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 arriving from a distant source along the fixed direction \(\hat{\mathbf k}\), \[ \mathbf E_{\rm inc}(\mathbf x,t)=E_0\,\hat{\boldsymbol\epsilon}\,\cos(\omega t-\mathbf k\cdot\mathbf x), \qquad \mathbf k\equiv\frac{\omega}{c}\hat{\mathbf k}, \qquad \hat{\boldsymbol\epsilon}\perp\hat{\mathbf k}, \] which reduces, at the charge’s location \(\mathbf x=\mathbf 0\), to \(\mathbf E_{\rm inc}(t)=E_0\,\hat{\boldsymbol\epsilon}\,\cos(\omega t)\) used below. This is a solution of the source-free Maxwell equations. Since \(\hat{\boldsymbol\epsilon}\) is a fixed vector, only the scalar factor is differentiated, with \(\nabla\cos(\omega t-\mathbf k\cdot\mathbf x)=\mathbf k\,\sin(\omega t-\mathbf k\cdot\mathbf x)\); Gauss’s law \(\nabla\cdot\mathbf E_{\rm inc}=0\) then holds automatically, \[ \nabla\cdot\mathbf E_{\rm inc}=E_0\sin(\omega t-\mathbf k\cdot\mathbf x)\,(\hat{\boldsymbol\epsilon}\cdot\mathbf k)=0, \] by the transversality condition \(\hat{\boldsymbol\epsilon}\perp\hat{\mathbf k}\); and the wave equation \(\Box\mathbf E_{\rm inc}=0\) fixes \(|\mathbf k|=\omega/c\), i.e. exactly the dispersion relation used above to define \(\mathbf k\). The magnetic field follows from Faraday’s law, \(\nabla\times\mathbf E_{\rm inc}=-\partial_t\mathbf B_{\rm inc}\): using \(\nabla\times\big(g(\mathbf x)\hat{\boldsymbol\epsilon}\big)=(\nabla g)\times\hat{\boldsymbol\epsilon}\) for the same reason (any scalar function \(g\) times the fixed vector \(\hat{\boldsymbol\epsilon}\)), \[ \nabla\times\mathbf E_{\rm inc}=E_0\sin(\omega t-\mathbf k\cdot\mathbf x)\,\big(\mathbf k\times\hat{\boldsymbol\epsilon}\big) =-\partial_t\mathbf B_{\rm inc}, \] which integrates in \(t\) (dropping an irrelevant static constant) to \[ \mathbf B_{\rm inc}(\mathbf x,t)=\frac{E_0}{\omega}\cos(\omega t-\mathbf k\cdot\mathbf x)\,\big(\mathbf k\times\hat{\boldsymbol\epsilon}\big) =\frac1c\,\hat{\mathbf k}\times\mathbf E_{\rm inc}(\mathbf x,t), \] using \(\mathbf k=(\omega/c)\hat{\mathbf k}\) in the last step. As a consistency check, this is exactly the radiation-field structure found in Radiation from a Moving Charge, \(\mathbf B_{\rm rad}=\frac1c\hat{\mathbf n}\times\mathbf E_{\rm rad}\) with \(\hat{\mathbf n}\cdot\mathbf E_{\rm rad}=0\): a plane wave is simply what any radiation field looks like once its distant source is out of the picture, with \(\hat{\mathbf k}\) now playing the role of the incoming propagation direction, in place of the outgoing \(\hat{\mathbf n}\) used there to point from source to observer, and \(\hat{\boldsymbol\epsilon}\perp\hat{\mathbf k}\) the same transversality condition as \(\hat{\mathbf n}\cdot\mathbf E_{\rm rad}=0\). Either way, \(|\mathbf B_{\rm inc}|=|\mathbf E_{\rm inc}|/c\), exactly as for any radiation field. 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 charge’s motion is governed by the covariant Lorentz force law, \[ m\,a^\mu=q\,F^{\mu\nu}u_\nu, \qquad a^\mu\equiv\frac{\mathrm{d}u^\mu}{\mathrm{d}\tau}, \] with \(u^\mu\) the four-velocity fixed in the conventions. This is the same content as the standard \(\mathrm{d}p^\mu/\mathrm{d}\tau=qF^{\mu\nu}U_\nu\) with \(p^\mu=mU^\mu\) and the textbook (dimensionful) \(U^\mu=\mathrm{d}y^\mu/\mathrm{d}\tau=cu^\mu\); writing it in terms of this course’s dimensionless \(u^\mu=U^\mu/c\) multiplies both sides by \(c\), which cancels, leaving the equation above with no explicit factor of \(c\) in front.

