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Sandbox Physics
Microswimming · A course with experiments

L3 · Chapter 5

Distance per stroke is an area in shape space

The distance per cycle is a line integral around the stroke loop in the plane of the two arm lengths. Stokes’ theorem in that plane turns it into an area integral of a single field, which acts like a curvature. Across a small loop the field is nearly constant, so the distance is a coefficient times the signed area. Reverse the loop and the sign flips; a figure eight’s two lobes nearly cancel.

Sandbox Physics ·

Draw the stroke as a loop

Plot the left arm length horizontally and the right arm length vertically. Every shape of the three-sphere swimmer is a point in this plane, and every periodic stroke is a closed curve. Najafi and Golestanian’s four-stroke square runs from the corner at ten and ten micrometres to six and ten, then to six and six, then to ten and six, and back. That is a counter-clockwise loop enclosing sixteen square micrometres.

Purcell drew the same kind of picture for his two-hinged swimmer, with the two hinge angles as axes and the stroke as a loop. He posed the direction of travel as a puzzle for the reader and remarked that the answer is not obvious. The loop picture turns that puzzle into a calculation.

From a line integral to an area

Chapter 3 wrote the distance as a path integral. For a closed loop it becomes a line integral around the curve. Green’s theorem converts it into an integral, over the enclosed region, of one scalar field: the rate at which the second coefficient changes along the first arm, minus the rate at which the first coefficient changes along the second.

This field tells you how much distance each patch of shape space is worth. It is largest where both arms are short and the spheres couple strongly, and it fades as the arms lengthen. A loop’s distance is the field summed over everything the loop encloses, counted positive for a counter-clockwise loop and negative for a clockwise one.

The benchmarks check the conversion numerically. Around the default square, the line integral gives 0.172091 micrometres; summing the field on a 60 by 60 grid over the enclosed square gives 0.172088. They agree to two parts in a hundred thousand.

ΔX=∮∂S(A1 dL1+A2 dL2)=∬SF dL1 dL2,F=∂A2∂L1−∂A1∂L2\Delta X=\oint_{\partial S}\left(A_{1}\,dL_{1}+A_{2}\,dL_{2}\right)=\iint_{S}F\,dL_{1}\,dL_{2},\qquad F=\frac{\partial A_{2}}{\partial L_{1}}-\frac{\partial A_{1}}{\partial L_{2}}

Worked example

The area law for the default square

When a loop is small compared with the arms, the field barely changes across it, and the integral becomes the field at the loop’s centre times the signed area. Golestanian and Ajdari give the coefficient in their equation 13. For three equal spheres it reduces to one third of the radius times a bracket of inverse squared arm lengths.

At the centre of the default square both arms are eight micrometres long. The coefficient is 0.009115 per micrometre, and with sixteen square micrometres of area the estimate is 0.146 micrometres. The full calculation gives 0.172, so the area law falls about fifteen percent short for this loop.

Two approximations separate the numbers. The loop is large: its side is half the arm length, so the field varies across it. The spheres are also not points, and the area law is only the leading term in radius over arm length. The benchmarks isolate this second effect with tiny loops; there the relative error shrinks along with the radius-to-arm ratio, approaching five sevenths of it. The Lab’s small-loop preset circles the point where both arms are nine micrometres, with a radius of 1.2 micrometres and an area of 4.32 square micrometres. It moves 0.0344 micrometres per cycle against an area-law estimate of 0.0311.

ΔX≈KA,K=a3[1ℓ12+1ℓ22−1(ℓ1+ℓ2)2]K(8,8)=1 μm3[264−1256]μm−2≈0.009115 μm−1,KA=0.009115×16≈0.146 μm\begin{aligned}\Delta X&\approx K\mathcal{A},\qquad K=\frac{a}{3}\left[\frac{1}{\ell_{1}^{2}}+\frac{1}{\ell_{2}^{2}}-\frac{1}{(\ell_{1}+\ell_{2})^{2}}\right]\\K(8,8)&=\frac{1\,\mu\mathrm{m}}{3}\left[\frac{2}{64}-\frac{1}{256}\right]\mu\mathrm{m}^{-2}\approx0.009115\,\mu\mathrm{m}^{-1},\qquad K\mathcal{A}=0.009115\times16\approx0.146\,\mu\mathrm{m}\end{aligned}

Worked example

Smooth strokes and the phase lag

Real actuators often move smoothly. Let each arm oscillate around its mean length with its own amplitude, both at the same frequency, with a phase difference between them. The loop in shape space is then an ellipse, and its area is pi times the two amplitudes times the sine of the phase difference.

