5 ms·

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.
    - logicallee 6y ago
      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: https://www.physics.purdue.edu/demos/display_page.php?item=2B-03 https://www.physics.purdue.edu/demos/display_page.php?item=2... So is it not accurate to say that water does this by oscillating and throwing away energy as parasitic losses, until it equalizes? Are you saying in general does a system of connected tubes NOT throw off lots of energy as it gets into the equalized state shown? If it does, I think this fact should also be mentioned when teaching the "water seeks its own level" demonstrated above. (Called Pascal's vases.)
      - saagarjha 6y ago
        Here's a video showing water oscillating: https://www.youtube.com/watch?v=cVEbh_COcRY https://www.youtube.com/watch?v=cVEbh_COcRY
        logicallee 6y ago
        Cool! Thanks :)
    - lutorm 6y ago
      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.
  - mindslight 6y ago
    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.
    - logicallee 6y ago
      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=2B-03 https://www.physics.purdue.edu/demos/display_page.php?item=2... ), does the system lose a lot of potential energy as the water equalizes? The demonstration is called Pascal's vases.
  - MauranKilom 6y ago
    > 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?
fpoling 6y ago
The equations do reveal that! As Wikipedia pointed out in the idealized model the current is infinite and the time to settle is zero. Thus the energy loss is proportional to ∞*0 , that tells immediately that the model is not applicable.
jbay808 6y ago
You can set this up with superconductors so they have zero resistance, but any wire will still have a trace inductance. (Moving charges must create a magnetic field, so if you're moving charges from one capacitor to another, you need to have some inductance somewhere). Once you include that, you'll see your LC resonator is perfectly undamped and oscillates the charges back and forth forever, breaking the assumption that a "steady state" would equalize the charge on both capacitors. The equations will also reveal the problem to you when you try to calculate the current that flows from one capacitor to another, with no inductance or resistance in between. You might try putting in an inductor and looking at the circuit behaviour in the limit as the inductance goes to zero -- you'd see the frequency of oscillation climbs to infinity; the full charge essentially teleports back and forth from one capacitor to the other, but still never settles into a steady state of equal charge on both capacitors. Once you see this, it's like having a problem set up with a frictionless ball on a hilltop beside a valley, saying "in the steady state, the ball has rolled down and settled in the valley. But there was no friction! Where did the energy go?", and the answer is just that the ball doesn't settle in the valley, but rather continues back up the other side, carried along by the kinetic energy that had been neglected in the problem statement. (Before the well-actualies point out that an LC circuit will damp itself via radiation, let's just say it's also perfectly shielded).
flak48 6y ago
We make an intuitive assumption about the final state and then use the rules we know to apply equations to verify the final state. In this case, the equations not matching up proved that our initial assumption (steady state) was itself wrong.

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.
  - logicallee 6y ago
    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: https://www.physics.purdue.edu/demos/display_page.php?item=2B-03 https://www.physics.purdue.edu/demos/display_page.php?item=2... So is it not accurate to say that water does this by oscillating and throwing away energy as parasitic losses, until it equalizes? Are you saying in general does a system of connected tubes NOT throw off lots of energy as it gets into the equalized state shown? If it does, I think this fact should also be mentioned when teaching the "water seeks its own level" demonstrated above. (Called Pascal's vases.)
    - saagarjha 6y ago
      Here's a video showing water oscillating: https://www.youtube.com/watch?v=cVEbh_COcRY https://www.youtube.com/watch?v=cVEbh_COcRY
      - logicallee 6y ago
        Cool! Thanks :)
  - lutorm 6y ago
    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.
- mindslight 6y ago
  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.
  - logicallee 6y ago
    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=2B-03 https://www.physics.purdue.edu/demos/display_page.php?item=2... ), does the system lose a lot of potential energy as the water equalizes? The demonstration is called Pascal's vases.
- MauranKilom 6y ago
  > 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?

fpoling 6y ago

The equations do reveal that! As Wikipedia pointed out in the idealized model the current is infinite and the time to settle is zero. Thus the energy loss is proportional to ∞*0 , that tells immediately that the model is not applicable.

jbay808 6y ago

You can set this up with superconductors so they have zero resistance, but any wire will still have a trace inductance. (Moving charges must create a magnetic field, so if you're moving charges from one capacitor to another, you need to have some inductance somewhere). Once you include that, you'll see your LC resonator is perfectly undamped and oscillates the charges back and forth forever, breaking the assumption that a "steady state" would equalize the charge on both capacitors. The equations will also reveal the problem to you when you try to calculate the current that flows from one capacitor to another, with no inductance or resistance in between. You might try putting in an inductor and looking at the circuit behaviour in the limit as the inductance goes to zero -- you'd see the frequency of oscillation climbs to infinity; the full charge essentially teleports back and forth from one capacitor to the other, but still never settles into a steady state of equal charge on both capacitors. Once you see this, it's like having a problem set up with a frictionless ball on a hilltop beside a valley, saying "in the steady state, the ball has rolled down and settled in the valley. But there was no friction! Where did the energy go?", and the answer is just that the ball doesn't settle in the valley, but rather continues back up the other side, carried along by the kinetic energy that had been neglected in the problem statement. (Before the well-actualies point out that an LC circuit will damp itself via radiation, let's just say it's also perfectly shielded).

flak48 6y ago

We make an intuitive assumption about the final state and then use the rules we know to apply equations to verify the final state. In this case, the equations not matching up proved that our initial assumption (steady state) was itself wrong.