Radiation from Accelerating Charges

10.1 Larmor Formula

A non-relativistic charge $q$ undergoing acceleration $\mathbf{a}$ radiates power:

$P = \frac{q^2 a^2}{6\pi\varepsilon_0 c^3}$

For an oscillating dipole $\mathbf{p} = q\mathbf{d}\cos\omega t$ with acceleration $a = \omega^2 d$:

$P = \frac{q^2 \omega^4 d^2}{12\pi\varepsilon_0 c^3} = \frac{\omega^4 p_0^2}{12\pi\varepsilon_0 c^3}$

Where $p_0 = qd$ is the dipole moment amplitude.

Radiation resistance: Equating $P = \frac{1}{2}I_0^2 R_{\text{rad}}$ for an antenna of length $\ell$ carrying current $I_0$ at frequency $\omega$:

$R_{\text{rad} = \frac{\mu_0 c}{6\pi}\left(\frac{\omega \ell}{c}\right)^2 = \frac{\pi}{6}Z_0\left(\frac{\ell}{\lambda}\right)^2 \approx 197\left(\frac{\ell}{\lambda}\right)^2\ \Omega}$

10.2 Electric Dipole Radiation

The radiation fields from an oscillating electric dipole at distance $r \gg \lambda$:

$\mathbf{E} = -\frac{\mu_0 \omega^2 p_0}{4\pi r}\sin\theta\, e^{i(kr - \omega t)}\,\hat{\boldsymbol{\theta}}$

$\mathbf{B} = -\frac{\mu_0 \omega^2 p_0}{4\pi c\, r}\sin\theta\, e^{i(kr - \omega t)}\,\hat{\boldsymbol{\phi}}$

The angular distribution of radiated power:

$\frac{dP}{d\Omega} = \frac{\mu_0 p_0^2 \omega^4}{32\pi^2 c}\sin^2\theta$

The total power (integrating over solid angle):

$P = \frac{\mu_0 p_0^2 \omega^4}{12\pi c}$

The radiation pattern is toroidal (doughnut-shaped), with zero radiation along the dipole axis ($\theta = 0, \pi$) and maximum in the equatorial plane ($\theta = \pi/2$).

10.3 Relativistic Radiation: Liénard—Wiechert Potentials

For a relativistic charge with velocity $\boldsymbol{\beta} = \mathbf{v}/c$ and acceleration $\dot{\boldsymbol{\beta}}$:

$P = \frac{q^2}{6\pi\varepsilon_0 c}\gamma^6\left[(\dot{\boldsymbol{\beta}})^2 - (\boldsymbol{\beta} \times \dot{\boldsymbol{\beta}})^2\right]$

For linear acceleration ($\boldsymbol{\beta} \parallel \dot{\boldsymbol{\beta}}$):

$P = \frac{q^2}{6\pi\varepsilon_0 c}\gamma^6\dot{\beta}^2$

For circular acceleration ($\boldsymbol{\beta} \perp \dot{\boldsymbol{\beta}}$E.g., synchrotron):

$P = \frac{q^2}{6\pi\varepsilon_0 c}\gamma^4\dot{\beta}^2 = \frac{q^2 c}{6\pi\varepsilon_0}\frac{\gamma^4}{R^2}$

Where $R$ is the radius of curvature. The $\gamma^4$ factor (vs. $\gamma^6$ for linear) explains why synchrotron radiation is significant for relativistic electrons but negligible for protons at the same energy ($\gamma$ is $m_p/m_e \approx 1836$ times smaller).

Synchrotron radiation spectrum: The critical frequency is $\omega_c = \frac{3}{2}\gamma^3\frac{c}{R}$. The spectrum peaks near $\omega_c$ and extends to high harmonics, making synchrotron radiation a powerful broadband source from infrared to X-rays.