Question about "Another look at reflections" article.
 

"K1TTT" wrote
...
On May 30, 5:05 pm, "Szczepan Bialek" wrote:

In schools are all theories.

the only purpose for disproved theories in schools is for historical
context. sometimes it is useful to show students what doesn't work so
they don't waste time repeating past mistakes. aether theories are
one that is taught and then demonstrated in class as incorrect. once
the student makes the measurements they get a better feeling why
aethers are bogus.

Up to now we have agreed that Maxwell/Lorents aether is bogus and that in
the space is the plasma (ions and electrons) and the dust. They rotate with
the Sun in the form of a whirl.

You are the first radioman who admit this.
S*

Question about "Another look at reflections" article.
 

On May 31, 12:34*pm, K1TTT wrote:
what does happen to that last photon in the infinite series of smaller
and smaller reflections between discontinuities??

It doesn't matter. At HF wavelengths, the last photon is not going to
have an appreciable effect ( but maybe at gamma-ray wavelengths). The
collapse of the probability function indicates that if one runs a
large number of trials, n% of those very last photons will be
reflected and (100-n%) of them will be absorbed. That probability
function has never been proven to be wrong.

I have no idea why an otherwise knowledgeable individual could be
argumentative concerning this subject.
--
73, Cecil, w5dxp.com

Question about "Another look at reflections" article.
 

On May 31, 12:58*pm, "Szczepan Bialek" wrote:
You are the first radioman who admit this.

What does the fact that particles with rest mass are spiraling away
from the sun have to do with photons that propagate in (almost) a
straight line away from the sun? The only effect that the sun has on
photons emitted by the sun is to slow them down.
--
73, Cecil, w5dxp.com

Question about "Another look at reflections" article.
 

On May 31, 5:58*pm, "Szczepan Bialek" wrote:
*"K1TTT" ...
On May 30, 5:05 pm, "Szczepan Bialek" wrote:



*In schools are all theories.
the only purpose for disproved theories in schools is for historical

context. *sometimes it is useful to show students what doesn't work so
they don't waste time repeating past mistakes. *aether theories are
one that is taught and then demonstrated in class as incorrect. *once
the student makes the measurements they get a better feeling why
aethers are bogus.

Up to now we have agreed that Maxwell/Lorents aether is bogus and that in
the space is the plasma (ions and electrons) and the dust. They rotate with
the Sun in the form of a whirl.

You are the first radioman who admit this.
S*

the solar wind is well know and easily studies with the satellite data
available today. but it is not an aether, the interplanetary plasma
does not propagate the light from the sun, it just gets in the way.

Question about "Another look at reflections" article.
 

On May 31, 6:05*pm, Cecil Moore wrote:
On May 31, 12:34*pm, K1TTT wrote:

what does happen to that last photon in the infinite series of smaller
and smaller reflections between discontinuities??

It doesn't matter. At HF wavelengths, the last photon is not going to
have an appreciable effect ( but maybe at gamma-ray wavelengths). The
collapse of the probability function indicates that if one runs a
large number of trials, n% of those very last photons will be
reflected and (100-n%) of them will be absorbed. That probability
function has never been proven to be wrong.

I have no idea why an otherwise knowledgeable individual could be
argumentative concerning this subject.
--
73, Cecil, w5dxp.com

i hate mechanical analogies, they lead to too many misconceptions.
the last photon question was meant to be rhetorical.

Question about "Another look at reflections" article.
 

On 25 mayo, 03:35, Richard Clark wrote:
On Mon, 24 May 2010 17:17:41 -0700 (PDT), lu6etj
wrote:

PSE, with the due respect and consideration toward you an the
distinguished colleagues and friends, Would you mind return to the
original question? (sorry if it is not this the most polite form to
ask it)

Hi Miguel,

I presume by "original question" you mean:

On Mon, 24 May 2010 13:06:19 -0700 (PDT), lu6etj wrote:

¿Absorb the reflected power or amortiguate the effects of variyng load
impedance?

The answer is YES.

Now, if you mean by absorb that all absorbtion results in heat, then
the answer is NO.

If you mean by absorb that all energy is combined in a load, then the
answer is YES.

The difference between YES and NO is the PHASE differences of the two
energies that are combined.

73's
Richard Clark, KB7QHC

Hi all

First the first: Sorry Szczepan, do not feel upset I wanted to say
"Me" don't go off topic with my own answers, not you.

Very interesting comments!
I think we must agree on meaning of words and basement concepts
employed to discuss these matters, otherwise we end up talking about
different "things" (Babel curse what confuses our tongues). If we do
not agree with meaning of, for example, "interaction", how can we
reach an agreement on much more complex things that depend on this
word/concept?
This is not a criticism to anyone, is only my point of view about what
I think is a partial source of apparently irreductible positions in
the group.
I believe Richard, K1TTT and me, have sinchronized minds about
"Interaction" word/concept meaning. I suppose must be a more deep
underlying assumptios that make it possible. However I believe I
understand the "idea" underlying Cecil concept about "interaction" and
I believe I can "see" his point, To me, Cecil's interaction concept it
is a very common idea, I do not think Cecil be a "hard to die" man :)
I think we have to do our more honest efforts to sinchronize ideas.
I do not want to go off topic, but let me bring a couple of thinkings
to the table (they are not mine).

* Students come into our classrooms with an established world-view,
formed by years of prior experience and learning.
* Even as it evolves, a student's world-view filters all experiences
and affects their interpretation of observations.
* Students are emotionally attached to their world-views and will not
give up their world-views easily.
* Challenging, revising, and restructuring one's world-view requires
much effort.
(from http://srri.umass.edu/topics/constructivism)

Why these above things would not happen to me?
......
Returning: Cecil, given A+B=C, do you you see C as result of an
interaction (or mutual action) among A and B, or C a simply result of
A added to B?
Another question to clarify my undestanding of your propositions: do
you see waves interacting themselves in a discontinuity, or you see
them interacting with the discontinuity? or both phenomena at the same
time?. Change discontinuity for load and please tell me.

Richard, you said: -"Because" leads to superstition-; you are pointing
to causal relations?
I believe was you who said dislike representations, if it is yes, were
you aiming to create mental images of physical phenomena?
Here we often use the word "methafor" in figured sense instead
"analogy", for example, "my car it is as strong as a locomotive" it is
a true methaphor. In metaphor, there are two levels or terms: the real
(my car) and evoked or imaginary (locomotive). Coulored water is it a
true methaphor or an analogy?

