The drag that integrates a whole history

The theory with no memory in it sets three response kernels side by side and observes that the viscous one falls as the inverse square root of time and therefore has no time constant at all. This essay is that kernel, measured on the problem it matters most for.

A particle in a fluid is the case where the whole of the unsteady force is visible, because the particle is small enough for the flow round it to be a Stokes flow and the force is therefore known in closed form. It has three parts: a steady drag proportional to the present velocity, an added mass proportional to the present acceleration, and a history integral proportional to nothing present at all.

The three forces, and which of them is a memory #

The steady drag is Stokes’ and it is a function of the state: velocity now, force now. The added mass is a function of the state too — acceleration now — and the mass a body has to borrow computes it for a sphere at exactly a half of the displaced fluid.

The third is different in kind. It is

$$F_h = -\tfrac{3}{2}d^2\sqrt{\pi\rho\mu}\int_0^t \frac{\dot v(\tau)}{\sqrt{t-\tau}}\,d\tau,$$

an integral over the particle’s entire acceleration history with a weight that falls as the inverse square root of how long ago. The physics behind it is diffusion — the mechanism where vorticity comes from identifies at every wall: an acceleration puts vorticity into the fluid at the surface, that vorticity spreads outwards, and it goes on exerting a force for as long as it is anywhere near — which is for ever, because diffusion has no end.

So a particle’s drag is not a function of its velocity. It is a functional of its whole history, and the practical question is how much of that history still matters.

What it does to a decelerating particle #

The cleanest experiment is a release: give a particle a velocity in still fluid and watch.

Quasi-steady drag alone gives an exponential decay with a time constant of the particle’s mass over the drag coefficient — 77 milliseconds for a hundred-micron particle in air. The full unsteady force gives something else.

The history term’s share of the total force grows with time, from three per cent at twenty milliseconds to 39 per cent at four hundred. The reason is that the quasi-steady drag is proportional to a velocity that is collapsing, while the history term is an integral over everything already accumulated and collapses far more slowly.

So the memory does not fade into irrelevance; it takes over.

The direction, which is the opposite of the intuition #

The natural expectation is that an extra force means extra drag and therefore a shorter stopping distance. It is wrong, and getting it wrong is what this computation did first.

The history term opposes the acceleration, whichever sign the acceleration has. For a particle being accelerated by gravity it retards the approach to terminal velocity, which is the textbook statement. For a particle decelerating it retards the deceleration — the fluid around the particle is still moving at the velocity the particle had a moment ago, and it pushes.

The particle therefore travels 12 per cent further than the quasi-steady answer, and at four hundred milliseconds it is still moving 3.8 times as fast.

A power law where an exponential was expected #

The shape of the decay changes as well as its size. The quasi-steady velocity is an exponential and disappears off the bottom of any plot; the full one is a power law, with a local exponent of about 1.87 over this window and an asymptote at three halves.

That is the signature of a memory with no time constant. An exponential decay is what a system with a single relaxation time does; a power law is what a system does when it has a continuum of them, and a diffusive memory has exactly that — the vorticity shed at every earlier instant is at a different distance now, and each contributes on its own time scale.

A sphere released at unit speed into a fluid of kinematic viscosity 1.5·10⁻⁵ m²/s has a Stokes time of 77.18 ms. Without the history term it stops after 77.15 mm; with it, after 86.55 — 12.2 per cent further, from a term whose peak share of the force is 38.6 per cent. The history term is therefore worth more than a tenth of the answer on a quantity as blunt as a distance.

The practical consequence is stark. There is no time after which the particle has stopped. The quasi-steady model says it has essentially stopped after three or four time constants; the real one says it is still moving, more slowly, for ever, and that the distance it eventually covers is a convergent integral of a divergent-looking process.

Why the kernel is a square root, in one paragraph #

The exponent is not fitted and it is worth deriving in words, because knowing where it comes from says where it stops.

An acceleration of the particle puts a sheet of vorticity on its surface. That vorticity diffuses outwards, so a time $\tau$ later it occupies a shell of thickness $\sqrt{\nu\tau}$. The force it can still exert on the particle is proportional to the velocity gradient it maintains at the surface, which is its strength divided by its thickness — so it falls as $1/\sqrt{\tau}$.

