(from my email)

In case anyone cares, I think I worked out the irreversibility thing (it's been good revision trying to figure it out, so may be good revision for you too:P):

1. All flow is time-reversible in the theoretical sense that if you reverse all particle's trajectories, you get another physical flow. This is not in general true, though, if you include the viscosity term (although it can be), as this is a term that is there to account for the degrees of freedom we aren't accouning for, so it is in general not time-reversible, just like friction isn't (balls don't just start spountaneously from rest and cooling the floor slightly, this is just the 2nd law).

2. What we usually talk about in fluid mechanics, though, is whether the flow is time-reversible in practice (I think this is called kinematic reversibility). What this means is whether or not I can perform the above theoretical operation of reversing the particle's trajectories, by performing a practically reasonable action. Such a reasonable action is usually changing the boundary conditions of the flow, as that is easy to do. In the case of the die drops, this is what they control, the surfaces of the cylinder Now, the boundary conditions determine a certain steady-state flow. The important thing about viscous flows (low Re), is that they reach stead state very quickly, so that most of the particles trajectories is spent in steady state, and so most of the trajectory is determined by the boundary condition (b.c.) we control. So we can just turn the cylinder one way, and then the other, and they will have very nearly retraced their steps (I think they turn the cylinder slowly because that keeps Re=UL/nu small). In higher Reynolds number flows, however, the time the system takes to reach the steady state set by the b.c.s is very significant. Therefore a significant portion of the particles' trajectories is spent in these transient periods. Now, I think the reason these transient periods break (practical) time reversal is because they are not deterimed completely by the b.c.s you control. I think this is because turbulence will most probably set off in them, and as we know, turbulence is random, i.e., out of your control, and thus you can't reverse that (significant) part of the flow. The reason I think turbulence will set off is that when you start moving the cylinder (in the experiment with the die), the no-slip boundary condition will cause a boundary layer, which is very thin due to the low viscosity. This sharp gradient in velocity means high vorticity, which as usual in high Re flows, will spread around, before getting dissipated eventually. This is just the standard onset of turbulence, actually.

Another note: I changed my mind on the pressure thing. I think Chris was right that the gradient of the pressure (though not the pressure itself) will change sign, for Stokes flow. This is actually just because what you do when you time-reverse is change the flow, and the flow determines the pressure distribution, so you can calculate what will happen to the pressure. If you do that for Stoke's equation, its gradient must change sign, as the viscous term does. Examples:

• In the Stokes flow around a sphere, if you reverse the flow, you clearly reverse the pressure gradient, as the drag is in the other direction.
• The cylinder with the die is interesting. If you compute the viscous term, it turns out to be zero (due to symmetry, although not obvious really), so the pressure gradient is zero, so its sign doesn't matter. Actually, if you are more careful, you realize that it can't be zero, there should be a radial component to make the fluid go in a circle, but that radial component would be of the same order, and canceling the inertial term, we ignored.

However, in non-Stokes flow, like say in steady-state inviscid flow, for which Bernouille's theorem holds, the gradient of the pressure doesn't change, as the other terms in NS equation don't either! Example:

• Venturi tube. Clearly, if you reverse the flow, the pressure is still high in the wide parts, and low in the constriction.