Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:

On Jan 3, 1:07am, Mike Monett wrote:

Keith, your model is not realistic. As you know, any signal you
impose on a conductor will form an electromagnetic wave. This is
the combination of electrostatic and electromagnetic fields, and
it propagates at the normal velocity for that medium.

However, electromagnetic waves do not interact with each other,
and they cannot bounce off each other.

That is the standard description, but it seems to have some
weaknesses.

No, there are no weaknesses. Maxwell's equations have stood the test
of time.

Recall that light from stars is electromagnetic. It tra vels many
light-years before it reaches your eyes. If electromagnetic waves
interacted, you would not be able to see individual stars they
would merge into a blur.

This would seem to me to depend on the nature of the interaction.
Clearly the interaction represented by the term "bounce" (for lack
of a better word) would have to be such as to not violate any of
these observed behaviours.

The term "bounce" means they interact. Electromagnetic signals do
not interact. They superimpose. Each is completely unaware and
unaffected by the other.

Similarly, the signals reaching your antenna and traveling down
the coax to your receiver do not interact with each other. As
long as your receiver is not overloaded, the signals remain sep
arate no matter how many stations are on the air at the moment.

So the statement that like charges repel does not apply to
electromagnetic waves,

Q1. Are you saying that it is inappropriate to view a transmission
line as distributed capacitance and inductance and analyze its
behaviour using charge stored in the capacitance and moving in the
inducatance?

That is not what you are saying. You are ignoring the magnetic
field.

If such analysis is appropriate, then it seems to me that a pulse
can be viewed as a chunk of charge moving down the line.

Q2. Is this an appropriate view?

No. You need to include the associated magnetic field.

Q3. If so, then what happens when two such chunks of charge
collide in the middle of the line?

Electromagnetic signals do not collide. They superimpose.

The existing analysis techniques tell us that no current ever
flows at the mid-point of the line, this means no charge crosses
the mid-point.

Q4. Is this correct?

That statement has no meaning.

Q5. If no charge crosses the mid-point, then how do the pulses,
made up of chunks of charge. pass the mid-point?

The pulses are not chunks of charge. They are the combination of
electrostatic and electromagnetic fields. You cannot separate the
two.

Q6. If they do not pass the mid-point, then what happens to them?

That statement has no meaning.

I have offerred a somewhat intuitive explanation.

Your explanation does not work.

Other explanations are welcome.

Any explanation that does not involve charge will immediately
cause me to ask Q1 again.

Please study Maxwell's equations and how they are derived.

Keith

Regards,

Mike Monett

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
When the pulses are not identical, the energy that crosses
the point is exactly sufficient to turn one pulse
into the other. The remainder of the energy must bounce
because it does not cross the mid-point.

All you have proved is that you cannot tell one photon
from another. Your whole charge repulsion argument
falls apart when dealing with photons (which constitute
EM waves). I suggest you study and discover what is
possible with photons and what is not possible. You
seem to be concentrating on the carriers of the waves
rather than the EM waves themselves. Photons do NOT
and cannot bounce off of each other under ordinary
circumstances. You are simply illustrating the limitations
of ignoring the basic physics of the situation and wasting
a lot of time and effort in the process.

I have sat on a cliff overlooking the Pacific Ocean
at Fitzgerald's Marine Reserve north of Santa Cruz, CA
and have seen waves rolling in, reflecting off the beach,
and rolling back out to sea. Those waves pass through
each other as if the other wasn't there. The wave energy
is moving in both directions. The H2O carriers move
hardly at all. You can argue that the energy in the
waves is equal and therefore no average energy is being
transferred, but I still see the waves with people
riding on those waves. I do not see waves bouncing off
of each other although one could, as you have, delude
oneself into creating a mental illusion of such.

When I look out into my back yard, I am seeing reflections.
If there were a thousand people here, they would all be
seeing different reflections all passing through each other.
Photonic waves pass through each other unimpeded. It would
be a weird looking world if they bounced off each other.

In a wire, photons do not bounce off each other. However,
superposition can cause a redistribution of photon energy
at an impedance discontinuity. We call that redistribution
of energy a "reflections".
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On 3 Jan, 07:38, Keith Dysart wrote:
On Jan 3, 1:07*am, Mike Monett wrote:

* Keith, your *model *is not realistic. As you *know, *any *signal you
* impose on a conductor will form an electromagnetic wave. This is the
* combination of *electrostatic *and *electromagnetic *fields, *and it
* propagates at the normal velocity for that medium.

* However, electromagnetic waves do not interact with each *other, and
* they cannot bounce off each other.

