Radiation from an Accelerating Charge
For a point charge \(q\) with acceleration \(\vec{a}\) (non-relativistic, \(v\ll c\)),
the radiation fields (dominating at large \(r\), falling as \(1/r\)) are:
Radiation fields\begin{align}
\vec{E}_\text{rad} &= \frac{q}{4\pi\eps c^2 r}
\left[\hat{r}\times(\hat{r}\times\vec{a})\right]_{t_r}
\\[6pt]
\vec{B}_\text{rad} &= \frac{1}{c}\hat{r}\times\vec{E}_\text{rad}
\end{align}
Both fall as \(1/r\) (radiation field) and are transverse (\(\perp\hat{r}\)).
Larmor Formula — Total Radiated Power\begin{equation}
\boxed{P = \frac{q^2 a^2}{6\pi\eps c^3}
= \frac{\muz q^2 a^2}{6\pi c}}
\end{equation}
Angular distribution of radiated power\begin{equation}
\frac{dP}{d\Omega} = \frac{q^2 a^2\sin^2\theta}{16\pi^2\eps c^3}
\end{equation}
where \(\theta\) is the angle between \(\hat{r}\) and \(\vec{a}\).
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Maximum at \(\theta=90\ensuremath{^\circ}\) (equatorial plane \(\perp\vec{a}\))
Zero at \(\theta=0\ensuremath{^\circ}\) and \(180\ensuremath{^\circ}\) (along acceleration direction)
Pattern: donut (toroidal) shape centred on \(\vec{a}\)
Example 7.1 Ratio of radiated powersParticle A: charge \(q\), acceleration \(a\), \(P_A\propto q^2a^2\).
Particle B: charge \(2q\), acceleration \(4a\).
\(P_B/P_A=(2q)^2(4a)^2/(q^2a^2)=4\times16=64\).
\[\boxed{P_B/P_A=64}\]
Example 7.2 Power from a non-relativistic dipoleTwo particles: A (\(m\), \(q\), \(a\)) and B (\(m/2\), \(2q\), velocity \(3v\), acceleration \(4a\)).
\(P_A\propto q^2a^2\); \(P_B\propto(2q)^2(4a)^2=64q^2a^2\).
\[\boxed{P_B=64P_A}\]
Example 7.3 Energy radiated during decelerationElectron decelerates uniformly: initial \(v_0\), final \(0\), time \(\tau\).
\(a=v_0/\tau\); total energy: \(E_\text{rad}=P\tau=\frac{e^2(v_0/\tau)^2}{6\pi\eps c^3}\cdot\tau
=\frac{e^2v_0^2}{6\pi\eps c^3\tau}\).
\[\boxed{E_\text{rad}=\frac{e^2v_0^2}{6\pi\eps c^3\tau}}\]
Example 7.4 Far-field power from E fieldIn the far field, the electric field at a point is
\(\vec{E}=E_0\sin(\omega t)\,\xhat\).
Find average power per unit area.
\(\langle S\rangle=\langle E^2\rangle/(\muz c)=E_0^2/(2\muz c)\).
\[\boxed{\langle S\rangle=\frac{E_0^2}{2\muz c}}\]
Concept Checks — Larmor Formula & Dipole Radiation
Concept check.Far from source, \(\vec{E}=\dfrac{A}{r}\sin\theta\,e^{i(kr-\omega t)}\thetahat\). Total average radiated power:
Show answer
\(\langle S\rangle = |A|^2\sin^2\theta/(2\muz cr^2)\). \(P=\int\langle S\rangle r^2d\Omega=\frac{|A|^2}{2\muz c}\int\sin^2\theta\,d\Omega=\frac{|A|^2}{2\muz c}\cdot\frac{8\pi}{3}=\frac{4\pi|A|^2}{3\muz c}\).
Concept check.Quadrupole moment tensor \(Q_{ij}\): number of independent components:
(a) 9 (b) 6 (c) 5 (d) 3
Show answer
(c). Symmetric (6) and traceless (removes 1) \(\Rightarrow\) 5 independent components.
