Jim Kelley wrote:
Cecil Moore wrote:

Harry wrote:

Would someone please explain that for me?


DC steady-state does not cause electrons to emit photons.

Except in flashlights, apparently.

For RF photons to be emitted from a copper wire dipole, the
free electrons must be accelerated and decelerated. The DC
component cannot accomplish that feat.

Quantum mechanics is completely unnecessary here, Cecil. I think
Faraday still provides the best explanation.

ac6xg

With a ns pulse into a zero resistance conductor in free space all
energy should be radiated, so no DC component if you look at the input
power. However if you introduce the same pulse to a circuit that has
magnetic properties, i.e. a ferrite inductor as an example, there is
energy used to align the molecules in the material, not radiated, which
results in a DC component. In a flashlight, most of the energy is
heat, which will result in a big DC component with no radiation rf
wise.
That's really advanced stuff for a Ham Radio Newsgroup.
Gary N4AST