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eHam Forums => Elmers => Topic started by: W5KKC on November 11, 2013, 07:07:25 PM



Title: current and voltage along a non resonant dipole
Post by: W5KKC on November 11, 2013, 07:07:25 PM
When  dipole is resonant the voltage is highest at the ends and current is highest as the middle.
But what happens when the dipole is not resonant, say for example and OCFD or a random wire at any frequency where it is not a half wavelength?  Where is the max current and what is the voltage at the ends?

tks
Dennis


Title: RE: current and voltage along a non resonant dipole
Post by: WD4ELG on November 11, 2013, 09:05:54 PM
It depends upon the length of the wire and the frequency of operation.  EZNEC can give you a good detailed answer.


Title: RE: current and voltage along a non resonant dipole
Post by: W5KKC on November 11, 2013, 09:30:00 PM
I have heard it said that the ends have to be low or no current and high voltage.  How can that be when the antenna is not 1/2 wavelength.


Title: RE: current and voltage along a non resonant dipole
Post by: WB6BYU on November 11, 2013, 09:36:14 PM
Highest voltage is always at the ends, and current is minimum, because there is no
place for current to flow to.

To analyze the current flowing on a wire, start at the ends and mark it as a voltage
maximum.  Go 1/4 wavelength in from the end and that will be a voltage minimum
and current maximum.  Another wave puts you back at a voltage maximum, etc.

At adjacent current maxima the currents will be 180 degrees out of phase.

So, for example, consider the current on a 1 wavelength wire:  voltage will
be maximum at the ends, currents maximum at 1/4 wave in from each end,
and voltage maximum at the center.  The currents in the two half wave
sections are out of phase.  I often find it easiest to draw a picture of the
antenna and mark it, for example, with dots at the voltage maxima and
arrows denoting direction (phase) at the current maxima:  in the case
of the 1 wavelength wire there would be a dot in the middle and one at
each end, and between the dots one arrow would point right and the
other left.

Then from looking at the paper (broadside to the antenna, such that the
arrows are equidistant from you) you would see equal numbers of arrows
pointing right and left, the radiation from which would cancel each other
in your direction, so there is a null in the pattern broadside to the antenna.

The feedpoint does make a difference, however:  if fed in the center such
a wire would have maximum radiation broadside to the antenna, while
fed 1/4 wave from the end you would have an "X" pattern with a null
broadside to the wire.  How to tell the difference?  Start by marking
the voltage and current nodes along the wire from each end towards
the feedpoint.  At the feedpoint you know that current flowing out
of one side of the feedpoint flows into the other side at the same
phase.  This means there is no phase shift across the feedpoint, whether
it is at a current maximum (low impedance) or voltage maximum (high
impedance, where otherwise there would be a phase reversal in the
currents.)

Once you draw out the currents on a few wires (especially those several
half wavelengths long) then it becomes easier to imagine the current
distributions on them.


The same approach works with a closed loop antenna, but in that case you
start from the midpoint of the loop at the far side from the feedpoint with
maximum current and work back around the loop from there.


Title: RE: current and voltage along a non resonant dipole
Post by: KA4POL on November 11, 2013, 09:48:00 PM
Similar to DC resistance you need to consider impedance. Where there is a low impedance you have a higher current. At the antenna ends you find high impedance and therefore no current. Keep in mind that the impedance is not constant over the complete length of the antenna.


Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 12, 2013, 05:03:50 AM
A dipole is a standing wave antenna. The SWR on a typical 1/2WL dipole is in the ballpark of 20:1 at the feedpoint. The SWR is nearly infinite at the tip ends of the dipole. For a pure standing wave, the voltage envelope is always 90 degrees out of phase with the current envelope, i.e. where a voltage maximum exists, a current minimum will exist and vice versa. The maximum points are called "loops" and the minimum points are called "nodes". The voltage and current distribution on a dipole is essentially the same pattern as the voltage and current distribution on an open-circuit transmission line stub.


Title: RE: current and voltage along a non resonant dipole
Post by: G3RZP on November 12, 2013, 06:28:42 AM
Cecil,

Don't you mean 180 degrees out of phase, not 90?


Title: RE: current and voltage along a non resonant dipole
Post by: WB6BYU on November 12, 2013, 06:38:51 AM
No, 90 degrees is correct.

Current is maximum when voltage is minimum.  1/4 wave down the line the current is minimum while
the voltage is maximum.

So the current is offset from the voltage by 1/4 of a cycle, or 90 degrees.


Title: RE: current and voltage along a non resonant dipole
Post by: W5KKC on November 12, 2013, 06:50:34 AM
But what about when the dipole is not a multiple of 0.5 wavelengths because it is NOT resonant? What would be the current distribution when the dipole is 3/4 wavelength? How can both ends be a high voltage? If one end is a high voltage, the other end would need to be ZERO volts because it is 3/4 wavelength away?

tks
Dennis


Title: RE: current and voltage along a non resonant dipole
Post by: WB6BYU on November 12, 2013, 08:29:31 AM
Quote from: AE5EK

But what about when the dipole is not a multiple of 0.5 wavelengths because it is NOT resonant? What would be the current distribution when the dipole is 3/4 wavelength? How can both ends be a high voltage? If one end is a high voltage, the other end would need to be ZERO volts because it is 3/4 wavelength away?



But if you have any current flowing at the end of the wire, where is it flowing from/to?
If you don't have current flowing, then the impedance is high.

(Or, if you look at it another way, all the current flowing outwards at that point is reflected
from the open circuit at the end of the wire so it flows back the opposite direction.  The
net current is zero.)


The answer to your conundrum is what happens at the feedpoint.

But let's look at the antenna first.  Starting at the right end, we have high impedance at
the end, low impedance (maximum current) 1/4 wave in from the end, and 1/8 wave
later we hit the feedpoint at a point in the cycle where both the voltage and current
are at intermediate values.  If we repeat the process from the left end, we have the
same current distribution, and end at the same intermediate point in the cycle.

If we were to plot the voltage distribution we'd see it high at the ends, dropping to zero
1/4 wave in from the end, then rising part way back up to the maximum at the feedpoint.
The current would be minimum at the ends, maximum 1/4 wave back, then dropping back
somewhat by the time you hit the feedpoint.

What happens at the feedpoint?  We know that the current has to be in phase across
the feedpoint, so if we are plotting the current along the wire we can make the
sections in phase on each side of it.  Since there is no point of minimum current
between the feedpoint and the end of the wire (which is where the current changes
phase along the wire) then all the current on the wire will be in the same phase.


Basically, the fact that the current must be 0 at the open end of a wire antenna, and
the sine / cosine distribution along the wire, force a particular condition at the feedpoint
regardless of the length of the wire.  That's where you have any discontinuity in the
voltage or current distributions, and that's why the impedance can vary wildly in the
center of a doublet as you change frequency.  If the feedpoint is at one end of the
wire rather than in the middle the current distribution will be different, but the same
rules apply.  (Except that you wouldn't say that the current is zero there because
it isn't a free end of the wire - it is connected to something, namely the feedline.)


