\documentclass[reqno]{amsart}
\begin{document}
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{\noindent\small Mathematical Physics and Quantum Field Theory,\newline
Electronic Journal of Differential Equations, Conf. 04, 2000, pp. 37--50.\newline
http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu \quad ejde.math.unt.edu (login: ftp)}
\thanks{\copyright 2000 Southwest Texas State University and
University of North Texas.} \vspace{1cm}
\title[Angular Momenta of Electromagnetic Radiation]
{The measurable distinction between the spin and orbital angular
momenta of electromagnetic radiation}
\author{James H. Crichton \& Philip L. Marston }
\address{James H. Crichton (deceased) \hfill\break\indent
Department of Physics, Seattle Pacific University \hfill\break\indent
Seattle, Washington 98119, USA}
\address{Philip L. Marston \hfill\break\indent
Department of Physics, Washington State University \hfill\break\indent
Pullman, Washington 99164, USA}
\email{marston@wsu.edu}
\thanks{Published July 12, 2000.}
\subjclass{78A40, 78A46, 78A45, 81U40}
\keywords{Electromagnetic radiation, scattering, angular momenta}
\begin{abstract}
We show how the angular momentum of electromagnetic radiation may
be decomposed into spin and orbital parts, of which the spin part
is measurable in terms of Stokes parameters, thereby providing an
unambiguous, gauge-invariant, distinction between the two parts.
\end{abstract}
\maketitle
\newcommand{\re}{\,\text{Re}\,}
\newcommand{\im}{\,\text{Im}\,}
\section{Introduction}
\label{sec:intro}
We are concerned with two issues in this paper. First, to what extent
are those physical quantities which are transported by electromagnetic
radiation measurable in terms of Stokes parameters? These quantities
would include energy, momentum and angular momentum. We find that, in
general, the angular momentum is not measurable, but only its spin
part. Thus, our second issue: how can the spin and orbital angular
momentum of electromagnetic radiation be distinguished?
An example will illustrate these issues. Consider a
left-circularly-polarized mono\-chromatic plane wave incident on a
dielectric sphere. The problem is to determine the force and torque
exerted by the incoming wave on the sphere~\cite{Debye09,Marston84}.
It would seem reasonable (but it turns out to be unnecessary) to
expect that it would be important to know experimentally how much
momentum and angular momentum would be transported by the scattered
radiation. Detectors could be set up in the radiation zone, with
appropriate polarization filters, to measure Stokes parameters at all
scattering angles. The question is then, is such a detection scheme
adequate to determine the momentum and angular momentum flux of the
scattered radiation? The answer is yes for momentum, but no for
angular momentum. The detection of the right- and left-circular
polarizations of the scattered light permits only a partial
measurement -- the spin part -- of the angular momentum.
Formulas for the force and torque exerted by beams of light on
arbitrarily-shaped target objects depend on a precise knowledge of the
electric and magnetic multipoles induced in the
target~\cite{Marston84,Chang85,Barton89}. We are not assuming such
precise knowledge, but are, in effect, inquiring whether measurements
of the Stokes parameters in the radiation zone are sufficient to
specify such detailed information of the source.
Relating the Stokes parameters to the angular momentum of light,
particularly to the spin angular momentum, is not a new idea. Jauch
and Rohrlich developed a quantum-mechanical version of the Stokes
parameters to discuss the spin of photons~\cite{Jauch55}. Here we are
working with the classical fields and the classical Stokes parameters,
but we will be able to exploit the close relationship between
circularly-polarized light and the helicity states of photons in developing and
interpreting our classical results.
In Section \ref{sec:planewaves}, we define the Stokes parameters and
review the relationships among angular momentum, helicity, and the
Stokes parameters. In Section \ref{sec:transport}, we obtain
expressions for the rate of transport of the physical quantities of
interest from a finite source through a spherical surface far removed
from the source. We express these quantities in terms of Stokes
parameters insofar as possible. An understanding of the measurement
of the angular momentum consistent with the quantum nature of light is
presented in Section \ref{sec:tenative} and related to the Stokes
parameters. The well-known decomposition of the angular momentum of
electromagnetic radiation into spin and orbital parts is reviewed in
Section \ref{sec:Humblet} and applied to the problem raised in
Sections \ref{sec:transport} and \ref{sec:tenative}. The problem of
the gauge invariance of this decomposition is discussed in Section
\ref{sec:spin}. We give some specific examples of the problem of
non-measurability in Section \ref{sec:examples} and save some
concluding remarks for Section \ref{sec:conclusion}. In an appendix,
we present formulas for the rate of transport of physical quantities,
given the source multipole moments. These formulas are useful for the
examples of Section \ref{sec:examples}.
\section{Plane waves and Stokes parameters}
\label{sec:planewaves}
Consider a monochromatic electromagnetic plane wave propagating in the
$z$-direction. With real functions, we could express such a wave with
transverse components as follows:
\begin{subequations}
\begin{gather}
E_{x} = E_{0x} \cos \left( kz - \omega t + \alpha \right),\\
E_{y} = E_{0y} \cos \left( kz - \omega t + \beta \right),
\end{gather}
\end{subequations}
where the amplitudes $E_{0x}$, $E_{0y}$ are real and positive. In the
physically-realizable case, they must decrease to zero at some finite
distance from the $z$-axis. Such a variation in $x$ and $y$ gives a
non-vanishing longitudinal component to the field which, as we shall
see, is related to the angular momentum carried by the
wave~\cite{Simmons70}.
In taking time-averages, it will be convenient to work with complex
fields:
\begin{subequations}
\begin{gather}
E_{x}=|E_{0x}| e^{i(kz-\omega t + \alpha)},
\label{complexfielda} \\
E_{y}=|E_{0y}| e^{i(kz-\omega t + \beta)},
\label{complexfieldb}
\end{gather}
\end{subequations}
whose real parts are the physical fields.
The Stokes parameters are measures of the energy flux and the
polarization of the electromagnetic wave. Using the convention of
Bohren and Huffman, we take these to be~\cite{Bohren83}:
\begin{subequations}
\begin{gather}
I=|E_{0x}|^{2} + |E_{0y}|^{2}, \\
Q=|E_{0x}|^{2} - |E_{0y}|^{2}, \\
U=2 \re \left( E_{x}E_{y}^{*} \right) = 2|E_{0x}||E_{0y}|\cos
\left( \alpha - \beta \right), \\
V=2 \im \left( E_{x}E_{y}^{*} \right) = 2|E_{0x}||E_{0y}|\sin
\left( \alpha - \beta \right).
