Home |
Search |
Today's Posts |
#31
|
|||
|
|||
Followon to image theory
Isn't the whole question of whether it's a reflection as from a conducting
plate, or radiation from the radials based on misunderstanding what a reflection is? The reflection from a conducting plane is the sum effect of the radiation from the currents induced in the plane by the incident field, say from a quarter wave vertical. The reflection forms an image 'behind the mirror' of the source of the incident field, for example a quarter wave vertical 'under the plane'. If these currents are radial then the plane is well approximated by a large number of long radial wires. The high resistance at right angles to the radials doesn't matter because no current is flowing in that direction. The plane is not so well approximated by a small number, say 3 or 4, radials, especially if they are short, say a quarter wavelength. However the approximation may be good enough to allow us to regard the short radials as reflecting the incident field from the quarter wave vertical especially if we just need a rough picture of the situation and a rough estimate of the input impedance.We may need to keep in mind that the reflection is due to radiation from the currents in the radials or we may be able to forget it. If we want the antenna pattern in detail and the input impedance accurately we have to do the maths properly. That's not simple even for a half-wave dipole! Brian LA0DG |
#32
|
|||
|
|||
Followon to image theory
David wrote:
With a vertical monopole antenna, the field emitted by the radials forms a near field and interacts with the wave radiated by the vertical element. Is it mainly in the vertical direction that the radial fields interact with the field from vertical element? I would expect the antenna feedpoint impedance to be formed from the length of antenna conductors and the interaction of the fields. If a monopole is simulated in a NEC program above a perfect ground plane, is it possible to see the reflection? I would expect NEC program to show reflection of waves from metallic surfaces e.g reflection from perfect ground plane or parabolic dish. Can a NEC program allow the user to see whether the wave is reflected (as for perfect ground plane) or whether effect is due to wave interaction (as for radials)? Can the user then vary a finite size ground plane to see whether antenna impedance comes from reflection or wave interaction? What do you mean when you say that waves interact with each other? Explain what you mean by the word "reflection." 73, Tom Donaly, KA6RUH |
#33
|
|||
|
|||
Followon to image theory
Brian Anthony Farrelly wrote:
Isn't the whole question of whether it's a reflection as from a conducting plate, or radiation from the radials based on misunderstanding what a reflection is? There is no contradiction in quantum electrodynamics. A photon encounters an electron in a radial and is absorbed. Later, that same electron emits a photon. The original photon energy could have come from anywhere. If it came from the vertical element in the first place, it is a "reflection" but that is irrelevant. If it came from that same radial in the first place, it is not a "reflection" but that is also irrelevant. If it bounced off a flagpole before it was absorbed by the electron, it is a "reflection" but no rational person cares. The photon cloud surrounding a radial contains photons that came from that radial and from everywhere else. That some photons must be put in a basket labeled "reflected" and some put in a basket labeled "not reflected" is simply nonsense. -- 73, Cecil http://www.w5dxp.com |
#34
|
|||
|
|||
Followon to image theory
Cecil Moore wrote:
That some photons must be put in a basket labeled "reflected" and some put in a basket labeled "not reflected" is simply nonsense. The very concept of arguing whether a photon was reflected or not seems to violate the uncertainty principle. There is only a probability that it was reflected. -- 73, Cecil http://www.w5dxp.com |
#35
|
|||
|
|||
Followon to image theory
On Wed, 17 Jan 2007 20:22:27 -0000, "David" nospam@nospam wrote:
With a vertical monopole antenna, the field emitted by the radials forms a near field and interacts with the wave radiated by the vertical element. Is it mainly in the vertical direction that the radial fields interact with the field from vertical element? Hi David, No. I would expect the antenna feedpoint impedance to be formed from the length of antenna conductors and the interaction of the fields. Sure. There is nothing you've offered to suggest otherwise. If a monopole is simulated in a NEC program above a perfect ground plane, is it possible to see the reflection? Yes. Even above an imperfect ground. Plane has nothing to do with it except for the matter of geometry - radials even less so. I would expect NEC program to show reflection of waves from metallic surfaces e.g reflection from perfect ground plane or parabolic dish. You should then expect an NEC program to show that reflection in relation to wavelength. They do this to as near to what is observed as to show no difference, naturally. Can a NEC program allow the user to see whether the wave is reflected (as for perfect ground plane) or whether effect is due to wave interaction (as for radials)? NEC is not responsible for insight or interpretation. That only comes with experience. Can the user then vary a finite size ground plane to see whether antenna impedance comes from reflection or wave interaction? The user might, but the proof of an argument is through a chain of evidence. A simple test of "reflection from radials" in a real implementation (with ground): 1. Change the angle of the radials by 45°, does the lobe angle change by 45°? Explain why reflection does not follow convention. 2. As suggested by Roy, add a top hat with a structure that identical to the radials, does the lobe angle point into the ground? Explain why reflection does not exist in this case. 3. Alternative: invert the radials/vertical, it is now upside down, does the lobe angle invert with it? Explain why radials are special only in the conventional case. You might want to repeat these in real space, but then you have removed the reflector, ground. Without it, what would you offer as a proof of reflection that can be verified through any of the three simple tests above? Explain why those tests demonstrate reflection in free space when those tests above do not. 73's Richard Clark, KB7QHC |
