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\begin{document}
\title {A circularly polarized beam carries\\
the double angular momentum}
\author {Radi I. Khrapko\thanks{Email addresses:
khrapko\_ri@mai.ru, khrapko\_ri@hotmail.com}\\ {\it Moscow Aviation
Institute, 125993, Moscow, Russia}} \date{} \maketitle
\begin{abstract}
It is shown that, in reality, a circularly polarized light beam without
an azimuthal phase structure carries the double angular momentum
in comparison with the prediction of the standard Maxwell
electrodynamics, and an experimental check of the statement is
suggested.
The Maxwell electrodynamics does not explain a result of the Beth
experiment and solves the problem of the two-element absorber
incorrectly. A conception of classical electrodynamics spin is
presented. \medskip
\noindent PACS number: 03.05.De\\
Keywords: classical spin; Beth experiment; electrodynamics' torque
\end{abstract}
%1
\section{The standard point of view}
All physicists \cite{Dar,Hum,Hei,Jac,
Sim,Oha,Cri,All99,All00,All02} consider
$$J=\cal E/\omega\eqno(1)$$
where $\cal E$ is energy of a circularly polarized light beam without
an azimuthal phase structure as total angular momentum of the
beam, and the angular momentum is given by the formula
$${\bf J}=\int{\bf r}\times({\bf E}\times{\bf H})dV.\eqno(2)$$
Heitler wrote \cite[p.401]{Hei},
\begin{quote}
A plane wave traveling in the $z$-direction and with infinite
extension in the $xy$-directions can have no angular momentum
about
the $z$-axis, because ${\bf E}\times{\bf H}$ is in the $z$-direction
and $({\bf r}\times({\bf E}\times{\bf H}))_z=0.$ However, this
is no longer the case for a wave with finite extension in the $xy$-
plane. Consider a cylindrical wave with its axis in the $z$-direction
and traveling in this direction. At the wall of the cylinder $r=R,$ say,
we let the amplitude drop to zero. It can be shown that the wall
of such a wave packet gives a finite contribution to $J_z.$
\end{quote}
Ohanian wrote \cite[p.502]{Oha}
\begin{quote}
In a wave of finite transverse extent, the {\bf E} and {\bf B} fields
have a component parallel to the wave vector (the field lines are
closed loops) and the energy flow has components perpendicular to
the wave vector. \dots The circulating energy flow in the wave
implies the existence of angular momentum, whose direction is along
the direction of propagation. This angular momentum is the spin
of the wave.
\end{quote}
Simmonds and Guttman \cite[p.227]{Sim} wrote
\begin{quote}
The electric and magnetic field can have a nonzero $z$-component
only within the skin region of the wave. Having $z$-component
within this region implies the possibility of a nonzero $z$-component
of angular momentum within this region. Since the wave is
identically zero outside the skin and constant inside the skin region,
the skin region is the only one in which the $z$-component of
angular momentum does not vanish.
\end{quote}
All physicists think that the angular momentum has been measured by
Beth \cite{Bet} by letting the beam pass through a doubly
refracting plate which alters the polarization. The plate was allowed to
rotate about the axis of the beam. It was found that the plate,
indeed, begins to rotate as soon as the beam passes through, and the
angular momentum acquired by the plate was in agreement with
(1) \cite[p.404]{Hei}.
%2
\section{The Beth experiment is a puzzle}
The Beth classical experiment \cite{Bet} was made almost 70 years
ago.
\begin{quote}
A beam of circularly polarized light should exert a torque on a doubly
refracting plate which changes the state of polarization of the light
beam.
The apparatus used involves a torsional pendulum with about a ten-
minute
period consisting of a round quartz half-wave plate one inch in
diameter
suspended with its plane horizontal from a quartz fiber about 25
centimeters long. About 4 millimeters above this is mounted a fixed
quartz
quarter-wave plate. The {\it top} side of the upper plate was coated by
evaporation with a reflecting layer of aluminum. The rotation of the
pendulum is observed by a telescope.