\end{gather}
\end{subequations}
For the monochromatic radiation under consideration, these parameters
are not all independent and are related
by~\cite{Bohren83,Jackson75,Hecht98}: $I^{2} = Q^{2} + U^{2} + V^{2}$.
The energy, momentum and angular momentum densities of electromagnetic
fields are given by the following well-known formulas (in Gaussian
units)~\cite{Jackson75}:
\begin{subequations}
\begin{gather}
e=\frac{1}{8\pi} \left( E^{2}+B^{2} \right), \\
p_{z}=\frac{1}{4\pi c}\left( \mathbf{E} \times \mathbf{B}
\right)_{z},\\
j_{z}=\frac{1}{4\pi c}\left[\mathbf{r} \times \left(
\mathbf{E} \times \mathbf{B} \right) \right]_{z}.
\end{gather}
\end{subequations}
For the case of plane waves, we can obtain expressions for the
time-averaged flux of energy, momentum and angular momentum by
considering the amount of each in a slab of thickness $c\delta t$
perpendicular to the $z$-axis. Then:
\begin{subequations}
\begin{gather}
\langle \dot{\varepsilon} \rangle = \frac{c}{4\pi} \int dx\,
dy\, \langle E^{2}+B^{2} \rangle = \frac{c}{8\pi} \int dx\,
dy\, (\mathbf{E}^{*} \cdot \mathbf{E}), \\
\langle \dot{P_{z}} \rangle = \frac{1}{4\pi} \int dx\, dy\,
\langle \mathbf{E} \times \mathbf{B} \rangle_{z} =
\frac{1}{8\pi} \int dx\, dy\, (\mathbf{E}^{*} \cdot
\mathbf{E}), \\
\langle \dot{J_{z}} \rangle = \frac{1}{8\pi} \int dx\, dy\,
\left[ \mathbf{r} \times (\mathbf{E}^{*} \times \mathbf{B})
\right]_{z}.
\label{angmomflux}
\end{gather}
\end{subequations}
In these formulas, we are using the standard procedure for
time-averaging sinusoid\-ally-varying waves, where the fields are now
given by the complex quantities of Eqs.\ \eqref{complexfielda} and
\eqref{complexfieldb}. The brackets $\langle \dots \rangle$ denote
the time-average of the quantity inside. Thus:
\begin{subequations}
\begin{gather}
\langle \dot{\varepsilon} \rangle = \frac{c}{8\pi} \int dx\,
dy\, \left( |E_{0x}|^{2}+|E_{0y}|^{2} \right) = \frac{c}{8\pi}
\int dx\, dy\, I,
\label{energyflux} \\
\langle \dot{P_{z}} \rangle = \frac{1}{8\pi} \int dx\, dy\, I.
\label{momentumflux}
\end{gather}
We see in Eqs.\ \eqref{energyflux} and \eqref{momentumflux} that
the energy and momentum fluxes can be expressed in terms of Stokes
parameters integrated over the surface normal to the propagation
direction. The corresponding expression for angular momentum flux
requires some careful treatment. It is useful to go to a
cylindrical coordinate system. In the source-free region, the
divergence of $\mathbf{E}$ vanishes, giving:
\begin{equation*}
\boldsymbol{\nabla} \cdot \mathbf{E} = \frac{1}{\rho}
\frac{\partial(\rho E_{\rho})}{\partial \rho} + \frac{1}{\rho}
\frac{\partial E_{\phi}}{\partial \phi} + \frac{\partial
E_{z}}{\partial z} = 0.
\end{equation*}
Thus, for a $z$-dependence $e^{ikz}$, we have:
\begin{equation*}
-ikE_{z} = \frac{1}{\rho} \frac{\partial(\rho
E_{\rho})}{\partial \rho} + \frac{1}{\rho} \frac{\partial
E_{\phi}}{\partial \phi}.
\end{equation*}
Likewise, for a time-dependence $e^{-i\omega t}$, Faraday's law
gives:
$$ \gathered
\frac{1}{\rho} \frac{\partial(E_{z})}{\partial \phi} -
ikE_{\phi} = ikB_{\rho}, \\
ikE_{\rho} - \frac{\partial E_{z}}{\partial \rho} =
ikB_{\phi}, \\
\frac{1}{\rho} \frac{\partial(\rho E_{\phi})}{\partial \rho} -
\frac{1}{\rho} \frac{\partial(E_{\rho})}{\partial \phi} =
ikB_{z}.
\endgathered$$
Substituting these results in Eq.\ \eqref{angmomflux} (neglecting
small quantities) gives:
\begin{equation*}
\begin{split}
\langle \dot{J_{z}} \rangle & = \frac{1}{8\pi} \int \rho\,
d\rho\, d\phi\, \rho(E_{z}^{*}B_{\rho} -
E_{\rho}^{*}B_{z}) \\
& = \frac{1}{8\pi ik} \int \rho\, d\rho\, d\phi\, \left(
E_{\rho}^{*} \frac{\partial E_{\rho}}{\partial \phi} +
E_{\phi}^{*} \frac{\partial E_{\phi}}{\partial \phi}
\right),
\end{split}
\end{equation*}
neglecting ``small'' terms and integrating by parts, as necessary.
Returning to Cartesian coordinates, using:
$$ \gathered
E_{\rho} = E_{x}\cos \phi + E_{y}\sin \phi, \\
E_{\phi} = -E_{x}\sin \phi + E_{y}\cos \phi,
\endgathered $$
we have finally:
\begin{equation}
\begin{split}
\langle \dot{J_{z}} \rangle & = \frac{1}{8\pi ik} \int
dx\, dy\, \left[ E_{x}^{*} \frac{\partial E_{x}}{\partial
\phi} + E_{y}^{*}\frac{\partial E_{y}}{\partial \phi} -
2i\im(E_{x}E_{y}^{*}) \right] \\
& = \frac{1}{8\pi k} \int dx\, dy\, \left[ \frac{1}{2} I
\frac{\partial}{\partial \phi}(\alpha + \beta) +
\frac{1}{2} \frac{\partial}{\partial \phi}(\alpha - \beta)
- V \right].