That is the standard description, but it seems to have
some weaknesses.

* Recall that *light *from stars is *electromagnetic. *It *travels many
* light-years before *it reaches your eyes. *If *electromagnetic waves
* interacted, you *would *not be able to see individual *stars *- they
* would merge into a blur.

This would seem to me to depend on the nature of the
interaction. Clearly the interaction represented by
the term "bounce" (for lack of a better word) would
have to *be such as to not violate any of these
observed behaviours.

* Similarly, the signals reaching your antenna and traveling *down the
* coax to *your receiver do not interact with each other. *As *long as
* your receiver *is *not overloaded, the *signals *remain *separate no
* matter how many stations are on the air at the moment.

* So the *statement *that *like * charges * repel *does *not *apply to
* electromagnetic waves,

Q1. Are you saying that it is inappropriate to view
a transmission line as distributed capacitance and
inductance and analyze its behaviour using charge
stored in the capacitance and moving in the
inducatance?

If such analysis is appropriate, then it seems
to me that a pulse can be viewed as a chunk of
charge moving down the line.

Q2. Is this an appropriate view?

Q3. If so, then what happens when two such chunks
of charge collide in the middle of the line?

The existing analysis techniques tell us that
no current ever flows at the mid-point of the
line, this means no charge crosses the mid-point.

Q4. Is this correct?

Q5. If no charge crosses the mid-point, then how
do the pulses, made up of chunks of charge.
pass the mid-point?

Q6. If they do not pass the mid-point, then what
happens to them?

I have offerred a somewhat intuitive explanation.
Other explanations are welcome.

Any explanation that does not involve charge will
immediately cause me to ask Q1 again.

...Keith

The above is correct omly when the mid point you are refering to
is at the point where the time to complete half a peried is reached.
On a half period length that is at the end of a half wave radiator
the return path is the second half.The ant6enna only radiates
when the current travels on the surface towards the inductance
when the capacitors are discharging
Physical radiator length is half a cycle or period because half
the time it is travelling forward and the rest of the time backwards.
Hams must stop equating antenna length with a full wave length,
it equates to half a wave length. It is obvious then that the
completion of a cycle thus at no time has current moving
other than a single direction. Until hams get this distinction
into their heads they will never understand radiation.
Art

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Still can't find one? ?

I haven't looked for one because I don't want to waste
my time. If there was a reference, Mr. Maxwell or Dr.
Bruene would have reported it by now but their argument
continues to rage, just like yours.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 3, 11:01*am, Mike Monett wrote:
* Keith Dysart wrote:

* On Jan 3, 1:07am, Mike Monett wrote:

* Keith, your *model is not realistic. As you know, any *signal you
* impose on a conductor will form an electromagnetic wave. *This is
* the combination of electrostatic and electromagnetic *fields, and
* it propagates at the normal velocity for that medium.

* However, electromagnetic *waves do not interact with *each other,
* and they cannot bounce off each other.

* That is *the *standard *description, but *it *seems *to *have some
* weaknesses.

* No, there are no weaknesses. Maxwell's equations have stood the test
* of time.

* Recall that light from stars is electromagnetic. It tra vels many
* light-years before it reaches your eyes. If electromagnetic waves
* interacted, you *would not be able to see *individual *stars they
* would merge into a blur.

* This would seem to me to depend on the nature of *the interaction.
* Clearly the interaction represented by the term "bounce" (for lack
* of a *better word) would have to be such as to not violate *any of
* these observed behaviours.

* The term *"bounce" means they interact. *Electromagnetic *signals do
* not interact. *They *superimpose. *Each *is *completely *unaware and
* unaffected by the other.

* Similarly, the *signals reaching your antenna and *traveling down
* the coax *to *your receiver do not interact with *each *other. As
* long as *your receiver is not overloaded, the signals *remain sep
* arate no matter how many stations are on the air at the moment.

* So the *statement *that *like charges *repel *does *not *apply to
* electromagnetic waves,

* Q1. Are you saying that it is inappropriate to view a transmission
* line as *distributed *capacitance and inductance *and *analyze its
* behaviour using charge stored in the capacitance and moving in the
* inducatance?

* That is *not *what *you are saying. You *are *ignoring *the magnetic
* field.

* If such analysis is appropriate, then it seems to me that *a pulse
* can be viewed as a chunk of charge moving down the line.

* Q2. Is this an appropriate view?

* No. You need to include the associated magnetic field.

* Q3. If *so, *then *what happens when *two *such *chunks *of charge
* collide in the middle of the line?

* Electromagnetic signals do not collide. They superimpose.

