On May 31, 6:16*am, K1TTT wrote:
conclusion: Standing waves are a figment of your instrumentation, ...

I would say: Standing waves are a virtual image caused by the two
traveling waves (forward and reverse). I wonder if we could see a
visible light standing wave in a cloud chamber if we used a
microscope?
--
73, Cecil, w5dxp.com

On May 31, 12:42*pm, Cecil Moore wrote:

If you add a pint of water to a pint of water, the two pints of
identical molecules interact, analogous to one joule of coherent/
collimated photons added to another joule of the *identical* photons.
The results is two joules of identical photons, indistinguishable from
each other. If the superposition process is reversible, interaction
has not taken place. If the superposition process is irreversible,
interaction has taken place. Both outcomes are possible depending upon
the initial conditions.

ah, but in this case consider that you have a pint of blue water
moving to the right and a pint of red water moving to the left to be a
better analogy to currents of the waves moving forward and backward in
the coax. there are then 3 possibilities:

1. the two mix and cancel giving you 2 pints of purple water not going
anywhere.
2. the two bounce off of each other now giving you red water moving
right and blue water moving left.
3. the two pass by each other not mixing at all and continue on their
way.

if 1 happened you would indeed cancel the waves and end up with a
spare pile of electrons not going anywhere. while this may be
adequate for a mechanical analogy it doesn't say where the energy of
those two moving pints went so is obviously wrong.

number 2 conserves energy at least.

number 3 also conserves energy and gives the same energy distribution
as 2.

since electrons aren't colored it makes number 2 and 3
indistinguishable, so clearly the result is that both waves continue
on their way undisturbed by the other... which is what is observed in
all cases of em wave interaction in linear media.

On May 31, 1:04*pm, Cecil Moore wrote:
On May 31, 6:16*am, K1TTT wrote:

conclusion: Standing waves are a figment of your instrumentation, ...

I would say: Standing waves are a virtual image caused by the two
traveling waves (forward and reverse). I wonder if we could see a
visible light standing wave in a cloud chamber if we used a
microscope?
--
73, Cecil, w5dxp.com

no need, just look at the lines on a hologram, that is the map of the
interference patterns from the reflected and reference laser beams.

On May 31, 9:09*am, K1TTT wrote:
ah, but in this case consider that you have a pint of blue water
moving to the right and a pint of red water moving to the left to be a
better analogy to currents of the waves moving forward and backward in
the coax. *there are then 3 possibilities:

No, no, no. I am NOT talking about forward and reflected waves moving
in opposite directions. I am talking about two coherent, collimated
waves *MOVING IN THE SAME DIRECTION* in an RF transmission line away
from an impedance discontinuity - either two waves moving forward
toward the load or two waves moving backwards in the opposite
direction toward the source. Such multiple wavefronts occur at
impedance discontinuities because of multiple reflections.

Forward and reflected waves (waves moving in opposite directions in a
constant Z0 environment) do NOT interact. At an impedance
discontinuity, the component reflections and transmissive waves do
interact if they are coherent, collimated, and MOVING IN THE SAME
DIRECTION.
--
73, Cecil, w5dxp.com

On May 31, 9:12*am, K1TTT wrote:
no need, just look at the lines on a hologram, that is the map of the
interference patterns from the reflected and reference laser beams.

Oh yeah, duh ... brain fart - if one slants the partial mirror
detector, one can spread the interference patterns out to an optimum
pattern for viewing by a human eye.
--
73, Cecil, w5dxp.com

On Mon, 31 May 2010 07:09:49 -0700 (PDT), K1TTT
wrote:

ah, but in this case consider that you have a pint of blue water
moving to the right and a pint of red water moving to the left to be a
better analogy to currents of the waves moving forward and backward in
the coax.
....
the result is that both waves continue
on their way undisturbed by the other... which is what is observed in
all cases of em wave interaction in linear media.

Miguel, I want you to note how David clearly exposes the failure of a
metaphor, using the metaphor's own analogy.

This is the danger of trying to explain one system in terms of another
without knowing how either work.

It also reveals in shades of deep purple, how the visual system lies
to us, and we believe we see the Truth.

