The drop a no-slip wall would never let spread

The air a wing does not carry took the no-slip condition at its word and found that the usual picture of it — a blanket of air dragged along by a wing — gets both the amount carried and the drag wrong, while the condition itself stays exact. That is the normal situation. For a liquid against a solid the condition has been tested down to a few molecular diameters, and it holds to within slip lengths of a nanometre or so on wetting surfaces and tens of nanometres on water-repellent ones, which on any scale anyone can see is no slip at all.

There is one place where it cannot hold, and it is not exotic. A drop of water spreading on glass, a coffee stain growing, a plate being dipped into paint: each has a line where liquid, gas and solid meet, and the line moves along the solid. The liquid at the line is touching the solid, so no slip says it must move with the solid. It is also on the free surface, which moves with the line. It cannot do both. Taken literally, the condition says a drop could not spread — not slowly, but at all — because the force needed to move its edge would be infinite. What follows computes that infinity, the length that removes it, and the surprising fact that the answer barely depends on how long that length is.

At a moving line, the wall is asked for two velocities at once #

The conventions are these. The contact angle θ is measured through the liquid. Everything is seen from the contact line, so the line is still and the solid slides beneath it at a speed U; a line advancing over dry solid sees the solid moving away from it underneath the liquid. The capillary number $\mathrm{Ca} = \mu U/\gamma$ compares viscous stress with surface tension and is positive for an advancing line. Water is at 20 °C, with a viscosity of 1.002 mPa·s and a surface tension of 72.8 mN/m, so a line moving at a millimetre a second has a capillary number of 1.38 × 10⁻⁵. Within a millimetre of the line the Reynolds number is far below one, and the flow is Stokes flow, with no inertia in it.

Huh and Scriven solved that flow in 1971 for a wedge of liquid under a gas whose viscosity is ignored. The stream function has the form $\psi = r f(\varphi)$, with $r$ the distance from the line and $\varphi$ the angle from the solid, and four conditions fix it: no flow through the solid, the solid’s own velocity along it, no flow through the free surface, and no shear stress on it.

The liquid rolls. It arrives along the free surface, turns in the corner and leaves along the solid, like the tread of a caterpillar track, and every streamline has the same shape at every distance from the line. That self-similarity is the whole difficulty. A solution with no length in it has velocity gradients that grow as the distance shrinks, because the same change of velocity, from the solid’s speed to the free surface’s, is squeezed into a gap proportional to $r$. The wall shear stress is

$$\tau_w = \frac{2\sin^2\theta}{\theta - \sin\theta\cos\theta}\,\frac{\mu U}{r},$$

which is 2.44 μU/r in a 60° wedge, 5.52 μU/r at 30° and 1.27 μU/r at 90°. The pressure pushing on the free surface has the same one-over-r form, 2.82 μU/r at 60°. For a shallow wedge both go over to the thin-film forms $3\mu U/\theta r$ and $3\mu U/\theta^2 r$: at 30° the thin-film stress is within four per cent of the exact one, and at 90° it is fifty per cent too high.

The stress rises as one over the distance, and its integral has no end #

What that one-over-r means is clearest in numbers.

For a 30° line of water moving at a millimetre a second, the stress on the solid a millimetre from the line is 5.7 mPa, which is nothing. A micron away it is 5.7 Pa. A nanometre away it is 5,740 Pa, and the line on the logarithmic plot does not bend. The force the solid exerts on the liquid, per unit length of line, is the integral of that stress, and the integral of $1/r$ from zero is infinite however far out it stops. The rate at which the flow dissipates energy is that force times U — the same bill a velocity gradient always runs up — and it is infinite too.

That is not a small correction waiting to be tidied. It says that a solid obeying no slip exactly, with a liquid on it at any angle, could not be moved along the liquid’s edge by any finite force, and that a drop placed on it would never spread. Huh and Scriven put it memorably: not even Herakles could sink a solid. Since drops do spread and plates can be dipped, the condition fails somewhere in the corner, and the only questions are where and by how much.