Summing over all the earlier accelerations gives the convolution, and the kernel is that $1/\sqrt{\tau}$. Every step of that argument is diffusion and none of it is about spheres, which is why the same exponent appears at a wall, in a vortex core and in a heat-transfer problem.

It also says exactly where the argument fails: as soon as the shell is no longer thin compared with the particle, the geometry enters and the exponent is no longer a half. That happens at a time of order $d^2/\nu$, which for the particle here is 0.6 seconds — comparable with the run, which is why the measured exponent has not reached its asymptote.

What truncating the history costs #

Anybody computing this has to decide how much history to keep, and the cost of that decision is worth having as a number rather than as a worry.

Keeping one step of history is wrong by 7.9 per cent. Keeping a hundred is wrong by 5.2. Keeping three hundred is wrong by 3.0. Only at a thousand steps — which is most of the run — does the error fall below a tenth of a per cent.

Truncating the kernel costs more than truncating an exponential would. Seven windows on the same problem, integrated at a coarser step whose own full answer is 83.53 mm rather than the 86.55 above — the 3.5 per cent between the two steppings is the same slow convergence, seen from the other side:

Keeping a thousand times as much history removes only 99 per cent of the error, and the first three hundred steps — a window a quarter of the run long — still leave 3.0 per cent on the table. That is a much slower convergence than any exponential kernel would give, and the reason is the same one: the weight given to the oldest contribution inside the window is the largest one still being dropped when the window is shortened. Truncating an inverse square root throws away the part of the integral that converges most slowly.

The cost of not truncating is that the computation is quadratic in the number of steps, which is why particle-laden flow codes either drop the term, truncate it, or approximate the kernel by a sum of exponentials — the last of which is the honest option and turns a convolution into a set of extra state variables.

What the solver computed, and how it was checked #

The Basset-Boussinesq-Oseen equation for a sphere much denser than the fluid, integrated forward with the history evaluated by direct summation over the stored accelerations. Nothing is approximated except the quadrature, and the quadrature’s convergence is what the truncation study measures.

Three checks. That the history term is a substantial share of the force — at least twenty per cent somewhere in the run, and it reaches 38.6, since a term that never exceeded a few per cent would not support the essay. That the stopping distance is longer rather than shorter, which is the claim about the direction and is the check that caught the sign error. And that the truncation error falls monotonically as more history is kept, which would catch a bug in the windowing.

The sign is the part worth recording. The first version of this check required the stopping distance to be shorter — on the reasoning that an extra force is extra drag — and it failed. The failure was correct: the history term opposes the deceleration for exactly the same reason it opposes an acceleration, and the fluid near a decelerating particle is a flywheel rather than a brake.

When it matters and when it does not #

The term is large here because the particle is dense and the fluid is not, and the general rule follows from where the density ratio sits in the three forces.

A heavy particle in a gas. The steady drag dominates and the history term is a correction — a large one, as measured here, but a correction. The particle time constant is long compared with the fluid’s own time scales.

A neutrally buoyant particle in a liquid. The added mass and the history term are comparable with the particle’s own inertia, and the history term routinely carries a third or more of the total. This is the case where dropping it is indefensible, and it is the case in most sediment and bubble calculations.

A bubble. The density ratio reverses, the particle’s own inertia is negligible, and the motion is governed almost entirely by the fluid terms. A bubble’s rise is a history problem from the first instant.

And any oscillating flow. In an oscillation of frequency $\omega$ the three terms scale as $1$, $\omega$ and $\sqrt{\omega}$, so the history term is the intermediate one — never dominant, never negligible, and always present at the order where two effects are being compared.

What this does to a measurement #

There is a practical consequence for anybody inferring a particle property from its motion, and it is the sort of error that produces a consistent bias rather than scatter.

Inferring a size from a stopping distance. A particle’s stopping distance is used to infer its inertia in impactor and settling measurements. Interpreting the measured distance with the quasi-steady formula gives a particle 12 per cent more inertial than it is, in the conditions computed here — always in the same direction, so averaging more runs does not help.

Inferring a viscosity from a decay. Fitting an exponential to a decay that is a power law returns a time constant that depends on the window fitted over: fit early and the answer is close to the quasi-steady one, fit late and it is much longer. The fit will look good in both cases, because a power law over a factor of two in time is an exponential to within the noise.

