper surface from the nose. The speed peaks near the leading edge and then falls, which is the layer climbing back up to the pressure it started at, and the parameter crosses the separation value where that climb becomes too steep.

Where the straight line stops

Ideal flow will report a lift coefficient at forty degrees of incidence without complaint. What ends the lift curve is the boundary layer refusing to follow the surface, and the estimate of when that happens joins two solvers that have nothing else in common.

The cheapest way to stay up

There is a speed at which an aircraft's drag is least, and a different, slower speed at which its power is least. The ratio between them is the fourth root of a third — a number that does not depend on the aircraft, the altitude, or anything else about the flight.

Every wavelength grows, and the shortest grows fastest. Growth rate against wavenumber for a vortex sheet of zero thickness and for a layer of thickness δ. The sheet's curve rises without limit, which is the model announcing that it has no shortest scale in it; the layer's turns over and dies at kδ = 1, the neutral wavenumber computed from Rayleigh's equation rather than fitted.

The gradient that does both

One line of the boundary-layer equations at the wall says the profile's curvature there equals the pressure gradient. That single sign causes separation and causes instability, and it causes the instability a long way before it causes the separation.

Two drag laws, one derived and one fitted. Flat-plate drag coefficient against Reynolds number, laminar and turbulent, on log axes. The laminar curve is Blasius' similarity solution, solved by shooting; the turbulent one is the 1/7-power correlation, which was fitted to experiment and is drawn dashed-in-kind to keep the difference visible. Their slopes differ — −1/2 against −1/5 — so the gap between them widens rather than staying put.

The cost of going turbulent

A turbulent boundary layer costs several times the friction of a laminar one, and the multiple is not a constant — it rises with Reynolds number, because the two laws have different exponents. That is why laminar flow is worth more on a long fast surface than on a short slow one.

The wall that shakes

Slide a wall back and forth in its own plane and the fluid above it does not follow — a wave travels upwards into the fluid and dies within one wavelength. The depth it reaches is √(2ν/ω), it contains no length from the geometry at all, and the whole thing is one of the very few exact solutions the Navier–Stokes equations have.

$A boundary layer with no x in it. The velocity profile over a porous wall with uniform suction: U(1 − e^{−Vy/ν}), exactly, at every station along the wall. The displacement thickness is ν/V, the momentum thickness is half of it, and the shape factor is two — all of them constants, none of them a function of distance. It is the cleanest demonstration there is that a boundary layer's thickness is a balance rather than an accumulation.$

The layer that stops growing

Blasius' boundary layer thickens as the square root of distance and never stops. Suck fluid through the wall at a uniform rate and it stops immediately — the profile becomes a single exponential with no x anywhere in it, and the friction comes out exactly equal to the momentum of the fluid that was taken away.

20.45 MPa out of a film 25 µm thick. The pressure along a tapered pad, from the closed-form solution of Reynolds' equation, with the same equation's tridiagonal grid solve drawn over it as points. The peak is 20.45 MPa — enough to yield mild steel — and it sits at 69 per cent of the way along rather than in the middle, because the pressure gradient vanishes where the film equals the harmonic mean of its two ends and the harmonic mean is biased towards the thinner one. The pad's own shape is drawn along the top, to a vertical scale of its own. Nothing pumps this oil: the runner drags it into a narrowing gap and the gap does the rest.

Nothing but the shape of the gap

A machine that holds a steel shaft off its bearing with a film of oil twenty-five microns thick has no pump in it, and the pressure it generates would yield mild steel. The mechanism is not the oil and not the speed; it is that the gap narrows.

45° of corner, and an infinite number of eddies in it. The creeping flow in a corner of 45 degrees, drawn from Moffatt's similarity solution. Each eddy turns the opposite way to its neighbours and is 3.17 times smaller and 1.6e+3 times weaker than the one outside it. The contour levels are rescaled inside each eddy, because they differ in strength by three orders of magnitude per step and a single set of levels would show the first and nothing else — which is itself the reason nobody has seen the third. The dividing lines between eddies are drawn where the stream function changes sign, and the dots are the centres, both found from the solved field.

