Talk:Ivor Catt/Archive 14

Latest comment: 18 years ago by Alfred Centauri in topic A third player enters

Everyone seems to be Catt-napping! Why is this?--Light current 22:50, 15 April 2006 (UTC)

Well, the archiving struck, and your last comment seemed to suggest that you didn't want to discuss the several very serious flaws in the Dec '78 Catt et al article, which I believe are the key to understanding the stand-off between Catt and "orthodox science". -- Kevin Brunt 21:50, 19 April 2006 (UTC)

Just because stuff has been archived does not mean it cant be discussed. Just bring it up again on the current talk page (ie here).--Light current 23:01, 19 April 2006 (UTC)

Discussion restart

The opening

Copied from archive #13--Light current 00:50, 20 April 2006 (UTC)

LC, it will save a lot of argument if you would review the Dec78 Catt et al paper that I'm working against. This paper very specifically models a capacitor of the form of a sector of a circular plate fed at the point as a TL, with the source impedance explicitly much greater than that of the TL. I am positing the "squaring off" of the curve, so that the distance from the point of feed to the outer edge varies depending on the direction. The issue is not that a TL charges up stepwise, it is that in many ways the 1D "TL model" is less useful than the 0D "lumped capacitance model" as a description of the way that charge distributes on the 2D surface of the plate of a capacitor. What, indeed, are the characteristic impedance and velocity of propagation of a 4700μF electrolytic capacitor and how long is it? -- Kevin Brunt 01:37, 21 March 2006 (UTC)

I'm familiar with the article. I dont think your argument regarding angles is at all relevant to Catts analysis which in my view is substantially correct. If there is a mismatch between the feed and the Z0 of the cap, you get reflections. What's the problem? BTW a capacitor is sometimes made from separated plates wound ito a cylinder. They are all transmission lines!--Light current 01:57, 21 March 2006 (UTC)
Kevin. Its your turn!!--Light current 00:52, 20 April 2006 (UTC)

Kevins opening move

OK. Let's start by noting that I have no quarrel with the analysis of the voltage waveforms observed on a "charging transmission line." However, in spite of the assertion in the article, this does not prove that "a capacitor is a transmission line." In fact it only demonstrates the converse, that a transmission line is a capacitor; specifically a long capacitor that is so thin that it can be considered to be a one-dimensional line. The analysis cannot be extended to the two-dimensional case of a "real" capacitor, because it cannot explain how (for example) a circular wavefront spreading out across a rectangular plate would reflect off the far edge. The 1-D analysis can only have the voltage front returning the way it came; it is far more likely that what really happens is similar to an optical reflection with the angle of incidence being equal to the angle of reflection.

Basically, the analysis by Catt et al is incomplete. The leap from 1-D to 2-D requires additional work which they have not done, and which they undoubtedly will not do, because it would require the admission that the equations for the characteristic impedance and velocity of propagation for a TL are not derived experimentally (as they appear to imply), but are a theoretical solution of the "Laplacian" Telegraphers' Equations. Viewed from the "other direction", the Telegraphers' Equations provide the way to explain why the behaviour of a voltage step travelling along a TL is dictated by the inductance and capacitance of the TL. Furthermore it is relatively trivial to generalise the TE's, not only from a 1-D line to a 2-D surface, but also to situations where the inductance and capacitance are not uniform across the surface, or even are not constant over time. But Catt et al believe that the Telegraphers' Equations have a "spurious high-frequency cutoff". (Is this because they avoid considering the implications of the resistance of their conductors?)

