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Scattering from a Small Dielectric Sphere

This is a relatively simple, and hence very standard problem.























Now, we have no desire to ``reinvent the sphere''14.1 but it is important that you understand where our results come from. First of all, let us introduce dimensionless, scaled versions of the relative permeability and permittivity (a step that Jackson apparently performs in J10 but does not document or explain):

$\displaystyle \epsilon_r$ $\textstyle =$ $\displaystyle \epsilon(\omega)/\epsilon_0$ (14.14)
$\displaystyle \mu_r$ $\textstyle =$ $\displaystyle \mu(\omega)/\mu_0 \approx 1$ (14.15)

where we assume that we are not at a resonance so that the spheres have normal dispersion and that these numbers are basically real. The latter is a good approximation for non-magnetic, non-conducting scatterers e.g. oxygen or nitrogen molecules.

If you refer back to J4.4, equation J4.56 and the surrounding text, you will see that the induced dipole moment in a dielectric sphere in terms of the relative permittivity is:

\begin{displaymath}
\mbox{\boldmath$p$}= 4\pi\epsilon_0 \left( \frac{ \epsilon_...
...1}{\epsilon_r + 2} \right) a^3 \mbox{\boldmath$E$}_{\rm inc}
\end{displaymath} (14.16)

To recapitulate the derivation (useful since this is a common question on qualifiers and the like) we note that the sphere has azimuthal symmetry around the direction of $\mbox{\boldmath$E$}$, so we can express the scalar potential inside and outside the sphere as

$\displaystyle \phi_{\rm in}$ $\textstyle =$ $\displaystyle \sum_\ell A_\ell r^\ell P_\ell(\cos \theta)$ (14.17)
$\displaystyle \phi_{\rm out}$ $\textstyle =$ $\displaystyle \sum_\ell \left\{ B_\ell r^\ell + C_\ell
\frac{1}{r^{\ell + 1}} \right\} P_\ell(\cos \theta) .$ (14.18)

We need to evaluate this. At infinity we know that the field should be (to lowest order) undisturbed, so the potential must asymptotically go over to

\begin{displaymath}
\lim_{ r \to \infty } \phi_{\rm out} = -E_0 z = -E_0 r \cos \theta = -E_0
r P_1(\cos \theta)
\end{displaymath} (14.19)

so we conclude that $B_1 = -E_0$ and all other $B_{\ell > 1} = 0$. To proceed further, we must use the matching conditions of the tangential and normal fields at the surface of the sphere:
\begin{displaymath}
- \frac{1}{a} \left. \frac{\partial \phi_{\rm in}}{\partial...
...{\partial \phi_{\rm out}}{\partial \theta} \right\vert _{r=a}
\end{displaymath} (14.20)

(tangential component) and
\begin{displaymath}
- \epsilon \left. \frac{\partial \phi_{\rm in}}{\partial r}...
...frac{\partial \phi_{\rm out}}{\partial r} \right\vert _{r=a}
\end{displaymath} (14.21)

(normal $\mbox{\boldmath$D$}$ onto $\mbox{\boldmath$E$}$).

Since this is the surface of a sphere (!) we can project out each spherical component if we wish and cause these equations to be satisfied term by term. From the first (tangential) equation we just match $\phi$ itself:

\begin{displaymath}
\frac{1}{a} (A_\ell a^\ell) = \frac{1}{a} \left( B_\ell a^\ell + C_\ell
\frac{1}{a^{\ell+1}} \right)
\end{displaymath} (14.22)

or (using our knowledge of $B_\ell$)
$\displaystyle A_1$ $\textstyle =$ $\displaystyle -E_0 + \frac{C_1}{a^3} \mbox{\hspace{1in}} \ell = 1$ (14.23)
$\displaystyle A_\ell$ $\textstyle =$ $\displaystyle \frac{C_\ell}{a^{2\ell+1}} \mbox{\hspace{1.35in} \rm else}$ (14.24)

From the second (normal) equation we get

$\displaystyle \epsilon_r A_1$ $\textstyle =$ $\displaystyle -E_0 - 2\frac{C_1}{a^3} \mbox{\hspace{1in}} \ell = 1$ (14.25)
$\displaystyle \epsilon_r A_\ell$ $\textstyle =$ $\displaystyle - \frac{(\ell+1) C_\ell}{a^{2\ell+1}}
\mbox{\hspace{1in} \rm else}.$ (14.26)

The second equation of each pair are incompatible and have only the trivial

\begin{displaymath}
A_\ell = C_\ell = 0 \hspace{1in} \ell \ne 1 .
\end{displaymath} (14.27)