Title: RE: current and voltage along a non resonant dipole
Post by: W5WSS on November 12, 2013, 09:48:21 AM
Dale, you are very good at technical writings :)

Appreciate you!

73 :)


Title: RE: current and voltage along a non resonant dipole
Post by: G3TXQ on November 12, 2013, 10:54:18 AM
On a related issue ......

Don't get confused between: on the one hand the phase relationship between the time-varying voltage and current at a particular point on the wire; and on the other hand the phase relationship between the voltage and current standing waves along the wire - they are two quite different things!

For example, we know at any point along a resonant half-wave dipole the time-varying voltage and current would be in-phase with each other; at the same time we know there is a 90 degree phase shift between the voltage and current standing wave patterns along that wire. They are different things, and it's easy to confuse the two.

Steve G3TXQ


Title: RE: current and voltage along a non resonant dipole
Post by: G3RZP on November 12, 2013, 02:42:21 PM
Right Steve, because if the feed voltage and feed current at the feed point have a 90 degree phase difference, the impedance must be a pure reactance - which, by definition, is not resonance.


Title: RE: current and voltage along a non resonant dipole
Post by: KC8Y on November 12, 2013, 02:46:45 PM
AE5EK:

I use an OCF dipole, that seems to work on 80/75, 40, 20, 17, 12, 10 & 6-meters; also just use end-fed 1/2 wave for 30-meters.  Have a tuner that does fine tuning for each band.  I never go above 50-watts & operate digital modes.

Ken KC8Y


Title: RE: current and voltage along a non resonant dipole
Post by: WB6BYU on November 12, 2013, 02:58:49 PM
A further source of confusion is between the magnitude and polarity of the standing
wave distribution along the wire, where there are points of minimum and maximum
voltage and current, and the actual plot of voltage vs. time at any single point, where
the voltage (and current) will show the normal alternation between positive and negative
values over the course of a single cycle of RF.


This is a confusing and difficult concept for a lot of people, especially with all
of the potential misunderstandings such as those that we have noted.  Take your
time, read through the various resources again, try plotting some distributions on
a drawing of an antenna, or whatever works best for your preferred learning style.

And kudos to you, Dennis, for asking questions and jumping in to try to understand
it.  Keep asking questions until you get it:  at that point things will make much
better sense, though it may not feel like that while you're still muddling in the middle.


And your reward (or, perhaps, penance might be a better term) will be then
to explain it to others...


Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 12, 2013, 06:19:35 PM
What would be the current distribution when the dipole is 3/4 wavelength? If one end is a high voltage, the other end would need to be ZERO volts because it is 3/4 wavelength away?

That would be true for a straight wire but the feedpoint provides a phase shift that makes the current zero at both ends. Note that the current maximums are 1/4WL from each end. Here is the current pattern according to EZNEC:

(http://www.w5dxp.com/270dip.JPG)

This is a confusing and difficult concept for a lot of people, especially with all of the potential misunderstandings such as those that we have noted.

Many years ago on rec.radio.amateur.antenna, both W7EL and W8JI proved that they didn't understand the difference between standing wave current and traveling wave current. That prompted me to write the following article which explains the mathematical difference between the equation for standing wave current and the equation for traveling wave current:

http://w5dxp.com/current2.htm

The main point was that the phase of standing wave current on a standing wave antenna cannot be used to measure the velocity of propagation through a wire or through a 75m mobile loading coil as both W7EL and W8JI described and defended for their loading coil measurements.


Title: RE: current and voltage along a non resonant dipole
Post by: W5KKC on November 12, 2013, 09:47:16 PM
This is great stuff and I appreciate your willingness to explain
I found this great graphic of standing waves.  I had never considered the difference between standing waves and the reflected waves on an antenna.
http://www.youtube.com/watch?v=ic73oZoqr70

Dennis


Title: RE: current and voltage along a non resonant dipole
Post by: W5KKC on November 12, 2013, 10:08:10 PM
So it would seem that on any wire length with open ends that the voltage and current are 90 degrees out of phase. Is this correct? 

tks
Dennis


Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 13, 2013, 10:28:11 AM
So it would seem that on any wire length with open ends that the voltage and current are 90 degrees out of phase. Is this correct?

Note that my previous postings contained the words, "pure standing wave". A pure standing wave only exists when the amplitude of the forward wave is exactly equal to the amplitude of the reflected wave. That almost never happens in the real world so almost all real-world waves are a combination of traveling waves and standing waves. A standing wave is a human mathematical construct that doesn't meet the definition of an "EM wave" because it carries no energy, no momentum, and doesn't move at the speed of light in the medium. Standing waves cannot exist without the forward and reflected traveling waves which do meet the definition of "EM waves".

"90 degrees out of phase" is only true for pure standing waves which exist only at the open-circuit ends of the dipole wire because the reflected wave is the same magnitude as the forward wave and the SWR on the standing wave antenna is (nearly) infinite at the ends of the wire. At the feedpoint of a 1/2WL dipole, the SWR on the standing wave antenna is in the ballpark of 20:1 which means that the forward power is about 20% greater than the reflected power on the antenna at the feedpoint. So about 80% of the energy is contained in the standing waves and is not radiated (until key up). The 20% of the energy that is radiated during steady-state key-down is contained in the traveling waves. On a standing wave antenna, there is (almost) always a mixture of traveling waves and standing waves which means that the angle between the total voltage and total current cannot be 90 degrees except at the open-circuits at the tip ends of the dipole.

This subject is easier to understand if one first uses lossless transmission lines as the example which eliminates the radiated power. Then stepping up to lossy transmission lines allows the radiated power from an antenna to be considered as just another loss.


Title: RE: current and voltage along a non resonant dipole
Post by: WS3N on November 13, 2013, 08:50:58 PM
A standing wave is a human mathematical construct that doesn't meet the definition of an "EM wave" because it carries no energy, no momentum, and doesn't move at the speed of light in the medium. Standing waves cannot exist without the forward and reflected traveling waves which do meet the definition of "EM waves".

Nonsense. Why is a standing wave a mathematical construct while a traveling wave is not? Standing waves are perfectly good solutions of Maxwell's equations and are required to match certain boundary conditions. Their instantaneous energy density, energy flux and momentum flux are non-zero, while the time-averaged energy density is non-zero and the time-averaged fluxes are zero.

One can express standing waves as linear combinations of traveling waves or the other way round. The physical field is what it is and one is free to choose any decomposition that is convenient.


Title: RE: current and voltage along a non resonant dipole
Post by: RFRY on November 14, 2013, 04:55:49 AM
At the feedpoint of a 1/2WL dipole, the SWR on the standing wave antenna is in the ballpark of 20:1 which means that the forward power is about 20% greater than the reflected power on the antenna at the feedpoint. So about 80% of the energy is contained in the standing waves and is not radiated (until key up).The 20% of the energy that is radiated during steady-state key-down is contained in the traveling waves.

Just to note that the maximum, free-space, far field radiated by a center-fed, 1/2-wave dipole driven with 1000 watts of Z-matched power is about 221.7 mV/m at a distance of 1 km - which is the value predicted by theory for virtually 100% radiation efficiency of that dipole.