The article also claims that "This model does not require use of the concept of charge." This is more than a little difficult to support, since the model does use the concept of impedance which is a statement of the relationship between voltage and current, and so is inescapably based on the concept of charge itself! -- Kevin Brunt 19:26, 20 April 2006 (UTC)

LCs blocking tactics

1. A capacitor is a transmission line is a capacitor.

Im going to take your points one at a atime

this does not prove that "a capacitor is a transmission line." In fact it only demonstrates the converse, that a transmission line is a capacitor; specifically a long capacitor that is so thin that it can be considered to be a one-dimensional line. The analysis cannot be extended to the two-dimensional case of a "real" capacitor,

Take two long strips of metal foil with parallel edges and place a dielectric matl between them . Connect a wire to the end if each bit of foil. Roll up the foil into a cylinder and secure with tape. Now what do you have? Is this an o/c TL, or a capacitor, or both? Simple answer PLEASE!--Light current 22:14, 20 April 2006 (UTC)

This is not just a TL, because the description of a TL in terms of the impedance at its ends, and the time taken for a voltage change at one end to be "seen" at the other is insufficient to describe exactly how the voltage at some arbitrary point on the surface of the foil varies with time. -- Kevin Brunt 11:02, 21 April 2006 (UTC)

I really dont understand what you are saying here. THe telegrahers equations predict exactly what the instantanous voltage and current are at any point along the line for a sine wave input. Also this can easily be verified by using a VSWR meter to look at the standing wave ratio for sine waves.

For a step input supplied from a matched source, the step travels to the far end, is reflected, reaches the driving end, then its all over with the line being charged to the o/c voltage of the gen. This can be seen with a scope . So in what way is this set up not just a TL?--Light current 16:56, 21 April 2006 (UTC)

My point is that the Telegraphers' Equations, as extended to deal with a 2D surface, are partial differential equations of the functions V(x,y,t) and I(x,y,t), which describe the voltage at some point (x,y) on the surface at some arbitrary time t. In fact, in the fully general case, the inductance,capacitance, etc, can also of be functions of position (and even time) and if you look at something like a varicap diode, it is even possible to envisage the capacitance being proportional to the voltage! The mathematics of the fully general case would be horrific.
The TL model that Catt et al use has been bolted down with a set of reasonable assumptions to make the maths easier, and which yield a useful description for the very common special case of a uniform conductor that is long enough that the time taken for a voltage change to propagate along its length is significant. In particular it is assumed that the voltage and current are identical in the y-direction for any given x and t, so that the functions V(x,y,t) and I(x,y,t) can be simplified to V(x,t) and I(x,t). If you solve these simplified equations for a voltage function based on the "Heaviside Step Function", you end up with the impedance/propagation time model.
This "impedance" model distorts something important; it suggests the equation V = Zο I. This is misleading. A more more accurate statement is that dV/Dt = Zο dI/dt. Using the latter form clears up the difficulty about the characteristic impedance being an AC property. It is also the key to understanding the error in Catt's "energy currents"; when a voltage step is reflected off the unterminated end of a TL, the positive voltage step travelling "backwards" is the same as a negative voltage step travelling forwards. In Catt's endlessly-repeated voltage-step-on-a-transmission-line scenarios, the outbound step represents where the current starts flowing, and the returning step the point where current is brought to a stop! -- 193.61.22.92 19:54, 21 April 2006 (UTC)

THere is no variation in voltage across the 'width' of the TL (your y direction I think) - why should there be? So theres no problem is there?--Light current 20:09, 21 April 2006 (UTC)

Not in a TL (by definition, more or less). However, I recently dissected an electrolytic capacitor. It consisted of two foil strips about 1cm by 10cm, with the leads connected assymetrically, so that the 'y' dimension is larger than the 'x'. Not only that, I believe that in some parts of the capacitor the voltage steps on the two plates are moving in opposite directions..... The issue is that a full understanding of a "real" 2D capacitor requires the deployment of the "heavy guns" of the physicists' partial differential equations, rather than the electrical engineers' "lightweight" 1D simplification, and that in doing it properly, the concepts of current and charge re-emerge. -- Kevin Brunt 21:07, 21 April 2006 (UTC)

If a TL is a cap, why is a cap (constructed like a TL as most are) not a TL?--Light current 00:46, 21 April 2006 (UTC)

