Roger wrote:
As your argument is developed below, you begin using positive x. What
is the zero point it is referenced to? I will assume that it is leading
edge of original reference wave.

Not speaking for Roy, but x is usually 0 at the feedpoint
of a 1/2WL dipole, for instance. For any fixed time t=N,
the phase of the standing-wave signal is constant all up
and down the antenna. Please ask Roy how that fixed phase
can possibly be used to measure the phase shift through
a 75m bugcatcher loading coil.

Wow! This would happen if the re-reflection actually reverses. How
could this occur if we consider that voltage is a collection of positive
or negative particles? We would have a positive reflection meeting with
a positive outgoing wave (or negative meeting negative). The situation
would be the same as at the open circuit end immediately following
initial reversal, or current would simply stop flowing from the source.

RF engineering convention in phasor notation: If the wave
reflects from an open circuit, the current reverses phase
by 180 degrees such that If+Ir=0. The voltage does not
reverse phase so Vf+Vr=2Vf=2Vr.

If the wave reflects from a short circuit, the voltage reverses
phase by 180 degrees such that Vf+Vr=0. The current does not
reverse phase so If+Ir=2If=2Ir.

Incidentally, this is a different convention from the field
of optics.

I think you would agree that a steady state standing wave would form
immediately upon reflected wave reaching the initiating source if the
wave did not reverse.

This is a tricky subject covered by many discussions among
experts. My take is that since the impedance "seen" by the
reflections is usually unknown (possibly even unknowable) the
discussion is a moot point. The convention is that if reflected
energy enters the source, it was, by definition, never generated
in the first place. All that is important at the output of a
source is the *NET* power output. Energy flow back into the
source is, by definition, completely ignored.

The gif's certainly made it clear why you are skeptical of the power of
traveling wave analysis.

The Poynting vector yields the power density of any EM traveling
wave. The power density equations from the field of optics are
just as valid for RF waves as they are for light waves. If one
wants to understand the redistribution of energy, one will need
to understand constructive and destructive interference during
superposition. Roy is on record as not caring where the energy
goes.
--
73, Cecil http://www.w5dxp.com