Golestanian and Ajdari’s equation 16 gives the mean velocity of this swimmer. It is largest when the arms are a quarter cycle apart and vanishes when they move in step or in exact opposition, because the ellipse then collapses to a line.

Take arms with a mean length of eight micrometres, amplitudes of two micrometres, one cycle per second and a quarter-cycle lag. The loop is a circle of radius two micrometres, with an area of about 12.6 square micrometres, and the mean speed comes to about 0.11 micrometres per second.

ui(t)=dicos⁡(ωt+φi),V0=K2 d1d2 ω sin⁡(φ1−φ2)=0.0091152×2×2×2π  μm s−1≈0.11  μm s−1u_{i}(t)=d_{i}\cos(\omega t+\varphi_{i}),\qquad V_{0}=\frac{K}{2}\,d_{1}d_{2}\,\omega\,\sin(\varphi_{1}-\varphi_{2})=\frac{0.009115}{2}\times2\times2\times2\pi\;\mu\mathrm{m\,s^{-1}}\approx0.11\;\mu\mathrm{m\,s^{-1}}

Reversed loops and figure eights

Running the same loop clockwise flips the sign of every area element, so the distance flips exactly. The Lab’s reversed preset moves 0.172 micrometres to the left per cycle. The energy per cycle is 6.93 attojoules in both directions, since dissipation cares nothing for the sense of the loop.

A figure eight has two lobes traversed in opposite senses, so its signed area is zero and the area law predicts zero. The field, though, differs between the lobes. In the Lab’s figure-eight preset, the counter-clockwise lobe lies where both arms are between eight and eleven micrometres, and the clockwise lobe where they are between five and eight. The short-arm lobe is worth more, and the swimmer drifts 0.018 micrometres to the left per cycle, about a tenth of the square’s step, while dissipating 9.34 attojoules. Area alone predicts the distance well only when the loop is small enough for the field to be uniform across it.

An experiment to think through

Predict, change, explain

  1. Before choosing “A small loop follows the area law”, predict whether Distance per cycle will come out above or below the Small-loop area law readout.
  2. Run the small-loop preset, then “A figure eight nearly cancels”. Finally drag one corner of the square loop toward shorter arms and watch both readouts.
  3. Explain the gap between the two readouts using the brightness of the shape panel, which shows how much distance each patch of shape space is worth.
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. The default square is run clockwise. What are its signed area and its distance per cycle?

Read an explained answer

Minus sixteen square micrometres and minus 0.172 micrometres. Every area element changes sign, so the distance changes sign exactly.

2. A small square loop has its side doubled while its centre stays put. How does the distance per cycle change?

Read an explained answer

It grows about fourfold, because the area grows fourfold while the coefficient at the centre stays the same. The factor departs from four once the loop is large enough for the field to vary across it.

3. Two harmonic arms move in step, with zero phase difference. How far does the swimmer go per cycle?

Read an explained answer

Nowhere on average. The ellipse collapses to a line segment with zero area, equation 16 gives zero mean velocity, and the stroke is reciprocal.

4. Using the area law, estimate the distance per cycle for the default arms with a stroke of two micrometres instead of four.

Read an explained answer

The square now runs between ten and eight micrometres and is centred at nine. The coefficient there is about 0.00720 per micrometre and the area is four square micrometres, so the estimate is about 0.029 micrometres.

K(9,9)=1 μm3[281−1324]μm−2≈0.00720 μm−1,KA≈0.00720×4≈0.029 μmK(9,9)=\frac{1\,\mu\mathrm{m}}{3}\left[\frac{2}{81}-\frac{1}{324}\right]\mu\mathrm{m}^{-2}\approx0.00720\,\mu\mathrm{m}^{-1},\qquad K\mathcal{A}\approx0.00720\times4\approx0.029\,\mu\mathrm{m}

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. 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.

  3. 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.

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