Talk:Ivor Catt/Archive 5

message: Kevin: before the TEM step arrives at any part of the cable, there is zero voltage.  As it arrives, the voltage increases and with it the current.  Once it has peaked, it remains steady.  The only period of interest is while the voltage is rising.  Ivor ignores this, and also introduces other errors like the claim that displacement current flowing where the voltage is steady.  Your sine wave radio signal versus square wave logic step argument is vacuous, since radio emission occurs so long as di/dt is not zero in any part of the aerial.  Fourier analysis is useful to me for the purpose of translating a plot of current or field strength versus time into a frequency spectrum.  According to Fourier analysis, you can represent a square wave as a series of sine waves!  But that's not my main argument.
Consider the 2 conductor transmission lines as 2 parallel radio aerials.  If you feed one with a signal (of any type) and leave the other passive, the first transmits energy to the passive one which receives energy only as a result of di/dt in the first one.  This is indistinguishable from Maxwell's "displacement current" equation.  Maxwell says vacuum "displacement current" i = e.dE/dt = e.dv/(x.dt) where e is permittivity and x is the width over which the step rises (definition: x = ct, where t is the rise-time).  We see that if x = 0, then i = infinity.  This disproves the idea of a truly abrupt step.  Moreover, the current rises over the rise-time from 0 to its peak, and since radio emission occurs in proportion to di/dt, it becomes more intense as the step rise-time is made smaller.
Now here is the proof.  Taking the 2 parellel aerials or transmission line conductors.  Feed one with any signal, and feed the other with the inversion of that signal.  While the signal strength rises, electrons accelerate and radio emission occurs in a perpendicular direction.
I've done this experiment and proved it experimentally.  During the rise-time, each conductor transmits a radio signal that is the exact opposite of that emitted from the other conductor.  At a long distance (several times the distance of the gap between the two conductors) there is no observable radio transmission at all, because each radio emission cancels out that of the other: perfect interference.  (The same concept is often used as white noise to suppress sounds, but that is less effective.)
The point is that the entire radio energy emitted by each conductor during the step is transmitted to, and received by, the other conductor.  This is the process by which the TEM wave is allowed to propagate.  Catt, ironically, gives the conventional textbook slab of drivel on this point!  See  (that web version misses out the formulae, but they are widely known) where Catt calculates the inductance of a single wire and finds: "The self inductance of a long straight conductor is infinite. This is a recurrence of Kirchhoff's First Law, that electric current cannot be sent from A to B. It can only be sent from A to B and back to A."  I think it is unhelpful for Catt, having defined E and B in fixed ratio for a TEM wave (E=cB), then goes along with the unfruitful textbook treatment of inductance which considers inductance as a B field effect!  The magnetic field loops around each conductor instead of going from one conductor to the other line "displacement current" or in fact radio energy.  This is probably where the conventional theory went wrong!  It is clear that the entire energy needed to propagate the TEM wave is transmitted as radio from one conductor to the other during the step.  No loss occurs because the step in each is inverted with respect to the other in a TEM wave.  I'm going to do the calculations to demonstrate how this solves the Catt anomaly.  172.209.113.91 20:52, 20 January 2006 (UTC)
replies: text: Nigel, you cannot inject RF into one conductor independent of the other. The RF energy must flow in the dielectric. Or dont you believe in the Poynting vector?--Light current 22:59, 20 January 2006 (UTC)

text: So, Nigel you are saying that vector calculus has been wrong all these years and so has Mr Poynting? You must publish this ASAP! Im sure it will be of interest to everyone!--Light current 02:15, 22 January 2006 (UTC)
thread_title:  20 January 2006 

message: :If energy does flow in the direction of the electric (or magnetic) field in a coax cable ( ie radially from inner to outer ), where does this energy go when it hits the outer conductor? Also if this were true, you would be able to inject RF power into one end of a coax with a perfect termination at the other end, and find that some energy had been lost due to radial propagation. I dont believe any results of such an experiment have yet been published. But maybe you know different?--Light current 19:23, 21 January 2006 (UTC)
thread_title:  Sideways energy flow?