Talk:Ivor Catt/Archive 17

New information on TLs
Kevin and Nigel please have a look at this interesting paper! 8-)--Light current 23:21, 22 May 2006 (UTC)

Kevins response
Yes, this is extermely interesting - it blows Catt out of the water! This is a couple of people doing the job properly. The key to understanding where this paper differs from Catt's misconceptions lies in comparing figure 3 with one of Catt's equivalents. Specifically, there are graphs not only for the voltage, but also for current and displacement current. (Catt, of course, ignores both.) The graphs clearly show that between the "Return Wavefront" and the load, the ratio of the voltage and current is not equal to the TL's characteristic impedance. -- Kevin Brunt 12:04, 23 May 2006 (UTC)

The first sentence of section 2 is equally important.
 * In electrical engineering transmission-line propagation is traditionally considered to be a distributed process in which an applied voltage progressively charges up the line capacitance, though the series inductance of the line, as it moves towards the load.

Much of Catt's case rests his claim that everyone before him thought that a capacitor charges evenly across its surface - (which he bases on a misrepresentation of Maxwell's derivation of displacement current as the rate of change of electric field.) This quote asserts that "everyone" actually thinks that in a TL the charging is concentrated at the voltage front. -- Kevin Brunt 12:04, 23 May 2006 (UTC)


 * Now before you get too excited Kevin, I have to say that I dont necessarily agree with everything in that paper. I have not studied it in detail yet. The main disturbing point I have picked up is that the authors seem to be suggesting that there is something physical (current) passing from one wire to the other at the step front. I dont see them explaning this except to say that it is Displacement current which we all know (and have agreed) is not a real current. If they mean dD/dt and a magnetic field, I would agree with them, but they go on to assume that this disp current at the step forms the return path for the travelling current in the top wire! This plainly cannot be true as it is shown. Can you explain this assumption of theirs? 8-?--Light current 12:29, 23 May 2006 (UTC)


 * Also, what do you assume that the authors are saying in Sect 2 - that the traditional concept is wrong? Or what? (ie C charging via L). Your following para seems a non sequitur 8-?--Light current 13:35, 23 May 2006 (UTC)

LC, there is "something physical" passing between the wires - the electric field. What they are doing with the dotted line linking the longitudinal currents in the wires with the transverse displacement current between the wires is stressing the association between the two. In particular, by showing that the displacement current matched the longitudinal current, they can work out what the total dD/dt is without needing to know what the shape of the wavefront (thus avoiding Nigel's infinity question.) The authors appear to have been careful to avoid saying that the displacement current actually flows. -- Kevin Brunt 18:23, 23 May 2006 (UTC)


 * I dont agree that the electric field flows between the conductors. Do electric fields flow? An electric filed is a stess in the medium! 8-|--Light current 21:36, 23 May 2006 (UTC)

I read the paragraph that starts section 2 as an assertion by the authors that their approach provides a deeper insight. Specifically, whereas the "traditional view" merely notes that the process of charging the TL's capacitance sweeps along the TL, (with a zone where displacement current is "happening" sweeping along with it) the authors argue that this displacement current (the changing electric field and the associated magnetic field) are essential to a proper understanding of the process. -- Kevin Brunt 18:23, 23 May 2006 (UTC)


 * Ahh. now I would agree with your terminology here:
 * (the changing electric field and the associated magnetic field). But please do not mention current. :Current is something that flows! The changing electric and mag fields may seem to indicate a flow of something between the conductors, but what is that something- and is it really there? After all, dont you get changing electric and mag fields with em radiation? Is there a current in free space em radiation? 8-)--Light current 21:36, 23 May 2006 (UTC)

My point is that Catt believes that Maxwell's displacement current theory relies on the assumption that the charge on a capacitor is evenly spread across its surface. Hence his conviction that demonstrating that a TL charges in a stepwise manner is sufficient to destroy all of Maxwell's work. I picked out the quote as a counter to the claim of a "crucial realisation by Catt in the 1960s and 1970s that [...] since a capacitor has physical size, so charge entering [...] has to spread itself out [...]" -- Kevin Brunt 18:23, 23 May 2006 (UTC)