The eddies nobody stirs

A slow flow past the mouth of a sharp corner does not simply fail to get in. The corner fills with an infinite sequence of counter-rotating eddies, each about three times smaller and sixteen hundred times weaker than the last, and the whole structure is decided by one complex number.

$The jet has to be fed from the sides. The velocity field of the plane jet with streamlines integrated through it. The seven central streamlines run down the jet and spread; the ten started at the top and bottom edges bend inwards and join it, which is entrainment and is a consequence of the solution rather than an addition to it. The dashed lines are the half-speed edges, widening as x^{2/3}. The transverse velocity far from the axis is 5.70e-3 m/s at this station, inward on both sides — a jet is a sink as seen from a distance, which is why two parallel jets pull together.$

What a jet keeps, and what it collects

A jet leaving a nozzle into still fluid has no boundary anywhere and one conserved quantity. Its momentum flux is exactly the same at every station downstream; its mass flux is not conserved at all and grows without limit, because a jet is a machine for acquiring fluid it did not start with.

The core spreads and the outside never notices. The swirl velocity at four times a factor of four apart, with the free vortex Γ/2πr drawn behind them. Every curve leaves the free-vortex line at its own core radius and turns over into solid-body rotation inside it; outside the core all four are the same curve, to the precision of the plot. Viscosity has rounded off the singularity and changed nothing else. The peak swirl falls from 52.4 to 6.6 m/s across the four, and the circulation is identical for all of them.

What viscosity cannot take away

Leave a vortex alone in a viscous fluid and every local measure of it falls — the peak spin, the peak velocity, the enstrophy. The circulation round a large loop does not move at all, ever, and the far field is identical to the line vortex it started as.

The current at the surface is 45° from the wind, and nothing sets that angle. The Ekman spiral drawn as a hodograph: each point is the velocity at one depth, and depth runs along the curve. At the surface the flow is at exactly 45 degrees to the wind that drives it — not approximately, exactly, and independently of the wind, the viscosity and the latitude. By one Ekman depth the flow has turned another radian and lost 1/e of its speed; by three it is a hundredth of the surface value and pointing back the way it came. The angle is a property of the equation having two terms in it, and nothing else.

The layer that stops at a depth

Every other boundary layer grows. This one does not — rotation supplies a frequency, the balance against diffusion supplies a length, and the transport that comes out contains the stress on the surface and not the viscosity underneath it.

The wall's condition, on its way to the middle. Five profiles across the half-channel, from just inside the entrance to fully developed, each drawn at the station where it occurs. The march starts from a slab of uniform flow and never assumes a shape: what arrives at the far end is a parabola, with a centre-line speed of 1.4979 times the mean against the exact 3/2 and a momentum flux of 1.1995 against 6/5. Notice what the middle does while the edges are being slowed: it speeds up, because the flow rate is held, and that acceleration is what the entrance's extra pressure drop pays for.

How far before a duct forgets what was fed into it

A pipe is always drawn with its answer already in place. Getting there takes a distance proportional to the Reynolds number, which means a more viscous fluid is done sooner — and the entrance costs a fixed number of dynamic pressures however long the pipe is.

An oscillation with no mean, and the steady flow it drives. The steady second-order velocity through a Stokes layer, in units of U U′/ω. The first-order flow averages to zero at every height; the average of its own nonlinear term does not, and the pale curve is that forcing. Integrating it twice across the layer, with no slip at the wall and no stress at the top, gives a steady velocity that rises through the layer and settles at -0.749998 — Rayleigh's −3/4, which was not put in anywhere. Beyond about five layer thicknesses nothing more happens, which is why the number is a boundary condition for the flow outside.

An oscillation with somewhere to go

Shake a fluid back and forth over a body and it develops a steady circulation that never reverses. The driving flow has no mean at all; the mean of its own nonlinear term does, and integrating that twice across the oscillatory layer gives a slip velocity of exactly three-quarters of U dU/dx over the frequency.

