A ball that bounces in water and not in oil

A steel ball dropped into a tank of water bounces off the bottom much as it would off a table. The same ball dropped into the same tank filled with heavy oil does not bounce at all: it arrives, and stays. Nothing about the ball or the floor has changed, and the liquid’s density is within ten per cent of the water’s.

That is awkward, because the squeeze film between them has already been shown to be unclosable. The force resisting the approach goes as the inverse cube of the gap, so the work needed to squeeze the last of the liquid out diverges, and there is no finite energy that removes it. If that is true then nothing ever touches anything, and a bounce is as impossible as a landing.

Both statements are right. What reconciles them is that the argument for the first was made at a constant load, and a falling ball does not apply one — it arrives with a momentum, spends it, and has no more.

The number is a momentum divided by a resistance #

Within a gap small against the radius the lubrication force on a sphere approaching a plane is exactly

$$F = \frac{6\pi\mu R^2\dot x}{x},$$

so the sphere obeys $m\ddot x = -6\pi\mu R^2\dot x/x$. Dividing the momentum it arrives with by the resistance that constant gives produces the group the whole essay is about:

$$\mathrm{St} = \frac{m v_0}{6\pi\mu R^2} = \frac{2}{9}\,\frac{\rho_s R v_0}{\mu}.$$

Written the second way it is recognisably the Stokes number, in both of its usual senses: the ratio of a particle’s own relaxation time to a flow time when a droplet is asked to turn a corner, and the measure of how far a tracer lags the fluid it is meant to follow. Here the flow time is the time the sphere has left, and the Stokes number is how many decades of gap the sphere’s momentum will buy.

It is also, written a third way, one ninth of the density ratio times the particle Reynolds number. Glass in water at two hundred millimetres a second has a Reynolds number of 1200 and a Stokes number of 333; glass in glycerol at the same speed has a Reynolds number of 1.5 and a Stokes number of a third. The two numbers move together and mean different things, and the one that decides the bounce is the one with the solid’s density in it.

That last point is worth dwelling on, because the two groups are routinely quoted for the same experiment and a reader can be forgiven for treating them as interchangeable. The Reynolds number compares the liquid’s inertia with the liquid’s viscosity and says what the wake looks like. The Stokes number compares the solid’s inertia with the liquid’s viscosity and says what the sphere does. A hollow plastic sphere and a solid steel one of the same size, arriving at the same speed into the same tank, make identical wakes and behave completely differently at the floor: their Reynolds numbers are equal and their Stokes numbers differ by a factor of eight.

The group is also not the ratio of two times, which most dimensionless groups are. There is no natural time in a lubrication approach — the force depends on the gap and the gap is what is being solved for — so the momentum has to be measured against a force rather than a duration. What $6\pi\mu R^2$ is, dimensionally, is a viscosity times an area, which is a momentum per unit of logarithm: it is precisely the momentum a sphere must spend to close the gap by a factor of $e$. The Stokes number is therefore a count. It says how many factors of $e$ of gap the sphere’s momentum is good for, and everything below is arithmetic on that sentence.

The equation solves itself, in the wrong variable #

Reynolds’ equation for the gap gives a force law that looks as though it needs integrating in time and does not, because $\ddot x$ can be written $v\,dv/dx$ and the $v$ on the right cancels:

$$m\,\frac{dv}{dx} = -\frac{6\pi\mu R^2}{x} \qquad\Longrightarrow\qquad v(x) = v_0\left[1 - \frac{1}{\mathrm{St}}\ln\frac{x_0}{x}\right].$$

Two things follow immediately and neither is visible in the force law. The speed falls as a logarithm of the gap, so every decade of approach costs the same fixed fraction of the arrival speed — $2.303/\mathrm{St}$ of it — however thin the film has already become. And the speed reaches zero at a finite gap, not at contact.

The curves in that figure are not plots of the formula. They are a Runge–Kutta march of the pair $\dot x = -v$, $\dot v = -(v_0/\mathrm{St})(v/x)$ with the step held to a thousandth of the gap’s own timescale, which is a genuinely separate calculation from the algebra above and agrees with it to $4\times10^{-15}$. That matters because the cancellation that produced the closed form is exactly the kind of step that quietly loses a factor.