* The existing *analysis *techniques tell us *that *no *current ever
* flows at *the mid-point of the line, this means no *charge crosses
* the mid-point.

* Q4. Is this correct?

* That statement has no meaning.

* Q5. If *no charge crosses the mid-point, then how *do *the pulses,
* made up of chunks of charge. pass the mid-point?

* The pulses *are *not chunks of charge. They are *the *combination of
* electrostatic and *electromagnetic fields. You *cannot *separate the
* two.

* Q6. If they do not pass the mid-point, then what happens to them?

* That statement has no meaning.

* I have offerred a somewhat intuitive explanation.

* Your explanation does not work.

* Other explanations are welcome.

* Any explanation *that *does not *involve *charge *will immediately
* cause me to ask Q1 again.

* Please study Maxwell's equations and how they are derived.

* Keith

* Regards,

* Mike Monett

You did not directly answer Q1, but I take if from all
the other responses that you are saying the answer is
"no, it is not appropriate to view a transmission
line as distributed capacitance and inductance and
analyze its behaviour using charge stored in the
capacitance and moving in the inducatance?"

Taking this invalidates all the subsequent questions
since they are based on the premise that this kind
of analysis is appropriate.

Or have I misinterpreted and your only concern with
Q1 was that I did not mention that energy is stored
in the electric field created by charge in the
capacitance and the magnetic created by charge
flowing in the inductance?

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Mike Monett wrote:
However, electromagnetic waves do not interact with each other, and
they cannot bounce off each other.

Mike, can we add "do not interact with each other" in
*an unchanging medium*? Coherent waves can apparently
interact at a change in mediums (impedance discontinuity).

One of the s-parameter equations illustrates that apparent
fact.

b1 = s11*a1 + s12*a2 = 0

Assuming a1 and a2 are non-zero forward and non-zero reflected
voltages, the only way they could superpose to zero is to
interact at the impedance discontinuity.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Mike Monett wrote:

...
The term "bounce" means they interact. Electromagnetic signals do
not interact. They superimpose. Each is completely unaware and
unaffected by the other.
...
Regards,

Mike Monett

EM fields act that same as static magnetic fields.

Why not just get some iron filings, a paper and a couple of magnets?

Move the magnets about below the paper with the iron filings above and
actually get a visual on some magnetic fields and how they react to each
other?

I like things simple ... then the math can follow ...

Regards,
JS

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
On Jan 2, 2:48 pm, Cecil Moore wrote:
Keith Dysart wrote:
I assume that you have not provided a reference to support
this assertion because you have not been able to find one.
I provided the reference a number of times and you
chose to ignore it. The reference is the chapter on
interference in "Optics", by Hecht.

I am suprised that a book on optics would discuss the
output impedance of Thevenin equivalent circuits.

The "Optics", by Hecht reference is for destructive
and constructive interference, not Thevenin equivalent
circuits, but your attempt to confuse everyone is noted.
"Thevenin" is not even in the index of "Optics" so your
attempted diversion is ridiculous on the face of it.

I will repeat an earlier posting that you conveniently
chose to ignore. If we measure the forward power and
reflected power with a Bird wattmeter at the output of
your source during steady-state, it will tell you that:

forward power = reflected power

From that we can calculate the reflection coefficient.

rho = SQRT(Pref/Pfor) = plus or minus 1.0
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 3, 11:10*am, Cecil Moore wrote:
Keith Dysart wrote:
When the pulses are not identical, the energy that crosses
the point is exactly sufficient to turn one pulse
into the other. The remainder of the energy must bounce
because it does not cross the mid-point.

All you have proved is that you cannot tell one photon
from another. Your whole charge repulsion argument
falls apart when dealing with photons (which constitute
EM waves). I suggest you study and discover what is
possible with photons and what is not possible.

Can photons explain the state of a transmission
line driven with a step function after the line
has settled to a constant voltage?

If not, there would seem to be some difficulty
with the applicability.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 3, 11:13*am, Cecil Moore wrote:
Keith Dysart wrote:
Still can't find one? ?

I haven't looked for one because I don't want to waste
my time. If there was a reference, Mr. Maxwell or Dr.
Bruene would have reported it by now but their argument
continues to rage, just like yours.

Excellent. So there is NO reference that claims
that the output impedance can not be used to
compute the reflection coefficient.

There are many references that do.

So that settles it, then. Many references on one side,
none on the other. It is time for you to accept the
standard methodology for computing the reflection
coefficient at a generator. And no, I am not holding
my breath while I wait.

And the arguments that I have seen between Mr. Maxwell
and Dr. Bruene are on a completely different matter.

...Keith