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Comment by barbegal | Hacker News Reader

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).

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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.

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

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.

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.

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.

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

And the loss will be in the form of gravitational waves.

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"?

I think so yes, but the effect would be very small

Probably more simply in kinetic and sound energy with water as it rumbles through the pipes and vibrates the pipes and barrels a bit.

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.

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.

In electric circuits capacity and inductance make energy oscillate between the electric and the magnetic field. The closest analogy I can think of for water pipes would be the oscillation between the kinetic and the potential energy (or the flow and the pressure). As for better understanding of inductance and magnetic field, imho it's better to stick to Faraday, Gauss and Lenz laws than looking for analogies, as EM field has it's very specific properties.

There's a property called "inertance" which is analogous to inductance in electrical circuits: https://en.wikipedia.org/wiki/Inertance

and it can cause things like water hammers: https://en.wikipedia.org/wiki/Water_hammer which are analogous to inductor sparks: http://web.physics.ucsb.edu/~lecturedemonstrations/Composer/...

This video explains water hammer really well.

https://youtu.be/xoLmVFAFjn4

Yes it does! Watched, learned something, grateful.

It is the reason why domotica valves move so slow. If they were to move instantaneously then it would cause water-hammer to the point where it might even dislodge the valve from its connections if the run before it is long enough!

There is also a practical device which is basically a boost converter for water: https://en.wikipedia.org/wiki/Hydraulic_ram

The analogy I always used as a kid was a heavy, free-spinning water wheel in the path of the water. It would tend to resist changes in current.

Might be an unpopular opinion but my advice as an electronics engineer would be to try and move away from water analogies as quickly as possible. While they can be useful at the start for some basic concepts, I feel like trying to relate everything to hydraulic concepts (or springs) makes it harder to understand things in the long run when you start looking at concepts where the analogies don't work as well.

What would be a better visualization?

What I’m saying is that it’s best to get to the point where you can reason more directly about voltage, current, charge, inductance etc.

Part of that is getting comfortable with the maths, and the way that’s visualised is just equations, graphs and sometimes things like diagrams of fields or heatmaps. I work in RF so we do antenna simulations and things like that, and the software generates radiation pattern diagrams in 2D, or in 3D heatmaps showing the energy density. It just doesn’t work to, say, try and think of what would happen if an antenna was spraying water out or something.

I have only a passing understanding of both, but enough to appreciate the (limited as you say) intuitive resemblances - can anyone enlighten me as to what are the fundamental mathematical differences between modelling electronics and water?

If you keep the water slow enough to stay in laminar flow and the electronic scales large enough to avoid quantum effects, electronic and hydraulic components have an essentially 1:1 correspondence.

The biggest difference is in mechanical effects: in hydraulics, physical force is primarily a function of the system pressure and velocity of the flow rate. For electromagnetics, those are reversed: voltage controls the speed of a motor and current controls the force, for example.

They also deviate from the linear regime in different ways, which means the more interesting components have to be built completely differently to perform the same job. A one-way check valve and a diode are both governed by similar equations on a macro scale, but you’ll never be able to understand the internal structure of the valve by an electronic analogy or the design of the diode by a hydraulic one.

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

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!

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.

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=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.)

Here's a video showing water oscillating: https://www.youtube.com/watch?v=cVEbh_COcRY

Cool! Thanks :)

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.

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?

The demonstration is called Pascal's vases.

> 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?

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.

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).

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.

If 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. :)

I 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.

Yes, 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.

* where h = half the water level in the first barrel.

Good observation!

Another 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...

It's also intuitively obvious that energy must disappear, otherwise nothing would happen (the system would already be in its lowest energy state).

Very 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

Heat 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.h...

I 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.

Gravity waves ;) - if they were many orders of magnitude stronger.

And that is what is called an analogy. Gives a simple and intuitive understanding but is not complete at all.

If 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?

A 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?

The 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%".

Because in the examples you have two capacitors/barrels, if you had three or four, you would be looking at 33% or 25%.

To 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.

This is wrong. The energy is proportional to the height, not its square. You would get wrong units otherwise.

The 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

Yeah, I totally forgot that there is a height component in the mass as well.

I think it's probably easier to imagine as a pendulum.

And if no work was done that energy turned into heat?

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