Current in antenna loading coils controversy - new measurement
 

Oh well, back to my rubber room....:/ MK


and stay there :-)

Some people can see further than others. Stand by for more. More thorough work
and decent article will take a bit more time. I intend to do that, have
promised cooperation of others in our camp. We should have more precise
figures, arguments and 'splanations. I mentioned some things in my previous
post, but if you can't see the benefits for improvement then be happy with what
you have.


Yuri

Yuri Blanarovich wrote:
Some people can see further than others. Stand by for more. More thorough work
and decent article will take a bit more time. I intend to do that, have
promised cooperation of others in our camp. We should have more precise
figures, arguments and 'splanations. I mentioned some things in my previous
post, but if you can't see the benefits for improvement then be happy with what
you have.

One thing I was originally confused about until I sat down and drew some
phasor diagrams. The phase of the *net* current doesn't change much over
the length of the mobile antenna. It's the phase of the forward current
and reflected current that is changing. But those phases are changing in
opposite directions so the sum of those two phasors results in very little
phase change in the net current.

For a 1/4WL vertical, the net current is approximately equal to 90 degrees
of a cosine wave, 1.0 at zero degrees and zero at 90 degrees. The magnitude
of the net current is an indication of where between zero and 90 degrees
the current is being measured. Where = arc-cos(|net current|)
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:
One thing I was originally confused about until I sat down and drew some
phasor diagrams. The phase of the *net* current doesn't change much over
the length of the mobile antenna. It's the phase of the forward current
and reflected current that is changing. But those phases are changing in
opposite directions so the sum of those two phasors results in very little
phase change in the net current.

It might be helpful to indicate your point of reference when talking
about phase, i.e. phase is changing with respect to voltage, or phase is
changing with respect to the source, phase is changing with respect to
position, etc.

Take the sentence "it's the phase of the forward and reflected current
that is changing" for example. For a traveling wave it is certainly
true that at any given point along a waveguide, the phase of the wave is
constantly changing with time. But it isn't necessarily true that at
that point, the phase of the current relative the voltage is changing
with time.

In the problem we've been discussing both points of reference are
relevant. In some case we're addressing the phase shift between voltage
and current, and in other cases we're discussing the change in phase
along the length of a conductor. I think it's helpful to be specific
about this.


For a 1/4WL vertical, the net current is approximately equal to 90 degrees
of a cosine wave, 1.0 at zero degrees and zero at 90 degrees. The magnitude
of the net current is an indication of where between zero and 90 degrees
the current is being measured. Where = arc-cos(|net current|)

That angle is the 'phase' of the current standing wave as a function of
position, not to be confused with the phase of the current with respect
to voltage. Roger?

73, Jim AC6XG

Jim Kelley wrote:
That angle is the 'phase' of the current standing wave as a function of
position, not to be confused with the phase of the current with respect
to voltage. Roger?

Since any reference to voltage on an antenna seems to be verboten, I
have avoided any such reference. Otherwise, here is a quote from Kraus:

"It is generally assumed that the current distribution of an infinitesimally
thin antenna is sinusoidal, and that the phase is *constant* over a 1/2WL
interval, changing abruptly by 180 degrees between intervals."

Don't you just love the phrase, "It is generally assumed ..."? He doesn't
say the current is sinusoidal. He doesn't say the phase is constant over
a 1/2WL interval.

For that general assumption to be true, the reflected current would have
to equal the forward current on a standing-wave antenna. But we know it
doesn't. However, this implies that the reflected current arriving back
at the feedpoint is not extremely/severely attenuated. Here's a cute
ballpark analysis sure to drive the gurus crazy.

Assume the Z0 of a traveling-wave dipole is 600 ohms. The ratio of forward
voltage to forward current is 600 ohms. The ratio of reflected voltage to
reflected current is 600 ohms. Assume the feedpoint current is one amp and
the feedpoint voltage is 50 volts. Assume the forward current and the reflected
current are in phase at the feedpoint. Assume the forward voltage and reflected
voltage are 180 degrees out of phase at the feedpoint. This is enough information
to solve for the ratio of forward current to reflected current at the feedpoint.
Assuming the net current is one amp at the feedpoint, I get 0.542 amps for
the forward current and 0.458 amps for the reflected current, i.e. the reflected
current is 85% of the value of the forward current. Remember, that is a ballpark
estimate.

