Talk:Ivor Catt/Archive 11

The model identifies "current" and "voltage" as distinct, if interrelated, concepts. When you apply this to a distinct "unconstrained" pulse (ie a pulse with definite leading and trailing edges) you get three distinct regions - the bit between the leading and trailing edges, where there is (moving) "charge", and the portions ahead of and behind the pulse where there is no charge. If you look at the pulse in terms of voltage, current and charge, and inductance and capacitance, you will see that the voltage is defined by the the amount of charge held on the capacitor defined by the pulse edges, and that the current is flowing in the inductance that is likewise defined. If, as Catt did, you calculate the energy stored in the capacitance and the inductance you will find them to be equal. Catt extrapolated this equality to a sort of "universal truth", that the energy stored in the electric and magnetic fields are always equal. It is this assertion that underpins his reverberating ("contrapuntal") TEM waves. The "conventional" position is that the equality is merely a property of the TL. (Presumably slowing the current flow increases the charge at that point, which increases the voltage in the capacitance, which tends to increase the current, etc.) I don't think I've seen any of Catt's theorising actually come up with a result that differs from the "conventional" model - basically he is arguing that because he gets the right answer, his model is correct, but he's looked up the answer at the back of the book and has cobbled together the maths to fit. And when the maths and the physics get difficult, he just argues that he is right and everyone else is wrong. -- Kevin Brunt 23:11, 8 March 2006 (UTC)


 * Kevin, There is no electric current in the portion of the logic step where the voltage is constant with distance. You need to have a voltage gradient with distance to keep the electrons drifting.  Only energy current, not electric current, is possible if the voltage is constant with distance along a conductor.  There is current at the front where the voltage varies from 0 to 10 volts.  See the Catt Anomaly diagram.  The lie in the Catt Anomaly diagram is that displacement current flows from one conductor to the other in all places, allowing electric current behind the front.  Theory C holds true behind the front of a logic step, as well as in the steady, charged capacitor.  Unless, that is, the magnetic field from the opposite conductor causes the current in the conductor of interest, which is of course what happens.  Each conductor causes the current in the other one.  At the front it is due to electromagnetic radiation emission due to charge accelerated by the electric voltage varying from 0 to 10 volts, and where the voltage is steady at 10 volts, the electric current is caused by the magnetic field from the other conductor.  In each case the electromagnetic radiation emission and the magnetic field from each conductor causes the current in the other conductor, a mutual induction situation.  At a large distance from the transmission line, no electromagnetic radiation or magnetic field exists, because the contribution from each conductor cancels that from the other conductor exactly in perfect interference.  So there is no energy lost from the system by radiation, 100% is exchanged between the two conductors, causing the currents in each of them.  Nigel 172.213.66.114 14:56, 9 March 2006 (UTC)

Nigel, there is a magnetic field, so there must be a current flowing to create it! Don't forget that we are dealing with current flow in an inductor - if the voltage is constant, the current must be as well, otherwise we violate the relationship between current and voltage in an inductor. I think you've lost track of the first and second differentials! -- Kevin Brunt 17:00, 9 March 2006 (UTC)

An earlier exchange of thoughts-
Just come across another problem quoted by Nigel Cook. When a capacitor is in series with a load, how exactly does the energy get from one plate to the other. Does this need displacement current?. If so, how is this displaement current generated? --Light current 20:37, 19 December 2005 (UTC)


 * The energy is transmitted when the field goes across the places. And, the energy goes through the wires on either side of the plate.  Pfalstad 20:43, 19 December 2005 (UTC)

Ahh but there can be no energy flow across (perpendicular to) the plates beacuse the Poynting vector points in a direction parallel to the plates-- no?. --Light current 20:52, 19 December 2005 (UTC)


 * Right you are! Interesting.  Well, then the energy goes down the wires.  It doesn't go across the plate.  Pfalstad 20:58, 19 December 2005 (UTC)

What, and by passes the capacitors as if they werent there?--Light current 21:17, 19 December 2005 (UTC)


 * No, I meant it stops at the capacitors as if they were an open circuit. Pfalstad 21:53, 19 December 2005 (UTC)

If thats the case, how does energy get to the load?