The price of a gradient

Viscosity does not charge for motion. It charges for the rate at which a parcel is being deformed, and a fluid in solid-body rotation at any speed whatever destroys nothing at all. What is charged for is a sum of squares, which is why the bill can be computed twice.

$How far downstream the heat is still being made. The dissipation accumulated from a station ahead of a cylinder to a station behind it, as a fraction of the whole of what is made inside the frame, at five Reynolds numbers. At Reynolds number 1 the fluid has finished paying by about a diameter behind the body. At 100 it has not finished at five, and the curve is still climbing at the edge of the picture — the drag is a force on the body, and the heat it stands for is somewhere else.$

Where the heat of a drag is made

The power it takes to tow a body through a fluid becomes heat, all of it, eventually. None of the interesting words in that sentence are the first four. It is the "eventually" that decides how a wake behaves, how far a disturbance reaches, and why no box drawn round a body contains its own bill.

The cheapest shape the walls allow

The parabola in a pipe is usually presented as what the equations give. It is better understood the other way round — of every profile that could carry that flow between those walls, it is the one that destroys the least energy, and the equations give it for that reason.

A force without the flow that makes it

A sphere carried along by a flow does not travel at the speed of the fluid at its centre. It travels at the average of the flow over its own surface — and getting that result needs no solution of the flow around the sphere at all, only the answer to a completely different problem that everybody already knows.

$A film with nothing useful to show for itself. The pressure under two discs squeezed together, as a fraction of the pressure at the centre. It is a parabola, and it holds an enormous load — but nothing is sliding, so there is no output at all and every joule put in becomes heat in the oil. The two routes to that heat share no arithmetic: one integrates the dissipation function over the film, the other multiplies the force by the approach speed.$

The last of the oil

A wet plate on a table is held down by nothing but the difficulty of getting air in underneath it. The force required to separate two flat surfaces with a film between them goes as the inverse cube of the gap, which means there is no finite energy that removes the last of the film and no finite time in which it drains away.

How hot seven films get. The peak temperature rise inside the film itself, on a logarithmic axis, for seven lubricated contacts with ordinary engineering numbers. A journal bearing runs tens of kelvin above its own housing, which is why lubrication systems are designed around heat rather than around load; a water-lubricated bush runs at a hundredth of a kelvin, because water conducts six hundred times better per unit of viscosity. The air bearing is marked: its gap is smaller than air's mean free path, so its film is not a continuum at all.

The film that heats itself

The oil in a bearing is not at the temperature of the metal around it. It is tens of kelvin hotter, the rise contains no length whatever, and whether the heating runs away or settles down depends on something that is not a property of the oil at all — whether the machine driving it holds the speed or holds the force.

Twice as fast is not twice as thick. The film a plate carries out of water, against the speed it is withdrawn at, both logarithmic. The slope is exactly two-thirds, so doubling the speed thickens the film by 58.7 per cent and never by more. The open marks are outside the range the derivation holds in — above a capillary number of about a hundredth the film is no longer thin against the capillary length, and the measured thickness leaves this line.

What a plate takes with it

Pull a plate out of a bath and it comes out wet. How wet is not set by the plate, the bath or how much liquid there is, but by a competition in a region a fraction of a millimetre long that nobody looking at the plate can see — and the film goes as the two-thirds power of the speed, never as the first.

Seven fluids in one pipe at one pressure gradient. Velocity profiles of power-law fluids in a round pipe, all at the same pressure gradient and the same consistency, scaled to the fastest. A shear-thinning fluid (n below one) is flatter in the middle and steeper at the wall; a shear-thickening one is the reverse. The flattening is often called plug-like, which invites the reading that the fluid is moving more freely — what has actually happened is that all the shear has been pushed into a thin annulus at the wall, which is the expensive place to put it.

A viscosity that depends on the question

Blood, paint, molten polymer and drilling mud have no viscosity. They have a relation between stress and strain rate that is not proportional, so the ratio of the two depends on how hard they are being sheared — and an instrument that reports one number is reporting a property of itself.

