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Microswimming · A course with experiments

L2 · Chapter 2

Stokes flow answers every push at once

Remove the inertial terms from the Navier–Stokes equations and what remains, the Stokes equations, is linear and has no time derivative. The flow at any instant is fixed by how the boundaries move at that instant. A sphere one micrometre in radius pulled through water at thirty micrometres per second feels about 0.57 piconewtons, and the flow it stirs fades only as one over the distance.

Sandbox Physics ·

Dropping the inertial terms

Lauga and Powers begin their review with the Navier–Stokes equations for an incompressible fluid. One side holds the inertia, density times the acceleration of a small parcel of fluid. The other side holds the pressure gradient and the viscous force. Measure velocities in units of a typical speed and lengths in units of a typical size, and the ratio of the two sides is the Reynolds number.

When that ratio is a few ten-thousandths, the inertial side can be set to zero. What remains is a balance of forces at every point: pressure pushes, viscosity resists and nothing accelerates. Together with the condition that the fluid is incompressible, these are the Stokes equations.

The review also points to a force built from viscosity and density alone, about one nanonewton in water. The drag on a slow body is roughly viscosity times speed times size, so the body’s Reynolds number is roughly the pulling force divided by this nanonewton. Anything pulled by much less than a nanonewton moves at low Reynolds number, whatever its size. Every force in this course is at least a thousand times smaller.

ρ(∂u∂t+u⋅∇u)=−∇p+η∇2u    ⟶    0=−∇p+η∇2u,∇⋅u=0\rho\left(\frac{\partial\mathbf{u}}{\partial t}+\mathbf{u}\cdot\nabla\mathbf{u}\right)=-\nabla p+\eta\nabla^{2}\mathbf{u}\;\;\longrightarrow\;\;\mathbf{0}=-\nabla p+\eta\nabla^{2}\mathbf{u},\qquad\nabla\cdot\mathbf{u}=0

Two properties that do the work

The Stokes equations contain no time derivative. Time enters only through the boundaries: when a sphere moves, the surrounding fluid rearranges at once into the one flow that matches the sphere’s present velocity. Stop the sphere and the flow everywhere stops in the same instant. This is Purcell’s remark about the past, written as mathematics.

The equations are also linear. Double every boundary velocity and every fluid velocity, pressure and force doubles too. Reverse them all and the whole flow reverses. Lauga and Powers call this kinematic reversibility, and Chapter 3 builds the scallop theorem on it.

Linearity is also what a viscosity measurement relies on. In a Newtonian liquid, the stress needed to shear a thin gap grows in proportion to the shear rate, and the ratio is the viscosity. A rotational rheometer finds that ratio from a torque and a rotation speed. Viscosity is the only property of the fluid that appears in the Stokes equations, so it is the one number to measure before modelling a swimmer.

Worked example

A one-micrometre sphere in water

Stokes’ law gives the drag on a slowly moving sphere: six times pi, times the viscosity, the radius and the speed. Take a radius of one micrometre, a speed of thirty micrometres per second and the viscosity of water, one millipascal second.

The drag comes to about 0.57 piconewtons. The power spent against it is drag times speed, about seventeen attowatts, which is seventeen billionths of a billionth of a watt.

Lauga and Powers also write the law the other way round, as a mobility: the speed gained per unit of applied force. The same mobility sets the thermal jitter. Multiply it by the thermal energy at room temperature and the sphere’s diffusion coefficient comes out near 0.21 square micrometres per second.

F=6πηaU=6π (10−3 Pa s)(10−6 m)(3×10−5 m s−1)≈5.65×10−13 NP=FU≈1.7×10−17 W,D=kBT6πηa≈0.21 μm2 s−1\begin{aligned}F&=6\pi\eta aU=6\pi\,(10^{-3}\,\mathrm{Pa\,s})(10^{-6}\,\mathrm{m})(3\times10^{-5}\,\mathrm{m\,s^{-1}})\approx5.65\times10^{-13}\,\mathrm{N}\\P&=FU\approx1.7\times10^{-17}\,\mathrm{W},\qquad D=\frac{k_{\mathrm B}T}{6\pi\eta a}\approx0.21\,\mu\mathrm{m^{2}\,s^{-1}}\end{aligned}

Worked example

The flow a point force leaves behind

Far from the sphere, its flow looks like the flow of a single point force pushing on the liquid. Lauga and Powers call this solution the Stokeslet. Its speed falls off only as one over the distance, more slowly than the gravitational pull of a mass.

