Interesting discussion! but what if we consider a lossless solid state switch like a MOSFET or a FET? on similar lines as mentioned by Mik3.
Thanks

 

Interesting discussion! but what if we consider a lossless solid state switch like a MOSFET or a FET? on similar lines as mentioned by Mik3.
Thanks

I think this is valid too. Even if we do not have lossless MOSFETs now, we have seen the development of high temperature superconductors. Perhaps someday we will have materials that allow superconducting semiconductor type devices to be built.

Superconductivity brings up the issue of the loss of zero resistance if the magnetic field or current is too large. This is relevant since a huge current spike and a huge magnetic field will exist.

Still, it is possible to imagine a modernization of the paradox stating that the capacitors, wires and switch are superconducting devices and no arc occurs. Then add the condition that superconductivity is maintained and the end result is two capacitors with less energy than before the switch was closed.

The explanation for this problem would be that radiation resistance maintained the current to low enough value to prevent superconductivity collapse and also explains where the missing energy went.

I agree it's getting pretty convoluted and the physics becomes questionable, as we are mixing classical electrodynamics and quantum mechanics with reckless abandon. But, it's still educational to think about it.

 

Interesting discussion! but what if we consider a lossless solid state switch like a MOSFET or a FET? on similar lines as mentioned by Mik3.
Thanks

It was mentioned that it will be no arcing but there will be EMI.

 

What is a bit sketchy for me to understand is that, can Electromagnetic Radiation introduce power loss analogous to circuit resistance? Thanx.

 

Would this analysis be along the lines (presented without maths)

Following closure of the switch charge moves to redistribute itself in the system between the capacitors.

This charge has a finite velocity in the conductors, but moving charge generates EM radiation which dissipates.

Once the charge has equilibrated it stops moving so stops generating EM.

The math is well over my head, so I just look at the pictures and consider the models and authors' comments. Boykin et al. state that they ignore the time dependence of the current in the ideal model (zero resistance, zero inductance) and treat the radiating circuit as a magnetic dipole.

A later paper by TC Choy (Am J Physics, 72: 662-670 (2004)) starts with the results from Boykin et al. but reduces the wire loop to zero. It goes into a discussion of Poynting vectors and capacitor antennas. The message I get from Choy is that the capacitor itself radiates. Here is his statement on that point:

(in this thought experiment, we)...reduce the wire loop to zero length. In the long wavelength
(λ -->∞) limit, the two capacitors connected by zero length wires can be viewed as an oscillating electric dipole arising from time-varying charges on the parallel plates of C.

One interesting aspect discussed by Choy is the claimed application of this theory to antenna designs that have been patented in the US and GB. He apparently disagrees with some of the claims in those patents, but nevertheless feels capacitor radiation may eventually be used in an antenna design.

If you cannot get those papers and would like to take a closer look at the math , I would be happy to scan portions and send them to you as pdf's under the academic use exemption.

John

 

What is a bit sketchy for me to understand is that, can Electromagnetic Radiation introduce power loss analogous to circuit resistance? Thanx.

Yes. As I mentioned before, they talk about "radiation resistance" in antenna design. This is the effective resistance that accounts for the power loss in a circuit that drives an antenna.

See the following.

http://en.wikipedia.org/wiki/Radiation_resistance

 

There is much to think about in this one.

I accept that there are probably several available routes for energy dissipation, including EM radiation.

I don't see that the introduction of superconductors adds anything, we already have 'ideal' wires in the model.

It is of interest to enquire whether the loss takes place at or just before the closing of the switch, or some time after. Only arcing can provide the former, only EM radiation the latter.

As to a perfect switch I wonder if a magnetostrictive switch could not be devised that actually allowed or prohibited passage through a wire so not contacts would be involved.

There is no restriction to say the switching must be quick or instantaneous, or that the build up of charge on the second capacitor must be quick. The same result should occur over any reasonable timescale.