capacitor | |   -| ||                     | |               |--|  ac source                            |  | Load (resistor) | |                                       |--|    |

--Light current 21:58, 19 December 2005 (UTC)

The energy gets to the load by the bottom wire. And to anticipate your next question, if you put another cap at the bottom, then I don't know what happens. There must be some energy flow across the cap... I can't simulate the fields with the resistor present, so I don't know what the poynting vector looks like in that case. Pfalstad 22:12, 19 December 2005 (UTC)

Ok, consider the case where there is a capacitor on either side of the load. Straighten everything out so the wire, caps, and resistor are in a line. Consider the time when current is maximum (voltages across each cap = 0). There is a magnetic field around the wire/caps/resistor. The electric field lines lead from the top wire to the bottom wire; near the resistor, they are parallel to the wire. If you work out the poynting vector, you'll find that it's pointing to the resistor. Energy is flowing from the wires into the resistor. It's not flowing across the caps, though; it just goes around them. Weird. Pfalstad 22:24, 19 December 2005 (UTC)


 * Interesting! Im going to think about that for a while. Needless to say, I dont know the answer.--Light current 22:30, 19 December 2005 (UTC)


 * Do you mean arrange cct as below: ?

|                                                                 |___                                    ____  C (Zo is very low) |                   Energy flow   |--| ->      |  |                                  |  | R Load ac source(Zout=R)      |  | |--|                                   |___                                    ____  C (Zo is very low) |       | yes.

I've drawn the Cs as little transmission lines to help us in thinking about this cct. This arrangement of components would indicate that the energy gets shared between the 3 components initially but we know that the Cs do not charge up and can therefore hold no energy. If they do not hold energy, they must reflect it.

Ah ha! Maybe the energy stolen by the capacitors in the first few ns (2 way transit time of TL) is given back to the load or source after reflection. Probably given back to the source actually (assuming matched source and load resistances). (ie the capacitors introduce a small but positive mismatch to the source) Any reaction to my thoughts?--Light current 23:07, 19 December 2005 (UTC)


 * Well the capacitors definitely charge, and they hold energy when charged. I'm not sure I understand what you are saying. Pfalstad 23:16, 19 December 2005 (UTC)

No. The capacitors do not charge up because you have an ac source! (the voltage on each side of each capacitor is the same--roughly). Charging means holding separated charges on each plate and having a steady voltage difference between the plates (Q=CV)-- we dont have that here. In normal circuit theory, large capacitors act like a low impedance to the ac and so dont drop any voltage. In the em field representation, they dont drop any voltage because their characteristic impedance is so low compared with the resistance of the load. Understand so far?--Light current 23:43, 19 December 2005 (UTC)


 * There you go again, assuming the high-frequency limit without saying it. Ok, in that case the caps charge very little.  I don't think I understand enough about your cap-as-TL model to say whether the energy is reflected or what happens to it.  Pfalstad 23:52, 19 December 2005 (UTC)

Sorry. Yes I am assuming the capacitors heve low impedance compared to the resistor. But thats what you expect for coupling capacitors isnt it? OOPs looks tho I didnt mention that. But the argument still holds for any size capacitor. Anyway so far so good. What Im trying to show here is that energy can get to the resistor without the need for displacement current going from one plate to the other in the capacitors. EM energy can flow up and down the transmission line capacitors which are o/c at the far end of course. My propostion is that it all must happen by EM fields.


 * I don't think anyone disputes that the transmission of energy across the capacitor all happens by EM fields. I certainly don't.  I don't understand how the TL representation helps.  Of course I mentally picture a TL as a chain of caps and inductors.  I suppose you have lots of experience with TL's, so perhaps they are more comfortable for you to work with. User:Pfalstad

Now in one limit, if we make the capacitors smaller and smaller the energy transferred to the resistor will tend to zero. In this case all the energy must be reflected back to the source by the capacitors. In the other limit, where C-->oo, none on the energy is reflected back to source. So the only effect the capacitors have is to decide how much of the incident radiation is reflected! Current does NOT need to pass from one plate to another. Energy enters the resistor sideways. Energy enters the capacitors sideways and is reflected (to a greater or lesser degree). (See : Talk:displacement current to see what Nigel said).--Light current 00:19, 20 December 2005 (UTC)