That means the current only decreases by 15% in its round trip to the
end of the antenna and back. Same for the voltage. Oops, I mentioned
voltage - sorry. But please note that the power loss is a lot higher
than 15% since both the current and voltage are reduced by the same 15%.

The argument seems to occur due to the ignoring of the component waves
on a standing wave antenna. Such is the steady-state model seduction attended
by its sacred cows. Who wants to join me in a beer bust and barbecue?
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Jim Kelley wrote:
That angle is the 'phase' of the current standing wave as a function of
position, not to be confused with the phase of the current with respect
to voltage. Roger?

Since any reference to voltage on an antenna seems to be verboten, I
have avoided any such reference.

I can understand how at DC, references to current in a dipole might be
verboten. :-)

I wonder if voltage on a dipole could be roughly likened to transverse
velocity at various points along a whip.

Don't you just love the phrase, "It is generally assumed ..."?

I guess it allows: 'I'm not responsible should it turn out not to be a
good assumption'. :-)

For that general assumption to be true, the reflected current would have
to equal the forward current on a standing-wave antenna. But we know it
doesn't. However, this implies that the reflected current arriving back
at the feedpoint is not extremely/severely attenuated.

Seems to me it just implies that current at the end of a dipole isn't
really zero.

Assume the Z0 of a traveling-wave dipole is 600 ohms. The ratio of forward
voltage to forward current is 600 ohms. The ratio of reflected voltage to
reflected current is 600 ohms. Assume the feedpoint current is one amp and
the feedpoint voltage is 50 volts. Assume the forward current and the reflected
current are in phase at the feedpoint. Assume the forward voltage and reflected
voltage are 180 degrees out of phase at the feedpoint. This is enough information
to solve for the ratio of forward current to reflected current at the feedpoint.
Assuming the net current is one amp at the feedpoint, I get 0.542 amps for
the forward current and 0.458 amps for the reflected current, i.e. the reflected
current is 85% of the value of the forward current. Remember, that is a ballpark
estimate.

50 volts difference across a 600 ohm impedance, over 2, plus and minus
half an amp.

That means the current only decreases by 15% in its round trip to the
end of the antenna and back.

I think it may actually make many round trips. There may be multiple
reflections.

The argument seems to occur due to the ignoring of the component waves
on a standing wave antenna. Such is the steady-state model seduction attended
by its sacred cows.

Well, you'll either have to write out everything as a series, which is a
lot of busy work, or use the steady state equivalent.

So what does all this say about reflectivity at the end of the dipole?
;-)

73, Jim AC6XG

Jim Kelley wrote:
Seems to me it just implies that current at the end of a dipole isn't
really zero.

The net current is very close to zero because the forward and reflected
currents are very nearly equal and 180 degrees out of phase.

I think it may actually make many round trips. There may be multiple
reflections.

Of course, but like a transmission line, there is only one forward wave
and one reflected wave. All the multiple reflections are contained in
those two waves.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Jim Kelley wrote:
Seems to me it just implies that current at the end of a dipole isn't
really zero.

The net current is very close to zero because the forward and reflected
currents are very nearly equal and 180 degrees out of phase.

Is that what you meant by:

"For that general assumption to be true, the reflected current would
have
to equal the forward current on a standing-wave antenna. But we know it
doesn't"?

It's not clear whether you're making this point in recognition of the
fact that wires are not lossless, or whether you're claiming it's
somehow fundamental to the performance of a radiator.

I think it may actually make many round trips. There may be multiple
reflections.

Of course, but like a transmission line, there is only one forward wave
and one reflected wave. All the multiple reflections are contained in
those two waves.

Then apparently you've decided not to completely eschew making at least
some steady state assumptions. Seductive indeed. ;-)

73, Jim AC6XG

Jim Kelley wrote:
Cecil Moore wrote:
The net current is very close to zero because the forward and reflected
currents are very nearly equal and 180 degrees out of phase.

This is at the tip end of a dipole.

Is that what you meant by:

"For that general assumption to be true, the reflected current would
have
to equal the forward current on a standing-wave antenna. But we know it
doesn't"?

