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Circularly polarized beam in a dielectric, etc.
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Radi Khrapko
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Joined: 04 May 2005
Posts: 142

PostPosted: Fri Jul 07, 2006 8:46 pm    Post subject: Re: Circularly polarized beam in a dielectric, etc. Reply with quote

The Beth experiment was intensively discussed at this page. So, I
inform you that I have posted a new paper "Paradox of the classical
Beth optics experiment" at www.sciprint.
I show that the celebrated Beth's experiment contradicts the angular
momentum conservation law in the frame of Maxwell electrodynamics
because Beth's birefringent plate experienced a torque without an
angular momentum flux in the surrounding space. However, this paradox
can be removed by introducing a classical spin tensor.
At the same time I submitted this paper to Journal of Optics A: Pure
and Applied Optics through IOP.
Radi Khrapko
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Radi Khrapko
science forum Guru Wannabe


Joined: 04 May 2005
Posts: 142

PostPosted: Sat Jul 01, 2006 5:32 pm    Post subject: Re: Circularly polarized beam in a dielectric, etc. Reply with quote

Josef Matz wrote:
Quote:
Experimental-Check-of-Electrodynamics,
www.sciprint.org


Could you tell me how i get your article on this web page ?
It's very simple
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Josef Matz
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Joined: 08 May 2005
Posts: 255

PostPosted: Thu Jun 29, 2006 9:37 am    Post subject: Re: Circularly polarized beam in a dielectric, etc. Reply with quote

<khrapko_ri@hotmail.com> schrieb im Newsbeitrag
news:1150783376.143795.311010@i40g2000cwc.googlegroups.com...
Quote:

Josef Matz wrote:
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

2 P/w how that ?

Quote:

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

for circular polarized light only or ?

Quote:

therefore your theory seems to be wrong.

Ok we will see.

Quote:
Therefore my theory is worthy of the Nobel Prize


All right. You have a fine goal !

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


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.

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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Josef Matz
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Joined: 08 May 2005
Posts: 255

PostPosted: Thu Jun 29, 2006 9:30 am    Post subject: Re: Circularly polarized beam in a dielectric, etc. Reply with quote

<khrapko_ri@hotmail.com> schrieb im Newsbeitrag
news:1151364710.259696.187920@b68g2000cwa.googlegroups.com...
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:


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
Belinfante-Rosenfeld procedure. I wrote about this many times, see
Measurement Techniques, 46, No. 4, 317 (2003), or www.sciprint.org.
Radi Khrapko


Yes and E x A is zero in the time average isnt it ? But the formula of timo
i speak of
is a E x E* and a purely imaginary number. Thats Timo also.

And i remember a discussion with you short time ago that you - and not
Timo -
said that E x A is the right spin tensor, isnt it ?

E x A obviously is wrong !
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Josef Matz
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Joined: 08 May 2005
Posts: 255

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

<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 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
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Josef Matz
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Joined: 08 May 2005
Posts: 255

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

<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(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

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Radi Khrapko
science forum Guru Wannabe


Joined: 04 May 2005
Posts: 142

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

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
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Radi Khrapko
science forum Guru Wannabe


Joined: 04 May 2005
Posts: 142

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

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.
Radi Khrapko

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
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Radi Khrapko
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Joined: 04 May 2005
Posts: 142

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

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
Belinfante-Rosenfeld 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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Timo Nieminen
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Joined: 12 May 2005
Posts: 244

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

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).

--
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
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Timo Nieminen
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Joined: 12 May 2005
Posts: 244

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

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.

--
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
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Josef Matz
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Joined: 08 May 2005
Posts: 255

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

<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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Radi Khrapko
science forum Guru Wannabe


Joined: 04 May 2005
Posts: 142

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

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.
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 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


Quote:




Radi Khrapko
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Josef Matz
science forum Guru Wannabe


Joined: 08 May 2005
Posts: 255

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

<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 (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







Quote:
Radi Khrapko
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Radi Khrapko
science forum Guru Wannabe


Joined: 04 May 2005
Posts: 142

PostPosted: Sun Jun 25, 2006 6:03 am    Post subject: Re: Circularly polarized beam in a dielectric, etc. Reply with quote

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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