next up previous
Up: PHY531-PHY532 Classical Electrodynamics

Radiation from a localized current source

Kevin Schmidt
Department of Physics and Astronomy
Arizona State University
Tempe, AZ

1 General Results

If a localized current is given as a function of space and time, the vector potential in the Lorentz gauge is

$\displaystyle\mbox{\boldmath$ {A} $}$($\displaystyle\mbox{\boldmath$ {r} $}$,t) = $\displaystyle{\textstyle\frac{1}{c}}$$\displaystyle\int$d 3r'$\displaystyle{\frac{\mbox{\boldmath$ {J} $}(\mbox{\boldmath$ {r'} $}, t - c^{-1... ... {r'} $}\vert)}{ \vert\mbox{\boldmath$ {r} $} - \mbox{\boldmath$ {r'} $}\vert}}$ . (1)

The retardation is often simpler if the current is Fourier transformed in time,

$\displaystyle\mbox{\boldmath$ {J} $}$$\scriptstyle\omega$($\displaystyle\mbox{\boldmath$ {r} $}$) = $\displaystyle\int_{-\infty}^{\infty}$dt$\displaystyle\mbox{\boldmath$ {J} $}$($\displaystyle\mbox{\boldmath$ {r} $}$,t)e i$\scriptstyle\omega$t (2)

so that

$\displaystyle\mbox{\boldmath$ {J} $}$($\displaystyle\mbox{\boldmath$ {r} $}$,t) = $\displaystyle\int_{-\infty}^{\infty}$$\displaystyle{\frac{d\omega}{2\pi}}$$\displaystyle\mbox{\boldmath$ {J} $}$$\scriptstyle\omega$($\displaystyle\mbox{\boldmath$ {r} $}$)e - i$\scriptstyle\omega$t . (3)

Since the current is real, the complex conjugate of the last equation shows that $\mbox{\boldmath$\space {J} $}$$\scriptstyle\omega$($\mbox{\boldmath$\space {r} $}$) = $\mbox{\boldmath$\space {J} $}$*- $\scriptstyle\omega$($\mbox{\boldmath$\space {r} $}$) . The current can be written in terms of positive frequencies alone as

$\displaystyle\mbox{\boldmath$ {J} $}$($\displaystyle\mbox{\boldmath$ {r} $}$,t) = Re $\displaystyle\int_{0}^{\infty}$$\displaystyle{\frac{d\omega}{\pi}}$$\displaystyle\mbox{\boldmath$ {J} $}$$\scriptstyle\omega$($\displaystyle\mbox{\boldmath$ {r} $}$)e - i$\scriptstyle\omega$t (4)

which therefore simply looks like a sum (i.e. the integral) of terms each with an e - i$\scriptstyle\omega$t time dependence.

Plugging this form in, the vector potential is

$\displaystyle\mbox{\boldmath$ {A} $}$($\displaystyle\mbox{\boldmath$ {r} $}$,t) = Re $\displaystyle\int_{0}^{\infty}$$\displaystyle{\frac{d\omega}{\pi}}$e - i$\scriptstyle\omega$t$\displaystyle\left [ \int d^3r' \frac{\mbox{\boldmath$ {J} $}_\omega(\mbox{\bol... ...ert}}{c \vert\mbox{\boldmath$ {r} $} - \mbox{\boldmath$ {r'} $}\vert} \right]\,$. (5)

To extract the radiation fields, the distance from the source to observation point is expanded assuming the source is localized and the observation point goes to r $\rightarrow$ $\infty$ . The vector potential is

$\displaystyle\lim_{r \rightarrow \infty}^{}$$\displaystyle\mbox{\boldmath$ {A} $}$($\displaystyle\mbox{\boldmath$ {r} $}$,t) = Re $\displaystyle\int_{0}^{\infty}$$\displaystyle{\frac{d\omega}{\pi}}$$\displaystyle{\frac{e^{i \frac{\omega}{c} r - i \omega t}}{rc}}$$\displaystyle\int$d 3r'$\displaystyle\mbox{\boldmath$ {J} $}$$\scriptstyle\omega$($\displaystyle\mbox{\boldmath$ {r'} $}$)e - i$\scriptstyle{\frac{\omega}{c}}$$\scriptstyle\mbox{\boldmath$ {\hat {r}} $}$ $\scriptstyle\cdot$ $\scriptstyle\mbox{\boldmath$ {r'} $}$ (6)

Notice that the current integral is just the ``on energy shell'' space-time Fourier transform of the current. That is defining the Fourier transform as

