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Timo Nieminen science forum Guru Wannabe
Joined: 12 May 2005
Posts: 244

Posted: Thu Jun 15, 2006 8:22 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



On Fri, 15 Jun 2006, khrapko_ri@hotmail.com wrote:
Quote:  We calculate absorption of a circularly polarized light beam in a
dielectric in the frame of the standard electrodynamics. A transfer of
energy, momentum, and angular momentum from the beam to the dielectric
is calculated. The calculation shows that the angular momentum flux in
the beam equals to two power of the beam divided by frequency.

Now try this for an infinite plane wave and an infinite dielectric. Do you
not get an AM flux of P/w? Does not using your spin tensor agree?
OK, both the infinite plane wave and infinite dielectric are unphysical.
However, consider a finite but very wide flattopped beam, and a finite
dielectric somewhat wider than the beam.
Calculating the torque due to absorption gives the same result as before,
P/w (we might disagree on this point!) and also shows that this torque is
evenly spread over the beam. The J=rxS/c school of thought claims that the
AM flux is P/w, and is localised at the edges. Something is wrong here,
for a sufficiently wide beam would require superluminal transport of AM
from the edges of the beam to the middle of the dielectric. You've already
pointed out the problem of what happens when the outer portion of the
dielectric is fixed, and the inner portion is free to rotate  torque due
to absorption tells you it will rotate, J=rxS/c says no. Clearly J=rxS/c
is wrong.
On the other hand, Humblet showed that it is possible to reconcile the
two, by showing that the integral of J=rxS/c gives the correct total AM,
under practical conditions, and gave an expression for the spin flux.
Now, the torque exerted on an object due to a change in circular
polarisation has been measured. Holbourn and Beth both did it, and their
results showed that a circularly polarised beam carries P/w AM. You say
"Not so!" due to the doublepass nature of Beth's experiment. I don't know
the details of Holbourn's experiment  his paper is very brief. Our own
experiments are singlepass, and show P/w to hold within a few percent.
OK, those aren't absorption experiments, and I can't think of any
directmeasurement absorption experiments. But how about:
M. E. J. Friese, J. Enger, H. RubinszteinDunlop and N. R. Heckenberg
Optical angularmomentum transfer to trapped absorbing particles
Physical Review A 54, 15931596 (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, 5254 (1997)
where it was shown that the spin flux is P/w if the orbital angular
momentum flux of vortex beams is lP/w, where l is the "charge" of the
singularity.
For your prediction of total torque due to absorption to be 2P/w, this
would require the orbital AM flux of a vortex beam to be double as well,
equal to 2lP/w. Is this compatible with your theory?

Timo Nieminen  Home page: http://www.physics.uq.edu.au/people/nieminen/
Eprints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html 

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Josef Matz science forum Guru Wannabe
Joined: 08 May 2005
Posts: 255

Posted: Sun Jun 18, 2006 6:27 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



<khrapko_ri@hotmail.com> schrieb im Newsbeitrag
news:1150393958.691618.235710@y41g2000cwy.googlegroups.com...
Quote:  We calculate absorption of a circularly polarized light beam in a
dielectric in the frame of the standard electrodynamics. A transfer of
energy, momentum, and angular momentum from the beam to the dielectric
is calculated. The calculation shows that the angular momentum flux in
the beam equals to two power of the beam divided by frequency. This

No its just P/w for circular polarised light. Not 2P/w !
Analogon: Energy of photon hbar w, Spin = Energy / w for circular polarized
photons.
And also note: Natural light has spin zero !
Quote:  result contradicts another part of the electrodynamics, which predicts
the flux equals to power of the beam divided by frequency. In addition

therefore your theory seems to be wrong.
Quote:  we show that this part of the electrodynamics contradicts the classical
Beth's experiment. Our inference is: the electrodynamics is
incomplete.

That is true
Quote:  To correct the electrodynamics, we introduce a spin tensor

Spin tensor concept is nonsense ! As long as your dielectrics do not move
with velocities of at least
1% of c, which is the case because your dielectrics are at rest, for what do
you need a spin tensor ?
Quote:  into the electrodynamics. The corrected electrodynamics is in
accordance with our calculation and with the Beth's experiment.

This i doubt.
Quote:  These results are published at www.sciprint.org and
www.mai.ru/projects/mai_works/
Unfortunately, Mathias Schubert, Topical Editor of the Journal of the
Optical Society of America A, rejected the paper without a peerreview
process.
My reply is:
Dear Mathias Schubert:
Sorry, you criminally try to hush up serious mistakes of R. Loudon, A.
Bishop and others. You will share the disgrace of journals that deny
spin in the classical electrodynamics. I do not need your favorable
report. I need a report of a peerreview process.
The editor's message was:
"Dear Dr. Khrapko:
After editorial reviewing of your paper, I have decided that the
content of your paper is not suitable for publication in the Journal of
the Optical Society of America A. A paper is acceptable for publication
in JOSA A only if the reviewers are convinced that, in addition to
being correct technically, it also adds a new and important result to
the field. Considering the history of scientific journal disapprovals
of the materials of your paper it obviously does not meet the criterion
of JOSA A. Thank you for submitting your paper to JOSA A. I regret that
it is not possible to send you a more favorable report on this
manuscript, and I hope that you will continue to consider JOSA A for
future submissions."
Radi Khrapko



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Josef Matz science forum Guru Wannabe
Joined: 08 May 2005
Posts: 255

Posted: Mon Jun 19, 2006 7:02 pm Post subject:
How it works with the complex index



