McEwan's Snow Job
18jan00. McEwan, after four years incommunicando (he only ever wrote once, in 1996, under instruction from is boss, and then ignored all further communications from third parties and instructions from his boss to write again), now does a snow job, garnished with the 'confidential' card, and salted with grovels to Pepper FRS,a Southerner. [McEwan should have grovelled to Pepper FRS's boss Howie FRS, who, like McEwan, is a Westerner. I.C.1feb00]
Dear Mr Catt,
I am offering a reply to your recent correspondence. I do hope you will
accept that this is entirely friendly and disinterested, and that I have
honestly tried to explain the problem.
I'd just like to first make a few personal comments about myself.
[About 600 'confidential' words erased by I Catt.]
I hope you will understand therefore that I simply can't afford to get
involved in a lot more correspondence on this issue, but I offer below some
thoughts which I hope will help.
I must say that I don't think you are doing anything useful by stirring up
issues of north versus south, east etc
I will trust to your integrity to treat my above comments, especially about
my own circumstances, as totally confidential.
[See p55 of the book "The Catt Anomaly", on this website, quoting Catt's 10sep96 letter to McEwan's boss; "I promise that his [McEwan's] response, and my further comments on him, will appear in future issues of this book." Should I now break my promise? These 'scientists' always play the 'confidential' card.]
Now let me make a few comments for public consumption:
"I previously offered to Mr Catt a simple explanation of how the charge is
conveyed along the transmission line. I used an uniform array of N
electrons and N positive ions spaced out along a section of line of length
L. I then pointed out that if we push in one extra electron at the left of
this section, and redistribute the N + 1 electrons uniformly over that
section, there appears a net unbalanced charge of one unit which is
distributed nearly uniformly over that section, but none of the charges
involved had to move a distance greater than L/N within the time it took to
redistribute the charges. The large values of N actually involved explains
why the particle velocity really is so small. This is the gist of my
explanation which I won't repeat in detail as I assume Mr Catt has already
included it and will recap it as necessary.
I still stand by this as a basic explanation of how the charge is carried
along the line. As I explained before, I think the anomaly only appears to
exist because there is a confusion about the identity of the charges
involved. The charge which actually supports the line voltage is actually a
very slight unbalance between very large densities of positive and negative
charges which are already in any given section of line before the
propagaing wave reaches them. (Note the italics!)
My description shows that a pattern of unbalanced charge can move far more
rapidly than the individual charges involved. (I could make the obvious
analogy with sound waves; after 1 second I hear the sound from a lightning
stroke 340 metres away but it is perfectly obvious that none of the
atmospheric molecules that were around the original discharge have arrived
at my ears. Putting it a bit facetiously, I don't smell any ozone at the
same time as the sound arrives and there certainly aren't any 340 m/sec
winds blowing round my head. But surely the idea of particles transmitting
stress to other particles is already clear enough.)
I would like to emphasise that my description using N charges in a line was
a deliberately simplified one intended to get over the key concept without
a lot of detail. This leads me to my next point.
I am prepared to take slight issue with Prof Pepper - again in a completely
friendly way I hope - about the main component of the velocity of the
charges. My recollection is that he agreed with me that the required
charges are already in the section of line to start with, but I think he
implied that the charges move laterally outward to generate the surface
charge as the wave moves over them. I would assert that the main component
of particle velocity is longitudinal.
In fact it is easy to show that the current flow must have both lateral and
longitudinal components, so I agree with Prof Pepper that there are lateral
charge movements but I do assert that the longitudinal velocity components
are the larger ones. We can go into this in a little more detail:
The surface charges on the metallic conductors exist only in a very thin
surface layer. Classical theory doesn't give any indication of the
thickness of this layer. To do it properly means solving the wave
mechanical equations for the states of the electrons near the surface. This
I am not competent to do. However, this distance scale is obviously an
Within the conductor deeper than the surface charge layer, we will find
there is no unbalanced charge density. We now have to introduce the
concept of skin depth. The current flow along the conductor occurs within a
layer near the surface whose thickness is the skin depth. Because the skin
depth varies inversely as the square root of frequency, we are obliged to
consider individual frequency components in the propagating pulse. However
the skin depth is very much greater than the surface charge layer thickness
up to very high frequencies, as (for copper) it is about 9 mm at 50 Hz and
about 2 microns at 1 GHz.
