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\begin{document} \title { The Beth's experiment is under review }
\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 the Maxwell electrodynamics cannot explain the
result of the Beth's experiment. So, the electrodynamics is not
complete. A spin tensor is introduced into the electrodynamics.
A theory of the Beth's experiment is presented.\medskip
\noindent PACS number: 03.05.De\\
Keywords: classical spin; Belinfante's procedure; torque
\end{abstract}
%1
\section{The Beth's experiment}
The Beth's 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 agrees with
the formula
$$\tau=4{\cal P}/\omega.\eqno(1)$$
%2
\section{The standard explanation of the Beth's result}
According to Maxwell's theory (see, for example, \cite{Hei}), the
Poynting vector ${\bf E}\times{\bf H}$ is interpreted as the density of
momentum of the field (we put speed of light, $c=1$). We can then also
define an angular momentum relative to a given point $O$ or to a given
axis,
$${\bf J}=\int{\bf r}\times({\bf E}\times{\bf H})dV.\eqno(2)$$
A circularly polarized plane wave without an azimuthal phase structure
travelling 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 beam. Consider a
cylindrical beam. At the wall of the cylinder we let the field drop to
zero. It can then be shown that the wall of a beam gives a finite
contribution to $J_z$ \cite{Jac,Dar,Hum,Oha,Cri,Sim}.
For example, Ohanian wrote \cite{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}
The angular momentum (2) of a beam was repeatedly calculated,
$$J =\int E^2_0dV/\omega.$$
The energy of the beam is
$${\cal E}=\int E^ 2_0dV,$$
where $E_0$ is the amplitude of the {\bf E} field in the inner region of
the beam. So, the ratio ${\cal E}/J= \omega$ appears as the ratio ${\cal
E}/S,$ i.e. energy/spin, for a photon.
A circularly polarization of the Beth's beam is being reversed when the
beam passes through the half-wave plate. So, according to the paradigm,
the plate gets
$$J=2{\cal E}/\omega$$
when the beam passes through the plate.
The plate gets the same angular momentum from the reflected beam. So,
having divided by time we arrive to Eq. (1).
%3
\section{The Beth's experiment is a puzzle}
At the same time, it is evident that the Poynting vector in the Beth's
experiment equals zero because the passed beam interferes with the
reflected one. 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,
$${\bf E}_{\rm tot}=2[({\bf x}\cos z-{\bf y}\sin z)E_0
-{\bf z}(\sin z\:\partial_x E_0+\cos z\:\partial_y E_0)]\cos t,
\eqno(3)$$
$${\bf H}_{\rm tot}=-2[({\bf x}\cos z-{\bf y}\sin z)E_0
-{\bf z}(\sin z\:\partial_x E_0+\cos z\:\partial_y E_0)]\sin t.
\eqno(4)$$
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}.
%4
\section{Electrodynamics spin}
The Poynting vector, ${\bf E}\times{\bf H}$, is a component of
the Maxwell-Minkowski energy-momentum tensor,
$$T^\alpha_\mu=-F_{\mu\nu}F^{\alpha\nu}
+ \delta^\alpha_\mu F_{\sigma\nu}F^{\sigma\nu}/4,\eqno(5)$$
and the Eq. (2) looks as
$$J^{\mu\nu}=\int2r^{[\mu}T^{\nu]\alpha}dV_\alpha\eqno(6)$$
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}:
$$\mathop T\limits_c{}^{\mu\alpha}
=\partial^\mu A_\sigma\frac{\partial{\cal L}}
{\partial(\partial_\alpha A_\sigma)}
-g^{\mu\alpha}{\cal L}
=-\p^\mu A_\sigma F^{\alpha\sigma}
+g^{\mu\alpha}F_{\sigma\rho}F^{\sigma\rho}/4,\eqno(7)$$
$$\mathop\Upsilon\limits_c{}^{\mu\nu\alpha}
=-2 A^{[\mu}\delta^{\nu]}_\sigma
\frac{\partial{\cal L}}
{\partial(\partial_\alpha A_\sigma)}
=-2A^{[\mu}F^{\nu]\alpha}.\eqno(8)$$
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
$$\p_\beta A_\mu F^{\alpha\beta}\eqno(9)$$
to the canonical energy-momentum tensor (7).