This is apparently not at the tip end of a dipole.

It's not clear whether you're making this point in recognition of the
fact that wires are not lossless, or whether you're claiming it's
somehow fundamental to the performance of a radiator.

Both, radiation is a "loss".

Of course, but like a transmission line, there is only one forward wave
and one reflected wave. All the multiple reflections are contained in
those two waves.

Then apparently you've decided not to completely eschew making at least
some steady state assumptions. Seductive indeed. ;-)

There are only two possible directions, forward and reverse, in which
energy can flow. Multiple reflections do not create any more directions.
I am not opposed to the steady-state solution and use it all the time. I am
opposed to people being seduced by the steady-state solution into believing
there is not such thing as forward and reflected waves even though standing
waves require forward and reflected waves.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:
I am
opposed to people being seduced by the steady-state solution into believing
there is not such thing as forward and reflected waves even though standing
waves require forward and reflected waves.

I also apologize if this particular dose of reality brought this thread to
a screeching halt.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:
I am
opposed to people being seduced by the steady-state solution into believing
there is not such thing as forward and reflected waves even though standing
waves require forward and reflected waves.

Say that in Sacramento, and some knucklehead legislator will pass a law
against it.

73, Jim AC6XG

Cecil, W5DXP wrote:
"There are only two possible directions, forward and reverse in which
energy can flow. Multiple reflections do not create any more
directions."

True. Further, all the same-frequency, same-direction signals merge. So,
as Cecil said, there are only two same-frequency signals on a
transmission line, forward and reverse. The interference pattern these
signals produce does not represent another signal.

Trying to use ordinary circuit analysis on standing-wave antennas is
problematic, but it`s been tried in this thread. Here is what R.W.P.
King wrote in "Transmission Lines, Antennas, and Wave Guides", King,
Mimno, and Wing, 1945, on page 86:
"Inductance and capacitance as used in near-zone circuits with uniform
current cannot be defined, and ordinary circuit analysis does not
apply." This has not stopped efforts in this thread to analyze LC
circuits as if we were dealing with low frequencies.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Cecil, W5DXP wrote:
"There are only two possible directions, forward and reverse in which
energy can flow. Multiple reflections do not create any more
directions."

True. Further, all the same-frequency, same-direction signals merge. So,
as Cecil said, there are only two same-frequency signals on a
transmission line, forward and reverse. The interference pattern these
signals produce does not represent another signal.

So the claim is that the amplitude of the reflection being bandied about
is the sum of multiple reflections? I haven't seen any indication of
that. It appears to be simply the amplitude of the first reflection.
At least, that's the way it appears to me, the uninitiated.

73, Jim AC6XG

Richard Harrison wrote:
Trying to use ordinary circuit analysis on standing-wave antennas is
problematic, but it`s been tried in this thread. Here is what R.W.P.
King wrote in "Transmission Lines, Antennas, and Wave Guides", King,
Mimno, and Wing, 1945, on page 86:
"Inductance and capacitance as used in near-zone circuits with uniform
current cannot be defined, and ordinary circuit analysis does not
apply." This has not stopped efforts in this thread to analyze LC
circuits as if we were dealing with low frequencies.

Very true. The basic problem, as I see it, is in assuming that the standing-
wave current only has one component. For standing-wave antennas, the
standing-wave current must necessarily have two components, If and Ir, as
explained by Balanis in _Antenna_Theory_, (page 489, 2nd edition).

In general, when forward waves and reflected waves exist in the circuit,
lumped circuit analysis fails and distributed network analysis is the only
method that yields the correct result. So even if one allows that the forward
current magnitude is constant through a coil and the reflected current
magnitude is constant through a coil, the net current magnitude will not
be constant because of the phase differences in the two superposed currents
at each end of the coil.

I've been told by the gurus that I can ignore the forward and reflected
waves and still obtain the correct steady-state solution. A mobile bugcatcher
coil on 75m seems to disprove that assertion. But I have been called a
"Grasshopper" and thus apparently have a lot yet to learn.
--
73, Cecil http://www.qsl.net/w5dxp



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Jim Kelley wrote:
So the claim is that the amplitude of the reflection being bandied about
is the sum of multiple reflections? I haven't seen any indication of
that. It appears to be simply the amplitude of the first reflection.
At least, that's the way it appears to me, the uninitiated.

