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Scattering from a Small Conducting Sphere























Perfect conductors are not just dielectrics where the electric field is completely zero inside. The electric field is exactly cancelled on the interior by the induced surface charge. As we have seen, this cancellation occurs close to the surface (within a few times the skin depth). However, the induced currents also tend to expel the time dependent magnetic field. We therefore have two modification of our results from the previous section. The electric polarization will have a different form, and there will be a contribution from the induced magnetic moment of the sphere as well.

Recall (from J2.5) that the induced dipole moment on a conducting sphere is

\begin{displaymath}
\mbox{\boldmath$p$}= 4\pi\epsilon_0 a^3 \mbox{\boldmath$E$}_{\rm inc} .
\end{displaymath} (14.43)

This is indeed the generalization of the result for p last time, as you should be able to derive in a few minutes of work. Either review that section or solve the boundary value problem where $\mbox{\boldmath$E$}_\perp$ is discontinuous at the surface and $\mbox{\boldmath$E$}_{\vert\vert} = 0$ on the surface to obtain:
\begin{displaymath}
\phi = -E_0 \left(r - \frac{a^3}{r^2} \right) \cos \theta
\end{displaymath} (14.44)

from which we can easily extract this $\mbox{\boldmath$p$}$.

But, the magnetic field is also varying, and it induces an EMF that runs in loops around the magnetic field lines and opposes the change in magnetic flux. Assuming that no field lines were trapped in the sphere initially, the induced currents act to cancel component of the magnetic field normal to the surface. The sphere thus behaves like a magnetically permeable sphere (see e.g. section J5.10 and J5.11, equations J5.106, J5.107, J5.115):

\begin{displaymath}
\mbox{\boldmath$M$}= \frac{\mbox{\boldmath$m$}}{4\pi a^3/3} ...
...mu -
\mu_0}{\mu + 2\mu_0}\right)\mbox{\boldmath$H$}_{\rm inc}
\end{displaymath} (14.45)

with $\mu_r = \mu/\mu_0 = 0$ so that:
\begin{displaymath}
\mbox{\boldmath$m$}= - 2\pi a^3 \mbox{\boldmath$H$}_{\rm inc}.
\end{displaymath} (14.46)

The derivation is again very similar to the derivation we performed last time, with suitably chosen boundary conditions on $\mbox{\boldmath$B$}$ and $\mbox{\boldmath$H$}$.

If we then repeat the reasoning and algebra for this case of the conducting sphere (substituting this $\mbox{\boldmath$p$}$ and $\mbox{\boldmath$m$}$ into the expression we derived for the differential cross-section), we get:

\begin{displaymath}
\frac{d\sigma}{d\Omega} = k^4 a^6 \left\vert \hat{\mbox{\bo...
..._0 \times \hat{\mbox{\boldmath$\epsilon$}}_0) \right\vert^2 .
\end{displaymath} (14.47)

After much tedious but straightforward work, we can show (or rather you can show for homework) that:

$\displaystyle \frac{d \sigma_\parallel}{d \Omega}$ $\textstyle =$ $\displaystyle \frac{k^4a^6}{2} \left\vert \cos \theta -
\frac{1}{2} \right\vert^2$ (14.48)
$\displaystyle \frac{d \sigma_\perp}{d \Omega}$ $\textstyle =$ $\displaystyle \frac{k^4a^6}{2} \left\vert 1 - \frac{1}{2 }\cos
\theta \right\vert^2$ (14.49)

so that the total differential cross section is:
\begin{displaymath}
\frac{d\sigma}{d\Omega} = k^4 a^6 \left\{\frac{5}{8} (1+\cos^2 \theta) - \cos
\theta )\right\}
\end{displaymath} (14.50)

and the polarization is:
\begin{displaymath}
\Pi(\theta) = \frac{3 \sin^2 \theta}{5(1 + \cos^2 \theta) - 8 \cos \theta}
\end{displaymath} (14.51)

Finally, integrating the differential cross section yields the total cross-section:
\begin{displaymath}
\sigma = \frac{10 \pi k^4 a^6}{3} = (4\pi a^2)(ka)^4 \frac{2.5}{3}
\sim \sigma_{\rm dielectric}
\end{displaymath} (14.52)

for $\epsilon_r >> 1$ curiously enough.

What do these equations tell us? The cross-section is strongly peaked backwards. Waves are reflected backwards more than forwards (the sphere actually casts a ``shadow''. The scattered radiation is polarized qualitatively alike the radiation scattered from the dielectric sphere, but with a somewhat different angular distribution. It is completely polarized perpendicular to the scattering plane when scattered through an angle of 60$^\circ$, not 90$^\circ$.

Figure 14.1: Differential cross-section and polarization of a small conducting sphere.

We see that dipole scattering will always have a characteristic $k^4$ dependence. By know you should readily understand me when I say that this is the result of performing a multipolar expansion of the reaction field (essentially an expansion in powers of $kd$ where $d$ is the characteristic maximum extent of the system) and keeping the first (dipole) term.

If one wishes to consider scattering from objects where $kd \sim 1$ or greater, one simply has to consider higher order multipoles (and one must consider the proper multipoles instead of simple expansions in powers of $kd$). If $kd >> 1$ (which is the case for light scattering from macroscopic objects, radar scattering from airplanes and incoming nuclear missiles, etc) then a whole different apparatus must be brought to bear. I could spend a semester (or a least a couple of weeks) just lecturing on the scattering of electromagnetic waves from spheres, let alone other shapes.

However, no useful purpose would be so served, so I won't. If you ever need to figure it out, you have the tools and can find and understand the necessary references.


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