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,
$$ A_\mu j^\alpha,\eqno(10)$$
repairs the divergence of the canonical tensor, and the first part,
$$\partial_\beta( A_\mu F^{\alpha\beta}),\eqno(11)$$
symmetrizes the contravariant form of the tensor.
The only reason for adding the term (9) to the canonical tensor (7) is
to obtain the Maxwell-Minkowski tensor (5) which was known
beforehand. The term (9) is not even a divergence.
Theorists ignore the second part (10) of the term. They do not see it.
But Belinfante and Rosenfeld \cite{Bel,Ros} made a considerable
study of the first part (11).
They pointed out that an antisymmertization of the first part yields
the divergence of the canonical spin tensor (8) with the minus sign,
$$2\partial_\beta(A^{[\mu} F^{\nu]\beta})
=-\p_\beta\mathop\Upsilon\limits_c{}^{\mu\nu\beta}.\eqno(12)$$
We shall use the fact below, but now we have to emphasize that the
Belinfante's part (11) itself turns the canonical energy-momentum
tensor (7) not to the Maxwell-Minkowski tensor (5), 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 (5) by adding the Belinfante's divergence
(11) to the canonical tensor (7). Moreover, on the grounds of (12),
theorists add $(-\mathop\Upsilon\limits_c{}^{\mu\nu\beta})$ to the
canonical spin tensor (8) and arrive to a zero as the electrodynamics'
spin tensor. So, the Belinfante's
divergence (11) does not lead to the Maxwell-Minkowski tensor (5), 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 (9) must satisfy an equation of type (12)
which uses (9) instead of (11),
$$2\partial_\beta A^{[\mu} F^{\nu]\beta}
=\p_\beta\Delta\mathop\Upsilon\limits_c{}^{\mu\nu\beta}.\eqno(13)$$
A simple expression satisfies the Eq. (13),
$$\Delta\mathop\Upsilon\limits_c{}^{\mu\nu\alpha}
=2A^{[\mu}\p^{\nu]}A^\alpha.$$
So, we obtain \cite{GR10,084,031}
$$\Upsilon^{\mu\nu\alpha}
=\mathop\Upsilon\limits_c{}^{\mu\nu\alpha}
+\Delta\mathop\Upsilon\limits_c{}^{\mu\nu\alpha}
=2A^{[\mu}\partial^{\mid\alpha\mid}A^{\nu]}.\eqno(14)$$
This result was submitted to ``JETP Letters'' on May 14, 1998.
The spin tensor (14) 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 (14) 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. Using vector notations, it
can be introduced as follows.
\centerline{If div{\bf D} = 0, then {\bf D} = curl$\Pi$.
If also curl{\bf H} = $\partial${\bf D}/$\partial t$,
then $\bf H = \partial{\Pi}/\partial t$.}
This procedure is similar to an introduction of the magnetic vector
potential:
\centerline{If div{\bf B} = 0, then {\bf B} = curl{\bf A}.
If also curl{\bf E} = - $\partial${\bf B}/$\partial t$,
then ${\bf E} = - \partial{\bf A}/\partial t$.}
Scalar potentials may participate in both cases still, but we may
consider they are zero.
The symmetry of the electrodynamics forces us to offer a symmetric
expression for the spin tensor consisting of two parts, electric and
magnetic \cite{GR10,084}.
$$\Upsilon^{\mu\nu\alpha} =
{\mathop\Upsilon\limits_e}^{\mu\nu\alpha}+
{\mathop\Upsilon\limits_m}^{\mu\nu\alpha}=
A^{[\mu}\partial^{\mid\alpha\mid}A^{\nu]}
+\Pi^{[\mu}\partial^{\mid\alpha\mid} \Pi^{\nu]}.\eqno(15)$$
%5
\section{The theory of the Beth's experiment}
Now we have to replace Eqs. (2) and (6) by
$$J^{\mu\nu}=\int2r^{[\mu}T^{\nu]\alpha}dV_\alpha
+\int\Upsilon^{\mu\nu\alpha}dV_\alpha.\eqno(16)$$
We have to name the first term {\it orbital} angular momentum of
electromagnetic waves. It is zero in the Beth's experiment. The second
term is spin angular momentum of the waves. Ohanian et al. were not
right. The spin part of the total angular momentum of the light (16)
exerts a torque on the Beth's plate. The part is absent in the Maxwell
electrodynamics. So, the electrodynamics is incomplete.
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 with the absence
of an energy current!
\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}.
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\end{document}