To extract the familiar 3-vector force law, specialize first to the instant at which the charge is at rest, \(u^\mu=(1,\mathbf 0)\) — the same specialization used below for the radiation field — so that \(u_0=-1\), \(u_i=0\), and, exactly as there, \(a^\mu=(0,\dot{\mathbf v}/c)\). Then \[ F^{i\nu}u_\nu=F^{i0}u_0=-F^{i0}=F^{0i}=\frac{E^i_{\rm inc}}{c}, \] with no contribution from the \(F^{ij}u_j\) piece, since \(u_j=0\) at this instant: the magnetic force vanishes identically, not merely approximately, whenever \(\mathbf v=\mathbf 0\). The covariant equation of motion then reads \(m\dot v^i/c=qE^i_{\rm inc}/c\), i.e. \[ m\dot{\mathbf v}=q\mathbf E_{\rm inc}(t). \] This is exact only at the single instant used to define the rest frame; away from it \(\mathbf v\neq\mathbf 0\) and a magnetic force reappears. To see how small it is, repeat the computation without specializing \(u^\mu\), keeping the general lab-frame \(u^\mu=\gamma(1,\mathbf v/c)\), so that \(u_0=-\gamma\), \(u_j=\gamma v_j/c\). Then \[ F^{i\nu}u_\nu = F^{i0}u_0+F^{ij}u_j = \frac{\gamma}{c}\Big[E^i_{\rm inc}+(\mathbf v\times\mathbf B_{\rm inc})^i\Big], \] using \(F^{i0}=-E^i_{\rm inc}/c\) and \(F^{ij}v_j=\epsilon^{ijk}B_kv_j=(\mathbf v\times\mathbf B_{\rm inc})^i\), while \(a^i=\mathrm{d}u^i/\mathrm{d}\tau=\gamma\,\mathrm{d}(\gamma v^i/c)/\mathrm{d}t\) using \(\mathrm{d}/\mathrm{d}\tau=\gamma\,\mathrm{d}/\mathrm{d}t\). Substituting both into \(ma^\mu=qF^{\mu\nu}u_\nu\) and cancelling the common factor \(\gamma/c\) from both sides gives the standard relativistic 3-force law, \[ m\frac{\mathrm{d}(\gamma\mathbf v)}{\mathrm{d}t}=q\big(\mathbf E_{\rm inc}+\mathbf v\times\mathbf B_{\rm inc}\big). \] Since \(|\mathbf B_{\rm inc}|=|\mathbf E_{\rm inc}|/c\), the magnetic term here is smaller than the electric term by a factor \(v/c\); in the non-relativistic regime \(v\ll c\) assumed throughout (Setup) this holds at every instant along the trajectory, not just the one where the charge happens to be at rest, and \(\gamma\approx1\) to the same accuracy. Dropping both recovers exactly the rest-frame result found above, now understood to hold throughout the motion, \[ m\dot{\mathbf v}=q\mathbf E_{\rm inc}(t), \qquad \dot{\mathbf v}(t)=\frac{q}{m}E_0\,\hat{\boldsymbol\epsilon}\,\cos(\omega t). \] 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.

