In article ,
JC wrote:

In a lossy coax the lost energy is, I suppose, heating up the dielectric.

Depends on the frequency.

At lower (HF and VHF) frequencies, the loss is coax is dominated by
resistive losses in the conductors. The RF current is flowing through
copper (which has a non-zero resistance per foot), and isn't even
flowing through *all* of the conductor (the "skin effect" constrains
the current to flow through a thin layer on the surface of each
conductor).

The current flow through the resistive copper creates a voltage drop,
and dissipates power... which then heats up the conductors. The
dielectric and outer insulation are also heated, but indirectly...
heat flows into the warmed-up conductors.

There are direct losses in the dielectric (in addition to the
resistive losses in the conductors) but as I recall these don't become
significant until you're well up into the microwave regions.

Certainly not ( it is well known that all the PE food containers used in
such ovens are not heated ), but what is wrong in this test ? how does it
differ from the dielectric heated in an actual operating lossy cable ?

Your test omits the indirect heating of the dielectric, from
conductors which are themselves heated by resistive losses.

--
Dave Platt AE6EO
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