The vorticity a clean surface cannot refuse

How many things a flow must be told counted the conditions at a boundary: the Navier–Stokes equations take two, and at a solid wall the second is that the fluid does not slip. A wall that is not quite there weakened that second condition to a slip length — a wall that holds the fluid back less than a solid does. This essay takes the other end of the same scale. A clean interface between a liquid and a gas holds the liquid back not at all: the gas is too thin to push, so the liquid at the surface feels no tangential stress.

That is a condition of a different kind, and the natural reading of it is that it makes nothing. Where vorticity comes from showed that a no-slip wall hands vorticity to the fluid because it forces a shear; a surface that forces no shear should hand over none. The reading is wrong wherever the surface is curved, and a bubble is nothing but curvature.

The stress a curved surface cannot hold #

Write the shear stress at a surface in terms of the velocity along it, $u_s$, and across it, $u_n$. On a flat surface the stress is the viscosity times $\partial u_s/\partial n$, and setting it to zero sets the shear to zero. On a surface with curvature $\kappa$ along the flow, the rate of strain picks up a term because the directions themselves turn: the stress is proportional to $\partial u_s/\partial n - \kappa u_s$, once the surface is impermeable. Setting that to zero gives

$$\frac{\partial u_s}{\partial n} = \kappa\,u_s ,$$

and the vorticity at the surface, which is $\partial u_s/\partial n + \kappa u_s$ on the same surface, is then

$$\omega = 2\kappa\,u_s .$$

A clean curved surface cannot have zero vorticity on it. It must carry exactly twice its curvature times its speed, and nothing in that statement contains the viscosity, the size of the bubble in any other way, or a Reynolds number.

The ideal flow round a bubble is irrotational, so it has the wrong value. It has zero vorticity, which means $\partial u_s/\partial n = -\kappa u_s$, and a stress proportional to $-2\kappa u_s$: the ideal flow is pulling on the surface, and a clean surface cannot be pulled.

The figure computes the ideal flow’s surface stress without the formula: the second derivatives of the potential are evaluated on the surface and projected onto its normal and tangent. Round a sphere the stress is $-3\mu U\sin\theta/a$, round a cylinder $-4\mu U\sin\theta/a$, and the two routes agree to the last digit. What a clean surface requires of the real flow is the vorticity that cancels it, and that vorticity is three times the flow speed over the radius at a sphere’s equator.

Why the factor is two #

The two in $2\kappa u_s$ has a picture behind it, and the picture is the most familiar flow with vorticity in it.

A shear-free surface curving with radius $R = 1/\kappa$ does not let the liquid next to it shear along it. So, just below the surface, the liquid moves round the curve the way a thin rim of something rigid would: its speed does not change from one layer to the next in a way that would stretch the fluid tangentially. Locally, that is a rigid rotation about the centre of curvature, at an angular speed of $u_s/R$ — and a rigid rotation at angular speed $\Omega$ carries vorticity $2\Omega$, everywhere in it, which is the definition spin is not the same as going round takes apart. With $\Omega = \kappa u_s$, the vorticity is $2\kappa u_s$.

The ideal flow cannot move like that at the surface, because it may not rotate at all. Near a curved boundary the only way an irrotational flow can go round the curve is the way a potential vortex does, faster on the inside and slower on the outside, and that difference in speed between neighbouring layers is exactly the tangential strain a clean surface refuses. The real flow near a clean bubble is therefore a thin transition from the potential vortex’s pattern to a rigid rim’s — a change in vorticity from zero to $2\kappa u_s$ — and nothing else. The rigid wall’s transition is from the full ideal-flow speed to zero, which is why it is so much larger.

The factor of two does not depend on the surface being a sphere. On any smooth surface that holds no shear, the component of vorticity along the surface and across the flow is twice the curvature in the direction of the flow times the speed; on a sphere the curvature is the same in every direction and the formula needs no qualification. For the cylinder it is $4U\sin\theta/a$ because the ideal flow round a cylinder moves faster at its surface — twice the stream speed at the shoulder, against one and a half times for a sphere — and the curvature is the same.

Vorticity of order U/a, not of order U/δ #

The number that matters is how large that vorticity is compared with what a solid wall makes, because that decides whether the flow near the surface is a thin region of intense shear or barely different from the ideal one.

A rigid sphere must bring the fluid to rest across a boundary layer whose thickness shrinks as the flow speeds up, so its surface vorticity is the flow speed over that thickness — of order $U/\delta$, which is $U\sqrt{Re}/a$. Near the front stagnation point that can be computed exactly: the layer there is Homann’s axisymmetric stagnation-point flow, driven by the ideal flow’s strain of $3U/2a$, and its wall slope comes out of a shooting calculation at 1.3119.

