8 ms·
Its easier to think about this using the water pressure analogy. Imagine a large barrel full of water connected to another identical barrel. The first barrel is
by barbegal 6y ago
Its easier to think about this using the water pressure analogy. Imagine a large barrel full of water connected to another identical barrel. The first barrel is filled to the top whilst the second is empty. A valve is opened between the two barrels and they equalise with half of the water in each.
The initial energy is mgh and the final energy is 2 * (m/2 * g * h/2) = mgh/2 so half of then energy has disappeared. It is clear that work could have been done by the water moving between the two barrels (like in a hydro-electric power station).
- Karto 6y agoIf I think straight, the initial potential energy in a column of liquid would be the integration of mgh (actually integration of density * g * section * height), therefore mgh/2 at the beginning of the situation. Or said intuitively : the center of gravity of the column at half its height. But obviously, even with this 1/2, your relation after dividing between two tanks still holds. :)
- thomasahle 6y agoI was thinking that too, the initial potential energy should be proportional to h^2/2. However then we still have 2*(h/2)^2/2 = h^2/4.
- Karto 6y agoYes, one h is hidden inside m... Ending up with energy = something * h^2 actually makes the relationship show the same face as the one for capacitors, where energy = something * v^2.
- unkulunkulu 6y ago* where h = half the water level in the first barrel.
- sirusdas 6y agoGood observation!
- bobbylarrybobby 6y agoAnd 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 agoThat, along with the parent comment, clinched it for me. Thank you.
- CamperBob2 6y agoI 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 agoEnergy 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 agoYes, 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 agoWell, technically there would also be some inductance... which as an externally coupled field would radiate, and that inherently causes loss!
- contravariant 6y agoIt's also intuitively obvious that energy must disappear, otherwise nothing would happen (the system would already be in its lowest energy state).
- surewhynat 6y agoIf you have just a full barrel of water, then you can call the pressure just atmospheric pressure, so when you connect another barrel, and you get half the water it's still the same atmospheric pressure. Instead, you should think of it like the first barrel of water has fit twice it's volume under pressure. So it's pressure is 2x atmospheric pressure. When you connect the second empty barrel, they both become full, now both at atmospheric pressure. The energy without the pressure in the system is half. But why? Because it took energy to squeeze 2x the water into the first barrel in the first place!
- adharmad 6y agoThe secondary question though is: Why does the energy always dissipate by 50% in both the electric and hydrodynamic models? Is there some underlying physical explanation of this? Something that says that "in a dynamical system the maximum efficiency can be at most 50%".
- joenathanone 6y agoBecause in the examples you have two capacitors/barrels, if you had three or four, you would be looking at 33% or 25%.
- deleted 6y ago[deleted]
- tijsvd 6y agoTo say that the energy of the water is mgh is a simplification. It's really 1/2 * h^2 * A * mass density. The water at the top has more energy than the water on the bottom. The energy in the water is square to the water level, but the amount of water is linear. It's the same with a capacitor, the energy is square to the voltage, but the number of electrons is linear. It works out to the same 50% because the analogy is just very good. Dividing over two barrels/capacitors means halving the potential level (amount of water/load doesn't change), which means 1/4 of the energy in each barrel.
- agapon 6y agoThis is wrong. The energy is proportional to the height, not its square. You would get wrong units otherwise.
- tijsvd 6y agoThe energy is proportional to both the height and the mass. The mass is proportional to the height itself. Hence the energy is proportional to the square of the height. If the fill level is h, so the weighted average of the mass is at h/2,then E = m * g * h / 2 m = V * r V = A * h E = A * r * g * h^2 / 2 m2 * kg/m3 * m/s2 * m2 = kg m2 / s2 = J
- cousin_it 6y agoVery nice! The hydraulic analogy actually works for electric circuits in general, and can be made precise (you can calculate how much pressure corresponds to one volt etc), though the numbers involved can get a little silly: http://amasci.com/miscon/voltpres.html http://amasci.com/miscon/voltpres.html
- scrooched_moose 6y agoHeat transfer as well. At the first engineering firm I worked for, we had a very good heat transfer solver but no electrical solver since it was an infrequent need in our field. One of old timers was an expert at reframing the electrical problems into heat transfer, solving in the available tools, then converting back. As he said "it's all unit conversions". I never picked it up beyond simple resistance networks, but it was a cool way to abuse the tools. https://lpsa.swarthmore.edu/Systems/Thermal/SysThermalElem.html https://lpsa.swarthmore.edu/Systems/Thermal/SysThermalElem.h...
- Matumio 6y agoI don't think the water analogy is a good fit here. For me the revelation was that energy can also be radiated away. When we did basic circuit exercises we never talked about that. Energy was either stored or turned into heat. Not sure how you would explain electromagnetic radiation with water pipes.
- ahartmetz 6y agoGravity waves ;) - if they were many orders of magnitude stronger.
- pddpro 6y agoAnd that is what is called an analogy. Gives a simple and intuitive understanding but is not complete at all.
- LeifCarrotson 6y agoIf you actually build this on a PCB with typical ceramic or electrolytic capacitors and a MOSFET or mechanical switch, an insignificant amount of energy will be radiated. Most of the energy does go into heating the capacitors through their internal series resistance. Radiation could be the sound produced by the rushing water?
- Matumio 6y agoA steady current would also produce a rush of water, but I don't think you can get EM radiation with just a steady current. Well, but maybe... if you equalize two water tanks with a very fat pipe it will swap forth and back, causing the whole assembly, table, and room to rock?
- castratikron 6y agoAnother equivalent analogy is a pendulum. Hold a pendulum with mass m at height h, so its potential energy is mgh. Let it go, and after a long time of swinging back and forth the pendulum eventually stops at the bottom (h=0). Where'd the energy go? It's all lost to heat through friction from the air and tension on the string. The rate of energy lost is just a lot less than the water barrel example. The general name for this kind of system is a harmonic oscillator. Now I want to try the water barrel thing myself and see how many times the water goes back and forth before it finally stops...
- mrfusion 6y agoAnd if no work was done that energy turned into heat?
- ouid 6y agoI think it's probably easier to imagine as a pendulum.