Old Chestnut

studiot

Joined Nov 9, 2007
4,998
Arguing definitions or angels on pinheads is a mugs game.

The theory I am talking about is often called classical physics. It describes transformers, radio antenna, condensers, waveguides, magnetrons, doorbells, motors, generators, ionic action and general paraphenalia up to some semiconductor action.

Maxwells equations and Poynting's theorem come under this heading.

What I am excluding is theory that requires (relativistic) quantum mechanics.


If you were to tell me that the charge made one solitary round of the circuit after the switch was thrown and in doing so generated EM radiation, like in Prof Lewin's demo I would happily accept this.

If you were to tell me that some quantum tunneling or other effect was going on I wouldn't.

I am not sure I agree about the oscillatory argument as I can't see a mechanism to promote it with zero resistance and inductance in the circuit.

Have you no comment on my mechanical analogy?
 

steveb

Joined Jul 3, 2008
2,436
Arguing definitions or angels on pinheads is a mugs game.
Agreed! :) But, clarifying definitions avoids confusion.

The theory I am talking about is often called classical physics. It describes transformers, radio antenna, condensers, waveguides, magnetrons, doorbells, motors, generators, ionic action and general paraphenalia up to some semiconductor action.

Maxwells equations and Poynting's theorem come under this heading.

What I am excluding is theory that requires (relativistic) quantum mechanics.
OK, no problem. I just have a different interpretation of the term "circuit theory". The term "classical physics", and your definition of it I am comfortable with it.

I am not sure I agree about the oscillatory argument as I can't see a mechanism to promote it with zero resistance and inductance in the circuit.
Actually, I'm not making any oscillatory argument. I was just talking about radiation resistance as a nice way to include EM radiation loss in a circuit equation. The usual definition of radiation resistance assumes sinusoidal variation. I think I said that the behavior is NOT oscillatory, although I'll have to go back and look now.

Have you no comment on my mechanical analogy?
I'm not sure what to comment on. If the ball stops in the valley, it has less energy. So, we have to explain where the energy went if there is no friction. I don't know where it went. I would expect the ball to keep moving with kinetic energy, or there was some unknown force that stopped the ball.

I guess I'm missing the point. Please explain.
 

studiot

Joined Nov 9, 2007
4,998
Hi Steve, et al.

Actually, I'm not making any oscillatory argument
I wasn't commenting on one particular post, but several.

Actually I wasn't commenting, rather 'thinking out loud'. Trying to organise facts and thoughts.

About my analogy.

With the electrical example, energy is stored in a potential field (electric) when the first capacitor is charged. When the switch is thrown charge accelerates and some of the energy is converted to magnetic energy. The residue is left as lower potential energy.

With the ball, energy is held in a potential field (gravitational). When the gate is opened the ball accelerates and some of the energy is converted to kinetic energy, the residue is left as lower potential energy. When the ball come to rest all the KE has gone somewhere.
 

b.shahvir

Joined Jan 6, 2009
457
With the ball, energy is held in a potential field (gravitational). When the gate is opened the ball accelerates and some of the energy is converted to kinetic energy, the residue is left as lower potential energy. When the ball come to rest all the KE has gone somewhere.

with all due respect, Sir, this somehow seems unpalatable! :)
 

studiot

Joined Nov 9, 2007
4,998
What's the problem with this Shavir?

At the start the ball is stationary and the system posesses the following PE

PE= mass x h1

After the ball has rolled into the valley the PEis

PE= mass x h2

The system has lost m(h1-h2) potential energy units.

Where did it go?
 

Attachments

b.shahvir

Joined Jan 6, 2009
457
What's the problem with this Shavir?
At the start the ball is stationary and the system posesses the following PE

PE= mass x h1
After the ball has rolled into the valley the PEis

PE= mass x h2
The system has lost m(h1-h2) potential energy units.
Where did it go?

Thanx for the detailed reply. But in nature there has to be some explanation howsoever arbitrary. The loss of PE cannot be directly explained, but one cannot even claim it has gone 'somewhere'!

There is no dispute in what has been presented but in this abstract case your guess is as good as mine! :(

Best regards,
Shahvir
 

Tesla23

Joined May 10, 2009
542
I see the situation this way.

In an idealised physical world where we allow perfect conductors, then if you have something in a perfectly shielded box then no EM energy can get in or out. (Simple application of Poynting vector on box surface). If the system in the box consists of perfect conductors, a pair of plates forming a capacitor, charged with +/-Q, and a perfectly conducting wire that is connected across them at t=0. At t=0 there is a certain amount of energy in the electric field (1/2CV^2), there is no way to convert the electrical energy to heat (perfect conductors), no radiation loss from the box (perfect shield) and no allowable stable state that doesn't violate conservation of energy, the only conclusion is that the oscillation continues indefinitely.

