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

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

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

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

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

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

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

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

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

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