To turn \(F_{\mu\nu}^{\rm rad}\) into an electric field, use the same covariant contraction as in the equation of motion above: the field measured by an observer of four-velocity \(w^\mu\) is \[ E^\mu(w)\equiv c\,F^{\mu\nu}w_\nu, \] automatically orthogonal to the observer, \(w_\mu E^\mu(w)=0\) (by antisymmetry of \(F^{\mu\nu}\)), and equal to \((0,\mathbf E)\) in that observer’s own rest frame: at \(w^\mu=(1,\mathbf 0)\), \(E^0(w)=cF^{00}w_0=0\) and, using \(F^{i0}=-F^{0i}=-E^i/c\), \[ E^i(w)=cF^{i\nu}w_\nu=cF^{i0}w_0=-cF^{i0}=cF^{0i}=E^i. \] (The similar-looking contraction \(u_\nu F^{\nu\mu}=-F^{\mu\nu}u_\nu\) has the opposite sign and is missing this factor of \(c\); the correctly normalized object is \(E^\mu=cF^{\mu\nu}u_\nu\), not \(u_\nu F^{\nu\mu}\).)

Apply this with \(w^\mu=u^\mu\), the radiating charge’s own four-velocity, at the instant it 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 \(\Delta x^\mu=(R,R\hat{\mathbf n})\), \(R=|\mathbf x|\), \[ u\cdot\Delta x=-R, \qquad \Delta x\cdot a=R\,\frac{\hat{\mathbf n}\cdot\dot{\mathbf v}}{c}. \] Since \(u^0=1\), \(u^i=0\), only the \(\mu=i,\nu=0\) component of the bracket in \(F_{\mu\nu}^{\rm rad}\) contributes to \(E_i^{\rm rad}=cF_{i\nu}^{\rm rad}u^\nu=cF_{i0}^{\rm rad}\). Using \(a_0=0\), \(u_0=-1\), \(u_i=0\), \(\Delta x_0=-R\), \(\Delta x_i=R\hat n_i\), and \(\Delta x\cdot a/(u\cdot\Delta x)=-\hat{\mathbf n}\cdot\dot{\mathbf v}/c\), \[ F_{i0}^{\rm rad} = \frac{\mu_0q}{4\pi R^2} \left[ \Delta x_i a_0-\Delta x_0 a_i-\frac{\Delta x\cdot a}{u\cdot\Delta x}(\Delta x_i u_0-\Delta x_0 u_i) \right] = \frac{\mu_0q}{4\pi R^2} \left[ R\frac{\dot v_i}{c} + \frac{\hat{\mathbf n}\cdot\dot{\mathbf v}}{c}(R\hat n_i) \right], \] so that \[ E_i^{\rm rad}=cF_{i0}^{\rm rad}=\frac{\mu_0q}{4\pi R}\Big[\dot v_i-(\hat{\mathbf n}\cdot\dot{\mathbf v})\hat n_i\Big], \qquad \boxed{ \mathbf E_{\rm rad}(\mathbf x,t) = \frac{\mu_0q}{4\pi R} \Big[\dot{\mathbf v}-(\hat{\mathbf n}\cdot\dot{\mathbf v})\hat{\mathbf n}\Big]_{t_r}. } \]