Present index theory is wrong.
With the right theory it is possible to decribe the Nimtz double prism
experiment. Main result:
Light can pass forbiden regions with velocities faster than the speed of
light limit c. Nimtz double
prism experiment is nothing else than macroscopic electrodynamics. Many
former publicatios
on measurement of superluminal effects are true. These experimentators have
been shot out by
the velocity not faster than c dogma which is proven to be wrong for
inhomogene wave superpositions.
With the right theory it is possible to get energy fluxes and energy
densities in metal layers. Two
things in the old Drude metall optics are wrong. the concept of vacuumlayers
eliminates unlogics
in the old theory. New so called tunnel fluxes in layers make energy
conservation understandable
at any surface between conductive substances.
Vacuum layers and tunnel fluxes are the main elements of the new theory of
complex index.
Even hard theoreticans now have to accept that complex dielectricity
constants exist.
Makroskopic electrodynamics becomes a strongly theoretical fundament of base
physics.
A formula for the spin flux, a second flux density in electrodynamics has
been developed.
This makes understandable that theoretical ununderstood effects and forces
occur when light
is passing matter. This spin forces are experimentally known since mid of
the 30s of the last century.
Now they can be calculated in terms of a strong theory. Spin forces occur
next to angular momentum forces
and momentum forces. Fundamental principle is: Spin is generally not
conserved when passing surfaces. This
non conservation can be calculated from index theory. But there is existing
a conservation law for the complete
angular momentum transfer everywhere in substances. This principle makes it
possible to calculate the forces.
There are to be distinguished volume and surface forces (momenta).
Magnetism is electronically induced. A strong phänomenological theory for
dia  and paramagnetism has been found.
This complete bundle can be the base for a new chapter in classical
electrodynamics of electromagnetic
waves which is in accordands with fundamental quantum concepts.
Invitation to dicussions welcomed.
Josef Matz 

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Radi Khrapko science forum Guru Wannabe
Joined: 04 May 2005
Posts: 142

Posted: Tue Jun 20, 2006 5:05 am Post subject:
Re: Circularly polarized beam in a dielectric, etc.



Timo A. Nieminen wrote:
Quote:  On Fri, 15 Jun 2006, khrapko_ri@hotmail.com wrote:
We calculate absorption of a circularly polarized light beam in a
dielectric in the frame of the standard electrodynamics. A transfer of
energy, momentum, and angular momentum from the beam to the dielectric
is calculated. The calculation shows that the angular momentum flux in
the beam equals to two power of the beam divided by frequency.
Now try this for an infinite plane wave and an infinite dielectric. Do you
not get an AM flux of P/w? Does not using your spin tensor agree?
A beam carries two power/w, but central part of a beam and a plane wave 
carry one (power density)/w. It's clear.
Quote: 
OK, both the infinite plane wave and infinite dielectric are unphysical.
However, consider a finite but very wide flattopped beam, and a finite
dielectric somewhat wider than the beam.
Why do you not show this calculation to me?
Calculating the torque due to absorption gives the same result as before,
P/w (we might disagree on this point!) and also shows that this torque is
evenly spread over the beam. The J=rxS/c school of thought claims that the
AM flux is P/w, and is localised at the edges. Something is wrong here,
for a sufficiently wide beam would require superluminal transport of AM
from the edges of the beam to the middle of the dielectric. You've already
pointed out the problem of what happens when the outer portion of the
dielectric is fixed, and the inner portion is free to rotate  torque due
to absorption tells you it will rotate, J=rxS/c says no. Clearly J=rxS/c
is wrong.
On the other hand, Humblet showed that it is possible to reconcile the
two, by showing that the integral of J=rxS/c gives the correct total AM,
under practical conditions, and gave an expression for the spin flux.
Humblet and others showed that \int rxS equals \int ExA rather than rxS 
is spin flux
Quote: 
Now, the torque exerted on an object due to a change in circular
polarisation has been measured. Holbourn and Beth both did it, and their
results showed that a circularly polarised beam carries P/w AM. You say
"Not so!" due to the doublepass nature of Beth's experiment. I don't know
the details of Holbourn's experiment  his paper is very brief. Our own
experiments are singlepass, and show P/w to hold within a few percent.
Your experiment is a puzzle for me. But formula J=rxS gives zero in the 
Beth experiment because S=0 in the Beth experiment. Only formula
J=rxS+spintensor gives the Beth result.
Quote: 
OK, those aren't absorption experiments, and I can't think of any
directmeasurement absorption experiments. But how about:
M. E. J. Friese, J. Enger, H. RubinszteinDunlop and N. R. Heckenberg
Optical angularmomentum transfer to trapped absorbing particles
Physical Review A 54, 15931596 (1996)
This paper confirms my theory. The point is their particles occupied 
the hole of the donut and the region where \partial u^2 > 0 and thus
the torque acts opposite to the beam polarization handedness.
Therefore, in the absence of spintensor, an absorbing particle must
rotate slower if the beam is changed to circularly polarized of the
same sense as the helicity of the donut, contrary to their result. Only
spintensor forces the particle to rotate faster.
Quote: 
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, 5254 (1997)
Results of this paper is qualitative. The authors believe that their 
particles were insufficiently uniform, resulting in an unwanted mode
transformation and giving an additional exchange of orbital angular
momentum. In the experiment not all regions where \partial u^2 < 0 were
occupied by a particle as well.
Quote: 
where it was shown that the spin flux is P/w if the orbital angular
momentum flux of vortex beams is lP/w, where l is the "charge" of the
singularity.
For your prediction of total torque due to absorption to be 2P/w, this
would require the orbital AM flux of a vortex beam to be double as well,
equal to 2lP/w. Is this compatible with your theory?

Timo Nieminen

Radi Khrapko 

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Radi Khrapko science forum Guru Wannabe
Joined: 04 May 2005
Posts: 142

Posted: Tue Jun 20, 2006 6:02 am Post subject:
Re: Circularly polarized beam in a dielectric, etc.



Josef Matz wrote:
Quote:  khrapko_ri@hotmail.com> schrieb im Newsbeitrag
news:1150393958.691618.235710@y41g2000cwy.googlegroups.com...
We calculate absorption of a circularly polarized light beam in a
dielectric in the frame of the standard electrodynamics. A transfer of
energy, momentum, and angular momentum from the beam to the dielectric
is calculated. The calculation shows that the angular momentum flux in
the beam equals to two power of the beam divided by frequency. This
No its just P/w for circular polarised light. Not 2P/w !
Analogon: Energy of photon hbar w, Spin = Energy / w for circular polarized
photons.
P/w is for a plane wave, but 2P/w is for a beam
And also note: Natural light has spin zero !
That's not the point
result contradicts another part of the electrodynamics, which predicts
the flux equals to power of the beam divided by frequency. In addition
therefore your theory seems to be wrong.
Therefore my theory is worthy of the Nobel Prize
we show that this part of the electrodynamics contradicts the classical
Beth's experiment. Our inference is: the electrodynamics is
incomplete.
That is true
To correct the electrodynamics, we introduce a spin tensor
Spin tensor concept is nonsense ! As long as your dielectrics do not move
with velocities of at least
1% of c, which is the case because your dielectrics are at rest, for what do
you need a spin tensor ?
Dielectric is not the point. Electromagnetic waves carry spin, and this 
spin is described by the tensor.
Quote: 
into the electrodynamics. The corrected electrodynamics is in
accordance with our calculation and with the Beth's experiment.
This i doubt.
All doubt! 
Radi Khrapko 

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Timo Nieminen science forum Guru Wannabe
Joined: 12 May 2005
Posts: 244

Posted: Tue Jun 20, 2006 8:44 am Post subject:
Re: Circularly polarized beam in a dielectric, etc.



