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

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

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.

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.

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

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

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

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

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.

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