A force without the flow that makes it

Here is a question that sounds as though it needs a calculation. A sphere is released, free of any force, in a flow that is not uniform — a pipe flow, say, where the velocity varies across the section. How fast does it go?

The answer that everybody gives, and that is wrong, is: at the speed of the fluid where the sphere is. The answer that is right requires no solution of the flow around the sphere, no boundary condition applied on its surface, and no equation solved that has not already been solved in every introductory course.

It goes at the average of the undisturbed flow over the sphere’s own surface.

The identity that gets it #

The tool is the Lorentz reciprocal theorem, and it is the second thing linearity buys after uniqueness. For any two Stokes flows in the same region of fluid,

$$\oint_S \mathbf{u}\cdot(\hat{\boldsymbol\sigma}\cdot\mathbf{n})\,dS \;=\; \oint_S \hat{\mathbf{u}}\cdot(\boldsymbol\sigma\cdot\mathbf{n})\,dS.$$

Both integrals are over the same closed surface, and in neither of them does a field appear with its own stress. The content is that the work the second flow’s stresses would do on the first flow’s velocities equals the work the first flow’s stresses would do on the second’s.

It follows in three lines from two facts and nothing else: the Stokes equations are linear, and the viscous stress tensor is symmetric. Take the divergence of $\mathbf{u}\cdot\hat{\boldsymbol\sigma}$ and it comes out as $2\mu\,\mathbf{e}:\hat{\mathbf{e}}$, which is symmetric in the two flows. So the difference of the two products is divergence-free, and its integral over any closed surface enclosing no singularity vanishes.

The mistake that makes it look broken #

The first pair chosen for that check was a translating sphere and a sphere held in a straining flow. Both are Stokes flows, both are exactly known, and both sides of the identity came out at about $10^{-8}$ — differing from each other by sixty per cent.

Sixty per cent is a large number and it looks like a broken theorem. It was a degenerate pair. A translating sphere’s disturbance is a vector harmonic of degree one; a stresslet’s is of degree two; their contraction integrates to exactly zero over any sphere centred on both, by orthogonality. Both sides of the identity were zero, both were being computed as quadrature noise, and the ratio of two noises is meaningless.

The symptom is worth recognising because it looks exactly like a failure: a relative residual that is enormous and an absolute value that is tiny. The fix is to test on a pair that can do work on one another, and the magnitudes are now printed alongside so that the reader can see the test was not vacuous.

The version with no integral in it #

Before using the theorem, it is worth seeing what it says in the simplest case, because that version is intuitive and the general one is not.

Apply a force at one point in a viscous fluid and measure the velocity it produces at another. Then apply the second force at the second point and measure the velocity at the first. The theorem says those two are related by

$$\mathbf{F}_1\cdot\mathbf{u}_2(\mathbf{x}_1) = \mathbf{F}_2\cdot\mathbf{u}_1(\mathbf{x}_2).$$

Push here and measure there, and the answer is the one that pushing there and measuring here would have given. It holds for any pair of points, any pair of directions, and any bodies in between.

Faxén’s law, in one page #

Now the payoff. Take the two flows to be the problem of interest and the problem already solved.

The problem of interest: a force-free sphere sitting in an ambient flow $\mathbf{u}^\infty$, which satisfies the Stokes equations on its own and is whatever the pipe or the shear was doing before the sphere arrived.

The auxiliary problem: the same sphere translating at unit speed through fluid at rest, whose answer is Stokes’ $6\pi\mu a$ and is the first thing anybody computes.

Apply the identity on the sphere’s surface. On it, the real flow’s velocity is the sphere’s own rigid translation, and the auxiliary flow’s traction is uniform — Stokes drag is distributed uniformly over a sphere, which is the special fact that makes this work. Everything collapses, and what is left is

$$\mathbf{U} = \langle \mathbf{u}^\infty \rangle_S + \frac{\mathbf{F}}{6\pi\mu a},$$

the surface average plus whatever an applied force would contribute. For a force-free particle the second term is zero and the first is the whole answer.

The surface average has a closed form for anything with a Taylor series. Expanding the ambient about the sphere’s centre, the odd moments vanish by symmetry and the second moment of a sphere is $a^2\delta_{ij}/3$, so

$$\langle \mathbf{u}^\infty \rangle_S = \mathbf{u}^\infty(0) + \frac{a^2}{6}\nabla^2\mathbf{u}^\infty + O(a^4).$$

For an ambient flow that is quadratic — Poiseuille is — the remainder is identically zero, and the law is exact rather than asymptotic.

