Talk:Ivor Catt/Archive 6

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Contents 1 Antennas and Induction fields 1.1 Deletion of posts 1.2 Nigel to LC? 1.2.1 LC to Nigel 1.3 LC's comment on (Nigels reply to Kevin 25 Jan )

Antennas and Induction fields

I believe Nigel is confused about radio propagation and Poynting's vector and that he is only considering the so called 'induction field' (near field) antennas whose signal strength decreases as the square of the distance from the antenna as opposed the the radiation fields that I am talking about (these field strengths are inversely prop to distance). Induction fields can, in my opinion, pass no energy as they consist of magnetic fields only or electric fields only. Both fields combined together in an em wave are needed to transfer energy from one place to another. If you look at proper high frequency antennas, you find that they are shaped like horns. This is done in order to match the impedance of the transmitter to the impedance of free space (377 ohms). Nigels antenna model is merely an induction antenna transmitting short range electric of magnetic fields only. Thier efficiency at radiating proper em waves is very low.--Light current 01:07, 23 January 2006 (UTC)

Nigel have a look at this. See if you agree with it!.[1]--Light current 03:42, 23 January 2006 (UTC)

Hi Light Current:

A time-varying current results in radio emission. Neither Catt nor anyone else has measured the fields in the space between two conductors as a TEM wave passes: they have only measured induced currents in other conductors. The diagrams in that article you quote, ignores radio emission occurring at the front of a logic step! Catt got the "Catt anomaly" wrong by relying on a book published in 1893 which ignored the step effects at the front of the TEM wave. Asserting ignorance is wrong. At the front of a logic step, current rises (in accepted picture) and this results in radio emission. Since each conductor is oppositely charged and carries an opposite current, the radio emission from each conductor (acting as aerials) is exactly out of phase with the other and so completely cancels that from the other as seen at a large distance. So there is no energy radiated to large distances! The only radio emission of energy occurs from each conductor to the other.

Yes I tend to agree with the last part of the above paragraph. But are you not changing your story here?. I thought you said before that you can have radiation from one isolated conductor. I said you couldnt! However, your interpretation is I feel slightly wrong. Radiation does not travel from one conductor to the other, iit travels between the conductors as per well knoewn TL theory. Poyntings vector etc. Im ignoring the small near (induction) field effects in open wire lines. And, of course, in coax there are no leakage fields.--Light current 04:17, 24 January 2006 (UTC)

Maxwell wrote "displacement current" in terms of electric field strength. However, as the voltage rises at front of the logic step, current rises. Maxwell should have written the extra current (displacement current in vacuum) equation in terms of the ordinary (conductor based) current, which means "displacement current" is radio. Maxwell: displacement current D = e.dE/dt = e(v/ct^2), v is uniform voltage rise over time t. What I'm saying is that the mutual radio emission causes the front of the logic step (the rising part) to propagate. Each conductor induces current in the other! It is fact that the inverse-square law doesn't apply: there is no net radio transmission beyond the system because of perfect interference, as the current rise in each conductor is the exact opposite of that in the other one so the radio transmissions from each conductor exactly cancels the other outside the transmission line! Illustration: [2] Many thanks, Nigel. 172.203.245.250 18:20, 23 January 2006 (UTC)

Well Nigel, it looks like you are now 'stealing my clothes'. However, one place we do differ is in the mechanism of high frequency currents being caused to flow in the conductors. You say there is some sort of mutual induction. Well, sort of! I say any currents that can be detected in the conductors will be induced by the em energy flow between them. Think about waveguides! The walls only act to reflect the waves. The only currents in the walls are induced currents. Regarding Maxwell, I agree that in a capacitor, the so called displacement current is indentically equal to the conduction current in the wires. That is trivially obvious. However that does not mean that displacement current actually flows across the plates from one to the other. THe current gets to one side from the other by means if the electromagnetic energy that is liniking both conductors (or plates). My previous posts have a fuller explained this view.--Light current 04:32, 24 January 2006 (UTC)