The bracket is the component of \(\dot{\mathbf v}\) transverse to \(\hat{\mathbf n}\): writing \(\dot{\mathbf v}=(\hat{\mathbf n}\cdot\dot{\mathbf v})\hat{\mathbf n}+\dot{\mathbf v}_\perp\), the bracket is exactly \(\dot{\mathbf v}_\perp\), manifestly orthogonal to \(\hat{\mathbf n}\). The same projection has an equivalent double-cross-product form: the BAC-CAB rule \(\mathbf A\times(\mathbf B\times\mathbf C)=\mathbf B(\mathbf A\cdot\mathbf C)-\mathbf C(\mathbf A\cdot\mathbf B)\) with \(\mathbf A=\mathbf B=\hat{\mathbf n}\), \(\mathbf C=\dot{\mathbf v}\), gives \[ \hat{\mathbf n}\times(\hat{\mathbf n}\times\dot{\mathbf v}) = \hat{\mathbf n}(\hat{\mathbf n}\cdot\dot{\mathbf v})-\dot{\mathbf v}(\hat{\mathbf n}\cdot\hat{\mathbf n}) = (\hat{\mathbf n}\cdot\dot{\mathbf v})\hat{\mathbf n}-\dot{\mathbf v}, \] using \(\hat{\mathbf n}\cdot\hat{\mathbf n}=1\), so \(\dot{\mathbf v}-(\hat{\mathbf n}\cdot\dot{\mathbf v})\hat{\mathbf n}=-\hat{\mathbf n}\times(\hat{\mathbf n}\times\dot{\mathbf v})\) and, equivalently, \[ \mathbf E_{\rm rad}(\mathbf x,t) = -\frac{\mu_0q}{4\pi R}\,\hat{\mathbf n}\times\big(\hat{\mathbf n}\times\dot{\mathbf v}\big)\Big|_{t_r}. \] Both forms make the transversality of the dipole radiation field to \(\hat{\mathbf n}\) manifest: the first as an explicit orthogonal projection, the second because \(\hat{\mathbf n}\times(\text{anything})\) is automatically perpendicular to \(\hat{\mathbf n}\). 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 \[ \langle S_{\rm inc}\rangle=c\varepsilon_0\langle E_{\rm inc}^2\rangle=\tfrac12c\varepsilon_0E_0^2. \]

What a cross section measures. A distant detector at angle \((\theta,\phi)\) subtending solid angle \(\mathrm{d}\Omega\) records scattered power \(\langle\mathrm{d}P/\mathrm{d}\Omega\rangle\,\mathrm{d}\Omega\), independent of its distance (since \(\mathrm{d}P/\mathrm{d}\Omega\) was built from the \(R\)-independent radiation field found above). The incident beam is instead characterized by its flux \(\langle S_{\rm inc}\rangle\), a power per unit area crossing a surface perpendicular to \(\hat{\mathbf k}\). Dividing the two cancels the (arbitrary) incident intensity and turns “power per solid angle” into “area per solid angle” — hence the name cross section: \[ \frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} \equiv \frac{\langle \mathrm{d}P/\mathrm{d}\Omega\rangle}{\langle S_{\rm inc}\rangle}, \qquad \big[\mathrm{d}\sigma/\mathrm{d}\Omega\big]=\text{area}. \] Concretely, \(\mathrm{d}\sigma\) is the cross-sectional area of the incident beam that carries exactly the power the charge re-radiates into \(\mathrm{d}\Omega\); integrated over all directions (below), the total cross section \(\sigma\) is the area of a flat, perfectly absorbing target that would intercept, head-on, the same total power the charge actually scatters in every direction. It is the effective size the charge presents to the beam. Because both \(\langle\mathrm{d}P/\mathrm{d}\Omega\rangle\) and \(\langle S_{\rm inc}\rangle\) scale as \(E_0^2\), this ratio is independent of the incident intensity: it depends only on the scattering geometry and on \(q\) and \(m\), which is what makes a cross section a property of the charge (or process) rather than of any particular beam. 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.

9 Exercises

9.1 Exercise 1: Dropping the Magnetic Force

a) Using \(|\mathbf B_{\rm inc}|=|\mathbf E_{\rm inc}|/c\), estimate the ratio \(|q\mathbf v\times\mathbf B_{\rm inc}|/|q\mathbf E_{\rm inc}|\) in terms of \(v/c\), confirming that it is suppressed for non-relativistic motion.

b) Using \(v\sim a_0/\omega\) for the oscillation velocity scale, with \(a_0=qE_0/m\), express \(v/c\) in terms of \(q,E_0,m,\omega,c\), and check that it is indeed small for visible light (\(\omega\sim10^{15}\,\mathrm{s^{-1}}\)) incident on a free electron at ordinary laboratory intensities.