\end{quote}
A circularly polarized light beam (power, ${\cal P}=80$ mW;
$\lambda=1.2$ mu; $\omega=1.6\cdot 10^{15}$ 1/s) passes through
the
half-wave plate, then it passes twice through the quarter-wave plate,
and
then returns through the half-wave plate. The torque, $\tau,$ exerting
on
the half-wave plate is 20 dyne-cm.
This result seems to be agreed with (1)
$$\tau=4{\cal P}/\omega.\eqno(3)$$
At the same time, it is evident that the Poynting vector ${\bf
E}\times{\bf H}$ equals to zero in the Beth experiment because the
passed beam interferes with the reflected one. So, ${\cal P}=0.$
Indeed, let us start from the Jackson's expression for a circularly
polarized beam \cite{Jac}.
($\omega = k =1$).
$${\bf E}=[{\bf x}\cos(z-t)-{\bf y}\sin(z-t)
+{\bf z}(-\sin(z-t)\p_x-\cos(z-t)\p_y)]E_0(x,y),$$
$${\bf H}=-\int{\rm rot}{\bf E}dt
=[{\bf x}\sin(z-t)+{\bf y}\cos(z-t)
+{\bf z}(\cos(z-t)\p_x-\sin(z-t)\p_y)]E_0(x,y),$$
Here $E_0(x, y) = {\rm Const}$
inside the beam, and $E_0(x, y) =0$ outside the beam.
The reflected beam may be got by changing the sign of $z$ and
$y$. Adding up the passed and reflected beams, we get
interesting expressions,
$${\bf E}_{\rm tot}=2[({\bf x}\cos z-{\bf y}\sin z)E_0
-{\bf z}(\sin z\:\partial_x E_0+\cos z\:\partial_y E_0)]\cos
t,\eqno(4)$$
$${\bf H}_{\rm tot}=-2[({\bf x}\cos z-{\bf y}\sin z)E_0
-{\bf z}(\sin z\:\partial_x E_0+\cos z\:\partial_y E_0)]\sin
t.\eqno(5)$$
The ${\bf E}_{\rm tot}$ and ${\bf H}_{\rm tot}$ fields are parallel
to
each other everywhere. So, the Poynting vector is zero, and,
according to
(2), angular momentum of the light in the Beth's experiment is zero!
So,
the light should not exert a torque on a plate. Why did the torque exert
the
plate then?
As a matter of fact, the all-recognized equation (2) seems to be
incorrect.
As a matter of fact, the Maxwell electrodynamics is not complete
\cite{29MAI,37MAI,IT,307}.
%3
\section{The problem of the two-element absorber }
An important question was raised at the
V. L. Ginsburg Moscow Physical Seminar in the spring of 1999. The
question was about absorption of circularly polarized light by a round
flat target, which is divided concentrically into an inner disc and a
closely fitting outer annulus \cite{Khr}.
If the target absorbs a circularly polarized beam, the annulus absorbs
the surface of the beam, which carries the angular momentum,
and the disc absorbs the body of the beam, which has no angular
momentum. According to the Maxwell theory, the expression
$d{\cal F}^j =T^{ji}da_i $
gives an infinitesimal force acting on a surface element $da_i$
(here $T^{ji}$ is the Maxwell stress tensor, i.e. a space part of the
Maxwell-Minkowski energy-momentum tensor).
So, the disc does not perceive a torque when the target absorbs a
circularly polarized beam, since the Poynting vector is perpendicular
to the disc. There are no ponderomotive forces, which are capable to
twist the disc. Tangential forces act only on the annulus.
But it is clear that in reality the disc does perceive a torque from the
wave, since the disc gets spin which is not described by the Maxwell
theory. Beth \cite{Bet}, Feynman \cite{Fey} clearly showed how
circularly polarized wave transferred torque to medium. The disc will
be twisted in contradiction with the Maxwell paradigm.
Allen and Padgett \cite{All02} have attempted to explain the torque
acting on the disc within the scope of the standard electrodynamics.
They have written, ``Any form of aperture introduces an intensity
gradient \dots and a field component is induced in the propagation
direction and so the dilemma is potentially resolved.'' They
decompose a plane wave into three beams: the inner beam, the
annulus beam, and the remainder.