$\displaystyle\mbox{\boldmath$ {\tilde J} $}$($\displaystyle\mbox{\boldmath$ {k} $}$,$\displaystyle\omega$) = $\displaystyle\int$d 3rdt$\displaystyle\mbox{\boldmath$ {J} $}$($\displaystyle\mbox{\boldmath$ {r} $}$,t)e - i$\scriptstyle\mbox{\boldmath$ {k} $}$ $\scriptstyle\cdot$ $\scriptstyle\mbox{\boldmath$ {r} $}$ + i$\scriptstyle\omega$t (7)

the vector potential in the radiation zone is

$\displaystyle\lim_{r \rightarrow \infty}^{}$$\displaystyle\mbox{\boldmath$ {A} $}$($\displaystyle\mbox{\boldmath$ {r} $}$,t) = Re $\displaystyle\int_{0}^{\infty}$$\displaystyle{\frac{d\omega}{\pi}}$$\displaystyle{\frac{e^{i \frac{\omega}{c} r - i \omega t}}{rc}}$$\displaystyle\mbox{\boldmath$ {\tilde J} $}$$\displaystyle\left (\frac{\omega}{c} {\mbox{\boldmath$ {\hat {r}} $}}, \omega \right)$ (8)

There are two cases of interest. The first is where the current (or at least the radiating part) is localized in time. In that case the total radiated energy in a small frequency range is finite. This of course is the only physical situation. However, in many cases, the motion is periodic for a very long time, and it is the energy radiated per cycle or equivalently the power radiated that is desired. If the motion continues indefinitely, the total energy radiated is infinite, so the mathematics needs to be slightly modified as shown in the spectral resolution notes.

The magnetic field in the radiation zone is given by taking the curl of $\mbox{\boldmath$\space {A} $}$ , and to give a r - 1 dependence, the gradient must operate on the exponent. Similarly, the electric field is given for a harmonic time dependence as

$\displaystyle\mbox{\boldmath$ {\nabla} $}$ x $\displaystyle\mbox{\boldmath$ {B} $}$$\scriptstyle\omega$ = i$\displaystyle{\frac{\omega }{c}}$$\displaystyle\mbox{\boldmath$ {E} $}$$\scriptstyle\omega$ . (9)

The fields are therefore
$\displaystyle\lim_{r \rightarrow \infty}^{}$$\displaystyle\mbox{\boldmath$ {B} $}$($\displaystyle\mbox{\boldmath$ {r} $}$,t) = Re $\displaystyle\int_{0}^{\infty}$$\displaystyle{\frac{d\omega}{\pi}}$$\displaystyle{\frac{e^{i \frac{\omega}{c} r - i \omega t}}{rc}}$$\displaystyle\left [-i \frac{\omega}{c} {\mbox{\boldmath$ {\hat {r}} $}} \times... ...eft (\frac{\omega}{c} {\mbox{\boldmath$ {\hat {r}} $}}, \omega \right ) \right]$   
$\displaystyle\lim_{r \rightarrow \infty}^{}$$\displaystyle\mbox{\boldmath$ {E} $}$($\displaystyle\mbox{\boldmath$ {r} $}$,t) = Re $\displaystyle\int_{0}^{\infty}$$\displaystyle{\frac{d\omega}{\pi}}$$\displaystyle{\frac{e^{i \frac{\omega}{c} r - i \omega t}}{rc}}$$\displaystyle\left [-i \frac{\omega}{c} {\mbox{\boldmath$ {\hat {r}} $}} \times... ...c{\omega}{c} {\mbox{\boldmath$ {\hat {r}} $}}, \omega \right ) \right ) \right]$ (10)

The $\mbox{\boldmath$\space {\hat {r}} $}$ x ($\mbox{\boldmath$\space {\hat {r}} $}$ x $\mbox{\boldmath$\space {\tilde J} $}$) is the transform of the transverse component of the current. The Poynting vector is ${\frac{c}{4\pi}}$$\mbox{\boldmath$\space {E} $}$ x $\mbox{\boldmath$\space {B} $}$ , and with direction along $\mbox{\boldmath$\space {\hat {r}} $}$ and magnitude given by c|$\mbox{\boldmath$\space {E} $}$|2/4$\pi$ .