A new start (already)
Why not start from an entirely different standpoint. Consider a (very) long straight (single) wire. Apply a square negative voltage pulse to the end of the wire. Since the wire is effectively a TL, we get a nice square pulse travelling down the wire. Now consider this in terms of electrons. Before we start, the electrons are distributed "evenly" along the wire. (They are in an equilibrium state.) We push some electrons in at the end. We now have an out-of-equilibrium situation at the end, so the electrons tend to move towards places where there are fewer electrons per unit volume. Eventually the electrons regain their original spacing and a number of electrons equal to the number inserted are pushed out the far end (for some definition of "end".) The "dynamics" actually cause the volume where the excess number of electrons is found to move down the wire as a "nice square pulse", but this would require a lot of physics and mathematics to demonstrate. However, note that the electrons move because the repulsion due to the like charges of the electrons results in a strong tendency to an even distribution of the electrons. -- Kevin Brunt 19:32, 8 March 2006 (UTC)
 * You say
 * Apply a square negative voltage pulse to the end of the wire
 * I ask, with respect to what: anohter conductor close by or a gound plane...You must define the return conductor or this will be like Nigels capacitor with only one plate.!--Light current 15:02, 9 March 2006 (UTC)

{Response to Light Current} For the sake of argument, the wire is infinitely long, and the universe is closed, so the wire goes all the way round and connects to the other side of the battery from the opposite direction. Alternatively, we just take the "return" conductor out in a wide loop and just keep on making it bigger until a further increase in size has no detectable effect, at which point we can ignore it. -- Kevin Brunt 18:57, 9 March 2006 (UTC)
 * OK so you have an infinite impedance TL! (C=0, L=u/8pi per m)--Light current 02:12, 10 March 2006 (UTC)

Now cut the wire so that there is a gap with clean faces perpendicular to the length of the wire. This is a capacitor, but unlike a Catt TL capacitor, the charging current flow is perpendicuar to the capacitor plates. Consider our square pulse in this circumstance. The same principle applies; however, the conductor is now finite, so the pulse will be reflected at the gap (and presumably at the input and as well.) Eventually the pulse will "run out of steam" as energy is lost into the resistance of the wire, by which time the excess electrons will have become spread evenly along the wire. -- Kevin Brunt 19:32, 8 March 2006 (UTC)
 * If you have capaitor charging, you have, by definition, a (semi) permanent separataion of charge. How does this separation happen in your cct. Ans: Assuming an infinitely long step applied from the generator (battery), it starts with the electric field progressing at the speed of light in the medium, and ends with the (v slowly moving) actual charge carriers moving round the circuit via the generator. The electric filed is the advance guard (so to spea) and the charge carries are the backup troops and supplies etc.--Light current 15:11, 9 March 2006 (UTC)

However, consider the situation at the gap. We are arguing that there are repulsive forces operating between the electrons in the wire. But is there any difference between the gaps between the electrons in the wire, and the gap separating the electrons in the two halves of the wire. The difference is that while the fields can cross the gap, the electrons can't. -- Kevin Brunt 19:32, 8 March 2006 (UTC)

A little thought should show that as the pulse travels down the wire, the spacing of the electrons at the leading and trailing edges is changing, whilst "within" the pulse (as indeed ahead and behind the pulse) the spacing of the electrons is effectively constant. Thus in Maxwell's terminology there is displacement current at the edges, but not elsewhere. At the gap the fields (and so the repulsion) can cross, but the electrons can. Thus what crosses the gap is not "current", or "radiation" or even an "electromagnetic" anything, but purely electric field. Because there are two electric fields there is force, and because that force can cause the electrons on the other side to move, there can be transfer of energy. -- Kevin Brunt 19:32, 8 March 2006 (UTC)

Note that this is all in terms of motion and fields acting along the length of the wire. Once the electrons have "settled", there is no current (or even displacement currrent) so there is no magnetic field to consider. The stored energy is all capacitive and (for a negative pulse) is stored in the reduced spacings between the electrons, which has required an input of energy. -- Kevin Brunt 19:32, 8 March 2006 (UTC)

This is a simple description of a capacitor. It omits the obfuscation of Catt's TL description, which hides the action perpendicular to the plates in all the cockadoodledoo about the flow along the plates. -- Kevin Brunt 19:32, 8 March 2006 (UTC)


