On Jun 2, 2:12*pm, Cecil Moore wrote:
On Jun 2, 8:57*am, K1TTT wrote:

wave function solutions to maxwell's equations are enough to prove
that for me.

Not a loaded question: How do Maxwell's equations applied to a
standing wave prove that the component forward and reflected waves are
moving at the speed of light in the medium? If it can and if I can
understand it, I wouldn't need to use the photon argument.
--
73, Cecil, w5dxp.com

easy, maxwell's equations don't predict standing waves! they are a
product of superposition and the simplest instrumentation used since
they were first discovered.

On Jun 2, 9:31*am, K1TTT wrote:
easy, maxwell's equations don't predict standing waves! *they are a
product of *superposition and the simplest instrumentation used since
they were first discovered.

Please correct me if I am wrong: If one starts with the (superposed)
standing wave equation, V(x,t) = A*sin(kx)*sin(wt), Maxwell's
equations seem to provide a perfectly valid result (not that I can
recognize perfection). Therefore, how do Maxwell's equations prove
that the component traveling waves necessarily possess a separate
existence? Again, a serious question from an engineer who considers
anything except a logical '0' or logical '1' to be broken. :-)
--
73, Cecil, w5dxp.com

On Jun 2, 3:13*pm, Cecil Moore wrote:
On Jun 2, 9:31*am, K1TTT wrote:

easy, maxwell's equations don't predict standing waves! *they are a
product of *superposition and the simplest instrumentation used since
they were first discovered.

Please correct me if I am wrong: If one starts with the (superposed)
standing wave equation, V(x,t) = A*sin(kx)*sin(wt), Maxwell's
equations seem to provide a perfectly valid result (not that I can
recognize perfection). Therefore, how do Maxwell's equations prove
that the component traveling waves necessarily possess a separate
existence? Again, a serious question from an engineer who considers
anything except a logical '0' or logical '1' to be broken. :-)
--
73, Cecil, w5dxp.com

my differential calculus is a bit rusty, but i don't think that
equation satisfies the basic wave equation. from Fields and Waves in
Communication Electronics, section 1.14. a solution for the wave
equation in a transmission line (or other 1d form) must satisfy the
condition that the 2nd derivative wrt x and wrt t equal v-squared...
too hard to write in text. but basically take the above and
differentiate twice wrt time and twice wrt distance and i don't think
you'll get anything that comes out to the velocity squared. the
problem is that t and x must be part of the same sinusoid in the F(t-x/
v) so that the wave is truly traveling instead of stationary like your
equation provides.

in the basic maxwell equations in differential form you also run into
the same problem where the curl of the one field is proportional to
the time derivative of the other i think you'll end up with a
contradiction like k=w if you eliminate the dimensionless terms... i
may be wrong on that because its been so long since i've had to deal
with them on that level.

On 2 jun, 13:48, K1TTT wrote:
On Jun 2, 3:13*pm, Cecil Moore wrote:

On Jun 2, 9:31*am, K1TTT wrote:

easy, maxwell's equations don't predict standing waves! *they are a
product of *superposition and the simplest instrumentation used since
they were first discovered.

Please correct me if I am wrong: If one starts with the (superposed)
standing wave equation, V(x,t) = A*sin(kx)*sin(wt), Maxwell's
equations seem to provide a perfectly valid result (not that I can
recognize perfection). Therefore, how do Maxwell's equations prove
that the component traveling waves necessarily possess a separate
existence? Again, a serious question from an engineer who considers
anything except a logical '0' or logical '1' to be broken. :-)
--
73, Cecil, w5dxp.com

my differential calculus is a bit rusty, but i don't think that
equation satisfies the basic wave equation. *from Fields and Waves in
Communication Electronics, section 1.14. *a solution for the wave
equation in a transmission line (or other 1d form) must satisfy the
condition that the 2nd derivative wrt x and wrt t equal v-squared...
too hard to write in text. *but basically take the above and
differentiate twice wrt time and twice wrt distance and i don't think
you'll get anything that comes out to the velocity squared. *the
problem is that t and x must be part of the same sinusoid in the F(t-x/
v) so that the wave is truly traveling instead of stationary like your
equation provides.

in the basic maxwell equations in differential form you also run into
the same problem where the curl of the one field is proportional to
the time derivative of the other i think you'll end up with a
contradiction like k=w if you eliminate the dimensionless terms... i
may be wrong on that because its been so long since i've had to deal
with them on that level.

Oh!, I surrender :) The thread do not converge to a solution, it
seems run out toward a naked singularity! I stay tuned until this
good brainstorm calms down ..

Miguel

On Jun 2, 5:56*pm, lu6etj wrote:
On 2 jun, 13:48, K1TTT wrote:



On Jun 2, 3:13*pm, Cecil Moore wrote:

On Jun 2, 9:31*am, K1TTT wrote:

easy, maxwell's equations don't predict standing waves! *they are a
product of *superposition and the simplest instrumentation used since
they were first discovered.