One slip length makes the force finite #

The standard repair is the one Navier wrote down in 1823. The liquid at the wall is allowed to move relative to it by a velocity proportional to the shear there, $u_{\text{wall}} - U = \lambda\,\partial u/\partial z$, and the constant of proportionality λ is a length, the slip length. Far from the line the stress is small, the slip is tiny, and nothing changes. Near the line, where the stress would grow without limit, the slip grows instead.

In the thin-film wedge the effect has a closed form. With the film’s thickness $h = \theta x$ and no net flow through the line, the pressure gradient is $3\mu U/h(h + 3\lambda)$ and the wall stress is

$$\tau_w = \frac{3\mu U}{\theta x + 3\lambda},$$

which is the no-slip stress far from the line and levels off at $\mu U/\lambda$ within about a slip length of it: 1,000 Pa for a slip length of a nanometre, 100 Pa for ten. The force per unit length out to a distance $L$ becomes $(3\mu U/\theta)\ln(1 + \theta L/3\lambda)$ — finite, and computable.

A slip length that short does not come from the physics that sets the slip of a rarefied gas, where the slip length is a mean free path and follows from kinetic theory; in a liquid there is no free path to speak of. Molecular simulations of moving contact lines find the liquid slipping over a few molecular diameters near the line and sticking elsewhere. Nor is it the slip a porous wall presents, which comes from flow inside the wall. The contact line needs only that something, over some small length, lets the liquid at the solid move — and other repairs do that without Navier’s condition at all: a precursor film of molecular thickness running ahead of the line, a diffuse interface a few molecules wide, evaporation at the corner. Each introduces a small length, and each produces the same logarithm.

The force knows the molecular scale only through its logarithm #

That logarithm is the central result, and it is worth seeing how weak it is.

With a slip length of one nanometre the force is 6.93 × 10⁻⁵ N/m, about a thousandth of the surface tension. With the diameter of a water molecule, 0.3 nm, it is 7.62 × 10⁻⁵. With ten nanometres it is 5.61 × 10⁻⁵. Each tenfold change in the slip length adds or removes the same 1.32 × 10⁻⁵ N/m, which is $3\mu U\ln 10/\theta$, so the plausible range for a liquid on a smooth solid — a tenth of a nanometre to ten — runs from 8.25 to 5.61 × 10⁻⁵ N/m: a third of the answer, for two orders of magnitude in the least-known length in the problem. The exact wedge with a sharp cutoff at the slip length, in place of a thin film with Navier’s condition, runs parallel a little above, at 7.64 × 10⁻⁵ N/m for a nanometre. The angle factor changes the slope, and the details of the repair change only the constant beside the logarithm.

This is the counterpart of something already familiar. A curved vortex filament drives itself along at a speed containing $\ln(8R/\delta)$, where δ is the size of its core, because — as a ring moves because it is bent computed — the Biot–Savart integral along the filament diverges logarithmically without one. The two problems share nothing physical. They share the geometry of a singular line whose influence falls as one over the distance from it, and in both the inner length that removes the infinity survives only inside a logarithm. That is why a vortex ring’s speed can be predicted without knowing its core precisely, and a drop’s spreading without knowing its slip length precisely.

The insensitivity has a cost, and it runs the other way. Because the force changes by less than a fifth for a tenfold change in λ, measuring the force, or anything that depends on it, and inverting for λ turns a five-per-cent measurement into an uncertainty of a factor of several in the slip length. The macroscopic world is well protected from the molecular details of the contact line, and by the same token can learn very little about them.

The angle anyone measures depends on where they look #

The pressure in the wedge has consequences beyond the force. A pressure of order μU/r pushing on the free surface has to be carried by the surface’s curvature, and a surface whose curvature varies as one over the distance from the line has a slope that changes with the logarithm of that distance.

In the thin film that balance is the equation the coating calculation integrated for a plate leaving a bath — a third derivative of the thickness against a viscous term — now with the slip length in it: $h''' = -3\mathrm{Ca}/h(h + 3\lambda)$. Marched outward from the line, starting at a microscopic slope of one half, with the curvature adjusted so that far from the line the surface has none of its own, it gives the shape of the surface over thirteen decades of distance.