The number, in a tube #

In a round tube the ambient is $u = u_{\max}(1 - r^2/R^2)$ and its Laplacian is $-4u_{\max}/R^2$, everywhere, with no dependence on position. So the correction does not depend on where in the tube the particle is. Only on how big it is.

$$U = u(r_c) - \frac{2}{3}\,\frac{a^2}{R^2}\,u_{\max}.$$

That number is the difference between a tracer and a thing being measured, and it is a systematic error rather than a scatter. A velocimetry technique that seeds a pipe with particles of a known size can correct for it exactly; one that seeds with a distribution of sizes cannot, because the correction goes as the square of the radius and the mean of a square is not the square of the mean.

What it costs to be wrong about this #

The lag is a fraction of a per cent for the particles velocimetry actually uses, and it would be fair to ask why it is worth a page. Three reasons, and the third is the one that decides it.

It is systematic. Random error in a velocity measurement falls as the square root of the number of particles. This does not fall at all: every particle in the field is slow, all of them by the same fraction of their local velocity, and averaging a million of them gives the same wrong answer with a smaller error bar on it.

It is largest where the gradients are. The correction is the Laplacian of the ambient, so it vanishes in a uniform flow, vanishes in a simple shear — a shear is linear, and a linear field has no Laplacian — and is largest exactly where a measurement is most interesting, which is in a boundary layer or a shear layer. A technique validated in a free stream is validated where the effect is zero.

And it has the wrong sign for the thing it is usually used to find. Seeding a boundary layer with particles and reading their speeds gives a profile that is systematically slow near the wall, where the curvature is greatest. The wall shear stress inferred from it is therefore systematically low, and the drag it implies is low with it — an error in the conservative direction for a lift estimate and the dangerous direction for a drag one.

What else falls out of the same identity #

Faxén’s is the tidy application; the theorem’s real value is how many other things it does without any new solve.

A drag from a slightly different shape. Deform a sphere a little and the change in its drag can be had from the undeformed sphere’s traction integrated against the deformation, with no solution of the deformed problem at all. That is Brenner’s result and it is the standard way small departures from sphericity are priced.

A swimming speed without solving for the swimmer. A body that deforms its own surface moves at a velocity that is a surface integral of that deformation weighted by the traction of the towed problem. Taylor’s waving sheet can be got this way, and so can the general result that a time-reversible stroke goes nowhere.

And a suspension’s viscosity. The extra stress a rigid particle contributes to a sheared fluid is its stresslet, and the stresslet is what the reciprocal theorem extracts from the straining problem. Einstein’s five halves is one of these integrals.

Symmetry of the resistance matrix, and why a bacterium can swim #

The mobility statement above — push here, measure there — has a compact form for a whole body, and the compact form contains a result worth more than the identity that produced it.

A rigid body in Stokes flow relates the force and torque on it to its translation and rotation through a single matrix,

$$\begin{pmatrix}\mathbf{F}\\ \mathbf{T}\end{pmatrix} = -\mu \begin{pmatrix} \mathsf{A} & \mathsf{C}^{\mathsf{T}} \\ \mathsf{C} & \mathsf{B}\end{pmatrix} \begin{pmatrix}\mathbf{U}\\ \boldsymbol{\Omega}\end{pmatrix},$$

and the reciprocal theorem is exactly the statement that this matrix is symmetric. Six by six, twenty-one independent entries rather than thirty-six, and the symmetry is not an approximation.

The diagonal blocks are familiar: $\mathsf{A}$ is the drag, $6\pi a$ for a sphere, and $\mathsf{B}$ is the rotational drag, $8\pi a^3$. It is the off-diagonal block that carries the interesting content. $\mathsf{C}$ couples torque to translation and, by the symmetry, couples force to rotation with the same coefficient. Spin the body and it moves; push it and it spins; and the two effects have one number between them rather than two.

A body with any mirror symmetry has $\mathsf{C} = 0$ identically. A sphere, an ellipsoid, a disc, a rod: spin any of them about any axis and they stay where they are, however fast and however long. The coupling is non-zero only for a body that is chiral — a helix, a corkscrew, a body with a handed twist and no plane of reflection.

Which is the whole of how a bacterium swims. A flagellum is a rigid helix turned by a rotary motor in the cell wall, at a hundred revolutions a second or so, and the helix’s coupling coefficient converts that rotation into a thrust. Nothing about the motion is reciprocal in time, because rotation is not a reciprocal stroke: reverse it and the organism goes backwards, which is exactly what a stroke that undoes itself cannot do. The cell body counter-rotates to balance the torque, and the swimming speed follows from the two resistance matrices solved together, at a few tens of microns a second for the numbers a bacterium has.

