5.2. Low Reynolds number[[Student version, December 8, 2002]] 151
Unmixing The full equations of fluid mechanics are rather complicated, but it’s not hard to guess
the minimal response of a fluid to the shearing force applied in Figure 5.2b. Since everything is
uniform in they, zdirections, we can think of the fluid layer as a stack of thin parallel sheets, each of
thickness dx,and apply Equation 5.9 to each layer separately. Denoting the relative velocity of two
neighboring sheets by dvz,each sheet pulls its neighbor with a force per area off/A=−ηdvdzx(x).
In particular, the sheet of fluid immediately next to a solid wall must move with the same speed
as the wall (theno-slip boundary condition), since otherwisevwould have an infinite derivative at
that point, and the required viscous force would be infinite too.
Since every sheet of fluid moves uniformly (does not accelerate), Newton’s Law of motion says
the forces on each slab must balance. Thus each must exert on its neighbor above the same force
exerted on it by its neighbor below, ordvdzx(x)must be a constant, independent ofx.Afunction
with constant derivative must be alinearfunction. Sincevmust go fromv 0 on the top plate to
zero on the bottom plate, we findvz(x)=(x/d)v 0.
Thus a volume element of water initially at (x 0 ,z 0 )movesintimetto (x 0 ,z 0 +(x 0 /d)v 0 t). This
motion is what stretches out an initially spherical blob of ink (Figure 5.2b). If we reverse the force
pulling the top plate for an equal timet,though, we find that every fluid element returns to exactly
its original starting point. The blob reassembles; if it had originally been stretched so far as to
appear mixed, it now appears to “unmix” (Figure 5.1).
Suppose now that we don’t insist on steady motion, and instead apply a time-dependent force
f(t)tothe top plate. Now the forces on each slab needn’t balance; the net force instead equals
the mass of fluid in the slab times its acceleration, by Newton’s Law of motion. As long as the
force is well below the viscous critical force, though, this correction will be negligible and all the
same conclusions as before apply: Once the top plate has returned to its initial position, each fluid
element has also returned. It’s a bit like laying a deck of cards on the table and pushing the top
card sideways, then back.
It’s not even necessary to apply the exact time-reversed force in order to return to the starting
configuration. Regardless of whether the return stroke is hard and short, or gentle and long, as
soon as the top plate returns to its original position, so have all the fluid elements (apart from a
small amount of true, diffusive mixing).
Time reversal The “unmixing” phenomenon points up a key qualitative feature of low Reynolds-
number fluid flow. To understand this feature, let’s contrast such flows with the more familiar world
of Newtonian mechanics.
If we throw a rock up in the air, it goes up and then down in the familiar way:z(t)=v 0 t−^12 gt^2.
Now imagine a related process, in which the positionzr(t)isrelated to the original one by “time
reversal”; that is,zr(t)≡z(−t)=−v 0 t−^12 gt^2. The time-reversed process is also a legitimate
solution of Newton’s laws, albeit with a different initial velocity from the original process. Indeed
wecan see directly that Newton’s Law has this property, just by inspecting it: Writing the force as
the derivative of a potential energy gives−ddUx=md
(^2) x
dt^2 .This equation contains two time derivatives,
and so is unchanged under the substitutiont→−t.
Asecond example may reinforce the point. Suppose you’re stopped at a traffic light when
someone rear-ends you. Thus starting at timet=0,the positionx(t)ofyour head suddenly
accelerates forward. The force needed to make this happen comes from your headrest; it’s also
directed forward, according tof=mdd^2 tx 2 .Nowimagine another process, in which your head moves
along the time-reversed trajectoryxr(t)≡x(−t). Physically,xrdescribes a process where your