Simple analogies as useful things until one (or more) of they not
work... then, ciao analogy..!, not so bad :), however... notice!, our
ideas are not equal to the "out there" world. Concepts, models,
(mathematical models also, of course)... are not they our mind's
"analogies" "out there" sensorial/injstrumental perceived world?
These are not trivial epistemology issues, our "observer" leads
directly to the question about "Is the moon there when nobody
looks?" (N. D.Mermin). When we go out of our classic-simplistic-
realistic-traditional ham world... "we are in troubles, Houston",
slippery soil!, I make the sign of the cross! :)

K1TTT said: -Standing waves are a figment of your instrumentation-. My
dictionary translates "figment" as "product" or "chimera", please,
tell me what word should I use to correctly read the sentence?

73

Miguel Ghezzi - LU6ETJ

Question about "Another look at reflections" article.
 

"lu6etj" wrote
...

Here we often use the word "methafor" in figured sense instead
"analogy", for example, "my car it is as strong as a locomotive" it is
a true methaphor. In metaphor, there are two levels or terms: the real
(my car) and evoked or imaginary (locomotive). Coulored water is it a
true methaphor or an analogy?

Simple analogies as useful things until one (or more) of they not
work... then, ciao analogy..!,

Radio waves and sound are in full analogy.

K1TTT said: -Standing waves are a figment of your instrumentation-. My
dictionary translates "figment" as "product" or "chimera", please,
tell me what word should I use to correctly read the sentence?

Waves always travel (pressure or voltage pulses). If the wave interact with
the reflected one than the places where the pressures/voltages change are
standing.

The reflected wave cam be weaker if the mirror is partialy transparent (or
if an absorbtion take place).
S*

73

Miguel Ghezzi - LU6ETJ

Question about "Another look at reflections" article.
 

"K1TTT" wrote
...
On May 31, 5:58 pm, "Szczepan Bialek" wrote:


Up to now we have agreed that Maxwell/Lorents aether is bogus and that in
the space is the plasma (ions and electrons) and the dust. They rotate
with
the Sun in the form of a whirl.

You are the first radioman who admit this.

the solar wind is well know and easily studies with the satellite data
available today. but it is not an aether,

The aether in sense of medium for light propagation.

the interplanetary plasma
does not propagate the light from the sun,

In the interstellar medium (ISM) are ions, electrons and dust.
http://www.astronomynotes.com/ismnotes/s2.htm

The medium are the electrons ony. The rest are contaminations, like the fog
and dust in the air for the sound waves.
Do you see any sensible another solution?
S*

Question about "Another look at reflections" article.
 

On May 31, 4:36*pm, K1TTT wrote:
i hate mechanical analogies, they lead to too many misconceptions.
the last photon question was meant to be rhetorical.

I apologize for not recognizing that. Words don't always convey the
intended context.
--
73, Cecil, w5dxp.com

Question about "Another look at reflections" article.
 

On Jun 1, 5:52*am, lu6etj wrote:
On 25 mayo, 03:35, Richard Clark wrote:



On Mon, 24 May 2010 17:17:41 -0700 (PDT), lu6etj
wrote:

PSE, with the due respect and consideration toward you an the
distinguished colleagues and friends, Would you mind return to the
original question? (sorry if it is not this the most polite form to
ask it)

Hi Miguel,

I presume by "original question" you mean:

On Mon, 24 May 2010 13:06:19 -0700 (PDT), lu6etj wrote:

¿Absorb the reflected power or amortiguate the effects of variyng load
impedance?

The answer is YES.

Now, if you mean by absorb that all absorbtion results in heat, then
the answer is NO.

If you mean by absorb that all energy is combined in a load, then the
answer is YES.

The difference between YES and NO is the PHASE differences of the two
energies that are combined.

73's
Richard Clark, KB7QHC

Hi all

First the first: Sorry Szczepan, do not feel upset *I wanted to say
"Me" don't go off topic with my own answers, not you.

Very interesting comments!
I think we must agree on meaning of words and basement concepts
employed to discuss these matters, otherwise we end up talking about
different "things" (Babel curse what confuses our tongues). If we do
not agree with meaning of, for example, "interaction", how can we
reach an agreement on much more complex things that depend on this
word/concept?
This is not a criticism to anyone, is only my point of view about what
I think is a partial source of apparently irreductible positions in
the group.
I believe Richard, K1TTT and me, have sinchronized minds about
"Interaction" word/concept meaning. I suppose must be a more deep
underlying assumptios that make it possible. However I believe I
understand the "idea" underlying Cecil concept about "interaction" and
I believe I can "see" his point, To me, Cecil's interaction concept it
is a very common idea, I do not think Cecil be a "hard to die" man :)
I think we have to do our more honest efforts to sinchronize ideas.
I do not want to go off topic, but let me bring a couple of thinkings
to the table (they are not mine).

* Students come into our classrooms with an established world-view,
formed by years of prior experience and learning.
* Even as it evolves, a student's world-view filters all experiences
and affects their interpretation of observations.
* Students are emotionally attached to their world-views and will not
give up their world-views easily.
* Challenging, revising, and restructuring one's world-view requires
much effort.
(fromhttp://srri.umass.edu/topics/constructivism)

Why these above things would not happen to me?
.....
Returning: Cecil, given A+B=C, do you you see C as result of an
interaction (or mutual action) among A and B, or C a simply result of
A added to B?
Another question to clarify my undestanding of your propositions: do
you see waves interacting themselves in a discontinuity, or you see
them interacting with the discontinuity? or both phenomena at the same
time?. Change discontinuity for load and please tell me.

Richard, you said: -"Because" leads to superstition-; you are pointing
to causal relations?
I believe was you who said dislike representations, if it is yes, were
you aiming to create mental images of physical phenomena?
Here we often use the word "methafor" in figured sense instead
"analogy", for example, "my car it is as strong as a locomotive" it is
a true methaphor. In metaphor, there are two levels or terms: the real
(my car) and evoked or imaginary (locomotive). Coulored water is it a
true methaphor or an analogy?

