Gene Fuller wrote:
This is truly bizarre. The E-field and the H-field are always at right
angles to each other in the plane wave case. They are also perpendicular
to the propagation direction. The same is true for the TEM mode in a
coaxial waveguide, which is the usual case discussed on RRAA. There is
no 0 or 180 degree involvement. I don't know what you are so confused
about, but you really need to rethink what you are saying.

What you are saying is 100% correct for plane EM traveling waves.
What you are saying is 100% incorrect for standing waves. It just
shows that Hecht was right when he said standing waves probably
don't deserve to be called waves. Maybe you should try to understand
why Hecht would say such a thing.

There is no need to sketch or calculate anything. A diagram showing the
relationship of the E-field vector and the H-field vector is in every
E&M and optics book I have ever seen.

Yes, and that diagram is for a *TRAVELING WAVE*, not for a standing
wave. Please find a reference with the E-fields and H-fields diagrammed
for a standing wave and get back to us. Better yet, consider there is
a need to sketch the fields if for no other reason, just to prove me
wrong.

This is not rocket science, Gene. When the magnitude of the Poynting
vector for the forward wave equals the magnitude of the Poynting vector
for the reflected wave, the net Poynting vector is obviously zero.
Since the Poynting vector equals ExH, the only way that the standing
wave Poynting vector can be zero everywhere is for the cross product
of ExH to be zero. The cross product is E*H*sin(A) and we know that
E and H are not zero, so sin(A) must necessarily be zero. So A must
necessarily be 0 or 180 degrees. Nothing else is possible.
--
73, Cecil http://www.w5dxp.com