Skip to main content
Sandbox Physics
Microswimming · A course with experiments

L2 · Chapter 4

Three beads, two arms and a mobility matrix

Write each sphere’s velocity as its own Stokes drag response plus the point-force flow from the other two. Add the condition that the swimmer exerts no net force, and the two prescribed arm rates. That gives three linear equations for three unknown forces. For equal spheres, all of the forward motion comes from the coupling between spheres: switch it off and the centre stays exactly still.

Sandbox Physics ·

The swimmer on paper

Najafi and Golestanian proposed the swimmer in 2004: three spheres of equal radius on a line, joined by two arms whose lengths the swimmer controls. The arms are thin rods that feel no drag. Their paper calls the middle sphere sphere 1 and the end spheres 2 and 3. This course numbers the spheres 1, 2 and 3 from left to right, so the middle sphere is sphere 2.

Their stroke has four steps: shorten the left arm, shorten the right arm, lengthen the left arm, lengthen the right arm. In the plane of the two arm lengths this traces a square once per cycle. With arms of ten micrometres and a stroke of four micrometres, each arm moves between ten and six micrometres.

Positions along the line count from left to right. The swimmer’s position is the centroid of the three spheres. Over a full cycle every sphere moves by the same net amount, so the choice of reference point changes the motion within a cycle and leaves the distance per cycle alone.

Three equations for three forces

Each sphere pushes on the fluid with some force. Its velocity is that force times its own Stokes mobility, plus the flow created at its centre by the other two spheres. For spheres far apart compared with their size, that flow is the point-force flow along the line from Chapter 2, falling off as one over the separation. Golestanian and Ajdari write this as their equations 1 to 3, and equation 6 of Najafi and Golestanian has the same form.

Two more conditions close the problem. Nothing external pushes the swimmer, so the three forces add up to zero. The arm rates are prescribed, so the differences between neighbouring sphere velocities are known. That makes three linear equations for three unknown forces. The matrix that turns forces into velocities is the mobility matrix. It is symmetric, and while the spheres stay well apart it is positive definite, which keeps the dissipated power positive.

The point-force form fails when spheres come close. Once two spheres are closer than one and a half radii, the matrix stops being positive definite and the model would create energy from nothing. The Lab therefore keeps every arm at least three radii long.

vi=fi6πηa+∑j≠ifj4πη ∣xi−xj∣f1+f2+f3=0,v2−v1=L˙1,v3−v2=L˙2\begin{aligned}v_{i}&=\frac{f_{i}}{6\pi\eta a}+\sum_{j\neq i}\frac{f_{j}}{4\pi\eta\,|x_{i}-x_{j}|}\\f_{1}+f_{2}+f_{3}&=0,\qquad v_{2}-v_{1}=\dot{L}_{1},\qquad v_{3}-v_{2}=\dot{L}_{2}\end{aligned}

Worked example

Solve one instant of the first stroke

Take the start of the first stroke: both arms ten micrometres long, the left arm shortening at thirty micrometres per second and the right arm fixed. The spheres have a radius of one micrometre and sit in water.

Solving the three equations gives sphere velocities of plus 19.70, minus 10.30 and minus 10.30 micrometres per second. The middle and right spheres move together, as the fixed right arm requires, and the left sphere closes in on them at thirty. The forces the spheres exert on the fluid are about plus 420, minus 228 and minus 191 femtonewtons, which add up to zero within rounding.

The centroid velocity is the average of the three, minus 0.303 micrometres per second. Now switch off the coupling between spheres and solve again. Each force is then just its own sphere’s drag, so the zero-force condition makes the three velocities add up to zero: plus 20, minus 10 and minus 10. The centroid stays exactly still. For equal spheres, all of the swimming comes from the flow each sphere creates at the others.

(v1,v2,v3)=(+19.70, −10.30, −10.30)  μm s−1,(f1,f2,f3)≈(+420, −228, −191)  fNX˙=13(v1+v2+v3)=−0.303  μm s−1,X˙uncoupled=13(20−10−10)=0\begin{aligned}(v_{1},v_{2},v_{3})&=(+19.70,\,-10.30,\,-10.30)\;\mu\mathrm{m\,s^{-1}},\qquad(f_{1},f_{2},f_{3})\approx(+420,\,-228,\,-191)\;\mathrm{fN}\\\dot{X}&=\tfrac{1}{3}(v_{1}+v_{2}+v_{3})=-0.303\;\mu\mathrm{m\,s^{-1}},\qquad\dot{X}_{\mathrm{uncoupled}}=\tfrac{1}{3}(20-10-10)=0\end{aligned}

Worked example

From one instant to a full cycle

Repeat the solve at every instant and integrate. The middle sphere’s steps over the four strokes are minus 1.353, plus 1.439, plus 1.439 and minus 1.353 micrometres. The centroid’s steps are smaller and lopsided: minus 0.020, plus 0.106, plus 0.106 and minus 0.020. Both lists add up to the same 0.172 micrometres to the right, since every sphere ends the cycle displaced by the same amount.