Shirley, you jest. Reference pages 17-20 in _T-Lines & Networks_ by Johnson.
Consider a T-line with an SWR of 5.8284:1 and a constant 100W Z0-matched
source. The forward power will be 100W, 150W, 175W, 187.5W, 193.75W,
196.875W, 198.4375W, ..., 200W. After steady-state is reached, the
reflected power is a constant 100W (sans modulation and noise).
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Jim Kelley wrote:
So the claim is that the amplitude of the reflection being bandied about
is the sum of multiple reflections? I haven't seen any indication of
that. It appears to be simply the amplitude of the first reflection.
At least, that's the way it appears to me, the uninitiated.

Shirley, you jest. Reference pages 17-20 in _T-Lines & Networks_ by Johnson.
Consider a T-line with an SWR of 5.8284:1 and a constant 100W Z0-matched
source. The forward power will be 100W, 150W, 175W, 187.5W, 193.75W,
196.875W, 198.4375W, ..., 200W. After steady-state is reached, the
reflected power is a constant 100W (sans modulation and noise).

Interesting reference. Wish I had it. He's showing re-reflections from
a Z0-matched source? I don't think I understand all I know about that.
:-)

73, AC6XG

On Tue, 18 Nov 2003 13:52:04 -0800, Jim Kelley
wrote:

|
|
|Cecil Moore wrote:
|
| Jim Kelley wrote:
| So the claim is that the amplitude of the reflection being bandied about
| is the sum of multiple reflections? I haven't seen any indication of
| that. It appears to be simply the amplitude of the first reflection.
| At least, that's the way it appears to me, the uninitiated.
|
| Shirley, you jest. Reference pages 17-20 in _T-Lines & Networks_ by Johnson.
| Consider a T-line with an SWR of 5.8284:1 and a constant 100W Z0-matched
| source. The forward power will be 100W, 150W, 175W, 187.5W, 193.75W,
| 196.875W, 198.4375W, ..., 200W. After steady-state is reached, the
| reflected power is a constant 100W (sans modulation and noise).
|
|Interesting reference. Wish I had it.

I have it and it doesn't say anything of the sort.

|He's showing re-reflections from
|a Z0-matched source? I don't think I understand all I know about that.
|:-)
|
|73, AC6XG

Jim Kelley wrote:
Interesting reference. Wish I had it. He's showing re-reflections from
a Z0-matched source? I don't think I understand all I know about that.

Nope, the reference is a DC transient buildup to steady-state but the
same principles apply to RMS values.
--
73, Cecil http://www.qsl.net/w5dxp



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Wes Stewart wrote:

Jim Kelley wrote:

|Cecil Moore wrote:
| Shirley, you jest. Reference pages 17-20 in _T-Lines & Networks_ by Johnson.

Now change the context from a DC example to an RF example:

| Consider a T-line with an SWR of 5.8284:1 and a constant 100W Z0-matched
| source. The forward power will be 100W, 150W, 175W, 187.5W, 193.75W,
| 196.875W, 198.4375W, ..., 200W. After steady-state is reached, the
| reflected power is a constant 100W (sans modulation and noise).
|
|Interesting reference. Wish I had it.

I have it and it doesn't say anything of the sort.

I didn't quote anything it says, Wes, I just gave it as a reference. Then I
suggested a different example to be considered. When the SWR is 5.8284:1,
the reflected power is 1/2 of the forward power making mental calculations
easy. The RF example was out of my head, not out of the reference.
--
73, Cecil http://www.qsl.net/w5dxp



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On Tue, 18 Nov 2003 23:32:11 -0600, Cecil Moore
wrote:

|Wes Stewart wrote:
|
| Jim Kelley wrote:
|
| |Cecil Moore wrote:
| | Shirley, you jest. Reference pages 17-20 in _T-Lines & Networks_ by Johnson.
|
|Now change the context from a DC example to an RF example:
|
| | Consider a T-line with an SWR of 5.8284:1 and a constant 100W Z0-matched
| | source. The forward power will be 100W, 150W, 175W, 187.5W, 193.75W,
| | 196.875W, 198.4375W, ..., 200W. After steady-state is reached, the
| | reflected power is a constant 100W (sans modulation and noise).
| |
| |Interesting reference. Wish I had it.
|
| I have it and it doesn't say anything of the sort.
|
|I didn't quote anything it says, Wes, I just gave it as a reference. Then I
|suggested a different example to be considered. When the SWR is 5.8284:1,
|the reflected power is 1/2 of the forward power making mental calculations
|easy. The RF example was out of my head, not out of the reference.

