On Thu, 03 Jun 2004 16:18:36 -0500, Cecil Moore
wrote:

Walter Maxwell wrote:
Cecil, this is exactly what I've been trying to persuade you of, but always said
no, there is no short developed. But you must also agree that under this
condition the current doubles.

Nope, for complete destructive interference, both the E-field and
H-field associated with the two interfering waves collapse to zero.
For the resulting complete constructive interference, the ratio of
the E-field to the H-field equals the characteristic impedance of
the medium.

The two corresponding rearward-traveling superposing currents might be:
(2a at 180 degrees) superposed with (2a at zero degrees)
The superposed sum of the two rearward-traveling currents is zero.
This acts like an open where currents go to zero.

Of course, but the voltage doubles.

Nope, again here are the two sets of reflected waves.

#1 100v at zero degrees and 2a at 180 degrees = 200W

#2 100v at 180 degrees and 2a at zero degrees = 200W

Superposing those two reflected waves yields zero volts and
zero amps.

Well, Cecil, here's where we part company to a degree. Unlike voltage and
current that can go to zero simultaneously only in the rearward direction, E and
H fields can never go to zero simultaneously.

For "complete destructive interference" as explained in _Optics_, by
Hecht, the E-field and B-field (H-field) indeed do go to zero
simultaneously. That is what causes a completely dark ring in a
set of interference rings. Of course, a resulting corresponding
complete constructive interference causes the brightest of rings.

If Steve understands the action of the fields in the EM wave it's hard to
understand why he finds it so erroneous to associate voltage and current with
the their respective fields in impedance matching. Apparently he can't conceive
that the voltages and currents in reflected waves can be considered to have been
delivered by separate generators connected with opposing polarities.

He pretty much ignored current. His power equations are exact copies of
the light irradiance interference equations from optics, but he apparently
didn't realize it until I pointed it out to him.

Well, Cecil, perhaps I don't understand Melles-Griot. However, when an EM wave
encounters a short circuit the E field goes to zero, but in its change from
normal level to zero, that changing field develops a corresponding H field that
adds to its original value, causing it to double--the H field does NOT go to
zero. But as it returns to its normal value that change in turn develops a new E
field propagating in the opposite direction. From there on the EM field
reconstitutes itself with normal E and H fields of equal value, each supporting
one half of the propagating energy.

This sequence must also prevail in optics--are you sure you are interpreting
your double zero at the correct point in the circles?

Walt