As shown in the spectral resolution notes, if the time integral of the product of two functions of time

f (t) = Re $\displaystyle\int_{0}^{\infty}$$\displaystyle{\frac{d\omega}{\pi}}$$\displaystyle\tilde{f}$($\displaystyle\omega$)   
g(t) = Re $\displaystyle\int_{0}^{\infty}$$\displaystyle{\frac{d\omega}{\pi}}$$\displaystyle\tilde{g}$($\displaystyle\omega$) (11)
is called I , the differential I per frequency range d$\omega$ is

$\displaystyle{\frac{dI}{d\omega}}$ = $\displaystyle{\textstyle\frac{1}{\pi}}$Re $\displaystyle\tilde{f}$($\displaystyle\omega$)$\displaystyle\tilde{g}^{*}_{}$($\displaystyle\omega$) . (12)

so that the energy radiated in frequency interval d$\omega$ in solid angle increment d$\Omega$ around $\mbox{\boldmath$\space {\hat {r}} $}$ is

$\displaystyle{\frac{d^2I}{d\omega d\Omega}}$ = $\displaystyle{\frac{\omega^2 }{4\pi^2 c^3}}$$\displaystyle\left \vert {\mbox{\boldmath$ {\hat {\epsilon}} $}}^* \cdot \mbox{... ...\frac{\omega}{c} {\mbox{\boldmath$ {\hat {r}} $}}, \omega \right ) \right\vert^$2 (13)

where $\mbox{\boldmath$\space {\hat {\epsilon}} $}$ is the detected polarization which must be perpendicular to $\mbox{\boldmath$\space {\hat {r}} $}$. Summing over polarizations will gives the total energy radiated in the frequency interval and solid angle interval. Compare this result to Jackson Eq. 14.70.

Again applying results of the spectral resolution notes to the case where the motion is periodic with period T = ${\frac{2\pi}{\omega_0}}$ , the power radiated per solid angle at the frequency n$\omega_{0}^{}$ is

$\displaystyle{\frac{dP_n}{d\Omega}}$ = $\displaystyle{\frac{n^2 \omega^4_0}{(2\pi c)^3}}$$\displaystyle\left \vert {\mbox{\boldmath$ {\hat {\epsilon}} $}}^* \cdot \mbox{... ...ft (\frac{n \omega_0}{c} {\mbox{\boldmath$ {\hat {r}} $}} \right ) \right\vert^$2 (14)

where

$\displaystyle\tilde{J}_{n}^{}$($\displaystyle\mbox{\boldmath$ {k} $}$) = $\displaystyle\int_{0}^{2\pi/\omega_0}$dt$\displaystyle\int$d 3rJ($\displaystyle\mbox{\boldmath$ {r} $}$,t)e - i$\scriptstyle\mbox{\boldmath$ {k} $}$ $\scriptstyle\cdot$ $\scriptstyle\mbox{\boldmath$ {r} $}$ + in$\scriptstyle\omega_{0}$t . (15)

Compare this result to that contained in Jackson Problem 14.13.

For reference, the spectral resolution of signals localized in time is given in Jackson section 14.5. The relationship between dPn/d$\Omega$ and d 2I/d$\omega$d$\Omega$ is given a physical meaning on Jackson page 681 around Eq. 14.92. Essentially, if you imagine that you have a signal that repeats with fundamental frequency $\omega_{0}^{}$ , but in each period is localized to time much smaller than the period, then the intensity of radiation for one period would be given by d 2I/d$\omega$d$\Omega$ where the current integration would be over just one period. Since the motion repeats, the power radiated in a frequency n$\omega_{0}^{}$ is the energy radiated at that frequency $\omega$ = n$\omega_{0}^{}$ divided by the period 2$\pi$/$\omega_{0}^{}$ to get the power, and multiplied by $\omega_{0}^{}$ to convert from d$\omega$ to dn . The overall factor is $\omega_{0}^{2}$/2$\pi$ which is the difference in the prefactors of the two expressions above.

2 Long Wavelength Approximations

If the current is localized in a region or a set regions, each which has spatial extent much smaller than a wavelength, $\lambda$ = 2$\pi$c/$\omega$ , the origin of integration can be translated to the approximate center of a region, and the exponential expanded in a power series. The terms in the power series are roughly powers of the extent of the region of nonzero current divided by the wavelength. The exponential becomes

e - i$\scriptstyle{\frac{\omega}{c}}$$\scriptstyle\mbox{\boldmath$ {\hat {r}} $}$ $\scriptstyle\cdot$ $\scriptstyle\mbox{\boldmath$ {r'} $}$ = 1 - i$\displaystyle{\frac{\omega }{c}}$$\displaystyle\mbox{\boldmath$ {\hat {r}} $}$ $\displaystyle\cdot$ $\displaystyle\mbox{\boldmath$ {r'} $}$ + ... (16)