Where a sphere stops, if the film is all there is #

Setting the bracket to zero gives the whole of the first half of the argument:

$$x_{\text{stop}} = x_0\,e^{-\mathrm{St}}.$$

An exponential in a dimensionless group is a steep function, and this one covers the entire interesting range in about a factor of twenty in the group.

At a Stokes number of two the sphere stops 406 microns short. That is a real distance: it can be photographed, and it is a straightforward demonstration that the film is doing what the inverse cube says. At a Stokes number of twenty the sphere stops 6.2 picometres short, and no statement about a fluid is meaningful at that scale — the continuum has been gone for four decades of length, and the electron clouds of the two solids overlap long before.

So the model stops predicting a distance and starts predicting an event. Between those two readings of the same formula lies the threshold, and the threshold is therefore a question about a length rather than about a speed, which is not what the phrasing “critical Stokes number” suggests.

The distance that has to be put in by hand #

Contact requires a length the film equation does not contain. Three candidates are usually offered and it does not matter much which: the roughness of the surfaces, the elastic flattening under the film’s own pressure, and the range at which the solids’ molecules interact. All three are somewhere between a nanometre and tens of microns, and all three enter the answer the same way.

Call the length $x_c$ and let $L = \ln(x_0/x_c)$. A sphere that reaches $x_c$ with speed $v_c$ leaves it at $e_d v_c$, where $e_d$ is the restitution the same impact would have had dry, and then has to climb back out through the same film. Each leg costs $v_0 L/\mathrm{St}$, so

$$e = \frac{v_{\text{out}}}{v_0} = e_d - \frac{L}{\mathrm{St}}\,(1 + e_d), \qquad \frac{e}{e_d} = 1 - \frac{\mathrm{St}_c}{\mathrm{St}}, \qquad \mathrm{St}_c = \frac{L\,(1 + e_d)}{e_d}.$$

The middle expression is what is measured. The right-hand one is where it comes from, and it is worth noticing that the outward leg is the expensive one: the sphere leaves the cut-off with only $e_d$ of the speed it arrived with, and pays the same fixed $L/\mathrm{St}$ on the way out as on the way in. That is where the factor $(1 + e_d)$ comes from, and it is why a sphere of low dry restitution needs a much larger Stokes number than one of high — a perfectly elastic solid has $\mathrm{St}_c = 2L$ and a solid with $e_d = 0.5$ has $3L$.

Computing the three steps numerically and comparing with the hyperbola reproduces it to one part in $10^{16}$, which says only that the algebra was done correctly and is worth checking for that reason.

Why one number fits four materials #

The interesting property of that expression is not its value but its insensitivity.

A decade of cut-off costs $(1 + e_d)/e_d \times \ln 10 = 4.68$ on a threshold near ten. That is a large fraction of the answer for one decade and a very small one for five: moving the cut-off from a nanometre to a hundred microns — the entire range anybody has proposed — changes the threshold by a factor of twenty-nine, against a factor of $10^5$ in the length.

That is the whole reason the threshold is worth quoting. A quantity that depended on the cut-off linearly would be a different number for every pair of surfaces, and there would be no such thing as “the critical Stokes number”; there would be a critical Stokes number for polished steel, another for lapped glass, and no general statement at all. A logarithm flattens five decades of ignorance into a factor of thirty, and what survives is a number.

The same structure appears whenever a thin-film singularity is cut off by a length nobody knows. A moving contact line has exactly this shape: the microscopic physics is genuinely unsettled, three different mechanisms remove the singularity, and all three give the same logarithm with a different constant beside it — so the macroscopic answer is protected and the microscopic one is not measurable. A bounce is the same bargain, made in the other direction: the length cannot be extracted, and the threshold does not need it.

Run the expression backwards and the measured threshold of ten implies $L = 4.92$, a ratio $x_0/x_c = 138$, and a cut-off of 21.8 microns on a 3 mm sphere. That is far larger than any roughness a polished sphere has and is about the scale of elastic flattening under the pressures the film reaches — which is a check on the picture rather than a derivation of it.