$Where Einstein's line stops being the measurement. The viscosity of a suspension of rigid spheres relative to the liquid's, against the volume fraction. Einstein's 1 + 5φ/2 is exact for one sphere and holds while the spheres cannot feel one another, which is up to about five per cent by volume. Batchelor and Green's two-sphere term takes it a little further; beyond about a fifth nothing derived works and the curve drawn is a fit, which diverges at a maximum packing that is itself a measurement.$

A viscosity made of particles

Stir rigid spheres into a liquid and the mixture is thicker, by five halves of the volume fraction. Einstein's coefficient is not an empirical constant — it is a dissipation calculation on one sphere — and doing it as an energy rather than as a stress shows that four-fifths of it comes from somewhere nobody mentions.

$Two-thirds to one, and nothing outside it. The drag on a spherical drop as a fraction of Stokes' drag on a rigid sphere of the same size, against the ratio of the viscosities. It runs from exactly two-thirds for a clean gas bubble to exactly one for a rigid particle and takes no value outside that range, however light or heavy the drop is — the formula has no density in it anywhere. A drop of water in air and a drop of mercury in water sit at opposite ends of it.$

The surface that moves with the flow

A clean gas bubble feels two-thirds of the drag a rigid sphere of the same size would, and the formula has no density in it anywhere. What buys the third is that the bubble's surface is free to move — and real bubbles in ordinary water do not get it, for a reason that is a millionth of a per cent of the water by mass.

A fifth of the drag power is not heat yet. The energy account of towing a flat plate a metre long through air at thirty metres a second. The power it takes is the drag times the speed. The heat made inside the boundary layer is half the free-stream energy times the energy thickness, and it is less — 78.6 per cent of what was paid. The rest has not been destroyed: it is kinetic energy still in the wake, which will become heat somewhere downstream, in fluid that is no longer touching the plate.

The third thickness

A boundary layer has no edge, so every thickness quoted for it is an integral of the profile against some weight. Two of them are famous. The third answers a question the other two cannot — how much of the power spent towing a plate has actually become heat by the time the fluid leaves it — and the answer is 78.6 per cent.

A vortex has no energy. The kinetic energy of a Lamb–Oseen vortex inside a circle, per metre of its length, against the logarithm of that circle's radius. It is a straight line and it does not stop: outside the core the swirl is Γ/2πr, the energy density falls as 1/r², and the area grows as r², so every decade of radius adds the same amount. There is no such thing as the energy of a line vortex without a stated cutoff, and no cutoff is physical.

The energy a vortex cannot have

A line vortex has infinite kinetic energy. Not a large amount — infinite, growing without limit as the logarithm of however far out the counting stops. And it is losing that energy at a rate that is finite, exactly known, and contains no cutoff at all.

$A swimmer that dissipates the same everywhere. Taylor's waving sheet, with the wave drawn along the bottom and the dissipation drawn against height above it. The dissipation function works out at 4μb²c²k⁶y²e^{−2ky} — with no x in it at all, so the sheet is destroying energy at the same rate under every part of the wave and at every instant of the cycle. It peaks one radian of wavelength above the sheet and is gone within about three.$

A swimmer that cannot go backwards

Taylor's waving sheet is the simplest self-propelled object in a viscous fluid, and its arithmetic contains a result that reads like a mistake — the work it does to travel a metre does not depend on how big its waves are. Doubling the amplitude quadruples both the speed and the power, and changes the bill for the journey by nothing at all.

The coefficient Stokes threw away is not a correction. What absorbs sound in dry air, split into the three transport effects that add to make the diffusivity of sound. Shear viscosity is the largest single contribution and it is not most of it; the bulk viscosity — which Stokes set to zero in 1845, saying in the same paper that he had no argument for doing so — is a quarter, and thermal conduction is a fifth. Setting ζ to zero does not make a small error in the absorption; it makes a 24 per cent one.

The viscosity nobody uses

A fluid has two viscosities. One resists a change of shape and appears in every figure on this site; the other resists a change of volume, was set to zero by Stokes in 1845 in a paper that says he had no argument for doing so, and for carbon dioxide is fifteen hundred times larger than the one everybody quotes.