\end{split}
\end{equation}
\end{subequations}
We note here that the angular momentum flux cannot be expressed
completely in terms of Stokes parameters as explained below Eq.\
\eqref{fluxstokes}. This fact is the central issue of this paper and
which will be examined for outgoing spherical waves in more detail in
later sections.
To illustrate the use of these expressions, consider a standard case,
a left-circularly-polarized (LCP) plane wave~\cite{Hecht98}:
$$\gathered
E_{x} = \frac{E_{0}}{\sqrt{2}}\cos(kz-\omega t) = \re \left[
\frac{E_{0}}{\sqrt{2}} e^{i(kz-\omega t)} \right], \\
E_{y} = -\frac{E_{0}}{\sqrt{2}}\sin(kz-\omega t) = \re \left[
\frac{E_{0}}{\sqrt{2}} e^{i(kz-\omega t+ \pi/2)} \right].
\endgathered$$
We obtain:
$$\gathered
\langle \dot{\varepsilon} \rangle_{\mathit{LCP}} = \frac{c}{8\pi}
\int dx\, dy\, |E_{0}|^{2}, \\
\langle \dot{P_{z}} \rangle_{\mathit{LCP}} = \frac{1}{8\pi} \int
dx\, dy\, |E_{0}|^{2}, \\
\langle \dot{J_{z}} \rangle_{\mathit{LCP}} = \frac{1}{8\pi k} \int
dx\, dy\, |E_{0}|^{2}.
\endgathered $$
Thus: $\langle \dot{\varepsilon} \rangle_{\mathit{LCP}} = c \langle
\dot{P_{z}} \rangle_{\mathit{LCP}} = \omega \langle \dot{J_{z}}
\rangle_{\mathit{LCP}}$, with obvious quantum-mechanical
interpretation that $\hbar$ of angular momentum corresponds to $\hbar
\omega$ of energy for this wave. In the language of particle physics,
such a wave would be said to have positive helicity: the angular
momentum component in the direction of motion is $+\hbar$. Likewise,
the right-circularly-polarized (RCP) states have negative helicity.
For such waves:
\begin{equation*}
\langle \dot{J_{z}} \rangle_{\mathit{RCP}} = - \frac{\langle
\dot{P_{z}} \rangle_{\mathit{RCP}}}{k} = - \frac{\langle
\dot{\varepsilon} \rangle_{\mathit{RCP}}}{\omega}.
\end{equation*}
It will be useful later to have the transverse components of the
electric field decomposed into LCP and RCP components. This is done
as follows:
\begin{equation*}
\mathbf{e}_{L} \equiv -\frac{(\mathbf{i} +
i\mathbf{j})}{\sqrt{2}}, \quad \mathbf{e}_{R} \equiv
\frac{(\mathbf{i} - i\mathbf{j})}{\sqrt{2}}.
\end{equation*}
Then:
\begin{equation*}
\mathbf{E} = E_{x}\mathbf{i} + E_{y}\mathbf{j} =
E_{L}\mathbf{e}_{L} + E_{R}\mathbf{e}_{R},
\end{equation*}
where:
\begin{equation*}
E_{L} = -\frac{E_{x} - iE_{y}}{\sqrt{2}}, \quad E_{R} =
\frac{E_{x} + iE_{y}}{\sqrt{2}}.
\end{equation*}
(For the example given above: $E_{L}=-E_{0} e^{i(kz-\omega t)}$,
$E_{R}=0$.)
Expressing the Stokes parameters in terms of these helicity amplitudes
yields:
\begin{subequations}
\begin{gather}
I = |E_{L}|^{2} + |E_{R}|^{2}, \\
Q = -2\re (E_{L}E_{R}^{*}), \\
U = -2\im (E_{L}E_{R}^{*}), \\
V = -\left( |E_{L}|^{2} - |E_{R}|^{2}\right).
\end{gather}
\end{subequations}
\section{Transport of physical quantities by outgoing spherical waves}
\label{sec:transport}
In this section, we develop expressions for the fluxes of energy,
momentum and angular momentum carried by outing spherical waves in the
radiation zone. Far from the source of the radiation, the transverse
components of the electric field, given appropriately in a spherical
polar coordinate system, have a $1/r$ behavior:
\begin{equation*}
E_{\theta}, E_{\phi} \sim \frac{1}{r}e^{i(kr - \omega t)}
\end{equation*}
as do the transverse magnetic field components, which, through
Faraday's law, are related to the transverse electric field
components:
\begin{equation*}
B_{\theta} = - E_{\phi}, \quad B_{\phi} = E_{\theta},
\end{equation*}
the error in these equations being of order $1/r^{2}$. There is a
radial component for each field, of order $1/r^{2}$, obtained from the
divergence conditions. Thus:
$$\gathered
-ikE_{r} = \frac{1}{r\sin \theta} \left[ \frac{\partial}{\partial
\theta} (\sin \theta E_{\theta}) + \frac{\partial
E_{\phi}}{\partial \phi} \right] + O\left(1/r^{3}\right), \\
ikB_{r} = \frac{1}{r\sin \theta} \left[ \frac{\partial}{\partial
\theta} (\sin \theta E_{\phi}) - \frac{\partial
E_{\theta}}{\partial \phi} \right] + O\left(1/r^{3}\right).
\endgathered $$
In terms of the transverse $\mathbf{E}$ components, the time-averaged
fluxes of energy, momentum and angular momentum through a spherical
surface of radius $r$ are given by:
\begin{subequations}
\begin{gather}
\langle \dot{\varepsilon} \rangle = \frac{c}{8\pi} \int
r^{2}\, d\Omega\, \left( |E_{\theta}|^{2} + |E_{\phi}|^{2}
\right),
\label{energyflux1} \\
\langle \dot{P_{z}} \rangle = \frac{1}{8\pi} \int r^{2}\,
d\Omega\, \left( |E_{\theta}|^{2} + |E_{\phi}|^{2} \right)
\cos \theta, \\
\langle \dot{J_{z}} \rangle = \frac{1}{8\pi k} \int r^{2}\,
d\Omega\, \left[ E_{\theta}^{*} \left( -i \frac{\partial
E_{\theta}}{\partial \phi} \right) + E_{\phi}^{*} \left( -i
\frac{\partial E_{\phi}}{\partial \phi} \right) \right],
\label{angmomflux1}
\end{gather}
\end{subequations}
where the integration is over the solid angle $d\Omega = \sin\theta\,
d\theta\, d\phi$. Some integration by parts was required to bring the
expression for $\langle \dot{J_{z}} \rangle$ into its final form.