Simple analogies as useful things until one (or more) of they not
work... then, ciao analogy..!, not so bad :), however... notice!, our
ideas are not equal to the "out there" world. Concepts, models,
(mathematical models also, of course)... are not they our mind's
"analogies" "out there" sensorial/injstrumental perceived world?
These are not trivial epistemology issues, our "observer" leads
directly to the question about "Is the moon there when nobody
looks?" (N. D.Mermin). When we go out of our classic-simplistic-
realistic-traditional ham world... "we are in troubles, Houston",
slippery soil!, I make the sign of the cross! :)

K1TTT said: -Standing waves are a figment of your instrumentation-. My
dictionary translates "figment" as "product" or "chimera", please,
tell me what word should I use to correctly read the sentence?

73

Miguel Ghezzi - LU6ETJ

in this context figment=product. often used to describe something
that is a result of an over active imagination. the only reason you
can see standing waves is because a measurement or observation makes
them look like they are 'standing' when it is really the interaction
of two or more regular traveling waves.

Question about "Another look at reflections" article.
 

On Jun 1, 8:42*am, "Szczepan Bialek" wrote:
*"K1TTT" ...
On May 31, 5:58 pm, "Szczepan Bialek" wrote:



Up to now we have agreed that Maxwell/Lorents aether is bogus and that in
the space is the plasma (ions and electrons) and the dust. They rotate
with
the Sun in the form of a whirl.

You are the first radioman who admit this.
the solar wind is well know and easily studies with the satellite data

available today. *but it is not an aether,

The aether in sense of medium for light propagation.

the interplanetary plasma

does not propagate the light from the sun,

In the interstellar medium (ISM) are ions, electrons and dust.http://www.astronomynotes.com/ismnotes/s2.htm

The medium are the electrons ony. The rest are contaminations, like the fog
and dust in the air for the sound waves.
Do you see any sensible another solution?
S*

the electrons can not do it either. there are not enough of them and
they move too slowly in the plasma to propagate light. these are
easily measured, you can watch the density directly using the ace
satellite web page and calculate the speed of longitudinal waves if
you want, but it won't be anywhere near light speed.

Question about "Another look at reflections" article.
 

On Jun 1, 12:52*am, lu6etj wrote:
Returning: Cecil, given A+B=C, do you you see C as result of an
interaction (or mutual action) among A and B, or C a simply result of
A added to B?

Of course, there are all types of "addition", e.g. arithmetic,
algebraic, voltage phasor, Poynting vector, scalar power,
superposition, merging, mixing, ... Which type of "addition" is
appropriate depends upon the nature of what is being added.

Simply put, if A + B = C creates an irreversible result, then A has
(obviously?) interacted with B. If the result is reversible, then A
has not interacted with B. For the great majority of cases,
superposition does not result in interaction. For the great majority
of cases, interference does not result in interaction. For some
(special?) cases, superposition (plus associated interference and wave
cancellation) results in interaction and the result is irreversible.
Non reflective glass is an example of wave interaction. The internal
reflection cancels the external reflection and the energy in those two
reflections changes directions. A Z0-match at an impedance
discontinuity in an RF transmission line is another example. Again, if
two waves are coherent, collimated, and traveling in the same
direction, those two waves will interact, i.e. their superposition
result is not reversible.
--
73, Cecil, w5dxp.com

Question about "Another look at reflections" article.
 

On Jun 1, 3:42*am, "Szczepan Bialek" wrote:
Do you see any sensible another solution?

What about the "quantum soup", particles winking in and out of
existence throughout space? How do you explain the Casimir effect? How
do you explain the flow of photons through evacuated space that is
devoid of electrons?
--
73, Cecil, w5dxp.com

Question about "Another look at reflections" article.
 

On Jun 1, 11:23*am, Cecil Moore wrote:
On Jun 1, 12:52*am, lu6etj wrote:

Returning: Cecil, given A+B=C, do you you see C as result of an
interaction (or mutual action) among A and B, or C a simply result of
A added to B?

Of course, there are all types of "addition", e.g. arithmetic,
algebraic, voltage phasor, Poynting vector, scalar power,
superposition, merging, mixing, ... Which type of "addition" is
appropriate depends upon the nature of what is being added.

Simply put, if A + B = C creates an irreversible result, then A has
(obviously?) interacted with B. If the result is reversible, then A
has not interacted with B. For the great majority of cases,
superposition does not result in interaction. For the great majority
of cases, interference does not result in interaction. For some
(special?) cases, superposition (plus associated interference and wave
cancellation) results in interaction and the result is irreversible.
Non reflective glass is an example of wave interaction. The internal
reflection cancels the external reflection and the energy in those two
reflections changes directions. A Z0-match at an impedance
discontinuity in an RF transmission line is another example. Again, if
two waves are coherent, collimated, and traveling in the same
direction, those two waves will interact, i.e. their superposition
result is not reversible.
--
73, Cecil, w5dxp.com

what exactly is the 'interaction' and why is it unique to that
special coherent, collimated, etc, case? if two waves can 'interact'
in that case there should be other evidence that they can interact in
other situations. just because a+b=c doesn't mean that a and b have
magically disappeared for some reason, in mathematics c=a+b is just as
valid and implies that c is made up of a and b... or using your water
analogy, adding one pint to another pint doesn't magically cause them
to 'interact' in some way, the original water is still there, perhaps
indistinguishable from each other, but still there. it may be
convenient to represent the result of summing an infinite series of
reflections as a single number, but it is not necessary.

Question about "Another look at reflections" article.
 

On Jun 1, 8:10*am, K1TTT wrote:
what exactly is the 'interaction' *and why is it unique to that
special coherent, collimated, etc, case?

Is it not obvious that when two reflected waves cancel in one
direction, as at the surface of a 1/4WL thin-film coating on glass,
and their combined EM energy is redistributed in the opposite
direction, that those two waves have interacted, i.e. have suffered a
permanent change and have lost their original identities?

Is it not obvious that when two waves are traveling two different
paths where the incident angle is, e.g. two degrees, that those two
waves will superpose and interfere throughout a certain space after
which they emerge intact, unaffected, and have obviously not
interacted, i.e. they suffered no permanent change and have maintained
their original identities?

Did superposition occur in both cases? Yes. Did interference occur in
both cases? Yes. Did wave cancellation occur in both cases? No, just
in the non-reflective glass case.

Consider a transmission line with an SWR of 5.83:1. The sourced power
is 100 watts. The forward power is 200 watts. The reflected power is
100 watts. All of the reflected power is redistributed back toward the
load at a Z0-match through reflection and wave cancellation. Zero
reflected power is incident upon the source.