Only the $\ell = 1$ term survives. With a little work one can show that
$\displaystyle A_1$ $\textstyle =$ $\displaystyle -\frac{3 E_0}{2 + \epsilon_r}$ (14.28)
$\displaystyle C_1$ $\textstyle =$ $\displaystyle \left(\frac{\epsilon_r - 1}{\epsilon_r+2} \right) a^3 E_0$ (14.29)

so that
$\displaystyle \phi_{\rm in}$ $\textstyle =$ $\displaystyle - \left ( \frac{3}{\epsilon_r+2} \right) E_0 r \cos
\theta$ (14.30)
$\displaystyle \phi_{\rm out}$ $\textstyle =$ $\displaystyle - E_0 r \cos \theta + \left( \frac{\epsilon_r -
1}{\epsilon_r + 2} \right) E_0 \frac{a^3}{r^2} \cos \theta .$ (14.31)

When we identify the second term of the external field with the dipole potential and compare with the expansion of the dipole potential

\begin{displaymath}
\phi(\mbox{\boldmath$r$}) = \frac{1}{4\pi\epsilon_0} \frac{\mbox{\boldmath$p$}\cdot\mbox{\boldmath$r$}}{r^3}
\end{displaymath} (14.32)

we conclude that the induced dipole moment is:
\begin{displaymath}
\mbox{\boldmath$p$}= 4\pi\epsilon_0 \left( \frac{\epsilon_r...
...{\epsilon_r + 2} \right)
a^3 E_0 \hat{\mbox{\boldmath$z$}}.
\end{displaymath} (14.33)

as given above.

There is no magnetic dipole moment, because $\mu_r = 1$ and therefore the sphere behaves like a ``dipole antenna''. Thus $\mbox{\boldmath$m$}= 0$ and there is no magnetic scattering of radiation from this system. This one equation, therefore, (together with our original definitions of the fields) is sufficient to determine the differential cross-section:

\begin{displaymath}
\frac{d \sigma}{d \Omega} = k^4 a^6 \left\vert \frac{\epsil...
...}^\ast \cdot \hat{\mbox{\boldmath$\epsilon$}}_0 \right\vert^2
\end{displaymath} (14.34)

where remember that $\epsilon_r(\omega)$ (for dispersion) and hopefully everybody notes the difference between dielectric $\epsilon $ and polarization $\hat{\mbox{\boldmath$\epsilon$}}$ (sigh - we need more symbols). This equation can be used to find the explicit differential cross-sections given $(\hat{\mbox{\boldmath$n$}},\hat{\mbox{\boldmath$n$}}_0,\hat{\mbox{\boldmath$\epsilon$}},\hat{\mbox{\boldmath$\epsilon$}}_0)$, as desired.

However, the light incident on the sphere will generally be unpolarized. Then the question naturally arises of whether the various independent polarizations of the incident light beam will be scattered identically. Or, to put it another way, what is the angular distribution function of radiation with a definite polarization? To answer this, we need to consider a suitable decomposition of the possible polarization directions.

This decomposition is apparent from considering the following picture of the general geometry:














Let $\hat{\mbox{\boldmath$n$}}, \hat{\mbox{\boldmath$n$}}_0$ define the plane of scattering. We have to fix $\hat{\mbox{\boldmath$\epsilon$}}^{(1)}$ and $\hat{\mbox{\boldmath$\epsilon$}}^{(2)}$ relative to this scattering plane and average over the polarizations in the incident light, $\hat{\mbox{\boldmath$\epsilon$}}^{(1)}_0$ and $\hat{\mbox{\boldmath$\epsilon$}}^{(2)}_0$ (also fixed relative to this plane). We can always choose the directions of polarization such that $\hat{\mbox{\boldmath$\epsilon$}}^{(2)} = \hat{\mbox{\boldmath$\epsilon$}}_0^{(2)}$ is perpendicular to the scattering plane and $\hat{\mbox{\boldmath$\epsilon$}}^{(1)} = \hat{\mbox{\boldmath$\epsilon$}}_0^{(1)}$ are in it, and perpendicular to the directions $\hat{\mbox{\boldmath$n$}}$ and $\hat{\mbox{\boldmath$n$}}_0$ respectively. The dot products are thus