Stokes' drag stops being the answer almost at once. The damping on a millimetre sphere oscillating in air, divided by Stokes' steady drag on the same sphere, against frequency. The extra term is the sphere's radius over the layer thickness, so it takes over as soon as the layer is thinner than the body — which for a millimetre sphere in air is below one hertz. At a kilohertz the damping is fifteen times the steady value, and the exponent is a half rather than zero.

What a fluid takes out of a swing

The damping a body feels from the air around it is not Stokes' drag, and stops being it far earlier than anybody expects — a millimetre sphere in air is already forty-six per cent above the steady answer at one hertz. Past that the damping rises as the square root of the frequency, and the fluid it is fighting is a shell a fraction of its own size.

The only candidate the far field allows, and the wall it slips past. The general Stokes solution has four constants; the condition at infinity kills two of them and fixes a third, leaving one to satisfy two conditions at the wall. Setting the stream function to zero there uses it up, and the tangential velocity that remains is exactly twice the free stream — for every radius, every speed, and every fluid.

The flow with no solution

Creeping flow past a sphere has a solution and everybody knows it. Creeping flow past a cylinder has none — not a difficult one, not one needing a clever method. The equations, the no-slip condition and the uniform stream at infinity are inconsistent, and the residual is exactly twice the free stream.

The singularity a layer makes for itself

March Prandtl's equations into an adverse pressure gradient and the wall shear reaches zero with an infinite slope at a finite station, and the solution cannot be continued past it. The singularity is real, it is not a numerical difficulty, and it belongs to the boundary condition rather than to the equations.

The core that does not move

Toothpaste in a tube has a region in the middle that is not being sheared at all, and its edge is at a radius you can write down before solving anything. Four things about that core are exact, and one of them is that the fluid does not flow below a threshold — not slowly, at all.

A pure number with no fluid in it. The load a plane slider bearing carries, divided by everything dimensional in it. What is left is a function of the convergence ratio alone — no viscosity, no speed, no size, no material — so the tilt that carries the most load is decided before anything about the machine is known. It is the root of one transcendental equation, at 2.1887048.

The gap that carries the most

A bearing's most consequential dimension is decided by a number with no oil in it, no speed and no size — the root of one transcendental equation, at 2.1887048. And the maximum it sits on is flat enough that the value printed in every handbook is not it.

One shape, at every station and for every jet. The axial velocity of Schlichting's round jet at four distances, each scaled on its own centreline speed and its own half-width. The four curves are one curve: the shape function (1 + eta²/4)⁻² has no parameter in it at all, and every station of every jet — laminar or modelled, weak or strong — collapses onto it exactly.

Exactly similar, and one number short

The round jet has an exact solution of the Navier–Stokes equations, and a turbulent jet is modelled by the same formula with the viscosity replaced by a fitted constant. The shape is identical. Nothing a measurement of the profile can do will tell the two apart.

Four profiles, one drag

The momentum integral is exact and asks nothing about the shape of the velocity profile. Four guesses at that shape span twenty-three per cent in the drag they give — and the two that satisfy the conditions the true profile satisfies are within three, which says the freedom belongs to the family rather than to the constraint.

The stress a pipe knows

A capillary viscometer measures a pressure drop and a flow rate and reports a viscosity. The first half of that inference is a force balance and is exact for any fluid there is; the second needs the slope of a whole flow curve, which is the experiment the instrument was bought to avoid.

A particle released in still fluid, with the history and without. The velocity of a hundred-micron particle released at one metre a second in air, computed with the unsteady Stokes force and with the quasi-steady drag alone. After five particle time constants the one that remembers is still moving nearly four times as fast.

The drag that integrates a whole history

A particle released in still air does not stop exponentially. The unsteady drag on it carries a term that is an integral over everything the particle has already done, weighted by the inverse square root of how long ago — so there is no time constant, and five particle time constants later it is still moving four times faster than the quasi-steady answer allows.

What the fluid at one height is listening to. The weight the fluid two millimetres above a moving wall gives to the wall's velocity a given delay earlier, in water. It peaks at two thirds of a second and has a tail that falls as the delay to the power minus three halves — so the fluid is responding to a broad stretch of the wall's past rather than to a moment of it.

