"Dave" wrote
...

"Szczepan Bialek" wrote in message
...

"Dave" wrote
...

"Szczepan Bialek" wrote in message
...
In the Hertz apparatus the charges (electrons) have at the centre the
max velocity and the acceleration equal zero.
At ends the situation is opposite. So your answer should be: "the ends
radiate of course".

of course you are wrong. there is a smooth transition between the
center and the ends, that whole length radiates. you can't just look at
the boundry conditions, you have to consider the whole length.

Yes. But the radiation is not uniform. What radiate stronger: the centre
or the ends?

both. when the current is high in the center it is creating a stronger
magnetic field, and when that current reaches the end it creates the
highest voltage so makes more electric field... both are part of the
electro-magnetic wave.

It is not Maxwell model. In it current create magnetic field and THIS field
create the electric field. AND SO ON.


It is very funny that engineers use electrons and do not know that in
the "Maxwell's equations" no electrons, There is incompressible
massless fluid.
You here do not use the "Maxwell's equations". The teachers use them to
teach math.
Engineers use the empirical equations following the rule "accelerating
charges create radiation".

Gauss's law is about charged particles, the one art so much likes to
distort.. and don't forget that the 'i' term is also about charged
particles moving... if they can move they are not imcompressible, and
since the force on them can be measured and accelerations are not
infinite they are not massless.

We all know now that the electrons are "not imcompressible, and since
the force on them can be measured and accelerations are not infinite
they are not massless."
But do you know what the electricity was like in the Maxwell theory from
1865?

sure, its the same as today. since his equations still work the
electricity hasn't changed.

"1864 - Maxwell reads a memoir before the Royal Society in which the
mechanical model is stripped away and just the equations remain. He also
discusses the vector and scalar potentials, using the Coulomb gauge. He
attributes physical significance to both of these potentials. He wants to
present the predictions of his theory on the subjects of reflection and
refraction, but the requirements of his mechanical model keep him from
finding the correct boundary conditions, so he never does this calculation"

Your (engineering people) model is O.K. but it is quite different from the
Maxwell model. This is the reason that Art can wrote: " "For your
information you have never built an antenna that conforms in its
entirety to Maxwell';s laws thus you cannot possibly understand
radiation as presented by Maxwell."
S*