On Tue, 19 Jun 2006, khrapko_ri@hotmail.com wrote:
[I'll see about posting a calculation of absorption torque in the near
future unless my time disappears.]
Quote:  Timo A. Nieminen wrote:
On the other hand, Humblet showed that it is possible to reconcile the
two, by showing that the integral of J=rxS/c gives the correct total AM,
under practical conditions, and gave an expression for the spin flux.
Humblet and others showed that \int rxS equals \int ExA rather than rxS
is spin flux

Exactly. rxS/c is clearly not spin flux. ExA could be spin flux, since
it's independent of choice of origin about which to take moments.
Quote:  Now, the torque exerted on an object due to a change in circular
polarisation has been measured. Holbourn and Beth both did it, and their
results showed that a circularly polarised beam carries P/w AM. You say
"Not so!" due to the doublepass nature of Beth's experiment. I don't know
the details of Holbourn's experiment  his paper is very brief. Our own
experiments are singlepass, and show P/w to hold within a few percent.
Your experiment is a puzzle for me. But formula J=rxS gives zero in the
Beth experiment because S=0 in the Beth experiment.
Only formula
J=rxS+spintensor gives the Beth result.

Even considering edge effects? Consider a beam narrow than the waveplate.
The longitudinal part of the Poynting vector cancels due to the
counterpropagation, but the azimuthal parts do not  the handedness of
the polarisation about the direction of propagation is reversed on
reflection, so the direction of the angular momentum about an axis fixed
wrt the apparatus is the same, so the azimuthal parts _add_ due to the
counterpropagation.
Without edge effects, J=spintensor gives the Beth result.
Other than that, J=ExA also works, for a suitable choice of A. For any
choice of A, Humblet's general result (including the orbital term) works.
Quote:  OK, those aren't absorption experiments, and I can't think of any
directmeasurement absorption experiments. But how about:
M. E. J. Friese, J. Enger, H. RubinszteinDunlop and N. R. Heckenberg
Optical angularmomentum transfer to trapped absorbing particles
Physical Review A 54, 15931596 (1996)
This paper confirms my theory. The point is their particles occupied
the hole of the donut and the region where \partial u^2 > 0 and thus
the torque acts opposite to the beam polarization handedness.
Therefore, in the absence of spintensor, an absorbing particle must
rotate slower if the beam is changed to circularly polarized of the
same sense as the helicity of the donut, contrary to their result. Only
spintensor forces the particle to rotate faster.

Well, yes, it's a disproof of the J=rxS/c school of thought. The Humblet
decomposition also explains it. See below for further
[cut]
Quote:  where it was shown that the spin flux is P/w if the orbital angular
momentum flux of vortex beams is lP/w, where l is the "charge" of the
singularity.
For your prediction of total torque due to absorption to be 2P/w, this
would require the orbital AM flux of a vortex beam to be double as well,
equal to 2lP/w. Is this compatible with your theory?

I suspect that the exactness of the equivalence between "spin" and
"orbital" AM (according to the view in paraxial vortex beams where L=lP/w
and S=(degree of circular polarisation)P/w) depends on the radius of the
trapped particle in the context of your theory. It's worth checking.

Timo Nieminen  Home page: http://www.physics.uq.edu.au/people/nieminen/
Eprints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html 

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Josef Matz science forum Guru Wannabe
Joined: 08 May 2005
Posts: 255

Posted: Tue Jun 20, 2006 5:40 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



"Timo A. Nieminen" <timo@physics.uq.edu.au> schrieb im Newsbeitrag
news:Pine.WNT.4.64.0606201827440.1300@serene.st...
Quote:  On Tue, 19 Jun 2006, khrapko_ri@hotmail.com wrote:
[I'll see about posting a calculation of absorption torque in the near
future unless my time disappears.]
Timo A. Nieminen wrote:
On the other hand, Humblet showed that it is possible to reconcile the
two, by showing that the integral of J=rxS/c gives the correct total
AM,
under practical conditions, and gave an expression for the spin flux.
Humblet and others showed that \int rxS equals \int ExA rather than rxS
is spin flux
Exactly. rxS/c is clearly not spin flux. ExA could be spin flux, since
it's independent of choice of origin about which to take moments.

r x S/c is angular momentum flux
S/c impulse flux
E x A cant be spin flux. If at all E x A* a possible description. E x A is
zero in time average !!!!
Quote:  Now, the torque exerted on an object due to a change in circular
polarisation has been measured. Holbourn and Beth both did it, and
their
results showed that a circularly polarised beam carries P/w AM. You say
"Not so!" due to the doublepass nature of Beth's experiment. I don't
know
the details of Holbourn's experiment  his paper is very brief. Our own
experiments are singlepass, and show P/w to hold within a few percent.
Your experiment is a puzzle for me. But formula J=rxS gives zero in the
Beth experiment because S=0 in the Beth experiment.
Only formula
J=rxS+spintensor gives the Beth result.