Does the 100w source wave lose its identity when it merges with the
100w of redistributed reflected wave to become the 200w forward wave?
Is the steady-state load energy coming from the source wave or the
redistributed reflected wave or both? Seems to me, it is obvious that
the two original component waves have interacted and lost their
original identities for good if they are pure coherent sine waves
traveling in the same direction confined to a coaxial transmission
line.
--
73, Cecil, w5dxp.com

Question about "Another look at reflections" article.
 

On Mon, 31 May 2010 22:52:35 -0700 (PDT), lu6etj
wrote:

Richard, you said: -"Because" leads to superstition-; you are pointing
to causal relations?

Hi Miguel,

Language would have us believe that causal relations "be" from
"cause." Common language sometimes uses "because" as a bridge between
phrases without necessarily implying causality - in other words, the
word "because" is verbal noise when that happens. "Because" is often
a one word reply to a child's question "Why?"

I believe was you who said dislike representations, if it is yes, were
you aiming to create mental images of physical phenomena?

Oh, I do that all the time. However, I do not mistake representations
as being the real thing. In other contexts, they are more real than
the real thing. So, if there is dislike, it comes from seeing
inferior representation when better work requires so little more
effort.

Here we often use the word "methafor"
now there's a curious and suggestive spelling
in figured sense instead
"analogy", for example, "my car it is as strong as a locomotive" it is
a true methaphor. In metaphor, there are two levels or terms: the real
(my car) and evoked or imaginary (locomotive). Coulored water is it a
true methaphor or an analogy?

Both. But context should resolve that, or it could still be both.
That is why metaphor and analogy are so dangerous. That is also why
it is so useful in fiction. We get enough fiction here.

Simple analogies as useful things until one (or more) of they not
work... then, ciao analogy..!, not so bad :), however... notice!, our
ideas are not equal to the "out there" world. Concepts, models,
(mathematical models also, of course)... are not they our mind's
"analogies" "out there" sensorial/injstrumental perceived world?
These are not trivial epistemology issues, our "observer" leads
directly to the question about "Is the moon there when nobody
looks?" (N. D.Mermin). When we go out of our classic-simplistic-
realistic-traditional ham world... "we are in troubles, Houston",
slippery soil!, I make the sign of the cross! :)

Namaste.

73's
Richard Clark, KB7QHC

Question about "Another look at reflections" article.
 

3. character assignation.

Hi Miguel,

How well does this translate to Spanish?

As I've offered, native speakers are often the poorest communicators
in their own language. Imagine trying to visualize an optical
metaphor when this comet hits the windshield.

73's
Richard Clark, KB7QHC

Question about "Another look at reflections" article.
 

On Jun 1, 1:52*pm, Cecil Moore wrote:
On Jun 1, 8:10*am, K1TTT wrote:

what exactly is the 'interaction' *and why is it unique to that
special coherent, collimated, etc, case?

Is it not obvious that when two reflected waves cancel in one
direction, as at the surface of a 1/4WL thin-film coating on glass,
and their combined EM energy is redistributed in the opposite
direction, that those two waves have interacted, i.e. have suffered a
permanent change and have lost their original identities?

Is it not obvious that when two waves are traveling two different
paths where the incident angle is, e.g. two degrees, that those two
waves will superpose and interfere throughout a certain space after
which they emerge intact, unaffected, and have obviously not
interacted, i.e. they suffered no permanent change and have maintained
their original identities?

no, it is not obvious. where do you draw the line... 1 degree, .1
degree, .001 degree? at what point is the angle small enough to say
that they have 'interacted' and the energy is redistributed?


Did superposition occur in both cases? Yes. Did interference occur in
both cases? Yes. Did wave cancellation occur in both cases? No, just
in the non-reflective glass case.


i propose that 'cancellation' is just a special case of interference
where the waves are 'close enough' to collinear that you never see the
interference pattern. this would of course always apply in a
transmission line because they are confined.

closely analyze the transient response of your non-reflective glass in
the case where the wave is not incident perpendicular to the glass.
do each reflection from each interface separately as the wave travels
in the coating at an angle. then reduce the angle to very near
perpendicular and you should see that there are indeed reflections
that should very nearly 'cancel' each other out as the number of
reflections gets bigger.

Question about "Another look at reflections" article.
 

On Tue, 1 Jun 2010 13:41:03 -0700 (PDT), lu6etj
wrote:

Richard: OK to "representations". Do you think there are things
accesible (or "visible") to our intelect (objectivism) or all we have
are models of external world built by our minds with the help and
mediation of our biological and technical sensing/measurements
apparatus?

Hi Miguel,

This has gotten pretty metaphysical.

I insist in that because I think that two or more models could be
capable to "explain" observed phenomena and we could partially agree
on they instead dispute hard about the "right one" :)

Data explains - the rest is the ego of pride of authorship.

Metaphysical discussions about the "thing itself" was very strong and
sterile before Newton stop caring about the "whys" and decided dealts
with "howtos" :). I think you were thinking about this when you
equalize "because"="superstition". Am I right?

I don't know.

However, though causality is an epistemological issue, usually in the
macroscopic phenomena we accept "If A, then B" as a causal relation,
example: If a rock hits my feet I absolutelly think the pain is due to
the stone, not a simple acausal correlation. In this sense is it valid
to use "because the stone")? Thanks to all for your comments

Don't you think your pain is due to nerve sensations? Rocks hit rocks
all the time and there is no pain due to anything any where.

Or maybe there is an optical proof that refutes this.

73's
Richard Clark, KB7QHC

Question about "Another look at reflections" article.
 

On Jun 1, 3:41*pm, lu6etj wrote:
the only reason you can see standing waves is because a measurement or observation makes them look like they are 'standing' when it is really the interaction of two or more regular traveling waves.

If I do not bad understand your response to Cecil, In your sentence I
would change "interaction" by "manifestation". What do you think?

I don't remember writing that, Miguel. If I did write that, then I
misused the word "interaction". I wrote elsewhere that forward and
reflected waves don't interact as long as Z0 stays constant.
--
73, Cecil, w5dxp.com

Question about "Another look at reflections" article.
 

On Jun 1, 5:44*pm, K1TTT wrote:
no, it is not obvious. *where do you draw the line... 1 degree, .1
degree, .001 degree? *at what point is the angle small enough to say
that they have 'interacted' and the energy is redistributed?

I don't know the answer but zero degrees (perfect collimation) will
result in interaction.

i propose that 'cancellation' is just a special case of interference
where the waves are 'close enough' to collinear that you never see the
interference pattern.

