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

Posted: Sat Jun 24, 2006 7:46 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



Maybe you know the Maxwell eqation
rot E =  d/dt B
Because the room and time dependence for plane waves is exp (ikriwt) it
follows that
k x E = B and from this BE = 0
The last formula is the scalar product of the two vectors B and E which is
zero.
Elementar optics books you find that in the real valued interpretation E and
B are perpendicular.
You make elementar mistakes. See my notes downwards
<khrapko_ri@hotmail.com> schrieb im Newsbeitrag
news:1151142022.157885.163100@y41g2000cwy.googlegroups.com...
Quote: 
Josef Matz wrote:
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)].

What is x + iy and x  iy meaning ? Since when are the amplititudes depend
on the location this way.
Amplitudes are constant for plane waves! Here lies the worm. Your fields
never fulfill the four Maxwell equations.
You have to learn a lot of maths before making theories i think.
Or ist it just that i do not understand your notation. But this cant be
also.
Josef Matz
Quote:  You will get E is parallel to H.
Please read physics/0102084, as the minimum.
Radi Khrapko



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

Posted: Sat Jun 24, 2006 8:06 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



<khrapko_ri@hotmail.com> schrieb im Newsbeitrag
news:1151142785.943100.140170@m73g2000cwd.googlegroups.com...
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.
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.
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.
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.
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.

I would not say it. Timo once sent me a publication he already done a couple
of years ago.
In this publication there he published the right spin formula for an
elliptic polarized beam.
So Timo is a good man so or so and the first who did that. Even though he
uses special
coordinates and just describes the dependence of spin from electric field.
So he has no field theory but a good formula. But there still are missing
two steps at Timo.
First the formulation in a field theory. This is possible. And second: A
second formula is
possible which uses magnetic fields. The true field theory of spin must
contain both possibilities,
the arguments therefore only can be found in the spin theory for absortive
substances (metal optics).
So Timo also is not looking through as you and we all. But one thing you
should realize:
A field theory is not possible with real valued field considerations. But
with complex fields it is.
Thanks
Josef Matz
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."
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.
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]
Timo Nieminen
Radi Khrapko



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

Posted: Sat Jun 24, 2006 10:17 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



Josef Matz wrote:
Quote:  Maybe you know the Maxwell eqation
rot E =  d/dt B
Because the room and time dependence for plane waves is exp (ikriwt) it
follows that
k x E = B and from this BE = 0
The last formula is the scalar product of the two vectors B and E which is
zero.
Elementar optics books you find that in the real valued interpretation E and
B are perpendicular.
You make elementar mistakes. See my notes downwards
khrapko_ri@hotmail.com> schrieb im Newsbeitrag
news:1151142022.157885.163100@y41g2000cwy.googlegroups.com...
Josef Matz wrote:
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)].
What is x + iy and x  iy meaning ? Since when are the amplititudes depend
on the location this way.
Amplitudes are constant for plane waves! Here lies the worm. Your fields
never fulfill the four Maxwell equations.
You have to learn a lot of maths before making theories i think.
Or ist it just that i do not understand your notation. But this cant be
also.
Josef Matz
Dear Josef, sorry. Please use the superposition principle and sum two 
plane wave:
\vec E_1=(\vec x+i\vec y)exp[i(zt)], \vec H_1=(i\vec x+\vec
y)exp[i(zt)]
and
\vec E_2= (\vec xi\vec y)exp[i(zt)], \vec H_2=(i\vec x\vec
y)exp[i(zt)].
This notations mean: E_{1x}= exp[i(zt)], E_{1y}=i exp[i(zt)],
H_{1x}=i exp[i(zt)], etc.
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: Sun Jun 25, 2006 6:03 am Post subject:
Re: Circularly polarized beam in a dielectric, etc.