This steady-state (key down) performance has been measured and proven for U.S. FM/TV broadcast antenna systems, whose radiated fields must meet a certain minimum value acceptable to the FCC.



Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 14, 2013, 05:47:16 AM
Why is a standing wave a mathematical construct while a traveling wave is not?

Well for one thing, a standing wave doesn't meet the definition of a "wave". Here are a couple of quotes:

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."

In short, the physics definition of an EM wave requires that it transport energy and momentum and travel at the speed of light in the medium. A standing wave doesn't satisfy those terms of definition.

When we talk about standing waves, we have not broken things down to an elementary level. A pure standing wave consists of two equal amplitude coherent traveling waves propagating in opposite directions at the speed of light in the medium. They are commonly referred to as the forward wave and reflected wave and they indeed do satisfy the definition of EM waves. Since the standing wave cannot exist without the two elementary traveling waves and since it doesn't satisfy the definition of an EM wave at all and since it can be separated into its two elementary components, that's why I call it a mathematical construct.

Proof that a standing wave is not traveling at the speed of light in the medium is easy. Simply measure the phase shift from the feedpoint of a 64 ft 40m dipole to a point halfway to the tip end, i.e. 16 ft from the feedpoint. At the speed of light in the medium (about 0.94 ft per nanosecond) it would take about 15 ns to travel that 45 degrees of dipole. If we measure the total current group delay, we get about 2 degrees which is obviously unrelated to the speed of light. 16 ft in 0.7 ns is faster than the speed of light. Is the standing wave really traveling faster than the speed of light? The answer to that question is why I call it a mathematical construct because it fails to satisfy the laws of physics.

Indeed, between the nodes of a pure standing wave, there is zero phase shift. That's why a measurement of the phase shift through a loading coil in a standing wave antenna will never yield the propagation delay through the coil. A 3ns propagation delay through an 80m Texas Bugcatcher coil is impossible


Title: RE: current and voltage along a non resonant dipole
Post by: G8HQP on November 14, 2013, 06:33:52 AM
Quote from: W5DXP
Proof that a standing wave is not traveling at the speed of light in the medium is easy.
. . . and quite unnecessary. The hint is in the name: 'standing wave'. It is not travelling at all.

Any linear superposition of Maxwell solutions is itself a Maxwell solution; just as physical as any other. Any argument must be about what it is called, not what it is.


Title: RE: current and voltage along a non resonant dipole
Post by: WS3N on November 14, 2013, 06:35:08 AM
I see, it's the combination of "standing" and "wave" that you dislike. Call it a mode of oscillation or whatever you like, it's still the same thing.

Your example of the antenna, demonstrating the fallacy of faster-than-light propagation, is a straw man. You're the only one who suggested such a thing, immediately before calling it impossible.




sin(kx - wt) = sin(kx) cos(wt) - cos(kx) sin(wt)

Hmmm, looks like a pure traveling wave consists of two equal-amplitude standing waves, 90 degrees out of phase.  ;)


Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 14, 2013, 06:45:17 AM
Just to note that the maximum, free-space, far field radiated by a center-fed, 1/2-wave dipole driven with 1000 watts of Z-matched power is about 221.7 mV/m at a distance of 1 km - which is the value predicted by theory for virtually 100% radiation efficiency of that dipole.

Nothing I said disagrees with that statement. After the standing wave energy has reached steady-state on a standing-wave antenna, no additional energy is required for the standing waves and the antenna radiates essentially all of the steady-state energy being delivered to the antenna. During the transient key-down state, it takes energy to charge up the antenna's standing waves. That energy is not radiated until key-up and the energy in the standing waves is approximately 4-6 times the energy being delivered to the antenna. Hence my assertion that only 20% of the energy contained in the antenna is radiated. The other 80% exists during steady-state in order to support the standing waves.

The dipole is a standing wave antenna. From key-down, it takes a certain amount of nanosecond transient time to charge up the antenna to the steady-state conditions. After the standing-wave antenna has been charged to steady-state, it is radiating all of the energy being supplied to the antenna but the energy contained in the forward waves and reflected waves on the antenna is 4-6 times that amount of energy.

Here's how I arrived at those numbers. The Z0 of a single horizontal #14 wire at a height of 30 ft above ground is 600 ohms according to the formula for single-wire transmission lines. The feedpoint impedance of the antenna is 50 ohms. The antenna element is 1/4WL long and open-circuited. It's just like a lossy transmission line open-stub problem. Given 100 watts steady-state to the stub, what is the forward power on the stub and what is the reflected power on the stub at the 50 ohm feedpoint. I get 352w forward power and 252w reflected power. It takes energy to support those forward and reflected waves and the energy in those waves is about 4-6 times the energy being applied to the stub during steady-state.




Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 14, 2013, 06:55:48 AM
Your example of the antenna, demonstrating the fallacy of faster-than-light propagation, is a straw man. You're the only one who suggested such a thing, immediately before calling it impossible.

On the contrary, you apparently have not have seen W8JI's web page where he asserts a 3ns propagation delay through a 10" coil that has an axial propagation factor of 3 deg/inch on 80m. 30 deg in 3 ns on 80m is about seven times faster than the speed of light.

http://www.w8ji.com/inductor_current_time_delay.htm

Quote
Hmmm, looks like a pure traveling wave consists of two equal-amplitude standing waves, 90 degrees out of phase.  ;)

Thanks for supporting my assertion that it is only a mathematical construct.:)

The hint is in the name: 'standing wave'. It is not travelling at all.

Then, by definition, it is not a "wave" at all so the name is actually a misnomer which is what I said previously.


Title: RE: current and voltage along a non resonant dipole
Post by: WS3N on November 14, 2013, 07:48:02 AM
Your example of the antenna, demonstrating the fallacy of faster-than-light propagation, is a straw man. You're the only one who suggested such a thing, immediately before calling it impossible.

On the contrary, you apparently have not have seen W8JI's web page where he asserts a 3ns propagation delay through a 10" coil that has an axial propagation factor of 3 deg/inch on 80m. That is certainly faster than the speed of light.

http://www.w8ji.com/inductor_current_time_delay.htm

No, I haven't. I didn't realize you were grinding an old axe.

Quote
Quote
Hmmm, looks like a pure traveling wave consists of two equal-amplitude standing waves, 90 degrees out of phase.  ;)

Thanks for supporting my assertion that it is only a mathematical construct.

Yes, both equally good, just two different ways of looking at things. I will use the tool that is suited to the task at hand.

Let's look at an example. Say the current in a wire is a "standing wave." At each point, the current will simply oscillate with a fixed amplitude and phase that depend on the position. Using what you say is the more fundamental description of its behavior,

i = sin(kx - wt) + sin(kx + wt).

If we use my "mathematical construct" we find

i = [sin(kx) cos(wt) - cos(kx) sin(wt)] + [sin(kx) cos(wt) + cos(kx) sin(wt)] = 2 sin(kx) cos(wt).

Which form more clearly shows what is the current actually doing?


Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 14, 2013, 08:45:03 AM
Which form more clearly shows what is the current actually doing?