 * ....stepwise manner is sufficient to destroy all of Maxwell's work.
 * Sorry? Kevin 8-? sounds like youre agreeing with Catt on capacitor charging. If Catts stepwise result destroys Maxwell, what does Edwards and Sahas stepwise graph do? Yes Maxwells disp current theory was mistaken. Maxwells equations  are not wrong. Its the interpretation of dD/dt thats wrong!--Light current 15:43, 24 May 2006 (UTC)

I wrote "Catt believes..." - and I believe that Catt is totally wrong! I think the sequence is this:- Maxwell observed that for a given charge the voltage on a capacitor varies with the geometry of the capacitor, but that when you calculate the total field between the plates the geometry cancels out and you get left with Q = εE = D, so that only the properties of the dielectric is relevant. From Q=D, obviously dQ/dt = dD/dt follows, (and indeed the proper partial differential 3D form.) By proper regard to the signs Maxwell got to dQ/dt + dD/dt = 0; this equation includes Kirchoff's Current Law (dQ/dt = 0) as a special case, which lead Maxwell to call dD/dt "displacement current" (because it fitted in with his concepts of the Aether) and to propose that it is the sum of currents + change of field that is zero, rather than just Kirchoff's sum of currents.

Maxwell's argument progresses from a specific case to the more general. Catt's argument is entirely a quibble about the method generalisation; Maxwell seems to have calculated the field between the capacitor's plates for the specific (hypothetical) case where the charge is evenly spread across the plates and the field between the plates is even throughout and doesn't "bulge out" at the edges. No doubt Maxwell chose these conditions to make the calculations easier; in fact, the additional calculus required for a "real" capacitor would have severely obscured his argument. I would imagine that Maxwell also proved (if the proofs were not already in existence) that his simplifications did not invalidate his argument. It is clear from what Catt has put up on the web that the Dec 78 paper is founded on the entirely erroneous premise that Maxwell's line of reasoning breaks down when the charge is not evenly distributed or the field is not absolutely "straight". -- Kevin Brunt 16:40, 24 May 2006 (UTC)

Alfreds contribution
Or, more precisely, it is the change in the electric field (and thus, flux density) that passes between he wires. By the way, you two do realize that Maxwell's equations, as usually written, give the divergence and curl of the fields everywhere based on present values of the charge and current densities, right? Doesn't this seem odd? What happened to no influence can travel faster than light in a vacuum? How can a change in charge density here and now, instantaneously affect the fields everywhere? The fact that a change here and now cannot affect the fields everywhere is the 'source' of the displacement current term in Maxwell's equations.

For example, look at the curl H equation. If the charge current density changes here and now, we know that this change in the curl of H must propagate at the speed of light to places distant from here. Sure enough, we find that the displacement current term cancels any change in the charge current density in such a way that the H field at distant points from here do not change until just the right amount of time for light to propagate to that point has elapsed.

In fact, it is possible to recast Maxwell's equations in a form (retarded potential form) where the displacement current term disappears. Alfred Centauri 20:00, 23 May 2006 (UTC)


 * Yes I was going to say 'retarded potential' although I never understood it at university (still dont!) Anyway if the displacement cuurrent disappears, I am very interested,. Can you elaborate?--Light current 21:43, 23 May 2006 (UTC)

Sure. The vector potential in the vacuum can be expressed in two ways:


 * $$\vec A = \frac{\mu_0}{4\pi} \int_{V}^{} \frac{\vec J(t) + \frac{\partial \vec D(t)}{\partial t}}{r}\, dv = \frac{\mu_0}{4\pi} \int_{V}^{} \frac{\vec J(t - r/c)}{r}\, dv$$


 * The above assumes the gauge:


 * $$\nabla \cdot \vec A = 0$$

Clearly the 2nd integral expression does not contain a displacement current term but instead uses the value of the charge current density element at a time in the past determined by dividing the distance to the current element by the speed of light in the vacuum. Interesting, huh? Alfred Centauri 00:53, 24 May 2006 (UTC)


 * Errr (cough, cough) yes of course! So we dont need displacement current if we only wait a while. Is that what the equations mean?--Light current 01:41, 24 May 2006 (UTC)


 * I suspect that it'd be a bit like Catt explaining a TL using voltage and impedance and then saying that charge doesn't exist. Alfred is advancing a different formulation. Since it hasn't completely ousted the "traditional" version, I suspect that the "displacement current" issue turns up in another form, that it makes some parts of the physics less obvious, or simply that it makes the mathematics even more horrid. Presumably, for a TEM wave in a vacuum, (t - r/c) is 0 (or at least constant), so that the wave is receding from the charge current density at some time in the past. -- Kevin Brunt 09:45, 24 May 2006 (UTC)
 * I don't think that Catt would believe it, in any case. -- Kevin Brunt 09:45, 24 May 2006 (UTC)

