Answer to the Question 06/96
The question was:
Fish are swimming by waving their tail left-and-right, while the spermatozoa are spinning their tail. Why can't spermatozoa swim like fish?
(8/01) Y. Kantor: The problem has been solved (2/6/2001) by Nathanael Schaeffer a Ph.D. student at ENS Lyon (France) (e-mail nschaeff@ens-lyon.fr).
The solution: Navier Stokes equation which describes the motion of a body in a liquid contains nonlinear (inertia) term (u*grad)u and viscous term {nu}(grad)2u, where {nu} is the kinematic viscosity and u is the velocity of the fluid. The ratio between these to terms is called Reynolds number
Re=UL/{nu}
where U is the typical velocity, L are the typical dimensions of the object. For a fish L=0.1m, U=0.1m/s, {nu}=10-6, and consequently Re=104, while for spermatozoan L=10-6, U=10-3m/s, and cosequently Re=10-3. Thus in the case of fish, the non-linear term is very strong, while in the case of spermatozoa, viscosity (linear term) dominates. Navier Stokes equation without non-linear term is time-reversible! Consequently, a body cannot advance by executing a motion which is time reversible, like the wigling of the tail of a fish. (Imagine the motion of the fish tail filmed, and then viewed backwards - it will look the same!) Consequently, spematozoan must execute motion which is not time reversible, such as a screw-like motion of the tail.
Here is what Schaeffer wrote about the problem (slightly edited):
At very low Reynolds numbers, the Navier Stokes equation ... is linear. Thus Thus, (u,p) and (-u,-p) are solutions of this equation. (p is the pressure) If you add some forcing term f (through pressure) with < f > = 0 (mean value over a period), taking the mean of the Stokes equation gives you < u >= 0, and so no displacement.
So if you try to swim like a fish, it won't work, because when you're flapping up and then down, you're moving exactly in the opposite way, resulting in a zero net advance. If you do like a spermatozoid, you're "screwing" yourself in the fluid. never inverting your movement. In this case, you have a non zero < f >, and so a non zero < u > .
All this does not apply to the NS equation, because if you take the mean of this equation, you have the mean of a quadratic term... thus < f > = 0 does not necesserely lead to < u > = 0.
(8/05) Y. Kantor: An excellent reference on the subject was brought to our attention by A. Bhatia: it is a talk by E. M. Purcell published in the American Journal of Physics 45, 3 (1977); it can be found here in PDF format.
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