If you replace spintensor by the word spinvector or the word spinfluxdensity
then you are right.
but dont forget the /c factor in the first term for J. J is the flow density
of angular momentum incl. spin.
J(total) = J + J(mechanical) is conserved everywhere.
Quote:  Even considering edge effects? Consider a beam narrow than the waveplate.
The longitudinal part of the Poynting vector cancels due to the
counterpropagation, but the azimuthal parts do not  the handedness of
the polarisation about the direction of propagation is reversed on
reflection, so the direction of the angular momentum about an axis fixed
wrt the apparatus is the same, so the azimuthal parts _add_ due to the
counterpropagation.
Without edge effects, J=spintensor gives the Beth result.
Other than that, J=ExA also works, for a suitable choice of A. For any
choice of A, Humblet's general result (including the orbital term) works.
OK, those aren't absorption experiments, and I can't think of any
directmeasurement absorption experiments. But how about:
M. E. J. Friese, J. Enger, H. RubinszteinDunlop and N. R. Heckenberg
Optical angularmomentum transfer to trapped absorbing particles
Physical Review A 54, 15931596 (1996)
This paper confirms my theory. The point is their particles occupied
the hole of the donut and the region where \partial u^2 > 0 and thus
the torque acts opposite to the beam polarization handedness.
Therefore, in the absence of spintensor, an absorbing particle must
rotate slower if the beam is changed to circularly polarized of the
same sense as the helicity of the donut, contrary to their result. Only
spintensor forces the particle to rotate faster.
Well, yes, it's a disproof of the J=rxS/c school of thought. The Humblet
decomposition also explains it. See below for further
[cut]
where it was shown that the spin flux is P/w if the orbital angular
momentum flux of vortex beams is lP/w, where l is the "charge" of the
singularity.
For your prediction of total torque due to absorption to be 2P/w, this
would require the orbital AM flux of a vortex beam to be double as
well,
equal to 2lP/w. Is this compatible with your theory?
I suspect that the exactness of the equivalence between "spin" and
"orbital" AM (according to the view in paraxial vortex beams where L=lP/w
and S=(degree of circular polarisation)P/w) depends on the radius of the
trapped particle in the context of your theory. It's worth checking.

Timo Nieminen  Home page: http://www.physics.uq.edu.au/people/nieminen/
Eprints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html



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Radi Khrapko science forum Guru Wannabe
Joined: 04 May 2005
Posts: 142

Posted: Wed Jun 21, 2006 1:44 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



Quote:  Your experiment is a puzzle for me. But formula J=rxS gives zero in the
Beth experiment because S=0 in the Beth experiment.
Only formula
J=rxS+spintensor gives the Beth result.
Even considering edge effects? Consider a beam narrow than the waveplate.
The longitudinal part of the Poynting vector cancels due to the
counterpropagation, but the azimuthal parts do not  the handedness of
the polarisation about the direction of propagation is reversed on
reflection, so the direction of the angular momentum about an axis fixed
wrt the apparatus is the same, so the azimuthal parts _add_ due to the
counterpropagation.
Without edge effects, J=spintensor gives the Beth result.
Other than that, J=ExA also works, for a suitable choice of A. For any
choice of A, Humblet's general result (including the orbital term) works.
Oh, my ears and whiskers! Dear Timo, why are you not attentive as well 
as hundreds of editors and referees? Many years ago I showed and
published that all parts of the Poynting vector are zero in the Beth
experiment (physics/0102084; Measurement Techniques, 46, No. 4, 317;
mp_arc@mail.ma.utexas.edu 03307; mp_arc@mail.ma.utexas.edu 03311;
www.sciprint.org). There is no AM flux in the Beth experiment without
using my spin tensor. The formula J=rxS gives zero in the Beth
experiment even considering edge effects. Only formula J=rxS+spintensor
gives the Beth result. But the spintensor is my spin tensor submitted
for the first time to JETP on Jan. 27, 1999. ExA is not a spin tensor
(see: Exercisesinthecanonicalspintensor www.sciprint.org).
Quote:  This paper confirms my theory. The point is their particles occupied
the hole of the donut and the region where \partial u^2 > 0 and thus
the torque acts opposite to the beam polarization handedness.
Therefore, in the absence of spintensor, an absorbing particle must
rotate slower if the beam is changed to circularly polarized of the
same sense as the helicity of the donut, contrary to their result. Only
spintensor forces the particle to rotate faster.
Well, yes, it's a disproof of the J=rxS/c school of thought. The Humblet
decomposition also explains it. See below for further
I wrote many times that Humblet, Jackson, Ohanian, Crichton and 
Marston, Stewart and others do NOT decompose the orbital AM \int
rx(ExB) into "orbital" and "spin" parts. They transform \int rx(ExB)
into the form of \int ExA or \int FxB. They do not get two terms. One
of their terms is zero (see physics/0102084;
Exercisesinthecanonicalspintensor or Electrodynamics spin at
www.sciprint.org. Now this paper is considered by "Communications in
Mathematical Physics" since May 9, 2006)
Quote:  I suspect that the exactness of the equivalence between "spin" and
"orbital" AM (according to the view in paraxial vortex beams where L=lP/w
and S=(degree of circular polarisation)P/w) depends on the radius of the
trapped particle in the context of your theory. It's worth checking.
Well, L=lP/w and Spin=(degree of circular polarisation)P/w are two 
different beings. L=\int rx(ExB) and Spin = spintensor. The spintensor
is my electrodynamics spin tensor which is absent in the standard
electrodynamics. So, the total AM is the sum J=L+Spin. So, the beam
carries 2P/w.
Quote:  
Timo Nieminen 

Radi Khrapko 

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Radi Khrapko science forum Guru Wannabe
Joined: 04 May 2005
Posts: 142

Posted: Wed Jun 21, 2006 1:50 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



