There is the energy-momentum tensor in the electrodynamics.
But, there is no spin tensor.
So,
a circularly polarized plane wave can have no angular momentum.
It is possible to refute the corollary experimentally
by lighting a target by a circularly polarized beam.

Please, mail
REQUEST: send papers
NUMBER: 03-307, 03-311, 03-315
to


Sincerely, Radi Khrapko

"Radi Khrapko" wrote in message
om...
| There is the energy-momentum tensor in the electrodynamics.
| But, there is no spin tensor.
| So,
| a circularly polarized plane wave can have no angular momentum.
| It is possible to refute the corollary experimentally
| by lighting a target by a circularly polarized beam.

Have a look at the following link.

http://www.physics.gla.ac.uk/Optics/...nglePhotonOAM/

FrediFizzx

Dear Timo,
My question (AJP 69, 405) which I set on Sept 15, 1999 was
provocative.
I understood that the absence of a spin tensor meant that J = r x
(ExH) was the orbital AM, not spin.
I knew that circularly polarized plane waves were brimful of spin.
“A circularly polarized plane wave can have no angular
momentum” are Heitler’s words (W. Heitler, 1954).
Humblet did not decompose the AM into an orbital term and spin (see
physics/0102084, Sect 7).
He wrote the orbital AM density as canonical spin density:
\int (rx(DxB))dV = \int (DxA)dV.
Momentum density DxB = 0 in the Beth’s beam.
So, spin density DxA = 0 also.
I mentioned it (Izmeritel'naya Tekhnika
(Ru ISSN 0368-1025, US ISSN 0543-1972), No.4, 2003, p. 3).
So, the Beth’s beam had a spin current without an energy current
and a spin density.
It’s striking!
Please, let me know “various RF measurements”.
Radi Khrapko

Have a look at the following link.

http://www.physics.gla.ac.uk/Optics/...nglePhotonOAM/

FrediFizzx
Many, many times I emphasized that I considered circularly polarized
light without an azimuthal phase structure.
I cited L. Allen, M. J. Padgett, M. Babiker at al. many, many times.
Please, be attentive.
Radi Khrapko

On Thu, 10 Jul 2003, Radi Khrapko wrote:

I understood that the absence of a spin tensor meant that J = r x
(ExH) was the orbital AM, not spin.

I'm not sure how you come to this conclusion, since J = r x (ExH) does
include an origin-independent component, which can be identified as spin
(that is, Humblet's M_s = 1/c \int ExA dv term, which, if you have a
monochromatic field, you can rewrite (in Gaussian units) as

s_i = -i/(8\pi\omega) e_{ijk} E_j^\star E_k

where e_{ijk} is the Levi-Civita symbol, which can also be written as the
usual 3x3 spin matrices for monochromatic EM fields.

[rearranged]

Humblet did not decompose the AM into an orbital term and spin (see
physics/0102084, Sect 7).

I don't follow what you did there. Basically, are you saying that the
Humblet decomposition is wrong? That is, are you saying that Humblet's
eqns (2)-(3) are wrong, or are you disputing his identification of the
orbital and spin terms in his expression for the total angular momentum?

He wrote the orbital AM density as canonical spin density:
\int (rx(DxB))dV = \int (DxA)dV.

Well, Humblet writes the total angular momentum density
J = rx(ExH) = orbital term + (1/c) ExA. In the case of a plane wave the
orbital term is zero.

Momentum density DxB = 0 in the Beth’s beam.
So, spin density DxA = 0 also.

I knew that circularly polarized plane waves were brimful of spin.
“A circularly polarized plane wave can have no angular
momentum” are Heitler’s words (W. Heitler, 1954).

Yes, taking the plane wave limit does make it difficult to calculate the
angular momentum (which is the whole point of using the Humblet
decomposition, since it avoids the problem completely). I don't see that
the usual difficulty with plane waves in any way invalidates the classical
theory, since there are at least two different ways (as mentioned in
previous post) to avoid the difficulty, and the plane wave limit is an
unphysical case.

I usually just use the semi-classical result, very quick and easy in
comparison.

I mentioned it (Izmeritel'naya Tekhnika
(Ru ISSN 0368-1025, US ISSN 0543-1972), No.4, 2003, p. 3).
So, the Beth’s beam had a spin current without an energy current
and a spin density.
It’s striking!

I agree, a quite striking result. Brings to mind comparisons with non-zero
electric current density in a conductor with zero charge density. The cool
thing is that the difference between the incident AM and the returning AM
is 2P/\omega, while the torque on the waveplate is 4P/\omega, with the
difference being made up by the torque of 2P/\omega on the mirrored
waveplate.

Make it simpler. Just consider some of the modern Beth-like experiments,
which only use a single pass through the waveplate. The best measurements
have an experimental error of 10%, and the torque agrees with the usual
result given by classical theory.

Please, let me know “various RF measurements”.

I'll see if I can find the references. There was one, maybe about 1970,
published in one of the IEEE journals, and also a more recent one. I have
both papers somewhere - but I need to find them.

--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html

On Thu, 10 Jul 2003, Radi Khrapko wrote:

My question (AJP 69, 405) which I set on Sept 15, 1999 was
provocative.

Sure, and it's an excellent question, and trying to find the answer is
an educational experience. I see that the replies you got are not entirely
satisfactory. Direct calculation of the absorption of a circularly
polarised wave in a lossy medium does actually give the torque due to
absorption via the stress tensor, so this probably be the best way to
approach the problem.

It also occured to me that the best measurements for determining the
angular momentum due to circular polarisation are probably rotational
frequency shift measurements, in which the rotational work is measured,
rather than the actual torque. See references below.

Please, let me know “various RF measurements”.