The TL is a special case for which a much simpler mathematical model can be built. In particular, the impedance/propagation time description is easily adapted so that the end of a TL can be modelled as a voltage source with a defined impedance, as used by electrical and electronic engineers when they design circuits. In fact, a TL is really an "engineering" concept, rather than a "science" one. Although the same underlying physics applies to all capacitors, the TL "simplification" just isn't useful in the general case. -- Kevin Brunt 11:02, 21 April 2006 (UTC)

Well I just cant agree here, haiving designed transmission line cicuits to produce high voltage pulses. THe line charges up as a capacitor and discharges as a TL. I have seen it numerous times. The experiments agree exactly with the theory.THe engineering reults agree with the scientific predictions I promise you. I agree that some real caps are not simple TLs- but TLs they are or can be decomposed into TLs. But see my post below--Light current 18:44, 21 April 2006 (UTC)

2 Does current need charge?

You said:

The article also claims that "This model does not require use of the concept of charge." This is more than a little difficult to support, since the model does use the concept of impedance which is a statement of the relationship between voltage and current, and so is inescapably based on the concept of charge itself!

What about em radiation in free space? Where is the charge? (BTW Z of free space = 377 ohms!)--Light current 00:20, 21 April 2006 (UTC)

The impedance of free space is what you get if you take the equation for the impedance of a TL as presented in the Dec78 article, set f (the geometrical factor) equal to 1, and set μ and ε to their "free space" values, at which point you get a theoretical value derived entirely from the speed of light, the mathematical constant Π and a numerical constant to convert the result into SI electrical units. EM radiation does not, by definition, have any charge involved; it has an electric field and a magnetic field, which are mutually generated by each other. EM radiation is described by a solution of Maxwell's Equations in which the J term (ie current; flow of charge) is 0. Since Maxwell's Equations use μ and ε, there is a connection with the impedance of free space, but attempting to form an association through to the relationship between current and voltage in a TL merely leads back into the confusion of Catt's ideas. -- Kevin Brunt 12:10, 21 April 2006 (UTC)

You say:

EM radiation does not, by definition, have any charge involved; it has an electric field and a magnetic field, which are mutually generated by each other. EM radiation is described by a solution of Maxwell's Equations in which the J term (ie current; flow of charge) is 0.

I agree. So why do you need charge with a guided wave? As you yourself have just said

Since Maxwell's Equations use μ and ε, there is a connection with the impedance of free space,...

So if Catt had talked about electric and magnetic fields in the TL insted of curent and voltage, that would have been OK and you would have agreed?--Light current 18:36, 21 April 2006 (UTC)

Catt does talk about electromagnetic fields in the TL! The trouble is that he doesn't accept that they are the consequence of the charge in the conductor....
Repeat the "charging TL" analysis, but with the source resistance set equal to the TL's impedance. Note that there is only one outbound/inbound voltage step cycle, with the total voltage of the returning step being equal to the supply voltage. During the cycle, half the supply voltage appears across the source resistance, and a constant current flows through it. After the end of the cycle, there is no voltage difference across the resistance so no current flows. If you use the formulae for the characteristic impedance and velocity of propagation of the TL in terms of the unit capacitance and unit inductance to get the current that flowed through the resitance and time, you will find that the amount of charge that flowed is precisely the amount needed to charge the capacitance of the TL to the supply voltage. The next step is to replace the source resistance with a further length of TL. It should be obvious that this does not change the discussion at all, and that it actually applies to any arbitrary point along the length of the TL.
When a voltage pulse travels along a TL it conveys energy. As Catt has shown, half of that energy is stored in the TL's capacitance and half in its inductance. IE, half is due to the position of the charge and half due to its motion. This equal division is undoubtedly the reason why a TL behaves as it does. When the pulse runs into the open end of the TL, the motion of the charge is, perforce, brought to a halt. The energy has to go somewhere, so it is converted into capacitive energy. When you apply the equations that relate charge, voltage and energy in a capacitor, you have to convert the moving pulse on the TL into a static one that is half as long, but of twice the voltage. This is precisely what is observed; where it differs from Catt's picture is that the voltage edge travelling backwards marks where the charge travelling forwards is being brought to a stop.
This is why the charge pulse in a TL is not the same as a TEM wave in free space. The charge in the TL moves as it does because slowing the pulse (ie reducing the current) down will increase the voltage in the TL capacitance, which tends to increase current, and vice versa, so that the electric and magnetic fields remain constant in magnitude, although they are moving. In a TEM wave the fields are constantly regenerating each other. Since their relationship is through rate-of-change, the fields must of necessity be continually changing. -- Kevin Brunt 20:46, 21 April 2006 (UTC)