Since the total current doesn't have an independent existence outside of the two component traveling-wave currents, I would say that

i = sin(kx - wt) + sin(kx + wt)

conveys more valid information closer related to a picture of the physical reality of two phasor quantities rotating in opposite directions than does the non-wave "mashed potatoes" short-cut version of the equation, an abstract mathematical construct somewhat divorced from reality. Since a standing wave has no moving parts, i.e. zero phase shift, zero transfer of energy, zero momentum, and zero velocity, how can it possibly cause additional losses on a transmission line?:)

Question: If the north-bound traffic on the Golden Gate Bridge equals the south-bound traffic, can we say there is zero net traffic? And since there is zero net traffic, no repairs are necessary? That's the kind of logic I have seen associated with standing waves.


Title: RE: current and voltage along a non resonant dipole
Post by: WS3N on November 14, 2013, 09:19:04 AM
Since a standing wave has no moving parts, i.e. zero phase shift, zero transfer of energy, zero momentum, and zero velocity, how can it possibly cause additional losses on a transmission line?:)

The standing-wave current is non-zero, so there will clearly be losses if you choose to include them.

Quote
Question: If the north-bound traffic on the Golden Gate Bridge equals the south-bound traffic, can we say there is zero net traffic? And since there is zero net traffic, no repairs are necessary? That's the kind of logic I have seen associated with standing waves.

I'll agree, your logic is flawed. That analogy makes no sense.


Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 14, 2013, 09:43:45 AM
The standing-wave current is non-zero, so there will clearly be losses if you choose to include them.

But pure standing wave current has no direction and is standing still so how can something that is not moving possibly cause losses? Could it be that the losses are actually caused by the existence and superposition of the forward and reflected waves?


Title: RE: current and voltage along a non resonant dipole
Post by: WS3N on November 14, 2013, 10:08:01 AM
The standing-wave current is non-zero, so there will clearly be losses if you choose to include them.

But pure standing wave current has no direction and is standing still so how can something that is not moving possibly cause losses? Could it be that the losses are actually caused by the existence and superposition of the forward and reflected waves?


Why do you say it is standing still? Look at the equation. From the example,

i = 2 sin(kx) cos(wt).

At point x the current has an amplitude 2 sin(kx) and it oscillates at frequency w.




This is getting tiresome, but let's look at one more example. You claim that there is no energy flow in a standing-wave field. Consider a simple example of the EM field in a one-dimensional vacuum cavity.


E = y E0 sin(kx) sin(wt)

dB/dt = - curl(E) = - z k E0 cos(kx) sin(wt)

B = z (k E0/w) cos(kx) cos(wt)


The energy density is

U = 0.5 eps0 E2 + 0.5 mu0 H2

   = 0.5 eps0 E02 sin2(kx) sin2(wt) + 0.5 mu0 (k E0/ w mu0)2 cos2(kx) cos2(wt)

  = 0.5 eps0 E02 [ sin2(kx) sin2(wt) + cos2(kx) cos2(wt) ].


The Poynting flux is

S = E x H = x (k E02/ w mu0) sin(kx) cos(kx) sin(wt) cos(wt)

 = x (E02 / 4 Z0) sin(2kx) sin(2wt)


The time-averaged energy density is

<U> = 0.25 eps0 E02


and the time-averaged energy/momentum flux is zero.



While the averages are zero it is clearly wrong that there is no energy flow. The energy moves from the center of the cavity to the edges twice per cycle.


Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 14, 2013, 11:41:16 AM
Why do you say it is standing still? Look at the equation. From the example,

i = 2 sin(kx) cos(wt).

At point x the current has an amplitude 2 sin(kx) and it oscillates at frequency w.

The fact that the phase angle is the same (wt) at a particular time and at all possible 'x' points on the wire (for 0<kx<pi) indicates that the current 'i' is standing still with an oscillating-in-place amplitude. In fact, since the standing wave current is not moving in either direction, one can just as easily write the standing wave current equation as:

i = 2*sin(kx)*cos(-wt)

and get exactly the same results. Putting a negative sign on the (wt) of a traveling wave reverses the direction of flow. But since a pure standing wave has no direction of flow, putting a negative sign on the (wt) term has no effect at all because  cos(wt)=cos(-wt).

When i=2*sin(kx)*cos(wt)=2*sin(kx)*cos(-wt)

The standing wave current flow does not have a direction. This has been used by some to try to prove that reflected power doesn't actually exist.


Title: RE: current and voltage along a non resonant dipole
Post by: WS3N on November 14, 2013, 12:17:01 PM
So what? Wait a half cycle and it certainly reverses direction. Your argument is akin to saying that the current in a dipole has no direction because, if I measure distance from the center, the amplitude is cos(kx), and cos(-kx) = cos(kx). It's simply a matter of where we choose our zero.


This is conceptually the same as a vibrating string. If I tell you the equation for the string's lateral displacement is

y(x,t) = sin(kx) cos(wt)

would you say the string doesn't move?


More trivially, I could have just as easily started with the example

sin(kx + wt) - sin(kx - wt) = [sin(kx) cos(wt) + cos(kx) sin(wt)] - [sin(kx) cos(wt) - cos(kx) sin(wt)] = 2 cos(kx) sin(wt).

The fact that it now changes sign is meaningless.


Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 14, 2013, 02:32:57 PM
Wait a half cycle and it certainly reverses direction.

Sorry, that's not true. The wire is one-dimensional. The base of the pure standing wave current phasor is anchored at a point 'x' and doesn't move away from point 'x'. It is true that the mathematical model of a current phasor reverses direction in 'y''z' space but for pure standing wave current phasors, its base is always anchored to exactly the same point 'x' on that one-dimensional wire. The current phasor has a constant amplitude, A*sin(kx), is perpendicular to the wire, and anchored to one and only one point 'x'. If you hang a crescent wrench on a wire by the hole in its handle and spin it at a single point on the wire, you will get a feel for what the standing wave current phasor looks like, i.e. it doesn't have a direction up and down the wire - only perpendicular around one point on the wire. This is essentially the same conceptual problem that W8JI had when "measuring" the 3ns delay of the current through a very large air-core loading coil. I tried explaining it to him almost 10 years ago but to no avail.

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y(x,t) = sin(kx) cos(wt)

would you say the string doesn't move?

No molecule changes its 'x' position on the string. All vibrations are 'y' and 'z' excursions perpendicular to the unplucked string and are centered at 'x' on the string when the string is at rest. (An EM-wave/string analogy is riddled with conceptual problems.)
 
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The fact that it now changes sign is meaningless.

Exactly! Meaningless to its 'x' position which has absolutely nothing to do with that phase angle. For a pure standing wave, (wt) is the same over 180 degrees at any time 't' all up and down the line. Over that 180 degree span no point changes sign while another point keeps its sign. At any time 't' all points over that 180 degrees have exactly the same phase angle. All you have proved is my point that the standing wave current doesn't move in the 'x' dimension.


