14 May 2025
Radiation from Accelerated Charges and Antennas
Angular radiation from accelerated charges, oscillating dipoles, current elements, linear antennas, and coherent antenna arrays.
The radiation part of the Liénard-Wiechert field is the term that decreases as $R^{-1}$:
\[\mathbf E_{\rm rad}=\frac{q}{4\pi\epsilon_0c} \left[ \frac{\mathbf n\times\{(\mathbf n-\boldsymbol\beta) \times\dot{\boldsymbol\beta}\}} {(1-\mathbf n\cdot\boldsymbol\beta)^3R} \right]_{t_r}, \qquad \mathbf B_{\rm rad}=\frac{1}{c}\mathbf n\times\mathbf E_{\rm rad}.\]Both fields are transverse to $\mathbf n$. Their far-zone Poynting vector is
\[\mathbf S=\frac{1}{\mu_0}\mathbf E_{\rm rad}\times\mathbf B_{\rm rad} =\epsilon_0cE_{\rm rad}^2\mathbf n.\]Because $dt=(1-\mathbf n\cdot\boldsymbol\beta)\,dt_r$, the energy crossing $R^2d\Omega$ during observer time $dt$ was emitted during $dt_r$. Hence
\[\boxed{ \frac{dP}{d\Omega} =\frac{q^2}{16\pi^2\epsilon_0c} \frac{\left\lvert\mathbf n\times[(\mathbf n-\boldsymbol\beta) \times\dot{\boldsymbol\beta}]\right\rvert^2} {(1-\mathbf n\cdot\boldsymbol\beta)^5}. }\]Low- and high-velocity radiation
For $v\ll c$, put $\boldsymbol\beta=0$ and $\dot{\boldsymbol\beta}=\mathbf a/c$. If $\theta$ is the angle between $\mathbf a$ and $\mathbf n$,
\[\frac{dP}{d\Omega} =\frac{q^2a^2}{16\pi^2\epsilon_0c^3}\sin^2\theta.\]Integration uses $d\Omega=2\pi\sin\theta\,d\theta$:
\[\begin{aligned} P&=\frac{q^2a^2}{16\pi^2\epsilon_0c^3} 2\pi\int_0^\pi\sin^3\theta\,d\theta\\ &=\frac{q^2a^2}{16\pi^2\epsilon_0c^3} 2\pi\left[\frac43\right] =\boxed{\frac{q^2a^2}{6\pi\epsilon_0c^3}}. \end{aligned}\]This is the Larmor power. To perform the arbitrary-speed integral, choose $\boldsymbol\beta=\beta\hat{\mathbf z}$ and write $\mathbf a=a_\parallel\hat{\mathbf z}+a_\perp\hat{\mathbf x}$, $\mu=\cos\theta$, and $\mathbf n=(\sqrt{1-\mu^2}\cos\phi, \sqrt{1-\mu^2}\sin\phi,\mu)$. Since $\dot{\boldsymbol\beta}=\mathbf a/c$, the azimuthal average of the numerator is
\[\begin{aligned} &\frac1{2\pi}\int_0^{2\pi} \left\lvert\mathbf n\times[(\mathbf n-\boldsymbol\beta) \times\mathbf a]\right\rvert^2d\phi\\ &\qquad=a_\parallel^2(1-\mu^2) +\frac{a_\perp^2}{2} \left[(1+\beta^2)(1+\mu^2)-4\beta\mu\right]. \end{aligned}\]Thus
\[P=\frac{q^2}{8\pi\epsilon_0c^3} \left(a_\parallel^2I_\parallel+a_\perp^2I_\perp\right),\]where
\[I_\parallel=\int_{-1}^{1} \frac{1-\mu^2}{(1-\beta\mu)^5}\,d\mu,\] \[I_\perp=\frac12\int_{-1}^{1} \frac{(1+\beta^2)(1+\mu^2)-4\beta\mu} {(1-\beta\mu)^5}\,d\mu.\]With $u=1-\beta\mu$, the limits become $1-\beta$ and $1+\beta$ and