When b1 = s11*a1 + s12*a2 = 0 at an impedance discontinuity, wave
cancellation has taken place. s11*a1 and s12*a2 are coherent sine
waves, equal in magnitude, and 180 degrees out of phase. Have you read
the FSU web page where they describe wave cancellation? All these
concepts are old hat to optical physicists.

micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/
waveinteractions/index.html

"... when two waves of equal amplitude and wavelength that are 180-
degrees ... out of phase with each other meet, they are not actually
annihilated, ... All of the photon energy present in these waves must
somehow be recovered or redistributed in a new direction, according to
the law of energy conservation ... Instead, upon meeting, the photons
are redistributed to regions that permit constructive interference, so
the effect should be considered as a redistribution of light waves and
photon energy rather than the spontaneous construction or destruction
of light."
--
73, Cecil, w5dxp.com

Question about "Another look at reflections" article.
 

On Jun 1, 11:31*pm, Cecil Moore wrote:
"... All of the photon energy present in these waves must

I suggest that you immediately dump any reference that includes
a phrase like "photon energy present in a wave".

There is a wave theory of light, and there is a particle theory
of light, and these two theories do not play well together.

While in many situations they will yield the same answers, it
is not permissible to mix the concepts from each. Distrust
the conclusions of any exposition which does so.

....Keith

Question about "Another look at reflections" article.
 

On Jun 2, 3:31*am, Cecil Moore wrote:
On Jun 1, 5:44*pm, K1TTT wrote:

no, it is not obvious. *where do you draw the line... 1 degree, .1
degree, .001 degree? *at what point is the angle small enough to say
that they have 'interacted' and the energy is redistributed?

I don't know the answer but zero degrees (perfect collimation) will
result in interaction.

what is the physical mechanism for 'interaction' that requires perfect
collimation?



i propose that 'cancellation' is just a special case of interference
where the waves are 'close enough' to collinear that you never see the
interference pattern.

When b1 = s11*a1 + s12*a2 = 0 at an impedance discontinuity, wave
cancellation has taken place. s11*a1 and s12*a2 are coherent sine
waves, equal in magnitude, and 180 degrees out of phase. Have you read
the FSU web page where they describe wave cancellation? All these
concepts are old hat to optical physicists.

micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/
waveinteractions/index.html

"... when two waves of equal amplitude and wavelength that are 180-
degrees ... out of phase with each other meet, they are not actually
annihilated, ... All of the photon energy present in these waves must
somehow be recovered or redistributed in a new direction, according to
the law of energy conservation ... Instead, upon meeting, the photons
are redistributed to regions that permit constructive interference, so
the effect should be considered as a redistribution of light waves and
photon energy rather than the spontaneous construction or destruction
of light."
--
73, Cecil, w5dxp.com

coming from a 'primer' and containing words like 'somehow' doesn't
raise my confidence in the explanation.

Question about "Another look at reflections" article.
 

On Jun 2, 10:33*am, Keith Dysart wrote:
On Jun 1, 11:31*pm, Cecil Moore wrote:

"... All of the photon energy present in these waves must

I suggest that you immediately dump any reference that includes
a phrase like "photon energy present in a wave".

There is a wave theory of light, and there is a particle theory
of light, and these two theories do not play well together.

While in many situations they will yield the same answers, it
is not permissible to mix the concepts from each. Distrust
the conclusions of any exposition which does so.

...Keith

agreed. photons are good when working with other elementary particle
interactions to represent the em energy lost or transferred during
particle interactions. they are not that useful when studying wave
propagation or interaction with macroscopic object... including 1/4
wave coatings.

Question about "Another look at reflections" article.
 

On Jun 2, 6:39*am, K1TTT wrote:
On Jun 2, 10:33*am, Keith Dysart wrote:

On Jun 1, 11:31*pm, Cecil Moore wrote:

"... All of the photon energy present in these waves must

I suggest that you immediately dump any reference that includes
a phrase like "photon energy present in a wave".

There is a wave theory of light, and there is a particle theory
of light, and these two theories do not play well together.

While in many situations they will yield the same answers, it
is not permissible to mix the concepts from each. Distrust
the conclusions of any exposition which does so.

...Keith

agreed. *photons are good when working with other elementary particle
interactions to represent the em energy lost or transferred during
particle interactions. *they are not that useful when studying wave
propagation or interaction with macroscopic object... including 1/4
wave coatings.

For those who may be interested, Richard Feynman offers an
introductory
lecture on photons he http://vega.org.uk/video/subseries/8

It illustrates that attempting to compute 1/4 wave coating behaviour
with photons would be extremely tedious, though possible.

On the other hand, at low light levels, where individual photons
become
discrete events, the wave theory becomes completely inadequate.
Fortunately, for practical applications, power levels are much higher
than this and the wave aproach is quite useful.

....Keith

Question about "Another look at reflections" article.
 

On Jun 2, 5:33*am, Keith Dysart wrote:
I suggest that you immediately dump any reference that includes
a phrase like "photon energy present in a wave".

If you (and others) will give up on the ridiculous concept of EM wave
energy standing still in standing waves, I will not have to refer to
photons again. Honor the technical fact that EM forward waves (with an
associated ExH energy) and EM reflected waves (with an associated ExH
energy) are always present when standing waves are present and that
those underlying waves (that cannot exist without energy) are moving
at the speed of light in the medium back and forth between impedance
discontinuities. Standing waves are somewhat of an illusion and
according to two of my reference books, do not deserve to be called
waves at all because standing waves do not transfer net energy as
required by the definition of "wave". In short, it is impossible for
EM waves to stand still.

Quoting one of my college textbooks, "Electrical Communication", by
Albert:

"Such a plot of voltage is usually referred to as a *voltage standing
wave* or as a *stationary wave*. Neither of these terms is
particularly descriptive of the phenomenon. A plot of effective values
of voltage, appearing as in Fig. 6(e), *is not a wave* in the usual
sense. However, the term "standing wave" is in widespread use."

From "College Physics", by Bueche and Hecht:

"These ... patterns are called *standing waves*, as compared to the
propagating waves considered above. They might better not be called
waves at all, since they do not transport energy and momentum."

Technically, RF waves *are* light waves, just not *visible* light
waves. All the laws of physics that govern EM waves of light also
apply to RF waves. That you find it inconvenient for your "mashed-
potatoes" theory of energy arguments is not a good reason to abandon
the photonic nature of EM waves. It is actually a good reason to keep
it in mind and abandon the mashed-potatoes energy arguments as human
conceptual constructs that cannot exist in reality. Most of the energy
in an EM wave is kinetic energy. Therefore, it cannot stand still.

