John Popelish wrote:
To the center conductor, carrying the standing wave, the shield is the
outside world. If there is no shield, the outside world is the outside
world, as far as displacement current goes. Do you imagine this current
changes in some way other than magnitude and wave velocity when you wrap
a shield around a wire carrying a standing wave?

No, that is your point, not mine. My point is that displacement
current to real ground is non-existent outside of a coax shield
(unless common mode current exists) and that it is usually a
secondary effect if the coax shield doesn't exist. The primary
reason for the variation in standing wave current along the line
is the phasor sum of the forward and reflected wave phasors that
are rotating in opposite directions. Do you understand phasor
addition? 1 at zero + 1 at 180 deg = zero at a standing wave
node? Displacement current to real ground doesn't cause that.

I am explaining distributed network theory to you.

:-) How? By denying the existence of the individual H-fields
in forward and reflected EM waves? Now, that's really funny.

http://www.qsl.net/w5dxp/travstnd.GIF

And I have agreed with that. Why do you keep bringing it up?

Because that's the whole point of this discussion. If you
agree with that, there is no reason to continue. I just
don't care about instantaneous current, Brownian motion, or
the exact location and velocity of every electron carrier.
There's too much uncertainty involved.
--
73, Cecil http://www.qsl.net/w5dxp