$\displaystyle \hat{\mbox{\boldmath$\epsilon$}}^{(1) \ast} \cdot \hat{\mbox{\boldmath$\epsilon$}}_0^{(1)}$ $\textstyle =$ $\displaystyle \hat{\mbox{\boldmath$n$}}\cdot \hat{\mbox{\boldmath$n$}}_0 = \cos \theta$ (14.35)
$\displaystyle \hat{\mbox{\boldmath$\epsilon$}}^{(2) \ast} \cdot \hat{\mbox{\boldmath$\epsilon$}}_0^{(2)}$ $\textstyle =$ $\displaystyle 1 .$ (14.36)

We need the average of the squares of these quantities. This is essentially averaging $\sin^2 \phi$ and $\cos^2 \phi$ over $\phi \in (0,2\pi)$. Alternatively, we can meditate upon symmetry and conclude that the average is just $\frac{1}{2}$. Thus (for the polarization in the plane ($\parallel$) and perpendicular to the plane ($\perp$) of scattering, respectively) we have:

$\displaystyle \frac{d \sigma_\parallel}{d \Omega}$ $\textstyle =$ $\displaystyle k^4 a^6
\left\vert \frac{\epsilon_r - 1}{\epsilon_r + 2} \right\vert^2 \frac{\cos^2 \theta}{2}$ (14.37)
$\displaystyle \frac{d \sigma_\perp}{d \Omega}$ $\textstyle =$ $\displaystyle k^4 a^6 \left\vert \frac{\epsilon_r -
1} {\epsilon_r + 2} \right\vert^2 \frac{1}{2}$ (14.38)

We see that light polarized perpendicular to the plane of scattering has no $\theta$ dependence, while light polarized in that plane is not scattered parallel to the direction of propagation at all (along $\theta = 0$ or $\pi$). We will invert this statement in a moment so that it makes more sense. See the diagram below.

Unfortunately, everything thus far is expressed with respect to the plane of scattering, which varies with the direction of the scattered light. If we define the polarization $\Pi(\theta)$ of the scattered radiation to be

\begin{displaymath}
\Pi(\theta) = \frac{\frac{d\sigma_\perp}{d\Omega} -
\frac{...
...parallel}{d\Omega}} = \frac{\sin^2 \theta}{1 + \cos^2 \theta}
\end{displaymath} (14.39)

then we obtain a quantity that is in accord with our intuition. $\Pi(\theta)$ is maximum at $\theta = \pi/2$. The radiation scattered through an angle of 90 degrees is completely polarized in a plane perpendicular to the plane of scattering.

















Finally, we can add the two pieces of the differential cross-section together:

\begin{displaymath}
\frac{d\sigma}{d\Omega} = k^4a^6 \left( \frac{\epsilon - 1}{\epsilon + 2}
\right)^2 \frac{1}{2}(1 + \cos^2 \theta)
\end{displaymath} (14.40)

which is strongly and symmetrically peaked forward and backward. Finally, this is easy to integrate to obtain the total cross-section:
\begin{displaymath}
\sigma = \frac{8\pi}{3} k^4 a^6 \left( \frac{\epsilon_r - 1}{\epsilon_r
+ 2} \right)^2 .
\end{displaymath} (14.41)

At last, we can put it all together. Molecules in the atmosphere behave, far from resonance, like itty-bitty dielectric spheres to a remarkable approximation. Since blue light is scattered more strongly than red, light seen away from its direction of incidence (the sky and not the sun) is shifted in color from white to blue. When Mr. Sun is examined directly through a thick layer of atmosphere (at sunset) the blue is all scattered out and the remaining light looks red. Finally, light from directly overhead at sunup or sundown is polarized in a north-south direction; at noon the light from the horizon is polarized parallel to the horizon (and hence is filtered by vertical transmission axis polarized sunglasses. You should verify this at your next opportunity outdoors with a pair of polarized sunglasses, as this whole discussion is taught in elementary terms in second semester introductory physics courses.

Don't say I never taught you anything14.2.

The last remarks I would make concern the total cross-section. Note that if we factor out a $4 \pi a^2$ we get the ``area'' of the sphere times a pure (dimensionless) number $(ka)^4$ associated with the relative size of the sphere radius and the wavelength and a second pure number involving only the dielectric properties of the medium:

\begin{displaymath}
\sigma = (4\pi a^2) (ka)^4 \left\{\frac{2}{3} \left( \frac{\epsilon_r -
1}{\epsilon_r + 2} \right)^2 \right\}.
\end{displaymath} (14.42)

This expression isn't any more useful than the one above, but it does make the role of the different terms that contribute to the total scattering cross-section more clear.


next up previous contents
Next: Scattering from a Small Up: Optical Scattering Previous: Radiation Reaction of a   Contents
Robert G. Brown 2013-01-04