9.2 Exercise 2: The Local-Frame Radiation Field (Essential)

a) Starting from \(E_i^{\rm rad}=cF_{i\nu}^{\rm rad}u^\nu\) specialized to \(u^\mu=(1,\mathbf 0)\), with \(\Delta x^\mu=(R,R\hat{\mathbf n})\) and \(\Delta x\cdot a=R\,\hat{\mathbf n}\cdot\dot{\mathbf v}/c\), write out \(F_{i0}^{\rm rad}\) term by term and confirm \[ \mathbf E_{\rm rad}=\frac{\mu_0q}{4\pi R}\Big[\dot{\mathbf v}-(\hat{\mathbf n}\cdot\dot{\mathbf v})\hat{\mathbf n}\Big]. \]

b) Verify the equivalent form \(-\hat{\mathbf n}\times(\hat{\mathbf n}\times\dot{\mathbf v})=\dot{\mathbf v}-(\hat{\mathbf n}\cdot\dot{\mathbf v})\hat{\mathbf n}\) using the BAC-CAB rule.

9.3 Exercise 3: Larmor Consistency Check (Essential)

a) Verify \(\int\sin^2\chi\,\mathrm{d}\Omega=8\pi/3\) by direct integration in spherical coordinates, with \(\chi\) the polar angle measured from \(\dot{\mathbf v}\).

b) Substitute this into \(\mathrm{d}P/\mathrm{d}\Omega\) and confirm that the result reproduces exactly the Larmor formula \(P=\mu_0q^2\dot v^2/(6\pi c)\) derived covariantly in Radiation from a Moving Charge, matching the numerical prefactor as well as the power of \(\dot v\).

9.4 Exercise 4: The Differential Cross Section, Step by Step (Essential)

a) Using \(\dot{\mathbf v}=(q/m)\mathbf E_{\rm inc}\) and \(\langle E_{\rm inc}^2\rangle=E_0^2/2\), verify \(\langle\dot v^2\rangle=(q/m)^2E_0^2/2\) and substitute it into \(\langle\mathrm{d}P/\mathrm{d}\Omega\rangle\) to reproduce the boxed intermediate expression given in the notes.

b) Divide by \(\langle S_{\rm inc}\rangle\) and, using \(1/\varepsilon_0=\mu_0c^2\), verify explicitly that every factor of \(c\), \(\mu_0\), and \(m\) combines to leave \(\mathrm{d}\sigma/\mathrm{d}\Omega=r_q^2\sin^2\chi\), an area.

9.5 Exercise 5: Unpolarized Averaging in a General Direction

a) Starting from \(\cos\chi=\hat{\mathbf n}\cdot\hat{\boldsymbol\epsilon}\), verify \(\sin^2\chi=1-(\hat{\mathbf n}\cdot\hat{\boldsymbol\epsilon})^2\).

b) Using the rotational symmetry of the setup about \(\hat{\mathbf k}\), argue that the unpolarized \(\mathrm{d}\sigma/\mathrm{d}\Omega\) can only depend on the polar angle \(\theta\) between \(\hat{\mathbf n}\) and \(\hat{\mathbf k}\), never on the azimuthal angle of \(\hat{\mathbf n}\) — so it suffices to check the formula for \(\hat{\mathbf n}\) in a single plane containing \(\hat{\mathbf k}\), as the notes do.

9.6 Exercise 6: The Total Cross Section (Essential)

a) Verify \(\int_0^{2\pi}\mathrm{d}\phi\int_{-1}^1(1+x^2)\,\mathrm{d}x=8\pi/3\) step by step, and hence confirm \(\sigma_T=\frac{8\pi}3r_q^2\).

b) Using \(r_q=q^2/(4\pi\varepsilon_0mc^2)\) and \(e=1.6\times10^{-19}\,\mathrm C\), \(m_e=9.11\times10^{-31}\,\mathrm{kg}\), verify numerically that \(\sigma_T\approx6.65\times10^{-29}\,\mathrm m^2\) for an electron.

---

Feel free to create issues, ask questions, or suggest improvements in the GitHub repository.