Title: RE: current and voltage along a non resonant dipole
Post by: WS3N on November 14, 2013, 02:49:18 PM
Sorry, that's not true. The wire is one-dimensional. The base of the pure standing wave current phasor is anchored at a point 'x' and doesn't move away from point 'x'. It is true that the mathematical model of a current phasor reverses direction in 'y''z' space but for pure standing wave current phasors, its base is always anchored to exactly the same point 'x' on that one-dimensional wire. The current phasor has a constant amplitude, A*sin(kx), is perpendicular to the wire, and anchored to one and only one point 'x'. If you hang a crescent wrench on a wire by the hole in its handle and spin it at a single point on the wire, you will get a feel for what the standing wave current phasor looks like, i.e. it doesn't have a direction up and down the wire - only perpendicular around one point on the wire. This is essentially the same conceptual problem that W8JI had when "measuring" the 3ns delay of the current through a very large air-core loading coil. I tried explaining it to him almost 10 years ago but to no avail.

Exactly! Meaningless to its 'x' position which has absolutely nothing to do with that phase angle. For a pure standing wave, (wt) is the same over 180 degrees at any time 't' all up and down the line. Over that 180 degree span no point changes sign while another point keeps its sign. At any time 't' all points over that 180 degrees have exactly the same phase angle. All you have proved is my point that the standing wave current doesn't move in the 'x' dimension.

I've read this a couple of times but I must be missing something. Are you really trying to tell me that the current is not along the wire, and that your "crescent wrench" spinning around the wire is the current, and its vector components are perpendicular to the wire?


Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 14, 2013, 04:07:25 PM
I've read this a couple of times but I must be missing something. Are you really trying to tell me that the current is not along the wire, and that your "crescent wrench" spinning around the wire is the current, and its vector components are perpendicular to the wire?

Essentially correct. The standing wave current is not flowing along the wire. It is standing still. If it was flowing along the wire, it would be a traveling wave. The crescent wrench represents the phasor of the current at point 'x'. Remember phasors are math models that don't exist in reality. Phasors have an imaginary component that we cannot measure. What we can measure is the real component. Look at the equation. What is 'x'? It's the location of a single physical point on the line. Does 'x' change with (wt)? No, it doesn't.

The current that W7EL and W8JI used for their coil group delay measurements was mostly standing wave current and cannot be used to measure propagation delay. The only way to obtain a valid propagation delay measurement is to completely eliminate the standing waves and use a traveling wave for the phase measurement. Here's an article I wrote about 75m Bugcatcher Coil measurements made at Louisiana Tech.

http://w5dxp.com/coilmeas.htm

One of the best references I have is Optics, by Hecht. It has a very good explanation of traveling waves, standing waves, interference, and superposition.


Title: RE: current and voltage along a non resonant dipole
Post by: WS3N on November 14, 2013, 04:48:06 PM
I've read this a couple of times but I must be missing something. Are you really trying to tell me that the current is not along the wire, and that your "crescent wrench" spinning around the wire is the current, and its vector components are perpendicular to the wire?

Essentially correct. The standing wave current is not flowing along the wire. It is standing still. If it was flowing along the wire, it would be a traveling wave. The crescent wrench represents the phasor of the current at point 'x'. Remember phasors are math models that don't exist in reality. Phasors have an imaginary component that we cannot measure. What we can measure is the real component. Look at the equation. What is 'x'? It's the location of a single physical point on the line. Does 'x' change with (wt)? No, it doesn't.

OMG

I know this is pointless but, just for fun, let's try one more. An AC source is connected to a resistor with short wires. Now there is no x dependence. All the little wrenches are the same length and they're all spinning in unison. Is the current standing still?


Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 15, 2013, 09:56:04 AM
I know this is pointless but, just for fun, let's try one more.

I'm not trying to drive you away and would like for you to continue this discussion until we agree on the technical concepts.

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An AC source is connected to a resistor with short wires. Now there is no x dependence. All the little wrenches are the same length and they're all spinning in unison. Is the current standing still?

There is no x dependence because there is no x in a lumped circuit. Phase shifts and delays are not allowed in a lumped circuit. That is one of its shortcomings and is unfortunately what most people carry around in their heads as a model of reality.

In the real world, there will always be a phase shift on the short wires from the source to a load with a resistive component. If it is a matched load, i.e. SWR=1:1, then the phase shift will be proportional to the speed of light in the medium. As the SWR rises from 1:1, the phase shift decreases until it disappears completely when SWR=infinity, e.g. with a purely reactive load.

It is a mathematical fact that if three phasors at points x, x+1, and x+2 all have the same phase angle at the same time, then they are not moving along x and are simply oscillating in place at a fixed x location. Those three phasors have to have different phase angles at the same time for them to be moving and one can tell from those phase angles which way they are moving.

I'm going to try to illustrate the above with graphics from EZNEC. The system is a 60 ft. wire running horizontal above mininec ground. On one end we have a source installed in a 1 ft wire to ground. For the first case, we have no load, i.e. an open-circuit exists at the end of the wire so the SWR is close to infinity (pure standing wave). For the second case, we have a 415 ohm load installed in a 1 ft wire to ground to get the SWR on the wire close to 1:1 (pure traveling wave).

In the case of the open-circuit at the end of the wire, the current phase is constant up and down the wire as can be seen from the end view. The current is obeying the standing wave equation. In the case of the 415 ohm load, the current phase can be seen to have rotated about 90 degrees as seen from the end view. The current is obeying the traveling wave equation.

In the open-circuit case, the standing wave current phasors are not advancing along the wire. They are oscillating in fixed x positions. In the 415 ohm load case, the traveling wave current phasors are advancing toward the load at the speed of light in the medium.

(http://www.w5dxp.com/phase1.JPG)


Title: RE: current and voltage along a non resonant dipole
Post by: WS3N on November 15, 2013, 11:19:53 AM
I know this is pointless but, just for fun, let's try one more.

I'm not trying to drive you away and would like for you to continue this discussion until we agree on the technical concepts.

That doesn't seem possible because you continue to ascribe some kind of reality to what is only a visualization aid (your wrenches), and talk about current standing still.

A current that stands still is completely ridiculous. It's an oxymoron. Current cannot stand still. Charges can be stationary. When charge moves it's called current. A current at some point can be constant (i.e. the same amount of charge flowing, in the same direction, past the point in a given time), or it can have a constant amplitude that periodically reverses direction in the time-harmonic case, but it makes no sense to say it stands still.

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It is a mathematical fact that if three phasors at points x, x+1, and x+2 all have the same phase angle at the same time, then they are not moving along x and are simply oscillating in place at a fixed x location. Those three phasors have to have different phase angles at the same time for them to be moving and one can tell from those phase angles which way they are moving.

This also makes no sense. Your little wrenches are, as I said, a visualization aid used in antenna modeling software. They do not flow up and down the wire, even for a traveling wave. They aren't real. Your wrench will still stay at point x and spin around in the traveling-wave case. It represents the magnitude and phase of the current at a point, whether the wave is traveling or stationary.