There is a wave theory of light, and there is a particle theory
of light, and these two theories do not play well together.

If they are both correct, they should play well together. If there is
any conflict, quantum electrodynamics wins the argument every time.

While in many situations they will yield the same answers, it
is not permissible to mix the concepts from each. Distrust
the conclusions of any exposition which does so.

Actually, distrust the wave theory if it disagrees with QED. Quantum
ElectroDynamics has never been proven wrong.

So feel free to prove that standing waves can exist without the
underlying component traveling waves traveling at the speed of light
in the medium. Feel free to prove that EM wave cancellation does not
"redistribute energy to areas that permit constructive interference"
as the FSU web page explains. Feel free to prove the Melles-Groit web
page wrong when they say such has been proven experimentally. In fact,
the interferometer experiment described here proves that reflected EM
waves, traveling at the speed of light, exist along with the necessary
energy. Take a look at the "non-standard output to screen".

http://www.teachspin.com/instruments...eriments.shtml

I, personally, am not interested in getting the right answer using the
wrong concepts. And I am absolutely sure that your math models do not
dictate reality. It is supposed to be the exact opposite.
--
73, Cecil, w5dxp.com

Question about "Another look at reflections" article.
 

On Jun 2, 5:37*am, K1TTT wrote:
what is the physical mechanism for 'interaction' that requires perfect
collimation?

The impossibility of divergence. That collapsed probability function
can occur in an RF transmission line and in an interferometer.
Secondary effects, e.g. noise, are obviously ignored.

coming from a 'primer' and containing words like 'somehow' doesn't
raise my confidence in the explanation.

The "somehow" means that even if someone doesn't comprehend enough to
have "confidence in the explanation", that lack of confidence will
have zero effect on the real-world outcome - the energy involved in
destructive interference wave cancellation will be preserved through
constructive interference based on the conservation of energy
principle. Please feel free to prove them wrong.
--
73, Cecil, w5dxp.com

Question about "Another look at reflections" article.
 

On Jun 2, 5:39*am, K1TTT wrote:
photons are good when working with other elementary particle
interactions to represent the em energy lost or transferred during
particle interactions. *they are not that useful when studying wave
propagation or interaction with macroscopic object... including 1/4
wave coatings.

Photons are useful for proving that EM waves must necessarily travel
at the speed of light in the medium. Photons cannot stand still -
therefore, EM waves, known to consist of photons, cannot stand still -
therefore any overly-simplified mashed-potatoes version of energy
stored in an RF transmission line violates the laws of physics.
--
73, Cecil, w5dxp.com

Question about "Another look at reflections" article.
 

On Jun 2, 1:39*pm, Cecil Moore wrote:
On Jun 2, 5:39*am, K1TTT wrote:

photons are good when working with other elementary particle
interactions to represent the em energy lost or transferred during
particle interactions. *they are not that useful when studying wave
propagation or interaction with macroscopic object... including 1/4
wave coatings.

Photons are useful for proving that EM waves must necessarily travel
at the speed of light in the medium. Photons cannot stand still -
therefore, EM waves, known to consist of photons, cannot stand still -
therefore any overly-simplified mashed-potatoes version of energy
stored in an RF transmission line violates the laws of physics.
--
73, Cecil, w5dxp.com

wave function solutions to maxwell's equations are enough to prove
that for me.

Question about "Another look at reflections" article.
 

On Jun 2, 8:57*am, K1TTT wrote:
wave function solutions to maxwell's equations are enough to prove
that for me.

Not a loaded question: How do Maxwell's equations applied to a
standing wave prove that the component forward and reflected waves are
moving at the speed of light in the medium? If it can and if I can
understand it, I wouldn't need to use the photon argument.
--
73, Cecil, w5dxp.com

Question about "Another look at reflections" article.
 

On Jun 2, 2:12*pm, Cecil Moore wrote:
On Jun 2, 8:57*am, K1TTT wrote:

wave function solutions to maxwell's equations are enough to prove
that for me.

Not a loaded question: How do Maxwell's equations applied to a
standing wave prove that the component forward and reflected waves are
moving at the speed of light in the medium? If it can and if I can
understand it, I wouldn't need to use the photon argument.
--
73, Cecil, w5dxp.com

easy, maxwell's equations don't predict standing waves! they are a
product of superposition and the simplest instrumentation used since
they were first discovered.

Question about "Another look at reflections" article.
 

On Jun 2, 9:31*am, K1TTT wrote:
easy, maxwell's equations don't predict standing waves! *they are a
product of *superposition and the simplest instrumentation used since
they were first discovered.

Please correct me if I am wrong: If one starts with the (superposed)
standing wave equation, V(x,t) = A*sin(kx)*sin(wt), Maxwell's
equations seem to provide a perfectly valid result (not that I can
recognize perfection). Therefore, how do Maxwell's equations prove
that the component traveling waves necessarily possess a separate
existence? Again, a serious question from an engineer who considers
anything except a logical '0' or logical '1' to be broken. :-)
--
73, Cecil, w5dxp.com

Question about "Another look at reflections" article.
 

On Jun 2, 3:13*pm, Cecil Moore wrote:
On Jun 2, 9:31*am, K1TTT wrote:

easy, maxwell's equations don't predict standing waves! *they are a
product of *superposition and the simplest instrumentation used since
they were first discovered.

Please correct me if I am wrong: If one starts with the (superposed)
standing wave equation, V(x,t) = A*sin(kx)*sin(wt), Maxwell's
equations seem to provide a perfectly valid result (not that I can
recognize perfection). Therefore, how do Maxwell's equations prove
that the component traveling waves necessarily possess a separate
existence? Again, a serious question from an engineer who considers
anything except a logical '0' or logical '1' to be broken. :-)
--
73, Cecil, w5dxp.com

my differential calculus is a bit rusty, but i don't think that
equation satisfies the basic wave equation. from Fields and Waves in
Communication Electronics, section 1.14. a solution for the wave
equation in a transmission line (or other 1d form) must satisfy the
condition that the 2nd derivative wrt x and wrt t equal v-squared...
too hard to write in text. but basically take the above and
differentiate twice wrt time and twice wrt distance and i don't think
you'll get anything that comes out to the velocity squared. the
problem is that t and x must be part of the same sinusoid in the F(t-x/
v) so that the wave is truly traveling instead of stationary like your
equation provides.

in the basic maxwell equations in differential form you also run into
the same problem where the curl of the one field is proportional to
the time derivative of the other i think you'll end up with a
contradiction like k=w if you eliminate the dimensionless terms... i
may be wrong on that because its been so long since i've had to deal
with them on that level.