 * You make so many different points at one time that it is not easy to answer you quickly, and some points may be missed. I believe all these points have been dealt with before, but I will try to answer them in due course. I dont agree that Catt's description of a capacitor as a TL is deliberate obfuscation. I believe that that is what he believes. --Light current 02:15, 10 March 2006 (UTC)

I really can't decide whether Catt honestly believes in his arguments or whether he is just being contrary. He seems to latch on to a point of view and stick with it. The reason that the discussions in Wireless/Electronics World eventually get dropped is that they never progress. Catt makes a bare assertion without adequate supporting proof, someone points out a difficulty in Catt's argument, Catt ignores the issue, picks up on something irrelevant in the comment and goes off at a tangent. Eventually, the argument gets back to the original subject and Catt reiterates his original point as if it were established as fact.

Heaviside's energy current is entirely plausible when looking at a "loose pulse" travelling down a TL, but it gets more and more problematic. Somewhere on Catt's pages he compares the concept of charge with the Phlogistic theory, which is somewhat ironic, since the contortions to keep Phlogiston "in sync" with the observed facts far more closely mirrors Catt's attempts to break differential calculus and the claims that the "contralpuntal" energy flows cancel out the energy lost into the TL's resistance than any of the alleged problems with the standard theory. -- Kevin Brunt 12:20, 10 March 2006 (UTC)
 * Alas, this is the way with so many arguments as revision of these pages may show. This is why it is important to deal with one point at a time and try to come to some sort of consensus on it before moving off at a tangent.--Light current 17:31, 10 March 2006 (UTC)

Nigel to LC

 * In the March 05 issue of Electronics World, there's a letter from me dealing with what happens if you charge up one plate of a capacitor before the other. It concludes a capacitor plate behaves like an aerial/antenna, and the other (parallel) plate behaves like a receiver antenna.  The publication of that letter led to the major breakdown in communication between Catt and myself.  Catt wrote to the editor complaining, but had no scientific comment to make.  If Roger Penrose had written the letter, I suspect Catt would have been more flattered.  Nigel 172.189.107.149 13:02, 8 March 2006 (UTC)


 * OK Nigel, but now you know wher I am coming from, and I think I know wher you are coming from. BTW can I just ask if you do understand (not asking you to argee) the above argument?--Light current 16:21, 8 March 2006 (UTC)


 * Light Current: by above argument you are referring to the December discussion, since Kevin's comments above are dated four hours after yours. I see what you are saying, but I don't think it is sufficient to say "energy enters a capacitor sideways" and "no current flows from one plate to another".  What you should say is that when you charge up both plates in a capacitor with opposite charge simultaneously, currents flow along the plates.  Because the direction of the current in each plate is the opposite to that in the other, it is confusing and vaguely inadequate to say "energy enters a capacitor sideways".  The statement "no current flows from one plate to another", is correct but in a sense that was a feature of Maxwell's system, where "displacement current" was ethereal current, not real charge flow.


 * Getting back to the Catt Anomaly issue: there is no electric current where the voltage does not vary with distance along a conductor. So there is no electric current in the logic step behind the front (where the voltage varies from 0 to 10 volts).  This is because you have to have a variation in electric field strength to get electrons to accelerate.
 * In the Catt anomaly diagrams, Catt shows a current flowing in the conductors where the potential along the conductor is constant 10 volts. No electric current will actually flow under this condition, because you have to have the electric field varying with distance along a conductor to cause electron drift. The electrons accelerated briefly at the front will presumably soon slow down behind the front due to resistance, so "energy current" not "electric current" is possible where the voltage is constant with distance along a conductor) Catt of course fiddles it completely by drawing displacement current as flowing where the electric field is both constant with time and constant with respect to distance along the transmission line. The Catt Anomaly is an total, complete, and utter hoax from start to finish.


 * This justifies the idea that the Catt Anomaly has (1) disproved the competence of conventionally trained physicists who didn't spot the mistakes (names listed in the Catt Anomaly book), and (2) proves that Theory C does hold behind the rise (front) of the logic step, in addition to Theory C applying in a steady charged capacitor, of course.