Nigel enters discussion
Light current and Kevin, my initial feeling after a brief glance through the paper, it is worse than Catt's paper; it makes the same fundamental error of assuming a vertical front on the current (zero rise time), which is impossible because you get radio emission from accelerated electrons and electrons accelerated from a net current of 0 to i amps instantaneously implies the acceleration of electrons is (1 mm/s or so)/(0 seconds) = infinity. This is a lie because we don't get an infinite radio emission from a logic step. What happens is that all real logic steps are not vertical but have a rise time, and this is crucial rather than technical trivia. So the authors of this paper have made the wrong simplifying assumptions. In addition, they should cite and discuss previous work on this subject area at the beginning of their paper, which they don't. It is basically an unhelpful armwaving paper, but not as bad as this one: .. Contrary to what Kevin says, it doesn't (at first glance) say what the displacement current is. Nigel 172.143.55.86 10:33, 24 May 2006 (UTC)


 * Nigel, a zero-rise time front will pass a point on the line in zero time, so Δt = 0. In zero time there can be no change of charge, so ΔQ = 0, and consequentally, ΔD = 0 as well. ΔD/Δt is thus 0/0 which is not infinite, but indefinite, it can be any value at all, and you have to work out what it is by some other means. For any non-zero Δt, ΔQ over the whole line (or any part of it that entirely contains the wavefront) is IΔt, where I is the current flowing into the line. ΔQ and ΔD are thus independent of the rise time, and the paper makes no assumption thereto. In fact by explicitly using displacement current the authors sidestep the entire (irrelevant) issue of rise time. -- Kevin Brunt 12:27, 24 May 2006 (UTC)

Alfreds response to Nigel and Kevin
Reading the responses from Kevin and Nigel reveals just how tricky analyzing this 'thing' can be. I think the root of this particular disagreement is in the author's statement "When a potential Va is suddenly applied across the input of the cable...". How do you do that? Is this even possible? If a proper analysis is done of how a voltage source can be suddenly connected to the cable, I'll bet that one will find themselves analyzing yet another TL problem! In other words, this problem is not quite properly defined yet because we have specified something (suddenly connecting a potential) that may not be possible. In any case, it is my gut feeling that in reality, the impulse of dD/dt is 'low-pass filtered' so that we don't actually have a step in the field quantities or currents. Alfred Centauri 13:35, 24 May 2006 (UTC)


 * Alfred, I dont really think the risetime of the step is that important to the papers description of the mechanics. Agreed you cant get zero risetime and therfore the so called 'Disp Current' will not be a delta fuction but more smeared out. The authors are considering the ideal case to simplify- as you and I would. I dont think this invalidates their argument.
 * What does invalidate the argument, IMO is that they are implying a return parh for the input current by means of 'displacement current' crossing from one conductor to the other at the 'step'. OK this may work in a line with a dielectric- but not in a vacuum. Is it not better to say that the step defines the wavefront of the advancing electric and magnetic fields? This wave front can be as fast as you can make it, or as slow as you like. No prob with 'infinite currents' and such like 8-| --Light current 15:55, 24 May 2006 (UTC)


 * Definitely agree that the rise time is a red herring. Disagree on displacement current, however. The use of displacement current is a convenient way of finding the total dD/dt at the step; the dotted line linking the charge flows in the conductors via the displacement current is a statement of their relationship, to to justify the assertion that the current in the conductor is matched by the changing electric field that is the displacement current. The authors were careful not avoid saying that the displacement current flows. Don't agree on vacuum vs dielectric either. A dielectric is merely a vacuum with charges in it, as is a conductor. In both, when pushed by an electric field the charges move, but in a dielectric the charges will move back. The motion of the charge due to the force requires energy; in a conductor the energy is permanently transfered, in a dielectric, the energy is given back. -- Kevin Brunt 17:22, 24 May 2006 (UTC)

Sorry guys, it appeared to me that Nigel and Kevin were disagreeing on the rise time issue so I thought my comments were quite appropriate in addressing that perceived disagreement. However, do keep in mind that even though we are talking about classical EM theory here, we do need to be careful with our simplfying assumptions here. Don't forget the lesson of the Ultraviolet catastrophe.