Please correct me if I am wrong: If one starts with the (superposed)
standing wave equation, V(x,t) = A*sin(kx)*sin(wt), Maxwell's
equations seem to provide a perfectly valid result (not that I can
recognize perfection). Therefore, how do Maxwell's equations prove
that the component traveling waves necessarily possess a separate
existence? Again, a serious question from an engineer who considers
anything except a logical '0' or logical '1' to be broken. :-)
--
73, Cecil, w5dxp.com

my differential calculus is a bit rusty, but i don't think that
equation satisfies the basic wave equation. *from Fields and Waves in
Communication Electronics, section 1.14. *a solution for the wave
equation in a transmission line (or other 1d form) must satisfy the
condition that the 2nd derivative wrt x and wrt t equal v-squared...
too hard to write in text. *but basically take the above and
differentiate twice wrt time and twice wrt distance and i don't think
you'll get anything that comes out to the velocity squared. *the
problem is that t and x must be part of the same sinusoid in the F(t-x/
v) so that the wave is truly traveling instead of stationary like your
equation provides.

in the basic maxwell equations in differential form you also run into
the same problem where the curl of the one field is proportional to
the time derivative of the other i think you'll end up with a
contradiction like k=w if you eliminate the dimensionless terms... i
may be wrong on that because its been so long since i've had to deal
with them on that level.

Oh!, I surrender *:) The thread do not converge to a solution, it
seems run out toward a naked singularity! *I stay tuned until this
good brainstorm calms down ..

Miguel

it will never converge, these discussions always go on until the
namecalling starts, then die an ugly death.

On Jun 2, 8:00*am, Cecil Moore wrote:
On Jun 2, 5:33*am, Keith Dysart wrote:

I suggest that you immediately dump any reference that includes
a phrase like "photon energy present in a wave".

If you (and others) will give up on the ridiculous concept of EM wave
energy standing still in standing waves, I will not have to refer to
photons again.

Refering to photons is just fine. Just do not mix them in with wave
theory.

Honor the technical fact that EM forward waves (with an
associated ExH energy) and EM reflected waves (with an associated ExH
energy) are always present when standing waves are present and that
those underlying waves (that cannot exist without energy) are moving
at the speed of light in the medium back and forth between impedance
discontinuities. Standing waves are somewhat of an illusion and
according to two of my reference books, do not deserve to be called
waves at all because standing waves do not transfer net energy as
required by the definition of "wave". In short, it is impossible for
EM waves to stand still.

Quoting one of my college textbooks, "Electrical Communication", by
Albert:

"Such a plot of voltage is usually referred to as a *voltage standing
wave* or as a *stationary wave*. Neither of these terms is
particularly descriptive of the phenomenon. A plot of effective values
of voltage, appearing as in Fig. 6(e), *is not a wave* in the usual
sense. However, the term "standing wave" is in widespread use."

From "College Physics", by Bueche and Hecht:

"These ... patterns are called *standing waves*, as compared to the
propagating waves considered above. They might better not be called
waves at all, since they do not transport energy and momentum."

All quite orthogonal to the original point, but your point about
standing
waves is quite correct, they are not really waves at all. But your
need
for the reality of underlying waves is quite excessive. The voltage
and
current distribution on a transmission line can be solved with a set
of differential equations which satisfy some boundary conditions.
There
is no mention of forward and reverse waves in this solution. Turns out
though, that the solution can also be factored in to a forward and
reflected wave and this technique will provide the same answer. It
does
not make these waves any more real.

I bring you back to a previous question which you have never
answered...
On an ideal line with 100% reflection, there are points where the
current
and voltage is always 0. Knowing that if either current or voltage is
0,
power is also 0, how does energy cross these point?

And if I cut the line at all the places where the current is zero, it
does not alter the energy distribution on the line one iota. How can
this be if energy is travelling from end to end on the line?

Technically, RF waves *are* light waves, just not *visible* light
waves. All the laws of physics that govern EM waves of light also
apply to RF waves.

PHYSICS has long given up on the idea of waves being an explanation
for light. The wave theory fails miserably when illumination levels
drop to the level that individual photons are being detected.

Though of course the earlier approximate models (waves) are still
useful
when intensity is high enough, just as we still use Newtonian
mechanics
to solve many every-day problems.

You might like to try http://vega.org.uk/video/subseries/8 for an
exposition on the strangeness of photon.

That you find it inconvenient for your "mashed-
potatoes" theory of energy arguments is not a good reason to abandon
the photonic nature of EM waves.

There you go again... mixing up you models. EM waves are analog and in
no way encompass the quantum nature of photons.

It is actually a good reason to keep
it in mind and abandon the mashed-potatoes energy arguments as human
conceptual constructs that cannot exist in reality. Most of the energy
in an EM wave is kinetic energy. Therefore, it cannot stand still.

There seems to be some misapprehension here. No one has claimed that
EM
waves stand still, though you may have been confused by the word
'standing' in 'standing waves'. But then earlier in your post you
quote
'College Physics' about 'standing waves', so it is not clear where
your confusion originates.