\[\begin{aligned} I_\parallel &=\frac1{\beta^3}\int_{1-\beta}^{1+\beta} \frac{(\beta^2-1)+2u-u^2}{u^5}\,du =\frac4{3(1-\beta^2)^3}=\frac43\gamma^6,\\ I_\perp &=\frac1{2\beta^3}\int_{1-\beta}^{1+\beta} \frac{(1-\beta^2)^2-2(1-\beta^2)u+(1+\beta^2)u^2} {u^5}\,du\\ &=\frac4{3(1-\beta^2)^2}=\frac43\gamma^4. \end{aligned}\]The $\beta\to0$ values follow continuously. Since $a^2-\lvert\boldsymbol\beta\times\mathbf a\rvert^2 =a_\parallel^2+(1-\beta^2)a_\perp^2$, the result is
\[\boxed{ P=\frac{q^2\gamma^6}{6\pi\epsilon_0c^3} \left[a^2-\lvert\boldsymbol\beta\times\mathbf a\rvert^2\right]. }\]Thus $P=q^2\gamma^6a^2/(6\pi\epsilon_0c^3)$ for acceleration parallel to the velocity and $P=q^2\gamma^4a^2/(6\pi\epsilon_0c^3)$ for perpendicular acceleration.
Oscillating electric dipole
Let a localized source have dipole moment $\mathbf p(t)=p_0\cos\omega t\,\hat{\mathbf z}$ and size $d\ll\lambda$. In the radiation zone $r\gg\lambda$ its retarded vector potential is
\[\mathbf A(\mathbf r,t) =\frac{\mu_0}{4\pi r}\dot{\mathbf p}(t_r), \qquad t_r=t-\frac rc.\]Keeping only derivatives of the retarded time, which supply the leading $r^{-1}$ fields,
\[\mathbf B_{\rm rad}=\nabla\times\mathbf A =-\frac{\mu_0}{4\pi cr} \mathbf n\times\ddot{\mathbf p}(t_r),\] \[\mathbf E_{\rm rad}=c\mathbf B_{\rm rad}\times\mathbf n =\frac{1}{4\pi\epsilon_0c^2r} \mathbf n\times[\mathbf n\times\ddot{\mathbf p}(t_r)].\]Since $\lvert\ddot{\mathbf p}\rvert=\omega^2p_0\lvert\cos\omega t_r\rvert$,
\[\left\langle\frac{dP}{d\Omega}\right\rangle =\frac{p_0^2\omega^4}{32\pi^2\epsilon_0c^3}\sin^2\theta, \qquad \boxed{\langle P\rangle =\frac{p_0^2\omega^4}{12\pi\epsilon_0c^3}}.\]The approximation requires source size $d\ll\lambda$, observation distance $r\gg\lambda$, and nonrelativistic source motion.
Small current element and linear antenna
A short element of length $\ell\ll\lambda$ carrying $I(t)=I_0\cos\omega t$ has charge amplitude $q_0=I_0/\omega$ by $I=dq/dt$, and dipole amplitude $p_0=q_0\ell=I_0\ell/\omega$. Therefore
\[\left\langle\frac{dP}{d\Omega}\right\rangle =\frac{I_0^2\ell^2\omega^2}{32\pi^2\epsilon_0c^3}\sin^2\theta =\frac{Z_0I_0^2(k\ell)^2}{32\pi^2}\sin^2\theta,\]where $k=\omega/c$ and $Z_0=\sqrt{\mu_0/\epsilon_0}=\mu_0c$.
Integrating $\sin^2\theta$ over solid angle gives $\int\sin^2\theta\,d\Omega=8\pi/3$, so the average radiated power of the current element is
\[\langle P\rangle=\frac{Z_0I_0^2(k\ell)^2}{12\pi}.\]If $I_0$ is the peak current, define its radiation resistance by $\langle P\rangle=I_0^2R_{\rm r}/2$. Hence
\[\boxed{R_{\rm r}=\frac{Z_0(k\ell)^2}{6\pi} \simeq20(k\ell)^2\ \Omega.}\]This is the resistance that would dissipate the same average power; it is not an ohmic loss. The current-element approximation requires $k\ell\ll1$.
