Entropy decrease using Stoke's flow?

There is a video of an experiment from University of Mexico using corn syrup (highly viscous) and water. They are "mixed together" in a container by turning a crank but when the crank is turned in the opposite direction they neatly un-mix.

I understand that the two liquids actually do not blend, but how can the syrup droplets be returned to very close to their initial state. It looks as if the entropy is greatly reduced. Is the increase in temperate of the liquids (through the turning of the crank) the answer?

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It's actually the University of New Mexico, and I think they were using all corn syrup, no water. But good question. – David Z Jun 13 '11 at 19:24

Entropy never increases during this "mixing process" as it normally would, because it is not really a mixing process. The only thing that happens is a huge distortion of the distribution of the colors, but no information about it is lost. Compare it to performing a Fourier transform on some function: this also results in something that looks completely different, sometimes chaotic and high-entropy-like, but it's really just an invertible transformation.

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To add some slightly more technical language to this, if you take the Navier-Stokes equations, and make it so that the viscous terms are dominant, you can show that the flow is reverisible. This means that highly viscous flow has no arrow of time, and therefore, must have constant entropy. – Jerry Schirmer Jun 13 '11 at 23:36
@Jerry the NS equation in this case has no time derivative and the behavior is defined entirely by the boundary conditions. The laws of physics are reversible, but time is necessary for the evolution of a low entropy state to a high entropy state to happen. In the case of the behavior defined by the BCs, the BCs can simply be played in reverse and get back to where you started. – Alan Rominger Jun 13 '11 at 23:57

My simple explanation is that the entropy never increased in the first place, and like a good magic trick, they showed you a state where it looked like mixing had occurred, but it in fact had not. In the false "mixed" state the colors were extremely systematically layered, looping around the apparatus several times.

Wikipedia actually does a far better job than I could with this, but I'll reference their equations anyway. Here is the typical Navier-Stokes equation beautifully written from the Wikipedia article. You find in the article about Stokes flow that in the case of a very low Reynolds number, the effective derivative part, the left hand side, is all equal to zero. Also, I think that $\mathbf{f}=-\rho g$ downward, but I'd like someone to confirm that fact.

$$\overbrace{\rho \Big( \underbrace{\frac{\partial \mathbf{v}}{\partial t}}_{ \begin{smallmatrix} \text{Unsteady}\\ \text{acceleration} \end{smallmatrix}} + \underbrace{\mathbf{v} \cdot \nabla \mathbf{v}}_{ \begin{smallmatrix} \text{Convective} \\ \text{acceleration} \end{smallmatrix}}\Big)}^{\text{Inertia (per volume)}} = \overbrace{\underbrace{-\nabla p}_{ \begin{smallmatrix} \text{Pressure} \\ \text{gradient} \end{smallmatrix}} + \underbrace{\mu \nabla^2 \mathbf{v}}_{\text{Viscosity}}}^{\text{Divergence of stress}} + \underbrace{\mathbf{f}}_{ \begin{smallmatrix} \text{Other} \\ \text{body} \\ \text{forces} \end{smallmatrix}} = 0$$

In addition to this, the mass balance in an incompressible flow is the following.

$$\boldsymbol{\nabla}\cdot\mathbf{v}=0$$

The solution to these equations includes $\mathbf{v}$ (vector velocity) and $p$ (scalar pressure). The important part to understand is that the time derivative is gone. In this case, the second you stop putting torque into the machine, the fluid is stationary. So it's actually entirely the boundary conditions that define the movement of the fluid at any given time. I think it's fair to them make the following statement about the position and velocity.

$$\int_0^{t_1} \mathbf{v} dt = \mathbf{r}$$

This gives the "mapping" of one particle in the position to the position at the end of the experiment. The entire idea is that they can produce the conditions such that the following is also met.

$$\int_{t_1}^{t_2} \mathbf{v} dt = -\mathbf{r}$$

Ordinarily this would be the exact thing prevented by entropy. Basically, the equation leads to the creation of too much information to be exactly reversed by simple set of boundary conditions. I don't have any good way to write this, but the boundary conditions are sufficiently reversed in time and the rate at which it's turned also doesn't matter as a result of the form of the equations.

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