 

There is no restriction to say the switching must be quick or instantaneous, or that the build up of charge on the second capacitor must be quick.

The original post implies that switching must be instantaneous. Non-instantaneous switching means the use of a time-dependent resistor which would seem to violate the "no resistance assumption". One could even argue (note I said "one" and not "I") that the no-resistance assumption implies that there is no arcing, since arcing does not create a zero-resistance path.

I agree that there is no restriction to say that the second capacitor must build up charge quickly, but the initial expectation is that the discharge will be very fast without resistance. Radiation resistance (which resolves the paradox) will slow the discharge down to a non-instantaneous value, and the speed will depend on the circuit dimensions. At this point the classification of the discharge being fast or slow becomes a matter of judgement. We have to ask: fast or slow relative to what?

The same result should occur over any reasonable timescale.

I agree that the same result will occur on any timescale. But the slow discharge is less mysterious since we can argue that most of the energy is lost as heat in the resistance. The paradox arises when we take the no-resistance/instantaneous-switch assumption. Here we find that the circuit-theory approximation breaks down. With or without arcing, EM radiation then will be significant. With arcing, it's hard to say how significant radiation loss is, but without arcing, I can't think of any energy loss other than EM radiation.

 

I have seen many variations of this over the years. Here are my thoughts:

The first observation is that circuits are meant to be models of the real world, and physical laws (e.g. conservartion of energy) only apply to circuits if they accurately model the real world.

One thing that is definitely missing from the circuit that appears in any real system is some inductance. Current cannot flow from one capacitor to another without creating a magentic field. Said in another way, the capacitors cannot both be in the same place so it must take some time for the charge to get from one to another, htis delay should be in the model in some form.

Once you add some inductance the problem goes away as what you have is a resonator, the energy simply oscillates between the capacitor and inductor.

In the real world there are typically losses in the conductors and dielectrics so the oscillation eventually dies out. Also, unless the system is shielded, it will radiate, as to whether this is significant compared to the ohmic losses depends on the system.

As others have pointed out as well, we don't have a perfect switch, so there are practical losses here as well.

 

Once you add some inductance the problem goes away as what you have is a resonator, the energy simply oscillates between the capacitor and inductor.

Since the wires are "lossless," there would be no damping either? The oscillation would be eternal?

 

Since the wires are "lossless," there would be no damping either? The oscillation would be eternal?

With perfect conductors and no dielectric losses, enclosed in a perfectly conducting shield to avoid radiation losses, then the oscillation theoretically goes on for ever.

 

The concept of EM radiation is a bit unclear to me. What does EM radiation actually consist/compose of? Is 100% shielding of EM radiation possible (ideal world) if so, how? Also, how does EM radiation carry packets of energy? Pardon my marathon of questions, but i did be grateful for a reply

Thanks & best regards,
Shahvir

 

Since the wires are "lossless," there would be no damping either? The oscillation would be eternal?

As mentioned, oscillating current will produce electromagnetic radiation, so oscillations would decay under a classical electrodynamic assumption.

This is a really interesting comment because it reminds me of the old classical paradox about the hydrogen atom. According to classical EM theory, the electron should radiate energy and plummet into the proton. The calculated lifetime is incredible short (sub-picosecond I think), yet a real atom is stable for billions of years. Quantum mechanics must be used to get the right answer.

It's quite possible that the best answer to this question requires QM to describe superconductivity and radiation from accelerating electrons in a superconductor. A superconductor expels magnetic field which only exists near the surface and decays exponentially as it tries to penetrate. So it's not clear to me how well classical EM can apply here. Still there is an energy discrepancy that must get resolved under the proper analysis.

 

The concept of EM radiation is a bit unclear to me. What does EM radiation actually consist/compose of? Is 100% shielding of EM radiation possible (ideal world) if so, how? Also, how does EM radiation carry packets of energy? Pardon my marathon of questions, but i did be grateful for a reply

Thanks & best regards,
Shahvir

In the world of perfect conductors it is possible to perfectly shield a system to prevent radiation entering or leaving simply by enclosing it in a perfectly conducting enclosure. As the tangential electric field must vanish on the shield, so thenormal component of the Poynting vector vanishes and there is no EM power flow in or out.