There is a wave theory of light, and there is a particle theory
of light, and these two theories do not play well together.

If they are both correct, they should play well together. If there is
any conflict, quantum electrodynamics wins the argument every time.

They are not both correct. QED aligns with many more observations than
does the wave theory. Another reason not to mix them.

While in many situations they will yield the same answers, it
is not permissible to mix the concepts from each. Distrust
the conclusions of any exposition which does so.

Actually, distrust the wave theory if it disagrees with QED. Quantum
ElectroDynamics has never been proven wrong.

And the wave theory does disagree with QED at low levels, while at
higher illumination levels QED agrees with the wave theory.

So feel free to prove that standing waves can exist without the
underlying component traveling waves traveling at the speed of light
in the medium. Feel free to prove that EM wave cancellation does not
"redistribute energy to areas that permit constructive interference"
as the FSU web page explains. Feel free to prove the Melles-Groit web
page wrong when they say such has been proven experimentally. In fact,
the interferometer experiment described here proves that reflected EM
waves, traveling at the speed of light, exist along with the necessary
energy. Take a look at the "non-standard output to screen".

And yet known of this aligns with the photons being part of the wave
theory of light. They two theories remain distinct.

http://www.teachspin.com/instruments...eriments.shtml

I, personally, am not interested in getting the right answer using the
wrong concepts.

Well, (and I am sorry, I can not resist), there is some evidence to
the
contrary. See:
http://www.w5dxp.com/energy.htm
http://www.w5dxp.com/nointfr.htm

....Keith

On Jun 2, 6:33*pm, Keith Dysart wrote:
The voltage and
current distribution on a transmission line can be solved with a set
of differential equations which satisfy some boundary conditions.

I am not interested in voltage and current. Like optical physicists, I
am only interested in tracking the RF energy flow.

It does not make these waves any more real.

Agreed, but the point is that it does not make the traveling waves any
LESS real!

I bring you back to a previous question which you have never
answered...
On an ideal line with 100% reflection, there are points where the
current
and voltage is always 0. Knowing that if either current or voltage is
0,
power is also 0, how does energy cross these point?

Good grief, I have answered that question at least a dozen times. The
net current being zero is just an illusion caused by superposition of
two magnetic fields propagating in opposite directions that are equal
in magnitude and opposite in phase. The two traveling waves keep on
trucking at their relatively constant ExH energy levels in opposite
directions. If you want to account for the energy, when the current is
zero, all of the energy existing at that point is in the electric
field and, sure enough, the voltage is at a maximum at that point-
duuuhhhh again. There is absolutely no point on an active transmission
line where the net energy level is zero. Again, you guys are never
going to convince anyone that the Golden Gate Bridge doesn't need
maintenance because the net traffic on the bridge is zero. How the
heck can thousands of vehicles traveling one direction while the same
number of vehicles are traveling in the opposite direction add up to
zero effect in reality? Please get real.

And if I cut the line at all the places where the current is zero, it
does not alter the energy distribution on the line one iota. How can
this be if energy is travelling from end to end on the line?

As I told you years ago, when you cut the line, you radically change
the impedance and create a reflection that didn't exist before you cut
the line. Hint: There are no reflections at a point of constant Z0.
There are only reflections at an impedance discontinuity. The
alteration of the energy distribution (to which you seem to be blind)
can easily be observed in a TV system.

PHYSICS has long given up on the idea of waves being an explanation
for light.

And RF waves are technically light, just not visible light. When I was
rummaging through the Texas A&M library, I found a book entitled,
"Light". It covered the subject of RF waves.

EM waves are analog and in
no way encompass the quantum nature of photons.

You could become famous if you can prove that EM waves are not
quantized. Good luck on that one.

No one has claimed that EM waves stand still, ...

Please prove your ridiculous assertion that no one has ever claimed
that EM waves can stand still.
--
73, Cecil, w5dxp.com

On Jun 2, 11:48*am, K1TTT wrote:
my differential calculus is a bit rusty, but i don't think that
equation satisfies the basic wave equation.

My calculus is probably a lot rustier than yours but it would be very
important for this discussion if Maxwell's equations do not work for
the standing wave equation. That would essentially prove that the
mashed-potatoes theory of transmission line energy is bogus.
--
73, Cecil, w5dxp.com

lu6etj wrote:

Oh!, I surrender :) The thread do not converge to a solution, it
seems run out toward a naked singularity! I stay tuned until this
good brainstorm calms down ..

Good day Miguel, you must be new here! I don't think any threads
converge to a solution. Mostly they do that singularity thing after
about the third message.

But hang in there, often much entertainment is to be had!

On Jun 2, 12:56*pm, lu6etj wrote:
Oh!, I surrender *:) The thread do not converge to a solution, ...

Miguel, do you fully realize the importance of Maxwell's equations
being nonfunctional with the equation for standing waves? It means
that any analysis based entirely on standing waves is probably invalid
in some area of the analysis - probably including Keith Dysart's
latest postings about the standing wave's zero current being
meaningful in some way.
--
73, Cecil, w5dxp.com