I agree with your first paragraph. In your second para , you say:

half of that energy is stored in the TL's capacitance and half in its inductance.

We could equally well say: half stored in mag field, half stored in electric field. Charge does not need to enter into it! In your scenario, charge travelling at 'c' in the material meets an open end, and instantaneously undergoes an infinite acceleration so that it can return down the line. How likely is that?.

If you were to accept the logical conclusion of counter propagating waves in the TL, you would have no problem in understanding what actually is happening. You seem to be making problems for yourself!

Your last para reads like pure waffle.--Light current 21:00, 21 April 2006 (UTC)

But I simply don't believe in Catt's counter-propagating waves! In order to get them to work he has to allow them to dissipate energy into the resistance of the conductor when flowing "unpaired", yet having them cancel out the resistance, and so not lose energy, when they are flowing contrarily. Catt tries to liken the standard theories to the Phlogistic Theory of the 18th century; in truth, there is a much stronger comparison to the contortions and special pleading needed to get his own theories to fit! -- Kevin Brunt 22:56, 21 April 2006 (UTC)

3. Capacitor with disk type plates

It would appear that you have not read Catts article very closely. He descibes the disc capacitor as being made up of a number of sectors of the disc all in parallel. Each sector can be considered as a // plate capacitor of varying (decreasing)Zo as you travel toward the perimeter of the disc. A circular wave front can be split into a large (infinite) number of straight line wave fronts each travelling down its own tapering transmission line. I see no problemo!!--Light current 00:28, 21 April 2006 (UTC)

This is another special case - the circular disc is highly symmetric. At any point on the edge of the disc the tangent to the edge is perpendicular to the line between the point and the centre of the disc. Consequently, the issue of non-perpendicular reflection does not arise. This is just more of Catt et al avoiding the issue. -- Kevin Brunt 12:22, 21 April 2006 (UTC)

Yes, its a special case as not too many caps are made in exactly this way (exapt perhaps disk ceramics - but there the leads do not enter at the centre of the disc, but at the circumference). But Catt is showing that even if they were, they would still look like a TL.

Regardless of the physical makeup of a capacitor, Im pretty sure that it can be decomposed into a plurality of parallel transmission lines (albiet of varying Zo) in every case. Therefore every cap can be considered a TL (or a collection of TLs in // if you want to be pernickity). If you doubt this, try to think of a capacitor form that cannot be so analysed. If you are talking about a curved edge at the end of each of the TL segments, then this problem can be overcome by letting the number of segments tend to infinity. Then each arc becomes a straight line (ie a perfect o/c with its partner below it). When the effect of all these tls in // is consisered you get just one tapering transmission line thats o/c at the end. Have you heard of tapering TLs and do you understand what they do?--Light current 18:50, 21 April 2006 (UTC)

For all of Catt's assertions, the voltage anywhere on the plate of a capacitor is a measure of the amount of charge there; an advancing voltage front is equally an advancing charge front. In a TL, because the front is advancing evenly across its width, any tendency to flow "sideways" will be self-correcting. However, if the front encounters an "edge" obliquely, so that the interaction does not occur simultaneously across the width of the front, the "symmetry" is destroyed and a 'y' component to the motion of the front is introduced. In terms of the "TL model" the voltage step at the open end is not constrained to travel back the way it came, but can spread out into any adjacent area with less charge per unit area (and therefore is at a lower voltage.) At this point the adjacent TLs are interacting with each other and any analysis based on parallel TLs collapses in a heap. -- Kevin Brunt 23:21, 21 April 2006 (UTC)

A third player enters

Does EM radiation need charge?