 * Of course you could preserve electric current in the part of the logic step where the voltage is a constant 10 volts by a mutual exchange of magnetic field energy from motion of charge in the other conductor.


 * So at the front of the logic step, the varying electric field due to the finite rise portion from 0 to 10 volts accelerates electrons. Once the slow-moving electrons are within the constant 10 volts part of the electric field, the electric field is unable to continue to exert a force to overcome resistance.  But the magnetic field from the charge in the opposite conductor then causes the current.


 * "Displacement current" if it exists equals permittivity times dv/(dt.dx).


 * As I've said, Maxwell confused this for "radio" (electromagnetic radiation, I'm not talking of sine waves) where the transverse emission of radiation (radiated power) is directly proportional to the acceleration of the charge, i.e., to the rate of change of the current in the conductor (not displacement current).


 * SUMMARY:
 * (1) Maxwell: circuit is completed by something called "displacement current" which is not radiation as such but :flows when voltage varies with time and distance.


 * (2) TRUTH: circuit is completed by radiation which flows as the ENERGY current in the conductor varies, accelerating charges at the front of the logic step.  Nigel 172.213.66.114 15:14, 9 March 2006 (UTC)


 * Ok Nigel Ill reply to your paras in turn.
 * 1 I agree that when you have an equal an opposite electric field on both plates in a capacitor, currents flow along the plates and em energy flows between and // to the plates. I do not like to use the word charged up as this implies a permanent separation of charge carriers that we all know cannot happen this fast. I agree that no current flows from one plate to another.
 * Why do you think it is not sufficient to say "energy enters a capacitor sideways". THis seems to be a perfect description of what happens and has been demonstrated many times by the placing of microwave loads in waveguides for best matching.--Light current 16:03, 9 March 2006 (UTC)


 * To the initiated "energy enters a capacitor sideways" is meaningful, but it is not visually helpful for people who picture energy carried in or guided along a wire entering a capacitor plate. At the end of the road there is a T-junction into the main road.  I drive to the end of my road, and then have to turn left or right on to the main road.  So you'd say "Nigel enters the main road sideways."  Does that help anyone?  Is it a perfect description? Nigel 172.213.66.114 16:25, 9 March 2006 (UTC)


 * OK Lets assume that all of us here are initiated by virtue of our long previos discussions. AS I said the side ways picture when were talking em energy flowing down a TL is actually more intuitive rather than less IMO--Light current 16:46, 9 March 2006 (UTC)
 * Reply to para 2
 * If electric current only flows at the step transistion, how do you explain the fact that an ammeter connected betweeen the gen and the line would show continous current being pumped in? I=V/Z--Light current 16:50, 9 March 2006 (UTC)


 * Reply to para 3. A hoax is some action intended to to make some people believe somethin that is not true. THe Catt anomaly as written mrely asks a quesion about the origni of the charge. Therfore, by definition, it cannot be a hoax.--Light current 18:11, 9 March 2006 (UTC)

Kevin to Nigel
Nigel, you have definitely got a problem with your differentials! Firstly, let me point out that you are not dealing with the motion of the electrons, but the motion of the electrons after the random motion due to the thermal energy has been "averaged out". Secondly, if we are working within Catt's calculations, there is no electrical resistance. because Catt uses the version of the Telegrapher's Equations that assume that the resistance is zero!Kevin


 * Kevin, since you are interrupting so will I. We have discussed the fact that the conduction electrons are going at 1 % of the speed of light due to normal motions, and that the 1 mm/s drift is a net drift.  The random drifts don't affect my argument, and I don't need to mention all the details repeatedly for this reason.  As you say resistance is important and added to the characteristic impedance, it determines the current.  But I still want to know what is going on regards "displacement current", i.e., the mechanism of the transverse action in the medium between the conductors.  It is not all longitudinal push.  Nigel 172.213.66.114 19:08, 9 March 2006 (UTC)

