Keith Dysart wrote:
As I expected, you claim that the situations are
*completely* different. And yet the voltage, current
and energy distributions are identical. There are
no observable differences. And yet you claim they
are *completely* different. And yet there are no
observable differences. And yet....

I did NOT claim that the situations are *completely*
different. I said that some conditions are different
and some conditions are the same. Voltages and currents
are the same yet there is certainly a difference between
an open circuit and a short circuit. Besides, in the
real world, cutting the line would certainly cause
observable differences.

Tis a puzzle, isn't it.

Nope, if you were born without your five senses,
you would feel that way about everything in existence.
Why do you deliberately choose to remain handicapped
by ignorance?

A bit of modulation would cure up the mystery for
you. If any modulation crosses the node, it is a
good bet that wave energy is carrying the
modulation. If phase locked TV signal generators
equipped with circulator load resistors are
installed at each end of a transmission line,
the TV signals can be observed on normal TV
sets crossing the standing wave nodes as if
they didn't exist. Removing the modulation
is unlikely to reverse the laws of physics.

And we know from circuit theory that we can cut
a conductor carrying no current without affecting
the circuit. Why should it be different here?

Please prove that a short circuit and an open circuit
are identical.

Please present your new laws of physics that allow
EM waves to reflect off of EM waves in the complete
absence of a physical discontinuity.

Again, not my claim.

Seems your theory requires such. Please explain how
reflections can occur at a passive standing wave node
without EM waves bouncing off of each other.

Energy and momentum both must be conserved. A causeless
reversal of energy and momentum is impossible whether
it is a bullet or an EM wave.

But using your previous approach
for analysis, perhaps we should insert a zero length
line of the appropriate impedance to provide the cause
for the reflection, if you insist on a reflection.

Please produce an example of
a real world transmission line that would support your
100% reflection. Hint: what would be the Z02
characteristic impedance in the reflection
coefficient equation, (50-Z02)/(50+Z02) = 1.0 ???
--
73, Cecil http://www.w5dxp.com