For a thin, centre-fed linear antenna of total length $L$ on the $z$-axis, take the standing current
\[I(z)=I_0\sin[k(L/2-\lvert z\rvert)],\qquad -L/2\le z\le L/2.\]Far-zone contributions from $z$ acquire phase $e^{-ikz\cos\theta}$. Put $A=kL/2$, $\eta=\cos\theta$, and $u=kz$. Evenness of $I(z)$ cancels the imaginary part, so
\[\begin{aligned} \int_{-L/2}^{L/2}I(z)e^{-ikz\eta}\,dz &=\frac{2I_0}{k}\int_0^A\sin(A-u)\cos(\eta u)\,du\\ &=\frac{2I_0}{k}\int_0^A [\sin A\cos u-\cos A\sin u]\cos(\eta u)\,du\\ &=\frac{2I_0}{k} \frac{\cos(A\eta)-\cos A}{1-\eta^2}. \end{aligned}\]The transverse projection supplies a factor $\sin\theta$. Apart from the common radial factor, the pattern amplitude is therefore
\[\boxed{ F(\theta)=\frac{2I_0}{k} \frac{\cos[(kL/2)\cos\theta]-\cos(kL/2)}{\sin\theta}. }\]The displayed quotient has a removable axial limit. For $\theta\to0$, with $A=kL/2$,
\[\cos(A\cos\theta)-\cos A =\frac{A\sin A}{2}\theta^2+O(\theta^4), \qquad \sin\theta=\theta+O(\theta^3),\]so $F(\theta)=(I_0A\sin A/k)\theta+O(\theta^3)\to0$; the $\theta\to\pi$ limit is likewise zero.
The radiation intensity is proportional to $\lvert F(\theta)\rvert^2$. For $L=\lambda/2$, this becomes $F(\theta)\propto\cos[(\pi/2)\cos\theta]/\sin\theta$.
Antenna arrays
For $N$ identical elements separated by $d$ along $z$, with phase advance $\delta$ between adjacent feeds, the phase step in direction $\theta$ is
\[\psi=kd\cos\theta+\delta.\]The array factor is the geometric sum
\[\begin{aligned} AF&=\sum_{m=0}^{N-1}e^{im\psi} =\frac{1-e^{iN\psi}}{1-e^{i\psi}}\\ &=e^{i(N-1)\psi/2} \frac{\sin(N\psi/2)}{\sin(\psi/2)}. \end{aligned}\]Thus
\[\boxed{\lvert AF\rvert^2=\left[\frac{\sin(N\psi/2)}{\sin(\psi/2)}\right]^2},\]with the limiting value $N^2$ when $\psi=2m\pi$. The total pattern equals the single-element intensity multiplied by $\lvert AF\rvert^2$.
Solved Problems
1. Fraction of Larmor power in an equatorial belt
What fraction of the nonrelativistic radiation lies between $60^\circ\le\theta\le120^\circ$, where $\theta$ is measured from the acceleration axis?
Because $dP/d\Omega=C\sin^2\theta$ and $d\Omega=\sin\theta\,d\theta\,d\phi$, the required fraction is
\[f=\frac{2\pi\displaystyle\int_{\pi/3}^{2\pi/3}\sin^3\theta\,d\theta} {2\pi\displaystyle\int_0^\pi\sin^3\theta\,d\theta}.\]Use
\[\int\sin^3\theta\,d\theta =-\cos\theta+\frac13\cos^3\theta.\]The denominator integral is $4/3$. Symmetry about $\pi/2$ gives
\[\int_{\pi/3}^{2\pi/3}\sin^3\theta\,d\theta =2\left[0-\left(-\frac12+\frac1{24}\right)\right] =\frac{11}{12}.\]Therefore
\[\boxed{f=\frac{11/12}{4/3}=\frac{11}{16}=0.6875.}\]The result is dimensionless and lies between zero and one. It also expresses the physical suppression of radiation along the acceleration axis.
