Ok. You might ask me, "Why do you laugh at people discussing antennas
emitting photons?

And, I would answer:

Photon emissions from an antenna element(s) seems difficult, at best, to
visualize (no pun intended.)

Consider a 1/2 inch dia. single element antenna (monopole?) If the
thing is emitting photons, one would think the photons are being emitted
equally around the elements circumference.

Well, now flatten that 1/2 dia rod into a very thin ribbon--however, the
ribbon still has the same area of cross section, and equal to the cross
section of the round rod. If this conductor is emitting photons, one
would expect them, now, to be off the two flat sides of the element and
relative few off the sides--indeed, one would now expect this element to
be becoming directional in two favored directions--off the flat sides
.... to date, I have NOT been able to measure an acceptable difference to
reinforce the "illumination properties" of the element.

The photon/wave properties of rf still remains a mystery ... and proof
hard to come by.

Regards,
JS

On Wed, 29 Aug 2007 12:53:25 -0700, John Smith
wrote:

Ok. You might ask me, "Why do you laugh at people discussing antennas
emitting photons?

(You don't close quotes is one larf. Why people laugh is a condition
of creationist-like explanations that attend the topic.)

And, I would answer:

Photon emissions from an antenna element(s) seems difficult, at best, to
visualize (no pun intended.)

The explanation is simple. Because of the pun (intended or otherwise)
too many expect that the experience of "seeing" is sufficient to
understanding "visualization." Nothing could be further from the
truth. The quote that follows provides sufficient evidence to this:

Consider a 1/2 inch dia. single element antenna (monopole?) If the
thing is emitting photons, one would think the photons are being emitted
equally around the elements circumference.

This confuses thinking with visualization now. Unfortunately it
proceeds from a false premise. It is also a false premise if we
simply ignore "photon" and discuss this in the rather more prosaic
term of "fields."

All points of all surfaces are active emitters. Your "perception"
(visualization) is a far field response of the total contributions of
all sources and their phases. This perception creates an "illusion."
Illusions are fun and interesting, but they bear only on cognitive
issues, not Physics.

73's
Richard Clark, KB7QHC

Richard Clark wrote:

...
73's
Richard Clark, KB7QHC

Unless there is a clear, defined and model-able example of why a "rf
photon" would behave different than a "light photon"--I expect both to
obey current laws/actions/expected-behaviors.

However, everyone enjoys a good fairytale, now and then.

There it is == "

The missing double quote! :-)

Regards,
JS

"John Smith" wrote in message
...
Ok. You might ask me, "Why do you laugh at people discussing antennas
emitting photons?

And, I would answer:

Photon emissions from an antenna element(s) seems difficult, at best, to
visualize (no pun intended.)

Consider a 1/2 inch dia. single element antenna (monopole?) If the thing
is emitting photons, one would think the photons are being emitted equally
around the elements circumference.

Well, now flatten that 1/2 dia rod into a very thin ribbon--however, the
ribbon still has the same area of cross section, and equal to the cross
section of the round rod. If this conductor is emitting photons, one
would expect them, now, to be off the two flat sides of the element and
relative few off the sides--indeed, one would now expect this element to
be becoming directional in two favored directions--off the flat sides ...
to date, I have NOT been able to measure an acceptable difference to
reinforce the "illumination properties" of the element.

The photon/wave properties of rf still remains a mystery ... and proof
hard to come by.

Regards,
JS

John

Imagine your ribbon antena flattened to the thickness of a razor blade.
Instead of using RF, heat the antenna with a blow torch until it becomes
white hot.

It is only when looking at the exact edge of the antenna that any
appreciable drop in light out put will be noticed. At all broadside angles
an appreciable amount of light would be seen. The same effects can be
expected to occur at RF but the majority of amateur test equipment would not
have the resolution to measure the dip with the antenna edge on. The width
of the receiving antenna and diffraction effects would tend to hide this in
the far field, and alignment, reflection effects and manufacturing
tolerances in the near field.

An example from nature can be seen when looking at the planet Saturn. The
rings are clearly visible through even a small telescope except for a couple
of weeks when they are aligned exactly edge on to the earth. Even at very
oblique angles, enough light is reflected to make them quite visible.