The implication of this is that the moving electrons must have both
transverse and longitudinal components of velocity. They have to arrive at
the surface of the metal, yet flow within a much thicker region. To arrive
at the surface, they must, as Prof Pepper says, move sideways. However, if
they only moved sideways, there would still not be any net charge imbalance
in any small section of line. So here I am saying that Prof Pepper's
description is incomplete, there have to be longitudinal motions as well.
You can imagine the lines of the current flow field (at a single frequency)
as like semi-loops in which one end of the loop starts on a patch of
positive surface charge, bends round very sharply within the skin depth,
then goes longitudinally along and terminates on a negative surface charge
patch. I emphasise again, however, that no individual charge originally at
one end of the loop has to arrive at the other end; only small individual
velocities are involved.
(This can be put a bit more formally using some mathematics. Because there
can be no unbalanced charge density within the conductor, the current flow
field must have zero divergence, i.e. if we use an x - axis along the
cable axis and a y - axis normal to the conductor surface, then we must
have dUsubx)/dx + dUsuby/dy = 0. Here Usubx and Usuby are the x and y
components of the current density flow vector. Now the first term is
certainly non - zero because the velocity does exist on the left of the
wave front and not on the right of it. This implies that Usuby can't be
zero. I include this only as shorthand for the benefit of those who are
familiar with this kind of maths, but it isn't essential.)
For the high frequency components within the propagating pulse, the ratio
of the longitudinal velocity components to the transverse ones will be the
approximate ratio of the wavelength of the guided wave to the skin depth.
For components at sufficiently low frequencies where the skin depth becomes
larger than the conductor transverse dimensions , the corresponding ratio
will be of the order of the ratio of the wavelength of the wave to the
transverse dimension of the appropriate conductor. I believe that in all
virtually all practical cases this ratio is very much greater than unity.
I am sure Prof Pepper will not be in the least offended by my raising this
contention, and anyway I am quite prepared to be shot down about it if I
myself am wrong.
Within the approximations of the classical equations, the problem of the
step wave propagating along a line made of conductors of finite
conductivity can in principle be solved numerically using the
finite-difference time domain method. I am not certain that the software
that is actually around can cope well with the different length scales of
the skin depth and the inter-conductor spacings. I don't have time to look
into this, but if anyone else would like to have a go (or maybe even has
done it already and I am not aware of it) I believe they will be able to
demonstrate a current flow field similar to what I described: I think it
will show almost purely longitudinal velocity components, uniformly
distributed across the conductors, a long way behind the wave front, and
transverse components that increase as you approach the propagating wave
I have noted Mr Catt's comments where he says that one explanation of the
wave transmission (and I believe it is correct) is that the electrons
transmit the wave by each one "nudging" the next.
Very best wishes,
Neil McEwan [18jan00]
Co-author Dr. A. Lynch to Ivor Catt, 30jan00
My physics dates back to the 1940's, since when I have usually called myself an electrical engineer. But I think the spacing of atoms in a solid or liquid is about 0.3nm, and the size of an atomic nucleus is less by a factor of thousands, so that energetic particles are able to pass through a thin film of solid with few collisions. The size and shape of an electron are, I believe, unknown. McEwan discusses the "nudging" sensibly, but he appears to assume that the electron is spherical - otherwise why "the" radius? ….
…. There are, however, no doubts about J.J.'s discovery [which J.J. described to young Arnold Lynch, now aged 83, - I.C.]: electric charge is associated with inertia, and this is what matters for your Anomaly.
Yours sincerely, Arnold Lynch
Comment by Ivor Catt, 1feb00.
Lynch first pointed out in our joint IEE paper that electrons are too far apart to nudge each other. Here, he points out that they must be far apart, not only in diameter, but also in their power to influence events, with large unaffected spaces between, "so that energetic particles are able to pass through a thin film of solid with few collisions." He is moving towards the suggestion that if electrons nudged each other, then X-ray photography would not work. - I.C. 1feb00.