Well then, I'm citing Einstein's Theory of Relatively, but I'm going
to use another example.

How about a DC voltage source and a switch connected to a Zo
transmission line terminated in R = 3*Zo. Sound familiar? (For those
without the text, it's Johnson's example)

Johnson says in part, "The first reflected wave will in turn be
reflected when it reaches the sending end. The terminating impedance
is *zero* (emphasis added) at the end....."

It's quite a leap from a DC situation with zero source impedance to a
RF situation with a matched source.

"Cecil Moore" wrote in message
...
I didn't quote anything it says, Wes, I just gave it as a reference. Then
I
suggested a different example to be considered. When the SWR is 5.8284:1,
the reflected power is 1/2 of the forward power making mental calculations
easy. The RF example was out of my head, not out of the reference.

But how does that relate to the topic of multiple reflections? Perhaps if
you had just quoted the reference instead....?

73, Jim AC6XG

Wes Stewart wrote:
Johnson says in part, "The first reflected wave will in turn be
reflected when it reaches the sending end. The terminating impedance
is *zero* (emphasis added) at the end....."

Yep, that's why Walter Maxwell calls a Z0-match point a "virtual short"
i.e. a virtual *zero* impedance, because the results are similar. I
call that Z0-match a point of interference because IMO that is a
more accurate description of what is occurring than a "virtual short".

It's quite a leap from a DC situation with zero source impedance to a
RF situation with a matched source.

The source I chose for my example was a *constant power source*.

Assume the source is a 50 ohm signal generator with a circulator load
so Pfwd1 below is a constant 100 watts.

Source----50 ohm lossless coax--+---291.5 ohm lossless twinlead---50 ohm load
Pfwd1-- Pfwd2--
--Pref1 --Pref2

Let time t1 be when the first non-zero Pref2 value arrives back at the '+'
point. tn is steady-state.

Pfwd1 Pref1 Pfwd2 Pref2
t0 100w 0w 0w 0w
t1 100w 50w 50w 25w
tn 100w 0w 200w 100w

If one plots the points between t1 and tn, the diagram will look much like
Johnson's. Pfwd2 steps up to 200w and Pref2 steps up to 100w.
--
73, Cecil http://www.qsl.net/w5dxp



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Jim Kelley wrote:

"Cecil Moore" wrote:
The RF example was out of my head, not out of the reference.

But how does that relate to the topic of multiple reflections?

How does the transient state relate to the topic of multiple
reflections????? Maybe I don't understand your question.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Jim Kelley wrote:

"Cecil Moore" wrote:
The RF example was out of my head, not out of the reference.

But how does that relate to the topic of multiple reflections?

How does the transient state relate to the topic of multiple
reflections????? Maybe I don't understand your question.

Is your assertion that multiple reflections occur only during the
transient period?

73, Jim AC6XG

Jim Kelley wrote:

Cecil Moore wrote:
How does the transient state relate to the topic of multiple
reflections????? Maybe I don't understand your question.

Is your assertion that multiple reflections occur only during the
transient period?

No, but during the transient period, they are visible and measurable.
Sans modulation and noise, they are invisible and unmeasurable during
steady-state. Some people say because they are invisible and unmeasurable
during steady-state, they cease to exist. But that's not me.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Jim Kelley wrote:

Cecil Moore wrote:
How does the transient state relate to the topic of multiple
reflections????? Maybe I don't understand your question.

Is your assertion that multiple reflections occur only during the
transient period?

No, but during the transient period, they are visible and measurable.
Sans modulation and noise, they are invisible and unmeasurable during
steady-state. Some people say because they are invisible and unmeasurable
during steady-state, they cease to exist. But that's not me.

Would you assert that what happens during the transient period and what
happens during the steady state are even necessarily the same thing?