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.
I would not say it. Timo once sent me a publication he already done a couple
of years ago.
In this publication there he published the right spin formula for an
elliptic polarized beam.
Please, forward this publication to me. I am sure this formula is \int 
rx(ExH), and so this quantity is an orbital AM, not spin in the sense
of the word as used in the field theory.
Quote: 
So Timo is a good man so or so and the first who did that. Even though he
uses special
coordinates and just describes the dependence of spin from electric field.
So he has no field theory but a good formula. But there still are missing
two steps at Timo.
First the formulation in a field theory. This is possible. And second: A
second formula is
possible which uses magnetic fields. The true field theory of spin must
contain both possibilities,
the arguments therefore only can be found in the spin theory for absortive
substances (metal optics).
So Timo also is not looking through as you and we all. But one thing you
should realize:
A field theory is not possible with real valued field considerations. But
with complex fields it is.
Thanks
Josef Matz
Timo is a very good man. He is the only physicist who tries to 
understand new physics. And you is a very good man. You are the only
physicist who asks me.
Radi Khrapko 

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

Posted: Sun Jun 25, 2006 6:36 am Post subject:
Re: Circularly polarized beam in a dielectric, etc.



<khrapko_ri@hotmail.com> schrieb im Newsbeitrag
news:1151187472.752559.234760@i40g2000cwc.googlegroups.com...
Quote: 
Josef Matz wrote:
Maybe you know the Maxwell eqation
rot E =  d/dt B
Because the room and time dependence for plane waves is exp (ikriwt) it
follows that
k x E = B and from this BE = 0
The last formula is the scalar product of the two vectors B and E which
is
zero.
Elementar optics books you find that in the real valued interpretation E
and
B are perpendicular.
You make elementar mistakes. See my notes downwards
khrapko_ri@hotmail.com> schrieb im Newsbeitrag
news:1151142022.157885.163100@y41g2000cwy.googlegroups.com...
Josef Matz wrote:
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)].
What is x + iy and x  iy meaning ? Since when are the amplititudes
depend
on the location this way.
Amplitudes are constant for plane waves! Here lies the worm. Your
fields
never fulfill the four Maxwell equations.
You have to learn a lot of maths before making theories i think.
Or ist it just that i do not understand your notation. But this cant be
also.
Josef Matz
Dear Josef, sorry. Please use the superposition principle and sum two
plane wave:
\vec E_1=(\vec x+i\vec y)exp[i(zt)], \vec H_1=(i\vec x+\vec
y)exp[i(zt)]
and
\vec E_2= (\vec xi\vec y)exp[i(zt)], \vec H_2=(i\vec x\vec
y)exp[i(zt)].
This notations mean: E_{1x}= exp[i(zt)], E_{1y}=i exp[i(zt)],
H_{1x}=i exp[i(zt)], etc.
You will get E is parallel to H.
Please read physics/0102084, as the minimum.

Radi, in no light beam exept the dark where fields are all zero  in no
light beam E and H are parallel.
In an dielctric that is not absorbing they are perpendicular exactly.
And another note: A light beam with no energy flux also does not transport
spin.
If you are overtaking such mistakes from others this does not help you 
your theory is completely
s**t.
Josef Matz


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

Posted: Sun Jun 25, 2006 6:49 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



Josef Matz wrote:
Quote:  Radi, in no light beam exept the dark where fields are all zero  in no
light beam E and H are parallel.
Sorry, I cannot understand your thought. Please say it in other words 
Quote:  In an dielctric that is not absorbing they are perpendicular exactly.
This is not exactly. E may be parallel to H even in vacuum. Please use 
the superposition principle and sum two plane waves in vacuum:
\vec E_1=(\vec x+i\vec y)exp[i(zt)], \vec H_1=(i\vec x+\vec
y)exp[i(zt)]
and
\vec E_2= (\vec xi\vec y)exp[i(zt)], \vec H_2=(i\vec x\vec
y)exp[i(zt)].
These notations mean: E_{1x}= exp[i(zt)], E_{1y}=i exp[i(zt)],
H_{1x}=i exp[i(zt)], etc.
You will get E is parallel to H.
Please read physics/0102084, as the minimum.
Quote:  And another note: A light beam with no energy flux also does not transport
spin.
The virtue of my spin tensor is it provides for spin flux without 
energy flux. I wrote about the Beth experiment, "Thus, the total
density of the spin angular momentum received by the halfwave plate is
equal to 4 in the absenceof an energy flux!" [Measurement Techniques,
Vol. 46, No. 4, 2003; ExperimentalCheckofElectrodynamics,
www.sciprint.org]
Quote: 
If you are overtaking such mistakes from others this does not help you 
your theory is completely
s**t.
What is s**t? 
"s**t n. VULGAR.
s**t! also oh s**t! said when you are angry or annoyed about something:
Oh, s**t! I forgot my passport!"
Quote: 
Josef Matz
Radi Khrapko 