Alas! A small clearance between the inner disc and outer annulus
does not aperture a beam and does not induce longitudinal field
components. The imaginary decomposition of the plane wave is not
capable to create longitudinal field components and, correspondingly,
transverse momentum and torque acting on the disc. Maxwell stress
tensor cannot supply the disc with a torque. According to the Maxwell
theory, the disc absorbs energy and feels pressure only.
%4
\section{ Classical electrodynamics' spin}
To resolve the problem, in our opinion, we must use the conception
of classical electrodynamics' spin which is described by a four-spin
tensor $\Upsilon^{\mu\nu\alpha}$ \cite{GR10,084,031,IT}
(more precisely, tensor density).
Electrodynamics' spin tensor is not zero, and ponderomotive forces
acting on a surface element $da_i$ consist of both, the force itself and
a torque.
$$d{\cal F}^j= T^{ji}da_i, \qquad d\tau^{jk}
= \Upsilon^{jki}da_i \eqno(6).$$
So, we must recognize that the standard classical electrodynamics is
not
complete. Therefore, the angular momentum (2) is not spin. It is
orbital
angular momentum $L$. This is the main point of the paper.
So, if a target absorbs a circularly polarized electromagnetic beam,
tangential forces act on the target at the area of absorption of the
surface of the beam. Besides, a torque acts on the target at the all
territory of absorption of the beam. As it was shown, the both parts of
acting are equal to each other, $L=S$. Therefore the resultunt angular
momentum received by a target will be equal to
$$J=L+S=2{\cal E}/\omega,$$
instead of $J=L$, as it is predicted by the Maxwell theory.
So, we double the angular momentum of a beam.
This result was submitted to ``JETP'' on February 25, 1999.
%5
\section{ An experimental check of the electrodynamics }
An experimental check of this result in a terrestrial laboratory is
complicated by thermal streams which is related to absorption of
energy. However, these difficulties disappear if a target is in free
space. So, consider \cite{IT} an aluminum disk with mass $m$ =
27~g
and area $a$ = 1~m$^2$ (the thickness is 10 microns,
the rotational inertia is $I$ = 4.3~g$\cdot$m$^2$).
I offer to illuminate it by circularly polarization
electromagnetic radiation, the wavelength is $\lambda$ = 10 cm,
and power is $\cal P$ = 100 W.
According to the Maxwell theory, the torque is
$$\tau={\cal P}/\omega ={\cal P}\lambda/2\pi c
=5.3\cdot10^{-9}{\rm N}\cdot{\rm m}.$$
Therefore, the disk will rotate through 1 radian in a time of 27
minutes, and, according to my prediction, in a time of 19 minutes
($t=\sqrt{2I/\tau}$).
%6
\section{A theory of the Beth experiment}
Let us apply the conception of the classical electrodynamics' spin to
the Beth experiment. Consider the expression
\cite{GR10,084,031,IT}
$$\Upsilon^{\mu\nu\alpha} =
{\mathop\Upsilon\limits_e}^{\mu\nu\alpha}+
{\mathop\Upsilon\limits_m}^{\mu\nu\alpha}=
A^{[\mu}\partial^{\mid\alpha\mid}A^{\nu]}
+\Pi^{[\mu}\partial^{\mid\alpha\mid} \Pi^{\nu]}.\eqno(7)$$
as the electrodynamics' spin tensor (see Sect. 7). Here $A^\mu$ is
the magnetic vector potential, and $\Pi^\mu$ is the electric vector
potential,
$$2\p_{[\nu}A_{\mu]}=F_{\nu\mu}, \quad \Pi_\mu
=\epsilon_{\mu\alpha\beta\sigma}\Pi^{\alpha\beta \sigma}, \quad
\p_\sigma \Pi^{\alpha\beta \sigma}=F^{\alpha\beta}.$$
Using vector notations, the potentials can be introduced as follows.
\begin{center}
If div{\bf B} = 0, then {\bf B} = curl{\bf A}.
