The tracer that is not one

A particle carried by a flow is pulled towards the local fluid velocity with a time constant, and the whole of the subject rests on that constant. Divide it by the time the flow has to turn and the result is the Stokes number, which decides whether the particle turns too — and below a critical value of an eighth a body in a stream collects nothing at all, however many particles are thrown at it.

The time constant is quoted everywhere in the same form:

$$\tau = \frac{\rho_p d^2}{18\mu}.$$

It is derived by balancing the particle’s own inertia against Stokes drag, and it is right for the case it was derived for. That case is a particle much denser than the fluid it is in, and the derivation stops being right the moment the two densities are comparable.

The two terms that were left out #

Write the momentum equation for a small sphere moving through a flow and there are more forces on it than drag and weight. Two of them are not corrections at all when the densities are comparable.

Added mass. A body accelerating through a fluid has to accelerate some of the fluid with it, which is why getting going costs a force that steady motion does not. For a sphere the added mass is exactly half the displaced fluid, which is a quantity this collection computes rather than quotes. It appears in the particle’s equation as an extra $\tfrac12\rho_f$ alongside the particle’s own $\rho_p$.

The pressure gradient of the undisturbed flow. Wherever the fluid itself is accelerating there is a pressure gradient, and that gradient acts over the particle’s volume whether the particle is there or not. It is what makes a cork rise in accelerating water, and it contributes a force $\rho_f V\,\mathrm{D}u/\mathrm{D}t$.

Carrying both through gives the Maxey–Riley form:

$$\tau = \frac{(2\rho_p + \rho_f)\,d^2}{36\mu}, \qquad \beta = \frac{3\rho_f}{2\rho_p + \rho_f}.$$

The relaxation time now contains the fluid’s density as well as the particle’s, and $\beta$ — which is the factor multiplying the fluid’s acceleration in the particle’s equation — is three for a bubble, one for a neutrally buoyant particle and tends to zero for anything much denser.

Why the omission survived #

Because for the case everybody uses, the two formulas agree.

A water drop in air has $\rho_p/\rho_f = 830$. The usual relaxation time is $830\,d^2/18\nu$ and the full one is $1661\,d^2/36\nu$, which differ by six hundredths of a per cent. A dust grain in air is the same story with a larger ratio. So every calculation of droplet impaction, spray penetration and inertial separation done since the 1920s has used a formula that is right for its case, and the case is nearly all of the applications.

The omission shows up the moment the densities are comparable, and the two commonest such cases are exactly the two where people most want a tracer.

The neutrally buoyant particle, which is not perfect but is nearly #

Take $\rho_p = \rho_f$. The usual formula gives $\tau = d^2/18\nu$; the full one gives $3d^2/36\nu = d^2/12\nu$. The true relaxation time is exactly three halves of the advertised one, and the usual formula is a third low.

That matters for particle image velocimetry, where the whole measurement rests on the seeding particles following the flow. Choosing a particle size from a relaxation-time criterion using the usual formula produces particles fifty per cent slower to respond than the calculation says — which is a systematic under-estimate of every velocity gradient the technique measures, largest exactly where the gradients are largest.

But the density-matched particle has a compensating virtue that no other particle has. Its $\beta$ is exactly one, and the radial drift in a vortex goes as $(1 - \beta)$:

$$v_r = (1 - \beta)\,\tau\,\frac{U_\theta^2}{r}.$$

At $\beta = 1$ that vanishes identically. A neutrally buoyant particle does not drift at all — not approximately, exactly — and the reason is not that it has no inertia. It has a relaxation time half as long again as the usual formula gives. The reason is that its inertia, its added mass and the pressure gradient cancel to the last term.

The bubble, which is the opposite of a tracer #

Now take $\rho_p \to 0$. The usual formula gives $\tau = 0$: a bubble has no inertia, so it follows the flow perfectly and has no dynamics of its own.

The full formula gives $\tau = d^2/36\nu$, which is not zero, and $\beta = 3$, which is the largest value $\beta$ can take. So a bubble is the furthest thing from a tracer available: it responds to three times the fluid’s own acceleration, and its drift in a vortex is inward at $-2\tau U_\theta^2/r$.

The consequence is visible in any glass of fizzy water that has been stirred, and it is the reverse of what a vortex does to the fluid itself. Bubbles collect on the axis. Heavy particles do the opposite — a stirred cup of tea gathers its leaves at the centre for a different reason, involving the Ekman layer at the bottom, but a suspended heavy particle in a free vortex migrates outward.