A clean bubble’s surface vorticity has no layer thickness in it at all. It is $3U\sin\theta/a$ at every Reynolds number, set by the geometry.

The ratio grows as the square root of the Reynolds number and has no ceiling. At a Reynolds number of a thousand — a bubble about a millimetre and a half across rising in clean water — the rigid sphere’s surface vorticity is eighteen times the clean bubble’s.

So the two surfaces are not two versions of one boundary layer. At a rigid wall the ideal flow is wrong by a whole velocity across a layer that thins without limit; at a clean surface it is wrong only by a vorticity of order U/a, which a layer of the same thickness can supply with a velocity correction of order $U/\sqrt{Re}$. The ideal flow is the flow, to leading order, right up to the surface.

The drag is the ideal flow’s dissipation #

If the flow is irrotational to leading order, d’Alembert’s theorem would seem to say it has no drag. The exact theory says nothing has any drag is that theorem, and it is a statement about the pressure: round a closed body in steady ideal flow the pressure pushes as hard on the back as on the front. What it does not say is that an irrotational flow of a viscous fluid dissipates no energy.

It does dissipate. An irrotational flow has no rotation but plenty of strain — fluid elements are stretched in one direction and squeezed in another as they pass the bubble — and viscosity turns strain into heat at a rate of $2\mu$ times the sum of the squared strain rates. A bubble rising steadily must supply that heat, and the only thing it can supply it with is the work its drag does. So the drag times the speed is the dissipation, and the dissipation can be computed from the ideal flow alone.

The quadrature integrates the squared second derivatives of the potential over the whole of the liquid outside the sphere and returns $12\pi\mu aU^2$, to fourteen figures. Divided by the speed, the drag is

$$D = 12\pi\mu a U, \qquad C_D = \frac{48}{Re},$$

which is Levich’s result. And the figure shows where the drag is made: not at the surface, in a layer, but in a shell of liquid round the bubble, with the dissipation density falling as the eighth power of the distance, so that all but three per cent of it lies within one radius of the surface.

That is also why the law has the viscosity to the first power. A drag made in a boundary layer carries the square root of the viscosity, because the layer’s thickness does. A drag made by straining a flow whose shape does not depend on the viscosity at all carries the viscosity exactly once. The same kind of calculation, done on a sphere in a flow with no inertia, is what gives a suspension its extra viscosity: a dissipation integral, divided by what is driving it.

The same calculation round a cylinder gives a dissipation of $8\pi\mu U^2$ per unit length, so a drag of $8\pi\mu U$ per unit length and a drag coefficient, on the diameter, of $16\pi/Re$. That is worth having because the cylinder has no creeping-flow answer at all — a cylinder in an unbounded fluid with no inertia has no solution — and at high Reynolds number the irrotational flow supplies one directly, with no layer and no matching.

A wave travels a long way on the same small vorticity #

The argument does not need a bubble. Any clean liquid surface that curves and moves carries vorticity $2\kappa u_s$ and no more, so any flow bounded by one is irrotational to leading order and dissipates only by straining — and the most important such flow covers seven-tenths of the planet.

The surface of a deep-water wave is a clean interface that curves up over each crest and down through each trough. The orbital motion beneath it is irrotational, and Lamb computed its viscous dissipation the way this essay computes a bubble’s: the strain of the irrotational flow, integrated through the water. A wave’s amplitude decays as $e^{-2\nu k^2 t}$, with $k$ the wavenumber.

The numbers are what make the point. A wave a metre long in water decays by a factor of $e$ in about three and a half hours, in which it travels some sixteen kilometres at its own speed. A wave ten metres long decays in a hundred times the time — about two weeks — and travels about five thousand kilometres before its amplitude has fallen by that factor. Swell crosses oceans because its surface makes almost no vorticity, and a boundary layer at the surface as strong as a rigid wall’s would stop it within a few wavelengths. A wave pattern whose shape does not depend on the boat that made it is the same irrotational water, seen at the instant it is made rather than weeks later.

Three times the creeping-flow drag, and far below a rigid sphere’s #

Two comparisons are in the figure, and both are instructive.

The surface that moves with the flow computed the drag of a clean bubble in the opposite limit, where inertia is negligible: Hadamard and Rybczynski’s $4\pi\mu aU$, a coefficient of $16/Re$, two-thirds of a rigid sphere’s Stokes drag. Levich’s law is exactly three times that. A bubble moving fast has three times the drag coefficient it would have at the same Reynolds number in a flow without inertia, and the reason is where the dissipation is. In creeping flow the disturbance spreads a long way and the strain is gentle; at high Reynolds number the flow round the bubble is the ideal flow, whose strain is concentrated close to the surface, and concentrated strain dissipates more.