So if we now consider what we can make, this is different. I'm no expert on superconductors, but from what I know it seems that they exist for DC, but where I see them applied at RF there seem to be some resistive losses. Someone may have better info here, but if we could make the system from superconducting material that was truly superconducting for AC currents at the oscillation frequency then it would work, but I don't know if such materials exist.
 

studiot

Joined Nov 9, 2007
4,998
the only conclusion is that the oscillation continues indefinitely.
What oscillation?

then if you have something in a perfectly shielded box then no EM energy can get in or out.
True by definition of the word 'perfect'.
But
I don't see that it matters whether the EM energy dissapates to the end of the universe or is absorbed in the walls of a perfect shield. It is still lost to the system and not recoverable by the capacitors.
 

Tesla23

Joined May 10, 2009
542
What oscillation?
The oscillation in the resonator formed by the capacitance of the plates and the inductance of the wire that the charges have to flow through to get from one plate to the other. This is approximately an LC resonant circuit.

True by definition of the word 'perfect'.
But
I don't see that it matters whether the EM energy dissapates to the end of the universe or is absorbed in the walls of a perfect shield. It is still lost to the system and not recoverable by the capacitors.
Sorry, by perfectly shielded I meant enclosed in a perfectly conducting shield. This doesn't dissipate any power, simply reflects if back to the system.
 

studiot

Joined Nov 9, 2007
4,998
I am not sure I agree about the oscillatory argument as I can't see a mechanism to promote it with zero resistance and inductance in the circuit.
I see no inductance in the ideal wires.
They could be infinitesimally short. Or they could be very long.
The result should be the same, and not depend upon their length or orientation.
 

Tesla23

Joined May 10, 2009
542
I see no inductance in the ideal wires.
They could be infinitesimally short. Or they could be very long.
The result should be the same, and not depend upon their length or orientation.
The capacitor plates must be separated by some distance. So the wire must have some length. When the current flows it will create a magnetic field, the energy stored in this will be represented in the circuit as some sort of inductance.

Depending on the length and path of the conductors the inductance will vary, so will the oscillation frequency. I don't see any reason why the result should be independent of the length of the wires, in fact the opposite, the inductance will depend on the length.

Inductanceless wires are an artifice of circuit theory, they don't exist even in idealised systems where perfect conductors are allowed.
 

thingmaker3

Joined May 16, 2005
5,083
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?
Since the ball comes to rest, I suspect we ran out of that expensive frictionless paint.
 

studiot

Joined Nov 9, 2007
4,998
Tesla,
Firstly we must agree to differ about circuit theory approximations.

At the simplest form, which the original poster (OP) introduced the problem the model allows zero resistance, zero inductance connections. Also adjacent wiring or board tracks posess no self capacitance. The calculations introduced by the Op and reinforced in my textbook attachment in post#15 are based on this.

Yes some models of wiring and circuit board analysis take or need to take connection inductance and or resistance into account, eg striplines on pcbs.

It would also be possible to cancel any magnetic field by placing the capacitors adjacent to each other and looping any wire into a hairpin so producing adjacent flows in opposite directions.

However that does not alter the fact that the energy remaining on the capacitors must be in accordance with th OP's calculations, and that some has been lost to the capacitors.

His question was where did it go?

I think the general consensus is that it went into a magnetic field.

However there is no need to invoke inductance to generate this, the circuit is a circuit of one complete turn and until the charges have redistributed will generate a magnetic field while the redistribution current flows.

I see you are a new member, Welcome.
I thoroughly recommend viewing the videos of Professor Lewins demonstrations

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

and the associated thread.
 

Tesla23

Joined May 10, 2009
542
Tesla,
Firstly we must agree to differ about circuit theory approximations.

At the simplest form, which the original poster (OP) introduced the problem the model allows zero resistance, zero inductance connections. Also adjacent wiring or board tracks posess no self capacitance. The calculations introduced by the Op and reinforced in my textbook attachment in post#15 are based on this.

Yes some models of wiring and circuit board analysis take or need to take connection inductance and or resistance into account, eg striplines on pcbs.

It would also be possible to cancel any magnetic field by placing the capacitors adjacent to each other and looping any wire into a hairpin so producing adjacent flows in opposite directions.
You can agree but I don't! Conservation of energy applies to physical systems (including idealised ones with perfect conductors) but NOT to circuit theory. You can draw up a circuit diagram that involves delta functions of current and you lose energy - so what - unless it corresponds to a real physical system it doesn't really matter.

The issue of magnetic fields is fundamental. I challenge you to create any charge distribution and then move the charges without generating magnetic fields. Check out maxwell's fourth equation, even changing E generates a magnetic field. If you have a circuit diagram that involves moving charges that doesn't include something that accounts for energy stroage in the magnetic field then it is non-physical and can violate anything really.