Question about "Another look at reflections" article.
 

On 2 jun, 13:48, K1TTT wrote:
On Jun 2, 3:13*pm, Cecil Moore wrote:

On Jun 2, 9:31*am, K1TTT wrote:

easy, maxwell's equations don't predict standing waves! *they are a
product of *superposition and the simplest instrumentation used since
they were first discovered.

Please correct me if I am wrong: If one starts with the (superposed)
standing wave equation, V(x,t) = A*sin(kx)*sin(wt), Maxwell's
equations seem to provide a perfectly valid result (not that I can
recognize perfection). Therefore, how do Maxwell's equations prove
that the component traveling waves necessarily possess a separate
existence? Again, a serious question from an engineer who considers
anything except a logical '0' or logical '1' to be broken. :-)
--
73, Cecil, w5dxp.com

my differential calculus is a bit rusty, but i don't think that
equation satisfies the basic wave equation. *from Fields and Waves in
Communication Electronics, section 1.14. *a solution for the wave
equation in a transmission line (or other 1d form) must satisfy the
condition that the 2nd derivative wrt x and wrt t equal v-squared...
too hard to write in text. *but basically take the above and
differentiate twice wrt time and twice wrt distance and i don't think
you'll get anything that comes out to the velocity squared. *the
problem is that t and x must be part of the same sinusoid in the F(t-x/
v) so that the wave is truly traveling instead of stationary like your
equation provides.

in the basic maxwell equations in differential form you also run into
the same problem where the curl of the one field is proportional to
the time derivative of the other i think you'll end up with a
contradiction like k=w if you eliminate the dimensionless terms... i
may be wrong on that because its been so long since i've had to deal
with them on that level.

Oh!, I surrender :) The thread do not converge to a solution, it
seems run out toward a naked singularity! I stay tuned until this
good brainstorm calms down ..

Miguel

Question about "Another look at reflections" article.
 

On Jun 2, 5:56*pm, lu6etj wrote:
On 2 jun, 13:48, K1TTT wrote:



On Jun 2, 3:13*pm, Cecil Moore wrote:

On Jun 2, 9:31*am, K1TTT wrote:

easy, maxwell's equations don't predict standing waves! *they are a
product of *superposition and the simplest instrumentation used since
they were first discovered.

Please correct me if I am wrong: If one starts with the (superposed)
standing wave equation, V(x,t) = A*sin(kx)*sin(wt), Maxwell's
equations seem to provide a perfectly valid result (not that I can
recognize perfection). Therefore, how do Maxwell's equations prove
that the component traveling waves necessarily possess a separate
existence? Again, a serious question from an engineer who considers
anything except a logical '0' or logical '1' to be broken. :-)
--
73, Cecil, w5dxp.com

my differential calculus is a bit rusty, but i don't think that
equation satisfies the basic wave equation. *from Fields and Waves in
Communication Electronics, section 1.14. *a solution for the wave
equation in a transmission line (or other 1d form) must satisfy the
condition that the 2nd derivative wrt x and wrt t equal v-squared...
too hard to write in text. *but basically take the above and
differentiate twice wrt time and twice wrt distance and i don't think
you'll get anything that comes out to the velocity squared. *the
problem is that t and x must be part of the same sinusoid in the F(t-x/
v) so that the wave is truly traveling instead of stationary like your
equation provides.

in the basic maxwell equations in differential form you also run into
the same problem where the curl of the one field is proportional to
the time derivative of the other i think you'll end up with a
contradiction like k=w if you eliminate the dimensionless terms... i
may be wrong on that because its been so long since i've had to deal
with them on that level.

Oh!, I surrender *:) The thread do not converge to a solution, it
seems run out toward a naked singularity! *I stay tuned until this
good brainstorm calms down ..

Miguel

it will never converge, these discussions always go on until the
namecalling starts, then die an ugly death.

Question about "Another look at reflections" article.
 

On Jun 2, 8:00*am, Cecil Moore wrote:
On Jun 2, 5:33*am, Keith Dysart wrote:

I suggest that you immediately dump any reference that includes
a phrase like "photon energy present in a wave".

If you (and others) will give up on the ridiculous concept of EM wave
energy standing still in standing waves, I will not have to refer to
photons again.

Refering to photons is just fine. Just do not mix them in with wave
theory.

Honor the technical fact that EM forward waves (with an
associated ExH energy) and EM reflected waves (with an associated ExH
energy) are always present when standing waves are present and that
those underlying waves (that cannot exist without energy) are moving
at the speed of light in the medium back and forth between impedance
discontinuities. Standing waves are somewhat of an illusion and
according to two of my reference books, do not deserve to be called
waves at all because standing waves do not transfer net energy as
required by the definition of "wave". In short, it is impossible for
EM waves to stand still.

Quoting one of my college textbooks, "Electrical Communication", by
Albert:

"Such a plot of voltage is usually referred to as a *voltage standing
wave* or as a *stationary wave*. Neither of these terms is
particularly descriptive of the phenomenon. A plot of effective values
of voltage, appearing as in Fig. 6(e), *is not a wave* in the usual
sense. However, the term "standing wave" is in widespread use."

From "College Physics", by Bueche and Hecht:

"These ... patterns are called *standing waves*, as compared to the
propagating waves considered above. They might better not be called
waves at all, since they do not transport energy and momentum."

All quite orthogonal to the original point, but your point about
standing
waves is quite correct, they are not really waves at all. But your
need
for the reality of underlying waves is quite excessive. The voltage
and
current distribution on a transmission line can be solved with a set
of differential equations which satisfy some boundary conditions.
There
is no mention of forward and reverse waves in this solution. Turns out
though, that the solution can also be factored in to a forward and
reflected wave and this technique will provide the same answer. It
does
not make these waves any more real.

I bring you back to a previous question which you have never
answered...
On an ideal line with 100% reflection, there are points where the
current
and voltage is always 0. Knowing that if either current or voltage is
0,
power is also 0, how does energy cross these point?

And if I cut the line at all the places where the current is zero, it
does not alter the energy distribution on the line one iota. How can
this be if energy is travelling from end to end on the line?