Quote:  Your experiment is a puzzle for me. But formula J=rxS gives zero in the
Beth experiment because S=0 in the Beth experiment.
Only formula
J=rxS+spintensor gives the Beth result.
Even considering edge effects? Consider a beam narrow than the waveplate.
The longitudinal part of the Poynting vector cancels due to the
counterpropagation, but the azimuthal parts do not  the handedness of
the polarisation about the direction of propagation is reversed on
reflection, so the direction of the angular momentum about an axis fixed
wrt the apparatus is the same, so the azimuthal parts _add_ due to the
counterpropagation.
Without edge effects, J=spintensor gives the Beth result.
Other than that, J=ExA also works, for a suitable choice of A. For any
choice of A, Humblet's general result (including the orbital term) works.
Oh, my ears and whiskers! Dear Timo, why are you not attentive as well 
as hundreds of editors and referees? Many years ago I showed and
published that all parts of the Poynting vector are zero in the Beth
experiment (physics/0102084; Measurement Techniques, 46, No. 4, 317;
mp_arc@mail.ma.utexas.edu 03307; mp_arc@mail.ma.utexas.edu 03311;
www.sciprint.org). There is no AM flux in the Beth experiment without
using my spin tensor. The formula J=rxS gives zero in the Beth
experiment even considering edge effects. Only formula J=rxS+spintensor
gives the Beth result. But the spintensor is my spin tensor submitted
for the first time to JETP on Jan. 27, 1999. ExA is not a spin tensor
(see: Exercisesinthecanonicalspintensor www.sciprint.org).
Quote:  This paper confirms my theory. The point is their particles occupied
the hole of the donut and the region where \partial u^2 > 0 and thus
the torque acts opposite to the beam polarization handedness.
Therefore, in the absence of spintensor, an absorbing particle must
rotate slower if the beam is changed to circularly polarized of the
same sense as the helicity of the donut, contrary to their result. Only
spintensor forces the particle to rotate faster.
Well, yes, it's a disproof of the J=rxS/c school of thought. The Humblet
decomposition also explains it. See below for further
I wrote many times that Humblet, Jackson, Ohanian, Crichton and 
Marston, Stewart and others do NOT decompose the orbital AM \int
rx(ExB) into "orbital" and "spin" parts. They transform \int rx(ExB)
into the form of \int ExA or \int FxB. They do not get two terms. One
of their terms is zero (see physics/0102084;
Exercisesinthecanonicalspintensor or Electrodynamics spin at
www.sciprint.org. Now this paper is considered by "Communications in
Mathematical Physics" since May 9, 2006)
Quote:  I suspect that the exactness of the equivalence between "spin" and
"orbital" AM (according to the view in paraxial vortex beams where L=lP/w
and S=(degree of circular polarisation)P/w) depends on the radius of the
trapped particle in the context of your theory. It's worth checking.
Well, L=lP/w and Spin=(degree of circular polarisation)P/w are two 
different beings. L=\int rx(ExB) and Spin = spintensor. The spintensor
is my electrodynamics spin tensor which is absent in the standard
electrodynamics. So, the total AM is the sum J=L+Spin. So, the beam
carries 2P/w.
Quote:  
Timo Nieminen 

Radi Khrapko 

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Timo Nieminen science forum Guru Wannabe
Joined: 12 May 2005
Posts: 244

Posted: Wed Jun 21, 2006 9:20 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



On Wed, 21 Jun 2006, khrapko_ri@hotmail.com wrote:
Quote:  Your experiment is a puzzle for me. But formula J=rxS gives zero in the
Beth experiment because S=0 in the Beth experiment.
Only formula
J=rxS+spintensor gives the Beth result.
Even considering edge effects? Consider a beam narrow than the waveplate.
The longitudinal part of the Poynting vector cancels due to the
counterpropagation, but the azimuthal parts do not  the handedness of
the polarisation about the direction of propagation is reversed on
reflection, so the direction of the angular momentum about an axis fixed
wrt the apparatus is the same, so the azimuthal parts _add_ due to the
counterpropagation.
Without edge effects, J=spintensor gives the Beth result.
Other than that, J=ExA also works, for a suitable choice of A. For any
choice of A, Humblet's general result (including the orbital term) works.
Oh, my ears and whiskers! Dear Timo, why are you not attentive as well
as hundreds of editors and referees? Many years ago I showed and
published that all parts of the Poynting vector are zero in the Beth
experiment

No, the azimuthal edgeeffect component of the counterpropagating beams
_adds_, not cancels. The longitudinal component cancels. But it's the
angular component that gives you the angular momentum.
Look, the azimuthal component of S at the edge of the beam depends of the
handedness of the circular polarisation of the beam. The handedness is
changed, along with the direction of propagation, by reflection from a
mirror. Hold your hands in front of you, touch the tips of your thumbs
together and see if your fingers point in the same direction.
Quote:  This paper confirms my theory. The point is their particles occupied
the hole of the donut and the region where \partial u^2 > 0 and thus
the torque acts opposite to the beam polarization handedness.
Therefore, in the absence of spintensor, an absorbing particle must
rotate slower if the beam is changed to circularly polarized of the
same sense as the helicity of the donut, contrary to their result. Only
spintensor forces the particle to rotate faster.
Well, yes, it's a disproof of the J=rxS/c school of thought. The Humblet
decomposition also explains it. See below for further
I wrote many times that Humblet, Jackson, Ohanian, Crichton and
Marston, Stewart and others do NOT decompose the orbital AM \int
rx(ExB) into "orbital" and "spin" parts. They transform \int rx(ExB)
into the form of \int ExA or \int FxB. They do not get two terms. One
of their terms is zero (see physics/0102084;
Exercisesinthecanonicalspintensor or Electrodynamics spin at
www.sciprint.org. Now this paper is considered by "Communications in
Mathematical Physics" since May 9, 2006)

In special cases, one of their terms is zero (also the 3rd "surface" term
which is zero for any case where the procedure is valid). In general, EM
fields and light _can_ carry orbital AM.
Quote:  I suspect that the exactness of the equivalence between "spin" and
"orbital" AM (according to the view in paraxial vortex beams where L=lP/w
and S=(degree of circular polarisation)P/w) depends on the radius of the
trapped particle in the context of your theory. It's worth checking.
Well, L=lP/w and Spin=(degree of circular polarisation)P/w are two
different beings. L=\int rx(ExB) and Spin = spintensor. The spintensor
is my electrodynamics spin tensor which is absent in the standard
electrodynamics. So, the total AM is the sum J=L+Spin. So, the beam
carries 2P/w.

But does it agree with experiments? Humblet et al say J=\int rxS/c, while
you say L=rxS/c. All agree (except for the "circularly polarised plane
wave carries no AM" folks) that spin=P/w for a circularly polarised plane
wave. This makes an experimentally observable difference, I think.

Timo Nieminen  Home page: http://www.physics.uq.edu.au/people/nieminen/
Eprints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html 

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Radi Khrapko science forum Guru Wannabe
Joined: 04 May 2005
Posts: 142

Posted: Thu Jun 22, 2006 12:43 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



