Gene Fuller wrote:
As for the E- and H-fields, this just gets more amusing by the minute.
The second paragraph is even more curious. Do you have a reference
for this migration of energy from the H-field to supply the suffering
E-field?

Here's the reference: From "Fields and Waves" by Ramo and Whinnery.
Take a close look at the exponential transmission line equations
for flat lines (no reflections):

V = Vmax(e^-az)(e^wt-bz) (1)

I = Vmax(e^-az)(e^wt-bz)/Z0 (2)

'a' (alpha) is the attenuation factor. The two equations are identical
except for the Z0 term. If you divide equation (1) by equation (2),
you get Z0. In a flat transmission line (no reflections) the current
is ALWAYS equal to the voltage divided by the characteristic impedance
of the transmission line. The voltage and current are attenuated by
EXACTLY the same factor. If the voltage drops because of I^2*R losses,
the current must decrease by exactly the same percentage. (I have avoided
calling it a current drop so it wouldn't upset you.)

Since the attenuation factor is R/2*Z0 + G*Z0/2 and since, for most
transmission lines used on HF, R/2*Z0 G*Z0/2, the current attenuation
is caused by the series I^2*R drop in the voltage and the V/I=Z0 ratio
that must be maintained - pretty simple logic.

I must have missed that day in class.

Yep, you must have. But it's not too late to learn what you missed
that day. :-)
--
73, Cecil, W5DXP