Six pairings, and the two that share a solid #

The three entries in water differ only in the solid. Steel, glass and nylon have densities of 7800, 2500 and 1140, and their Stokes numbers fall in the same ratio — 1040, 333 and 152 — so all three rebound, at 96, 94 and 90 per cent of their dry values. The density enters the number linearly and the recovery above the threshold is $1 - \mathrm{St}_c/\mathrm{St}$, so a factor of seven in density buys very little once the threshold is comfortably passed.

The three entries in steel differ only in the liquid. Water, light oil and heavy oil at $10^{-3}$, $2\times10^{-2}$ and $2\times10^{-1}$ pascal seconds give Stokes numbers of 1040, 52 and 5.2 — and the last of those is below the threshold. A single sphere crosses from bouncing to not bouncing on a change of nothing but the fluid’s viscosity, and the crossing is sharp rather than gradual, because the restitution reaches zero at the threshold rather than tailing off through it.

Glass in glycerol, at a Stokes number of a third, is the extreme: it stops 3 mm above the floor and descends the rest of the way under gravity, slowly, as the load problem in the earlier figure describes.

What the list does not contain is a single combination that rebounds at a low Stokes number or fails to at a high one, and that absence is the content of the threshold. Six pairings spanning a factor of a thousand in viscosity, seven in density and three thousand in the Stokes number sort themselves onto the two sides of one line with nothing left over. A rule that sorted five out of six would be a correlation; this one is a criterion, and the reason it can be is that the quantity it sorts on is the only one the equation of motion contains.

Where the energy goes, which is not at the end #

The restitution is a ratio of speeds and the loss is a fraction of energy, and converting between them says something the speed curve does not.

At a Stokes number of 1040 — steel in water — the sphere leaves at 96.1 per cent of its arrival speed, so it keeps 92.4 per cent of its kinetic energy. Of the 7.6 per cent it loses, 5.9 belongs to the dry impact and would have been lost against a table in air; the film takes 1.7. At a Stokes number of 52 — the same steel in light oil — it leaves at 78.0 per cent, keeps 60.8 per cent of its energy, and the film has taken 33 of the 39 per cent lost. The film goes from a rounding error to the dominant loss over a factor of twenty in one number, while the solids do exactly the same thing in both cases.

The distribution of that loss across the approach is the part that contradicts the intuition. Because the speed falls by a fixed fraction per decade of gap, and energy goes as the square of speed, the energy lost in successive decades is a falling but not a vanishing sequence. For a sphere at $\mathrm{St} = 20$ closing from 3 mm to 20 microns, the first decade of approach costs 21.7 per cent of the arrival energy, the second costs 19.0, and the last fifth of a decade costs 3.1. The bill is spread almost evenly over the approach.

That is the opposite of what a force going as $1/x$ suggests, and the reconciliation is that the sphere is travelling fastest where the force is weakest. The power delivered to the film, $6\pi\mu R^2 v^2/x$, does rise as the gap closes — but only because $v$ falls more slowly than $x$ does, and the time spent in each decade falls in step. Integrating force over distance rather than power over time is what flattens it, and it is why a sphere cannot be saved by a thin film near the end: by then most of the momentum has already gone somewhere else.

What the far field is doing, and why it is not in the answer #

Everything above treats the liquid as a lubrication film and nothing else, which for a sphere arriving at a Reynolds number of a thousand deserves an explanation.

The resolution is that the lubrication force and the far-field drag are different objects. The lubrication result comes from the thin gap alone: it needs the gap to be small against the radius and the local Reynolds number built on the gap to be small, and both are satisfied long before the sphere arrives whatever the far-field Reynolds number is. The far-field drag is a constant-order force; the lubrication force diverges as $1/x$. So in the last decades of approach, which is where all the decisions happen, the film is the only thing in the problem by a wide margin.

What the far field does decide is $v_0$ — the speed the sphere arrives with — and $x_0$, the gap at which the lubrication force overtakes everything else. The second of those sits inside the logarithm and has already been shown not to matter much. The far field sets the initial condition and the film does the rest, which is why an account of a bounce contains no wake.