Gains and losses differ because the coupling depends on the arm lengths at the moment of each stroke. In the first and fourth strokes the arm that stays fixed is ten micrometres long; in the second and third it is six. The cycle takes 0.533 seconds, so the average speed is 0.323 micrometres per second, and each cycle dissipates 6.93 attojoules.

What scales and what drops out

Viscosity multiplies every entry of the mobility matrix by the same factor, so it cancels from the velocities. The arm rates enter linearly, so the pace cancels from the distance, as Chapter 3 showed. The numerical benchmarks confirm both. Changing the viscosity leaves the distance per cycle unchanged to rounding, and scaling the radius, arms and stroke together by 2.5 multiplies the distance by 2.5.

Energy follows a different rule. Dissipated power is force times velocity, and both grow with the pace, so the energy per cycle grows in proportion to the arm speed and to the viscosity. Running the arms at forty micrometres per second instead of ten costs four times the energy for the same distance.

ΔX(sa, sL1, sL2)=s ΔX(a, L1, L2),Ecycle∝η W\Delta X(sa,\,sL_{1},\,sL_{2})=s\,\Delta X(a,\,L_{1},\,L_{2}),\qquad E_{\mathrm{cycle}}\propto\eta\,W

An experiment to think through

Predict, change, explain

  1. With “Four strokes swim forward” running, predict whether a smaller Sphere radius makes Distance per cycle longer or shorter.
  2. Lower Sphere radius, then raise Arm speed. Finally drag an end sphere along the axis to make a stroke of your own.
  3. Explain which change alters Distance per cycle and which alters only the period and the energy, using the coupling between spheres.
Stroke a three-sphere swimmer

The three-sphere Lab couples three equal spheres through point-force Stokes flow in unbounded water. It leaves out walls, neighbours and thermal noise.

Try explaining it yourself

1. Show that three equal spheres with no hydrodynamic coupling stay in place on average.

Read an explained answer

Without coupling, each force is the drag coefficient times that sphere’s velocity. The forces add up to zero, so the velocities add up to zero, and the centroid velocity is zero at every instant. Whatever the arms do, the centroid stays where it started.

2. In the first stroke the middle sphere moves minus 1.353 micrometres, while the centroid moves only minus 0.020. Why are they so different?

Read an explained answer

Within a stroke the spheres move by different amounts. The sphere on the moving arm travels about four micrometres relative to the other two, so it moves about plus 2.65 while the others move minus 1.35. The centroid averages the three, and only after a full cycle do all spheres share the same net step.

3. Double the sphere radius from one to two micrometres with the arms and stroke unchanged. Roughly how does the distance per cycle change?

Read an explained answer

It roughly doubles. For spheres small compared with the arms, the swimming comes from coupling terms proportional to the radius. The factor is only approximate, because the next correction grows with the ratio of radius to arm length.

4. Triple the viscosity. What happens to the forces, the distance and the energy per cycle?

Read an explained answer

Every force triples, since the same velocities now need three times the push. The distance per cycle stays the same, and the energy per cycle triples.

Sources to read alongside

  1. Najafi & Golestanian, A simplest swimmer at low Reynolds number: three linked spheres, arXiv:cond-mat/0402070v1 (2004); Physical Review E 69, 062901 (2004)

    Preprint version 1 read; the journal version was not checked. Equations 1 to 11, Figure 3 and the in-text example below Equation 10 are used. The curve of Figure 4 was not digitised.

  2. Golestanian & Ajdari, Three-sphere low Reynolds number swimmer revisited, arXiv:0711.3700v1 (2007); Physical Review E 77, 036308 (2008)

    Preprint version 1 read; the journal version was not checked. Equations 1 to 6, 9, 12, 13 and 16, Section IV on the far-field flow and footnote 23 are used.

FeedbackSomething broken, confusing, or missing on this page?Tell us
What kind of feedback?

Sent with this page’s address, its Lab ID, the site language and your device type. Please leave out personal details. Feedback is kept for up to 12 months.Privacy policy