Gene Fuller wrote:
... your long exposition above
is not what you said previously, E*H*sin(A).

I think you know that I meant the magnitudes only,
but to be entirely technically correct it should
have been:

Poynting vector = ExH = |E|*|H|*sin(A)

The same question remains: When E and H are both
not zero, how can the Poynting vector be zero?
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Gene Fuller wrote:
Therefore, your proof that my "Incorrect" assertion is incorrect is
itself incorrect.

Please answer the question. It is known that
Re(ExH*)/2 = 0 for pure standing waves all up and
down the line. If E and H* are both non-zero, how
can Re(ExH*)/2 be zero?

Known by whom?

If you are talking about time and/or spatial averages then yes, the
Poynting vector will be zero. If you are talking about instantaneous
calculations at any point and time, then no, the Poynting vector will
not be zero all up and down the line.

Once again you have misinterpreted something you read in a book.

73,
Gene
W4SZ

Cecil Moore wrote:
Gene Fuller wrote:
Cecil Moore wrote:
The E-field and H-field of an EM wave can certainly
be represented as an exponential function when the
EM wave is normal to the reference plane. Think
about it. Hecht used that feature in his book.

Totally ducking the issue, as usual.

No, I'm waiting for an answer from you. It is known
that Re(ExH*)/2=0 for pure standing waves all up
and down the line. Since E and H* are not zero all
up and down the line, how is that possible?

Anything is possible in your Fractured Fairytale Physics.

Cecil Moore wrote:
Gene Fuller wrote:
The phasor has nothing to do with spatial direction of the field vector.

Sorry, but The IEEE Dictionary disagrees with you.
In the description of the Poynting vector, it says:
"E and H are the electric and magnetic field vectors
in phasor notation ..."

I have prepared a graph of a snapshot in time of
a forward wave and reflected wave of equal magnitudes.
When the waves are superposed at the reference plane,
the total E-field and total H-field are 180 degrees
out of phase. The graph is at:

http://www.w5dxp.com/EHSuper.JPG

The graph is complete nonsense. There is no rotation of the fields when
they undergo reflection. Any ordinary text on E&M or optics will show
you the equations and the correct sketches.

You are obviously still confusing phasors and fields. More FFP.

73,
Gene
W4SZ

On Fri, 18 Jan 2008 07:22:48 -0500, Chuck
wrote:

Thanks for the response, Richard.

Hi Chuck,

You're welcome.

I had consulted Feynman's Volumes I and II prior to posting and saw
nothing there to support the idea that an EM wave performs work in
moving through a region of free space. Hence, my original post.

It is also noteworthy that electrons don't exhibit much friction
either. (Feynman is particularly interesting in regard to friction
too.)

73's
Richard Clark, KB7QHC

Gene Fuller wrote:
Once again you have misinterpreted something you read in a book.

Once again, you have refused to answer a simple question.
If E and H are not zero, how can ExH be zero?
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Gene Fuller wrote:
Once again you have misinterpreted something you read in a book.

Once again, you have refused to answer a simple question.
If E and H are not zero, how can ExH be zero?

What is this? The RRAA version of "When did you stop beating your wife?"

Cecil Moore wrote:

Jim Kelley wrote:


Cecil Moore wrote:

Since the standing wave energy is being used to transform
impedances, it cannot also be used for radiation purposes -
like the energy in a standing wave on a transmission line
cannot be used both to transform impedances and to heat
up a load.


Would you care to elaborate on the idea that "energy is being used to
transform impedances"?


If reflections are nonexistent, no transformation takes
place, i.e. there is no SWR circle, just a point at the
center of the Smith Chart and the system is flat.

If reflections exist, then the superposition of the
forward wave and reflected wave transforms the
load impedance to some other impedance on the SWR
circle.

A 1/4WL transformer, for instance, will not transform
unless there are reflections present.

Cecil Moore wrote:

Jim Kelley wrote:


Cecil Moore wrote:

Since the standing wave energy is being used to transform
impedances, it cannot also be used for radiation purposes -
like the energy in a standing wave on a transmission line
cannot be used both to transform impedances and to heat
up a load.


Would you care to elaborate on the idea that "energy is being used to
transform impedances"?


If reflections are nonexistent, no transformation takes
place, i.e. there is no SWR circle, just a point at the
center of the Smith Chart and the system is flat.

If reflections exist, then the superposition of the
forward wave and reflected wave transforms the
load impedance to some other impedance on the SWR
circle.

A 1/4WL transformer, for instance, will not transform
unless there are reflections present.

You fail to explain how any of that makes energy unavailable for
"radiation purposes".

ac6xg

Gene Fuller wrote:
The graph is complete nonsense. There is no rotation of the fields when
they undergo reflection. Any ordinary text on E&M or optics will show
you the equations and the correct sketches.

Good grief, Gene. There is a 360 degree rotation in the fields
every wavelength. The direction of rotation is associated with
the direction of travel of the wave and is displayed by EZNEC
when the current phase option is turned on. All you have to
do to see the rotation of the traveling wave is to download
http://www.w5dxp.com/rhombicT.EZ

Those simplified sketches are making you simple-minded.
Assuming you are a member of the IEEE, look up this paper:
"Rotation in electromagnetic field equations". Or Google
"Rotation and the Electromagnetic Field".

"Optics", by Hecht, is one of your ordinary texts. That is
where the material for that graph comes from. The IEEE Dictionary
says: "E and H are the electric and magnetic field vectors in
phasor notation". If that graph is nonsense to you, it is your
fault, not mine.

Hecht says "Optics", 4th edition, page 289, about standing waves:

"The composite disturbance is then:

E = Eo[sin(kx+wt) + sin(kx-wt)]

Applying the identity:

sin A + sin B = 2 sin 1/2(A+B)*cos 1/2(A-B)

yields:

E(x,t) = 2*Eo*sin(kx)*cos(wt)"

"This is the equation for a STANDING or STATIONARY WAVE, as opposed
to a traveling wave. Its profile does not move through space; it is
clearly not of the form Func(x +/- vt)."

"... a phasor rotating counterclockwise at a rate omega is equivalent
to a wave traveling to the left (decreasing x), and similarly, one
rotating clockwise corresponds to a wave traveling to the right
(increasing x)."

Hecht uses phasors to represent EM waves all through his book.
He explains the standing wave E-field based on the two traveling
waves, E1-field and E2-field, thusly:

"The resultant phasor is E1 + E2 = E ... Keeping the two [traveling
wave] phasors tip-to-tail and having E1 rotate counterclockwise as
E2 rotates (at the same rate) clockwise, generates E [total] as a
function of 't'."

[Standing wave phase] "doesn't rotate at all, and the resultant
wave it represents doesn't progress through space - its a standing
wave."

Traveling wave phase rotates. Standing wave phase doesn't.

Speaking of "... net transfer of energy, for the pure standing
wave there is none."

The forward wave and reflected wave E-field and H-field vectors
are represented by phasors just as indicated in the IEEE Dictionary.
One is rotating clockwise and the other is rotating counterclockwise.
The Poynting vector for a pure standing wave is equal to zero just
as illustrated in my graph at: http://www.w5dxp.com/EHSuper.JPG
Given those boundary conditions and solving for the angle between
the standing wave E-field and H-field yields 0 or 180 degrees.

Are you really more interested in presenting false information
and saving face than you are in valid technical facts?
--
73, Cecil http://www.w5dxp.com