Mike G0ULI

On 29 Aug, 14:57, John Smith wrote:
Richard Clark wrote:

...

73's
Richard Clark, KB7QHC

Unless there is a clear, defined and model-able example of why a "rf
photon" would behave different than a "light photon"--I expect both to
obey current laws/actions/expected-behaviors.

However, everyone enjoys a good fairytale, now and then.

There it is == "

The missing double quote! :-)

Regards,
JS

Intersting debating point as to whether it is a fairy tail!
May I point out that a H bomb explosion creats via the velocity of the
explosion
radiation with out the use of a resonant element! In the old days we
used a spark gap to create
radiation again without a resonant element! Tho scientists would say
that it is a consequence
of time varient current and thus settle on the current change of
velocity ie accelleration, They could have easily stated that
radiation is pulsed form after all current does go thru zero and
Newton himself phrased it as "packets"
of radiation in his day. But keeping to the same rules of the Masters
of what creats radiation one could easily see from the explosion
theory that radiation is created by the exchange of energy between
capapaciters and inductance
each of which provides a explosion when shorted at the end of each
cycle which promotes the particulate theorem without conflict with the
Masters. I would remind you that "Gausses theorem" with respect to
equilibrium supports the particulate theorem which to the surprise of
many is supported by Maxwells equations. I read the reply that you got
but I could not determine his position if any or what points he was
trying to make...if any.
A point to ponder on, capacitance and inductance are both capable of
energy storage and shorting the terminals is time varient to eject
particulates at a high velocity thus creatin two pulses per cycle and
where in their absence scientists have only time varient current to
hang their hats on for the wave theorem.
Regards
Art

On Aug 29, 12:53 pm, John Smith wrote:
Ok. You might ask me, "Why do you laugh at people discussing antennas
emitting photons?

And, I would answer:

Photon emissions from an antenna element(s) seems difficult, at best, to
visualize (no pun intended.)

Consider a 1/2 inch dia. single element antenna (monopole?) If the
thing is emitting photons, one would think the photons are being emitted
equally around the elements circumference.

Well, now flatten that 1/2 dia rod into a very thin ribbon--however, the
ribbon still has the same area of cross section, and equal to the cross
section of the round rod. If this conductor is emitting photons, one
would expect them, now, to be off the two flat sides of the element and
relative few off the sides--indeed, one would now expect this element to
be becoming directional in two favored directions--off the flat sides
... to date, I have NOT been able to measure an acceptable difference to
reinforce the "illumination properties" of the element.

The photon/wave properties of rf still remains a mystery ... and proof
hard to come by.

Regards,
JS

You'd have just as much trouble understanding the behaviour of visible-
light photons, given your desire to view them, apparently, as you
would billiard balls or some other macro-size physical object. You
might enjoy reading how Feynmann described the behaviour in his
physics lectures at Cal Tech. It's something along the lines of,
"They behave differently than anything you have any experience with.
Much differently."

On the other hand, there's probably not much utility in discussing
photons of, say, a 14MHz signal, simply because the energy contained
in one quantum at that frequency is so small that you won't be able to
detect it: a little less than 10^-26 joules per photon. At one
photon per second, that's under 10^-26 watts, if you collect all the
energy. At 50 ohms, that's less than a picovolt. Noise in a 1Hz
bandwidth in a 50 ohm resistor at room temperature is about a
thousand times that much. -- Yes, the energy is quantized. But the
quanta are going to be _very_ difficult to distinguish.

Cheers,
Tom

John Smith wrote:
Ok. You might ask me, "Why do you laugh at people discussing antennas
emitting photons?

And, I would answer:

Photon emissions from an antenna element(s) seems difficult, at best, to
visualize (no pun intended.)

Consider a 1/2 inch dia. single element antenna (monopole?) If the
thing is emitting photons, one would think the photons are being emitted
equally around the elements circumference.

Well, now flatten that 1/2 dia rod into a very thin ribbon--however, the
ribbon still has the same area of cross section, and equal to the cross
section of the round rod. If this conductor is emitting photons, one
would expect them, now, to be off the two flat sides of the element and
relative few off the sides--indeed, one would now expect this element to
be becoming directional in two favored directions--off the flat sides
... to date, I have NOT been able to measure an acceptable difference to
reinforce the "illumination properties" of the element.

