154 Chapter 5. Life in the slow lane: the low Reynolds-number world[[Student version, December 8, 2002]]
Figure 5.6:(Schematic.) A hypothetical microscopic swimmer trying to make progress by cycling between forward
and backward strokes of its paddles. On the first stroke the paddles move backward relative to the body at relative
speedv,propelling the body through the fluid at speedu.Onthe second stroke the paddles move forward at relative
speedv′,propelling the body backward at speedu′.Then the cycle repeats. The progress made on the first stroke
is all lost on the second stroke; reciprocal motion like this cannot give net progress in low Reynolds-number fluid
mechanics.
the body through the fluid requires a force determined by a viscous friction coefficientζ 0 .Moving
the paddles through the fluid requires a force determined by a different constantζ 1.
Initially the body’s center is located atx=0.Then it pushes its paddles backward (toward
negativex)relative to its body at a relative speedvfor a timet.Next it pushes the paddles forward
at a different relative speedv′to return them to their original location. The cycle repeats. Your
friend suggests that by making the “recovery” stroke slower the “power” stroke (that is, taking
v′<v), the creature can make net progress, just as in a rowboat.
Example a. The actual speed at which the paddles move through the water depends both on
the givenvand on the speeduof the body, which you don’t know yet. Findufor
the first half of the cycle.
b. How far and in what direction does the body move in the first stroke?
c. Repeat (a,b) for the second (return) stroke.
d. Your friend proposes to choosevandv′to optimize this process. How do you
advise him?
Solution:
a. The velocity of the paddles relative to the surrounding fluid is the relative velocity,
−v,plusu.Balancing the resulting drag force on the paddles against the drag force
on the body givesu=ζ 0 ζ+^1 ζ 1 v.
b. ∆x=tu,forward, whereuis the quantity found in (a).
c.u′=−ζ 0 ζ+^1 ζ 1 v′,∆x′=t′u′.Wemust taket′u′=tuif we want the paddles to
return to their original positions on the body. Thus ∆x′=(tv/v′)u′=−∆x.
d. It won’t work. The answers to (b) and (c) always cancel, regardless of what we
take forvandv′.For example, if the “recovery” stroke is half as fast as the “power”
stroke, the corresponding net motion is also half as fast. But such a recovery stroke