Technically, RF waves *are* light waves, just not *visible* light
waves. All the laws of physics that govern EM waves of light also
apply to RF waves.

PHYSICS has long given up on the idea of waves being an explanation
for light. The wave theory fails miserably when illumination levels
drop to the level that individual photons are being detected.

Though of course the earlier approximate models (waves) are still
useful
when intensity is high enough, just as we still use Newtonian
mechanics
to solve many every-day problems.

You might like to try http://vega.org.uk/video/subseries/8 for an
exposition on the strangeness of photon.

That you find it inconvenient for your "mashed-
potatoes" theory of energy arguments is not a good reason to abandon
the photonic nature of EM waves.

There you go again... mixing up you models. EM waves are analog and in
no way encompass the quantum nature of photons.

It is actually a good reason to keep
it in mind and abandon the mashed-potatoes energy arguments as human
conceptual constructs that cannot exist in reality. Most of the energy
in an EM wave is kinetic energy. Therefore, it cannot stand still.

There seems to be some misapprehension here. No one has claimed that
EM
waves stand still, though you may have been confused by the word
'standing' in 'standing waves'. But then earlier in your post you
quote
'College Physics' about 'standing waves', so it is not clear where
your confusion originates.

There is a wave theory of light, and there is a particle theory
of light, and these two theories do not play well together.

If they are both correct, they should play well together. If there is
any conflict, quantum electrodynamics wins the argument every time.

They are not both correct. QED aligns with many more observations than
does the wave theory. Another reason not to mix them.

While in many situations they will yield the same answers, it
is not permissible to mix the concepts from each. Distrust
the conclusions of any exposition which does so.

Actually, distrust the wave theory if it disagrees with QED. Quantum
ElectroDynamics has never been proven wrong.

And the wave theory does disagree with QED at low levels, while at
higher illumination levels QED agrees with the wave theory.

So feel free to prove that standing waves can exist without the
underlying component traveling waves traveling at the speed of light
in the medium. Feel free to prove that EM wave cancellation does not
"redistribute energy to areas that permit constructive interference"
as the FSU web page explains. Feel free to prove the Melles-Groit web
page wrong when they say such has been proven experimentally. In fact,
the interferometer experiment described here proves that reflected EM
waves, traveling at the speed of light, exist along with the necessary
energy. Take a look at the "non-standard output to screen".

And yet known of this aligns with the photons being part of the wave
theory of light. They two theories remain distinct.

http://www.teachspin.com/instruments...eriments.shtml

I, personally, am not interested in getting the right answer using the
wrong concepts.

Well, (and I am sorry, I can not resist), there is some evidence to
the
contrary. See:
http://www.w5dxp.com/energy.htm
http://www.w5dxp.com/nointfr.htm

....Keith

Question about "Another look at reflections" article.
 

On Jun 2, 6:33*pm, Keith Dysart wrote:
The voltage and
current distribution on a transmission line can be solved with a set
of differential equations which satisfy some boundary conditions.

I am not interested in voltage and current. Like optical physicists, I
am only interested in tracking the RF energy flow.

It does not make these waves any more real.

Agreed, but the point is that it does not make the traveling waves any
LESS real!

I bring you back to a previous question which you have never
answered...
On an ideal line with 100% reflection, there are points where the
current
and voltage is always 0. Knowing that if either current or voltage is
0,
power is also 0, how does energy cross these point?

Good grief, I have answered that question at least a dozen times. The
net current being zero is just an illusion caused by superposition of
two magnetic fields propagating in opposite directions that are equal
in magnitude and opposite in phase. The two traveling waves keep on
trucking at their relatively constant ExH energy levels in opposite
directions. If you want to account for the energy, when the current is
zero, all of the energy existing at that point is in the electric
field and, sure enough, the voltage is at a maximum at that point-
duuuhhhh again. There is absolutely no point on an active transmission
line where the net energy level is zero. Again, you guys are never
going to convince anyone that the Golden Gate Bridge doesn't need
maintenance because the net traffic on the bridge is zero. How the
heck can thousands of vehicles traveling one direction while the same
number of vehicles are traveling in the opposite direction add up to
zero effect in reality? Please get real.

And if I cut the line at all the places where the current is zero, it
does not alter the energy distribution on the line one iota. How can
this be if energy is travelling from end to end on the line?

As I told you years ago, when you cut the line, you radically change
the impedance and create a reflection that didn't exist before you cut
the line. Hint: There are no reflections at a point of constant Z0.
There are only reflections at an impedance discontinuity. The
alteration of the energy distribution (to which you seem to be blind)
can easily be observed in a TV system.

PHYSICS has long given up on the idea of waves being an explanation
for light.

And RF waves are technically light, just not visible light. When I was
rummaging through the Texas A&M library, I found a book entitled,
"Light". It covered the subject of RF waves.

EM waves are analog and in
no way encompass the quantum nature of photons.

You could become famous if you can prove that EM waves are not
quantized. Good luck on that one.

No one has claimed that EM waves stand still, ...

Please prove your ridiculous assertion that no one has ever claimed
that EM waves can stand still.
--
73, Cecil, w5dxp.com

Question about "Another look at reflections" article.
 

On Jun 2, 11:48*am, K1TTT wrote:
my differential calculus is a bit rusty, but i don't think that
equation satisfies the basic wave equation.

My calculus is probably a lot rustier than yours but it would be very
important for this discussion if Maxwell's equations do not work for
the standing wave equation. That would essentially prove that the
mashed-potatoes theory of transmission line energy is bogus.
--
73, Cecil, w5dxp.com

Question about "Another look at reflections" article.
 

lu6etj wrote:

Oh!, I surrender :) The thread do not converge to a solution, it
seems run out toward a naked singularity! I stay tuned until this
good brainstorm calms down ..

Good day Miguel, you must be new here! I don't think any threads
converge to a solution. Mostly they do that singularity thing after
about the third message.

But hang in there, often much entertainment is to be had!

Question about "Another look at reflections" article.
 

On Jun 2, 12:56*pm, lu6etj wrote:
Oh!, I surrender *:) The thread do not converge to a solution, ...

Miguel, do you fully realize the importance of Maxwell's equations
being nonfunctional with the equation for standing waves? It means
that any analysis based entirely on standing waves is probably invalid
in some area of the analysis - probably including Keith Dysart's
latest postings about the standing wave's zero current being
meaningful in some way.
--
73, Cecil, w5dxp.com