And the same effect decides where cavitation starts. Nuclei — microscopic bubbles that seed the cavity — are drawn into the core of a tip vortex by exactly this drift, so the region with the lowest pressure is also the region that has concentrated its own nuclei. Tip-vortex cavitation appears earlier than the inception threshold for the blade surface would suggest, and this is one of the reasons.

The Stokes number of a bubble is not zero #

The practical statement is about the group rather than about the drift. The Stokes number of a bubble computed from the usual relaxation time is zero, so a bubble is predicted to follow every streamline exactly and to be uncollectable by any body in any flow.

Computed properly, a 100 µm air bubble in water has $\tau = d^2/36\nu = 2.8\times10^{-4}$ s, and in a flow with a turning time of a millisecond that is a Stokes number of 0.28 — well above the critical eighth. A body in that flow collects bubbles, which the usual formula says is impossible, and which is the basis of froth flotation, of bubble-column contactors and of every gas-liquid separator that works by inertia.

That is the sharpest form of the error. It is not a quantitative correction; it converts an answer of “never” into an answer of “routinely”.

The arithmetic, laid out #

The three cases in a table, because the pattern is easier to see than to describe:

Reading down the ratio column: the correction is invisible at the top, nineteen per cent for sand, fifty per cent for a tracer, and infinite for a bubble. Reading down the $\beta$ column: the drift reverses between sand and a tracer, and there is no configuration in which a bubble drifts outward.

The one row that has no entry in the usual formula at all is the last. A model that returns zero for a case is not being slightly inaccurate about it; it is declining to describe it, and the decline is silent.

The same two terms, weighed rather than spun #

The drift above is the corrections acting in a rotating flow. Put them in a stationary fluid with gravity instead and they produce a result that is easier to check and just as surprising, and it turns out to be the same expression.

Release a particle from rest. Before it has any speed there is no drag, so the only forces are its weight and the buoyancy, and the mass being accelerated is its own plus the added mass:

$$\left(\rho_p + \tfrac12\rho_f\right)V\,\frac{dv}{dt} = \left(\rho_p - \rho_f\right)V g .$$

Rearranged, the initial acceleration is

$$\frac{dv}{dt}\bigg|_{0} = g\,\frac{\rho_p - \rho_f}{\rho_p + \tfrac12\rho_f} = (1 - \beta)\,g ,$$

exactly the same factor $(1-\beta)$ that governs the drift in a vortex. That is not a coincidence: in both cases the particle is responding to the difference between what its own inertia demands and what the surrounding fluid’s acceleration supplies, and gravity is simply the fluid acceleration a hydrostatic pressure gradient corresponds to.

Read the three cases off it. A heavy particle has $\beta \to 0$ and starts falling at $g$, as expected. A neutrally buoyant one has $\beta = 1$ and does not start at all, which is the same exact cancellation as its zero drift. And a bubble has $\beta = 3$, so it starts upward at $2g$ — twice gravity, from rest, in a fluid where nothing is pushing it but its own buoyancy.

That factor of two is the added mass made visible, and it is the cleanest demonstration of the term available. Without it a bubble would be predicted to accelerate at $g\rho_f/\rho_p$, which for air in water is eight hundred times gravity — an absurdity that the simple formula produces and that the half-a-displaced-volume of added mass removes.

The correction also decides the rest of the trajectory, not merely its start. A bubble’s rise is a balance between buoyancy, drag and the added-mass term, and because the effective inertia is dominated by the fluid rather than by the bubble, the approach to terminal velocity is fast and the transient is governed entirely by the fluid’s properties. A bubble’s dynamics contains almost nothing about the bubble — its mass has dropped out of every term that matters — which is exactly why treating it as a massless tracer is such an appealing mistake.

The missing history term makes the same point one level further down. The Basset integral decays only as $t^{-1/2}$, so a particle never quite forgets its own acceleration, and its size relative to the others is set by the density ratio in the same way: negligible for a dust grain in air, comparable with everything else for a bubble in water. The corrections this essay restores are exactly the ones that matter when the particle stops being much heavier than what it is in, and the term still missing is the third member of the same family.

What is still missing from the equation #

The two terms restored above are not the whole of the Maxey–Riley equation, and it is worth naming what is still absent.

The Basset history term is an integral over the particle’s whole past, arising because the vorticity a particle sheds while accelerating diffuses away slowly and keeps acting on it. It has the awkward property of decaying only as $t^{-1/2}$, so it never quite goes away, and it is comparable with the added-mass term whenever the densities are comparable — which is to say in every case this essay is about.

