Thomson Scattering

These notes use the conventions fixed earlier in the course, and build directly on the covariant radiation field derived in Radiation from a Moving Charge and the energy flux introduced in Energy of the Electromagnetic Field. The goal is to go from the covariant field of an accelerated charge, through its non-relativistic (local rest-frame) limit, to the classical differential and total cross sections for scattering of light by a free charge.

1 Setup

A free charge \(q\) of mass \(m\) (an electron, in practice) sits at rest at the origin and is illuminated by a monochromatic plane wave, \[ \mathbf E_{\rm inc}(t)=E_0\,\hat{\boldsymbol\epsilon}\,\cos(\omega t), \qquad \mathbf B_{\rm inc}(t)=\frac1c\,\hat{\mathbf k}\times\mathbf E_{\rm inc}(t), \] with \(\hat{\boldsymbol\epsilon}\perp\hat{\mathbf k}\), evaluated at the charge’s location. Two approximations define the classical (Thomson) regime:

• non-relativistic motion: the charge’s response velocity \(v\ll c\), so terms of order \(v/c\) are dropped;
• no radiation reaction: the energy radiated per cycle is a small fraction of the charge’s kinetic energy, so the incident field alone drives the motion.

2 Equation of motion

The Lorentz force on the charge is \(q(\mathbf E_{\rm inc}+\mathbf v\times\mathbf B_{\rm inc})\). Since \(|\mathbf B_{\rm inc}|=|\mathbf E_{\rm inc}|/c\), the magnetic term is smaller than the electric term by a factor \(v/c\ll1\) and is dropped, leaving \[ m\dot{\mathbf v}=q\mathbf E_{\rm inc}(t), \qquad \dot{\mathbf v}(t)=\frac{q}{m}E_0\,\hat{\boldsymbol\epsilon}\,\cos(\omega t). \] The charge oscillates along the fixed direction \(\hat{\boldsymbol\epsilon}\) with acceleration amplitude \(a_0\equiv qE_0/m\); since \(a_0/\omega\ll c\) in the non-relativistic regime, the charge’s excursion is tiny compared to the wavelength, and it radiates essentially from a fixed point (the dipole approximation).

3 From the covariant field to the local frame

The radiation field of an accelerated charge, valid for any velocity, was written covariantly as \[ F_{\mu\nu}^{\rm rad} = \frac{\mu_0q}{4\pi(u\cdot\Delta x)^2} \left[ \Delta x_\mu a_\nu-\Delta x_\nu a_\mu - \frac{\Delta x\cdot a}{u\cdot\Delta x} \big(\Delta x_\mu u_\nu-\Delta x_\nu u_\mu\big) \right]_{\tau=\tau_r}, \] with \(a^\mu=\mathrm{d}u^\mu/\mathrm{d}\tau\). Here this is specialized to a charge that is instantaneously at rest, as is the case for the non-relativistic oscillation driven by the incident wave.

At the instant the charge is at rest, \(u^\mu=(1,\mathbf 0)\) and, since \(u_\mu a^\mu=0\) forces \(a^0=0\) there, \(a^\mu=(0,\dot{\mathbf v}/c)\) exactly (this is the same identification used for the Larmor formula). With \(\Delta x^\mu=(R,R\hat{\mathbf n})\), \(R=|\mathbf x|\), \[ u\cdot\Delta x=-R, \qquad \Delta x\cdot a=R\,\frac{\hat{\mathbf n}\cdot\dot{\mathbf v}}{c}. \] Substituting into \(F_{0i}|_{\rm rad}=-E_i/c\) and simplifying (the two \(1/R\)-carrying terms combine; the \(\Delta x_\mu u_\nu\) term contributes \(R\hat n_i\) using \(u_i=0,u_0=-1\)) gives \[ \boxed{ \mathbf E_{\rm rad}(\mathbf x,t) = \frac{\mu_0q}{4\pi R} \Big[\dot{\mathbf v}-(\hat{\mathbf n}\cdot\dot{\mathbf v})\hat{\mathbf n}\Big]_{t_r} = -\frac{\mu_0q}{4\pi R}\,\hat{\mathbf n}\times\big(\hat{\mathbf n}\times\dot{\mathbf v}\big)\Big|_{t_r}, } \] the familiar non-relativistic dipole radiation field, transverse to \(\hat{\mathbf n}\) by construction. Writing \(\chi\) for the angle between \(\hat{\mathbf n}\) and \(\dot{\mathbf v}\), its magnitude is \[ |\mathbf E_{\rm rad}|=\frac{\mu_0q\,|\dot{\mathbf v}|}{4\pi R}\sin\chi. \]

4 Radiated power and a consistency check

In the radiation zone, \(u=\varepsilon_0E^2\) and \(\mathbf S=cu\,\hat{\mathbf n}\) (from the Radiation zone section), so the power radiated per solid angle is \[ \frac{\mathrm{d}P}{\mathrm{d}\Omega}=R^2\,\mathbf S\cdot\hat{\mathbf n}=R^2c\varepsilon_0|\mathbf E_{\rm rad}|^2 = \frac{\mu_0q^2\dot v^2}{16\pi^2c}\sin^2\chi, \] using \(1/(4\pi\varepsilon_0)=\mu_0c^2/(4\pi)\). As a check, integrating over solid angle with \(\int\sin^2\chi\,\mathrm{d}\Omega=8\pi/3\) recovers exactly the non-relativistic Larmor formula already derived, \[ P=\int\frac{\mathrm{d}P}{\mathrm{d}\Omega}\,\mathrm{d}\Omega=\frac{\mu_0q^2\dot v^2}{16\pi^2c}\cdot\frac{8\pi}{3}=\frac{\mu_0q^2\dot v^2}{6\pi c}, \] matching the boxed result \(P=\mu_0q^2\dot v^2/(6\pi c)\) obtained independently from the covariant radiation tensor.

