The actual equations do capture it. It's just the simplified minimal equations that result in 'paradox'.
A full model is complicated by the fact that your capacitors have an internal series resistance and leakage resistance, and that the leads and circuit board traces have resistance and inductance. Just like the pipes and valve has some resistance to flow, and some water might leak out or evaporate, and the water has inertia and nonzero viscosity, and turbulence will turn some of the motion to heat, and depending on the phase of the moon, the time of day, and the compass orientation of the barrels, the water may be pulled into a picometers-higher tide in one barrel. When you say "they equalize with half the water in each" you don't typically mention that the phase of the moon may be a factor.
Still, picometers are very different from "half the energy in the system".
We do store energy in water towers for example, so it is pretty surprising and unintuitive that if you put two large water tanks, one full and one empty, right next to each other, open a valve between them allowing their levels to equalize, then assuming there is no distance and you used teflon coated valves you lose... half of the energy as they equalize!
I certainly wouldn't have thought so. I'd have thought you keep 70%-95+% of the energy.
Actually the oscillation explanation didn't match my intuition at all, because I would have thought the water flows from high to low until the point of equalization and then stops flowing, without oscillation.
I get that this doesn't happen, but I would have thought it would!
> Actually the oscillation explanation didn't match my intuition at all, because I would have thought the water flows from high to low until the point of equalization and then stops flowing, without oscillation.
It might very well do that. You will just have lost half the energy already to heat from friction with the piping and due to water's viscosity.
> it is pretty surprising and unintuitive that if you put two large water tanks, one full and one empty, right next to each other, open a valve between them allowing their levels to equalize, then assuming there is no distance and you used teflon coated valves you lose... half of the energy as they equalize!
If you did that, they would equalize - very briefly - and then the second tank would fill higher and higher, until it's (nearly) full. Then the reverse begins.
This is very similar to a pendulum. Just that our intuition about pendulums is better than for near-frictionless transfer of fluids in connected systems.
> I'd have thought you keep 70%-95+% of the energy.
Well, the potential energy being zero at the bottom of each tank is arbitrary. If both tanks are inside a water tower, you might keep 99% of the "useful" energy even if you let them equalize, because the height above ground is greater than the height above "tank bottom". Maybe this is the source of some confusion here?
You lose 50% of the energy relative to the bottom of the water tanks (actually, the level you're using the water to perform work). If you did this with two 20 foot high tanks, but on towers 40 feet off the ground, then you retain 45 / 50 (the ratio of the changed center of mass) == 90% of the energy.
The oscillation bit happens, but it doesn't dominate. Water analogies can be misleading because water has intrinsic properties (eg turbulence ~= resistance) which aren't always significant in an electronic circuit. To make the equivalent of an LC-dominant circuit with (open) water tanks, you'd need something like a high-momentum turbine in the transfer pipe.
When water equalizes in the "water seeks its own level" demonstration (in a series of connected glass tubes of various shapes, the water is at the same level in the whole system, as here: https://www.physics.purdue.edu/demos/display_page.php?item=2... ), does the system lose a lot of potential energy as the water equalizes?
In normal situations, you would not lose half the energy, or hydropower would be totally bunk (actually maybe not, considering the efficiency of burning fossil fuels…) But in this degenerate case, you cannot ignore the losses.
In a hydro power plant, the water turbine constitutes the "friction" so, if there were no other losses, the half of the energy "lost" from the system is precisely what your generator gave you.
Are you sure? I was going to add that the intuition comes from demonstrations we might have seen, where in a series of tubes, regardless of shape, water will be at one level. As in this picture:
A full model is complicated by the fact that your capacitors have an internal series resistance and leakage resistance, and that the leads and circuit board traces have resistance and inductance. Just like the pipes and valve has some resistance to flow, and some water might leak out or evaporate, and the water has inertia and nonzero viscosity, and turbulence will turn some of the motion to heat, and depending on the phase of the moon, the time of day, and the compass orientation of the barrels, the water may be pulled into a picometers-higher tide in one barrel. When you say "they equalize with half the water in each" you don't typically mention that the phase of the moon may be a factor.