Inside a slip length the slope barely changes. Outside, its cube grows in a straight line against the logarithm of the distance, by 9.0025 times the capillary number for each factor of e. Voinov’s law says nine exactly:

$$\theta^3 = \theta_m^3 + 9\,\mathrm{Ca}\ln(x/x^*),$$

with $\theta_m$ the slope at the solid and $x^*$ a length of the order of the slip length, which the march puts at 2.21 slip lengths. At a capillary number of 10⁻³, a surface that leaves the solid at 26.6° stands at 33.9° a millimetre away, which is 10⁹ slip lengths of a nanometre. There is no single dynamic contact angle. There is an angle at each distance, and the one a microscope reports is set by where the microscope is focused.

For angles that are not small, Cox’s treatment of 1986 replaces $\theta^3/9$ by a function $g(\theta)$ whose slope is the inverse of the wedge’s pressure factor, so that $g(\theta_{\text{app}}) = g(\theta_m) + \mathrm{Ca}\ln(L/\lambda)$. The function stays within two per cent of $\theta^3/9$ up to 60°, and it carries a viscosity ratio for the gas. It is the same law, in a form that runs to 180°.

Two speeds a contact line cannot pass #

Cox’s form makes a prediction the small-angle law hides, and it is one of the most practical results in the subject.

An advancing line steepens and a receding one flattens. But the receding angle cannot fall below zero, and $g(\theta_m) - \mathrm{Ca}\ln(L/\lambda)$ reaches zero at a finite capillary number. Past it there is no solution with a contact line at all. Taking $L$ as the capillary length of water, 2.7 mm, the receding line fails at 0.29 cm/s for a microscopic angle of 10°, at 7.7 cm/s for 30° and at 61 cm/s for 60°. A plate withdrawn from water faster than that cannot drag its contact line down with it, and it comes out carrying a film — the regime where the Landau–Levich film takes over. That calculation assumed perfect wetting and noted that a partially wetting plate comes out dry below a critical speed; this is that speed, to the accuracy the logarithm allows. The slip length shows through here, faintly: at ten nanometres rather than one, the critical speed at 30° is 9.2 cm/s rather than 7.7.

The advancing side has a subtler limit. Under a gas with no viscosity, $g(\theta)$ grows without bound as the angle approaches 180°, so an advancing line can always steepen a little more and never fails. Air has a viscosity of 1.8 per cent of water’s, and Cox’s function with that ratio is finite at 180°, at 1.748. The advancing angle therefore reaches 180° at a capillary number near 0.12, and air rather than liquid is drawn into the corner — the entrainment that sets the ceiling on how fast a coating line can run. For water that capillary number is 8.5 m/s, a Reynolds number of twenty thousand on the capillary length, long after the Stokes-flow wedge has stopped describing the flow. For a 100 cSt silicone oil the same balance gives 6 cm/s at a Reynolds number near one, where the estimate is at least inside its own assumptions. The advancing ceiling belongs to the gas; the receding one belongs to the liquid.

A drop spreads as the tenth root of time #

The logarithm’s weakness has its most striking consequence in a drop left to spread on a solid it wets completely.

The drop is small enough that gravity does not matter, so it is a spherical cap, and its volume $V$ fixes its angle once its radius $R$ is known: at small angles $\theta \approx 4V/\pi R^3$. The microscopic angle is zero, so Voinov’s law has the edge moving at a capillary number of $\theta^3/9\ell$, with ℓ the logarithm of the drop’s radius over the slip length. Putting the two together makes $R^9\,dR/dt$ a constant, and

$$R = \left(\frac{640\,\gamma V^3}{9\pi^3\mu\ell}\,t\right)^{1/10}.$$

That is Tanner’s law. The marched version drawn here keeps the exact cap, Cox’s function and the logarithm’s slow drift as the drop grows, in place of every small-angle form.

The drop starts as a hemisphere 0.34 mm across. It reaches a radius of 0.36 mm at one second with an angle of 15.6°, 0.54 mm at a minute with 4.5°, and 0.82 mm at an hour with 1.3°, when its edge is creeping at 23 nanometres a second. Between a hundred seconds and ten thousand the exponent is 0.0994 — the tenth, less a little, because the logarithm grows as the drop does.