2. Half-power beamwidth of a half-wave antenna
For $L=\lambda/2$, find the half-power directions and the half-power beamwidth in the plane containing the antenna.
After normalization to its broadside value, the power pattern is
\[G(\theta)=\left[ \frac{\cos\!\left((\pi/2)\cos\theta\right)}{\sin\theta} \right]^2, \qquad G\!\left(\frac\pi2\right)=1.\]In the first half-plane the amplitude is positive, so $G=1/2$ requires
\[g(\theta)\equiv \frac{\cos\!\left((\pi/2)\cos\theta\right)}{\sin\theta} =\frac1{\sqrt2}.\]Writing $a=\pi/2$, the derivative needed for Newton iteration is
\[g^{\prime}(\theta)= \frac{a\sin(a\cos\theta)\sin^2\theta -\cos(a\cos\theta)\cos\theta}{\sin^2\theta}.\]Starting with $\theta_0=0.900\ \mathrm{rad}$ and using $\theta_{n+1}=\theta_n-[g(\theta_n)-1/\sqrt2]/g^{\prime}(\theta_n)$ gives
\[\theta_1=0.8894022,\qquad \theta_2=0.8894397,\qquad \theta_3=0.8894397\ \mathrm{rad}.\]Thus symmetry gives
\[\theta_{\rm HP}=50.961^\circ,\ 129.039^\circ, \qquad \boxed{\mathrm{HPBW}=78.078^\circ.}\]Substitution gives $G=0.500000$ at either boundary. The result concerns the far-zone power pattern; it does not specify near-field structure.
Descriptive Questions
- Derive the angular power distribution of an arbitrarily moving point charge, including the conversion from observation time to retarded emission time.
- Obtain the Larmor and Liénard powers and explain why parallel and perpendicular accelerations carry different powers of $\gamma$.
- Derive the radiation fields and average power of an oscillating electric dipole, stating the source-size, speed, and radiation-zone approximations.
- Starting from the current distribution of a centre-fed linear antenna, derive its pattern factor and then obtain the array factor for equally spaced phased elements.
Numerical Problems
- Find the Larmor power of an electron whose nonrelativistic acceleration is $2.00\times10^{17}\ \mathrm{m\,s^{-2}}$.
- A singly charged particle has $\beta=0.800$ and acceleration $5.00\times10^{15}\ \mathrm{m\,s^{-2}}$ perpendicular to its velocity. Find its instantaneous Liénard power.
- An electric dipole has peak moment $p_0=3.00\times10^{-12}\ \mathrm{C\,m}$ and frequency $50.0\ \mathrm{MHz}$. Find its average radiated power.
- A current element has peak current $I_0=2.00\ \mathrm A$, length $\ell=2.00\ \mathrm{cm}$, and frequency $300\ \mathrm{MHz}$. Find its average radiated power and radiation resistance.
- Six identical elements form a linear array with $d=\lambda/2$ and feed phase $\delta=-\pi/3$. Find the principal maximum, its two adjacent nulls, and the peak value of $\lvert AF\rvert^2$.
Answers: 1. $2.283\times10^{-19}\ \mathrm W$; 2. $1.101\times10^{-21}\ \mathrm W$; 3. $9.748\times10^{-6}\ \mathrm W$; 4. $0.6321\ \mathrm W$, $R_{\rm r}=0.3160\ \Omega$; 5. principal maximum at $\theta=70.529^\circ$, adjacent nulls at $48.190^\circ$ and $90.000^\circ$, and $\lvert AF\rvert^2_{\max}=36$.
The angular integrals, half-power root, radiation-resistance relation, array factor, and all printed numerical answers are checked in the Maxima worksheet; every printed residual is zero.
References
- Larmor formula
- David J. Griffiths, Introduction to Electrodynamics, 4th ed., Chapter 11, “Radiation.”
- John D. Jackson, Classical Electrodynamics, 3rd ed., Chapter 14, “Radiation by Moving Charges.”
- Constantine A. Balanis, Antenna Theory: Analysis and Design, 4th ed., Chapter 4, “Linear Wire Antennas.”
Discussion