Quote
I'm going to try to illustrate the above with graphics from EZNEC. The system is a 60 ft. wire running horizontal above mininec ground. On one end we have a source installed in a 1 ft wire to ground. For the first case, we have no load, i.e. an open-circuit exists at the end of the wire so the SWR is close to infinity (pure standing wave). For the second case, we have a 415 ohm load installed in a 1 ft wire to ground to get the SWR on the wire close to 1:1 (pure traveling wave).

In the case of the open-circuit at the end of the wire, the current phase is constant up and down the wire as can be seen from the end view. The current is obeying the standing wave equation. In the case of the 415 ohm load, the current phase can be seen to have rotated about 90 degrees as seen from the end view. The current is obeying the traveling wave equation.

In the open-circuit case, the standing wave current phasors are not advancing along the wire. They are oscillating in fixed x positions. In the 415 ohm load case, the traveling wave current phasors are advancing toward the load at the speed of light in the medium.

You say there is a current in the standing-wave case. Please describe its behavior as a function of time. If you are going to tell me it's standing still then please describe this phenomenon more fully. What does it mean for a current to stand still?


Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 15, 2013, 01:14:51 PM
A current that stands still is completely ridiculous. It's an oxymoron. Current cannot stand still.

I agree, that's my point exactly, and it means there is something very wrong with the standing wave model. When one is dealing with standing waves, one is dealing with an abstract mathematical construct that does not correspond to the real world. Just because one may come up with the correct answers doesn't mean that the math model matches reality. What is an oxymoron is a "standing wave" because EM waves cannot stand still. A standing wave is a human abstraction and is simply not a wave at all. Waves cannot stand still. Current cannot stand still. Seems we agree.

In a pure standing wave, the total current is the superposition of two currents traveling in opposite directions at the speed of light in the medium. When we perform the following trig function, we are divorcing ourselves from reality by merging those two waves and this thread is exposing the consequences of such actions.

sin(kx + wt) - sin(kx - wt) =  2 cos(kx) sin(wt)

The left side of the equation correctly represents the forward wave and reflected wave traveling in opposite directions at the speed of light in the medium while obeying the laws of physics for EM waves. The right side of the equation does not represent the real world and cannot exist in the real world of EM waves because, as you have pointed out, that equation leads to contradictions that violate the laws of physics governing EM waves. The right side of the equation is standing still, something that is impossible for EM waves to do.

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You say there is a current in the standing-wave case. Please describe its behavior as a function of time. If you are going to tell me it's standing still then please describe this phenomenon more fully. What does it mean for a current to stand still?

You missed my point. I did not say the current is standing still. I said that the standing wave model/equation tells us that the current is standing still. Therefore it is invalid because it doesn't agree with reality. I believe that any math expert will verify that's what the standing wave equation is telling us.

I believe that there are two currents moving at the speed of light in opposite directions just as they are supposed to. I believe that the equation:

i=2 cos(kx) sin(wt)

does not represent the real world because it turns the two underlying EM waves into standing wave mashed potatoes and violates the laws/definitions of physics. It indicates that the current is not moving and that is impossible. As Bueche and Hecht say in College Physics:

"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."


Title: RE: current and voltage along a non resonant dipole
Post by: WS3N on November 15, 2013, 01:51:17 PM
A current that stands still is completely ridiculous. It's an oxymoron. Current cannot stand still.

I agree, that's my point exactly, and it means there is something very wrong with the standing wave model. When one is dealing with standing waves, one is dealing with an abstract mathematical construct that does not correspond to the real world. Just because one may come up with the correct answers doesn't mean that the math model matches reality. What is an oxymoron is a "standing wave" because EM waves cannot stand still. A standing wave is a human abstraction and is simply not a wave at all. Waves cannot stand still. Current cannot stand still. Seems we agree.

In a pure standing wave, the total current is the superposition of two currents traveling in opposite directions at the speed of light in the medium. When we perform the following trig function, we are divorcing ourselves from reality by merging those two waves and this thread is exposing the consequences of such actions.

sin(kx + wt) - sin(kx - wt) =  2 cos(kx) sin(wt)

The left side of the equation correctly represents the forward wave and reflected wave traveling in opposite directions at the speed of light in the medium while obeying the laws of physics for EM waves. The right side of the equation does not represent the real world and cannot exist in the real world of EM waves because, as you have pointed out, that equation leads to contradictions that violate the laws of physics governing EM waves. The right side of the equation is standing still, something that is impossible for EM waves to do.

The left and right sides are completely equivalent. There is only one field. It is what it is. It can be described in two ways but that doesn't change the field or the math.

Quote
Quote
You say there is a current in the standing-wave case. Please describe its behavior as a function of time. If you are going to tell me it's standing still then please describe this phenomenon more fully. What does it mean for a current to stand still?

You missed my point. I did not say the current is standing still. I said that the standing wave model/equation tells us that the current is standing still. Therefore it is invalid because it doesn't agree with reality. I believe that any math expert will verify that's what the standing wave equation is telling us.

It doesn't take a math expert to tell you that you are wrong. The right side says no such thing.

i(x,t) = [2 cos(kx)] sin(wt) = I(x) sin(wt)

The amplitude I(x) varies with position but the current is most certainly not standing still.

Look at this link. Someone has done us the favor of creating animated gifs of the current on a transmission line in both cases (traveling wave at the top, standing waves at about the middle of the page).

http://en.wikipedia.org/wiki/Transmission_line

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"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."

Whether one chooses to call them standing waves (a term which is, of course, widely used and understood) or something else is irrelevant. As I showed in my worked example of the cavity oscillator, to which you failed to respond, these fields do transport energy locally, but there is no net transport.


Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 15, 2013, 04:53:27 PM
The left and right sides are completely equivalent. There is only one field.

The left and right sides are mathematically equivalent. That doesn't mean that there is only one current. There is, in fact, two currents and without either one that field could not exist as a standing wave. So in reality, there exist two superposed fields, not just one. Each field retains its own independent existence without interacting with the other until they encounter an impedance discontinuity at the same place at the same time. People who ignore that fact have come up with some really far out ideas, e.g. reflected power doesn't exist.

Have you ever tried to create a standing wave using only one field generator? No fair causing reflections.:)

On the subject of trusting our math models to represent reality, quite often our math models result in a negative resistance solution. Does that mean that one can go out and buy a negative resistor?

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It doesn't take a math expert to tell you that you are wrong. The right side says no such thing.

I suggest you consult with a math expert as I have. Any time this identity exists,

i=2 cos(kx) cos(wt)=2 cos(kx) cos(-wt)

there is no movement of the current because at any particular time, there is no difference in the phase on each side of any particular point 'x'. When that condition is true, nothing is moving in the 'x' direction. Check it out if you don't believe me.

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http://en.wikipedia.org/wiki/Transmission_line

Unfortunately, those plots don't represent RF reality. At RF frequencies, the electrons move hardly at all. They are better visualized by assuming they oscillate in place. The wikipedia "drift velocity" web page indicates that the movement of an electron at 60 MHz would be about 2x10-12 meter. Here's a better example. Assuming the blue trace is standing wave current, in which direction is it flowing?

http://www.csupomona.edu/~ajm/materials/animations/stwaves/swr11.mov


Title: RE: current and voltage along a non resonant dipole
Post by: WS3N on November 15, 2013, 06:20:48 PM
The left and right sides are completely equivalent. There is only one field.