#36
|
|||
|
|||
Followon to image theory
Cecil Moore wrote:
There is no contradiction in quantum electrodynamics. There's certainly a paradox. A photon encounters an electron in a radial and is absorbed. We know that electrons are inspired to move in a particular fashion when they are irradiated. But how does a photon tell the electron which way it should move? Kinematics would seem to have little to do with this phenomenon. Later, that same electron emits a photon. A photon which, if the direction in which it is emitted is unknown, can be made to create a diffraction pattern - implying that a wave was emitted rather than a particle. Putting us right back where we started. The photon cloud surrounding a radial contains photons that came from that radial and from everywhere else. But what is it about this "cloud" that is actually cloud-like? Isn't it really more like a wave of photons? :-) That some photons must be put in a basket labeled "reflected" and some put in a basket labeled "not reflected" is simply nonsense. I think Feynman uses almost that exact analogy, with probabilities assigned to each, in the first few pages of QED. 73, ac6xg |
#37
|
|||
|
|||
Followon to image theory
Jim Kelley wrote:
But how does a photon tell the electron which way it should move? Energy and momentum must be conserved. A photon which, if the direction in which it is emitted is unknown, can be made to create a diffraction pattern - implying that a wave was emitted rather than a particle. Putting us right back where we started. Allow me to paraphrase Feynman: Only particles exist. But what is it about this "cloud" that is actually cloud-like? Isn't it really more like a wave of photons? :-) If you prefer liquid analogies to gaseous analogies, feel free. I think Feynman uses almost that exact analogy, with probabilities assigned to each, in the first few pages of QED. You missed the point. That an individual photon must be put in a basket labeled "reflected" and some other individual photon must be put in a basket labeled "not reflected" is simply nonsense. Talking about exactly where an individual photon goes is nonsense. -- 73, Cecil, http://www.qsl.net/w5dxp |
#38
|
|||
|
|||
Followon to image theory
On Thu, 18 Jan 2007 12:51:40 -0800, Jim Kelley
wrote: A photon encounters an electron in a radial and is absorbed. We know that electrons are inspired to move in a particular fashion when they are irradiated. But how does a photon tell the electron which way it should move? Kinematics would seem to have little to do with this phenomenon. And photon/electron interaction is not so indiscriminate (it doesn't always interact in the first place). Later, that same electron emits a photon. A photon which, if the direction in which it is emitted is unknown, can be made to create a diffraction pattern - implying that a wave was emitted rather than a particle. Putting us right back where we started. An electron does not always give rise to just one photon, or any photon, or a photon of the same energy. The photon cloud surrounding a radial contains photons that came from that radial and from everywhere else. But what is it about this "cloud" that is actually cloud-like? Isn't it really more like a wave of photons? :-) Cloud indeed. This is like a sack of magic beans in exchange for a cow. Is this ill-discussion of photons the legacy of image theory? It is like preening in front of a conjugate mirror. 73's Richard Clark, KB7QHC |
#39
|
|||
|
|||
Followon to image theory
Richard Clark wrote:
An electron does not always give rise to just one photon, or any photon, or a photon of the same energy. This is true. But in an amateur transmitter driven antenna system, the great majority of radiated photons are coherent with the transmitting frequency. Coherent photons are required to have the same energy level. The photons coherent with the transmitting frequency are first order effects. Assuming one is not close to another transmitter, the photons not coherent with the transmitting signal are Nth order effects. -- 73, Cecil http://www.w5dxp.com |
#40
|
|||
|
|||
Followon to image theory
Cecil Moore wrote:
Jim Kelley wrote: But how does a photon tell the electron which way it should move? Energy and momentum must be conserved. Do you really want to go there? :-) According to QED, you first need to find some negative energy photons. Then you need to get them to propagate backward in time and subsequently interact with electrons. Then, the electrons can then move forward in time and emit positive energy photons while conserving momentum. A photon which, if the direction in which it is emitted is unknown, can be made to create a diffraction pattern - implying that a wave was emitted rather than a particle. Putting us right back where we started. Allow me to paraphrase Feynman: Only particles exist. Dear Merrium Webster, On behalf of a friend of mine, please consider adding this definition to your dictionary: paraphrase - to restate text in different form such that it conveys unintended meaning But what is it about this "cloud" that is actually cloud-like? Isn't it really more like a wave of photons? :-) If you prefer liquid analogies to gaseous analogies, feel free. A wave of photons. It's a physics joke, Cecil. Laugh already. 73, ac6xg |
Reply |
Thread Tools | Search this Thread |
Display Modes | |
|
|
Similar Threads | ||||
Thread | Forum | |||
balun and image | Shortwave | |||
A "single conversion" question | Shortwave | |||
And Incase Lennie Doubted that MARS and Amateur Radio are a "Service to the Nation..." MARS Chief Says Otherwise | Policy | |||
Rare Books on Electronics and Radio and Commmunications | Equipment | |||
Rare Books on Electronics and Radio and Commmunications | Equipment |