# Dipole Radiation and Rayleigh Scattering ## Hertzian Dipole A short current element $I\,l\,\hat z$ (equivalently a dipole moment with $Il=j\omega p$) at the origin. **Far field** ($kr\gg1$), with $e^{+j\omega t}$: $ E_\theta=\frac{j\eta k\,(Il)\,\sin\theta}{4\pi r}\,e^{-jkr}=-\frac{k^2p\,\sin\theta}{4\pi\epsilon_0 r}\,e^{-jkr},\qquad H_\phi=\frac{E_\theta}{\eta} $ - The field locally looks like a plane wave moving radially: $\vec E\perp\vec H\perp\hat r$. - The pattern goes as $\sin^2\theta$: no radiation along the dipole axis. **Radiated power** (peak phasors): $ \langle S_r\rangle=\frac{|E_\theta|^2}{2\eta},\qquad P=\frac{\eta k^2|Il|^2}{12\pi}=\frac{\eta_0k^2\omega^2|p|^2}{12\pi} $ using $ \oint\sin^2\theta\,d\Omega=\frac{8\pi}{3} $ Near field ($kr\ll1$): the electrostatic dipole field. A small scatterer's boundary problem is therefore solved **quasistatically**. ## Rayleigh Scattering from a Small Sphere ($ka\ll1$) A sphere of radius $a$ and permittivity $\epsilon=\epsilon_r\epsilon_0$ in free space, with incident field $E_0\hat z$ propagating along $\hat y$. 1. Since $a\ll\lambda$, the incident field is uniform over the sphere. Solve the electrostatic problem: a uniform field inside, and uniform plus dipole outside. 2. Boundary conditions at $r=a$: tangential $E_\theta$ continuous and normal $D_r$ continuous. Use $\hat z=\hat r\cos\theta-\hat\theta\sin\theta$. 3. Result: $ \vec E_\text{in}=\frac{3}{\epsilon_r+2}\,E_0\hat z,\qquad \vec p=4\pi\epsilon_0a^3\,\frac{\epsilon_r-1}{\epsilon_r+2}\,E_0\hat z $ 4. This induced dipole oscillates at $\omega$ and radiates like the Hertzian dipole above. That radiation is the scattered field: $ E_{\theta,\text{scat}}=-\frac{k^2a^3\sin\theta}{r}\,\frac{\epsilon_r-1}{\epsilon_r+2}\,E_0\,e^{-jkr} $ **Scattering cross section** = scattered power ÷ incident intensity $\dfrac{|E_0|^2}{2\eta_0}$: $ \sigma_s=\frac{8\pi}{3}\,k^4a^6\left|\frac{\epsilon_r-1}{\epsilon_r+2}\right|^2 $ - $\sigma_s\propto\lambda^{-4}$: blue scatters about 5× more than red. That's why the sky is blue (and not black). - Compare with the geometric area: $ \frac{\sigma_s}{\pi a^2}=\frac{8}{3}(ka)^4\left|\frac{\epsilon_r-1}{\epsilon_r+2}\right|^2\ll1\quad\text{when } ka\ll1 $ The pset 2 P2(b) wording suggests $\sigma_s>\pi a^2$, but in the Rayleigh limit that only happens near $\epsilon_r\approx-2$ (the plasmon resonance of a metal particle, where the denominator vanishes). For the sky, each molecule scatters weakly, but there are an enormous number of them. - The light scattered at 90° to the sun is **linearly polarized**, because the dipole can't radiate along its own axis. - A free electron (no restoring force) gives the Thomson cross section instead, which does not depend on $\lambda$.