If also $\p{\bf B}/\p t=-$curl{\bf E}
then $\p{\bf A}/\p t=-{\bf E}$.\\
If div{\bf D} = 0, then {\bf D} = curl$\Pi$.
If also $\p{\bf D}/\p t$ = curl{\bf H},
then $\p{\Pi}/\p t={\bf H}.$
\end{center}
We consider the Beth's light beams as plane waves because the
Poynting
vector is zero, and the wall effects are of no importance.
In front of the half-wave plate we have (3) and (4) with $E_0=1$,
$${\bf E}=2({\bf x}\cos z-{\bf y}\sin z)\cos t,$$
$${\bf A}=-\int{\bf E}dt
=2({\bf x}\cos z-{\bf y}\sin z)(-\sin t).$$
So, the electrical part of spin current density is uniform, but pulses,
$$\mathop\Upsilon\limits_e{}^{xyz}=2\sin^2t.$$
The magnetic part is
$${\bf H}=2(-{\bf x}\cos z+{\bf y}\sin z)\sin t,$$
$$\Pi=\int{\bf H}dt=2({\bf x}\cos z-{\bf y}\sin z)\cos t,\quad
\mathop\Upsilon\limits_m{}^{xyz}=2\cos^2t.$$
So, the total spin current density is constant,
$\Upsilon^{xyz}=\mathop\Upsilon\limits_e{}^{xyz}
+\mathop\Upsilon\limits_m{}^{xyz}=2.$
The same calculation for the domain behind the plate gives
$\Upsilon^{xyz}=-2.$
This means that the $S^{xy}$-component of the spin moves opposite
the $z$-direction, i.e.\ towards the plate also. In result, the plate
receives the spin current density of 4, as in (3), with the absence of
an energy current!
\section{Appendix. The Belinfante-Rosenfeld mistake }
The Poynting vector, ${\bf E}\times{\bf H}$, is a component of
the Maxwell-Minkowski energy-momentum tensor,
$$T^\alpha_\mu=-F_{\mu\nu}F^{\alpha\nu}
+ \delta^\alpha_\mu F_{\sigma\nu}F^{\sigma\nu}/4,\eqno(8)$$
and the Eq. (2) looks as
$$J^{\mu\nu}=\int2r^{[\mu}T^{\nu]\alpha}dV_\alpha\eqno(9)$$
in the Minkowski space.
There are no simple theoretical method to obtain the
Maxwell-Minkowski tensor. The canonical Lagrangian,
$${\cal L}=-F_{\mu\nu}F^{\mu\nu}/4,\quad
F_{\mu\nu}=2\p_{[\mu}A_{\nu]},$$
according to Noether's theorem, gives the canonical energy-
momentum
and spin tensors \cite[Sec.\ 7g]{Sch}:
$$\mathop T\limits_c{}^{\mu\alpha}
=\partial^\mu A_\sigma\frac{\partial{\cal L}}
{\partial(\partial_\alpha A_\sigma)}
-g^{\mu\alpha}{\cal L}
=-\p^\mu A_\sigma F^{\alpha\sigma}
+g^{\mu\alpha}F_{\sigma\rho}F^{\sigma\rho}/4,\eqno(10)$$
$$\mathop\Upsilon\limits_c{}^{\mu\nu\alpha}
=-2 A^{[\mu}\delta^{\nu]}_\sigma
\frac{\partial{\cal L}}{\partial(\partial_\alpha A_\sigma)}
=-2A^{[\mu}F^{\nu]\alpha}.\eqno(11)$$
But the canonical energy-momentum tensor is asymmetric and has an
incorrect divergence,
$$\p_\alpha \mathop T\limits_c{}^{\mu\alpha}
=-\p^\mu A_\nu j^\nu.$$
The divergence of a true energy-momentum tensor must be equal to
$-
F_{\mu\nu}j^\nu.$
To symmetrize the canonical energy-momentum tensor and to turn it
to
the Maxwell-Minkowski tensor, theorists has to add a term
$$\p_\beta A_\mu F^{\alpha\beta}\eqno(12)$$
to the canonical energy-momentum tensor (10).