The left and right sides are mathematically equivalent.

They are completely equivalent. The charges in the medium cannot do two things at the same time. There is only one total current. As long as we are in a linear medium the two cases are indistinguishable. The total current is the linear superposition of the two traveling waves, physically and mathematically.

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Quote
It doesn't take a math expert to tell you that you are wrong. The right side says no such thing.

I suggest you consult with a math expert as I have. Any time this identity exists,

i=2 cos(kx) cos(wt)=2 cos(kx) cos(-wt)

there is no movement of the current because at any particular time, there is no difference in the phase on each side of any particular point 'x'. When that condition is true, nothing is moving in the 'x' direction. Check it out if you don't believe me.

How can you keep insisting that there is no current? At, say, x = 0,

i(0,t) = 2 cos(wt)

The current is obviously oscillating. It is positive and flows in the +x direction until it reaches a maximum value of +2, then reverses, passes through zero, and goes negative, flowing in the -x direction until it equals -2, then back again.

Please tell me how this means there is no current.

Quote
Quote
http://en.wikipedia.org/wiki/Transmission_line

Unfortunately, those plots don't represent RF reality. At RF frequencies, the electrons move hardly at all. They are better visualized by assuming they oscillate in place. The wikipedia "drift velocity" web page indicates that the movement of an electron at 60 MHz would be about 2x10-12 meter.

Now you are purposely being obstinate. Would it be satisfactory if the dots were the size of electrons and they only moved 10-12 m? Maybe it would be better if you viewed it on a very tiny screen. If you really object to this then all of your plots are similarly incorrect because they only demonstrate concepts and are not at the true scale. That is just silly.

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Here's a better example. Assuming the blue trace is standing wave current, in which direction is it flowing?

http://www.csupomona.edu/~ajm/materials/animations/stwaves/swr11.mov

Again, OMG. Now I finally see the problem. I can't believe I have been wasting my time because of this. It's no wonder you didn't respond to my worked example of the field in the cavity.

You said, "Assuming the blue trace is standing wave current, in which direction is it flowing?"

Unbelievable. You have no idea what the blue trace means. The direction of current flow has nothing to do with whether or not the trace appears to move left or right. Current is a vector quantity. In one dimension, the direction of the current is indicated by its sign. Positive means it flows in the +x direction, negative means it flows in the -x direction, as I wrote a few lines above this. When the blue trace is above the axis the current is flowing to the right, and when it's below the axis the current is flowing to the left.

I don't know what your background is but you seem to be lacking a fundamental appreciation of the relationship between the mathematics and the physical concepts it represents.


Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 16, 2013, 07:57:10 AM
The current is obviously oscillating.

Sorry, I thought it was obvious that I was asking you to take a look at the current NODES. I'll rephrase the question: "Assuming the blue trace is standing wave current, in which direction is it flowing at the current nodes?"

How is it possible for the current to just slosh back and forth between current nodes? Exactly what happens when that alleged flowing standing wave current encounters a current node? The mathematics says the conditions are identical to an open circuit. But we know it is not an open circuit because if we cut the wire, things change.

What is actually happening in reality is easy to understand when a laser is used to create the standing waves in free space without the encumbrance of wires and currents.

Quote
I don't know what your background is but you seem to be lacking a fundamental appreciation of the relationship between the mathematics and the physical concepts it represents.

My credentials are listed on my QRZ bio page. I don't respond well to ad hominem attacks so I will bow out.


Title: RE: current and voltage along a non resonant dipole
Post by: WS3N on November 16, 2013, 08:24:29 AM
The current is obviously oscillating.

Sorry, I thought it was obvious that I was asking you to take a look at the current NODES. I'll rephrase the question: "Assuming the blue trace is standing wave current, in which direction is it flowing at the current nodes?"

Why does the current just slosh back and forth between current nodes? Exactly what happens when that alleged flowing standing wave current encounters a current node? The mathematics says the conditions are identical to an open circuit. Why can't we just cut the wire?

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I don't know what your background is but you seem to be lacking a fundamental appreciation of the relationship between the mathematics and the physical concepts it represents.

My credentials are listed on my QRZ bio page. I don't respond well to ad hominem attacks so I will bow out.

You never said anything about nodes, not that it matters. That is just more smoke.

You have listed you credentials, so I'll give you mine. I have BS degrees in math and physics from MIT and a PhD in theoretical physics from Penn State. I have worked at one of the Navy's research facilities for 25 years, where I was originally hired to fill a slot in EM theory. I have worked on many problems in EM propagation and scattering, and written my own FEM, BEM and MoM (what's used in NEC) codes.

My comment was hardly an attack but this is also my last post in this thread. This is a waste of time because you have demonstrated that you know just enough to be dangerous.


Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 16, 2013, 10:31:47 AM
I previously posted standing wave current graphics from EZNEC. Here is another set covering multiple wavelengths. Please note that, unlike the standing wave current phasor, the magnitude of the traveling wave current phasor is constant and never zero. For real world EM waves, the electric and/or magnetic energy cannot fall to zero while the wave exists. Just more evidence that a standing wave is not an EM wave.

(http://www.w5dxp.com/StndTrav.JPG)

Here's some previous information on the difference between a standing wave and a traveling wave.

(http://www.w5dxp.com/travstnd.gif)


Title: RE: current and voltage along a non resonant dipole
Post by: WS3N on November 16, 2013, 12:36:55 PM
Boy, you just keep digging a deeper hole.

I previously posted standing wave current graphics from EZNEC. Here is another set covering multiple wavelengths. Please note that, unlike the standing wave current phasor, the magnitude of the traveling wave current phasor is constant and never zero. For real world EM waves, the electric and/or magnetic energy cannot fall to zero while the wave exists. Just more evidence that a standing wave is not an EM wave.

Just more evidence that you don't know what you're talking about. I guess you won't be satisfied until that's clear to everyone.

EZNEC is like a power tool. Anyone can pick it up but it's dangerous in the wrong hands.

Look at figure 6 on page 11 of this document, http://www.people.fas.harvard.edu/~djmorin/waves/electromagnetic.pdf

It sure looks like the electric and magnetic fields, which are in-phase in your traveling wave, do periodically go to zero as they reverse direction. BTW, a discussion of standing waves (!!!) follows.

Or, if you don't like that one, this link has a really pretty picture. http://physics.info/em-waves/

You can skip all the math stuff, much too complicated.



Let me make an edit and add something, before we have to go through another round of this nonsense.

I'm going to explain to you what your phasor means. The radius of the helix (it's a helix since this is a traveling wave) is the amplitude of the wave (E0) and the angle around the axis is its phase (phi(z,t)). I'm using z for the position variable since that's what the document uses.

The electric field, for example, is E(z,t) = E0 sin(phi(z,t)). The amplitude (helix radius) doesn't change but the phase phi, and so the field, is a function of position and time. As the text above fig. 6 says, the whole figure slides along the z axis as time progresses. Your phasor helix is a snapshot in time, so its phase only depends on position. If we could see the time dependence it would turn like a screw.

