"W5DXP" napisal w wiadomosci
...
On Friday, July 27, 2012 11:57:54 AM UTC-5, Szczepan Bialek wrote:
Why you do not like the electrons?

I don't dislike electrons and have a bunch of them in my body. Without free
electrons, antennas wouldn't be able to radiate photons.

For what you need the next " the equivalent of current flow".

Technically, current is defined as the movement of charged particles. Since
photons don't carry an electric charge, they technically don't meet the
definition. But since current flow implies energy transfer and photons are
capable of energy transfer, one can come up with the concept of "equivalent
current" corresponding to photon movement. Thus the "current flow" through
a capacitor is explained by photon flow rather than displacement current.

"Energy transfer": "Umov was the first who introduced in physics such basic
concepts as speed and direction of movement of energy,".
" In his first works of this period, Umov considered potential energy as
kinetic energy of some environments "imperceptible for us"."
From: http://en.wikipedia.org/wiki/Nikolay_Umov

Photons as the wave packet transfer the energy in the same way as the sound
waves.
In radars are "pulses". Light is not coherent. It consists of pulses or
photons. The same is in sonars.

For what you need the new "photons".
S*

"Szczepan Bialek" wrote in message
...

Photons as the wave packet transfer the energy in the same way as the
sound waves.
S*
Good morning Szczepan.
Sound waves propagation through a gaseous medium. No medium - no sound wave.

Light waves and radio waves can propagate without a gaseous medium.
Please get a textbook and read some theory. If you choose to say, once
again, that textbooks are for children then do remember that children can
learn and understand.

Regards, Ian.

"W5DXP" napisal w wiadomosci
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On Friday, July 27, 2012 12:50:32 PM UTC-5, Szczepan Bialek wrote:
Are the voltages doubled at the ends?

The voltage doubling at the ends of a dipole is simple to understand. A
dipole is a standing wave antenna. When the forward wave from the feedpoint
encounters the open-circuit at the end of the dipole wire, a reflection
takes place where the reflected voltage is in phase with the forward
voltage and the reflected current is 180 degrees out of phase with the
forward current.

Exactly like with the sound waves.

At the end of the dipole, |Vfor|=|Vref| and |Ifor|=|Iref|. So the total
voltage and total current at the ends of a dipole a

Vtot = Vfor + Vref = 2*Vfor = 2*Vref

Itot = Ifor - Iref = 0

And of course, no current is going to flow into the open-circuit at the end
of the dipole so the total energy in the magnetic field at that point is
zero.

You assume that there no field electron emission. Why?

All of the energy in the EM waves migrates to the electric field and that's
why we get a standing wave voltage maximum at the ends of a 1/2WL dipole.

Nobody know what the EM waves are. Write about electrons.

However, under certain corona conditions, when the impedance at the end of
the dipole conductor is much less than infinite, electrons actually migrate
from the antenna into the conductive air, e.g. salty fog on the coast. But
this is undesirable non-coherent behavior unless you are building a Tesla
coil.

They migrate if the voltage is:
"Field emission was explained by quantum tunneling of electrons in the late
1920s. This was one of the triumphs of the nascent quantum mechanics. The
theory of field emission from bulk metals was proposed by Ralph H. Fowler
and Lothar Wolfgang Nordheim.[1] A family of approximate equations,
"Fowler-Nordheim equations", is named after them."

I once saw a mobile antenna emitting a red corona glow in the fog on HWY 1
near Monterrey, CA. A traffic cop stopped him for having a "red light" that
could be seen by oncoming traffic.

Each your antennas are the electron emitter:
"How would the ideal field emitter look like? It should be very long and
very thin, made of conductive material with high mechanical strength, be
robust, and cheap and easy to process". From:
http://ipn2.epfl.ch/CHBU/NTfieldemission1.htm

The voltage doubling is only theoretical. In reality the voltage is rising
to the level necessary for effective field emissions.
If no proper voltage no radiation.
S*

Szczepan Bialek wrote:
And of course, no current is going to flow into the open-circuit at the end
of the dipole so the total energy in the magnetic field at that point is
zero.

You assume that there no field electron emission. Why?

Because if it was there, you would see the sparks and corona effects.

Each your antennas are the electron emitter:
"How would the ideal field emitter look like? It should be very long and
very thin, made of conductive material with high mechanical strength, be
robust, and cheap and easy to process". From:

Of course we all know that antennas exist that are not long and thin,
and still work very well. Well, we all know that except you the stubborn
Pole.

On Saturday, July 28, 2012 2:59:06 AM UTC-5, Szczepan Bialek wrote:
Light is not coherent.

On the contrary, any single frequency light, i.e. single color light, is coherent. Most of the laser light in an interferometer is coherent. Otherwise, steady-state interference could not be achieved.
--
73, Cecil, w5dxp.com

On Saturday, July 28, 2012 3:58:40 AM UTC-5, Szczepan Bialek wrote:
You assume that there no field electron emission. Why?