How EM radiation carries energy is tied up with the Poynting vector. How the energy occurs in packets does not have a simple explanation, try looking up quantum electrodynamics.

 

As mentioned, oscillating current will produce electromagnetic radiation, so oscillations would decay under a classical electrodynamic assumption.

This is a really interesting comment because it reminds me of the old classical paradox about the hydrogen atom. According to classical EM theory, the electron should radiate energy and plummet into the proton. The calculated lifetime is incredible short (sub-picosecond I think), yet a real atom is stable for billions of years. Quantum mechanics must be used to get the right answer.

It's quite possible that the best answer to this question requires QM to describe superconductivity and radiation from accelerating electrons in a superconductor. A superconductor expels magnetic field which only exists near the surface and decays exponentially as it tries to penetrate. So it's not clear to me how well classical EM can apply here. Still there is an energy discrepancy that must get resolved under the proper analysis.

I don't think you need QM here, you did with the hydrogen atom as we knew what was in it (a positive nucleus and electrons outside) but the problem was to find a stable configuration, and there was no such configuration available in classical mechanics. There were no conductors involved.

Here we can come up with an arrangement of conductors and an initial charge distribution (contrived somehow), that if we shield the electromagnetic radiation there will be no energy loss (perfect conductors) and the oscillation will go on forever.

 

I don't think you need QM here.

Maybe not. I'm not saying we do, but just that the physics of superconductors is not clear to me. Classical EM theory is not exact and breaks down in some quantum mechanical situations.

If we want to talk about lossless conductors and reality at the same time, then quantum theory of superconductors can't be regarded as irrelevent in my opinion.

As an example/analogy, consider Hawking Radiation. Classical general relativity predicts black-holes and says that nothing escapes from them. Yet, Hawking combined QM and GR and showed that black-holes radiate energy and evaporate.

Can we really be sure that the situation you described will radiate NO energy. I agree we can talk about perfect conductors under classical EM theory assumptions, but this does not necessarily correspond with reality.

 

I think we are getting way too complicated here.

It should be possible to explain and quantify this situation within the bounds of classical circuit theory.

Nor do I see how you can contain the EM radiation within a 'shield' . That would contravene energy conservation. If energy is withdrawn form the circuit by EM radiation, it doesn't matter whether it radiates away to space or gets lost in a metal box, it is lost to the 'system'.

Nor do I see any Entropy change to figure in.

Consider a mechanical analogy (not exact).
A ball is mounted on the top of a hill, but prevented from rolling down by a gate.
the system contains a certain amount of potential energy by virtue of the ball's elevation.
The gate is opened and the ball released.
It rolls down a frictionless hill, coming to rest in the valley below.
The system now has less energy than before.

Where did this energy go?

 

I think we are getting way too complicated here.

It should be possible to explain and quantify this situation within the bounds of classical circuit theory.

I probably am getting too complicated, but I like throwing in some ideas to think about. However, staying within the bounds of classical circuit theory is getting way too simple. At the very least we need to consider classical electrodynamics.

 

Do we not have another example like Prof Lewin's excellent demo?

http://forum.allaboutcircuits.com/showthread.php?t=16150&highlight=disingenuous

 

Do we not have another example like Prof Lewin's excellent demo?

Perhaps.

Here is a reference that conforms to my interpretation of the difference between EM theory and circuit theory. If you use a different definition and include wave-propagation effects within the confines of circuit theory then there is no point worrying about the distinction.

One can include radiation loss as a resistive term, as I mentioned before, but usually this is done with sinusoidal signals. Personally, I would approach this problem directly with Maxwell's equations, not circuit equations.

http://books.google.com/books?id=VR...6cjjCw&sa=X&oi=book_result&ct=result&resnum=7