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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: 2003-07-13 15:41:02 PST
The AM flux can be found by many methods, including, but not limited
to:
(1) the semi-classical 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 2003-07-13:
"A profile of beam (distribution on a cross-section) 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 radius-vector 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 Laguerre-Gauss 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 Laguerre-Gauss beam
carrying orbital AM" is senseless because Humblet and others did NOT
consider Laguerre beams. Nevertheless, a Laguerre-Gauss 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 flat-top beam. I was just commenting that the term can be non-zero. OK! You agree that the term is zero about the beam axis for a Gaussian

or flat-top 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 anti-establishment. 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 quarter-wave 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]

 Quote: Timo Nieminen 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 (ikr-iwt) 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
 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(z-t)]+ (x-iy)exp[i(-z-t)] H=(-ix+y)exp[i(z-t)]+ (-ix-y)exp[i(-z-t)].

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 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

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 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 (ikr-iwt) 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(z-t)]+ (x-iy)exp[i(-z-t)] H=(-ix+y)exp[i(z-t)]+ (-ix-y)exp[i(-z-t)]. 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(z-t)], \vec H_1=(-i\vec x+\vec
y)exp[i(z-t)]
and
\vec E_2= (\vec x-i\vec y)exp[i(-z-t)], \vec H_2=(-i\vec x-\vec
y)exp[i(-z-t)].
This notations mean: E_{1x}= exp[i(z-t)], E_{1y}=i exp[i(z-t)],
H_{1x}=-i exp[i(z-t)], etc.
You will get E is parallel to H. 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. 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
 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 (ikr-iwt) 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(z-t)]+ (x-iy)exp[i(-z-t)] H=(-ix+y)exp[i(z-t)]+ (-ix-y)exp[i(-z-t)]. 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(z-t)], \vec H_1=(-i\vec x+\vec y)exp[i(z-t)] and \vec E_2= (\vec x-i\vec y)exp[i(-z-t)], \vec H_2=(-i\vec x-\vec y)exp[i(-z-t)]. This notations mean: E_{1x}= exp[i(z-t)], E_{1y}=i exp[i(z-t)], H_{1x}=-i exp[i(z-t)], 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 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(z-t)], \vec H_1=(-i\vec x+\vec
y)exp[i(z-t)]
and
\vec E_2= (\vec x-i\vec y)exp[i(-z-t)], \vec H_2=(-i\vec x-\vec
y)exp[i(-z-t)].
These notations mean: E_{1x}= exp[i(z-t)], E_{1y}=i exp[i(z-t)],
H_{1x}=-i exp[i(z-t)], etc.
You will get E is parallel to H.

 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 half-wave plate is
equal to 4 in the absenceof an energy flux!" [Measurement Techniques,
Vol. 46, No. 4, 2003; Experimental-Check-of-Electrodynamics,
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 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
 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 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(z-t)], \vec H_1=(-i\vec x+\vec y)exp[i(z-t)] and \vec E_2= (\vec x-i\vec y)exp[i(-z-t)], \vec H_2=(-i\vec x-\vec y)exp[i(-z-t)]. This notations mean: E_{1x}= exp[i(z-t)], E_{1y}=i exp[i(z-t)], H_{1x}=-i exp[i(z-t)], 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 non-zero. This is basically the Beth
experiment.

--
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html 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
Lorentz-invariantly monochromatic (except for plane waves).

--
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html 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(z-t)], 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
Measurement Techniques, 46, No. 4, 317 (2003), or www.sciprint.org. 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 non-zero. 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.

 Quote: -- Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/ E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html 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 Lorentz-invariantly 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
 Quote: -- Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/ E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html Radi Khrapko 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
 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(z-t)], \vec H_1=(-i\vec x+\vec y)exp[i(z-t)] and \vec E_2= (\vec x-i\vec y)exp[i(-z-t)], \vec H_2=(-i\vec x-\vec y)exp[i(-z-t)]. These notations mean: E_{1x}= exp[i(z-t)], E_{1y}=i exp[i(z-t)], H_{1x}=-i exp[i(z-t)], 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 half-wave plate is equal to 4 in the absenceof an energy flux!" [Measurement Techniques, Vol. 46, No. 4, 2003; Experimental-Check-of-Electrodynamics, 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 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
 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 non-zero. 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

 Quote: Radi Khrapko -- Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/ E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html  Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First
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