This term divides in two parts:
$$T^\alpha_\mu =\mathop T\limits_c{}^\alpha_\mu
+\p_\beta A_\mu F^{\alpha\beta}
=\mathop T\limits_c{}^\alpha_\mu + \partial_\beta( A_\mu
F^{\alpha\beta})+ A_\mu j^\alpha.$$
The second part,
$$ A_\mu j^\alpha,\eqno(13)$$
repairs the divergence of the canonical tensor, and the first part,
$$\partial_\beta( A_\mu F^{\alpha\beta}),\eqno(14)$$
symmetrizes the contravariant form of the tensor.
The only reason for adding the term (12) to the canonical tensor (10)
is to
obtain the Maxwell-Minkowski tensor (8) which was known
beforehand. The term (12) is not even a divergence.
Theorists ignore the second part (13) of the term. They do not see it.
But Belinfante and Rosenfeld \cite{Bel,Ros} made a considerable
study of the first part (14).
They pointed out that an antisymmertization of the first part yields the
divergence of the canonical spin tensor (11) with the minus sign,
$$2\partial_\beta(A^{[\mu} F^{\nu]\beta})
=-\p_\beta\mathop\Upsilon\limits_c{}^{\mu\nu\beta}.\eqno(15)$$
We shall use the fact below, but now we have to emphasize that the
Belinfante's part (14) itself turns the canonical energy-momentum
tensor
(10) not to the Maxwell-Minkowski tensor (8), but to a tensor that
may
be named the Belinfante's tensor,
$$\mathop T\limits_{B}{}^{\mu\alpha}
=\mathop T\limits_c{}^{\mu\alpha}
+\partial_\beta(A^\mu F^{\alpha\beta})
=T^{\mu\alpha}-A^\mu j^\alpha.$$
Nevertheless, theorists believe that they obtain the
Maxwell-Minkowski tensor (8) by adding the Belinfante's divergence
(14) to the canonical tensor (10). Moreover, on the grounds of (15),
theorists
add $(-\mathop\Upsilon\limits_c{}^{\mu\nu\beta})$ to the canonical
spin tensor (11) and arrive to a zero as the electrodynamics' spin
tensor.
So, the Belinfante's
divergence (14) does not lead to the Maxwell-Minkowski tensor (8),
but it
eliminates spin tensor. That is why classical spin is absent in the
Maxwell electrodynamics. The classical spin tensor is considered
zero.
That is why they consider a circularly polarized plane wave has no
angular momentum \cite{Khr,
Jac,Dar,Hum,Oha,Cri,Sim,All99,All00}.
Here a problem arises: what is an electrodynamics' true spin tensor.
What
must we add to the canonical spin tensor to get the true spin tensor?
Our answer is as follows: a spin addition,
$\Delta\mathop\Upsilon\limits_c{}^{\mu\nu\alpha}$, and the
energy-momentum addition (12) must satisfy an equation of type (15)
which uses (12) instead of (14),
$$2\partial_\beta A^{[\mu} F^{\nu]\beta}
=\p_\beta\Delta\mathop\Upsilon\limits_c{}^{\mu\nu\beta}.\eqno(16)
$$
A simple expression satisfies the Eq. (16),
$$\Delta\mathop\Upsilon\limits_c{}^{\mu\nu\alpha}
=2A^{[\mu}\p^{\nu]}A^\alpha.$$
So, we obtain \cite{GR10,084,031}
$$\Upsilon^{\mu\nu\alpha}
=\mathop\Upsilon\limits_c{}^{\mu\nu\alpha}
+\Delta\mathop\Upsilon\limits_c{}^{\mu\nu\alpha}
=2A^{[\mu}\partial^{\mid\alpha\mid}A^{\nu]}.\eqno(17)$$
This result was submitted to ``JETP Letters'' on May 14, 1998.
The spin tensor (17) is a function of the vector potential $A_\mu$ and
is
not gauge invariant. We greet this fact \cite{031}. As is shown,
$A^\mu$
must satisfy the Lorentz condition, $\partial_\mu A^\mu=0$.