73, Jim AC6XG

Jim Kelley wrote:
Would you assert that what happens during the transient period and what
happens during the steady state are even necessarily the same thing?

Does some new particle of physics manifest itself during steady-state?
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Jim Kelley wrote:
Would you assert that what happens during the transient period and what
happens during the steady state are even necessarily the same thing?

Does some new particle of physics manifest itself during steady-state?

The two questions are not equivalent. Yours is ludicrous. Sorta like
this:
"Measured near field photons may simply recombine with the antenna's
free electrons and not contribute to far field radiation."

Onward through the fog.

73, Jim AC6XG

Jim Kelley wrote:
Cecil Moore wrote:

Jim Kelley wrote:
Would you assert that what happens during the transient period and what
happens during the steady state are even necessarily the same thing?

Does some new particle of physics manifest itself during steady-state?

The two questions are not equivalent. Yours is ludicrous.

Well, if no new particles manifest during steady-state, why wouldn't
the transient state and the steady-state follow exactly the same laws
of physics? Does something supernatural happen at the transient-state
to steady-state threshold? Or not? (Hint: rhetorical question)

Sorta like this:
"Measured near field photons may simply recombine with the antenna's
free electrons and not contribute to far field radiation."

Hmmmmm, I doubt that Feynman would find that statement to be "ludicrous".
I wonder if "ludicrous" is the term the priests used when they condemned
Galileo to house arrest for agreeing with Copernicus? :-) Photons re-
combining with electrons in the near field is a really simple concept.

On second thought, maybe you are inferring that the measured photons cannot
recombine? I would agree with that but the measured photons are negligible
compared to the total number of photons involved in the near field.
--
73, Cecil http://www.qsl.net/w5dxp



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I started this thread late on, so I don't know what the original issue was.
However, that bit about the reflections from the matched source really threw
me. Its been a long time since I was deep into this stuff in school (it all
made good sense then) & then at work, so I have to really strain the 'ole
brain here. Some of the comments do seem to take a turn off to the side.
With a 5.8:1 with half the power reflected at the load means that the
source'll get warmer to the tune of half the power. The use of a "constant
power" source seems strange to me.
And By the Way, you CAN use a DC circuit to show some of the same
concepts.

Steve, k\9\d\c\i




"Cecil Moore" wrote in message
...
Jim Kelley wrote:
Interesting reference. Wish I had it. He's showing re-reflections from
a Z0-matched source? I don't think I understand all I know about that.

Nope, the reference is a DC transient buildup to steady-state but the
same principles apply to RMS values.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:
Well, if no new particles manifest during steady-state, why wouldn't
the transient state and the steady-state follow exactly the same laws
of physics? Does something supernatural happen at the transient-state
to steady-state threshold? Or not? (Hint: rhetorical question)

Given your propensity for hyperbole, if we can't agree that there are
differences between the transient and steady states, I don't think we'll
have too much luck discussing the subject further.

Hmmmmm, I doubt that Feynman would find that statement to be "ludicrous".

Do you really think you're in any postion to be able to speak for
Feynman?

On second thought, maybe you are inferring that the measured photons cannot
recombine? I would agree with that but the measured photons are negligible
compared to the total number of photons involved in the near field.

The photon and the electron were never really "combined" to begin with.
Therefore, the notion that they "recombine" is somewhat off the mark.
Things don't work that way. A photon could on the other hand impart
some or all of its energy to an electron. Certainly the near field can
be seen as affecting the fields within the conductor, thus having an
effect on the charges within that conductor.

73, Jim AC6XG

Jim Kelley wrote:
Cecil Moore wrote:

Given your propensity for hyperbole, if we can't agree that there are
differences between the transient and steady states, I don't think we'll
have too much luck discussing the subject further.

Maybe you had better list those differences, one by one, so we can agree
or disagree.

Do you really think you're in any postion to be able to speak for
Feynman?

Just read one of his books. He says, "An electron emits or absorbs an
electron." "Absorbs" and "combines" sure seems like the same thing to me.

Things don't work that way. A photon could on the other hand impart
some or all of its energy to an electron.

