On Mon, 10 Apr 2006 20:52:06 GMT, "Tom Donaly"
wrote:

Not everyone is happy with the term "displacement current." Albert
Shadowitz, in his book _The Electromagnetic Field_, has a chapter
entitled "The So-called Displacement Current." The term isn't in
the index to Feynman's _Lectures on Physics_. (At least I couldn't
find it.) All that is academic to the fact that AC current seems to
be able to make its way through a capacitor with no more opposition
than the capacitive reactance. Fortunately, no one on this
newsgroup has any objection to the way the term is commonly used.

Hi Tom, and others,

The "labeled" currents span a much too small arena. There are also
the induced currents (no, not necessarily from flux linkage) and
convection currents (which IS the primary correlative to the induced
current).

The convection currents are possibly the only current that attain the
speed of light velocity. The others are so astronomically slow, that
it is arguable to say that any current (electron/hole transport) in a
wire is any more significant than that that is supposed to never cross
through the dielectric of a capacitor.

In other words, the displacement current is labeled fictitious because
no electron ever moves from one plate to the other. Now, if we simply
substitute solid gold for that dielectric (still maintaining the same
plates); then no electron ever makes it from one plate to the other -
and yet current flows in the entire AC circuit by proportion to the
impedance presented to it by either the dielectric capacitor, or the
gold capacitor.

This, of course, illustrates the corruption of usage in the term
"current."

73's
Richard Clark, KB7QHC

Gene Fuller wrote:
My point is in complete agreement with Tom, W8JI. The only thing that
allows "current taper" is displacement current.

True, but it doesn't happen as W8JI describes.

The distributed capacitance in a coil causes a transmission
line effect. The displacement currents cause delays (phase
shifts) in traveling wave currents. The traveling wave
currents can be considered to have constant magnitude, i.e.
*negligible current taper* in the traveling wave in spite
of the known displacement currents.

The displacement current effect on traveling waves is in
the phase, not the magnitude. Such is illustrated as an
EZNEC result in the left hand graphic at:

http://www.qsl.net/w5dxp/travstnd.GIF

Please note that in spite of the distributed capacitance,
the magnitude is fixed and flat, i.e. no taper. The displacement
currents cause phase shift delays in traveling waves but has
virtually no effect on the magnitude of the traveling wave.

The distributed capacitance is the same in the transmission
line whether a single traveling wave is present or standing
waves present. So displacement currents don't necessarily
result in current taper. How do you explain that one?

Now take a look at the right hand graph involving standing
wave current. The *phase is fixed and unchanging*. The magnitude
of the standing wave current is *tapered as a cosine function
of distance from the source*. Displacement current indeed does
cause this effect but it is a transmission line effect of
superposition of forward and reflected waves, not the effect
of some imagined sideways third path for current to earth ground.
--
73, Cecil http://www.qsl.net/w5dxp

Richard Harrison wrote:
"Displacement current which is the a-c current through a capacitor, that
has no a-c conduction, is not the "ONLY" thing that allows a conductor
to have a current taper." It was Tom, W8JI who shouted: "The ONLY thing
etc." I just said displacement current is NOT the only thing. Energy
level often declines between ends of a wire or coil due to losses from
radiation or dissipation in the wire or coil. Tom is mistaken.

Sorry Richard, that is not correct.

Radiation does not cause current taper. Dissipation does not either.

Consider dissipation first. If dissipation caused current reduction,
the return to a battery from a light bulb would have less current than
the outgoing terminal. There has to be a third path to allow current to
divide, but the totals of the division equal the initial amount. That's
a rule we learn way back in basic electricity. Current or charges are
not converted into heat.

Radiation is no different. Radiation is not conversion of charges into
a force that allows action at a distance. Radiation is a force on other
charges at a distance caused by charge acceleration.

The only thing that allows an antenna to have current taper or current
change along the length of a wire suspended in space is displacement
current. It is not standing waves, it is not radiation, it is not
resistance.

Of course we could add a shunt resistance or inductance to provide a
path, but when there is no leakage resistance or shunting inductance
the path can only be what is called displacement current.

A series impedance or resistance by itself, even if the cause is
radiation or loss resistance, cannot cause current reduction with
distance along a conductor.

A model that only considers reflected and forward "waves" is fine, if
applied correctly. Cecil doesn't even seem to understand current, and
appears to think there is a forward current and reflected current
moving in opposite directions at the same instant of time in the very
same location in a conductor.

Wave theory is just fine, but it has to be understood it is just a
modelling shortcut and the results cannot conflit with basic laws of
physics. The current we measure with a clamp on meter IS the current
that causes radiation, standing waves or not. It is also the current
that causes all of the heating. We cannot really have two opposite
directions of charge movement at the same time in a single conductor.

73 Tom

Cecil,

All I can say is why don't you write this magic tale into a technical
article and submit it to your favorite IEEE journal or AIP journal.

