6 ms·

And the resolutions are the same as in the capacitor example: either friction in the pipes damps the oscillation, or the inertia of the water keeps it oscillati

And the resolutions are the same as in the capacitor example: either friction in the pipes damps the oscillation, or the inertia of the water keeps it oscillating between the two barrels indefinitely and half the energy is kinetic.

sopooneo 6y ago
That, along with the parent comment, clinched it for me. Thank you.
- CamperBob2 6y ago
  I think the Wikipedia article's final sentence is probably the easiest way to visualize the question if you want to stick to electrical terms. If there's just one capacitor and you short it out by closing the switch, it's not hard to see how all of the energy vanishes, rather than complicating the question by referring to half of it.
  - wrycoder 6y ago
    Energy doesn’t vanish. Part of it is dissipated in the wire’s resistance and the rest is radiated as EM energy in a ring-down oscillation.
    - CamperBob2 6y ago
      Yes, and that's the point; to answer the paradox you have to explain where the seemingly-"vanished" energy goes, and that's relatively straightforward in the single-capacitor case. It's then equally straightforward to see how the same explanation applies to the more contrived case of two capacitors.
kurthr 6y ago
Well, technically there would also be some inductance... which as an externally coupled field would radiate, and that inherently causes loss!
- chronolitus 6y ago
  tomorrow's popsci headline: "Do barrels of water emit harmful 5G radiation? The answer may surprise you!"
- red75prime 6y ago
  And the loss will be in the form of gravitational waves.
  - throwphoton 6y ago
    Argh is this real or not? Would energy radiated away as a result of inductance be analogous to energy radiated away by gravitational waves in the "water pressure analogy"?
    - adammunich 6y ago
      I think so yes, but the effect would be very small
  - majikandy 6y ago
    Probably more simply in kinetic and sound energy with water as it rumbles through the pipes and vibrates the pipes and barrels a bit.
- Nextgrid 6y ago
  I’m curious, how would you describe inductance in the context of the water analogy? I’m not very experienced in electronics so I’m hoping the “water version” of it might help me understand electric inductance better.
  - jimmyswimmy 6y ago
    It is the inertia of the water in the connecting pipes. In electrical systems, the inductance is that property of a component which tries to keep current constant. In both forms of the thought experiment, if you shut off the connection instantly, the voltage or pressure build to destroy the thing which stopped the connection. Water analogies only go so far. You would do well to pick up a book on basic electrical networks to learn this material further. Don't shy away from the math; it's really the only way to build the understanding that leads to intuition.
Yajirobe 6y ago
So how do we capture this 'resolution' in actual equations? Why aren't the equations revealing this to us?
- LeifCarrotson 6y ago
  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.
  - logicallee 6y ago
    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!
    - saagarjha 6y ago
      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.

sopooneo 6y ago

That, along with the parent comment, clinched it for me. Thank you.

CamperBob2 6y ago
I think the Wikipedia article's final sentence is probably the easiest way to visualize the question if you want to stick to electrical terms. If there's just one capacitor and you short it out by closing the switch, it's not hard to see how all of the energy vanishes, rather than complicating the question by referring to half of it.
- wrycoder 6y ago
  Energy doesn’t vanish. Part of it is dissipated in the wire’s resistance and the rest is radiated as EM energy in a ring-down oscillation.
  - CamperBob2 6y ago
    Yes, and that's the point; to answer the paradox you have to explain where the seemingly-"vanished" energy goes, and that's relatively straightforward in the single-capacitor case. It's then equally straightforward to see how the same explanation applies to the more contrived case of two capacitors.

kurthr 6y ago

Well, technically there would also be some inductance... which as an externally coupled field would radiate, and that inherently causes loss!

chronolitus 6y ago
tomorrow's popsci headline: "Do barrels of water emit harmful 5G radiation? The answer may surprise you!"
red75prime 6y ago
And the loss will be in the form of gravitational waves.
- throwphoton 6y ago
  Argh is this real or not? Would energy radiated away as a result of inductance be analogous to energy radiated away by gravitational waves in the "water pressure analogy"?
  - adammunich 6y ago
    I think so yes, but the effect would be very small
- majikandy 6y ago
  Probably more simply in kinetic and sound energy with water as it rumbles through the pipes and vibrates the pipes and barrels a bit.
Nextgrid 6y ago
I’m curious, how would you describe inductance in the context of the water analogy? I’m not very experienced in electronics so I’m hoping the “water version” of it might help me understand electric inductance better.
- jimmyswimmy 6y ago
  It is the inertia of the water in the connecting pipes. In electrical systems, the inductance is that property of a component which tries to keep current constant. In both forms of the thought experiment, if you shut off the connection instantly, the voltage or pressure build to destroy the thing which stopped the connection. Water analogies only go so far. You would do well to pick up a book on basic electrical networks to learn this material further. Don't shy away from the math; it's really the only way to build the understanding that leads to intuition.

Yajirobe 6y ago

So how do we capture this 'resolution' in actual equations? Why aren't the equations revealing this to us?

LeifCarrotson 6y ago
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.
- logicallee 6y ago
  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!
  - saagarjha 6y ago
    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.