The expression (17) is not final. As a matter of fact, the
electrodynamics is asymmetric. Magnetic induction is closed, but
magnetic field strength has electric current as a source:
$$\partial_{[\alpha}F_{\beta\gamma]}=0,\quad
\partial_\nu F^{\mu\nu} =j^\mu.$$
So, a magnetic vector potential exists, but, generally speaking, an
electric vector potential does not exist. However, when currents are
absent the symmetry is restored, and a possibility to introduce an
electric multivector potential $\Pi^{\mu\nu\sigma}$ appears. The
electric multivector potential satisfies the equation
$$\partial_\sigma\Pi^{\mu\nu\sigma}=F^{\mu\nu}.$$
A covariant vector, dual relative to the multivector potential,
$$\Pi_\alpha=\epsilon_{\alpha\mu\nu\sigma}\Pi^{\mu\nu\sigma},$$
is an analog of the magnetic vector potential $A_\alpha$.
We name it the electric vector potential.
The symmetry of the electrodynamics forces us to offer a symmetric
expression for the spin tensor consisting of two parts, electric and
magnetic (7).
%
\section*{Conclusion}
The Maxwell electrodynamics is not complete. We have to put a spin
tensor in the modern electrodynamics. Theorists did not catch sight of
the classical spin because of ignoring a localization of
energy-momentum and due to the concept of the ambiguity of an
energy-momentum tensor. A significance of the Lagrange formalism
was overestimated.\medskip
\noindent I am deeply grateful to Professor Robert H. Romer for
publishing my question \cite{Khr}.
%
\section*{Notes}
The material of this paper was rejected or ignored by the journals
(dates
of the submissions are in brackets):\\
Phys. Rev. D (25 Sep 2001, 23 Sept 2002), \\
Foundation of Physics (28 May 2001, 03 May 2002, 08 Oct 2002, 23
Oct 2002, 16 Apr 2003, 17 Apr 2003), \\
American J. of Physics (15 Sep 1999, 10 Sep 2001, 26 Nov 2001, 28
Mar 2002, 03 Jun 2002, 13 Mar 2003), \\
Acta Physica Polonica B (28 Jan 2002, 09 May 2002, 02 Jun 2002, 12
Mar 2003, 13 Mar 2003), \\
Phys. Lett. A (22 July 2002, 28 Aug 2002, 14 Nov 2002, 12 Mar
2003, 21 Apr 2003), \\
Journal of Physics A (23 Jun 2002, 02 Mar 2003, 12 Mar 2003),\\
Journal of Mathematical Physics (28 Nov 2002, 13 Mar 2003, 28 Mar
2003),\\
Optics Communications (22 Sept 2002),\\
Europhysics Letters (15 Oct 2002, 15 Apr 2003),\\
Experimental \& Theor. Phys. Lett. ({\bf14 May 1998}, 17 June 2002, 05
Aug 2002, 20 Nov 2002), \\
J. Experimental \& Theor. Phys. (27 Jan 1999, {\bf25 Feb 1999}, 13 Apr
2000, 25 May 2000, 16 May 2001, 26 Nov 2001, 05 Jul 2002, 12 Dec
2002), \\
Theor. Math. Phys. (29 Apr 1999, 17 Feb 2000, 29 May 2000, 18 Oct
2000, 05 Jul 2002, 19 Dec 2002), \\
Physics - Uspekhi (25 Feb 1999, 12 Jan 2000, 31 May 2000, 04 Jul
2002, 31 Jul 2002), \\
Rus. Phys. J. (18 May 1999, 15 Oct 1999, 1 March 2000, 25 May
2000, 31 May 2001, 24 Nov 2001, 05 Jul 2002).
The Editors, V. F. Gantmakher, A. F. Andreev, A.~A.~Logunov,
V.~N.~Detinko, ignored my objections.
The arXiv blacklisted all computers of the Moscow Aviation Institute
and rejected my submissions 21 Jan 2002, 18 Feb 2002, 02 June 2002,
13 June 2002, 12 Apr 2003, 15 May 2003.
Prof. Ginsparg ignored my objections.
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\end{document}