next up previous contents
Next: Scattering from a Small Up: Optical Scattering Previous: Optical Scattering   Contents

Radiation Reaction of a Polarizable Medium

Usually, when we consider optical scattering, we imagine that we have a monochromatic plane wave incident upon a polarizable medium embedded in (for the sake of argument) free space. The target we imagine is a ``particle'' of some shape and hence is mathematically a (simply) connected domain with compact support. The picture we must describe is thus




















The incident wave (in the absence of the target) is thus a pure plane wave:

$\displaystyle \mbox{\boldmath$E$}_{\rm inc}$ $\textstyle =$ $\displaystyle \hat{\mbox{\boldmath$\epsilon$}}_0 E_0 e^{ik\hat{\mbox{\boldmath$n$}}_0 \cdot {\bf r}}$ (14.1)
$\displaystyle \mbox{\boldmath$H$}_{\rm inc}$ $\textstyle =$ $\displaystyle \hat{\mbox{\boldmath$n$}}_0 \times \mbox{\boldmath$E$}_{\rm inc}/Z_0 .$ (14.2)

The incident wave induces a time dependent polarization density into the medium. If we imagine (not unreasonably) that the target is a particle or atom much smaller than a wavelength, then we can describe the field radiated from its induced dipole moment in the far zone and dipole approximation (see e.g. 4.122):
$\displaystyle \mbox{\boldmath$E$}_{\rm sc}$ $\textstyle =$ $\displaystyle \frac{1}{4\pi\epsilon_0} k^2 \frac{e^{ikr}}{r}
\left\{ (\hat{\mb...
...{\boldmath$n$}}- \hat{\mbox{\boldmath$n$}}\times \mbox{\boldmath$m$}/c \right\}$ (14.3)
$\displaystyle \mbox{\boldmath$H$}_{\rm sc}$ $\textstyle =$ $\displaystyle \hat{\mbox{\boldmath$n$}}\times \mbox{\boldmath$E$}_{\rm sc}/Z_0.$ (14.4)

In these expressions, $\hat{\mbox{\boldmath$n$}}_0 = \frac{\mbox{\boldmath$k$}_0}{k_0}$ and $\hat{\mbox{\boldmath$n$}}=
\frac{\mbox{\boldmath$k$}}{k}$, while $\hat{\mbox{\boldmath$\epsilon$}}_0, \hat{\mbox{\boldmath$\epsilon$}}$ are the polarization of the incident and scattered waves, respectively.

We are interested in the relative power distribution in the scattered field (which should be proportional to the incident field in a way that can be made independent of its magnitude in a linear response/susceptibility approximation). The power radiated in direction $\hat{\mbox{\boldmath$n$}}$ with polarization $\hat{\mbox{\boldmath$\epsilon$}}$ is needed per unit intensity in the incident wave with $\hat{\mbox{\boldmath$n$}}_0,
\hat{\mbox{\boldmath$\epsilon$}}_0$. This quantity is expressed as

\begin{displaymath}
\frac{d\sigma}{d\Omega}(\hat{\mbox{\boldmath$n$}},\hat{\mbo...
...}}_0^\ast \cdot \mbox{\boldmath$E$}_{\rm inc} \right\vert^2}
\end{displaymath} (14.5)

[One gets this by considering the power distribution:

$\displaystyle \frac{dP}{d\Omega}$ $\textstyle =$ $\displaystyle \frac{1}{2} {\rm Re} \left\{ r^2 \hat{\mbox{\boldmath$n$}}\cdot
\left( \mbox{\boldmath$E$}\times \mbox{\boldmath$H$}^\ast \right) \right \}$  
  $\textstyle =$ $\displaystyle \frac{1}{2Z_0} \left\vert \mbox{\boldmath$E$}\times (\hat{\mbox{\boldmath$n$}}\times \mbox{\boldmath$E$}) \right\vert$  
  $\textstyle =$ $\displaystyle \frac{1}{2Z_0} \left\vert \mbox{\boldmath$E$} \right\vert^2$ (14.6)

as usual, where the latter relation steps hold for transverse EM fields 7.1 and 7.2 only and where we've projected out a single polarization from the incident and scattered waves so we can discuss polarization later.]

This quantity has the units of area ($r^2$) and is called the differential cross-section:

\begin{displaymath}
\frac{d\sigma}{d\Omega} = \frac{dP/d\Omega}{dP_0/dA} \propto
\frac{dA}{d\Omega} \sim A.
\end{displaymath} (14.7)

In quantum theory a scattering cross-section one would substitute ``intensity'' (number of particles/second) for ``power'' in this definition but it still holds. Since the units of angles, solid or not, are dimensionless, a cross-section always has the units of area. If one integrates the cross-section around the $4\pi$ solid angle, the resulting area is the ``effective'' cross-sectional area of the scatterer, that is, the integrated are of its effective ``shadow''. This is the basis of the optical theorem, which I will mention but we will not study (derive) for lack of time.

The point in defining it is that it is generally a property of the scattering target that linearly determines the scattered power:

\begin{displaymath}
\frac{dP}{d\Omega} = \frac{d\sigma}{d\Omega} \times I_0
\end{displaymath} (14.8)

where the last quantity is the intensity of the incident plane wave beam. The cross-section is independent (within reason) of the incident intensity and can be calculated or measured ``once and for all'' and then used to predict the power distribution for a given beam intensity.

We need to use the apparatus of chapter 7 to handle the vector polarization correctly. That is, technically we need to use the Stokes parameters or something similar to help us project out of E a particular polarization component. Then (as can easily be shown by meditating on:

\begin{displaymath}
\hat{\mbox{\boldmath$\epsilon$}}^\ast \cdot \mbox{\boldmath...
...{\boldmath$n$}}\times \mbox{\boldmath$m$}/c \right\} \right\}
\end{displaymath} (14.9)

for a transverse field):
\begin{displaymath}
\frac{d\sigma}{d\Omega} = r^2 \frac{\frac{1}{2Z_0}
\left\v...
...psilon$}}^\ast)
\times \mbox{\boldmath$m$}/c \right\vert^2 .
\end{displaymath} (14.10)

To get this result, we had to evaluate (using vector identities)
\begin{displaymath}
\hat{\mbox{\boldmath$\epsilon$}}^\ast \cdot (\hat{\mbox{\bo...
...at{\mbox{\boldmath$\epsilon$}}^\ast \cdot \mbox{\boldmath$p$}
\end{displaymath} (14.11)

and
\begin{displaymath}
\hat{\mbox{\boldmath$\epsilon$}}^\ast \cdot (\hat{\mbox{\bo...
...boldmath$n$}}\times \hat{\mbox{\boldmath$\epsilon$}}^\ast) .
\end{displaymath} (14.12)

From this we immediately see one important result:

\begin{displaymath}
\frac{d \sigma}{d \Omega} \propto k^4 \propto \frac{1}{\lambda^4}.
\end{displaymath} (14.13)

This is called Rayleigh's Law; the scattering cross-section (and hence proportion of the power scattered from a given incident beam) by a polarizable medium is proportional to the inverse fourth power of the wavelength. Or, if you prefer, short wavelengths (still long with respect to the size of the scatterer and only if the dipole term in the scattering dominates) are scattered more strongly than long wavelengths. This is the original ``blue sky'' theory and probably the origin of the phrase!

To go further in our understanding, and to gain some useful practice against the day you have to use this theory or teach it to someone who might use it, we must consider some specific cases.


next up previous contents
Next: Scattering from a Small Up: Optical Scattering Previous: Optical Scattering   Contents
Robert G. Brown 2014-08-19