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, 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, 5:39*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

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.

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, 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