Richard Clark wrote:
On Mon, 10 Jul 2006 17:57:42 -0400, John Popelish
wrote:


I meant that half as many photons
are produced, compared to the full dipole antenna that produces the
same fields above the center line.


Hi John,

So, proceeding along your avowed lines of Photons, one of several
questions:
Presuming 100W radiated, how many photons would that be so that we
can talk about them by halves.

Yes, that is perhaps unfair, however it demonstrates how easily the
discussion can tumble for lack of quantifiables such as that original
offering of 100W.

Should we discuss how infinitesimal the energy is in a 40M photon?
(Easily accounts for why so many are needed for that same 100W.)

No, I suppose not.

Want to get into the problems of diffraction with object lenses that
measure less than a wavelength of the photon?

Hard to escape, and makes a mess of describing mirrors too, especially
when they are skeletal approximations as well.

I can offer more thread-busters when it comes to photonics, but that
is a slam dunk. Get us rolling on one ace proposition, and I will get
back to you in a couple of hours.

73's
Richard Clark, KB7QHC

How many photons does it take to make a Watt?
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:

How many photons does it take to make a Watt?

1/(Hz*6.63*10^-34).

The lower the frequency the less energy per photon.

John Popelish wrote:

Tom Donaly wrote:

How many photons does it take to make a Watt?


1/(Hz*6.63*10^-34).

The lower the frequency the less energy per photon.

That's joules per second, is it?
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
John Popelish wrote:

Tom Donaly wrote:

How many photons does it take to make a Watt?



1/(Hz*6.63*10^-34).

The lower the frequency the less energy per photon.


That's joules per second, is it?

A watt is a joule per second. The formula gives the number of photons
per second that carry a watt (or a joule per second) once you provide
the Hz (frequency).

By the way, I am having second thoughts as to whether or not there
should be a 2*pi factor in there, since most physics formulas deal
with frequency in radians per second, not cycles per second. But the
photon energy formulas usually deal with wavelength, and I have never
seen one that assumes a wavelength is a radian of a cycle, rather that
a full cycle, so, perhaps Hz is the correct unit.

If anyone can clear this up for me, I would appreciate it.

On Tue, 11 Jul 2006 11:06:19 -0400, John Popelish
wrote:

By the way, I am having second thoughts as to whether or not there
should be a 2*pi factor in there, since most physics formulas deal
with frequency in radians per second, not cycles per second. But the
photon energy formulas usually deal with wavelength, and I have never
seen one that assumes a wavelength is a radian of a cycle, rather that
a full cycle, so, perhaps Hz is the correct unit.

If anyone can clear this up for me, I would appreciate it.

Hi John,

That would be 2 pi radians per second as frequency - same thing as a
cycle. For photonic interactions the classic treatment is usually
with wavenumber as frequency not cycles nor radians. However, the 2
pi difference is the difference between the Planck constant
represented as h, and its rational equivalent (with 2 pi divided out)
of h-bar.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
On Tue, 11 Jul 2006 11:06:19 -0400, John Popelish
wrote:


By the way, I am having second thoughts as to whether or not there
should be a 2*pi factor in there, since most physics formulas deal
with frequency in radians per second, not cycles per second. But the
photon energy formulas usually deal with wavelength, and I have never
seen one that assumes a wavelength is a radian of a cycle, rather that
a full cycle, so, perhaps Hz is the correct unit.

If anyone can clear this up for me, I would appreciate it.


Hi John,

That would be 2 pi radians per second as frequency - same thing as a
cycle. For photonic interactions the classic treatment is usually
with wavenumber as frequency not cycles nor radians. However, the 2
pi difference is the difference between the Planck constant
represented as h, and its rational equivalent (with 2 pi divided out)
of h-bar.

Thank you. Makes good sense.