5 Differential cross section

The incident wave carries energy density \(u_{\rm inc}=\varepsilon_0E_{\rm inc}^2\) and flux \(S_{\rm inc}=cu_{\rm inc}\) (same radiation-zone relations, since a plane wave is locally the same kind of transverse field); its time average over a cycle is \[ \langle S_{\rm inc}\rangle=c\varepsilon_0\langle E_{\rm inc}^2\rangle=\tfrac12c\varepsilon_0E_0^2. \] The differential cross section is defined as scattered power per solid angle, divided by incident flux: \[ \frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} \equiv \frac{\langle \mathrm{d}P/\mathrm{d}\Omega\rangle}{\langle S_{\rm inc}\rangle}. \] Using \(\dot{\mathbf v}=(q/m)\mathbf E_{\rm inc}\) from the equation of motion, \(\langle\dot v^2\rangle=(q/m)^2\langle E_{\rm inc}^2\rangle=(q/m)^2E_0^2/2\), so \[ \left\langle\frac{\mathrm{d}P}{\mathrm{d}\Omega}\right\rangle = \frac{\mu_0q^2}{16\pi^2c}\cdot\frac{q^2E_0^2}{2m^2}\sin^2\chi, \] and dividing by \(\langle S_{\rm inc}\rangle=c\varepsilon_0E_0^2/2\), with \(1/\varepsilon_0=\mu_0c^2\), \[ \frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \left(\frac{\mu_0q^2}{4\pi m}\right)^{\!2}\sin^2\chi. \] Defining the classical radius of the charge, \[ r_q\equiv\frac{\mu_0q^2}{4\pi m}=\frac{q^2}{4\pi\varepsilon_0mc^2}, \] this is \[ \boxed{ \frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}=r_q^2\sin^2\chi, } \] with \(\chi\) the angle between the observation direction \(\hat{\mathbf n}\) and the (fixed) oscillation direction \(\hat{\boldsymbol\epsilon}\): scattering is strongest broadside to the oscillation and vanishes along it, exactly like a driven dipole antenna.

6 Unpolarized incident light

For unpolarized incident light, average \(\sin^2\chi\) over the two independent transverse polarizations. Take \(\hat{\mathbf k}=\hat{\mathbf z}\) and, without loss of generality, the observation direction in the \(x\)-\(z\) plane at scattering angle \(\theta\) from \(\hat{\mathbf k}\), \(\hat{\mathbf n}=\sin\theta\,\hat{\mathbf x}+\cos\theta\,\hat{\mathbf z}\). For \(\hat{\boldsymbol\epsilon}=\hat{\mathbf x}\), \(\sin^2\chi=1-(\hat{\mathbf n}\cdot\hat{\mathbf x})^2=\cos^2\theta\); for \(\hat{\boldsymbol\epsilon}=\hat{\mathbf y}\), \(\hat{\mathbf n}\cdot\hat{\mathbf y}=0\) so \(\sin^2\chi=1\). Averaging the two, \[ \langle\sin^2\chi\rangle=\frac{\cos^2\theta+1}{2}, \] so that \[ \boxed{ \left.\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}\right|_{\rm unpol} = \frac{r_q^2}{2}\left(1+\cos^2\theta\right), } \] the classical Thomson formula, with \(\theta\) now the angle between the scattered and incident directions.

7 Total cross section

Integrating over solid angle, with \(x=\cos\theta\), \[ \sigma_T = \frac{r_q^2}{2}\int_0^{2\pi}\!\!\mathrm{d}\phi\int_{-1}^{1}(1+x^2)\,\mathrm{d}x = \pi r_q^2\left[x+\frac{x^3}{3}\right]_{-1}^{1} = \pi r_q^2\cdot\frac83, \] \[ \boxed{ \sigma_T=\frac{8\pi}{3}r_q^2. } \] For an electron (\(q=e\), \(m=m_e\)), \(r_q\) is the classical electron radius and \(\sigma_T\approx6.65\times10^{-29}\,\mathrm m^2\) is the Thomson cross section.

8 Validity

This is a purely classical, elastic calculation: the scattered light has the same frequency \(\omega\) as the incident light, with no recoil. It holds provided:

• \(v\ll c\) throughout the motion (already used to drop the magnetic force and to evaluate \(a^\mu\) at \(u^\mu=(1,\mathbf 0)\));
• the photon energy \(\hbar\omega\) is much smaller than \(mc^2\), so that the quantum recoil corrections of Compton scattering are negligible;
• the charge is genuinely free (or driven far from any resonance), so that no restoring force competes with the incident field in the equation of motion.

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