Scattering & Diffraction

Step 1: Equation of motion for a free electron

An incident wave $\mathbf{E} = E_0\hat{\epsilon}\,e^{-i\omega t}$ drives the electron: $m_e\ddot{\mathbf{x}} = -eE_0\hat{\epsilon}\,e^{-i\omega t}$. The acceleration is $\mathbf{a} = -(eE_0/m_e)\hat{\epsilon}\,e^{-i\omega t}$.

Step 2: Apply the Larmor formula

The time-averaged radiated power is: $P = e^2|a|^2/(12\pi\epsilon_0 c^3) = e^4 E_0^2/(12\pi\epsilon_0 m_e^2 c^3)$.

Step 3: Define the cross section

$\sigma = P/\langle S_{\text{inc}}\rangle$ where $\langle S_{\text{inc}}\rangle = \epsilon_0 c E_0^2/2$ is the incident intensity.

Step 4: Compute

$$\sigma_T = \frac{e^4}{6\pi\epsilon_0^2 m_e^2 c^4} = \frac{8\pi}{3}\left(\frac{e^2}{4\pi\epsilon_0 m_e c^2}\right)^2 = \frac{8\pi}{3}r_e^2$$

Step 5: Differential cross section

For polarized incident light with $\hat{\epsilon}$ perpendicular to the scattering plane: $d\sigma/d\Omega = r_e^2$ (isotropic). For unpolarized light: $$\boxed{\frac{d\sigma}{d\Omega} = \frac{r_e^2}{2}(1+\cos^2\theta)}$$

Thomson scattering is the high-frequency limit ($\omega \gg \omega_0$) of the general scattering formula and is the dominant opacity mechanism in stellar interiors and the early universe.

Step 1: Driven oscillator model

$$m_e\ddot{x} + m_e\gamma\dot{x} + m_e\omega_0^2 x = -eE_0 e^{-i\omega t}$$

where $\omega_0$ is the natural frequency and $\gamma$ is the radiation damping rate.

Step 2: Steady-state solution

$$x_0 = \frac{-eE_0/m_e}{\omega_0^2 - \omega^2 - i\gamma\omega}$$

The acceleration amplitude is $|a_0| = \omega^2|x_0|$.

Step 3: General cross section

$$\sigma(\omega) = \sigma_T\frac{\omega^4}{(\omega^2-\omega_0^2)^2 + \gamma^2\omega^2}$$

Step 4: Low-frequency limit (Rayleigh)

For $\omega \ll \omega_0$: $\sigma \approx \sigma_T(\omega/\omega_0)^4$. Blue light (400 nm) scatters $(700/400)^4 \approx 9.4$ times more than red light (700 nm).

Step 5: High-frequency limit (Thomson)

For $\omega \gg \omega_0$: $\sigma \to \sigma_T$, recovering the free-electron result. Near resonance ($\omega \approx \omega_0$), the cross section peaks at $\sigma_{\max} \approx \sigma_T(\omega_0/\gamma)^2 \gg \sigma_T$, explaining resonance fluorescence.

Step 1: Kirchhoff's diffraction integral

The field at a point $\mathbf{r}$ behind a screen with aperture $\mathcal{A}$ is given by integrating over the aperture:

$$\mathbf{E}_1(\mathbf{r}) = \int_\mathcal{A} [\text{Kirchhoff integrand}]\,da'$$

Step 2: Complementary screen

The complement has aperture $\bar{\mathcal{A}}$ (where the original screen was opaque). Its diffracted field integrates over $\bar{\mathcal{A}}$:

$$\mathbf{E}_2(\mathbf{r}) = \int_{\bar{\mathcal{A}}} [\text{Kirchhoff integrand}]\,da'$$

Step 3: Sum of complementary fields

Since $\mathcal{A} \cup \bar{\mathcal{A}}$ covers the entire plane: $\mathbf{E}_1 + \mathbf{E}_2 = \int_{\text{entire plane}}[\ldots]\,da' = \mathbf{E}_0$.

Step 4: Consequence for intensities

Away from the forward direction ($\mathbf{E}_0 \approx 0$): $\mathbf{E}_1 \approx -\mathbf{E}_2$, so $|\mathbf{E}_1|^2 = |\mathbf{E}_2|^2$. The diffraction patterns are identical.

Step 5: Electromagnetic generalization

For vector fields, Babinet's principle requires swapping E and B between complementary screens: $\mathbf{E}_2 = c\mathbf{B}_1$ (with appropriate sign changes). This means the polarization of the diffracted field rotates by 90 degrees between a screen and its complement.

Step 1: Fraunhofer approximation

For $r \gg a^2/\lambda$, the phase across the aperture is $e^{-iky'\sin\theta}$ and the amplitude variation is negligible.

Step 2: Fourier transform integral

$$E(\theta) \propto \int_{-a/2}^{a/2} e^{-iky'\sin\theta}\,dy' = a\,\text{sinc}\!\left(\frac{ka\sin\theta}{2}\right)$$

Step 3: Intensity pattern

$$I(\theta) = I_0\left(\frac{\sin u}{u}\right)^2, \quad u = \frac{\pi a\sin\theta}{\lambda}$$

Step 4: Minima and subsidiary maxima

Zeros occur at $a\sin\theta = m\lambda$ ($m = \pm 1, \pm 2, \ldots$). The first subsidiary maximum has intensity 4.7% of the central maximum.

Step 5: Angular width and resolution

The central maximum has angular half-width $\Delta\theta \approx \lambda/a$. This sets the diffraction limit of optical instruments: $$\boxed{\theta_{\min} = 1.22\frac{\lambda}{D}}$$

(Rayleigh criterion for a circular aperture of diameter $D$).

Step 1: Total field

The total field is $\mathbf{E} = \mathbf{E}_i + \mathbf{E}_s$, where $\mathbf{E}_i = E_0 e^{ikz}\hat{x}$ and $\mathbf{E}_s = E_0 f(\theta,\phi)\frac{e^{ikr}}{r}\hat{\epsilon}_s$.

Step 2: Energy conservation

The power removed from the beam equals the scattered power plus absorbed power: $P_{\text{ext}} = P_{\text{scat}} + P_{\text{abs}}$. This is computed from the Poynting vector integrated over a large sphere.

Step 3: Interference term

The cross term $\mathbf{E}_i \times \mathbf{B}_s^* + \mathbf{E}_s \times \mathbf{B}_i^*$ gives the extinction. Using the stationary phase approximation on the large sphere, only the forward direction contributes.

Step 4: Extract the forward amplitude

The extinction cross section is: $\sigma_{\text{ext}} = (4\pi/k)\,\text{Im}[f(0)]$.

Step 5: Interpretation

$$\boxed{\sigma_{\text{ext}} = \frac{4\pi}{k}\,\text{Im}[f(\theta=0)]}$$

This remarkable result connects the total cross section (an integral over all angles) to the forward scattering amplitude alone. It is a consequence of unitarity (energy conservation) and holds in quantum mechanics as well.