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.(adsbygoogle = window.adsbygoogle || []).push({});

##\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?

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