Standing morphing to travelling waves, and other stupid notions
 

Jim Kelley wrote:
Cecil Moore wrote:

Standing waves do not meet the requirements for an EM wave!

Here's a perfect example of what you're saying:

http://id.mind.net/~zona/mstm/physic.../StandingWaves

A standing wave, yet presumably not an electromagnetic wave! ;-)

ac6xg

Sorry. Truncated URL. There's more. The correct URL is:

http://id.mind.net/~zona/mstm/physic...ingWaves1.html

jk

Standing morphing to travelling waves, and other stupid notions
 

Cecil Moore wrote:
On Jan 14, 10:53 am, Gene Fuller wrote:
Is your position that
anyone who disagrees with you must be a spoiled brat?

The day you decided that you already know everything there is to know
is the day you started getting more and more ignorant and started
acting like a spoiled brat. Instead of screaming in anger that I am
wrong, I challenge you to produce the math that proves me wrong.

It is
not reasonable to substitute current and voltage for E-field and H-field
when considering the Poynting vector.

Please prove that strange assertion. Is the voltage not proportional
to the E-field? Is the current not proportional to the H-field? For
traveling waves, is not Z0 = V/I and proportional to E/H?

Go take a look at the
actual field configurations for a TEM wave, which includes standing
waves, and report back to us.

That's what I did already, Gene. You should take your own advice. It's
a little harder to visualize than voltage and current but follows
exactly the same concepts plus the right hand rule. Let's take two
equal coherent TEM plane waves forming standing waves in free space as
described by Hecht in "Optics". At one point the forward wave E-field
is at 135 degrees which puts the H-field at -135 degrees. The
reflected wave E-field is at 45 degrees which puts the H-field at -45
degrees.

Efor at 135 deg + Eref at 45 deg = |Efor|+|Eref| at 90 degrees

Hfor at -135 deg + Href at -45 deg = |Hfor|+|Href| at - 90 degrees

For standing waves, the H-field is 180 degrees offset from the E-
field. The power in the standing wave is ExH = E*H*sin(180) = ZERO.
Your assertion that the E-field and H-field are 90 degrees apart for
standing waves is simply FALSE! Please prove fact of physics that for
yourself.

Standing waves do not meet the requirements for an EM wave! Hecht was
right about that.

Feel free to do something besides wave your hands, mount ad hominem
attacks, and engage in obfuscation.
--
73, Cecil, w5dxp.com



c


Cecil,

Enjoy your superiority! When you finally figure out that phasor angles
(such as those used above) have no connection with the vector directions
considered in the Poynting analysis then you will have found even more
wisdom.

73,
Gene
W4SZ

Standing morphing to travelling waves, and other stupid notions
 

On Jan 14, 3:40 pm, Keith Dysart wrote:
So energy does move within the line, though no energy crosses a point
where the voltage or current is always 0.

Zero *NET* energy moves in a standing wave pattern whether it be at a
node or anywhere else. Forward energy and reflected energy moves
across a node and everywhere else because there are no reflections in
a homogeneous medium.

Please site a reference that says that reflections are possible in a
homogeneous medium. Until you do that, you are preaching religion, not
science.
--
73, Cecil, w5dxp.com

Standing morphing to travelling waves, and other stupid notions
 

On Jan 14, 3:09 pm, Jim Kelley wrote:
Cecil Moore wrote:
I'm glad you finally agree.

I'm glad you finally understand.

I've always known that a graph drawn with a pencil on a piece of paper
is not an EM standing wave but, for awhile, you seem to have forgotten
that simple fact. The envelope is also not the wave.
--
73, Cecil, w5dxp.com

Standing morphing to travelling waves, and other stupid notions
 

On Jan 14, 4:22 pm, Jim Kelley wrote:
A standing wave, yet presumably not an electromagnetic wave! ;-)

EM waves necessarily travel at the speed of light in a medium. Does a
standing wave travel at the speed of light? If your answer is 'yes',
where does it go? If your answer is 'no', it is not an EM wave.
--
73, Cecil, w5dxp.com

Standing morphing to travelling waves, and other stupid notions
 

On Jan 14, 8:10 pm, wrote:
Jim Kelley wrote:
Sorry. Truncated URL. There's more. The correct URL is:
http://id.mind.net/~zona/mstm/physic...es/standingWav...

