# Radiation far zone expansion

1. May 10, 2015

### decerto

So using the lorenz gauge and maxwells equations we find that both the scalar and vector potentials obey the wave equation with a charge/current density source.

$\Box \phi= \frac{\rho}{\epsilon}$
$\Box\vec{A} = \mu \vec{J}$

So using the greens function for the wave equation we can compute the scalar or vector potential given either a charge or current density.

$\phi(x,t)=\frac{1}{4\pi \epsilon}\int \frac{\delta(t'-[t-\frac{|x-x'|}{c}])}{|x-x'|}\rho(x,t) dt'\ d^3x'$
$\vec{A}(x,t)=\frac{\mu}{4\pi}\int \frac{\delta(t'-[t-\frac{|x-x'|}{c}])}{|x-x'|}\vec{J}(x,t) dt'\ d^3x'$

If the charge/current densities are sinusoidally time dependent and using the retarded condition from the greens function we get a spatially oscillating term in the integral solution for the potentials, this term can then be expanded in the 'far zone'.

$\rho(x,t)=\rho(x)e^{-i\omega t}$
$\vec{A}(x,t)=\vec{A}(x)e^{-i\omega t}$

$\phi(x,t)=\frac{1}{4\pi \epsilon}\int \frac{\delta(t'-[t-\frac{|x-x'|}{c}])}{|x-x'|}\rho(x')e^{-i\omega t'} dt'\ d^3x'$
$\vec{A}(x,t)=\frac{\mu}{4\pi}\int \frac{\delta(t'-[t-\frac{|x-x'|}{c}])}{|x-x'|}\vec{J}(x')e^{-i\omega t'}dt'\ d^3x'$

$\phi(x,t)=\frac{1}{4\pi \epsilon}\int \frac{1}{|x-x'|}\rho(x')e^{-i\omega( t-\frac{|x-x'|}{c})} d^3x'$
$\vec{A}(x,t)=\frac{\mu}{4\pi}\int \frac{1}{|x-x'|}\vec{J}(x')e^{-i\omega (t-\frac{|x-x'|}{c})} d^3x'$

Making the time dependence implicit

$\phi(x)=\frac{1}{4\pi \epsilon}\int \frac{e^{ik|x-x'|}}{|x-x'|}\rho(x') d^3x'$
$\vec{A}(x)=\frac{\mu}{4\pi}\int \frac{e^{ik|x-x'|}}{|x-x'|}\vec{J}(x') d^3x'$

Far zone $|x-x'| \approx |x|-\frac{x \cdot x'}{|x|}=r-\hat{n}\cdot x'$ for exponential and $|x-x'|=|x|=r$ for the denominator

$\phi(x)=\frac{1}{4\pi \epsilon}\frac{e^{ikr}}{r}\int e^{ik\hat{n}\cdot x'}\rho(x') d^3x'$
$\vec{A}(x)=\frac{\mu}{4\pi}\frac{e^{ikr}}{r}\int e^{ik\hat{n}\cdot x'}\vec{J}(x') d^3x'$

$\phi(x)=\frac{1}{4\pi \epsilon}\frac{e^{ikr}}{r}\sum_{n=0}^\infty\frac{(-ik)^n}{n!}\int (\hat{n} \cdot x')^n \rho(x') d^3x'$
$\vec{A}(x)=\frac{\mu}{4\pi}\frac{e^{ikr}}{r}\sum_{n=0}^\infty\frac{(-ik)^n}{n!}\int (\hat{n} \cdot x')^n \vec{J}(x') d^3x'$

My questions are about these expansions, for the scalar potential expansion the 1st term gives

$\phi(x)=\frac{1}{4\pi \epsilon}\frac{e^{ikr}}{r}\int \rho(x') d^3x'$

To me this looks like a monopole wave but jackson in section 9.1 for anyone who has the book explicitly states the monopole part of the the potential is necessarily static(spatially static) by considering $|x-x'|=r$

$\phi(x,t)=\frac{1}{4\pi \epsilon}\int \frac{\delta(t'-[t-\frac{r}{c}])}{r}\rho(x,t) dt'\ d^3x'$

$\phi(x,t)=\frac{q(t'=t-\frac{r}{c})}{4\pi \epsilon}$

I don't understand where this comes from and it appears to contradict the 1st term in the expansion.

My other question is about the the second term in the scalar expansion and the first term in the vector expansion given by

$\phi(x)=\frac{-ik}{4\pi \epsilon}\frac{e^{ikr}}{r}\hat{n} \cdot \int x' \rho(x') d^3x'$
$\vec{A}(x)=\frac{\mu}{4\pi}\frac{e^{ikr}}{r}\int\vec{J}(x') d^3x'$

By the continuity equation and integration by parts this can be shown to be

$\vec{A}(x)=-\frac{i\omega\mu}{4\pi}\frac{e^{ikr}}{r}\vec{p}$
$\phi(x)=-\frac{ik}{4\pi \epsilon}\frac{e^{ikr}}{r}\hat{n} \cdot \vec{p}$

with $\vec{p}=\int x' \rho(x') d^3x'$

So there is obviously a lot of symmetry between these two equations $\omega$ and $k$, $\vec{p}$ and $\hat{n} \cdot \vec{p}$, $\mu$ and $\frac{1}{\epsilon}$

Is this symmetry perserved for all paired terms in both expansions like this, what do the higher order expansion terms look like and if I did this type of expansion in covariant notation would these be the part of the same term essentially?

2. May 11, 2015

### fzero

There is indeed a $1/r$ missing in the text (http://www-theory.lbl.gov/jdj/Errata(2010).pdf). It does not change the discussion about the static nature, since that just refers to the time independence.

The symmetry between the numerical coefficients is indeed a property of the Lorentz invariance of electrodynamics. The fact that both terms are dipoles is more a property of the multipole expansion of the Green function (9.98). The fact that the dipole comes in as the first term of the expansion of the vector potential vs the 2nd term for the scalar potential is due to the vector vs scalar nature of the objects. At higher orders, the structure is dictated by the polar symmetry of the spherical harmonics. I believe the quadropole moment is given somewhere in Ch 9. Some explicit forms for higher mulipole moments appears in http://zon8.physd.amu.edu.pl/historia/kielich-publ/129.pdf

3. May 11, 2015

### decerto

What does the first term of the scalar expansion correspond to then?

4. May 11, 2015

### fzero

It's the same static monopolar potential that you'd find if the charge configuration was static and spherically symmetric.

5. May 12, 2015

### decerto

But it has a spatially oscillating term $e^{ikr}$ and a temporal term $e^{i\omega t}$?