We again follow Bohren and Huffman in defining Stokes parameters for
the outgoing radiation at angles $\theta$ and $\phi$:
\begin{subequations}
\begin{gather}
I = |E_{\theta}|^{2} + |E_{\phi}|^{2}, \\
Q = |E_{\theta}|^{2} - |E_{\phi}|^{2}, \\
U = -2\re (E_{\theta}E_{\phi}^{*}), \\
V = +2\im (E_{\theta}E_{\phi}^{*}).
\end{gather}
\end{subequations}
The expressions for the fluxes in terms of these Stokes parameters are
then given by:
\begin{subequations}
\label{fluxstokes}
\begin{gather}
\langle \dot{\varepsilon} \rangle = \frac{c}{8\pi} \int
r^{2}\, d\Omega\, I, \\
\langle \dot{P_{z}} \rangle = \frac{1}{8\pi} \int r^{2}\,
d\Omega\, I \cos \theta, \\
\langle \dot{J_{z}} \rangle = \frac{1}{16\pi k} \int r^{2}\,
d\Omega\, \left[ I \frac{\partial}{\partial \phi} (\alpha +
\beta) + Q \frac{\partial}{\partial \phi} (\alpha - \beta)
\right].
\label{fluxstokesc}
\end{gather}
\end{subequations}
with $E_{\theta} = |E_{\theta}|e^{i\alpha}$ and $E_{\phi} =
|E_{\phi}|e^{i\beta}$. For outgoing spherical waves, as for plane
waves, the angular momentum flux cannot be expressed completely in
terms of Stokes parameters, hence is not measurable. The unmeasurable
quantity is the sum of the phases, $\alpha + \beta$, the difference
being given by the inverse tangent of $-V/U$.
\section{A tentative measurement of the angular momentum of outgoing
spherical radiation}
\label{sec:tenative}
We have seen that the theoretical expression for the angular momentum
flux derived in the previous section does not give much hope for
measuring angular momentum in the radiation zone. But certainly we
can measure something. If we think of the scattered light as
consisting of photons, each of which has a probability for being found
to have positive or negative helicity upon measurement, then we can
count the contributions of all the photons in carrying a certain
component of angular momentum.
Here is the measurement that we propose: set a detector with an
appropriate set of polarization filters at given angles $\theta$ and
$\phi$. The detector can be used to measure the irradiance of
left-circularly-polarized light (energy carried by photons found to
have positive helicity) and that of right-circularly-polarized light.
The resulting contribution to angular momentum transport over the
total solid angle is (again, the $z$-component for convenience) we
denote by $\langle \dot{J_{z}} \rangle_{M}$:
\begin{equation}
\langle \dot{J_{z}} \rangle_{M} = \int r^{2}\, d\Omega\, \left[
\frac{I_{L}}{h\nu}(+\hbar) + \frac{I_{R}}{h\nu}(-\hbar) \right]
\cos\theta.
\end{equation}
In this expression, the irradiances, $I_{L}$ and $I_{R}$, or left- and
right-circularly-polarized light, respectively, are divided by
$\hbar\omega$ to give the number of photons per unit area per unit
time. An angular momentum $+\hbar$ is attributed to the LCP states
and $-\hbar$ to the RCP states. The irradiances called for are
obtained by decomposing the electric field into LCP and RCP components
as in Section \ref{sec:planewaves} with
\begin{equation}
\mathbf{e}_{L} = -\frac{(\mathbf{e}_{\theta} +
i\mathbf{e}_{\phi})}{\sqrt{2}}, \quad \mathbf{e}_{R} =
\frac{(\mathbf{e}_{\theta} - i\mathbf{e}_{\phi})}{\sqrt{2}}.
\end{equation}
Here the LCP and RCP field components are:
\begin{equation}
E_{L} = -\frac{(E_{\theta} - iE_{\phi})}{\sqrt{2}}, \quad E_{R} =
\frac{(E_{\theta} + iE_{\phi})}{\sqrt{2}}.
\label{fieldcomponents}
\end{equation}
The irradiances are then given by $I_{L,R} = (c/8\pi)|E_{L,R}|^{2}$,
so that
\begin{equation}
\begin{split}
\langle \dot{J_{z}} \rangle_{M} & = \frac{c}{8\pi\omega} \int
r^{2}\, d\Omega\, \left( |E_{L}|^{2} - |E_{R}|^{2} \right)
\cos \theta \\
& = - \frac{1}{8\pi k} \int r^{2}\, d\Omega\, V\cos\theta.
\end{split}
\end{equation}
Thus, the angular momentum flux that we could measure is expressible
in terms of a Stokes parameter, but nevertheless there must be some
part of the angular momentum that we have missed, since $\langle
\dot{J_{z}} \rangle_{M}$ is clearly not equal to $\langle \dot{J_{z}}
\rangle$ as given by Eq.\ \eqref{fluxstokesc}.
\section{The Humblet decomposition: orbital and spin angular momenta
of light}
\label{sec:Humblet}
It is a common feature of quantum field theories that expressions for
total field angular momentum can be naturally decomposed into orbital
and spin parts~\cite{Jauch55,Bjorken65}. Humblet applied this idea to
the classical electromagnetic field~\cite{Humblet43}. For the
time-averaged angular momentum density ($i$-th Cartesian component),
\begin{equation}
j_{i} = \frac{1}{8\pi c} \left[ \mathbf{r} \times (\mathbf{E}^{*}
\times \mathbf{B}) \right]_{i},
\label{angmomdensity}
\end{equation}
a simple rearrangement of terms gives the desired result. We start
with Faraday's law (for monochromatic fields with time-dependence
$e^{-i\omega t}$):
\begin{equation*}
\boldsymbol{\nabla} \times \mathbf{E} = ik\mathbf{B}.
\end{equation*}
Then:
\begin{equation}
j_{i} =\frac{1}{8\pi i \omega} \left\{ \mathbf{r} \times \left[
\mathbf{E}^{*} \times (\boldsymbol{\nabla} \times \mathbf{E})
\right] \right\}_{i} = l_{i} + s_{i} + j_{i}^{\mathit{surf}};
\end{equation}
with
\begin{subequations}
\begin{gather}
l_{i} = \frac{1}{8\pi\omega} E_{j}^{*} \left[ -i(\mathbf{r}
\times \boldsymbol{\nabla}) \right]_{i} E_{j}, \\
s_{i} = \frac{1}{8\pi\omega} E_{j}^{*} (-i \varepsilon_{ijk})
E_{k}, \\
j_{i}^{\mathit{surf}} = - \frac{1}{8\pi\omega} \nabla_{j}
(E_{j}^{*}(\mathbf{r} \times \mathbf{E})_{i}).
