Where Does the Power Go?
 

Cecil Moore wrote:
Jim Kelley wrote:

The problem with your paper, Cecil, is the part where you try to
invent the "4th Mechanism of Reflection".


I wish I had invented it, Jim, but the mechanism of wave
reflection due to interference was well known and under-
stood by optical engineers long before I was born. It's
how non-reflective glass works. Ideally, interference
at the thin-film coating reflects all of the light back
toward the picture behind the glass.

What has been known since long before you were born is that only
direct interaction with matter causes EM waves to reflect. You're far
too modest Cecil. The "4th mechanism of reflection" is completely
your idea! ;-)


73, Jim, AC6XG

Where Does the Power Go?
 

Jim Kelley wrote:
What has been known since long before you were born is that only direct
interaction with matter causes EM waves to reflect.

Would you say the changing characteristic impedance between
two waveguides in outer space is a direct interaction
with matter? There is no matter inside the waveguide with
which to interact.

However, I am wondering if I am using the wrong word when
I say interference can cause reflections. Since the same
thing happens with scattering S-parameters, it may be
a 180 degree refraction instead of a 180 degree reflection.
I'll have to take a look at the math.

But no matter what it is called, the results are the same.
"A rose by any other name ..."
--
73, Cecil http://www.w5dxp.com

Where Does the Power Go?
 

Jim Kelley wrote:
What has been known since long before you were born is that only direct
interaction with matter causes EM waves to reflect. You're far too
modest Cecil. The "4th mechanism of reflection" is completely your
idea! ;-)

Nope, it's not. The interference is caused
by the interaction of the forward and reflected waves
*at a physical impedance discontinuity*. The impedance
discontinuity is the primary cause of everything and is
certainly a point of "direct interaction with matter".
The reflection coefficients based on RF impedance
discontinuities are similar in kind to the reflection
coefficients based on differing light indices of
refraction between materials.

The forward reflection of light from a non-reflective
thin-film glass surface was well understood before I was
born. Optical engineers have no problem with that change
of EM wave momentum due to interference between the internal
reflection and the external reflection. That's the same
mechanism that happens at a Z0-match point in a transmission
line.
--
73, Cecil http://www.w5dxp.com

Where Does the Power Go?
 

Cecil Moore wrote:

[snip]

However, I am wondering if I am using the wrong word when
I say interference can cause reflections. Since the same
thing happens with scattering S-parameters, it may be
a 180 degree refraction instead of a 180 degree reflection.
I'll have to take a look at the math.

Cecil,

You may have taken the first step along the path to enlightenment. It
has been explained previously on RRAA that interference is a result, not
a cause. There are no primary equations related to interference that are
useful for analyzing the exact behavior of a system. Sure, there are
lots of handwaving explanations, but nothing that can actually give real
numbers for fields, currents, or whatever.

If you start with the proper equations for the fields, or voltage and
current if you desire, add in the correct boundary conditions on the
interfaces, and then find the numerical solution, any interference will
appear. No need to make a special case.

If you do this then two positive things will occur.

1. You will be in accord with virtually every mathematician, physical
scientist, optical scientist, and even engineer in applying standard
analysis techniques.

2. All of the worry about missing energy, canceling waves at the
interfaces, etc. simply disappears. It all pops right out from the
proper application of the math, automatically.

73,
Gene
W4SZ

Where Does the Power Go?
 

Gene Fuller wrote:
You may have taken the first step along the path to enlightenment. It
has been explained previously on RRAA that interference is a result, not
a cause.

Of course, the Big Bang is the cause of everything,
but what caused the Big Bang?

When one says A causes B which causes C which causes
D, etc., it is not false to say C causes D. There may
be a long line of causes and effects. One step's cause
is the previous step's effect. Interference causes
visible interference rings in a light experiment. And
of course, interference is just one event in a long
line of cause and effect. Without interference, there
would be no interference rings. Without beams of light,
there would be no interference. Without a Big Bang,
there would be no light.

In a line of events, interference is an event that has
a cause and has an effect.
--
73, Cecil http://www.w5dxp.com

Where Does the Power Go?
 

Cecil Moore wrote:
Gene Fuller wrote:
You may have taken the first step along the path to enlightenment. It
has been explained previously on RRAA that interference is a result,
not a cause.

Of course, the Big Bang is the cause of everything,
but what caused the Big Bang?

When one says A causes B which causes C which causes
D, etc., it is not false to say C causes D. There may
be a long line of causes and effects. One step's cause
is the previous step's effect. Interference causes
visible interference rings in a light experiment. And
of course, interference is just one event in a long
line of cause and effect. Without interference, there
would be no interference rings. Without beams of light,
there would be no interference. Without a Big Bang,
there would be no light.

In a line of events, interference is an event that has
a cause and has an effect.

Cecil,

You just proved the point perfectly. I did not say that interference is
imaginary or that it is not a useful description. I said that
interference is not a primary tool for achieving detailed numerical
solutions. It is a result from such calculations.