I am loathe to respond to Nigel's latest post, since he appears to have reposted it following its deletion by an administrator! I am certain that over the years an enormous number of careful and well-designed experiments have been performed on every conceivable configuration of conductor and means of excitation. Nigel's experimental work, on the other hand, would appear to consist entirely of an afternoon's fiddling with a pair of hand-held transceivers, borrowed when his dad wasn't looking. He seems not to realise that in most cases a transmitter antenna is driven by the best approximation to a sinusoid waveform that can be reasonably achieved , precisely in order to avoid the spurious generation of RF energy at uncontrolled and inappropriate frequencies. -- Kevin Brunt 23:19, 23 January 2006 (UTC)

No, Kevin, it involved measurements of radiated field strength with an oscillscope: the radio emission is proportional to di/dt fed into the aerial. You are the one who appears to not have a grasp of this fact. Sine waves are indeed an ideal case because it maximises the overall emission (a square wave input doesn't provide any radio emission at all during the flat parts, just a series of spaced pulses during the rises and falls of the wave. Thanks, Nigel 172.203.223.44 11:50, 24 January 2006 (UTC)

Deletion of posts

As far as I am aware, no one has the authority on WP to delete posts on talk pages. If this has happened, it is what we term as vandalism (or an error). The perpetrator will be warned not to repeat this action. --Light current 00:36, 24 January 2006 (UTC)

As Light Current notes, Nigel is clearly observing a "near field effect". I would have said that there was some sort of inductive or capacitive "coupling", but it amounts to much the same thing. I would disagree with LC to the extent that there has to be some transfer of energy, otherwise there could be no signal to be detected, but a radio receiver is specifically designed to detect small signals. -- Kevin Brunt 23:19, 23 January 2006 (UTC)

The external fields around the open wire TL are, in my opinion, near field effects. If any energy were absorbed from these fields, it must affect the transmission of energy along the conductors. However, I wish to restrict the argument to coaxial cable as this simplifies matters greatly in that there are no leakage fields to worry about.--Light current 04:40, 24 January 2006 (UTC)

LC's link is extremely informative. The author might possibly be overasserting his case slightly. It might perhaps be slightly more accurate to say that the electric and magnetic fields surround the moving charge and the fact that some of that surrounding space might be a conductor is merely detail. The important point is that the fields are entirely bound to the charge and its motion, and in the case of a charge moving in a conductor (ie current flow) the motion of the charge is constrained by the conductor. If energy from the charge's motion is lost into the resistance, the energy stored in the fields is reduced. If energy is abstracted from the fields in some way the motion of the charge is affected. (And an antenna is designed specifically to maximise the transfer of energy out of the fields into a "coinciding" transverse electromagnetic wave, which "detachs" itself from the antenna and carries the energy off into "free space".) -- Kevin Brunt 23:19, 23 January 2006 (UTC)

Yes, well, the only problem I have with Bill Beaty's description is that he seems to be restricting the cause of fields to charge and charge movements. We all know that charges cannot move at light speed, so the description does not answer all the questions. However, the em energy (energy current) induction theory does appear to explain displacement current (ie that it doesnt exist) and how 'conduction' current can appear to get from one side of a capacitor to the other without actually passing between the plates thro the (vacuum) dielectric. (Thats a subtle one!)--Light current 04:50, 24 January 2006 (UTC)

This is where Catt's "electromagnetism" diverges from orthodoxy. Catt views current flow as "TEM" which is necessarily both electric field and magnetic field. The magnetic field is due to motion (as in a sense is the electric field in a TEM wave,) so he has to argue for continuous motion. The orthodox view regards the electric and magnetic fields that exist when current flows in a conductor as separate. Orthodoxy has no problem with the idea that in a charging capacitor, charge flows into the capacitor and comes to a stop, and in stopping the magnetic energy associated with the motion is transferred into the electric field. -- Kevin Brunt 23:19, 23 January 2006 (UTC)

Kevin, the orthodox view on capacitor charging has been completely blown away by read Catts article on @diplacement current (and how to get rid of it). Even Nigel agrees with that!! So I dont see what point you are making in this post except that you realise that Catt is proposing reciprocating energy currents. Yes! I agree that reciprocating energy currents are one answer to DC situations as I have been saying for about 3 months now!--Light current 04:58, 24 January 2006 (UTC)

Hi Light Current, it is hard to reply to your response where you put comments into the original reply instead of afterwards. You say: "Yes I tend to agree with the last part of the above paragraph. But are you not changing your story here?. I thought you said before that you can have radiation from one isolated conductor. I said you couldnt! However, your interpretation is I feel slightly wrong. Radiation does not travel from one conductor to the other, iit travels between the conductors as per well knoewn TL theory."