\end{gather}
\end{subequations}
(In these expressions, summation over repeated indices is assumed and
$\varepsilon_{ijk}$ is the Levi--Civita symbol, which is $+1$ $(-1)$
when the indices are an even (odd) permutation of $123$ and zero
otherwise.)
The ``orbital'' angular momentum density, $l_i$ , has a striking
similarity to a quantum-mechanical density with the orbital
angular-momentum operator (except for a factor of $h/2\pi$) sandwiched
between a wave function and its complex conjugate. (Elsewhere, it is
explained why the electric field components do not make good wave
functions for photons~\cite{Cohen89}.)
The ``spin'' angular momentum density, $s_{i}$, is appropriately named
in that there is no moment arm. Its relation to spin is even more
strongly suggested when we form matrices $\Sigma_{i}$ according to the
following rule:
\begin{equation}
(\Sigma_{i})_{jk} = - i \varepsilon_{ijk}.
\end{equation}
The set of three $3 \times 3$ matrices $\{\Sigma_{i}\}$ satisfies the
angular momentum commutation relations and further:
\begin{equation}
\Sigma_{1}^{2} + \Sigma_{2}^{2} + \Sigma_{3}^{2} = 2 I_{3},
\end{equation}
as would be appropriate for spin-one matrices. The ``surface'' term,
$j_{i}^{\mathit{surf}}$ is a three-divergence: integration of this
term throughout a volume is thus equivalent to integration of
$E_{n}^{*} (\mathbf{r} \times \mathbf{E})_{i}$ over a surface ($E_{n}$
being the component of $\mathbf{E}$ normal to that surface) which
could be indefinitely far removed from the source. Such an integral
can be safely taken to vanish, although this should be checked
whenever the Humblet decomposition is used. Humblet showed that, in
calculating the time-averaged flux of angular momentum through a
spherical surface, the surface term makes no contribution.
Thus, for the angular momentum flux we have:
\begin{subequations}
\begin{gather}
\langle \dot{J_{z}} \rangle = \langle \dot{L_{z}} \rangle +
\langle \dot{S_{z}} \rangle, \\
\langle \dot{L_{z}} \rangle = \frac{1}{8\pi k} \int r^{2}\,
d\Omega\, E_{j}^{*} (-i) \frac{\partial E_{j}}{\partial \phi},
\label{angmomflux2b} \\
\langle \dot{S_{z}} \rangle = \frac{1}{8\pi k} \int r^{2}\,
d\Omega\, \left( |E_{L}|^{2} - |E_{R}|^{2} \right) \cos
\theta,
\label{angmomflux2c}
\end{gather}
\end{subequations}
where in Eq.\ \eqref{angmomflux2c}, we have made use of Eq.\
\eqref{fieldcomponents}. We see that $\langle \dot{S_{z}} \rangle$ is
exactly what we suggested could be measured in the hypothetical
experiment of Sec.\ \ref{sec:tenative}, and the orbital part is that
which cannot be measured.
Our result shows that there is a measurable distinction between the
spin and orbital parts of the angular momentum of electromagnetic
radiation. Yet, it is often stated that there is no unique,
gauge-invariant, separation of these two kinds of angular
momentum~\cite{Jauch55,Humblet43,Cohen89}. To this problem we turn
next.
\section{Spin and gauge invariance}
\label{sec:spin}
The standard decomposition of the angular momentum density uses
$\mathbf{B} = \boldsymbol{\nabla} \times \mathbf{A}$ in the
development of Eq. \eqref{angmomdensity}. There results the
following:
\begin{equation}
j_{i} = l_{i} + s_{i} + j_{i}^{\mathit{surf}};
\end{equation}
with
\begin{subequations}
\begin{gather}
l_{i} = \frac{1}{8\pi} E_{j}^{*} \left[(\mathbf{r} \times
\boldsymbol{\nabla})\right]_{i} A_{j}, \\
s_{i} = \frac{1}{8\pi} E_{j}^{*} \varepsilon_{ijk} A_{k} \\
j_{i}^{\mathit{surf}} = - \frac{i}{8\pi} \nabla_{j} \left[
E_{j}^{*}(\mathbf{r} \times \mathbf{A})_{i} \right].
\end{gather}
\end{subequations}
Under a gauge transformation,
\begin{equation}
\mathbf{A} \rightarrow \mathbf{A}' = \mathbf{A} +
\boldsymbol{\nabla} \chi,
\end{equation}
the spin angular momentum density transforms as:
\begin{equation}
s_{i} \rightarrow s_{i}' = s_{i} + \frac{1}{8\pi} (\mathbf{E}^{*}
\times \boldsymbol{\nabla} \chi)_{i}.
\end{equation}
Because no measurement can distinguish between $\mathbf{A}$ and
$\mathbf{A}'$, it would seem that the spin density is not unique. We
suggest below a number of ways out of this difficulty.
First of all, if we restrict the physical situation to one of
monochromatic radiation, the decomposition performed in Sec.\
\ref{sec:Humblet} is valid, and the decomposition there is expressed
in terms of gauge-invariant quantities, i.e., the transverse
electric-field components in the radiation zone.
Second, we point out that the change in $s_{i}$, to leading order in
powers of $1/r$, is:
\begin{equation}
\delta \mathbf{s} = \frac{1}{8\pi} (\mathbf{E}^{*} \times
\boldsymbol{\nabla} \chi) = -\frac{ik}{8\pi r}(\mathbf{r} \times
\mathbf{E}^{*}\chi),
\end{equation}
i.e., it has a moment arm. It appears then that to the spin density
has been added an orbital part. But what is true of the orbital
angular momentum density is true here also: the change in the spin
density is transverse to the propagation direction and cannot be
measured. Our gauge-invariant decomposition gives gauge-invariant
quantities, one of which is measurable.