and the current integrals needed to this order are
I (0)k = $\displaystyle\int$d 3r'J$\scriptstyle\omega$ k($\displaystyle\mbox{\boldmath$ {r'} $}$)   
I (1)jk = $\displaystyle\int$d 3r'x'jJ$\scriptstyle\omega$ k($\displaystyle\mbox{\boldmath$ {r'} $}$) . (17)
Just as in magnetostatics, some integrations by parts using the divergence theorem are useful,
$\displaystyle\mbox{\boldmath$ {\nabla} $}$ $\displaystyle\cdot$ (xk$\displaystyle\mbox{\boldmath$ {J} $}$$\scriptstyle\omega$) = xk$\displaystyle\mbox{\boldmath$ {\nabla} $}$ $\displaystyle\cdot$ $\displaystyle\mbox{\boldmath$ {J} $}$$\scriptstyle\omega$ + J$\scriptstyle\omega$ k   
  = i$\displaystyle\omega$xk$\displaystyle\rho_{\omega}^{}$ + J$\scriptstyle\omega$ k   
$\displaystyle\mbox{\boldmath$ {\nabla} $}$ $\displaystyle\cdot$ (xjxk$\displaystyle\mbox{\boldmath$ {J} $}$$\scriptstyle\omega$) = xjxk$\displaystyle\mbox{\boldmath$ {\nabla} $}$ $\displaystyle\cdot$ $\displaystyle\mbox{\boldmath$ {J} $}$$\scriptstyle\omega$ + [xjJ$\scriptstyle\omega$ k + xkJ$\scriptstyle\omega$ j]   
  = i$\displaystyle\omega$xjxk$\displaystyle\rho_{\omega}^{}$ + [xjJ$\scriptstyle\omega$ k + xkJ$\scriptstyle\omega$ j] . (18)
The symmetry in j and k of the last term suggests writing I (1) as

I (1)jk = $\displaystyle{\textstyle\frac{1}{2}}$$\displaystyle\left \{ \int d^3r'[ x'_j J_{\omega ~k}(\mbox{\boldmath$ {r'} $}) ... ...ox{\boldmath$ {r'} $}) - x'_k J_{\omega ~j}(\mbox{\boldmath$ {r'} $})] \right\}$ (19)

and the integrals become
I (0)k = - i$\displaystyle\omega$$\displaystyle\int$d 3r'x'k$\displaystyle\rho_{\omega}^{}$($\displaystyle\mbox{\boldmath$ {r'} $}$) = - i$\displaystyle\omega$d$\scriptstyle\omega$ k   
I (1)jk = $\displaystyle{\textstyle\frac{1}{2}}$$\displaystyle\left \{ -i \omega \int d^3r' x'_j x'_k \rho_\omega(\mbox{\boldmat... ...ox{\boldmath$ {r'} $}) - x'_j J_{\omega ~k}(\mbox{\boldmath$ {r'} $})] \right\}$ . (20)
The I (0)k integral is proportional to the k component of the electric dipole moment. The I (1)jk has two terms. The second looks like the magnetic dipole moment

$\displaystyle\int$d 3r'[x'jJ$\scriptstyle\omega$ k($\displaystyle\mbox{\boldmath$ {r'} $}$) - x'jJ$\scriptstyle\omega$ k($\displaystyle\mbox{\boldmath$ {r'} $}$)] = 2c$\displaystyle\epsilon_{jk\ell}^{}$m$\scriptstyle\omega$$\scriptstyle\ell$ . (21)

Noting that in the field equation, one component is dotted with $\mbox{\boldmath$\space {\hat {r}} $}$, while the other is crossed with $\mbox{\boldmath$\space {\hat {r}} $}$, a constant diagonal element can be added to it without changing the result. That is

$\displaystyle\sum_{ji}^{}$rjri$\displaystyle\epsilon_{ik\ell}^{}$$\displaystyle\mbox{\boldmath$ {\hat {x}} $}$$\scriptstyle\ell$$\displaystyle\delta_{jk}^{}$ = 0 . (22)

so that the replacement

$\displaystyle\int$d 3r'x'jx'k$\displaystyle\rho_{\omega}^{}$($\displaystyle\mbox{\boldmath$ {r'} $}$) $\displaystyle\rightarrow$ $\displaystyle{\textstyle\frac{1}{3}}$$\displaystyle\int$d 3r'$\displaystyle\left (3 x'_j x'_k - \delta_{jk} \right)\rho$$\scriptstyle\omega$($\displaystyle\mbox{\boldmath$ {r'} $}$) = $\displaystyle{\textstyle\frac{1}{3}}$Q$\scriptstyle\omega$ jk (23)