Please don't tell me what I am assuming because you are invariably wrong. As I (and Richard Feynman) have told you before, the fields emitted by electrons consist of photons. Electrons escaping a transmitting antenna are a non-coherent corona discharge. Photons escaping a transmitting antenna are coherent RF energy.

Back in the early 20th century when spark gap transmitters were being used, the spark gap function, like lightning, generated non-coherent photons which filled the RF spectrum. Nowadays, a CW signal is mostly coherent with a small amount of non-coherence associated with the transient rise and fall times.
--
73, Cecil, w5dxp.com

"Rob" napisa³ w wiadomo¶ci
...
Szczepan Bialek wrote:
And of course, no current is going to flow into the open-circuit at the
end
of the dipole so the total energy in the magnetic field at that point is
zero.

You assume that there no field electron emission. Why?

Because if it was there, you would see the sparks and corona effects.

It is better to measure it:
"Ian White GM3SEK wrote:

Coming back to the effectiveness of baluns, the final decider is the
amount of unwanted RF current on the outside of the coax shield,
compared to the wanted current in the antenna element. The only way to
find that out for sure is to *measure* it, in the system as installed.

I'm a big fan of RF current meters based on simple snap-on ferrite
beads. Add a few turns of wire, one resistor, a diode detector, and you
have a real measuring instrument. It's a real eye-opener to be able to
snap the meter over any cable and *see* the common mode RF current.

Have you the result of counting of electrons which go forwards and backwards
at the feed point?
S*

Szczepan Bialek wrote:

"Rob" napisa³ w wiadomo¶ci
...
Szczepan Bialek wrote:
And of course, no current is going to flow into the open-circuit at the
end
of the dipole so the total energy in the magnetic field at that point is
zero.

You assume that there no field electron emission. Why?

Because if it was there, you would see the sparks and corona effects.

It is better to measure it:

Why don't you try that?

"Ian White GM3SEK wrote:

Coming back to the effectiveness of baluns, the final decider is the
amount of unwanted RF current on the outside of the coax shield,
compared to the wanted current in the antenna element. The only way to
find that out for sure is to *measure* it, in the system as installed.

I'm a big fan of RF current meters based on simple snap-on ferrite
beads. Add a few turns of wire, one resistor, a diode detector, and you
have a real measuring instrument. It's a real eye-opener to be able to
snap the meter over any cable and *see* the common mode RF current.

Have you the result of counting of electrons which go forwards and backwards
at the feed point?

You have the misunderstanding that the current on the outside of the
coax cable is the result of electron emission by the antenna. This
is of course not true. It is the result of imperfect symmetry or
imperfect balun operation.

"Ian" napisa³ w wiadomo¶ci
...
"Szczepan Bialek" wrote in message
...

Photons as the wave packet transfer the energy in the same way as the
sound waves.
S*
Good morning Szczepan.
Sound waves propagation through a gaseous medium. No medium - no sound
wave.

Light waves and radio waves can propagate without a gaseous medium.
Please get a textbook and read some theory. If you choose to say, once
again, that textbooks are for children then do remember that children can
learn and understand.

"Electric currents that oscillate at radio frequencies have special
properties not shared by direct current or alternating current of lower
frequencies. The energy in an RF current can radiate off a conductor into
space as electromagnetic waves (radio waves); this is the basis of radio
technology."

As you see something radiate off a conductor into space. So the something
must be in the space.
Are such issue (special properties) for children?

S*

"Rob" napisa³ w wiadomo¶ci
...
Szczepan Bialek wrote:

"Rob" napisa3 w wiadomo?ci
...
Szczepan Bialek wrote:
And of course, no current is going to flow into the open-circuit at the
end
of the dipole so the total energy in the magnetic field at that point
is
zero.

You assume that there no field electron emission. Why?

Because if it was there, you would see the sparks and corona effects.

It is better to measure it:

Why don't you try that?

"Ian White GM3SEK wrote:

Coming back to the effectiveness of baluns, the final decider is the
amount of unwanted RF current on the outside of the coax shield,
compared to the wanted current in the antenna element. The only way to
find that out for sure is to *measure* it, in the system as installed.

I'm a big fan of RF current meters based on simple snap-on ferrite
beads. Add a few turns of wire, one resistor, a diode detector, and you
have a real measuring instrument. It's a real eye-opener to be able to
snap the meter over any cable and *see* the common mode RF current.

Have you the result of counting of electrons which go forwards and
backwards
at the feed point?

You have the misunderstanding that the current on the outside of the
coax cable is the result of electron emission by the antenna. This
is of course not true. It is the result of imperfect symmetry or
imperfect balun operation.

But now is possible to measure the currents. Somebody wrote: "I'm a big fan
of RF current meters".
Do you know the results?
S*