An isolated radio aerial can radiate energy. An isolated conductor connected to a charge (battery terminal) radiates energy as it charges up. It behaves like a radio aerial, and as the current varies in it, during charging, radio emission occurs. The current falls off rapidly along the wire because of this emission of energy. Hence you cannot transfer significant energy with a single wire. For a pair of conductors connected to the two terminals of a battery (Catt's anomaly situation) the current in each conductor is the opposite of the other, so the radio emission cancels out beyond the system, and no energy is wasted by radio -it all goes into the opposite conductor, so the two conductors help each out out. This is during the rise time portion of the TEM wave, see the illustration: http://electrogravity.blogspot.com/2006/01/solution-to-problem-with-maxwells.html

What you are talking about, Light Current, is the part of the TEM wave behind the rise at the front, and what you say about that part is fine! I'm concerned with Catt's anomaly, which is the front part of the TEM wave, where the current increase occurs. Thanks, Nigel 172.203.223.44 11:41, 24 January 2006 (UTC)

OK Nigel I think everyone should reply outside the body of others' posts. I'll start now. It would help though if the posts were kept shorter and deal with few points at one time. Rgarding your last post, yuo say

an isolated conductor connected to a charge (battery terminal) radiates energy as it charges up. However, this plainly cannot be the case as there is no return path for the current- unless of course the other end of the capaciance is connected to the other battery terminal. In this case you have a two conductor system and the capacitance must be taken into account in the analysis. BTW, charging a capacitor does not radiate energy, unless youre very unlucky! You then say For a pair of conductors connected to the two terminals of a battery (Catt's anomaly situation) the current in each conductor is the opposite of the other, so the radio emission cancels out beyond the system,

I agree with this mostly. In a coax of coarse there is no radiation at all-- Im not sure how much radiation ther is from a pair of open wires.

BTW you keep talking about a step as only having effect at that instant. Do a Fourier analyisis of the step to see what sinusodal components it contains! --Light current 14:27, 24 January 2006 (UTC)

Hi Light current; (1) an aerial is a single conductor and does radiate radio as the current applied varies, we know a single conductor can't propagate a constant current because its inductance is infinite (which is a mechanism for Kirchoff's law), (2) if the capacitor is a transmission line, as stated before, the radio emission due to each conductor (capacitor plate) is the exact opposite of the other, and cancels out as seen from a distance. What I'm saying is that to resolve the Catt anomaly the TEM wave step needs to be analysed in two parts, first where the current is increasing (which is omitted from today's treatment), and second where the current is constant(which the current treatment does describe, using steady magnetic and electric fields). If the current rise (step front) was vertical, "displacement current" there (however you think of it) would be infinite, and since "displacement current" is an invention by Maxwell to retain continuity of current flow across the vacuum, you would then have the paradox of a finite current flowing along one wire turning into an infinite "displacement current" across the vacuum and then returning to a finite displacement current in the other wire. The true rise is not vertical, because the current does not rise from 0 to i instantly at any point on a conductor as the step passes by. It can be very great. The standard treatment of radio shows that radio emission is proportional to the variation rate of the net current di/dt in a conductor. "Displacement current" is the radio exchange process where the front of the TEM wave in each conductor swaps energy by radio (or electromagnetic pulse, if you prefer to reserve "radio" for sine wave shaped electromagnetic waves, as Kevin does). The wires must swap energy across the vacuum to propagate; each one is inducing the current in the other one. This is why the TEM wave goes at the speed of light in the vacuum between the conductors. Best wishes, Nigel 172.202.0.92 16:17, 24 January 2006 (UTC)