Some RF measurements:

F. S. Chute
"The reaction torque on an axial multipole radiator"
IEEE Transactions on Antennas and Propagation 15, 585-587 (1967)

M. Kristensen and J. P. Woerdman
"Is photon angular momentum conserved in a dielectric medium?"
Physical Review Letters 72, 2171-2174 (1994)

Some optical measurements:

M. E. J. Friese, J. Enger, H. Rubinsztein Dunlop, and N. R. Heckenberg
"Optical angular-momentum transfer to trapped absorbing particles"
Physical Review A 54, 1593-1596 (1996)

N. B. Simpson, K. Dholakia, L. Allen, and M. J. Padgett
"Mechanical equivalence of spin and orbital angular momentum of light: an optical spanner"
Optics Letters 22, 52-54 (1997)

And some stuff on rotational Doppler shift, as noted above. Some of these
are RF, some are optical, and some are theory.

M. J. Padgett and L. Allen
"The angular momentum of light: optical spanners and the rotational frequency shift"
Optical and Quantum Electronics 31, 1-12 (1999)

Iwo Bialynicki Birula and Zofia Bialynicki Birula
"Rotational frequency shift"
Physical Review Letters 78, 2539-2542 (1997)

J. Courtial, K. Dholakia, D. A. Robertson, L. Allen, and M. J. Padgett
"Measurement of the rotational frequency shift imparted to a rotating light beam possessing orbital angular momentum"
Physical Review Letters 80, 3217-3219 (1998)

J. Courtial, D. A. Robertson, K. Dholakia, L. Allen, and M. J. Padgett
"Rotational frequency shift of a light beam"
Physical Review Letters 81, 4828-4830 (1998)

V. Bagini, F. Gori, M. Santarsiero, F. Frezza, G. Schettini, and G. Schirripa Spagnolo
"Change of energy of photons passing through rotating anisotropic elements"
European Journal of Physic 15, 71-78 (1994)

A. B. Pippard
"Change of energy of photons passing through rotating anisotropic elements"
European Journal of Physics 15, 79-80 (1994)

F. Bretenaker and A. Le Floch
"Energy exchanges between a rotating retardation plate and a laser beam"
Physical Review Letters 65, 2316 (1990)

And some general theory on spin and orbital AM of EM fields which might be
of interest to you:

James H. Crichton and Philip L. Marston
"The measurable distinction between the spin and orbital angular momenta of electromagnetic radiation"
Electronic Journal of Differential Equations Conf. 04, 37-50 (2000)

S. J. van Enk and G. Nienhuis
"Commutation rules and eigenvalues of spin and orbital angular momentum of radiation fields"
Journal of Modern Optics 41(5), 963-977 (1994)

--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html

"Radi Khrapko" wrote in message
om...
| Have a look at the following link.
|
| http://www.physics.gla.ac.uk/Optics/...nglePhotonOAM/
|
| FrediFizzx
| Many, many times I emphasized that I considered circularly polarized
| light without an azimuthal phase structure.
| I cited L. Allen, M. J. Padgett, M. Babiker at al. many, many times.
| Please, be attentive.
| Radi Khrapko

What the heck is a "circularly polarized plane wave"? From the link above,
it only looks like a plane wave can be only linearly polarized.

Where are these citations you are talking about? They are not in this
thread.

FrediFizzx

http://www.flashrock.com/upload/photong/photong.html
http://www.flashrock.com/upload/photong.ps

On Thu, 10 Jul 2003, FrediFizzx wrote:

What the heck is a "circularly polarized plane wave"? From the link above,
it only looks like a plane wave can be only linearly polarized.

Easy to get a circularly polarised plane wave (apart from the problem in
getting a plane wave in the first place; approximately plane will do).
Just add two plane waves with perpendicular linear polarisation a
quarter wave out of phase. Any half-way decent optics book will cover it.

Where are these citations you are talking about? They are not in this
thread.

In his papers, so you didn't just miss seeing them in this thread.

--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html

"Timo Nieminen" wrote in message
..
..
| On Thu, 10 Jul 2003, FrediFizzx wrote:
|
| What the heck is a "circularly polarized plane wave"? From the link
above,
| it only looks like a plane wave can be only linearly polarized.
|
| Easy to get a circularly polarised plane wave (apart from the problem in
| getting a plane wave in the first place; approximately plane will do).
| Just add two plane waves with perpendicular linear polarisation a
| quarter wave out of phase. Any half-way decent optics book will cover it.

OK, right. But I wasn't talking about mixing *two* plane waves.

FrediFizzx

On Thu, 10 Jul 2003, FrediFizzx wrote:

"Timo Nieminen" wrote in message
..
.
| On Thu, 10 Jul 2003, FrediFizzx wrote:
|
| What the heck is a "circularly polarized plane wave"? From the link
above,
| it only looks like a plane wave can be only linearly polarized.
|
| Easy to get a circularly polarised plane wave (apart from the problem in
| getting a plane wave in the first place; approximately plane will do).
| Just add two plane waves with perpendicular linear polarisation a
| quarter wave out of phase. Any half-way decent optics book will cover it.

OK, right. But I wasn't talking about mixing *two* plane waves.

Mix them, and then you have one plane wave.

Or, if you prefer, take a left-circularly polarised plane wave, and add a
right-circularly polarised plane wave, and you get a plane-polarised plane
wave.

Take a plane wave of arbitrary polarisation, and you can represent it as
either (1) a sum of two plane polarised (x and y) plane waves, or (2) the
sum of two circularly polarised (left and right) plane waves. Classically,
there isn't any particular reason to prefer one representation over the
other in general (although there will be in specific cases), but
non-classically, photon spin is a pretty good argument that the circular
basis is actually fundamental.

--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html