The second comparison is with a rigid sphere, and it goes the other way. A rigid sphere’s boundary layer separates and leaves a wake, and its drag coefficient levels off near a half while the clean bubble’s keeps falling. The borrowed correction for the clean bubble’s weak boundary layer, Moore’s $(1 - 2.211/\sqrt{Re})$, is the size a correction of order $1/\sqrt{Re}$ should be, and it trims Levich’s value by a fifth at a Reynolds number of a hundred and by a fourteenth at a thousand.

What a trace of surfactant costs #

A real bubble in ordinary water does not keep its clean surface. Surface-active molecules — in tap water, in anything that has touched skin or a plastic container — collect on the interface, are swept to the back of the bubble by the flow, and pile up there, and the gradient in surface tension they create pulls the surface forwards against the flow. The surface stops moving, and the bubble behaves as a solid sphere of the same size. The mechanism and its thresholds belong to the creeping-flow version of this argument, where the stakes are a third of the drag.

At the Reynolds numbers of a bubble rising through a glass of water the stakes are much larger. The ratio of the two drags is three at a Reynolds number of a hundred and ten at a thousand, and the condition that decides which curve a bubble is on is invisible: a concentration of surfactant far below anything that changes the water’s appearance, its taste or its bulk surface tension.

Balanced against buoyancy, the two drag laws give two rise speeds for the same bubble. At half a millimetre in radius a clean bubble rises at 31 centimetres a second and a contaminated one at 11. Nothing about the bubble differs between the two — not its size, not the gas, not the liquid — except whether its surface can move, and the difference is nearly a factor of three.

The shaded region marks where the calculation stops describing either bubble. A clean bubble rising fast enough deforms: its Weber number, inertia against surface tension, passes one at a radius of 0.47 millimetres, and a flattened bubble has a larger frontal area and a different ideal flow. The spherical result is a statement about bubbles smaller than that, and the size at which a drop’s own weight flattens it is the static version of the same competition.

What the shear-free model leaves out #

The shape. Every calculation here is for a sphere or a circular cylinder. A deformed bubble has curvature that varies over its surface, so its required vorticity varies with it, and the ellipsoidal ideal flow dissipates differently.

The gas. The gas inside is taken to transmit no stress at all. Its viscosity is fifty times smaller than water’s, so the correction is small, and it drives a slow circulation inside the bubble that the calculation ignores.

The boundary layer’s own dissipation, and the wake. Moore’s correction for the weak layer is borrowed rather than derived. A clean spherical bubble’s layer does not separate, which is what keeps the ideal flow valid at the back as well as the front; at high enough Reynolds number, and for deformed bubbles, a small separated region appears, and the drag rises above Moore’s curve.

The path. A bubble larger than about a millimetre does not rise straight. Its wake becomes unstable and it zigzags or spirals, and the drag of a bubble on a spiral path is not the drag of one rising steadily.

Partial contamination. A surface is either clean or immobilised here. A real one can be clean at the front and stagnant at the back, with a drag between the two curves and a rise speed that changes as the surfactant accumulates.

Levich’s argument #

Levich published the dissipation argument in 1949, and it is one of the few results in viscous flow at high Reynolds number obtained without solving for a boundary layer at all. Moore supplied the boundary-layer correction in 1963 and showed that the layer on a clean bubble does not separate. The surface condition itself — that a stress-free curved surface carries twice its curvature times its speed in vorticity — had been written down by Longuet-Higgins a decade earlier for the surface of water waves, where the same small vorticity is what lets a wave lose energy to viscosity slowly enough to travel far.

What joins the three is the reading that runs through this essay. A clean surface makes vorticity because it is curved, not because it is sticky, and it makes so little that the irrotational flow survives — and a flow that survives unrotated can still dissipate, and pay for its dissipation with a drag.

Still open: a surface that slips in one direction only #

A clean bubble slips equally in every direction along its surface, and a solid wall slips in none. Between them is a surface that slips differently along one direction and across it: a patterned surface of gas pockets and solid ridges, where each gas stripe is a small clean interface and each ridge is a small wall. The condition such a surface imposes on a flow far above it is not a single slip length but a pair of them, and their ratio is fixed by the geometry rather than measured.

That is twice as slippery along as across: Philip’s exact slip lengths for stripes, the factor of exactly two between them, and the limit a logarithm puts on how much slip any gas fraction can buy. Beside it sits the moving solid wall’s own budget, which puts into the fluid exactly its own speed, and which a shear-free surface — making vorticity by curvature rather than by acceleration — does not obey.