The Faxén corrections account for the flow’s curvature over the particle’s finite size, and they matter when the particle is not small compared with the velocity gradients — near a wall, or in a vortex whose core is a few particle diameters across.

And Stokes drag itself is wrong by a per cent at a particle Reynolds number of 0.054 — the threshold a group of order one does not supply, which is a threshold the drifting particles above cross whenever they are doing anything interesting.

So the corrected relaxation time is a better model and not a complete one. What it does establish — and what no amount of further correction will undo — is the sign structure: $\beta$ is greater than one for anything lighter than the fluid and less for anything heavier, and the drift reverses at exactly the density match.

What the picture cannot show #

Every particle here is a rigid sphere. A bubble is not: it deforms, it has internal circulation, and its drag law is different — Hadamard and Rybczynski’s, giving two thirds of Stokes’ drag for a clean bubble. Surfactant contamination makes it behave as a rigid sphere again, which is why bubble experiments are notoriously irreproducible and why the clean-bubble result is the exception rather than the rule.

The drift is a leading-order result. It is the first term in an expansion in the relaxation time, so it holds when the particle nearly follows the flow — exactly the regime where the drift is smallest and takes longest to observe.

Nothing here is turbulent. The vortex the drift is computed in is steady and axisymmetric. In a turbulent flow the same mechanism produces preferential concentration — heavy particles clustering in the strain regions between vortices, bubbles clustering in the cores — and quantifying it needs the statistics rather than a single vortex.

And $\beta$ assumes a uniform particle. A hollow glass sphere, a droplet with a solid inclusion and a bubble with a surfactant shell all have effective densities that differ from their apparent ones, and $\beta$ is sensitive to the difference exactly where the drift is smallest.

The measurement that has to trust this #

Every optical flow measurement — particle image velocimetry, laser Doppler anemometry, particle tracking — works by measuring the velocity of something that is not the fluid and calling it the fluid’s. The whole technique rests on the relaxation time being short compared with the flow’s own time scales, and the criterion is invariably quoted with the simple formula.

Two consequences follow from doing it properly.

The particle size limit is more restrictive than advertised. Requiring $\mathrm{St} < 0.01$ for a one-per-cent tracking error, on a flow with a millisecond time scale, gives a maximum diameter with $\sqrt{18}$ in it on the usual formula and $\sqrt{12}$ on the true one for a density-matched particle — a maximum size smaller by twenty per cent, and a factor of one and a half in the relaxation time it was chosen to bound.

And bubbles are not usable as tracers at all, which is worth saying because bubble seeding is attractive: bubbles are easy to make, cheap, and highly visible. Their $\beta$ of three means they report three times the fluid’s acceleration and drift into every vortex core in the field, so a velocity map made from bubbles is systematically wrong in exactly the regions a reader is looking at the map to find.

Who found it, and when #

Basset, Boussinesq and Oseen each wrote a version of the equation of motion for a sphere in an unsteady flow between 1885 and 1927; the added-mass and pressure-gradient terms are in all of them. Maxey and Riley assembled the modern form in 1983 and put the Faxén corrections on it, and the equation carries their names because they wrote the version that could be used.

What is odd is that the simplified relaxation time survived alongside it. Every textbook derives $\tau = \rho_p d^2/18\mu$ from a force balance, notes in a footnote that added mass may be included, and then uses the simple form for the rest of the chapter. The footnote is correct and the sentence that would matter — that the simple form gives a bubble no dynamics at all — is not usually there.

The surprising connection is that the perfect tracer is a cancellation. A density-matched particle does not fail to drift because it has no inertia; it has a longer relaxation time than the usual formula gives it. It fails to drift because three separate forces — its own inertia, the added mass it drags, and the pressure gradient it sits in — sum to zero at exactly that density. A quantity that is zero by cancellation behaves quite differently from one that is zero by absence, and the difference shows up as soon as anything perturbs the cancellation: a slightly non-spherical particle, a slightly mismatched density, or a Basset term.

Where the ladder goes next #

Above this rung is preferential concentration in turbulence, where the same $(1-\beta)$ drift acting in a field of vortices sorts particles by density into sheets and cores. It is one of the most studied effects in two-phase flow and it needs the statistics of a turbulent field, which this collection can synthesise but cannot solve.

Beside it sits cavitation inception, which the drift moves, and the critical Stokes number, which the corrected relaxation time changes by fifty per cent for a tracer and by everything for a bubble. Below it is what a parcel does in the first instant, which is the fluid’s own version of the same question.