Quote:  Oh, my ears and whiskers! Dear Timo, why are you not attentive as well
as hundreds of editors and referees? Many years ago I showed and
published that all parts of the Poynting vector are zero in the Beth
experiment
No, the azimuthal edgeeffect component of the counterpropagating beams
_adds_, not cancels. The longitudinal component cancels. But it's the
angular component that gives you the angular momentum.
Look, the azimuthal component of S at the edge of the beam depends of the
handedness of the circular polarisation of the beam. The handedness is
changed, along with the direction of propagation, by reflection from a
mirror. Hold your hands in front of you, touch the tips of your thumbs
together and see if your fingers point in the same direction.
Dear Timo, do not be obstinate. Please take a pen in your hand and sum 
E & H fields of incident and reflected Beth's beams as well as I did it
for physics/0102084, for example. I wrote there, "Let us start from the
Jackson's expression for a circularly polarized beam.. .. Here E_0(x;
y) is the electric field of the beam. E_0(x; y) = Const inside the
beam, and E_0(x; y) = 0 outside the beam. The returning light beam may
be got by changing the sign of z and y. Adding up the passing and
returning light beams we get interesting expressions,.. .. The E and H
fields are parallel everywhere. So, the Poynting vector is zero."
Quote: 
I wrote many times that Humblet, Jackson, Ohanian, Crichton and
Marston, Stewart and others do NOT decompose the orbital AM \int
rx(ExB) into "orbital" and "spin" parts. They transform \int rx(ExB)
into the form of \int ExA or \int FxB. They do not get two terms. One
of their terms is zero (see physics/0102084;
Exercisesinthecanonicalspintensor or Electrodynamics spin at
www.sciprint.org. Now this paper is considered by "Communications in
Mathematical Physics" since May 9, 2006)
In special cases, one of their terms is zero (also the 3rd "surface" term
which is zero for any case where the procedure is valid). In general, EM
fields and light _can_ carry orbital AM.
Sorry, I am forced to quote myself. I wrote about the term in 
physics/0102084, "But the derivative .. .. has only z component inside
the beam, and, in any case, the term .. .. is z directed. So, the term
is an integral moment of a longitudinal component of the momentum and
equals zero if the origin of the radiusvector is at the axis of
symmetry. The term bears no relation to the angular momentum of the
beam."
Quote: 
Well, L=lP/w and Spin=(degree of circular polarisation)P/w are two
different beings. L=\int rx(ExB) and Spin = spintensor. The spintensor
is my electrodynamics spin tensor which is absent in the standard
electrodynamics. So, the total AM is the sum J=L+Spin. So, the beam
carries 2P/w.
But does it agree with experiments? Humblet et al say J=\int rxS/c, while
you say L=rxS/c. All agree (except for the "circularly polarised plane
wave carries no AM" folks) that spin=P/w for a circularly polarised plane
wave. This makes an experimentally observable difference, I think.
What we see?! You write, "All agree that spin=P/w for a circularly 
polarised plane wave."! And what about your formula J=\int rxS/c?
Quote: 

Timo Nieminen 

Radi Khrapko 

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Josef Matz science forum Guru Wannabe
Joined: 08 May 2005
Posts: 255

Posted: Thu Jun 22, 2006 8:16 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



<khrapko_ri@hotmail.com> schrieb im Newsbeitrag
news:1150980215.077601.53930@p79g2000cwp.googlegroups.com...
Quote: 
Oh, my ears and whiskers! Dear Timo, why are you not attentive as well
as hundreds of editors and referees? Many years ago I showed and
published that all parts of the Poynting vector are zero in the Beth
experiment
No, the azimuthal edgeeffect component of the counterpropagating beams
_adds_, not cancels. The longitudinal component cancels. But it's the
angular component that gives you the angular momentum.
Look, the azimuthal component of S at the edge of the beam depends of
the
handedness of the circular polarisation of the beam. The handedness is
changed, along with the direction of propagation, by reflection from a
mirror. Hold your hands in front of you, touch the tips of your thumbs
together and see if your fingers point in the same direction.
Dear Timo, do not be obstinate. Please take a pen in your hand and sum
E & H fields of incident and reflected Beth's beams as well as I did it
for physics/0102084, for example. I wrote there, "Let us start from the
Jackson's expression for a circularly polarized beam.. .. Here E_0(x;
y) is the electric field of the beam. E_0(x; y) = Const inside the
beam, and E_0(x; y) = 0 outside the beam. The returning light beam may
be got by changing the sign of z and y. Adding up the passing and
returning light beams we get interesting expressions,.. .. The E and H
fields are parallel everywhere. So, the Poynting vector is zero."

Such a nonsense !!!!
E and H are perpendicular in light beam ! Maxwell equation greet !
Quote: 
I wrote many times that Humblet, Jackson, Ohanian, Crichton and
Marston, Stewart and others do NOT decompose the orbital AM \int
rx(ExB) into "orbital" and "spin" parts. They transform \int rx(ExB)
into the form of \int ExA or \int FxB. They do not get two terms. One
of their terms is zero (see physics/0102084;
Exercisesinthecanonicalspintensor or Electrodynamics spin at
www.sciprint.org. Now this paper is considered by "Communications in
Mathematical Physics" since May 9, 2006)
In special cases, one of their terms is zero (also the 3rd "surface"
term
which is zero for any case where the procedure is valid). In general, EM
fields and light _can_ carry orbital AM.
Sorry, I am forced to quote myself. I wrote about the term in
physics/0102084, "But the derivative .. .. has only z component inside
the beam, and, in any case, the term .. .. is z directed. So, the term
is an integral moment of a longitudinal component of the momentum and
equals zero if the origin of the radiusvector is at the axis of
symmetry. The term bears no relation to the angular momentum of the
beam."
Well, L=lP/w and Spin=(degree of circular polarisation)P/w are two
different beings. L=\int rx(ExB) and Spin = spintensor. The spintensor
is my electrodynamics spin tensor which is absent in the standard
electrodynamics. So, the total AM is the sum J=L+Spin. So, the beam
carries 2P/w.
But does it agree with experiments? Humblet et al say J=\int rxS/c,
while
you say L=rxS/c. All agree (except for the "circularly polarised plane
wave carries no AM" folks) that spin=P/w for a circularly polarised
plane
wave. This makes an experimentally observable difference, I think.
What we see?! You write, "All agree that spin=P/w for a circularly
polarised plane wave."! And what about your formula J=\int rxS/c?