The symmetry then makes a prediction that is easy to test and slightly startling. Because $\mathsf{C}$ is shared, towing the same helix through the fluid must make it rotate, at a rate fixed by the same coefficient that governs its propulsion. A corkscrew dragged through syrup turns itself, and how fast it turns per unit towing speed equals how fast it advances per unit imposed spin. Two experiments, one number, and the equality is a theorem rather than a coincidence.

And the handedness matters exactly as it should. A left-handed helix rotated the same way as a right-handed one swims in the opposite direction, because $\mathsf{C}$ changes sign under reflection — which is why artificial magnetic microswimmers are manufactured with a specified handedness, and why their direction is reversed by reversing the field’s rotation rather than by anything about the fluid. Propulsion at zero Reynolds number is a property of a body’s chirality, and the reason one number describes both of its faces is the symmetry this essay’s identity establishes.

Two spheres, and why suspensions are hard #

There is one more thing the theorem gives cheaply and it is the reason a suspension of particles is not a simple problem.

Two spheres settling side by side in a viscous fluid fall faster than either would alone, because each one is falling through the downwash of the other. The reciprocal theorem, applied to the pair, gives the leading interaction without solving the two-sphere problem: it is the Stokeslet of one evaluated at the other, and it decays as $1/r$ — the slowest decay of any interaction in physics that is not electrostatic.

That slowness is the whole difficulty. A pair correction that fell as $1/r^3$ would be a neighbour effect and a suspension would be a gas of independent particles with a small correction. A correction that falls as $1/r$ is not a neighbour effect at all: summing it over a uniform distribution of particles gives an integral that diverges, and making it converge takes an argument about the suspension as a whole rather than about pairs.

So the dilute limit of a suspension is not the limit of few neighbours; it is the limit of few particles per volume, however far apart they are. That is why Einstein’s coefficient stops being the measurement at a few per cent by volume rather than at a few tens, and why the second term in the expansion took another sixty years to compute.

Why it stops the moment there is inertia #

Every step of the proof used the linearity of the Stokes equations, and the nonlinear term destroys all of it. There is no reciprocal theorem for the Navier–Stokes equations, and the reason is easy to state: reciprocity is a symmetry, and the convective term breaks it by picking a direction — the direction the fluid is going.

The practical boundary is the same one creeping flow itself sits inside, which is a good deal lower than a Reynolds number of one. A particle in a tube at Re = 0.1 is well inside it; the same particle at Re = 10 is not, and its lag is then a different and much harder problem with a lift force in it as well.

What the picture cannot show #

The tube wall is outside the theorem. Faxén’s law is derived for a sphere in an unbounded fluid, and the ambient it uses is whatever the flow was doing before the sphere arrived. Putting a wall a few radii away adds a correction of the opposite sign and of order $(a/R)^3$ on the axis — separable from the $(a/R)^2$ term over the range drawn here, and not separable beyond about a third.

The particle is rigid and spherical. A deformable drop obeys a different law with its viscosity ratio in it, and a non-spherical particle rotates as well as translating, which brings in a second Faxén law for the angular velocity that is not drawn here.

And nothing here is Brownian. A particle small enough for the surface-average correction to be negligible is usually small enough for thermal motion to dominate its trajectory entirely, which is a different subject with a different mathematics.

Who found it, and when #

Lorentz gave the theorem in 1896, in a paper largely about something else, and it sat almost unused for fifty years. Faxén derived his laws in 1922 by a direct expansion rather than by reciprocity — the reciprocal derivation, which is the one above and is a page rather than a chapter, came later. The modern use of the theorem as a general-purpose tool for getting forces without flows dates from the 1960s and is largely Brenner’s.

The surprising connection is with a piece of applied mathematics that is not fluid mechanics at all. The result that a sphere samples the average of a field over its surface rather than the field at its centre, with the correction being the Laplacian times $a^2/6$, is the same statement as the mean value property of harmonic functions and its failure for functions that are not harmonic. A sphere in a viscous fluid is, to that order, an instrument that measures the spherical mean — and the correction it makes is exactly the amount by which the ambient fails to be harmonic.

Where the ladder goes next #

Beside this rung is the minimum-dissipation theorem, which is the other thing linearity buys, and which fails at the same place for the same reason. Above it are the two applications this collection now makes of the identity: what a suspension does to a viscosity and a swimmer that cannot go backwards.

Below it is the world with no inertia, which is the regime all of this holds in and only in, and how many things a flow must be told, which is where the boundary conditions the theorem contracts against come from.