At the leading edge the (averaged) electrons are accelerated to the electron drift velocity. Before they started moving the forces acting on them were in balance. The acceleration is due to the increased force as the electrons "behind" approach. The accelerating electrons approach the electrons in front, so an increased force acts against the accelerating electrons. (Newton's 3rd Law also requires that there is a force acting to push the electrons ahead forward.) A new equilibrium is established where the electrons have been compressed together and are moving forwards, all at the same speed (on average, of course.) Behind the trailing edge, there are no longer electrons pushing forwards. At the edge itself, the electrons experience a net force against the direction of motion and decelerate. Between the edges the net force on an electron is zero, which means that the electron neither accelerates not decelerates. This does not mean that it is stationary. It means that its velocity is constant. Note that in compressing the electrons an input of energy is needed as the electrons move against the repulsive forces, and that in accelerating the electrons more energy is needed to supply the magnetic field due to the moving charge. These inputs of energy are observed on the macroscopic level as the capacitance and the inductance of the conductor. (There is also an input of energy needed to supply the kinetic energy of the moving electrons, but this is so small that we don't need to worry about it.) Note also that although the forces on an individual electron are in balance, it is moving against the force acting backwards, and with the force acting forwards; there is a continual transfer of energy from the electric field behind the electron accelerating the electron forward at exactly the same rate that energy is being transferred into the field ahead of the electron as it is being decelerated by the backwards force. The motion of the electrons forward transfers energy with it. -- Kevin Brunt 18:43, 9 March 2006 (UTC)

It is becoming impossible to maintain the structure of this discussion! -- Kevin Brunt 21:12, 9 March 2006 (UTC)

Nigel maintains that because he can't get the motion of the electrons in a conductor to propagate at the required velocity, that all the physicists who say that it does must be wrong. I can't get it to work either, but I am willing to believe that I'm wrong rather than the physicists! I have a background in chemistry, and I suspect that if I were to ask a chemist about this (and there's a whole department of them a couple of floors above my desk...) that I'm quite likely to get an answer along the lines that it doesn't matter what the electrons are doing, because what is moving is the probability of where each electron is to be found. In other words, somewhere down at the quantum mechanical level, an electron finds that the volume in which it is supposed to be found 95% of the time has shifted down the conductor, so it has to move to suit. As long as it doesn't have to move faster than light there's no problem. At this level, the idea that an electron has a known position and velocity becomes severely entangled with Heisenberg's Uncertainty Principle. I was careful in my discussion above about the motion of the electrons, the forces acting on them and the transfer of energy, to stress the word average. At the "average" level, the electrons behave like that. An appeal to the motion of the electrons at a lower level is irrelevant, because the average explains what is happening, not why it is doing it. -- Kevin Brunt 21:12, 9 March 2006 (UTC)

My comment about Nigel's derivatives is because he is showing distinct signs of confusion about the difference between velocity and acceleration. It ought to be clear that when a free-standing pulse travels down a TL it moves the conduction electrons along the conductor by a distance such that the distance moved by the electrons times the cross-sectional area of the conductor contains precisely the number of conduction electrons needed to match the quantity of change transfered along the conductor by the pulse. It ought also to be clear that the electrons move while  the pulse is passing, and that they start moving as the leading edge passes, that they stop as the trailing edge passes and that in between they are moving with a constant velocity. A litle more thought will show that because of the precise relationship between voltage and current in a TL, the electrons move at a velocity proportional to the voltage. -- Kevin Brunt 21:33, 9 March 2006 (UTC)

Nigel asked about the displacement current in my description. Obviously, before and after the pulse the electrons are "stationary" so their separation is unchanged and the electric fields likewise are unchanged. The same applies during the "body" of the pulse; no change in separation, no change of field. However, at the leading and trailing edges the separation is changing, so the fields are changing too. These changing fields are along the conductor, not perpendicular to it. Because the electrons are compressed together (or stretched apart for a positive pulse), there is a net imbalance of charge between the pulse edges. This can create a field perpendicular to the conductor as well as the one along it, which is where the interaction between the conductors in Catt's TLs comes in - it is in addition to the action along the conductor, not instead of it!


 * Gentlemen! please try to the follow the simple rules I outlined at the beginning and dont interrupt each others posts. I know its very tempting to do so- but you must resist that temptation!.Also try to keep your posts short and to the point without bringing in too many different threads in one post. THanks--Light current 23:22, 9 March 2006 (UTC)


 * Kevin please try to only post one msg at a time. It will be easier for everyone. Thanks--Light current 02:56, 10 March 2006 (UTC)