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

Posted: Mon Jun 26, 2006 7:23 am Post subject:
Re: Circularly polarized beam in a dielectric, etc.



<khrapko_ri@hotmail.com> schrieb im Newsbeitrag
news:1151215409.238543.106360@r2g2000cwb.googlegroups.com...
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.
I would not say it. Timo once sent me a publication he already done a
couple
of years ago.
In this publication there he published the right spin formula for an
elliptic polarized beam.
Please, forward this publication to me. I am sure this formula is \int
rx(ExH), and so this quantity is an orbital AM, not spin in the sense
of the word as used in the field theory.

No this formula is a E x E* but expressed in (x,y,z) coordinate system.
Quote: 
So Timo is a good man so or so and the first who did that. Even though
he
uses special
coordinates and just describes the dependence of spin from electric
field.
So he has no field theory but a good formula. But there still are
missing
two steps at Timo.
First the formulation in a field theory. This is possible. And second: A
second formula is
possible which uses magnetic fields. The true field theory of spin must
contain both possibilities,
the arguments therefore only can be found in the spin theory for
absortive
substances (metal optics).
So Timo also is not looking through as you and we all. But one thing you
should realize:
A field theory is not possible with real valued field considerations.
But
with complex fields it is.
Thanks
Josef Matz
Timo is a very good man. He is the only physicist who tries to
understand new physics. And you is a very good man. You are the only
physicist who asks me.
Radi Khrapko



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

Posted: Mon Jun 26, 2006 12:19 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



On Sun, 25 Jun 2006, Josef Matz wrote:
Quote:  khrapko_ri@hotmail.com> schrieb:
Dear Josef, sorry. Please use the superposition principle and sum two
plane wave:
\vec E_1=(\vec x+i\vec y)exp[i(zt)], \vec H_1=(i\vec x+\vec
y)exp[i(zt)]
and
\vec E_2= (\vec xi\vec y)exp[i(zt)], \vec H_2=(i\vec x\vec
y)exp[i(zt)].
This notations mean: E_{1x}= exp[i(zt)], E_{1y}=i exp[i(zt)],
H_{1x}=i exp[i(zt)], etc.
You will get E is parallel to H.
Please read physics/0102084, as the minimum.
Radi, in no light beam exept the dark where fields are all zero  in no
light beam E and H are parallel.
In an dielctric that is not absorbing they are perpendicular exactly.
And another note: A light beam with no energy flux also does not transport
spin.

This is two beams propagating in opposite directions. With equal power
each, the total energy flux is zero. With opposite handedness of circular
polarisation, the AM flux is nonzero. This is basically the Beth
experiment.

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

Posted: Mon Jun 26, 2006 12:27 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



On Sun, 24 Jun 2006, khrapko_ri@hotmail.com wrote:
Quote:  Please, forward this publication to me. I am sure this formula is \int
rx(ExH), and so this quantity is an orbital AM, not spin in the sense
of the word as used in the field theory.

This would be the AM of a focussed beam paper, on arxiv, or via the
eprints link in my sig below, or email me and I can send you a copy.
This is just the same formula for spin that appears in Crichton or (iirc)
van Enk, which is Humblet's result for a monochromatic wave, given a
particular choice of gauge.
Quote:  This is possible. And second: A
second formula is
possible which uses magnetic fields.

With monochromatic fields, you know H if you know E. There is stuff
written about the division of AM into spin and orbital parts not being
Lorentz invariant  note that monochromatic fields are not
Lorentzinvariantly monochromatic (except for plane waves).