When a photon imparts its energy to an electron, doesn't that "combine"
the two energies? What am I missing? Also, please describe how a photon
could impart half its energy to an electron. Is the result half a photon?
--
73, Cecil http://www.qsl.net/w5dxp



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Severely SNIPPED


When a photon imparts its energy to an electron, doesn't that "combine"
the two energies? What am I missing? Also, please describe how a photon
could impart half its energy to an electron. Is the result half a photon?

A photon is modeled as energy as a function of wavelength.

There are high energy photons, gamma rays. [energies at 0.5 MeV and higher]

There are medium energy photons, X rays. [energies at 0.5 KeV to 0.5 MeV]

There are low energy photons, radio [EM] waves. [Energies below 0.05 KeV]

Electrons are bound in their atomic orbits with energies from a few tens
of eVs to about 300 KeV. [My cross section tables were retired when I
was.][And these energy levels are different for all materials.]

Apply conservation of energy, under the poor assumption of a 100%
absorption factor [actually a probability distribution is more
appropriate] and it's easy to see that gamma rays surrender ONLY a
portion of their energy as they interact with electrons. Elementary
Example: 1 MeV photon interacts with and 100% ionizes a H atom releasing
2 electrons totaling about 40 KeV. So, the forward traveling photon's
energy changed from 1 MeV to 960 KeV. [i.e. a change in wavelength
occurred for the photon ... in radio terms it retuned to a lower
frequency band.] Conservation of energy applies to the photon.
Conservation of energy also applies to the ionized electron. When it
recombines with an ionized H atom the energy may be released in various
forms including a 40 KeV photon, or heat, or both.

Medium energy photons also conserve energy. The Absorptions cross
sections for most materials are several orders of magnitude larger than
for gamma rays but the principle is still the same. There is an
additional case where a photon may be absorbed but the energy is
insufficient to fully ionized the atom. In these cases there is a
transient increase to an excited state for the electrons and a
subsequent release of energy, mostly as heat but also a lower energy
photon [EM noise source] is created.

So, Cecil to answer your basic question: "The energy in the photon
changes [decreases] when there is an interaction with an electron."

Corollary: "The energy level in the electron changes [increases] when it
absorbs energy from a photon."

Deacon Dave

Cecil Moore wrote:
Just read one of his books. He says, "An electron emits or absorbs an
electron." "Absorbs" and "combines" sure seems like the same thing to me.

Note that he did not use the word combine. Do you think that was an
oversight on his part?

73, Jim AC6XG

On Thu, 20 Nov 2003 15:20:18 GMT, Dave Shrader
wrote:
Medium energy photons also conserve energy.

Hi Dave,

Such responses wholly lack the context of RF, where the energy of a
Photon is NOT imparted to ANY single atom as there is absolutely no
elemental atom that supports the necessary de Broglie wavelength for
this to occur. Nor does it occur for any molecule at the atomic
scale. It must then be a product of the super-Macro scale (of course,
at the wavelength involved) in the far far conduction bands.

One may argue the conservation of energy, but the energy of such
photons are incredibly weak, they in no sense qualify as "medium
energy" (qualifiers are not necessary).

73's
Richard Clark, KB7QHC

Jim Kelley wrote:
Onward through the fog.

I've added some information concerning this subject to my web page.
--
73, Cecil http://www.qsl.net/w5dxp



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I've added some information concerning this subject to my web page.
--
73, Cecil http://www.qsl.net/w5dxp

================================

What's new Cec?
It's fairly obvious stuff.
Of what use is it?
Can you describe it in numbers?

A coil change current distibution in an antenna but then again so does the
length of antenna that the coil replaces.
lumped inductance = lumped change in current. distributed inductance =
distributed change in current. Is there more to it than that?


"Cecil Moore" wrote in message
...
Jim Kelley wrote:
Onward through the fog.

I've added some information concerning this subject to my web page.
--
73, Cecil http://www.qsl.net/w5dxp



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Reg Edwards wrote:
It's fairly obvious stuff.

Not to everyone, Reg.
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Jimmy wrote:
lumped inductance = lumped change in current.

Actually, I think the assertion was that
lumped inductance = no change in current.
--
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Reg Edwards wrote:
What's new Cec?
It's fairly obvious stuff.

Reg, just for you, I have added some background information to that
web page. The underlying physics are obviously not so obvious to
the other side of the argument.
--
73, Cecil http://www.qsl.net/w5dxp/current.htm



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