Finally, we note that the electromagnetic fields can be uniquely
decomposed into longitudinal ($\parallel$) and transverse ($\perp$)
parts~\cite{Cohen89}:
\begin{equation}
\mathbf{E} = \mathbf{E}_{\parallel} + \mathbf{E}_{\perp}, \quad
\mathbf{B} = \mathbf{B}_{\parallel} + \mathbf{B}_{\perp},
\end{equation}
with
\begin{equation}
\boldsymbol{\nabla} \times \mathbf{E}_{\parallel} =
\boldsymbol{\nabla} \times \mathbf{B}_{\parallel} = 0; \quad
\boldsymbol{\nabla} \cdot \mathbf{E}_{\perp} = \boldsymbol{\nabla}
\cdot \mathbf{B}_{\perp} = 0.
\end{equation}
The vector potential $\mathbf{A}$ can be similarly decomposed. A
gauge transformation is then represented as:
\begin{subequations}
\begin{gather}
\mathbf{A}_{\parallel} \rightarrow \mathbf{A}_{\parallel}' =
\mathbf{A}_{\parallel} + \boldsymbol{\nabla} \chi, \\
\mathbf{A}_{\perp} \rightarrow \mathbf{A}_{\perp}' =
\mathbf{A}_{\perp}.
\end{gather}
\end{subequations}
The angular momentum density can also be expressed in terms of
longitudinal and transverse parts:
\begin{subequations}
\begin{gather}
\mathbf{j}_{\parallel} = \frac{1}{4\pi} \mathbf{r} \times
(\mathbf{E}_{\parallel} \times \mathbf{B}), \\
\mathbf{j}_{\perp} = \frac{1}{4\pi} \mathbf{r} \times
(\mathbf{E}_{\perp} \times \mathbf{B}).
\end{gather}
\end{subequations}
Making the substitution $\mathbf{B} = \boldsymbol{\nabla} \times
\mathbf{A}$ as before, it can be shown that the longitudinal part can
be distributed to the moment of (canonical) momentum of the charged
particles constituting the source of the fields. The longitudinal
part is not gauge-invariant. However, the transverse part, which can
be expressed as:
\begin{equation}
\mathbf{j}_{\perp} = \frac{1}{4\pi} \left[ E_{\perp i} (\mathbf{r}
\times \boldsymbol{\nabla}) A_{\perp i} + \mathbf{E}_{\perp}
\times \mathbf{A}_{\perp} \right],
\end{equation}
is obviously gauge-invariant. It is this transverse part with which
we are dealing in the radiation zone.
Our conclusion, then, is that the spin flux $\langle \dot{S_{z}}
\rangle$ is measurable and certainly satisfies gauge invariance as
much as any other measurable quantity.
\section{Examples of measurable spin and unmeasurable orbital angular
momentum}
\label{sec:examples}
We now present some examples to illustrate this measurability problem
with the angular momentum of outgoing spherical electromagnetic waves.
First, we consider the field of an electric dipole rotating in the
$x$-$y$ plane with angular frequency $\omega$:
\begin{subequations}
\begin{equation}
\mathbf{D} = D ( \mathbf{i} \cos \omega t + \mathbf{j} \sin
\omega t),
\end{equation}
or, in terms of complex quantities:
\begin{equation}
\mathbf{D} = D (\mathbf{i} + i\mathbf{j}) e^{-i\omega t}.
\end{equation}
\end{subequations}
The electric-field components in the radiation zone
are~\cite{JacksonP395}:
\begin{subequations}
\begin{gather}
E_{\theta} = \frac{k^{2}D}{r} \cos\theta e^{i\phi}
e^{i(kr-\omega t)}, \\
E_{\phi} = i \frac{k^{2}D}{r} e^{i\phi} e^{i(kr-\omega t)}.
\end{gather}
\end{subequations}
Using Eqs.\ \eqref{energyflux1} -- \eqref{angmomflux1}, we obtain the
fluxes of interest:
\begin{subequations}
\begin{gather}
\langle \dot{\varepsilon} \rangle = \frac{2}{3}k^{3}
D^{2}\omega, \\
\langle \dot{P_{z}} \rangle = 0, \\
\langle \dot{J_{z}} \rangle = \frac{2}{3} k^{3} D^{2},
\end{gather}
and, from Eqs.\ \eqref{angmomflux2b} and \eqref{angmomflux2c}:
\begin{equation}
\langle \dot{L_{z}} \rangle = \langle \dot{S_{z}} \rangle =
\frac{1}{3} k^{3}D^{2}.
\end{equation}
\end{subequations}
In this example, in which we know what the radiated angular momentum
will be because we know the source of the fields, we see that half of
the total angular momentum, in the orbital form, will escape
unmeasured: it cannot in principle be measured from Stokes parameters.
Consider next the example given in the introduction: the scattering of
LCP light from a dielectric sphere. The measurement of spin angular
momentum is predicted to be:
\begin{equation}
\langle \dot{S_{z}} \rangle = \frac{1}{2k^{3}} \sum_{n} \left[
\frac{2n+1}{n(n+1)} \left( |a_{n}|^{2} + |b_{n}|^{2} \right) +
\frac{2n(n+2)}{n+1} \re \left( a_{n}b_{n+1}^{*} + b_{n}a_{n+1}^{*}
\right) \right]
\end{equation}
where the $a_{n}$, $b_{n}$ are the Mie scattering
coefficients~\cite{Marston84b}. (This expression is derived in the
Appendix.) This is another case where we know, theoretically, the
source of the fields, so the measurement of the spin would be a test
of the theory. The total angular momentum, however, is given
theoretically by a much simpler expression. Matching boundary
conditions of incoming and scattered waves on the surface of the
sphere gives to the spherical polar components of the scattered field
a simple $\phi$-dependence: $e^{i\phi}$. The expression for the total
angular momentum is then:
\begin{equation}
\langle \dot{J_{z}} \rangle = \frac{1}{8\pi k} \int r^{2}\,
d\Omega\, I = \frac{\langle \dot{\varepsilon} \rangle}{\omega}.
\end{equation}
We must emphasize that this simple result, whose expression in terms
of Mie coefficients is given in the Appendix, is a theoretical one
which cannot be verified by measurements in the radiation zone.