73,
Gene
W4SZ

Cecil Moore wrote:
Gene Fuller wrote:

My point is in complete agreement with Tom, W8JI. The only thing that
allows "current taper" is displacement current.


True, but it doesn't happen as W8JI describes.

The distributed capacitance in a coil causes a transmission
line effect. The displacement currents cause delays (phase
shifts) in traveling wave currents. The traveling wave
currents can be considered to have constant magnitude, i.e.
*negligible current taper* in the traveling wave in spite
of the known displacement currents.

The displacement current effect on traveling waves is in
the phase, not the magnitude. Such is illustrated as an
EZNEC result in the left hand graphic at:

http://www.qsl.net/w5dxp/travstnd.GIF

Please note that in spite of the distributed capacitance,
the magnitude is fixed and flat, i.e. no taper. The displacement
currents cause phase shift delays in traveling waves but has
virtually no effect on the magnitude of the traveling wave.

The distributed capacitance is the same in the transmission
line whether a single traveling wave is present or standing
waves present. So displacement currents don't necessarily
result in current taper. How do you explain that one?

Now take a look at the right hand graph involving standing
wave current. The *phase is fixed and unchanging*. The magnitude
of the standing wave current is *tapered as a cosine function
of distance from the source*. Displacement current indeed does
cause this effect but it is a transmission line effect of
superposition of forward and reflected waves, not the effect
of some imagined sideways third path for current to earth ground.

Gene Fuller wrote:
I believe after a long series of EZNEC models and RRAA messages you came
to the conclusion that the 75 meter bugcatcher coil at 4 MHz had a
traveling wave phase shift of around 10 degrees.

Note that is not a measurement - that is what EZNEC reports but
let's assume, for the sake of discussion, that it is correct. W8JI
measured a 3nS, 4 degree phase shift in a coil twice as long with
43% more inductance. A bigger coil would obviously have a bigger
phase shift because of less current field linkage between the end
coils. So even if the phase shift through the coil is 10 degrees
as reported by EZNEC, W8JI's phase shift measurements were probably
off by *MORE THAN 200%* and that's why Tom is wrong.

This same coil resonated an antenna with a whip length of 10 feet or so.
A quarter wavelength at 4 MHz is around 60 feet. The phase shift that
could be attributed to the whip is therefore around 15 degrees. The
phase shift of the missing 50 feet of wire for a plain quarter wave
antenna would be around 75 degrees.

You are confused. Some time ago, I explained why a mobile antenna
may not be 90 degrees long at all. Did you understand that posting?
All we can say is that (Vfor+Vref)/(Ifor+Iref) is purely resistive.
We don't know how many degrees the reflected wave has traveled in
its round trip because there are too many variables.

So please stop the diversions. I have always said that the delay
through a coil *IS WHAT IT IS* but it is NOT zero and it is not the
3 nS measured by W8JI for that 100 uH coil. It is also not the near-zero
phase shift measured by W7EL using standing wave current phase as the
reference. You, yourself, implied that is an invalid measurement when
you told us there is no phase information in standing wave phase.

Seems to me you are making my argument for me and that your real
argument is with the other side. Have you told W7EL that standing
wave current phase cannot be used to measure the delay through a
coil?
--
73, Cecil http://www.qsl.net/w5dxp

wrote:
Radiation does not cause current taper. Dissipation does not either.

Radiation and dissipation are considered to be losses in a
transmission line covered by the attenuation factor. All that
is needed to prove your above assertions to be false is to
quote a transmission line equation. It can even be the more
simple flat form where the SWR is 1:1. Here it is in ASCII:

I = Im*e^(ax)*e^j(wt-bx)

Note this is the equation for *CURRENT* where 'a' is the attenuation
factor. The attenuation factor includes radiation and dissipation.

Your statements indicate a high level of ignorance. Assuming a flat
transmission line with an SWR of 1:1, if the loss in the transmission
line is 3 dB, we can put 200 watts into 50 ohm coax at the source end
and get 100 watts out at the 50 ohm load end. The current out of the
source is SQRT(200w/50) = 2 amps. The current through the 50 ohm load
is SQRT(200w/50) = 1.414 amps. The current has dropped from the source
to the load by exactly the same percentage that the voltage has dropped.

What you seem to be missing is that the H-field is attenuated by the
same amount as the E-field while the ratio of E-field to H-field
remains constant and equal to Z0. Current is proportional to the
H-field and voltage is proportional to the E-field.
--
73, Cecil http://www.qsl.net/w5dxp

Gene Fuller wrote:
All I can say is why don't you write this magic tale into a technical
article and submit it to your favorite IEEE journal or AIP journal.

I'm sure there are hundreds of such papers already, Gene.
Much of this stuff is in the Corum paper. But you rejected
that Corum IEEE paper that I presented as evidence so why
would me writing one make any difference to your fixed
preconceptions?
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:

wrote:

Radiation does not cause current taper. Dissipation does not either.


