Radiation Reaction

Step 1: Energy balance

The power radiated by a non-relativistic charge is $P = \mu_0 q^2 a^2/(6\pi c)$. The work done by the radiation reaction force is $\mathbf{F}_{\text{rad}}\cdot\mathbf{v}$. Energy conservation requires:

$$\int_{t_1}^{t_2}\mathbf{F}_{\text{rad}}\cdot\mathbf{v}\,dt = -\int_{t_1}^{t_2}\frac{\mu_0 q^2}{6\pi c}a^2\,dt$$

Step 2: Integrate by parts

$\int a^2\,dt = \int \dot{v}\cdot\dot{v}\,dt = [\dot{v}\cdot v]_{t_1}^{t_2} - \int \ddot{v}\cdot v\,dt$. If the motion is periodic or the boundary terms vanish:

$$\int a^2\,dt = -\int\dot{\mathbf{a}}\cdot\mathbf{v}\,dt$$

Step 3: Identify the force

$$\int\mathbf{F}_{\text{rad}}\cdot\mathbf{v}\,dt = \frac{\mu_0 q^2}{6\pi c}\int\dot{\mathbf{a}}\cdot\mathbf{v}\,dt$$

Step 4: Read off the force

$$\boxed{\mathbf{F}_{\text{rad}} = \frac{\mu_0 q^2}{6\pi c}\dot{\mathbf{a}} = m\tau_0\dot{\mathbf{a}}}$$

Step 5: Equation of motion

The full equation becomes: $m\mathbf{a} = \mathbf{F}_{\text{ext}} + m\tau_0\dot{\mathbf{a}}$. This is a third-order ODE, requiring three initial conditions. The extra degree of freedom leads to the notorious runaway and preacceleration problems.

Step 1: Rewrite the equation

$m(\mathbf{a} - \tau_0\dot{\mathbf{a}}) = \mathbf{F}_{\text{ext}}$. Multiply by the integrating factor $e^{-t/\tau_0}$:

$$\frac{d}{dt}(\mathbf{a}\,e^{-t/\tau_0}) = -\frac{1}{m\tau_0}\mathbf{F}_{\text{ext}}\,e^{-t/\tau_0}$$

Step 2: Integrate from t to infinity

Requiring $\mathbf{a}(t)\to 0$ as $t\to\infty$ (no runaways):

$$\mathbf{a}(t) = \frac{1}{m\tau_0}\int_t^\infty \mathbf{F}_{\text{ext}}(t')\,e^{-(t'-t)/\tau_0}\,dt'$$

Step 3: Preacceleration

The acceleration at time $t$ depends on the force at future times $t' > t$. For a force that turns on at $t = 0$, the particle begins accelerating at $t < 0$.

Step 4: Timescale of acausality

The preacceleration decays with time constant $\tau_0 \approx 6\times 10^{-24}$ s, far smaller than any classical measurement timescale. Quantum effects (pair creation, Compton scattering) become important at this scale.

Step 5: Interpretation

The preacceleration is not a true violation of causality but an indication that classical electrodynamics is incomplete at the scale $\tau_0 \sim r_e/c$. Quantum electrodynamics resolves these issues consistently, with radiation reaction arising naturally from the interaction of the electron with its own quantized field.

Step 1: Covariant power balance

The 4-vector radiation reaction force $\Gamma^\mu$ must satisfy $\Gamma^\mu u_\mu = 0$ (orthogonal to 4-velocity) and reproduce the Larmor power: $\Gamma^\mu u_\mu/c = 0$.

Step 2: Most general form

The most general 4-vector quadratic in the motion is $\Gamma^\mu = \alpha\dot{a}^\mu + \beta a_\nu a^\nu u^\mu$. Requiring $u_\mu\Gamma^\mu = 0$ and using $u_\mu\dot{a}^\mu = -a_\mu a^\mu$ fixes $\beta = \alpha/c^2$.

Step 3: Fix the coefficient

Integrating the energy component $\Gamma^0 v$ over a period and comparing with the integrated Larmor power fixes $\alpha = m\tau_0$.

Step 4: Full equation

$$m a^\mu = F_{\text{ext}}^\mu + m\tau_0\left(\dot{a}^\mu - \frac{a_\nu a^\nu}{c^2}u^\mu\right)$$

The term $-(a_\nu a^\nu/c^2)u^\mu$ is the Schott term, responsible for the relativistic correction to radiation reaction.

Step 5: Non-relativistic limit

For $v \ll c$: $a^\mu a_\mu \approx -a^2$ and $\dot{a}^i \approx \dot{a}_i$. The spatial part reduces to $m\mathbf{a} = \mathbf{F}_{\text{ext}} + m\tau_0\dot{\mathbf{a}}$, recovering the Abraham-Lorentz equation.

Step 1: Self-energy of a charged sphere

A uniformly charged sphere of radius $a$ has electrostatic self-energy $W = e^2/(8\pi\epsilon_0 a) \times 3/5$ (for a uniform volume distribution) or $W = e^2/(8\pi\epsilon_0 a)$ (surface charge).

Step 2: Electromagnetic mass

Identifying $W = m_{\text{em}}c^2$: $m_{\text{em}} = e^2/(8\pi\epsilon_0 ac^2)$. If all the electron mass is electromagnetic, then $a = r_e/2$ where $r_e = e^2/(4\pi\epsilon_0 m_e c^2)$.

Step 3: Momentum of the field

The electromagnetic momentum of a moving charged sphere is $\mathbf{p}_{\text{em}} = (4/3)m_{\text{em}}\gamma\mathbf{v}$. The factor 4/3 (not 1) means $E \neq pc$ for the field alone.

Step 4: Poincare stresses

Poincare showed that non-electromagnetic cohesive forces (Poincare stresses) are needed to hold the electron together. These stresses contribute $-W/3$ to the total momentum, restoring the correct relativistic relation $p = \gamma mv$.

Step 5: Resolution in QED

In quantum electrodynamics, the self-energy divergence is handled by renormalization: the bare mass $m_0$ and the electromagnetic self-energy $\delta m$ combine to give the observed physical mass $m = m_0 + \delta m$. Only the physical mass is observable, and the theory makes finite, testable predictions (e.g., the anomalous magnetic moment to 10 decimal places).

Step 1: Damped oscillator solution

For $\Gamma \ll \omega_0$: $x(t) = x_0 e^{-\Gamma t/2}\cos(\omega_0 t)$. The energy decays as $e^{-\Gamma t}$.

Step 2: Fourier transform of the signal

The Fourier transform of $e^{-\Gamma t/2}e^{-i\omega_0 t}$ for $t > 0$ is a Lorentzian:

$$|\hat{x}(\omega)|^2 \propto \frac{1}{(\omega-\omega_0)^2 + (\Gamma/2)^2}$$

Step 3: Full width at half maximum

$$\Delta\omega = \Gamma = \tau_0\omega_0^2$$

Step 4: Classical lifetime

The $e$-folding time for energy decay is $\tau = 1/\Gamma$. For the Lyman-alpha line ($\lambda = 121.6$ nm): $\Gamma \approx 6.3\times 10^8$ s$^{-1}$, giving $\tau \approx 1.6$ ns.

Step 5: Comparison with quantum result

The classical result gives the correct order of magnitude. The quantum mechanical linewidth for hydrogen Lyman-alpha is $\Gamma_{\text{QM}} = \omega_0^3 r_e/(3c) \times f_{12}$, where $f_{12}$ is the oscillator strength. The classical and quantum widths agree to within factors of order unity, one of the remarkable successes of the classical oscillator model.