As a final example, we use multipole sources which are nonvanishing
only for $m = 0$ components (in the notation of the Appendix):
\begin{equation}
a_{E,M}(n,m) = a_{E,M}(n,0) \delta_{m0},
\end{equation}
Then, setting $a_{E}(n,0) = c_{n}$ and $a_{M}(n,0) = d_{n}$, we have:
\begin{subequations}
\begin{gather}
\langle \dot{\varepsilon} \rangle = \frac{c}{\pi k^{2}}
\sum_{n} \left( |c_{n}|^{2} + |d_{n}|^{2} \right), \\
\langle \dot{P_{z}} \rangle = -\frac{1}{4\pi k^{2}} \sum_{n}
\left[ \frac{n(n+2)}{(2n+1)(2n+3)}\right]^{1/2} \im \left(
c_{n}c_{n+1}^{*} + d_{n}d_{n+1}^{*} \right), \\
\langle \dot{J_{z}} \rangle = 0, \\
\langle \dot{S_{z}} \rangle = \frac{1}{4\pi k^{3}} \sum_{n}
\left[ \frac{n(n+2)}{(2n+1)(2n+3)}\right]^{1/2} \re \left(
d_{n}c_{n+1}^{*} - c_{n}d_{n+1}^{*} \right).
\end{gather}
\end{subequations}
Here, without further specification of the multipole sources, we have
an example in which the unmeasurable orbital angular momentum just
cancels the measurable spin to give zero ($z$-component) total angular
momentum. As in the previous cases, measurement of the spin provides
no useful information concerning the total angular momentum.
\section{Conclusion}
\label{sec:conclusion}
It should be noted that restricting our development to the
$z$-components of momentum, angular momentum and spin does not limit
our conclusions. It is simply an artifact of the spherical polar
coordinate system that the corresponding expressions for $x$- and
$y$-components are somewhat more complicated. Nevertheless, $\langle
\dot{P_{x}} \rangle$, $\langle \dot{P_{y}} \rangle$, $\langle
\dot{S_{x}} \rangle$ and $\langle \dot{S_{y}} \rangle$ can be
expressed completely in terms of Stokes parameters but $\langle
\dot{J_{x}} \rangle$ and $\langle \dot{J_{y}} \rangle$ cannot.
We have shown above, on the basis of classical electromagnetic theory,
that it is possible to measure energy, momentum and spin angular
momentum transported by electromagnetic radiation from a finite source
region \emph{and} that it is \emph{not} possible to measure total
angular momentum using Stokes parameters.
Such a finding is surprising on the one hand because of the lack of
gauge invariance when decomposing the total angular momentum into spin
and orbital parts. On the other hand, it is not surprising when we
consider these measurements from the standpoint of quantum theory.
The detectors used to measure the Stokes parameters determine photon
momentum, all three components. Physical quantities whose
corresponding operators do not commute with the momentum operator are
left undetermined, unmeasurable. The orbital angular momentum is one
such quantity. The helicity, however, is the projection of angular
momentum along the momentum direction and the corresponding operator
commutes with the momentum operator. Hence, helicity is measurable
along with momentum. We have shown then that the classical theory is
consonant with the quantum theory in this regard.
\appendix
\section{Rate of transport of energy, momentum, angular momentum and
spin from a given source of electromagnetic radiation}
To obtain the formulas for energy, momentum, angular momentum and spin
transport in terms of sources, we need the transverse components of
the electric field in the radiation zone. The most general solution
of Maxwell's equations corresponding to outgoing spherical waves in
the radiation zone is, according to Jackson:
\begin{equation}
\mathbf{E} = \frac{1}{kr} e^{i(kr-\omega t)} \sum_{n,m} (-i)^{n+1}
\left[ a_{E}(n, m) \mathbf{X}_{nm} + a_{M}(n, m) \mathbf{n} \times
\mathbf{X}_{nm}\right] \times \mathbf{n},
\end{equation}
where the $\mathbf{X}_{nm}$ are vector spherical harmonics,
\begin{equation*}
\mathbf{X}_{nm} = \left[n(n+1)\right]^{-1/2} \mathbf{n} \times
\boldsymbol{\nabla} Y_{nm},
\end{equation*}
$\mathbf{n}$ is a unit vector radially outward, and the $a_{E}(n,m)$,
$a_{M}(n,m)$ are the electric and magnetic multipole moments of the
source. Helicity components $E_{L}$, $E_{R}$ are found to be:
\begin{multline}
E_{L,R} = \pm \frac{1}{kr} e^{i(kr-\omega t)} \sum_{n,m}
(-i)^{n+1} (-1)^{m} \left[\frac{2n+1}{8\pi}\right]^{1/2} d_{\pm
1,m}^{n}(\theta) e^{im\phi} \\
\times \left[a_{M}(n, m) \pm ia_{E}(n, m)\right].
\label{helicitycomps}
\end{multline}
In Eq.\ \eqref{helicitycomps} the $d_{mm'}^{n}$ are the reduced Wigner
rotation matrices. Various properties of the vector spherical
harmonics, $d$-matrices and Clebsch--Gordan coefficients needed in
these derivations can be found in the monograph of Varshalovich,
Moskalev and Khersonkii~\cite{Varshalovich88}.
The time-averaged energy flux at distance $r$ from the source in terms
of helicity components is given by:
\begin{subequations}
\label{energyfluxeqns}
\begin{equation}
\langle \dot{\varepsilon} \rangle = \frac{c}{8\pi} \int
r^{2}\, d\Omega\, \left( |E_{L}|^{2} + |E_{R}|^{2} \right).
\end{equation}
Substitution of $E_{L}$ and $E_{R}$ from Eq.\
\eqref{helicitycomps} and integration yield:
\begin{equation}
\langle \dot{\varepsilon} \rangle = \frac{c}{8\pi k^{2}}
\sum_{n,m} \left( |a_{E}(n, m)|^{2} + |a_{M}(n, m)|^{2}
\right).
\end{equation}
\end{subequations}
The time-averaged angular momentum flux ($z$-component for
convenience) is:
\begin{equation}
\begin{split}
\langle \dot{J_{z}} \rangle & = \frac{1}{8\pi k} \int r^{2}\,
d\Omega\, \left[ E_{L}^{*} (-i) \frac{\partial}{\partial \phi}
E_{L} + E_{R}^{*} (-i) \frac{\partial}{\partial \phi} E_{R}
\right] \\
& = \frac{1}{8\pi k^{3}} \sum_{n,m} m \left( |a_{E}(n, m)|^{2}
+ |a_{M}(n, m)|^{2} \right).