It is the magnetic field that prevents the delta function from arising in practice.

Even at a circuit level, cancelling out inductance with capacitance only works in the frequency domain at discrete frequencies - not in the time domain like we are working here.

However that does not alter the fact that the energy remaining on the capacitors must be in accordance with th OP's calculations, and that some has been lost to the capacitors.

His question was where did it go?

I think the general consensus is that it went into a magnetic field.

However there is no need to invoke inductance to generate this, the circuit is a circuit of one complete turn and until the charges have redistributed will generate a magnetic field while the redistribution current flows.

I see you are a new member, Welcome.
I thoroughly recommend viewing the videos of Professor Lewins demonstrations

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

and the associated thread.
I haven't had time to check out the videos, I will, but I have had many discussions on this over time. The fundamental objection remains though, the OP's circuit does not correspond to any physical circuit. Heck, it even violates relativity - the charge gets from one capacitor to another instantaneously (really the same issue - model the real path that the charges flow along and you will have some inductance, no delta functions)
 

studiot

Joined Nov 9, 2007
4,998
You can agree but I don't
Sorry mate I meant disagree!

However to return to the question you are so carefully sidestepping what happened to the energy lost from the capacitors?


Even at a circuit level, cancelling out inductance with capacitance only works
You must have misunderstood I wasn't cancelling inductantance with capacitance, merely routing a connecting wire so half was going in one direction and half in the other - a not uncommon non-inductive winding method.

Nor did I ever suggest that moving charge does not generate a magnetic field. However my understanding is that it is accelerating charge, not charge moving at a steady velocity per se, which
Correction produces EM radiation.
Obviously in this case charge must accelerate from stationary to some speed to redistribute, so we can accept the production of EM radiation, as opposed to some ghostly magnetic field, which must cease as soon as the charges stop moving.

Nor did I ever suggest that the redistribution is instantaneous. In fact in earlier posts I was trying to draw out the fact that the result is independant of time. It depends solely in the intitial and final states, not how long, or how far it takes to get there.

You surely can't be seriously suggesting that the final energy states of the capacitor depends upon the length of connecting wires?
 
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Tesla23

Joined May 10, 2009
542
However to return to the question you are so carefully sidestepping what happened to the energy lost from the capacitors?
In the circuit theory model, there is a delta function of current and half the energy in the circuit disappears. Who cares - it doesn't correspond to any physical system. There are other problems, simply shorting a charged capacitor produces a similar loss of energy - again a non-physical circuit so who cares?


You must have misunderstood I wasn't cancelling inductantance with capacitance, merely routing a connecting wire so half was going in one direction and half in the other - a not uncommon non-inductive winding method.

Nor did I ever suggest that moving charge does not generate a magnetic field. However my understanding is that it is accelerating charge, not charge moving at a steady velocity per se, which does this. Obviously in this case charge must accelerate from stationary to some speed to redistribute.
The challenge is to come up with a physical system where the charge can move from one capacitor to another without creating a magnetic field. I don't believe you can so there is no point applying physical laws to the circuit that doesn't correspond to reality.

Moving charges are sufficient to creat a magnetic field, just look at a wire carrying a steady current. Mind you the charges are initially at rest, so to move anywhere they have to accelerate, so even with your understanding there are magnetic fields.

Nor did I ever suggest that the redistribution is instantaneous. In fact in earlier posts I was trying to draw out the fact that the result is independant of time. It depends solely in the intitial and final states, not how long, or how far it takes to get there.

You surely can't be seriously suggesting that the final energy states of the capacitor depends upon the length of connecting wires?
If you are not suggesting that the redistribution is instantaneous then why are you worried about applying physical laws to the circuit diagram which has as a solution an instantaneous charge distribution? Why aren't you complaining about the violation of relativity as well as the violation of conservation of energy.

You try to brush off the non-physical nature of the circuit saying that it can be fixed without changing the outcome. It is fundamental that fixing it will change the outcome. The result of adding inductance (which is fundamental) is to avoid the delta function and the circuit acts as a resonator, which if not shielded will radiate and the oscillation die out. The missing power is dissipated at the end of the universe. If it is shielded it oscillates forever and there is no missing power.
 

studiot

Joined Nov 9, 2007
4,998
Given good quality real world components I could charge up one capacitor to 100volts, connect through a series of wires of different lengths and diameters, some wound as formal inductors, some straight to a second similar capacitor and then disconnect.

In each and every case I would walk away with two capacitors charged to pretty close to 50 volts.

Are you suggesting that somehow the energy is not 5000C before the second capacitor is connected and 2500C after ?
 
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