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Re: Series resonance/Was: Waveguide TC



Original poster: "Paul Nicholson by way of Terry Fritz <twftesla-at-qwest-dot-net>" <paul-at-abelian.demon.co.uk>

John Couture wrote:

 > I take that to mean [Tesla] was going to use currents instead of
 > wave radiation. Aren't these two different methods of transferring
 > electrical energy?

No, they're the same.  Take a conductor or two, carrying power to
a load.  You'll come along and measure the current, and the voltage
between the conductor and its return path, and multiply the two to
obtain a measure of the power flow.   Instead, I'll measure the
electric and magnetic fields, and integrate their vector product over
a surface separating source and load, and I'll arrive at my figure for
power flow into the load.

Our two answers will be the same.  Think of the conductor(s) as a
guide for the EM waves.  The fields are a consequence of the current
flow, or the current flow is a consequence of the fields. Take your
pick.

 > The energy via current has to have a return conductive circuit.

Not necessarily.  The return path can be a displacement current
rather than a conduction current.  So too can be the outgoing path.
However, if you take these options, the coupling between source and
load reduces as the frequency is lowered, reaching zero for DC.

Charge conservation prevents the launching of an EM wave of zero
frequency from an isolated transmitter: the symmetry of the system
prevents you from raising a sustained (static) EM field with a
non-zero integral of the vector product of E and H over a Gaussian
surface surrounding it.  However hard you try, using whatever
fiendish mix of static charges and magnets you can devise, you'll
find at some point you need to create and destroy individual charges,
or teleport them from A to B to avoid creating the H field that would
occur if they were moved through normal space.

A conductor allows an EM wave to be conveniently guided.  A pair of
conductors also allows use of an EM wave of zero frequency (ie a DC
circuit) by providing a charge path for both the transmitter and
receiver which breaks the symmetry and enables an EM wave of zero
frequency to be set up between source and load, carrying EM energy
continuously in the desired direction - this despite the fact that
the phase of the wave never moves off the zero mark.  My opening
statement equating the VI product with the ExH integral still applies,
even at zero frequency.

It's as well to remember that an E field or an H field alone cannot
convey energy, since ExH is zero everywhere. They can each apply a
force to a charge at a distance, but that's all.  If a distant charge
responds to the force applied by a pure E or H field (by a component
of movement parallel to the force) then work is done, energy is
transferred to or from the distant charge, and a transient field of
the opposite kind, H or E respectively, exists for a time. Then, the
integral of the cross product becomes non-zero, indicating a real
energy exchange.

Tesla thought that by feeding a current into the ground, it would
set up a uniform charge distribution over the earth.  Instead, the
charge distribution follows the strength of the field established by
the transmitter that drives the current.  Thus you are free to say
that the energy is distributed by the current, or by the field,
because these are two aspects of the same phenomena.

As for Hertzian waves, well, you can minimise the radiation lost to
the far field by using a low frequency, so that the wavelength is of
the same order or larger, than the region containing the transmitter
and receivers.  You can further minimise it by transmitting with an
E/H ratio very different from the impedance of free space.  But, it
does not follow from this that the energy is no longer carried by the
field. On the contrary, the energy picked up by the receivers is
still represented correctly by the integral of ExH, and although we
are immersed in the reactive near field of the transmitter, the
components of the field that are actually carrying the energy (the
real in-phase E and H perpendicular to the direction of flow) are
indistinguishable from a far field Hertzian wave except perhaps for
the ratio of E to H.
--
Paul Nicholson
Manchester, UK.
--