Now put your hands down, solve the real equations, and stop all that
mumbo-jumbo about canceling waves and reversing momentum.

8-)

73,
Gene
W4SZ

Where Does the Power Go?
 

On Tue, 03 Oct 2006 16:56:32 GMT, Gene Fuller
wrote:

reversing momentum

Imagine the G forces of that for an infinitesmal 1/10¹² second or so.

Where Does the Power Go?
 

Gene Fuller wrote:
Now put your hands down, solve the real equations, ...

Do you mean the mashed-potatoes energy equations where
the wave energy components are completely ignored? That's
how some arrived at such strange unreal concepts in
the first place. When one assumes that standing waves
have an existence separate from the necessary components
of the standing wave, one loses touch with reality. There
should be a rehab clinic for sufferers of the mashed
potatoes steady-state addiction.
--
73, Cecil http://www.w5dxp.com

Where Does the Power Go?
 

Do you mean the mashed-potatoes energy equations where
the wave energy components are completely ignored?

Calling the steady state solution "mashed-potatoes" is ridiculous. All
other terms besides the steady-state ones have left the picture in the
steady state. This is true of the waves, their amplitudes, the energy
in the line, everything. That's what makes steady-state analysis
useful... you can do it and be right.

If you want to know what happens during the transient, you MUST start
from the original differential equation that describes how the fields
were set up in the line in the first place, but it doesn't matter in
the steady state. The answer is the same. If you start with the full,
time dependent equations for turning the source on into some line,
whatever it is that happens with the power and where it goes will
become clear from the transient solution, which I know you haven't
worked out. Neither have I.

I don't want to have to do it but one of these days I think it might be
necessary.

This is not a problem that will be solved in with Logic and English,
Cecil. There is no argument if you start with a full , time dependent
mathematical description of the waves. That is what will answer the
question "Where Does the Power Go" You'll end up with a time dependent
Poynting vector that will tell you. I hypothesize, for no particularly
good reason, that such a description will lead to the excess power
having been delivered to the load through the interaction of the
transient solutions on the line.

- - - - - -

Copied and pasted from the rest of my post at eHam:

In a matched line, none of us would disagree that the power flux out of
one end of the line is the power flux into the other end.

From what I understand, the electromagnetic energy contained in that
line is related to the Poynting vector. The integral of the Poynting
vector over the cross sectional area of the line gives you the time
averaged power flowing in the line. The power flows at the group
velocity of the waves in the line, for normal transmission lines, this
is somewhere between 0.6 - 1.0 times the speed of light.

The power flow divided by the group velocity gives you the energy
density per unit length in the line.

The power flow is the Poynting vector integrated over the cross
sectional area of the line.

The Poynting vector in a mismatched line has been worked out by me,
here, and I'd appreciate comments on the MATH.

http://en.wikipedia.org/wiki/User:Dan_Zimmerman/Sandbox

My claim is that that's it. That's all. No handwaving, no setting up
steady state startup and claiming that that energy remains in the line.
The real part of the Poynting vector represents real power flux and
the imaginary part represents reactive power flux. To get the energy
that exists in the line you just add it all up...

Take the real part, divide it by the propagation velocity on the line,
integrate it over the area of the line (I think for TEM, the fields
are uniform and so multiplying by the area of the line is sufficient)

This gives you the energy density per unit length for the power that's
flowing from source to load. Integrate that over the length of the
line. Set aside.

Take the imaginary part, do the same thing. This gives you the energy
contained in the line of the reactive, circulating power. Add in the
previously calculated energy of the flowing power, and you're done.

Am I wrong, Cecil et. al? If so, Cecil, could you please write down
the description mathematically or get someone to help you out, and
could you please look at and check my math?

I started with the assumption that there are two counterpropagating
waves with some arbitrary electric field amplitudes and an arbitrary
phase relationship. I didn't use anything about Thevenin. I didn't
use anything about virtual open circuits.

The forward and reverse waves have been included, from the beginning.
They're both there.

When the SWR is 1:1, there is no energy in the line associated with
reactive stored power. None at all. The reflection coefficient is
zero, and all energy in the line is associated with flowing power.

When the SWR is infinite, there is no energy in the line associated
with flowing power. None at all. The reflection coefficient is 1, and
all energy in the line is associated with circulating power (reactance
only, as you would expect from a stub)


Dan

Where Does the Power Go?
 

Richard Clark wrote:
Gene Fuller wrote:

reversing momentum

Imagine the G forces of that for an infinitesmal 1/10¹² second or so.

The pressure of such a momentum change can be calculated.
From "Optics", by Hecht, 4th edition, page 57:
"When the surface under illumination is *perfectly
reflecting*, the beam that entered with a velocity
of +c will emerge with a velocity of -c. This
corresponds to twice the change in momentum that
occurs on absorption, and hence,"

Pressure of a absorbed wave = S(t)/c

Pressure of a reflected wave = 2*S(t)/c
--
73, Cecil http://www.w5dxp.com