And the slip length has almost vanished from the answer. At one hour the radius is 0.804 mm with a slip length of a tenth of a nanometre, 0.817 mm with one nanometre and 0.832 mm with ten: a hundredfold change in the only molecular length in the problem moves the radius by 3.5 per cent, because it enters as the tenth root of a logarithm. Tanner measured this law with silicone oils in 1979, and it remains the cleanest macroscopic evidence that the contact-line singularity is cut off by something very small — evidence that says nothing about what.

The calculation stops being right after about an hour for this drop. By then its radius is half the oil’s capillary length of 1.49 mm, its Bond number has passed 0.3, and gravity — which fixes how large a drop is allowed to be before it slumps — begins to flatten it faster than surface tension alone would, turning the exponent towards an eighth.

Each closed form against a calculation that does not use it #

The wedge’s stream function meets its four boundary conditions to 3.6 × 10⁻¹⁵, and its stress and pressure factors, read off its own coefficients, match the closed forms to the same precision. The thin-film force with slip matches a quadrature of the stress to 2.8 × 10⁻⁷. Inverting Cox’s function recovers a chosen angle to 5 × 10⁻¹³. The nine in Voinov’s law came out of a march that was never told it, at 9.0025, and the tenth in Tanner’s law out of a drop marched in time, at 0.0994. Two rows are limits rather than errors: the exact stress factor sits 1.3 × 10⁻⁷ from its thin-film form at an angle of a thousandth of a radian, and Cox’s function 5 × 10⁻⁵ from $\theta^3/9$ at 0.05 radians, which is the size the square of the angle says each should be. The calculation refuses a moving line on a wall with no slip, a wedge of no angle or of 180°, an advancing angle of 180° under a gas with no viscosity, an outer length inside the slip length, and a spreading drop that cannot slip.

What the corner cannot show #

The slip length’s value. Every result here needs one, and nothing in continuum mechanics supplies it. The calculation shows how little it matters, which is not the same as knowing it.

Pinning. Real solids are rough and chemically patchy, and a contact line on them does not move until the force on it passes a threshold. The difference between the angles at which a line starts to advance and to recede — the hysteresis — is often tens of degrees, and at slow speeds it swamps the logarithm.

The microscopic angle. It is taken as the static equilibrium angle, independent of speed. The molecular-kinetic picture, in which a line advances by molecules hopping between sites on the solid, makes it depend on speed as well; the two mechanisms add, and experiments rarely separate them cleanly.

The receding threshold’s precision. The critical speeds come from the dynamic-angle law pushed to an angle of zero, where it is least reliable. Eggers’s more careful matching to the static meniscus puts the true threshold somewhat lower, so those numbers are estimates of the right size rather than predictions to a per cent.

Inertia, evaporation and surfactants. The wedge is Stokes flow. A fast line, a volatile liquid or a surface carrying a gradient of surfactant each adds a stress this calculation does not have.

Still open: what actually slips at a moving contact line #

The logarithm protects the macroscopic answer and hides the microscopic one, and that leaves the physics of the corner genuinely unsettled. Navier slip, a precursor film and a diffuse interface all remove the singularity, all give the same logarithm, and differ only in the constant beside it — a constant that shifts a measured angle by a degree or two and cannot be pulled cleanly out of an experiment. Molecular simulations find the slip concentrated within a few molecular diameters of the line, which fits a Navier slip length whose value no one has derived.

The advancing side has an open question of its own. Coating lines run faster before entraining air when the gas is at reduced pressure, and the reason may be a second corner: the gas’s own wedge, whose mean free path — 68 nm at atmospheric pressure and ten times that at a tenth of an atmosphere — is not small against the gas gap within a few microns of the line, so that the gas may slip where the liquid does not. Whether the advancing ceiling is set by the gas’s viscosity, as Cox’s function has it, or by where the gas stops being a continuum in the last microns of the wedge, is the next calculation: the same wedge, with slip on the gas side and a Knudsen number that grows towards the line.