Radiation and dissipation are considered to be losses in a
transmission line covered by the attenuation factor. All that
is needed to prove your above assertions to be false is to
quote a transmission line equation. It can even be the more
simple flat form where the SWR is 1:1. Here it is in ASCII:

I = Im*e^(ax)*e^j(wt-bx)

Note this is the equation for *CURRENT* where 'a' is the attenuation
factor. The attenuation factor includes radiation and dissipation.

Your statements indicate a high level of ignorance. Assuming a flat
transmission line with an SWR of 1:1, if the loss in the transmission
line is 3 dB, we can put 200 watts into 50 ohm coax at the source end
and get 100 watts out at the 50 ohm load end. The current out of the
source is SQRT(200w/50) = 2 amps. The current through the 50 ohm load
is SQRT(200w/50) = 1.414 amps. The current has dropped from the source
^^^
Obviously, should be 100w. Sorry for the typo.

to the load by exactly the same percentage that the voltage has dropped.

What you seem to be missing is that the H-field is attenuated by the
same amount as the E-field while the ratio of E-field to H-field
remains constant and equal to Z0. Current is proportional to the
H-field and voltage is proportional to the E-field.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil,

As usual, you evaded the question, but this time you did not even do a
very slick job of evasion.

The question is what happens to the 75 degrees that was formerly
represented by the now-replaced wire. The coil may offer about 10 degrees.

I believe that Tom is stating that 75 is not equal to 10. Sounds like a
reasonable statement to me.

I don't know why you are so worried about the precise details of phase
measurements. Even your standard of precision, +/- 59%, won't make 10
equal to 75.

73,
Gene
W4SZ

Cecil Moore wrote:
Gene Fuller wrote:
I believe after a long series of EZNEC models and RRAA messages you came
to the conclusion that the 75 meter bugcatcher coil at 4 MHz had a
traveling wave phase shift of around 10 degrees.

Note that is not a measurement - that is what EZNEC reports but
let's assume, for the sake of discussion, that it is correct. W8JI
measured a 3nS, 4 degree phase shift in a coil twice as long with
43% more inductance. A bigger coil would obviously have a bigger
phase shift because of less current field linkage between the end
coils. So even if the phase shift through the coil is 10 degrees
as reported by EZNEC, W8JI's phase shift measurements were probably
off by *MORE THAN 200%* and that's why Tom is wrong.

This same coil resonated an antenna with a whip length of 10 feet or so.
A quarter wavelength at 4 MHz is around 60 feet. The phase shift that
could be attributed to the whip is therefore around 15 degrees. The
phase shift of the missing 50 feet of wire for a plain quarter wave
antenna would be around 75 degrees.

You are confused. Some time ago, I explained why a mobile antenna
may not be 90 degrees long at all. Did you understand that posting?
All we can say is that (Vfor+Vref)/(Ifor+Iref) is purely resistive.
We don't know how many degrees the reflected wave has traveled in
its round trip because there are too many variables.

So please stop the diversions. I have always said that the delay
through a coil *IS WHAT IT IS* but it is NOT zero and it is not the
3 nS measured by W8JI for that 100 uH coil. It is also not the near-zero
phase shift measured by W7EL using standing wave current phase as the
reference. You, yourself, implied that is an invalid measurement when
you told us there is no phase information in standing wave phase.

Seems to me you are making my argument for me and that your real
argument is with the other side. Have you told W7EL that standing
wave current phase cannot be used to measure the delay through a
coil?

Cecil,

Wow! I think you may have set a new world record for the most irrelevant
concepts per word dragged into an RRAA posting.

We got transmission lines, attenuation factors, H-fields, E-fields, and
even SWR. Not to mention watts, dB, and Zo.

It is truly unfortunate that none of this is connected to the subject at
hand, displacement current, but it makes for a colorful message.

73,
Gene
W4SZ

Cecil Moore wrote:
wrote:

Radiation does not cause current taper. Dissipation does not either.


Radiation and dissipation are considered to be losses in a
transmission line covered by the attenuation factor. All that
is needed to prove your above assertions to be false is to
quote a transmission line equation. It can even be the more
simple flat form where the SWR is 1:1. Here it is in ASCII:

I = Im*e^(ax)*e^j(wt-bx)

Note this is the equation for *CURRENT* where 'a' is the attenuation
factor. The attenuation factor includes radiation and dissipation.

Your statements indicate a high level of ignorance. Assuming a flat
transmission line with an SWR of 1:1, if the loss in the transmission
line is 3 dB, we can put 200 watts into 50 ohm coax at the source end
and get 100 watts out at the 50 ohm load end. The current out of the
source is SQRT(200w/50) = 2 amps. The current through the 50 ohm load
is SQRT(200w/50) = 1.414 amps. The current has dropped from the source
to the load by exactly the same percentage that the voltage has dropped.

What you seem to be missing is that the H-field is attenuated by the
same amount as the E-field while the ratio of E-field to H-field
remains constant and equal to Z0. Current is proportional to the
H-field and voltage is proportional to the E-field.