Now, to address LC's statement about the return path of the input current. The displacement current term, by design, closes currents so that Ampere's law is satisfied when field values there and now are determined by changes here and now. As I stated above, when we express field values there and now in terms of changes here some time ago, we see that the displacement current term (in present value formulation) is actually the result of changes in 'real' current at some time in the past - see Jefimenko's equations.

Yes, the math is uglier when expressed this way in the usual 3 space + time notation but I'll bet the math looks much nicer in 4-vector notation. Alfred Centauri 18:30, 24 May 2006 (UTC)

Zero-rise time step
Alfred, the zero-rise time step is one of Nigel's particular bêtes-noires; he also thinks that displacement current is "radio transmission". I must agree about simplifying assumptions - some of Catt's writing is based on quote-mining introductory textbooks, and misusing the over-simplifications he finds. (I'd mention electron drift velocity in this context, except that Nigel and I always get into crossed purposes about it!) -- Kevin Brunt 19:45, 24 May 2006 (UTC)


 * Kevin, wrong! (1) The zero rise time for a current implies acceleration from a mean drift velocity of 0 to a mean drift velocity of whatever (1 mm/s) over zero time, which is infinite acceleration.  This can't occur for the reasons already given, such as the fact that infinite acceleration of charge would produce an infinitly intense burst of radio emission.  You are getting yourself tangled up.  (2) The 'displacement current' is no more radio emission than modern thermodynamics (kinetic theory and radiation) are synonymous with caloric and phlogiston.  Your sneer on this point is therefore bunk.  I'm not saying displacement current is radiation, I'm saying the phenomena normally attributed to aethereal charge motion are actually accounted for by radiation.  (Any effect from vacuum displacement current is due to charge polarisation in response to the the electric field set up by light speed radiation, not vice-versa.)  The difference has profound consequences in physics, changing the nature of displacement current into something equivalent to the time-dependent Schroedinger equation and Dirac's equation; energy transfer by time-varying fields .  Nigel 172.189.116.131 10:49, 25 May 2006 (UTC)

Nigel, I said that we always end up at crossed purposes over electron drift. There's no need to demonstrate.... Try visualising the charge as a rigid rod moving at the velocity of propagation. The rise time of the wavefront is effectively the profile of the end of the rod and there is no mechanical objection to that being flat. There is similarly no conceptual barrier to the front of the wavefront being flat as well; it can't happen in practice, but the infinities in your theoretical argument is merely a case of using the wrong mathematics and/or the wrong physical picture. Your zero divided by zero is because you're not moving from ΔQ/Δt to dQ/dt as Δt→0. Your instantaneous acceleration to drift velocity ignores the fact that the electrons are continuously changing velocity as they exchange energy with each other. -- Kevin Brunt 18:49, 25 May 2006 (UTC)


 * Kevin you make the arm-waving and false statement: "Your instantaneous acceleration to drift velocity ignores the fact that the electrons are continuously changing velocity as they exchange energy with each other." If current changes instantaneously, electrons have to move instantly.  If you drop a white feather, Kevin, and it instantly moves at drift speed with zero acceleration time, the mean acceleration equals [change in velocity]/[zero time] = infinity.  This breaks the laws of motion, since Newton's F=ma implies infinite force for infinite acceleration.  The proper explanation I've given time and again here is the only one which works.  (What's the difference between self-confidence and arrogance in asserting an explanation?  Answer: having factual evidence you are right!  I remember an argument with a physics teacher where I won.  He didn't thank me for teaching him physics but was instead rude.  Don't become a physicist if you expect thanks.)  Nigel 172.143.208.206 11:05, 29 May 2006 (UTC)

Nigel, $$Force = mass \times acceleration$$ is almost irrelevant here. The important equation is $$Work \ done = Force \times Distance \  moved$$. The work needed to supply the kinetic energy of the accelerated mass of the electron is trivial compared with the energy needed to supply the magnetic field of the accelerated charge and the energy needed to "fill" the increasing electric field as the electron moves towards the electrons ahead of it. The finite acceleration of the electron does not explain why the wavefront has a finite slope; instead it explains why the wavefront propagates at a velocity less than that of the speed of light in vacuo - it is advancing at a rate limited by the amount of energy available. From this it is obvious why the wavefront is not "flat" - the area between the "perfect" square step and the observed ramp represents lost energy - the energy that is lost (mostly at least) into the resistance! -- Kevin Brunt 18:53, 30 May 2006 (UTC)