I must comment on this although I'm sure I'll regret sticking my nose once again into this ongoing debate. While it is true that Maxwell's equations have travelling wave solutions in a region free of charge, one must be careful to not forget boundary conditions when solving a system of differential equations. Bottom line, charge is the source of the electric field and EM waves are propagating disturbances in this field. No charge anywhere, no electric field anywhere and so no EM waves anywhere (the 'trivial' solution).

Consider the solution for an EM wave propagating in a parallel plate waveguide. This solution is based on the boundary condition provided by the mobile charge within the conductive plates. If we let the distance between the plates go to infinity, we still have a plane wave solution but there is apparently no charge in this space. Yet, we know that there is charge, but it is 'at infinity'. Alfred Centauri 21:17, 21 April 2006 (UTC)

Welcome Alfred. Im sure you'll regret it :-), but I know you enjoy it also!
You seem to be saying that even for em waves in free space, charges are involved, albeit at infinity. I must disagree with you here. For instance, if one suddenly decided to create some em radiation (v. easy), here on earth, then by your reckoning, this must somehow induce charges at infinity. This assumes information can travel at infinite speed. How do you suggest this happens? (I know Ill regret asking this!)--Light current 21:38, 21 April 2006 (UTC)
Hi from me as well! I suspect that bringing in a reference to a waveguide might prove to be a mistake, as a major part of the "Catt Problem" is that he is totally confused about the distinction between an EM wave in free space and a current flowing in a conductor, and it is impossible to discuss waveguides, where the two things interact without getting completely tied up in Catt's misconceptions! LC has picked up on the infinity in the second paragraph. I'm reading the first paragraph as saying that it is not possible to create an EM wave in the absence of charge. Is it not possible to create the electric field with a varying magnetic field? -- Kevin Brunt 22:27, 21 April 2006 (UTC)

LC, please don't jump to such silly conclusions so quickly. Now, think carefully about what I said. Better yet, allow me to clarify and amplify my earlier statements.

EM radiation is a propagating disturbance in the electric field. Thus, for EM radiation to exist, there must be an electric field to disturb. That implies the existence of electric charge somewhere at sometime. That is, even if the region you are considering is free of electric charge, this fact does not imply that there is no electric charge somewhere outside that region. This charge outside the region of consideration places constraints (boundary conditions) on the solutions to the source free Maxwell equations inside the region. Got it?

If boundary conditions are not considered when solving the source free equations, the solutions are plane waves over all space and time. However, these solutions cannot be physical for the very reason that they imply infinite planes of charge at infinity. If you are considering only a limited region of space, plane wave solutions do not exist without the appropriate boundary conditions (e.g., perfectly conducting planes in the case of a planar wave guide).

To sum up: the statement that EM waves do not require electric charge allows for the possibility that EM waves can exist without any charge existing anywhere for at least some finite time and that is simply wrong. To say that EM waves in a bounded region do not require electric charge in that region is fine as long as it is understood that somewhere (sometime) outside or on the boundary of that region, there exists electric charge. Alfred Centauri 22:50, 21 April 2006 (UTC)

Kevin, where did the magnetic field that you are varying come from? What are the sources of the magnetic field? An electric current (requires electric charge) is one. The other is a changing electric field which of course, implies the presence of an electric field to change. Of course, one could try to argue that the source of the changing electric field is non-uniformly changing magnetic field but then we find we are in an infinite loop. Alfred Centauri 22:50, 21 April 2006 (UTC)

Well first of all, Alfred, I was not jumping to any conclusions that were not implied by your original statement! I take issue with your statement that em must be a disturbance in the electric field only. I would have said that it is a disturbance in the electromagnetic field. The propagating electric and magnetic fields are symbiotic and need no external source of charge or current or even magnets etc. I'm sure you will agree with this from Maxwells equations. Since we know that changing magnetic fields can create electric fields, I dispute your argument that real charge must exist for the propagation of em waves anymore than real magnets or real current are required.--Light current 23:05, 21 April 2006 (UTC)

I'll ask you the same question I asked Kevin: what is the source of this changing magnetic field that creates this electric field in the absence of any charge anywhere at anytime? Alfred Centauri 23:39, 21 April 2006 (UTC)

No one said there was no charge anywhere at any time. Thats silly!--Light current 23:44, 21 April 2006 (UTC)

Yes, but that is exactly what you disputed. In case you don't believe it, let me refresh your memory.