Timo Nieminen 
Radi Khrapko



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Timo Nieminen science forum Guru Wannabe
Joined: 12 May 2005
Posts: 244

Posted: Thu Jun 22, 2006 8:31 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



On Thu, 22 Jun 2006, khrapko_ri@hotmail.com wrote:
Quote:  Oh, my ears and whiskers! Dear Timo, why are you not attentive as well
as hundreds of editors and referees? Many years ago I showed and
published that all parts of the Poynting vector are zero in the Beth
experiment
No, the azimuthal edgeeffect component of the counterpropagating beams
_adds_, not cancels. The longitudinal component cancels. But it's the
angular component that gives you the angular momentum.
Look, the azimuthal component of S at the edge of the beam depends of the
handedness of the circular polarisation of the beam. The handedness is
changed, along with the direction of propagation, by reflection from a
mirror. Hold your hands in front of you, touch the tips of your thumbs
together and see if your fingers point in the same direction.
Dear Timo, do not be obstinate. Please take a pen in your hand and sum
E & H fields of incident and reflected Beth's beams as well as I did it
for physics/0102084, for example. I wrote there, "Let us start from the
Jackson's expression for a circularly polarized beam.. .. Here E_0(x;
y) is the electric field of the beam. E_0(x; y) = Const inside the
beam, and E_0(x; y) = 0 outside the beam. The returning light beam may
be got by changing the sign of z and y. Adding up the passing and
returning light beams we get interesting expressions,.. .. The E and H
fields are parallel everywhere. So, the Poynting vector is zero."

Ah! You're hiding stuff in the singularity at the edge! Try it again for
two counterpropagating Gaussian beams of opposite handedness (ie same
angular momentum).
Quote:  I wrote many times that Humblet, Jackson, Ohanian, Crichton and
Marston, Stewart and others do NOT decompose the orbital AM \int
rx(ExB) into "orbital" and "spin" parts. They transform \int rx(ExB)
into the form of \int ExA or \int FxB. They do not get two terms. One
of their terms is zero (see physics/0102084;
Exercisesinthecanonicalspintensor or Electrodynamics spin at
www.sciprint.org. Now this paper is considered by "Communications in
Mathematical Physics" since May 9, 2006)
In special cases, one of their terms is zero (also the 3rd "surface" term
which is zero for any case where the procedure is valid). In general, EM
fields and light _can_ carry orbital AM.
Sorry, I am forced to quote myself. I wrote about the term in
physics/0102084, "But the derivative .. .. has only z component inside
the beam, and, in any case, the term .. .. is z directed. So, the term
is an integral moment of a longitudinal component of the momentum and
equals zero if the origin of the radiusvector is at the axis of
symmetry. The term bears no relation to the angular momentum of the
beam."

Yes, this is for one of the special cases. Try it for a LaguerreGauss
beam carrying orbital AM.
I don't disagree that the term is zero about the beam axis for a Gaussian
or flattop beam. I was just commenting that the term can be nonzero.
Quote:  Well, L=lP/w and Spin=(degree of circular polarisation)P/w are two
different beings. L=\int rx(ExB) and Spin = spintensor. The spintensor
is my electrodynamics spin tensor which is absent in the standard
electrodynamics. So, the total AM is the sum J=L+Spin. So, the beam
carries 2P/w.
But does it agree with experiments? Humblet et al say J=\int rxS/c, while
you say L=rxS/c. All agree (except for the "circularly polarised plane
wave carries no AM" folks) that spin=P/w for a circularly polarised plane
wave. This makes an experimentally observable difference, I think.
What we see?! You write, "All agree that spin=P/w for a circularly
polarised plane wave."! And what about your formula J=\int rxS/c?

J = \int rxS/c = P/w. Humblet et al say that J=S in this case (ie L=0,
since Humblet's "orbital" term is zero, as you note above). You are saying
the orbital AM density is rxS/c, yes? Thus, that L = \int rxS/c = P/w.
This is why you say J=L+S=2P/w, while others say J=S=P/w. For a finite
beam, even those who say that the _only_ AM flux density is rxS/c agree,
but they'd write J=L=P/w.
I think the important point is that your spin tensor gives the correct
result for the spin. Thus useful, and publishable. No, I don't think it
gets rejected because it's antiestablishment. Your papers can be hard to
read, you run into referees who say "the only AM density is rxS/c, so it
must be wrong", claiming classical EM theory is wrong without clear
experimental evidence is touchy, etc. Seriously, try to write as simple
and clear as possible a paper _only_ about spin and your spin tensor.
Nothing about "mistakes" etc.
Anyway, beams that carry orbital AM about the beam axis might be a good
test of whether you are correct about the _total_ AM.

Timo Nieminen  Home page: http://www.physics.uq.edu.au/people/nieminen/
Eprints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html 

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Radi Khrapko science forum Guru Wannabe
Joined: 04 May 2005
Posts: 142

Posted: Sat Jun 24, 2006 9:40 am Post subject:
Re: Circularly polarized beam in a dielectric, etc.



Josef Matz wrote:
Quote:  Such a nonsense !!!!
E and H are perpendicular in light beam ! Maxwell equation greet !
Please compare the real parts of two sums: 
E=(x+iy)exp[i(zt)]+ (xiy)exp[i(zt)]
H=(ix+y)exp[i(zt)]+ (ixy)exp[i(zt)].
You will get E is parallel to H.
Please read physics/0102084, as the minimum.
Radi Khrapko 

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Radi Khrapko science forum Guru Wannabe
Joined: 04 May 2005
Posts: 142

Posted: Sat Jun 24, 2006 9:53 am Post subject:
Re: Circularly polarized beam in a dielectric, etc.