And inferring a drag law from a trajectory. A measured trajectory carries the history term whether the analysis contains one or not, so a drag coefficient fitted from an unsteady trajectory is a drag coefficient plus whatever the memory contributed — which is exactly the situation the instrument in the answer is about.

The way out in all three cases is the same: measure at a condition where the term is small, or model it, and say which was done.

The same kernel elsewhere #

The inverse square root is not the particle’s; it is diffusion’s, and it turns up wherever a diffusing quantity exerts a force back on what produced it.

At a wall. The wall the fluid is listening to is the same kernel written for a plane surface, where the delay a given height responds at is a height squared over a viscosity.

In a vortex core. What viscosity cannot take away has the same diffusion spreading a core as the square root of the elapsed time, which is the same statement about the same equation.

And in heat transfer. A body suddenly heated in a fluid loses heat at a rate falling as the inverse square root of time, for the same reason and with the same lack of a time constant.

What they share is that the memory is stored in a field rather than in a state variable, and a field with a diffusion equation on it has no characteristic time until something supplies a length.

Three memories in one small object #

It is worth separating the three terms by what kind of memory each is, because they are usually written as one force and they behave completely differently.

The steady drag has none. Force now, velocity now, and a particle whose velocity is restored to an earlier value feels exactly the force it felt then.

The added mass has an instantaneous one. Its force depends on the acceleration, which is a derivative, so it responds to the immediate past — but only to it. It is the particle’s version of the memoryless ideal flow in everything about the start, except one vector: its impulse over a manoeuvre depends only on the endpoints.

And the history term has an unbounded one. No time constant, no forgetting, and a contribution from every instant since the motion began.

A particle is therefore a small laboratory containing all three of kinds of response collected here at once, in a problem with an exact closed-form force — which is why it is the standard test case for anybody building a model of unsteady drag, and why the three terms are almost never all kept.

What the picture cannot show #

The figures draw a velocity and a force, and the object doing the remembering is neither: it is the vorticity in the fluid around the particle, spreading outwards. A picture of that would show a shell of rotation growing as the square root of time, and the force being drawn here is that shell’s effect integrated over the particle’s surface.

The truncation figure also cannot show the thing it is about, which is that the error is dominated by the oldest contributions dropped rather than by the number of them. Two windows keeping the same number of steps at different ages give different errors, and only one age is plotted.

Who found it, and when #

The history term is Basset’s and Boussinesq’s, independently, in the 1880s, in work on a sphere oscillating in a viscous fluid. It was regarded as a mathematical curiosity for most of a century because the flows people computed were steady.

Particle-laden flow made it practical. The Maxey-Riley equation of 1983 assembled the terms into the form used for a small particle in a non-uniform unsteady flow, and the numerical difficulty of the convolution has been an active problem since — the sum-of-exponentials approximations that make it affordable are from the 2000s.

Limits recorded rather than smoothed over #

Stokes flow only. The whole force is the unsteady Stokes solution, valid at particle Reynolds numbers well below one. At larger Reynolds number the wake is not a diffusing shell and the kernel is not an inverse square root; the finite-Reynolds-number history kernel falls off faster, and how much faster is a research question rather than a formula.

The particle is a rigid sphere. A drop or a bubble has an internal circulation and a mobile interface, which changes the steady drag by up to a factor of 3/2 — the surface that moves with the flow — and changes the history kernel too, in a way not computed anywhere here.

A uniform fluid at rest. The Maxey-Riley form has extra terms for a fluid that is itself moving and accelerating, and they are not small in a turbulent flow.

The history is summed directly, which is O(N²) and is why the step counts here are modest. The error from the quadrature is bounded by the truncation study rather than estimated independently.

One particle, one condition. The 39 per cent, the 12 per cent and the 3.8 are for a hundred-micron particle at 2500 kg/m³ in air released at one metre a second. Every one of them moves with the density ratio and with how the particle is being driven; the structure of the result does not.

And the 1.87 is a local exponent over this window, not the asymptote. The theory gives three halves at long times; reaching it takes far longer than the run, and quoting the measured number as the theoretical one would be exactly the mistake the range a real Reynolds number does not have is about.