Even the first of those is less of a handle than it looks. Doubling the arrival speed doubles the Stokes number, which halves $\mathrm{St}_c/\mathrm{St}$ and so moves the restitution ratio from, say, 0.5 to 0.75 — a real change, but one that saturates: no arrival speed produces a restitution above the dry value, and the curve is within ten per cent of it by four times the threshold. So the far field can decide whether a bounce happens and cannot decide how good it is, which is a division of labour worth stating because it is the reason a dropped-ball experiment is reproducible at all. Everything that is hard to control — the release, the wake, the tank’s walls — acts through $v_0$, and $v_0$ acts through one number that the film then spends in a way nothing upstream can influence.

What the picture cannot show #

The solids are rigid everywhere except at the cut-off. They are not: the film pressure at a micron reaches a hundred megapascals, the surfaces flatten, and the correct treatment is elastohydrodynamic — the gap shape and the pressure are solved together. That calculation replaces the sharp cut-off with a smooth minimum and changes the coefficient in front of the logarithm, which is the same insensitivity working in the model’s favour.

The film is Newtonian and isothermal. A film squeezed to a micron at a metre a second is being sheared at $10^6$ per second, which takes a mineral oil out of its Newtonian range, and it is heating itself while it does so.

The liquid cannot take tension, and this calculation lets it. The rebound leg asks the film to pull the sphere back — the pressure there is below ambient — and a real film cavitates instead, which removes part of the outward cost and raises the measured restitution above this model’s. The correction goes the right way and is not small.

Gravity is absent throughout. For the low-Stokes cases it should not be: a sphere that stops 400 microns above the floor is still heavy, and what happens next is the constant-load drainage problem, over minutes.

And the sphere does not rotate or wobble. A real sphere arrives slightly off-centre, the film is then asymmetric, and it acquires a spin on the way out which carries away energy this accounting gives back to it.

Who found it, and when #

Davis, Serayssol and Hinch gave the elastohydrodynamic treatment of the collision in 1986 and are the source of the modern picture: the approach, the elastic flattening, and the criterion in the Stokes number. Barnocky and Davis measured it shortly afterwards, and Gondret, Lance and Petit published in 2002 the measurements that made the form $e/e_d = 1 - \mathrm{St}_c/\mathrm{St}$ with $\mathrm{St}_c \approx 10$ the standard statement — over spheres of glass, steel, nylon and Teflon in water, silicone oils and glycerol mixtures.

The surprising connection is with a threshold in a completely different subject. A critical Stokes number also decides whether a droplet in an air stream hits an obstacle or goes round it, and that one is exactly one eighth — an algebraic number, out of a discriminant, with no tolerance in it anywhere. This one is about ten, is not algebraic, and has a logarithm of an unmeasured length in it. Two thresholds in the same dimensionless group, and only one of them is a number the mathematics knows. The difference is that the droplet problem is a solved trajectory in a solved flow, with nothing put in by hand, and the impact problem has a singularity in it that has to be stopped by physics from outside. A sharp threshold is not evidence of an exact one.

Still open: whether the threshold moves when the film cavitates #

The outward leg is the weak point of the three-step model, and it is weak in a way that can be computed rather than argued.

A film being pulled apart goes into tension, and liquids sustain very little of it, so the real film ruptures at some gap on the way out and the sphere leaves with less resistance than this accounting charges it. That would lower the cost of the outward leg from $L/\mathrm{St}$ towards something like $\ln(x_{\text{cav}}/x_c)/\mathrm{St}$, and therefore lower the threshold — by $(1/e_d)$ times $\ln(x_0/x_{\text{cav}})/\ln(x_0/x_c)$ of its value, which for a rupture at a tenth of a millimetre is a reduction of about a third.

The calculation that would settle it is the same march with a pressure floor at the liquid’s vapour pressure, following the ruptured region’s radius outwards and stopping the film’s contribution where it has opened. The measurement that would settle it is the asymmetry: a cavitating film should make the restitution depend on the ambient pressure, which nothing else in this problem does, and an impact run in a partially evacuated tank should rebound differently from the same impact at an atmosphere. Neither appears to have been done, and the threshold’s stubborn insensitivity to everything else suggests the effect would be visible.