On Sun, 12 Jul 2015 19:54:44 -0700, Jeff Liebermann
wrote:

Oops. I goofed in typing in several places. The last part should be:

(d) Energy_of_Photon = hf = 6.63*10^-34 * 2*10^9 J = 1.3*10^-24 J
where h=6.63*10^-34 Js (Plank's constant)
This is the energy of a 2.5 MHz photon.
From (a), PV=5.5*10-^18 Js^-1 m^-2
Therefore, number of photons =
(5.5*10^-18 / 1.3*^10^-24) = 4.2*10^6 m^-2 s^-1

Hmm... I have no idea where the "2.5 MHz" came from or the strange
units for the "number of photons".



--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

Jeff Liebermann wrote:

On Sun, 12 Jul 2015 19:54:44 -0700, Jeff Liebermann
wrote:

Oops. I goofed in typing in several places. The last part should be:

(d) Energy_of_Photon = hf = 6.63*10^-34 * 2*10^9 J = 1.3*10^-24 J
where h=6.63*10^-34 Js (Plank's constant)
This is the energy of a 2.5 MHz photon.
From (a), PV=5.5*10-^18 Js^-1 m^-2
Therefore, number of photons =
(5.5*10^-18 / 1.3*^10^-24) = 4.2*10^6 m^-2 s^-1

Hmm... I have no idea where the "2.5 MHz" came from or the strange
units for the "number of photons".




Poltergeists?

They share a simply amazing number of similarities to "rf photons" ...
yanno'?

In article , Jeff Liebermann writes:
Photon (RF or light) pressure have been measured in the laboratory by
using two pressure gauges, blocking RF and light from one gauge, and
measuring the differential pressure. The differential measurement
cancels external influences, such as gravity, wind, earth movement,
etc.

Maxwell's equations - classical field theory - predict light
pressure even without photons and quantum theory. Double slit
experiments show interference patterns are followed even by
single photons allowed to to pass - exactly as if each photon
converted to a wave and portions passed through each slit and
thus _the photon interfered with itself_.

You really have to observe quantum effects before you can
register individual photons.

And, with e = h nu, nu being frequency, quantum effects at UHF
and below are much harder to see because each photon has such
low energy.

George

On 7/13/2015 3:16 AM, George Cornelius wrote:
In article , Jeff Liebermann writes:
Photon (RF or light) pressure have been measured in the laboratory by
using two pressure gauges, blocking RF and light from one gauge, and
measuring the differential pressure. The differential measurement
cancels external influences, such as gravity, wind, earth movement,
etc.

Maxwell's equations - classical field theory - predict light
pressure even without photons and quantum theory. Double slit
experiments show interference patterns are followed even by
single photons allowed to to pass - exactly as if each photon
converted to a wave and portions passed through each slit and
thus _the photon interfered with itself_.

You really have to observe quantum effects before you can
register individual photons.

And, with e = h nu, nu being frequency, quantum effects at UHF
and below are much harder to see because each photon has such
low energy.

That is the real problem with observing EM photons. IR which has a much
shorter wavelength and higher frequency stimulates molecular motion,
vibration and spinning which is heat. To see even microwave quanta the
apparatus would have to be cooled to very low temperatures to eliminate
the interference.

Do Josephson junctions work at the level of EM quanta? It has been a
long time since I've seen much about them. I don't even remember what
they do except that they are QM phenomenon.

--

Rick

"Jeff Liebermann" wrote in message
...

Yep, antennas radiate photons.
+1
There is not any proof that RF behaves differently than light.
Things are already quite complicated without it :-)

On Mon, 13 Jul 2015 13:45:43 +0200, "bilou" wrote:

"Jeff Liebermann" wrote in message
.. .

Yep, antennas radiate photons.
+1
There is not any proof that RF behaves differently than light.
Things are already quite complicated without it :-)

One of my not so great ideas was to devise a contraption that would
let me "see" RF. It certainly would make troubleshooting RF devices
much easier. Essentially, it would be a human eye analog implimented
with RF components. According to theory, if it works for light, it
should also work for RF. At the time, I was working at about 1GHz.
Light is about 400 THz. So, all I need is an eyeball that's 400,000
times larger than the human eye. I'll give myself a -1 for the idea.

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

"Jeff Liebermann" wrote in message
...

On Mon, 13 Jul 2015 13:45:43 +0200, "bilou" wrote:

"Jeff Liebermann" wrote in message
. ..

Yep, antennas radiate photons.
+1
There is not any proof that RF behaves differently than light.
Things are already quite complicated without it :-)

One of my not so great ideas was to devise a contraption that would
let me "see" RF. It certainly would make troubleshooting RF devices
much easier. Essentially, it would be a human eye analog implimented
with RF components. According to theory, if it works for light, it
should also work for RF. At the time, I was working at about 1GHz.
Light is about 400 THz. So, all I need is an eyeball that's 400,000
times larger than the human eye. I'll give myself a -1 for the idea.