The wall the fluid is listening to

Water two millimetres above a moving wall is responding to what the wall did two thirds of a second ago — most likely. Half of its response is older than four and a half seconds, a tenth is older than two minutes, and the average age of what it is responding to does not exist at all.

The flow stops and the stress does not. Shear stress against time for a fluid sheared at a constant rate for two seconds and then left alone. Nothing is moving after the second mark and the fluid is still stressed: 37 per cent of the peak one relaxation time later, and 0.7 per cent after five.

The fluid that has not finished its last deformation

Shear a polymer solution for two seconds and stop. Nothing is moving and the fluid is still stressed — 37 per cent of the peak one relaxation time later, and still measurably stressed after five. Two histories imposing exactly the same total strain leave it in states differing by a factor of 3.7.

Two external flows that agree where it matters and nowhere else. The velocity just outside the boundary layer, for two pressure distributions, against distance along the surface. They cross at the half-way station with the same value and the same gradient, and they have nothing else in common: one accelerates steadily and the other does most of its accelerating at once.

A layer that is an integral of everything upstream

Two surfaces are given external velocity distributions that agree exactly at one station — the same speed and the same gradient. The boundary layers there differ by 38 per cent in momentum thickness, and the two surfaces separate five per cent of their length apart.

$How long a fluid takes to forget it was not rotating. The fraction of solid-body rotation a container's interior has reached, against time, by the two available routes. The Ekman layers on the end walls pump fluid radially and carry angular momentum inwards in a hundred seconds; diffusion alone would need ten thousand.$

How long a fluid takes to forget it was not rotating

Spin a container of water and the fluid inside reaches solid-body rotation in a hundred seconds rather than the three hours diffusion would need. The shortcut is the thin layers on the end walls, and the advantage they give is exactly the reciprocal of the square root of the Ekman number.

A duct that forgets everything but one number

Whatever is fed into a pipe, what survives a little way down it is one shape. The disturbance's higher modes decay as the square of their mode number, so the sixth is gone in thirteen centimetres where the first survives four and a half metres — and the entrance length is that one mode's decay rate and an arbitrary threshold.

Three profiles that do not depend on the radius. The radial, azimuthal and axial velocities of the flow above a rotating disc, as functions of one similarity variable. The radial one is a jet: fluid thrown outward by the swirl it has picked up, peaking at 0.181 of the local disc speed a fifth of the way through the layer. The azimuthal one falls from the disc's own speed to nothing. And the axial one is the surprise — it does not vanish far from the disc but tends to a constant, so the disc draws fluid down onto itself at 0.8845 times the square root of the viscosity times the rotation rate, at every radius and for ever.

The solution that keeps its nonlinear term

Every exact solution before this one has been exact because the nonlinear term vanished. A rotating disc's does not vanish — at the wall it is the whole of the balance — and the reduction is exact anyway, because the radius divides out of all three momentum equations at once.

One channel, one flux, two flows

Flow between two plane walls meeting at a line has an exact solution. Past a threshold that turns out to be a ratio of gamma functions, it has two — the same wedge carrying the same flux, once outward everywhere and once with the fluid running backwards along both walls, and nothing in the equations chooses.

Six that are symmetries and five that look like them. Each transformation applied to an exact solution, with the Navier-Stokes residual recomputed from the transformed field by finite differences — nothing differentiated by hand. The six symmetries leave the residual at the differencing floor, a few parts in 10^8. The five near-misses leave between 0.048 and 4.3, which is six to nine orders of magnitude larger. The gap is what makes this a test rather than an illustration: a transformation that is nearly a symmetry does not exist here, and every one of the five is something a reader might reasonably believe.

Why the list is this long

Every textbook list of exact solutions of the Navier–Stokes equations is about a dozen long, and the usual explanation is that the equations are hard. It is not the reason. A similarity reduction is a solution invariant under a subgroup of the equations' own symmetries, so the catalogue of possible reductions is the catalogue of subgroups — and that is a finite, countable object.

All essays