Now the good part. The fact that the phasor helix doesn't go to zero does not mean that the electric field is always non-zero. Whenever phi passes through a multiple of pi (since I used sin in this example) the E field will be zero.


Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 16, 2013, 07:27:02 PM
It sure looks like the electric and magnetic fields, which are in-phase in your traveling wave, do periodically go to zero as they reverse direction.

I've lost count of the straw men that you have attempted to set up for torture.

I was not talking about instantaneous values of E and H. I was talking about the ExH Poynting vector for plane waves which has a constant term indicating that the wave carries an average power. So let me upgrade my sentence:

"For real world EM waves, the average ExH value cannot fall to zero while the wave exists."

Question: What is the value of the Poynting vector for a pure standing wave?


Title: RE: current and voltage along a non resonant dipole
Post by: WS3N on November 16, 2013, 11:39:47 PM
It sure looks like the electric and magnetic fields, which are in-phase in your traveling wave, do periodically go to zero as they reverse direction.

I've lost count of the straw men that you have attempted to set up for torture.

I was not talking about instantaneous values of E and H. I was talking about the ExH Poynting vector for plane waves which has a constant term indicating that the wave carries an average power.

Bull. You never said anything about averages. You keep trying to get yourself out of this mess of your own creation.

You said, "For real world EM waves, the electric and/or magnetic energy cannot fall to zero while the wave exists."


I'm not going to work out another example because 1) I don't want to take the time to do it, 2) I don't think you'll understand it anyway, and 3) you'll just come up with another lame excuse.

So I'll again refer to the document I found today when I was looking for a picture of a traveling wave, http://www.people.fas.harvard.edu/~djmorin/waves/electromagnetic.pdf

Page 15, equation 52.

The energy density of a plane wave in free space is

E = eps0 E02 cos2(kz - wt)

and the Poynting vector is

S = c eps0 E02 cos2(kz - wt)k.


There is no constant term. For a traveling wave in free space, the energy density E and energy flux S do go to zero whenever E and B are zero. The energy travels in bunches.

You're wrong, again.

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So let me upgrade my sentence:

"For real world EM waves, the average ExH value cannot fall to zero while the wave exists."

That's a statement entirely of your own making, based on your distinction between "real-world EM waves" (whatever that means) and something else.

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Question: What is the value of the Poynting vector for a pure standing wave?

This one I did work out for you (post #30). The energy density is

U  = 0.5 eps0 E02 [ sin2(kx) sin2(wt) + cos2(kx) cos2(wt) ]

and the Poynting vector is

S = E x H = x (k E02/ w mu0) sin(kx) cos(kx) sin(wt) cos(wt)

 = x (E02 / 4 Z0) sin(2kx) sin(2wt).

The behavior of the lowest mode, where the cavity walls are 1/2 wavelength apart, is the easiest to picture. During one cycle of the field (0 < wt < 2 pi), the energy density (given by the function U) starts at the walls (none in the middle), then flows to the middle (none at the walls), then back to the walls, two times. The energy oscillates twice as fast (frequency 2w) as the fields themselves.

The Poynting vector S describes the flow of the energy. It vanishes at the times of maximum energy density (all at the walls or all in the middle) since there is momentarily no flow at these peak times (like waves coming onto a beach, stopping, and then retreating). The energy flux reaches its peak values at the times between the energy peaks, as the energy moves back and forth. S is a vector quantity so its sign indicates its direction, positive means right, negative means left. S is positive between the left wall and the center as the energy there flows to the right, from the wall to the center, and, at the same time, negative between the right wall and the center as the energy there flows to the left, from that wall to the center. A quarter of a cycle later the sign of S reverses, as the energy flows back out to the walls.

As I also said before, the average value of S is zero and there is no net energy flow. That does not mean S is always zero.

Section 8.4.3 of the same document, also starting on page 15, describes the energy in standing waves, and gives essentially the same expressions that I derived. You can see that he switches freely between the two representations of a standing wave, namely as a single expression or as two opposing traveling waves (also in 8.3.3 where he deals with E and B). As I have always said, there is no physical or mathematical difference between the two viewpoints. Note the last paragraph of that section, at the top of page 16, where he says the energy flow in a standing wave is only zero on average, not identically zero.

Strange that this fellow also fails to mention "real-world EM waves." He just merrily switches back and forth between the two representations. Poor devil, he just doesn't realize the error of his ways.



BTW, you never thanked me for setting you straight about your phasors in my last post, when I was responding to your statement

"Please note that, unlike the standing wave current phasor, the magnitude of the traveling wave current phasor is constant and never zero. For real world EM waves, the electric and/or magnetic energy cannot fall to zero while the wave exists."

I clearly showed that the fields (and energy) do go to zero, and your assumption that a constant-magnitude phasor somehow prevented this was completely wrong. You're welcome. I'm sure the fact that you failed to mention this was just an oversight.



This is absolutely my last post on this thread. You can follow this with any crazy ideas about "real-world EM waves" or anything else that you like. I will never respond to another of your posts. To the other readers of this forum I will only say buyer beware.


Title: RE: current and voltage along a non resonant dipole
Post by: W5DXP on November 17, 2013, 07:04:16 AM
Bull. You never said anything about averages. You keep trying to get yourself out of this mess of your own creation.

OTOH, using that same logic, I never said anything about instantaneous values either so why did you "mistakenly" presume I was talking about instantaneous energy? If I refer to scalar values, like "energy" and "power", 99+% of the time I am talking about average energy and power. I don't find instantaneous energy and instantaneous power to be very useful concepts.

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You said, "For real world EM waves, the electric and/or magnetic energy cannot fall to zero while the wave exists."

That's exactly what I said where "the electric and/or magnetic energy" was intended to imply the same meaning as the "average ExH energy". I used those colloquial words because a lot of readers here don't know what ExH means.

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There is no constant term. You're wrong, again.

Then your argument is with Ramo and Whinnery, not me. Reference: Fields and Waves in Communications Electronics; 3rd edition; Ramo, Whinnery, Van Duzer; page 143, Example 3.12c, "Poynting Flow in a Plane Wave". I'm going to use K for a constant involving permeability, permittivity, and the maximum envelope value.

Pz = K[1/2 + 1/2 cos 2(wt-kz)]    (equation 18)

"Note that there is a constant term showing that the wave carries an average power, as expected. There is also a time-varying portion representing the redistribution of stored energy in space as maxima and minima of fields pass through a given region." (my emphasis)

Are you going to contact the publisher and demand that they correct the gross error that has been in that book for more than half a century?:)

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BTW, you never thanked me for setting you straight about your phasors in my last post, when I was responding to your statement

You want me to thank you for setting up a straw man and then knocking it down??? Virtually nothing of what I was trying to say disagrees with what you have said. Are you prone to tilting at windmills?:)

That EZNEC current envelope plot that I posted is a graphical indication that a standing wave is transporting zero average power while the traveling wave is indeed transporting some average power. (I'm using the IEEE definition of "power".) We'll just have to agree to disagree on that subject.