An aerial/antenna is specifically optimised to transfer energy from the current flow in the conductor into a self-sustaining transverse electromagnetic wave outside the conductor. Capacitors and transmission lines are usually intended to avoid this sort of energy transfer. The error that creates the Catt Anomaly is Catt's assumption that the magnetic and electric fields associated with the flow of current in a conductor constitute a "TEM wave"; they do not for any definition of a "TEM wave" that conforms to normal physics usage. -- Kevin Brunt 20:12, 24 January 2006 (UTC)

"Displacement current" is the label used by physicsts to refer to Maxwell's correction to Ampere's equation which is needed to yield consistent results when the equation is applied to a volume of space containing only one of the two plates of a charging capacitor. It predicts that a changing electric field causes a magnetic field. Asserting that the "displacement current" somehow "completes the cicuit" is stretching a metaphor too far. -- Kevin Brunt 20:12, 24 January 2006 (UTC)

LC - Catt et al, Displacement current (and how to get rid of it) does not "blow away" the orthodox view. The article does not prove its claims. It merely demonstrates that when an open-ended TL is fed from a constant voltage through a resistance, the voltage at the input end of the TL approximates to that expected if the TL were replaced with a "point" capacitance equal to the total capacitance of the line, and that approximation is increasingly accurate as the resistance is increased, or the TL's length is decreased. The article does not dispose of charge because it explicitly uses the unit capacitance of the TL which is, by definition, the ratio of the charge to the voltage. Equally, displacement current is not disproved, because the authors are using a set of equations derived for the convenience of electrical engineers who don't have the time (and possibly the mathematical competence) to handle the partial differential equations that describe the voltage and current at an arbitrary point on a TL. The authors seem to want to pretend that they are using a set of equations based on experimental observation. In fact they can show that a transmission line behaves like a capacitor precisely because they are using the results of an analysis that starts from the assertion that a TL is a capacitor (or rather many capacitors in parallel.) -- Kevin Brunt 20:12, 24 January 2006 (UTC)

Nigel to LC?

Kevin???(dont you mean LC), coax or a waveguide (for microwave frequencies) is used in many practical cases, but let's keep to the simple physics of a DC TEM wave step propagated by two straight aerial-like conductors, as the Catt anomaly uses. We get electromagnetic radiation (radio emission) from a net time-varying current in a conductor. If you have two such conductors, with each having an inverted form of the signal in the other, they exchange energy which induces the current in the other. But there is no long distance propagation of this energy due to exact interference, so the coupling is perfect. This is how the front of the logic step propagates: each conductor causes the current in the opposite conductor by simple electromagnetic radiation due to the time-varying current as it rises. Catt simply missed out this mechanism, and is now too prejudiced against the mainstream to back these facts. Best wishes, Nigel 172.203.152.63 23:18, 24 January 2006 (UTC)

LC to Nigel

Both Catts conductors are connected to the battery. Only one of yours is. Therfore yours dont do nothin man!(IMO) (Unless youre talking about electrostatic charge buildup on the lone conductor)- but there again there must be a return path for the current thro' the stray capacitance to the other end of the battery.--Light current 05:24, 25 January 2006 (UTC)

Light current: you say "there must be a return path for the current". Radio is the return path. Radio is not just a sine wave phenomena. You cannot stop radio (self-propagating emission of energy) if the current changes with time. The current in a single wire falls rapidly, the energy being propagated away as radio. Since the self-inductance is infinite, the current continues falling rapidly, although the whole wire still "charges" up to the battery terminal voltage at light speed. Thanks, Nigel 172.212.87.209 15:40, 27 January 2006 (UTC)

Nigel,I dont really understand your reply here. If you say radio is the return path, you must mean it returns thro' the earth, in which case the other end of the gen must be connected to earth. Is that what you mean?--Light current 22:49, 28 January 2006 (UTC)

Nigel is, I think, trying to replace "displacement current" with "radio". In this he is trying to follow on from Catt, while at the same time trying to avoid some of the known problems with Catt's arguments. However, he's not managed to see through some of Catt's misconceptions. I have come to the conclusion that Catt has either picked up a very loose and misleading statement from an introductory textbook, or has completely misinterpreted how Maxwell came to derive the additional term to Ampere's equation that is now referred to as "displacement current". -- Kevin Brunt 22:16, 29 January 2006 (UTC)