does not change the fields. The latter expression is the quadrupole moment Qjk . The three terms are then identified as the electric dipole, magnetic dipole and quadrupole terms. Explicitly we can make the replacements
$\displaystyle\mbox{\boldmath$ {\tilde J} $}$$\displaystyle\left (\frac{\omega}{c} {\mbox{\boldmath$ {\hat {r}} $}}, \omega \right)$ = - i$\displaystyle\omega$$\displaystyle\mbox{\boldmath$ {d} $}$$\scriptstyle\omega$ + i$\displaystyle\omega$$\displaystyle\mbox{\boldmath$ {\hat {r}} $}$ x $\displaystyle\mbox{\boldmath$ {m} $}$$\scriptstyle\omega$ - $\displaystyle{\frac{\omega^2}{6c}}$$\displaystyle\;\stackrel{\leftrightarrow}{Q}\;$$\scriptstyle\omega$ $\displaystyle\cdot$ $\displaystyle\mbox{\boldmath$ {\hat {r}} $}$ + ... (24)
or
$\displaystyle\mbox{\boldmath$ {\tilde J} $}$n$\displaystyle\left (\frac{n \omega_0}{c} {\mbox{\boldmath$ {\hat {r}} $}}\right)$ = - in$\displaystyle\omega_{0}^{}$$\displaystyle\mbox{\boldmath$ {d} $}$n + in$\displaystyle\omega_{0}^{}$$\displaystyle\mbox{\boldmath$ {\hat {r}} $}$ x $\displaystyle\mbox{\boldmath$ {m} $}$n - $\displaystyle{\frac{n^2 \omega^2_0}{6c}}$$\displaystyle\;\stackrel{\leftrightarrow}{Q}\;$n $\displaystyle\cdot$ $\displaystyle\mbox{\boldmath$ {\hat {r}} $}$ + ... (25)
where the subscript n indicates the Fourier integral over one period of $\omega_{0}^{}$ of e in$\scriptstyle\omega_{0}$t.

As a check against Jackson's results, look at the case where only the diagonal quadrupole terms contribute with Q33 = Q0 , and Q11 = Q22 = - Q0/2 each oscillating as cos($\omega_{0}^{}$t) , so that only the n = 1 term survives, and the time integration gives

$\displaystyle\;\stackrel{\leftrightarrow}{Q}\;$1 = $\displaystyle{\frac{\pi Q_0}{2 \omega_0}}$$\displaystyle\left [2{\mbox{\boldmath$ {\hat {z}} $}} {\mbox{\boldmath$ {\hat {... ...}} -{\mbox{\boldmath$ {\hat {y}} $}} {\mbox{\boldmath$ {\hat {y}} $}} \right]\,$. (26)

The current transform is then
$\displaystyle\mbox{\boldmath$ {\tilde J} $}$1$\displaystyle\left (\frac{\omega_0}{c} {\mbox{\boldmath$ {\hat {r}} $}}\right)$ = - $\displaystyle{\frac{\pi \omega Q_0}{12 c}}$$\displaystyle\left [2 \cos \theta {\mbox{\boldmath$ {\hat {z}} $}} - \sin \thet... ...{\hat {x}} $}} - \sin \theta \sin \phi {\mbox{\boldmath$ {\hat {y}} $}} \right]$ (27)
The power radiated per solid angle is
$\displaystyle{\frac{dP_1}{d\Omega}}$ = $\displaystyle{\frac{\omega_0^6 Q_0^2}{1152 \pi c^5}}$$\displaystyle\left \vert {\mbox{\boldmath$ {\hat {r}} $}} \times \left [2 \cos ... ... - \sin \theta \sin \phi {\mbox{\boldmath$ {\hat {y}} $}} \right ] \right\vert^$2   
  = $\displaystyle{\frac{\omega_0^6 Q_0^2}{1152 \pi c^5}}$$\displaystyle\left [ 4 \cos^2 \theta + \sin^2 \theta - \left ( 2 \cos^2 \theta - \sin^2 \theta \right )^2 \right]$   
  = $\displaystyle{\frac{\omega_0^6 Q_0^2}{128 \pi c^5}}$$\displaystyle\cos^{2}_{}$$\displaystyle\theta$$\displaystyle\sin^{2}_{}$$\displaystyle\theta$ (28)
which agrees with Jackson Eq. 9.51 after unit conversion.

next up previous
Up: PHY531-PHY532 Classical Electrodynamics