Sorry, it didn't change and is truncated again. Maybe you should just
use two lines?
--
73, Cecil, w5dxp.com

Standing morphing to travelling waves, and other stupid notions
 

On Jan 14, 10:07 pm, Gene Fuller wrote:
Enjoy your superiority! When you finally figure out that phasor angles
(such as those used above) have no connection with the vector directions
considered in the Poynting analysis then you will have found even more
wisdom.

I challenged you to post the math, Gene, and you are apparently afraid
to do so. I'll bet you $100 that I am right and can prove it. I showed
that two plane waves in free space form standing waves with the total
E-field either 0 or 180 degrees apart from the H-field, contrary to
what you asserted. Please prove me wrong. If you cannot, please send
me $100.

If you would put aside your delusions of grandeur and simply sketch
those fields on a piece of paper, you would see for yourself that I am
right. When I get back home, I'll knock out a 3D graphic that will
leave no doubt.
--
73, Cecil. w5dxp.com

Standing morphing to travelling waves, and other stupid notions
 

Cecil Moore wrote:
On Jan 14, 10:07 pm, Gene Fuller wrote:
Enjoy your superiority! When you finally figure out that phasor angles
(such as those used above) have no connection with the vector directions
considered in the Poynting analysis then you will have found even more
wisdom.

I challenged you to post the math, Gene, and you are apparently afraid
to do so. I'll bet you $100 that I am right and can prove it. I showed
that two plane waves in free space form standing waves with the total
E-field either 0 or 180 degrees apart from the H-field, contrary to
what you asserted. Please prove me wrong. If you cannot, please send
me $100.

If you would put aside your delusions of grandeur and simply sketch
those fields on a piece of paper, you would see for yourself that I am
right. When I get back home, I'll knock out a 3D graphic that will
leave no doubt.
--
73, Cecil. w5dxp.com


Cecil,

This is truly bizarre. The E-field and the H-field are always at right
angles to each other in the plane wave case. They are also perpendicular
to the propagation direction. The same is true for the TEM mode in a
coaxial waveguide, which is the usual case discussed on RRAA. There is
no 0 or 180 degree involvement. I don't know what you are so confused
about, but you really need to rethink what you are saying.

There is no need to sketch or calculate anything. A diagram showing the
relationship of the E-field vector and the H-field vector is in every
E&M and optics book I have ever seen. I presume that Hecht has the same
diagram.

73,
Gene
W4SZ

Standing morphing to travelling waves, and other stupid notions
 

Gene Fuller wrote:

...
There is no need to sketch or calculate anything. A diagram showing the
relationship of the E-field vector and the H-field vector is in every
E&M and optics book I have ever seen. I presume that Hecht has the same
diagram.

73,
Gene
W4SZ

Gesus, e fields/h fields at right angles to each other? These are
things which make our graphs make "sense" to us--as far as being "real",
gawd--who knows?

IDIOTS ALL!

Regards,
JS

Standing morphing to travelling waves, and other stupid notions
 

On Jan 14, 7:19 pm, Roger Sparks wrote:
On Mon, 14 Jan 2008 12:40:39 -0800 (PST)

Keith Dysart wrote:

snip...............

As you say, the energy moves between the E-field and the H-field,
but the locations of maximum energy along the line for each of these
fields is different, so the energy changes position on the line with
each cycle. The energy at any point on the line is not constant.

E-field energy will peak at the voltage maximums.
H-field energy will peak at the current maximums.
These are at different places (90 degrees apart).
So energy does move within the line, though no energy crosses a point
where the voltage or current is always 0.

I can understand no energy crossing a zero current point, but how do you justify no energy crossing a zero voltage point when current IS observed?