\end{split}
\label{angmomflux3}
\end{equation}
The time-averaged momentum flux ($z$-component) is given by:
\begin{equation}
\langle \dot{P_{z}} \rangle = \frac{1}{8\pi} \int r^{2}\,
d\Omega\, \left( |E_{L}|^{2} + |E_{R}|^{2} \right) \cos\theta,
\end{equation}
while the time-averaged spin angular momentum flux ($z$-component) is
given by:
\begin{equation}
\langle \dot{S_{z}} \rangle = \frac{1}{8\pi k} \int r^{2}\,
d\Omega\, \left( |E_{L}|^{2} - |E_{R}|^{2} \right) \cos\theta.
\end{equation}
Thus, we must calculate the integral
\begin{equation*}
I_{\pm} = \int r^{2}\, d\Omega\, \left( |E_{L}^{2} \pm |E_{R}|^{2}
\right) \cos\theta.
\end{equation*}
Exploiting the fact that $d_{00}^{1}(\theta) = \cos\theta$, and using
various symmetry properties of the $d$ matrices, we obtain:
\begin{multline*}
I_{\pm} = \frac{1}{6k^{2}} \sum_{n,m} \sum_{n'} \left[
(2n+1)(2n'+1) \right]^{1/2} (-i)^{n-n'} (-1)^{m-1} \\
\times \left\{ \left[ a_{M}(n, m) + i a_{E}(n, m) \right] \left[
a_{M}^{*}(n', m) - i a_{E}^{*}(n', m)\right] \right. \\
\times (n1n'-1|nn'10) (nmn'-m|nn'10) \\
\pm \left[ a_{M}(n, m) - i a_{E}(n, m) \right] \left[
a_{M}^{*}(n', m) + i a_{E}^{*}(n', m)\right] \\
\left. \times (n1n'1|nn'10) (nmn'-m|nn'10) \right\},
\end{multline*}
where the $(j_{1}m_{1}j_{2}m_{2}|j_{1}j_{2}JM)$ are the
Clebsch--Gordan coefficients in a standard notation~\cite{PDG98}.
Explicit evaluation of the C--G coefficients leads to:
\begin{multline}
I_{\pm} = \frac{1}{2k^{2}} \sum_{n,m} \left\{ \frac{m}{n+1} \left[
|a_{M}(n, m) + i a_{E}(n, m)|^{2} \mp |a_{M}(n, m) - i a_{E}(n,
m)|^{2} \right] \right. \\
- \frac{2}{n+1} \left[ \frac{n(n+2)(n-m+1)(n+m+1)}{(2n+1)(2n+3)}
\right]^{1/2} \\
\times \left\{ \im \left[ \left( a_{M}(n, m) + i a_{E}(n, m)
\right) \left(a_{M}^{*}(n+1, m) - i a_{E}^{*}(n+1, m) \right)
\right] \right. \\
\left. \left. \pm \im \left[ \left( a_{M}(n, m) - i a_{E}(n, m)
\right) \left( a_{M}^{*}(n+1, m) + i a_{E}^{*}(n+1, m) \right)
\right] \right\} \right\}.
\end{multline}
Taking the upper signs, we have:
\begin{multline}
\langle \dot{P_{z}} \rangle = - \frac{1}{4\pi k^{2}} \sum_{n,m}
\left\{ \frac{m}{n(n+1)} \im \left[ a_{E}(n,m) a_{M}^{*}(n,m)
\right] \right. \\
+ \frac{1}{(n+1)} \left[ \frac{n(n+2)(n-m+1)(n+m+1)}{(2n+1)(2n+3)}
\right]^{1/2} \\
\left. \times \im \left[ a_{M}(n, m) a_{M}^{*}(n+1, m) + a_{E}(n,
m) a_{E}^{*}(n+1, m) \right] \right\}.
\label{uppersigns}
\end{multline}
The lower signs give:
\begin{multline}
\langle \dot{S_{z}} \rangle = \frac{1}{8\pi k^{3}} \sum_{n,m}
\left\{ \frac{m}{n(n+1)} \left[ |a_{E}(n, m)|^{2} + |a_{M}(n,
m)|^{2} \right] \right. \\
+ \frac{2}{(n+1)} \left[ \frac{n(n+2)(n-m+1)(n+m+1)}{(2n+1)(2n+3)}
\right]^{1/2} \\
\left. \times \re \left[ a_{M}(n, m) a_{E}^{*}(n+1, m) - a_{E}(n,
m) a_{M}^{*}(n, m) \right] \right\}.
\label{lowersigns}
\end{multline}
For the case of a left-circularly-polarized plane wave scattering from
a dielectric sphere, we have, in the standard notation:
$$\gathered
a_{E}(n, m) = (i)^{n+1} \left [4\pi (2n + 1)\right]^{1/2} a_{n}
\delta_{m1}, \\
a_{M}(n, m) = (i)^{n+2} \left[ 4\pi (2n + 1)\right]^{1/2} b_{n}
\delta_{m1}.
\endgathered$$
With this restriction, Eqs.\ \eqref{energyfluxeqns},
\eqref{angmomflux3}, \eqref{uppersigns} and \eqref{lowersigns} reduce
to~\cite{BohrenMarston}:
\begin{gather}
\langle \dot{\varepsilon} \rangle_{\mathit{MS}} = \frac{c}{2k^{2}}
\sum_{n} (2n + 1) \left( |a_{n}|^{2} + |b_{n}|^{2} \right), \\
\langle \dot{J_{z}} \rangle_{\mathit{MS}} = \frac{1}{2k^{3}}
\sum_{n} (2n + 1) \left( |a_{n}|^{2} + |b_{n}|^{2} \right), \\
\langle \dot{P_{z}} \rangle_{\mathit{MS}} = \frac{1}{k^{2}}
\sum_{n} \left[ \frac{2n+1}{n(n+1)} \re(a_{n}b_{n}^{*}) +
\frac{n(n+2)}{n+1} \re \left( a_{n}a_{n+1}^{*} + b_{n}b_{n+1}^{*}
\right) \right] \\
\langle \dot{S_{z}} \rangle_{\mathit{MS}} = \frac{1}{2k^{3}}
\sum_{n} \left[ \frac{2n + 1}{n(n + 1)} \left( |a_{n}|^{2} +
|b_{n}|^{2} \right) + \frac{2n(n + 2)}{n + 1} \re \left(
a_{n}b_{n+1}^{*} + b_{n}a_{n+1}^{*} \right) \right],
\end{gather}
with subscripts ``MS'' to denote Mie scattering.
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\end{document}