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: Mon Jun 26, 2006 11:31 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



Quote:  Please, forward this publication to me. I am sure this formula is \int
rx(ExH), and so this quantity is an orbital AM, not spin in the sense
of the word as used in the field theory.
Josef Matz wrote: 
Quote: 
No this formula is a E x E* but expressed in (x,y,z) coordinate system.
Dear Josef, I taught you many times that they used a wrong formula 
spin=ExA.
If E=exp[i(zt)], A=iE, and spin=Re(ExA*)/2 =Re(iExE*)/2 =Im(ExE*)/2.
This formula is presented by Timo et al. in Nature 394, 348 (1998), and
physics/0408080 with small mistakes.
This formula is wrong because ExA is a component of the wrong canonical
spin tensor. Physicists eliminate this tensor by the use of the wrong
BelinfanteRosenfeld procedure. I wrote about this many times, see
Measurement Techniques, 46, No. 4, 317 (2003), or www.sciprint.org.
Radi Khrapko 

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

Posted: Mon Jun 26, 2006 11:37 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



Quote:  And another note: A light beam with no energy flux also does not transport
spin.
Timo A. Nieminen wrote: 
Quote: 
This is two beams propagating in opposite directions. With equal power
each, the total energy flux is zero. With opposite handedness of circular
polarisation, the AM flux is nonzero. This is basically the Beth
experiment.
Dear Timo, no prompting! Josef must recognize his mistakes himself. He 
claim, "Such a nonsense !!!! E and H are perpendicular in light beam !
Maxwell equation greet !" I presented a disproving example and waited
for his recognition. In my example the parallel E & H fields satisfy
Maxwell equations. You have prevented his recognition.
On the other side, it is not good that you confuse Josef. He is right
that, according to Maxwell electrodynamics, according to the formula:
total AM flux density =rx(ExH), a light beam with no energy flux also
does not transport spin.
In my simple example, two plane wave propagate in opposite directions,
and ExH=0. But I repeat, you can easy make sure that ExH=0 for beams of
the Beth experiment. So, you must admit that AM flux is zero for the
Beth experiment, according to the formula j=rx(ExH). The expression ExA
cannot help you. Thank God, you do not claim j=rx(ExH)+ExA.
Radi Khrapko


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

Posted: Mon Jun 26, 2006 11:40 pm Post subject:
Re: Circularly polarized beam in a dielectric, etc.



Timo A. Nieminen wrote:
Quote:  With monochromatic fields, you know H if you know E. There is stuff
written about the division of AM into spin and orbital parts not being
Lorentz invariant  note that monochromatic fields are not
Lorentzinvariantly monochromatic (except for plane waves).
Dear Timo, please be attentive. I repeated many times that the division 
of AM flux=rx(ExH) into spin orbital parts is an illusion. AM
flux=rx(ExH) does not contain spin in the sense of the word as used in
the field theory. The Humblet et al. procedure leads to the senseless
identity \int rx(ExH)=\int ExA in very simple cases only


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

Posted: Thu Jun 29, 2006 9:20 am Post subject:
Re: Circularly polarized beam in a dielectric, etc.



<khrapko_ri@hotmail.com> schrieb im Newsbeitrag
news:1151261343.386221.129440@p79g2000cwp.googlegroups.com...
Quote: 
Josef Matz wrote:
Radi, in no light beam exept the dark where fields are all zero  in no
light beam E and H are parallel.
Sorry, I cannot understand your thought. Please say it in other words
In an dielctric that is not absorbing they are perpendicular exactly.
This is not exactly. E may be parallel to H even in vacuum. Please use
the superposition principle and sum two plane waves in vacuum:
\vec E_1=(\vec x+i\vec y)exp[i(zt)], \vec H_1=(i\vec x+\vec
y)exp[i(zt)]
and
\vec E_2= (\vec xi\vec y)exp[i(zt)], \vec H_2=(i\vec x\vec
y)exp[i(zt)].
These notations mean: E_{1x}= exp[i(zt)], E_{1y}=i exp[i(zt)],
H_{1x}=i exp[i(zt)], etc.
You will get E is parallel to H.
Please read physics/0102084, as the minimum.

