Angular momentum transfer in a bottle vortex A colleague and I were discussing the following fluid mechanics experiment, which most people have probably tried in their kitchen: take a bottle full of water (a 2-litre transparent soda pop bottle works well), invert it, and swirl it around to impart some angular momentum to the fluid. It will quickly form a vortex, which allows the bottle to drain rapidly.
The question that neither of us knows the answer to is this: is there some mechanism by which additional angular momentum is transferred into the system from outside as the water drains, or is it just that the initial angular momentum (which you add by swirling) becomes more concentrated?
Probably equivalently, if I were to cut the bottom off the bottle and then do the same experiment while continually adding water from the tap, would the vortex continue to spin indefinitely, or would the angular momentum in the system eventually be depleted, killing the vortex?*
If angular momentum is transferred to the system from outside the bottle, what is the mechanism for this? I assume that the Coriolis force is not relevant on such a small scale (the fact that a bottle vortex can be formed with either chirality would seem to back this up) so I would imagine that any transfer of angular momentum would be due to the fluid exerting a torque on the bottle it's in. Obviously frictional forces can only exert a torque in the direction that would slow the rotation down, but perhaps there are pressure gradient effects that can exert one in the opposite direction.
* this is obviously a rather easy experiment, and my only excuse for not doing it is that I don't have a pair of scissors handy.
 A: I think the answer to the question is that there is no mechanism by which angular momentum is transferred to the water from outside, after you stop the initial swirling. Rather, the vortex takes angular momentum from the draining water and
transfers it to that remaining in the bottle. I give a fuller explanation below -
it talks about water draining from a bath but the principles are the same.
Three basic physical principles apply: 
First, angular momentum is conserved: at any time during the draining of the 
bath the total angular momentum of the water (inside and outside the bath) 
stays the same. 
Second, the only energy input to the system is that available from gravity
as the water drains.
Third, the water is viscous, so that any rotating volume of water within the 
bath will tend to transfer its angular momentum to the body of water as a whole.
The presence of the vortex tube with a water/air surface shows that the force 
experienced by the water at that interface points radially outward and must be balanced by the normal pressure of the water in the bath. As water exits the bottom of the vortex, pressure from the body of bath water outside pushes more water inward towards the vortex centre. 
Conservation of angular momentum means that as this water moves in, its angular 
velocity and consequent rotational energy increase. At the same time viscosity 
acts to decrease the shear in water velocities, transferring angular momentum 
from the fast rotating water to the main body of water away from the vortex. 
The two effects combine to build rotational energy near the vortex but 
transfer angular momentum away from it.
What appears to the watcher to be the adding of angular momentum to the water is, 
if this model is right, rather its transfer from one part of the water volume to 
another. The liquid that falls out of the bottom end of the vortex has high angular energy but low angular momentum. Most of its original angular momentum remains in the liquid still in the vessel; as the amount of liquid in the vessel decreases so its angular speed increases.
The increase in angular energy is fuelled by the decrease in gravitational potential energy as the water falls down the vortex.
Some numbers. If the air tube at the centre of the vortex in a draining sink has a radius of 3mm, and the gradient of the vortex tube surface is 9, then the angular velocity of the water must be:
$\omega^2 \cdot 3\times 10^{-3} = 9g$
that is, $\omega=171$ rad/s; quite rapid. Conversely, a speck of water rotating with the same angular momentum at a distance of 20cm from the vortex centre would have 
$\omega=3.9\times 10^{-2}$ rad/s (2.2 deg/s); almost too slow to notice. Though the two specks 
of water have the same angular momentum, their rotational energies differ by a 
factor of 4325.
A: I know it's been ages but I wanted to add this.
It's more about the bottles geometry then anything. Your soda bottle is defining the path for the vortex. The more perfect this container is to a hyperbolic egg the easier it is for the vortex to do its thing while conserving angular momentum. 
If you could add water and let it flow out the bottom you could get it to spin indefinitely so long as the water addition didn't impact the current spin. It's like a kid on a swing, you only need to add a small force in the right place to conserve momentum but adding it in the wrong place with destroy momentum. So if you pour water in adding to the spin it will reinforce it combined with gravity. 
The vortex is truly fasinating! It seems to disobey entropy. it's behavior has everything to do with the geometry in the closed system or external influences in an open system . Look up viktor shuburger for some interesting reading.
