On Sat, 11 Jul 2015 08:33:24 -0700, "Wayne"
wrote:

“The antenna, like the eye, is a transformation device converting
electromagnetic
photons into circuit currents; but, unlike the eye, the antenna can also
convert energy
from a circuit into photons radiated into space. In simplest terms an
antenna converts
photons to currents or vice versa.” Antennas, Second Edition, 1988, by John
D. Kraus. Page 19.

Yep, antennas radiate photons.

Quantum theory treats RF and light as both a wave and a particle. If
you put a pressure gauge behind a flat plate in front of your antenna,
you would be able to detect the tiny pressure produced by your RF
emissions. The problem is that it's very very very small because the
energy decreases linearly with the frequency. When calculating
orbital mechanics, light pressure is a major consideration. When
building interstellar space craft, "light sails" are a common idea.

This should explain it:
https://www.khanacademy.org/test-prep/mcat/physical-processes/light-and-electromagnetic-radiation-questions/v/photon-energy#

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.


In Kraus 3rd edition (2002), the term "photons" appears in questions
2-16-1.

2-16-1. Spaceship near moon.
A spaceship at lunar distance from the earth transmits 2 GHz waves. If
a power of 10 W is radiated isotropically, find (a) the average
Poynting vector at the earth, (b) the rms electric field E at the
earth and (c) the time it takes for the radio waves to travel from the
spaceship to the earth. (Take the earth-moon distance as 380 Mm.) (d)
How many photons per unit area per second fall on the earth from the
spaceship transmitter?

I couldn't paste the answer because the original, in MS Word format,
used characters in the formulas that don't translate to ASCII very
gracefully. I'll try to trasnscribble and annotate for clarity:

(d) Energy_of_Photon = hf = 6.63*10^-34 * 2*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 (or 1/J)

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