 * F=ma disproves your false statement, so it certainly is not trivial for your model. I told you before that the work energy in the electron motion of drift is trivial.  It seems that you just want to be silly and tell me what I've already stated as if I wasn't the one to tell you.  Your statement that the wavefront goes slower than light in vacuo is false; I'm discussing the case where the two conductors are separated by vacuum.  In this case the wavefront goes at light velocity for vacuum.  The reason is that electrons at the front are being accelerated by light velocity electromagnetic radiation coming from the other conductor, as already explained to you (illustration at top of page).  This is a matter of fact, not of opinion, and you're just plain wrong.  Admit it!  Nigel 172.209.184.108 09:59, 31 May 2006 (UTC)

Nigel, a voltage step propagates down a transmission line at a velocity determined by the line's characteristic inductance and capacitance. These parameters can be changed by varying the spacing between the conductors, the surface area of the conductors and the material that the conductors are made of (eg use of iron rather than copper). Changing the parameters will modify the propagation velocity. It will not modify the velocity of electromagnetic radiation travelling in the gap between the conductors, except in the special case that the wavelength of the radiation matches the size of some constituent part of the TL, when transfer of energy into the structure of the TL can occur.

Regardless of your assertions, since an electron possesses charge as well as mass, accelerating it requires an input of energy to supply the created magnetic field due to the moving charge as well as that required to supply the kinetic energy of the moving mass. In addition, since an electron in a conductor is moving relative to other charges, it is necessary to take account of the energy associated with the forces resulting from the interaction of the fields due to these charges. On the macroscopic scale the stored energy due to the magnetic and electric fields is observed as the inductance and capacitance of the conductor. Note that the fields are due to the charge on the electron; they are bound to the current flow in the conductor. Because of this they are not "transverse electromagnetic waves", as physicists define the term, because they use "TEM" to refer to sinusoidally-varying waves in free space, where the electric and magnetic fields are explicitly not bound to moving charge. -- Kevin Brunt 19:25, 31 May 2006 (UTC)


 * Kevin said "since an electron possesses charge as well as mass, accelerating it requires an input of energy to supply the created magnetic field due to the moving charge as well as that required to supply the kinetic energy of the moving mass." I don't quite buy that, Kevin.  The magnetic field that is, as you say, bound to the moving charge, can be viewed as the result of the charge in motion or I, as the observer, in motion with respect to the charge.  That is, I can transform the magnetic field away by changing my frame of reference.  However, I cannot transform EM radiation away so I'm thinking you must be referring to the the energy that is carried away by the EM radiation.  Alfred Centauri 22:28, 31 May 2006 (UTC)

Kevins reply to Alfred
Alfred, I'm not sure that we're in the same scenario! Nigel and I are arguing over the flow of current in a conductor. The issue is that Nigel thinks that the electric and magnetic fields are separate from the motion of the electrons, and has had an article published in Electronics World which purports to prove his case by demonstrating that the kinetic energy of a lot of electrons moving at 1mm/second is not very much. I'm not sure how much of Catt's work Nigel currently agrees with; Catt thinks that the "Heaviside energy current" ($$E \times H$$) view of a DC current in a conductor is the same thing as a "TEM wave" in free space, and I'm far from clear how Nigel's "light velocity electromagnetic radiation coming from the other conductor" fits in. I get the impression that Nigel is not distinguishing between "field" and "radiation" and that he's really talking about displacement current. -- Kevin Brunt 00:15, 1 June 2006 (UTC)


 * Kevin: your statement that you can change the speed of electricity by altering the shape of the transmission line shows you are talking nonsense ... I've explained to you that QED Yang-Mills component of the Standard Model which accurately describes existing data and has predicted major experimental discoveries at CERN and elsewhere, explains that the electromagnetic field is composed of exchange boson radiation: electric fields are caused by this as are magnetic ones.  ...  All the best, Nigel 172.143.209.196 11:05, 1 June 2006 (UTC)  - OK, I've removed the reactionary comments.  Nigel 172.215.63.55 16:43, 10 July 2006 (UTC)