So, for metaphors and analogies, when the reader knows better, its a
lark (an amusing adventure or escapade); or when the writer knows
better, its a lark (behavior in a mischievous way).

If you both don't know better, its a lark (activity regarded as
foolish or a waste of time by the rest of us).

Carefully parse the following:
"Truth is a lark."

73's
Richard Clark, KB7QHC

On May 31, 11:57*am, Richard Clark wrote:
Miguel, I want you to note how David clearly exposes the failure of a
metaphor, using the metaphor's own analogy.

Actually Miguel, I want you to note how a simple semantic
misunderstanding can lead to false assumptions that propagate to: 1.
false assertions, 2. accusations, and 3. character assignation. I said
earlier that I was referring only to coherent/collimated waves that
are TRAVELING IN THE SAME DIRECTION. K1TTT apparently missed that
caveat and was simply confused about what I had said. Based on that
false assumption, Richard, as usual, jumped on the "kill the
messenger" bandwagon. He doesn't seem to realize that he has to prove
quantum electrodynamics to be wrong to prove me wrong, a feat that no
mortal has yet accomplished.
--
73, Cecil, w5dxp.com

On May 31, 2:42*pm, Cecil Moore wrote:
On May 31, 9:09*am, K1TTT wrote:

ah, but in this case consider that you have a pint of blue water
moving to the right and a pint of red water moving to the left to be a
better analogy to currents of the waves moving forward and backward in
the coax. *there are then 3 possibilities:

No, no, no. I am NOT talking about forward and reflected waves moving
in opposite directions. I am talking about two coherent, collimated
waves *MOVING IN THE SAME DIRECTION* in an RF transmission line away
from an impedance discontinuity - either two waves moving forward
toward the load or two waves moving backwards in the opposite
direction toward the source. Such multiple wavefronts occur at
impedance discontinuities because of multiple reflections.

Forward and reflected waves (waves moving in opposite directions in a
constant Z0 environment) do NOT interact. At an impedance
discontinuity, the component reflections and transmissive waves do
interact if they are coherent, collimated, and MOVING IN THE SAME
DIRECTION.
--
73, Cecil, w5dxp.com

add another condition and i might buy it... their polarization must be
the same... if you satisfy ALL those conditions then i believe you
would not be able to separate the waves and you could combine their
amplitudes. but that still doesn't mean they are interacting, just
that their fields always happen to be aligned... now if you are an
engineer like i am and deal with macroscopic processes i would
consider it perfectly logical to add the fields in a linear medium and
carry on with a single wave in each direction created by an infinite
series of reflections... HOWEVER, if i switch my hat to the scientist
part of my job title and i was working in photons I would come to a
point where it would be impossible to divide the last photon and
things would fall apart.... fortunately the ham/engineer side usually
outvotes the scientist part and i take the infinite summation and call
it a day.

On May 31, 5:20*pm, Cecil Moore wrote:
On May 31, 11:57*am, Richard Clark wrote:

Miguel, I want you to note how David clearly exposes the failure of a
metaphor, using the metaphor's own analogy.

Actually Miguel, I want you to note how a simple semantic
misunderstanding can lead to false assumptions that propagate to: 1.
false assertions, 2. accusations, and 3. character assignation. I said
earlier that I was referring only to coherent/collimated waves that
are TRAVELING IN THE SAME DIRECTION. K1TTT apparently missed that
caveat and was simply confused about what I had said. Based on that
false assumption, Richard, as usual, jumped on the "kill the
messenger" bandwagon. He doesn't seem to realize that he has to prove
quantum electrodynamics to be wrong to prove me wrong, a feat that no
mortal has yet accomplished.
--
73, Cecil, w5dxp.com

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

On May 31, 12:32*pm, K1TTT wrote:
add another condition and i might buy it... their polarization must be
the same...

If they are coherent, their polarization must necessarily be the same
at the same point at the same time. I specified that they are
coherent, collimated, and traveling in the same direction. I'm not
omniscient so if that's not enough boundary conditions, please
enlighten me.
--
73, Cecil, w5dxp.com