The photon/wave properties of rf still remains a mystery ... and proof
hard to come by.

Regards,
JS

Get a copy of Richard Feynman's "QED". It's 4 of his lectures on the
subject.

jk

K7ITM wrote:

...
On the other hand, there's probably not much utility in discussing
photons of, say, a 14MHz signal, simply because the energy contained
in one quantum at that frequency is so small that you won't be able to
detect it: a little less than 10^-26 joules per photon. At one
photon per second, that's under 10^-26 watts, if you collect all the
energy. At 50 ohms, that's less than a picovolt. Noise in a 1Hz
bandwidth in a 50 ohm resistor at room temperature is about a
thousand times that much. -- Yes, the energy is quantized. But the
quanta are going to be _very_ difficult to distinguish.

Cheers,
Tom


If there are, indeed, as many photons being emitted by the thin edge of
the ribbon, as by the broad edges, what law/effect/affect is being
demonstrated here?

Or. why are the photons "drawn" to the thin edge with such magnitude of
force?

If this ribbon was white hot (even infrared) a meter would indicate more
energy from the greatest surface area. Occams' razor is wrong, again?

I have never read of the phenomenon you seem to be suggesting here ...

Regards,
JS

On Aug 29, 4:11 pm, "Mike Kaliski" wrote:
"John Smith" wrote in message

...



Ok. You might ask me, "Why do you laugh at people discussing antennas
emitting photons?

And, I would answer:

Photon emissions from an antenna element(s) seems difficult, at best, to
visualize (no pun intended.)

Consider a 1/2 inch dia. single element antenna (monopole?) If the thing
is emitting photons, one would think the photons are being emitted equally
around the elements circumference.

Well, now flatten that 1/2 dia rod into a very thin ribbon--however, the
ribbon still has the same area of cross section, and equal to the cross
section of the round rod. If this conductor is emitting photons, one
would expect them, now, to be off the two flat sides of the element and
relative few off the sides--indeed, one would now expect this element to
be becoming directional in two favored directions--off the flat sides ...
to date, I have NOT been able to measure an acceptable difference to
reinforce the "illumination properties" of the element.

The photon/wave properties of rf still remains a mystery ... and proof
hard to come by.

Regards,
JS

John

Imagine your ribbon antena flattened to the thickness of a razor blade.
Instead of using RF, heat the antenna with a blow torch until it becomes
white hot.

It is only when looking at the exact edge of the antenna that any
appreciable drop in light out put will be noticed. At all broadside angles
an appreciable amount of light would be seen. The same effects can be
expected to occur at RF but the majority of amateur test equipment would not
have the resolution to measure the dip with the antenna edge on. The width
of the receiving antenna and diffraction effects would tend to hide this in
the far field, and alignment, reflection effects and manufacturing
tolerances in the near field.

Or perhaps more appropriately, with visible light being around 500
nanometers wavelength, imagine your antenna wire being about 0.01
nanometers thick and 1 nanometer wide (and 250 nanometers long, if you
wish) ... Now does you intuition tell you anything useful about the
angular distribution of emitted photons? I suppose not.

Cheers,
Tom

Mike Kaliski wrote:

...
It is only when looking at the exact edge of the antenna that any
appreciable drop in light out put will be noticed. At all broadside
angles an appreciable amount of light would be seen. The same effects
can be expected to occur at RF but the majority of amateur test
equipment would not have the resolution to measure the dip with the
antenna edge on. The width of the receiving antenna and diffraction
effects would tend to hide this in the far field, and alignment,
reflection effects and manufacturing tolerances in the near field.
...
Mike G0ULI

The eye, like the ear, has defects, in the fact it is not linear.

However, if a ribbon the width and depth of a razor blade is white hot,
a light meter available and rotated around this ribbon--the least energy
would come from the side, the most from the flat. There would be
something of a linear graph in the 90 degree rotation between thinnest
to broadest ... please, don't attempt to kid a kidder.

Regards,
JS