Superposition principles  interesting thing. An infall wave and reflected
outfall wave is indeed
not zero flux for infall and outfall. But the resulting flux is zero i
agree.
Thus int r x S(total) is zero obviously.
Your example is a example that is not good because if at all a very special
case to add fields and
using trigonometric relations for the cos and sin of the sum and difference
of two angles. I have not
proven if everywhere everytime resulting E and H are parallel. This
treatment would be one connected
with what is called standing wave. That could be right. But i think there
are further conditions necessary
to form standing waves as for example a exact parallel boundary within the
light emitting source or
similar things. The normal back  reflection at small angles of incidence
towards the source does not have
standing waves involved. And i think Beths experiment does not have standing
waves and brings
probably almost same result at small angles of incidence or ?
I have also a nice contribution to superposition principles in metals.
Superposition of fields is wrong in
general complex situations.
The energy flux for E and H waves must be calculated seperately from each
other.
So you have two superposition principles. And a third flux, the tunnel flux
must be calculated seperately.
This flux is the dominant 100% flux in the gap of Nimtz double prism
experiment. The tunneling flux results from mixture of reflected and infall
wave. Just that you see that superposition is nothing new to me.
Quote: 
And another note: A light beam with no energy flux also does not
transport
spin.
The virtue of my spin tensor is it provides for spin flux without
energy flux. I wrote about the Beth experiment, "Thus, the total
density of the spin angular momentum received by the halfwave plate is
equal to 4 in the absenceof an energy flux!" [Measurement Techniques,
Vol. 46, No. 4, 2003; ExperimentalCheckofElectrodynamics,
www.sciprint.org]

Could you tell me how i get your article on this web page ? Or please send
copy
to josefmatz@arcor.de. I will tell you how i think of it, promised.
Quote: 
If you are overtaking such mistakes from others this does not help you 
your theory is completely
s**t.
What is s**t?
"s**t n. VULGAR.
s**t! also oh s**t! said when you are angry or annoyed about something:
Oh, s**t! I forgot my passport!"

Ok thats right. I normally do not do that, sorry for that lapsus linguae.
Quote: 
Josef Matz
Radi Khrapko
Radi Khrapko



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

Posted: Thu Jun 29, 2006 9:26 am Post subject:
Re: Circularly polarized beam in a dielectric, etc.



<khrapko_ri@hotmail.com> schrieb im Newsbeitrag
news:1151365043.946801.10340@p79g2000cwp.googlegroups.com...
Quote: 
And another note: A light beam with no energy flux also does not
transport
spin.
Timo A. Nieminen wrote:
This is two beams propagating in opposite directions. With equal power
each, the total energy flux is zero. With opposite handedness of
circular
polarisation, the AM flux is nonzero. This is basically the Beth
experiment.
Dear Timo, no prompting! Josef must recognize his mistakes himself. He
claim, "Such a nonsense !!!! E and H are perpendicular in light beam !
Maxwell equation greet !" I presented a disproving example and waited
for his recognition. In my example the parallel E & H fields satisfy
Maxwell equations. You have prevented his recognition.

Are we doing a game ?
Quote:  On the other side, it is not good that you confuse Josef. He is right
that, according to Maxwell electrodynamics, according to the formula:
total AM flux density =rx(ExH), a light beam with no energy flux also
does not transport spin.

ok.
Quote:  In my simple example, two plane wave propagate in opposite directions,
and ExH=0. But I repeat, you can easy make sure that ExH=0 for beams of
the Beth experiment. So, you must admit that AM flux is zero for the
Beth experiment, according to the formula j=rx(ExH). The expression ExA
cannot help you. Thank God, you do not claim j=rx(ExH)+ExA.

Thats right. But was E x A that not the expression you claimed to be the
spin tensor ?
That is only a short time ago when you said so !!
Josef Matz


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