"Tom Donaly" wrote:

When you pluck a string, you are exciting the whole string at
once.

If a sound wave of the right frequency impinges on a string
perpendicular to the string's axis, the string will vibrate
sympathetically. In that case, it's hard to justify saying that
two waves are traveling in opposite directions up and down the
string.

OK, lets change the string. Now it's the top guy wire for a 1/4 wave
vertical at 560KHz. When you pluck it, you can hear it pinging as
the waves are reflected. Maybe it would be difficult to take that to
a symphony performance, but hey, true art is art no matter where you
find it

Nevertheless, the solution of the partial differential equation
describing the motion of the string, as proposed and solved by the
French mathematician D'Alembert, in 1747, is consistent with the
idea of two waves of arbitrary function traveling in opposite
directions on the string. If I were you, I'd find a copy of the
differential equation of a wave on a string and compare it to the
same equation describing an electromagnetic wave on a transmission
line. How similar are the two?

We may have lost the validity of the comparison to EM waves.

73,
Tom Donaly, KA6RUH

Regards,

Mike Monett

Mike Monett wrote:
Since photons do not
interact, EM waves also do not interact with each other.

The following quote sounds like an interaction of photons to me.

http://micro.magnet.fsu.edu/primer/j...ons/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."

I suspect that coherent photons can interact at an impedance
discontinuity which causes reflections.
--
73, Cecil http://www.w5dxp.com

Mike Monett wrote:
Roy Lewallen wrote:
. . .
Again my apology. You do indeed have it right. Incidentally, it's
not possible for a medium to have a purely reactive (imaginary) Z0
at any non-zero frequency.

Roy Lewallen, W7EL

Thanks very much, Roy. It was probably my mistake, using the word
"Orthogonal" when quadrature would probably have worked better.

Can you explain your last sentence? Why does this happen?

The "non-zero" was unnecessary, and a result of a too-quick evaluation
of an equation, although the meaning of Z0 at DC isn't clear anyway.

There are at least two related ways to show that a medium can't have a
purely imaginary Z0 (more correctly, intrinsic impedance). One is to use
the telegrapher's equation for a transmission line immersed in the medium:

Z0 = sqrt((R + jwL)/(G + jwC)) (w = omega, the rotational frequency)

For Z0 to be purely imaginary, the quantity under the radical has to be
purely real and negative. A little algebraic manipulation shows that
this requires that RG + w^2LC 0. All the quantities are positive, so
it can't happen.

You can also use

Zc = sqrt(mu/ceps) where ceps = the complex permittivity, mu = the
permeability of the medium, and Zc the intrinsic impedance.

The complex permittivity ceps = eps - j*sigma/w

where eps = the real (DC) permittivity
sigma = the conductivity of the material

You end up with the same situation, where for Zc to be purely imaginary,
the quantity under the radical has to be purely real and negative, which
requires that mu * eps 0. Remember that mu and eps here are the actual
permeability and permittivity, not the relative values we often use.

A little further research reveals that there are some fairly recently
created man-made materials which have a negative permeability. Those
could presumably have a purely imaginary intrinsic impedance, provided
that they have a positive permittivity. So there might be an exception
to my statement, although it isn't something you're likely to encounter
for some time to come.

I have been following these threads with some interest, and I very
much appreciate your analysis, as it adds greatly to my
understanding. Thank you very much for taking the time to write so
clearly.

There is one point I still have trouble with. The concept of power
flowing in standing waves where the superposition goes to zero, and
yet the energy flow is unaffected and continues in opposite
directions on either side of the null point.

Anyway, I have googled until my fingers get sore, and I haven't
found a good explanation of why this happens. Everyone says it is
well understood from basic undergraduate theory, but the only
references I can find are from graduate studies in Quantum
Electrodynamics. This is not much help.

Please exclude me from the "everyone" in "everyone says". I don't say
that power flows, period. We've seen the serious traps people have
fallen into by making this assumption and trying to build from it.
That's why you won't find it in texts.

So I have to form some image in my mind of why these waves do not
interact. Here is a partial pictu

1. Electromagnetic waves travel at the speed of light in whatever
medium they are in. For them to interact, there must be some advance
information they are about to collide. But that would require
transferring information faster than the speed of light, which is
forbidden.

You don't need a reason for them to not interact, you need a reason for
them to do so. In a linear medium, there is none.

2. The fields in electromagnetic waves are at right angles to the
direction of propagation. There is no longitudinal component, and
therefore the waves have no advance warning they are about to
collide. There is no vector component that is common to both that
would allow any interaction, so there is no way this can happen.

3. Photons carry no charge. They are not deflected by electrostatic
or electromagnetic fields, and do not interact with other photons.
Electromagnetic waves are made up of photons. Since photons do not
interact, EM waves also do not interact with each other.

The above concepts seem to make sense, and allow me to get some
sleep at night. Can you tell me if they are valid, and if there are
other ways of explaining this phenomenon?

I'm glad they work for you. I'll have to leave it to others to comment
on their validity, since I don't buy into the notion of flowing power in
the first place.

Roy Lewallen, W7EL