Wouldn't such a gadget be awesome for adjusting antennas!

On Mon, 13 Jul 2015 08:33:34 -0700, "Wayne"
wrote:
"Jeff Liebermann" wrote in message
.. .

On Mon, 13 Jul 2015 13:45:43 +0200, "bilou" wrote:

"Jeff Liebermann" wrote in message
...

Yep, antennas radiate photons.
+1
There is not any proof that RF behaves differently than light.
Things are already quite complicated without it :-)

One of my not so great ideas was to devise a contraption that would
let me "see" RF. It certainly would make troubleshooting RF devices
much easier. Essentially, it would be a human eye analog implimented
with RF components. According to theory, if it works for light, it
should also work for RF. At the time, I was working at about 1GHz.
Light is about 400 THz. So, all I need is an eyeball that's 400,000
times larger than the human eye. I'll give myself a -1 for the idea.

Wouldn't such a gadget be awesome for adjusting antennas!

Yep. I later realized that it would be marginal for RF circuits
because I could only see the components and traces that radiate RF. If
the circuit was any good, it wouldn't radiate anything.

I also burned some time trying to make an RF equivalent to a liquid
crystal sheet.
http://www.edmundoptics.com/testing-targets/calibration-standards/temperature-sensitive-liquid-crystal-sheets/1642/
Before thermal imagers became relatively inexpensive, I would place a
sheet over the power amplifier or whatever, and be able to see the hot
spots. I was also somewhat successful at creating a blurry thermal
image, using a small germanium lens and one of these sheets.

However, the ideal would be to have a liquid crystal sheet that was
sensitive to RF instead of heat. I couldn't find anything that
detected low frequency RF directly, but did get some interesting
effects by screen printing carbon squares on the thermal sensitive
liquid crystal sheets. The carbon would get slightly warm from the
RF, and cause the color to change. You can also use thermal crayons
to get a similar color change with temperatu
http://www.tiptemp.com/Products/Color-Changing-Thermal-Paint-Crayons/TLCSEN464-245-Color-Change-Crayon-Kit

Long ago, in High Skool, the instructor waved a neon lamp (NE-2) over
a transmission line, so that we could see standing waves. I thought
that was cool, but would be even better if a had a row of neon lamps
so that I didn't need to move the lamp. So, I built one with about
100 NE-2 lamps. Not only could I see the standing waves, but I could
also tune the load for minimum SWR. Today, I could probably built
something similar out of the LED strip lighting on rolls:
http://www.amazon.com/Triangle-Bulbs-T93007-Waterproof-Flexible/dp/B005EHHLD8
However, it would take more power to light up than the NE-2. At 4.8
watts/meter of LED strip, a 20 meter half wave dipole would require 48
watts to fully light at 10 meter long strip.

There are admittedly many things wrong with the aforementioned ideas.
None of them will work because of obvious (and not-so-obvious)
reasons. That's not the point. One has to start somewhere, and
started at "close, but not quite" is as good a place as any.


--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

"Jeff Liebermann" wrote in message
...

One of my not so great ideas was to devise a contraption that would
let me "see" RF. It certainly would make troubleshooting RF devices
much easier. Essentially, it would be a human eye analog implimented
with RF components. According to theory, if it works for light, it
should also work for RF. At the time, I was working at about 1GHz.
Light is about 400 THz. So, all I need is an eyeball that's 400,000
times larger than the human eye. I'll give myself a -1 for the idea.
Yes it is a question of scale.
There is the trick to use a fluorescent light bulb close to an aerial.
Energy saving lamps can be quite small .
Puting the glass part of one in a microwave oven can be instructive.
Don't forget the cup of water. :-)

In article , Jeff Liebermann writes:
let me "see" RF. It certainly would make troubleshooting RF devices
much easier. Essentially, it would be a human eye analog implimented
with RF components. According to theory, if it works for light, it
should also work for RF. At the time, I was working at about 1GHz.
Light is about 400 THz. So, all I need is an eyeball that's 400,000
times larger than the human eye. I'll give myself a -1 for the idea.

A word: synthetic aperture. Remember the dish arrays in
the Jodie Foster movie Contact? You still need the same
scale factor - many times the wavelength - but most of a
dish array can be air.

So with the eyeball analogy, I would first reduce to the
size of the pupil - the aperture - and that is perhaps
5 mm. Times 400K gives 2000m for the same theoretical
resolution. Of course, for a 2D image you would need
an array of antennas spread over a disk of that radius.

Or just calculate directly. I think the angular
resolution of an array or a telescope in radians is
something like

0.22 * wavelength / aperture .

Multiply by about 60 to get degrees.

So for 1 Ghz (.3m) it's 0.22 * .3m / 2000m, or
33 x 10^-6 radians. About 7 seconds of arc.

George