The flow is lopsided. Along the line of the force the fluid moves twice as fast as at the same distance off to the side. For our sphere, ten micrometres ahead on its path the fluid moves at about 4.5 micrometres per second, fifteen percent of the sphere’s own speed. Even 150 micrometres ahead it still moves at one percent.

The real sphere’s flow is the Stokeslet plus a correction that falls off as the cube of the distance and makes the fluid stick to the sphere’s surface. The three-sphere model in the next chapters keeps only the point-force part. That works when the spheres are small compared with the arms between them, and the size of the shortcut can be checked: adding the leading finite-size correction moves the default swimming step by about four percent.

u(r)=18πηr(I+rrr2)⋅F,u∥=F4πηr=3a2r U\mathbf{u}(\mathbf{r})=\frac{1}{8\pi\eta r}\left(\mathbf{I}+\frac{\mathbf{r}\mathbf{r}}{r^{2}}\right)\cdot\mathbf{F},\qquad u_{\parallel}=\frac{F}{4\pi\eta r}=\frac{3a}{2r}\,U

Thermal jitter is larger than one stroke

The diffusion coefficient from the worked example has a sobering consequence. During the 0.53 seconds of one default three-sphere stroke cycle, a lone sphere of this size wanders about half a micrometre by thermal motion along any one direction. The swimmer’s deterministic step in that cycle is 0.17 micrometres.

Steps add up in proportion to the number of cycles, while the wandering grows only as the square root of that number. After roughly eight cycles the drift overtakes the jitter, and from then on the swimming shows. The swimmer models in this course compute that deterministic mean, which a real micrometre-sized swimmer reveals after averaging over many cycles or many swimmers.

An experiment to think through

Predict, change, explain

  1. In the three-sphere Lab, predict how Distance per cycle and Energy dissipated per cycle change when you triple the fluid viscosity.
  2. Open the Fluid group and raise Fluid viscosity, then raise Arm speed. In the viscosity Lab, double the rotation speed with the Newtonian sample and compare the torque.
  3. Explain why viscosity and arm speed change the energy and the time per cycle, while the distance per cycle stays where it was.
Stroke a three-sphere swimmer Measure viscosity in a sheared gap

The three-sphere Lab couples three equal spheres through point-force Stokes flow in unbounded water. It leaves out walls, neighbours and thermal noise. The rheometer Lab models steady flow of Newtonian and shear-thinning liquids in a sheared gap.

Try explaining it yourself

1. What is the drag on a sphere twice as large, moving at the same speed?

Read an explained answer

The drag grows in proportion to the radius, so it doubles to about 1.13 piconewtons. The power spent against it doubles as well.

2. Estimate the Reynolds number of the default three-sphere swimmer, with arms ten micrometres long changing at thirty micrometres per second.

Read an explained answer

Density times speed times length, divided by viscosity, gives about 0.0003. Inertia is smaller than viscosity by more than three orders of magnitude.

Re=(103 kg m−3)(3×10−5 m s−1)(10−5 m)10−3 Pa s=3×10−4\mathrm{Re}=\frac{(10^{3}\,\mathrm{kg\,m^{-3}})(3\times10^{-5}\,\mathrm{m\,s^{-1}})(10^{-5}\,\mathrm{m})}{10^{-3}\,\mathrm{Pa\,s}}=3\times10^{-4}

3. How far ahead of the moving sphere does the flow along its path drop to one percent of the sphere’s speed?

Read an explained answer

Set three halves of the radius over the distance equal to one hundredth. The distance is 150 radii, 150 micrometres here, so a neighbour several arm lengths away still feels the push.

4. In the rheometer, the torque doubles when the rotation speed doubles. What does that tell you about the liquid?

Read an explained answer

The stress grows in proportion to the shear rate, so the viscosity is constant over that range. That is a Newtonian liquid, the kind the Stokes equations assume.

Sources to read alongside

  1. Purcell, Life at low Reynolds number, American Journal of Physics 45, 3 (1977)

    Pages 3 to 6 were read: the Reynolds-number estimates of Figure 3, the coasting example of Figure 4, the scallop theorem of Figure 6, the two-hinge swimmer of Figure 7 and the flexible oar and corkscrew of Figure 9. Pages 7 to 11 were not read for this course.

  2. Lauga & Powers, The hydrodynamics of swimming microorganisms, arXiv:0812.2887v1 (2008); Reports on Progress in Physics 72, 096601 (2009)

    Preprint version read. Sections 3 and 4 supply the Stokes equations, the point-force flow and the proofs of the scallop theorem; Sections 5 to 7 cover resistive-force theory, flexible filaments, walls and synchronisation.

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