Re: Circularly polarized beam in a dielectric, etc.


<khrapko_ri@xxxxxxxxxxx> schrieb im Newsbeitrag
news:1150980215.077601.53930@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Oh, my ears and whiskers! Dear Timo, why are you not attentive as well
as hundreds of editors and referees? Many years ago I showed and
published that all parts of the Poynting vector are zero in the Beth
experiment

No, the azimuthal edge-effect component of the counterpropagating beams
_adds_, not cancels. The longitudinal component cancels. But it's the
angular component that gives you the angular momentum.

Look, the azimuthal component of S at the edge of the beam depends of
the
handedness of the circular polarisation of the beam. The handedness is
changed, along with the direction of propagation, by reflection from a
mirror. Hold your hands in front of you, touch the tips of your thumbs
together and see if your fingers point in the same direction.
Dear Timo, do not be obstinate. Please take a pen in your hand and sum
E & H fields of incident and reflected Beth's beams as well as I did it
for physics/0102084, for example. I wrote there, "Let us start from the
Jackson's expression for a circularly polarized beam.. .. Here E_0(x;
y) is the electric field of the beam. E_0(x; y) = Const inside the
beam, and E_0(x; y) = 0 outside the beam. The returning light beam may
be got by changing the sign of z and y. Adding up the passing and
returning light beams we get interesting expressions,.. .. The E and H
fields are parallel everywhere. So, the Poynting vector is zero."

Such a nonsense !!!!

E and H are perpendicular in light beam ! Maxwell equation greet !



I wrote many times that Humblet, Jackson, Ohanian, Crichton and
Marston, Stewart and others do NOT decompose the orbital AM \int
rx(ExB) into "orbital" and "spin" parts. They transform \int rx(ExB)
into the form of \int ExA or \int FxB. They do not get two terms. One
of their terms is zero (see physics/0102084;
Exercises-in-the-canonical-spin-tensor or Electrodynamics spin at
www.sciprint.org. Now this paper is considered by "Communications in
Mathematical Physics" since May 9, 2006)

In special cases, one of their terms is zero (also the 3rd "surface"
term
which is zero for any case where the procedure is valid). In general, EM
fields and light _can_ carry orbital AM.
Sorry, I am forced to quote myself. I wrote about the term in
physics/0102084, "But the derivative .. .. has only z component inside
the beam, and, in any case, the term .. .. is z directed. So, the term
is an integral moment of a longitudinal component of the momentum and
equals zero if the origin of the radius-vector is at the axis of
symmetry. The term bears no relation to the angular momentum of the
beam."

Well, L=lP/w and Spin=(degree of circular polarisation)P/w are two
different beings. L=\int rx(ExB) and Spin = spintensor. The spintensor
is my electrodynamics spin tensor which is absent in the standard
electrodynamics. So, the total AM is the sum J=L+Spin. So, the beam
carries 2P/w.

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

--
Timo Nieminen -

Radi Khrapko



.

Relevant Pages

  • Re: 1c+1c Light and matter
    ... >> In a situation where this is less than zero, ... and verifies the beam operating or blocked. ... > If one does not like blockers, ... set up a lab 2 disc light velocity measurer. ...
    (sci.physics)
  • Re: 1c+1c Light and matter
    ... >> In a situation where this is less than zero, ... and verifies the beam operating or blocked. ... > If one does not like blockers, ... set up a lab 2 disc light velocity measurer. ...
    (sci.physics.relativity)
  • Re: Circularly polarized beam in a dielectric, etc.
    ... published that all parts of the Poynting vector are zero in the Beth ... the azimuthal component of S at the edge of the beam depends of the ... handedness of the circular polarisation of the beam. ...
    (sci.physics.electromag)
  • Re: 19v2000 vector monitor HV adjustment
    ... should be 14.5KV at the anode with the beam current at zero. ... "Beam current at zero" that has me confused. ... and then adjust away. ...
    (rec.games.video.arcade.collecting)
  • Re: The Physics Behind Contractions.
    ... A     A' ... clock in the frame of a moving observer?'. ... does not represent a continuous beam. ... The only light beam in your diagram is represented  by the ...
    (sci.physics.relativity)