Quote:  Dear Timo, do not be obstinate. Please take a pen in your hand and sum
E & H fields of incident and reflected Beth's beams as well as I did it
for physics/0102084, for example. I wrote there, "Let us start from the
Jackson's expression for a circularly polarized beam.. .. Here E_0(x;
y) is the electric field of the beam. E_0(x; y) = Const inside the
beam, and E_0(x; y) = 0 outside the beam. The returning light beam may
be got by changing the sign of z and y. Adding up the passing and
returning light beams we get interesting expressions,.. .. The E and H
fields are parallel everywhere. So, the Poynting vector is zero."
Ah! You're hiding stuff in the singularity at the edge! Try it again for
two counterpropagating Gaussian beams of opposite handedness (ie same
angular momentum).
My dear Timo. I hide nothing. Contrary, I show, show, show during 7 
years, but without a result. If you read my papers or "Classical
Electrodynamics" by J. D. Jackson, you will see that there is NO
singularity in our calculations. Our result is independent of the
profile of the beam.
You wrote me
"From: Timo Nieminen (timo@physics.uq.edu.au)
Date: 20030713 15:41:02 PST
The AM flux can be found by many methods, including, but not limited
to:
(1) the semiclassical approximation
(2) finding the AM flux of, for example, a Gaussian beam and taking the
limit as the width infinity"
I answered to you 3 years ago, on 20030713:
"A profile of beam (distribution on a crosssection) is of no
importance"
You can easily see that an explicit form of the function E_0(x,y) is
not in the use and the E and H fields are parallel everywhere. So, the
Poynting vector is zero and Maxwell electrodynamics is incomplete
Nevertheless I used the Gaussian beam in a paper "The Beth experiment
is under review" (www.scoprint.org,
www.mai.ru/projects/mai_works/articles/num21/article6/auther.htm) and
got the result. Unfortunately, this paper was rejected by APP, EPL, FP,
JMP, JOPA, JPA, NJP, OL, PLA, PRA, PRE. Now this paper is considered by
JMO (Professor P. L. Knight) since February 22, 2006.
Quote:  In special cases, one of their terms is zero (also the 3rd "surface" term
which is zero for any case where the procedure is valid). In general, EM
fields and light _can_ carry orbital AM.
Sorry, I am forced to quote myself. I wrote about the term in
physics/0102084, "But the derivative .. .. has only z component inside
the beam, and, in any case, the term .. .. is z directed. So, the term
is an integral moment of a longitudinal component of the momentum and
equals zero if the origin of the radiusvector is at the axis of
symmetry. The term bears no relation to the angular momentum of the
beam."
Yes, this is for one of the special cases. Try it for a LaguerreGauss
beam carrying orbital AM.
My dear Timo. This is NOT for one of the special cases. One of their 
terms is zero for their common case, and other term, \int (ExA), equals
the whole expression, \int rx(ExH), and represents orbital AM, L = \int
rx(ExH), because spin in the sense of the word as used in the field
theory is absent in Maxwell electrodynamics. If you read my papers or
"Classical Electrodynamics" by J. D. Jackson, you will see that our
results are independent of the profile of the beam. You can easily see
that an explicit form of the function E_0(x,y) or u(x,y) is not in the
use.
Nevertheless I used the Gaussian beam in a paper "Defects of the
general field theory" (V2). (www.scoprint.org). Unfortunately, this
paper was rejected by MPEJ, JMP, IJTP, NJP, JPA, PRL, PLA.
However, your recommendation, "Try it for a LaguerreGauss beam
carrying orbital AM" is senseless because Humblet and others did NOT
consider Laguerre beams. Nevertheless, a LaguerreGauss beam obviously
carries two different orbital AM.
Quote: 
I don't disagree that the term is zero about the beam axis for a Gaussian
or flattop beam. I was just commenting that the term can be nonzero.
OK! You agree that the term is zero about the beam axis for a Gaussian 
or flattop beam and thus you confirm the fictitious division of the
orbital angular momentum L = \int rx(ExH). I showed it in a paper
"Electrodynamics' spin" (www.sciprint.org) rejected by AJP, PLA, JPA,
PRA, JMP, NJP, OC, PRL, IJTP, CMP.
Quote:  What we see?! You write, "All agree that spin=P/w for a circularly
polarised plane wave."! And what about your formula J=\int rxS/c?
J = \int rxS/c = P/w. Humblet et al say that J=S in this case (ie L=0,
since Humblet's "orbital" term is zero, as you note above). You are saying
the orbital AM density is rxS/c, yes? Thus, that L = \int rxS/c = P/w.
This is why you say J=L+S=2P/w, while others say J=S=P/w. For a finite
beam, even those who say that the _only_ AM flux density is rxS/c agree,
but they'd write J=L=P/w.
J = \int rx(ExH)/c = P/w holds only in Maxwell electrodynamics and for 
a circularly polarized beam. But Maxwell electrodynamics is not
complete. In reality, J = \int rx(ExH)/c + spintensor = 2P/w for a
circularly polarized beam. Let us agree that moment of the Poynting
vector is an orbital AM, i.e. \int rx(ExH) = L forever. You must also
admit that Maxwell electrodynamics does not contain conception of spin
in the sense of the word as used in the field theory, i.e. S = 0, and J
= L in Maxwell electrodynamics. So, we must recognize that ExA is not
spin, and the Humblet's equality \int rx(ExH) = \int ExA holds, but is
senseless. Humblet et al. say that J=S by mistake in any case. Any
plane wave has neither spin nor orbital AM, and the beam has L = P/w, S
= 0, J = L in Maxwell electrodynamics. However, in reality, circularly
polarized plane wave has spin because the spintesor. The use a word
"spin" is unscientific in Maxwell electrodynamics. Nobody can agree
that spin=P/w for a circularly polarized plane wave in Maxwell
electrodynamics because of two reasons: the concept of spin is absent,
J = L =\int rx(ExH) = 0.
Quote: 
I think the important point is that your spin tensor gives the correct
result for the spin.
You must not write a word "spin". You do not know spin in the sense of 
the word as used in the field theory. You know only "spin" of a bike
wheel. You are an incorrigible Maxwellian physicist.
Quote:  Thus useful, and publishable. No, I don't think it
gets rejected because it's antiestablishment. Your papers can be hard to
read,
They claim, "A strict criterion for acceptance in this journal is that 
manuscripts must convey new physics". If they cannot read new physics,
they must change their criterion.
Honest people inquire unintelligible things. I had no questions.
Btw, Editors hide favorable reports. For example, Christian Brosseau
hid Reviewer 1 of "Optics Letters" who wrote: "The result, dealing with
matters at the heart of the rather confused matter of electromagnetic
angular momentum, is interesting and merits publication. The paper is
concise, and likely as clear as possible, given the subject matter."
Quote:  you run into referees who say "the only AM density is rxS/c, so it
must be wrong", claiming classical EM theory is wrong without clear
experimental evidence is touchy, etc. Seriously, try to write as simple
and clear as possible a paper _only_ about spin and your spin tensor.
Nothing about "mistakes" etc.
Not only my referee, but also you say "the only AM density is rxS/c" 
because J = \int rx(ExH)/c = P/w is your formula.
Quote: 
Anyway, beams that carry orbital AM about the beam axis might be a good
test of whether you are correct about the _total_ AM.
I submitted a paper "Beth's experiment modification" to OL on 29 Jul 
2003. I offered to replace the Beth's quarterwave plate by a mirror
that will reflect the beam away from the apparatus. I predict that the
replacement will not change the result of the experiment. But Robert A.
Fisher rejected the paper.
Earlier I offered to illuminate an aluminum black disc in free space by
circularly polarized electromagnetic radiation. [Measurement
Techniques, 46, No. 4, 317]
Radi Khrapko 

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