Well I start with P = V * I, so whenever the current or the voltage
is zero, there is no power. Specifically, V and I are measured at
the terminals of a network and P will be the power flowing into or
out of the network.

In the case under discussion, there are two networks, one to
the left of the point on the line and one to the right and
we are measuring the power flowing between these two networks.

For an example of current without power, consider a loop of
superconductor with a current flowing in it. No voltage, no
power, but there is current.

Current is defined as movement of charges, and charges have energy by definition (how can they be charges without energy?).

Consider an object flying through space. No work is being done
(and therefore there is no power), but the object still has
kinetic energy.

Another point, the current is observed to change directions during the cycles, polarity also changes on each side of the zero voltage point. Where might the polarized energy come from if it does not cross the zero voltage point?

A thought experiment I have found useful is to consider a simple
resonant circuit made of an ideal capacitor and inductor. Charge
the capacitor to 10 volts and then connect the inductor. A sinusoidal
voltage and current will appear in the circuit.

Just as the inductor is connected:
- all the energy is stored in the capacitor
- the voltage on the capacitor is maximum
- there is no current in the inductor
After connecting the inductor:
- energy starts to transfer to the inductor
- the voltage on the capacitor is dropping
- the current in the inductor is increasing
Some time later:
- the voltage on the capacitor is 0
- the current in the inductor is maximum
- there is no energy stored in the capacitor
- all the energy is stored in the inductor
- no energy is moving from the capacitor to
the inductor
But the inductor insists that current continue
to flow:
- the capacitor begins to charge with a negarive
voltage
- energy begins to transfer from the inductor
back to the capacitor (note the change in the
direction of energy flow)
- the voltage on the capacitor is increasing
negatively
- the current in the inductor is dropping
Sometime later:
- the current in the inductor has dropped to
zero
- the capacitor has a maximum negative voltage
- all the energy is in the capacitor
And this continues forever at the resonant
frequency of the capacitor and inductor circuit.

But no energy is moving from the capacitor to
the inductor when the voltage on the capacitor
is zero and the current in the inductor is
maximum. It is at these times that the direction
of energy flow is changing, as well as when the
voltage in the capacitor is maximum and the
current is zero.

I can kinda see how like charges could repell so that waves of like polarity might "bounce" but I can't see how waves of opposite polarity might "bounce". If waves of opposite polarity "bounced", why would the polarity change during the cycle on each side of the "bounce" point?

An excellent counter-example. I may have fallen into
the trap of looking at the examples that support the
argument rather than looking for the ones that don't.
This will take some cogitating. Maybe its the end
of the line for the "bounce hypothesis".

To me, it is much more rewarding to work with traveling waves that pass through one another, interacting to create standing waves.

I don't object to this view, as long as the waves are
viewed as having voltages or currents but no power.

The difficulty is that some waves definitely transport
energy while others do not and I do not see a good
explanation for what turns the former into the latter,
as happens, for example, when the pulses collide.

Would it help your visualization process to observe that when two waves of SAME POLARITY but traveling in opposite directions cross, the currents accompaning the waves are moving in opposite directions both before and after crossing? When two waves of OPPOSITE POLARITY but traveling in opposite directions cross, the currents are moving in the same direction both before and after crossing.

I do understand (at least I think I do) the methodology
for superposing the waves of voltage and current,
computing the results and deriving the power from the
result.

I just am not happy that this results in waves sometimes
transporting energy and sometimes not, without a good
explanation of the transition.

If we were talking about water, water behind a dam is like voltage, with the height of the water the potential energy, measured in head (feet), or PSI if measured at the bottom of the dam. A pipe to the bottom of the dam will squirt water at a high velocity but no head or PSI. The potential energy of the water behind the dam has been converted to kinetic energy measued in velocity of a moving mass. The moving water can be stopped, and if carefully done, the static head reached by stopping the water will nearly reach the original water level behind the dam. It would reach the same level if it were not for friction losses. Electrical current is something like that moving water.

Agreed. I have used this analogy as an aid to understaning
though it becomes challenging at RF.

....Keith