I said: "...the statement that EM waves do not require electric charge allows for the possibility that EM waves can exist without any charge existing anywhere for at least some finite time and that is simply wrong."

You said: "...I dispute your argument that real charge must exist for the propagation of em waves anymore than real magnets or real current are required."

So, either you are not comprehending my comments or you are contradicting yourself. Let me state my argument one more time. The existence of EM waves in a source free region of space requires that electric charge exists (or has existed in the past) outside that region. Otherwise, you get plane wave solutions that extend over all space and time and, if you bother to check the literature, you'll find that these are not physical solutions. Go ahead, look it up. Alfred Centauri 00:46, 22 April 2006 (UTC)

You are not reading between my lines! The fact that real charges exist in the TX on earth does not mean that real charges exist in outer space!(Or does it). Thats where the only real charges are: this side of the antenna!--Light current 00:50, 22 April 2006 (UTC)

I confess that I am guilty of not reading between your lines (in fact, I don't know to do this). On the other hand, you apparently read something into my arguments that I never put in. So, if you think that I have said that EM waves need charge to propagate through some region, then I understand why you dispute this. Problem is, that's not what I said. I hope that if you carefully re-read my comments, you'll come to understand this. EM waves are globally inseparable from electric charge but not locally so. Alfred Centauri 01:05, 22 April 2006 (UTC)

Do you mean to say you dont know how to read between my lines. I'll give you a clue. Its when I write very short posts or questions- then you must be vigilant.
No, you implied charge must magically come into existence at infinity every time an em wave is launched. Did you not? Once an em wave is launched, it is my belief that charge, whether near of far, has no part to play in its propagation. Do you disagree with this fundamental staement. It seems you do , but Im not sure!--Light current 01:25, 22 April 2006 (UTC)

No, I did not imply such a thing and further, I cannot imagine how you could construe (or contort or distort) any of my statements above to imply the magical appearance of electric charge at infinity. What do you think I mean when I talk about boundary conditions in the context of a system of differential equations such as Maxwell's equations? Are you familiar with boundary value problems? Perhaps this is where our disconnect is? Alfred Centauri 02:25, 22 April 2006 (UTC)

I fear this conversation is diverging. Are you saying that the solution of any free space wave propagation requires some boundary conditions to be satisfied and that at these boundaries (say the antenna) you have charges? If so I would agree. But what im talking about is no charge necessary in space for the propagation of em thro' space. I thought I was making a very simple statement about the nature of em waves. To generate an em wave, you probably need charges at some point in the process. --Light current 02:34, 22 April 2006 (UTC)
I think I understand what you are driving at now, but you have not been explicit. You're sayin at some point in time you need charges to generate em waves. Is that correct?--Light current 02:39, 22 April 2006 (UTC)

Correct. Allow me to quote myself: "Thus, for EM radiation to exist...implies the existence of electric charge somewhere at sometime." Are we good now? Alfred Centauri 02:54, 22 April 2006 (UTC)

Yes well Alfred, you would make a good lawyer. It was the 'somewhere at sometime' (in the dim and distant past)? that seemed to be a throwaway phrase. Any way, Im going to agree for now (only) that you need charges unless you have a secret store of electromagnetic waves that can be called on by closing a switch! Hey hang on ;-) -- I know of one of those-- its a TL charged up with dc (or counter propagating waves - whichever you fancy). Does that involve charges when the energy is let out?--Light current 03:03, 22 April 2006 (UTC)

I'll save that conversation for another day. Good night! Alfred Centauri 03:13, 22 April 2006 (UTC)