Maxwell realised that there was a mathematical inconsistency that showed up when Ampere's equation was applied to a volume of space that enclosed one of the two plates of a charging capacitor. In particular Maxwell saw that the inconsistency could be fixed by inserting the term we now know as "displacement current", which he derived from exising work (by Kelvin) - the "continuity equation". ([www.cmmp.ucl.ac.uk/~drb/Teaching/PHAS3201_MaxwellsEquationsFull.pdf See this PDF from UCL]) Maxwell, of course, went on to try to interpret displacement current in terms of his "aether". I think that Catt has picked up on the words "current" and "continuity" and convinced himself that physicists claim that the "flow" of displacement current "completes the circuit" - a position that physics has certainly not held since Maxwell's "aether" was abandoned by the start of the 20th Century. Nigel is just following on from this. -- Kevin Brunt 22:16, 29 January 2006 (UTC)

LC's comment on (Nigels reply to Kevin 25 Jan )

Nigel, you appear not to have heard of Mr Fourier and his famous Fourier analysis which says that any repetitive waveform can be represented by a sum of sine and /or cosine waveforms of appropriate amplitude. Now in your experiment, the square wave that you used would have a fundamental plus all the odd harmonics decreasing in amplitude proportional to the harmonic. So when you say 'the flat part doesn't do anything', strictly speaking, you are incorrect or your statement is careless.--Light current 05:36, 25 January 2006 (UTC)

Actually, Catt's "TEM step" isn't a "repetitive waveform", which is part of the problem, since it confused what is going on at the step and what is going on either side of it. Nigel has muddled it further by introducing a radio transmitter, so we now have too many things going on at once. -- Kevin Brunt 22:32, 25 January 2006 (UTC)

The issue of one conductor versus two has some relevance, because when you stop to consider it properly a TL is merely a conductor with regular properties to make it more convenience to do theory with. A single isolated conductor is a TL, and a lot of what we've been discussing does not really require a TL at all. In fact, if you read the "Lynch IEE version" of the Catt Anomaly, it is clear that the inclusion of the vacuum-spaced transmission line is merely noise, and that all of Catt's "Southern" and "Western" obfuscation is swept away, reducing the Question to "How can current propagate at nearly the speed of light when the individual electrons only move at less 1mm/second?" -- Kevin Brunt 22:32, 25 January 2006 (UTC)

When you connect a conductor to a battery, a voltage step propagates down the conductor. Behind the voltage step there is movement of charge - current flows and because charge has moved into a space that was previously uncharged, the charge with the volume of the conductor increases. There is a magnetic field and an electric field associated with the charge. For a TL it regularity means that behind the voltage step the charge remains constant at least until the step reaches the far end. Thus for a TL, ahead of and behind the step the fields are constant. At the step itself the fields are changing so, by Maxwell's equations, the changing magnetic field created an additional electric field, and vice versa. In a normal radio transmitter the antenna is excited by a sinusoidal signal, so that the fields around the antenna are deliberately not constant. Furthermore the antenna is "tuned" to maximise the field strengths at the right times to transfer energy into the desired TEM wave. The more the exciting waveform diverges from a pure sinusoid, the more energy is wasted. -- Kevin Brunt 22:32, 25 January 2006 (UTC)

Light current: I use Fourier analysis to convert waveforms (signal strength versus time) into frequency spectra! The flat part has constant voltage and thus current variation, di/dt = 0. Therefore there is no radio emission from the flat part, as I said. Thanks, Nigel 172.212.87.209 15:32, 27 January 2006 (UTC)

Nigel: If only a single step, you may be correct. Is that what you are referring to in your arguments? However, to say that no energy flows to the load after the initial step I think is not correct.--Light current 15:44, 27 January 2006 (UTC)

Light current: yes! I was breaking down the Catt question into two stages, where the current is rising and where it is steady. Of course current continues when the rate of increase of current is zero. The current rises to a peak at ant location along a conductor, then continues flowing at a steady rate, not varying with time. So once the front with the rising current has passed, the situation is then covered by the ordinary TEM wave treatment. Nigel 172.212.87.209 18:10, 27 January 2006 (UTC)