Changes

::<math>=\frac{-1}{(2\pi)^2}\left(\frac{1}{|\mathbf{r}-\mathbf{r'|}}\right)2\int_0 dk \sin(ck(t-t')) \sin(k|\mathbf{r}-\mathbf{r'}|)\Theta</math><br><br>

::<math>=\frac{-1}{(2\pi)^2}\left(\frac{1}{|\mathbf{r}-\mathbf{r'|}}\right)2\int_0 dk \sin(ck(t-t')) \sin(k|\mathbf{r}-\mathbf{r'}|)\Theta</math><br><br>

::<math>=\frac{1}{(2\pi)^2}\frac{2}{|\mathbf{r}-\mathbf{r}'|}\frac{2\pi}{4} \left[\delta(|\mathbf{r}-\mathbf{r}'|+c(t-t'))-\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))\right]\Theta</math><br>

::<math>=\frac{1}{(2\pi)^2}\frac{2}{|\mathbf{r}-\mathbf{r}'|}\frac{2\pi}{4} \left[\delta(|\mathbf{r}-\mathbf{r}'|+c(t-t'))-\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))\right]\Theta</math><br>

 

But the term <math>\delta(|\mathbf{r}-\mathbf{r}'|+c(t-t'))\rightarrow 0 \quad\forall\quad t>t'</math> so that

But the term <math>\delta(|\mathbf{r}-\mathbf{r}'|+c(t-t'))\rightarrow 0 \quad\forall\quad t>t'</math> so that

:<math>  G(r,r')=\frac{-1}{4\pi}\quad \frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}</math><br>

:<math>  G(r,r')=\frac{-1}{4\pi}\quad \frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}</math><br>

Now use Green's theorem, with the generating function

Now use Green's theorem, with the generating function

:<math>F^\mu=A(r)\part_\mu G_1(r,r')-G_1(r,r')\part_\mu A(r)</math><br><br>

:<math>F^\mu=A(r)\part_\mu G_1(r,r')-G_1(r,r')\part_\mu A(r)</math><br><br>

<math>\int \part_\mu F_\mu d^4r= \int cdt \int d^3r[\part_\mu A \part^\mu G+A\part_\mu \part^\mu G_1-\part_\mu G \part^\mu A -G_1\part_\mu \part^\mu A]</math><br><br>

:<math>\int \part_\mu F_\mu d^4r= \int cdt \int d^3r[\part_\mu A \part^\mu G+A\part_\mu \part^\mu G_1-\part_\mu G \part^\mu A -G_1\part_\mu \part^\mu A]</math><br><br>

:<math>\part_\mu \part^\mu G_1(r,r')=\delta^4(r-r')</math><br><br>

:<math>\part_\mu \part^\mu G_1(r,r')=\delta^4(r-r')</math><br><br>

<math>\part_\mu \part^\mu A(r)= \mu j(r)</math>, let  <math>j(r)=0 \quad</math><br><br>

:<math>\part_\mu \part^\mu A(r)= \mu j(r)</math>, let  <math>j(r)=0 \quad</math><br><br>

<math>\int \part_\mu F_\mu d^4r=A(r')</math><br><br>

:<math>\int \part_\mu F_\mu d^4r=A(r')</math><br><br>

Now invoke the divergence theorem on the half space <math>z>0 \quad</math>:<br><br>

Now invoke the divergence theorem on the half space <math>z>0 \quad</math>:<br><br>

<math>A(r')=-\int d^2r\int cdt\left[A(r)\frac{\part}{\part z}G_1(r,r')-G_1(r,r')\frac{\part}{\part z}A(r)\right]</math>, where the last term is zero by the condition of<math>G_1(z=0,r')=0 \quad</math><br><br>

:<math>A(r')=-\int d^2r\int cdt\left[A(r)\frac{\part}{\part z}G_1(r,r')-G_1(r,r')\frac{\part}{\part z}A(r)\right]</math>, where the last term is zero by the condition of<math>G_1(z=0,r')=0 \quad</math><br><br>

<math>A(r')=-c\int dt\int d^2rA(r)\frac{\part}{\part z}G_1(r,r')</math><br><br>

:<math>A(r')=-c\int dt\int d^2rA(r)\frac{\part}{\part z}G_1(r,r')</math><br><br>

To do the t integral, I need to bring out the z derivative.  To do this, I first turn it into a z' derivative, using the relation: <br><br><br>

To do the t integral, I need to bring out the z derivative.  To do this, I first turn it into a z' derivative, using the relation: <br><br><br>

<math>G_1(r,r')=\frac{-1}{4\pi}\left(\frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}-\frac{\delta(|\mathbf{r}-\mathbf{r}''|-c(t-t'))}{|\mathbf{r}-\mathbf{r}''|}\right)</math>, where <math>\mathbf{r}''=\mathbf{r}'-2z'\hat{e_3}</math><br><br>

:<math>G_1(r,r')=\frac{-1}{4\pi}\left(\frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}-\frac{\delta(|\mathbf{r}-\mathbf{r}''|-c(t-t'))}{|\mathbf{r}-\mathbf{r}''|}\right)</math>, where :<math>\mathbf{r}''=\mathbf{r}'-2z'\hat{e_3}</math><br><br>

<math>\frac{\part}{\part z}G_1(r,r')=\frac{-1}{4\pi}\left(\frac{\part}{\part z}\left(\frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}-\frac{\delta(|\mathbf{r}-\mathbf{r}''|-c(t-t'))}{|\mathbf{r}-\mathbf{r}''|}\right)\right)</math><br><br>

:<math>\frac{\part}{\part z}G_1(r,r')=\frac{-1}{4\pi}\left(\frac{\part}{\part z}\left(\frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}-\frac{\delta(|\mathbf{r}-\mathbf{r}''|-c(t-t'))}{|\mathbf{r}-\mathbf{r}''|}\right)\right)</math><br><br>

&there4; <math>A(r')=\frac{-1}{4\pi}\frac{\part}{\part z'}\int_{z=0} d^2r\left(2\frac{A(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c})}{|\mathbf{r}-\mathbf{r}'|}\right)</math><br><br>

&there4; <math>A(r')=\frac{-1}{4\pi}\frac{\part}{\part z'}\int_{z=0} d^2r\left(2\frac{A(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c})}{|\mathbf{r}-\mathbf{r}'|}\right)</math><br><br>

At <math>z=0 \quad </math>, <math>|\mathbf{r}-\mathbf{r}'|=\sqrt{r^2+z'^2}=S, dS=\frac{rdr}{\sqrt{r^2+z'^2}}</math><br><br>

At <math>z=0 \quad </math>, <math>|\mathbf{r}-\mathbf{r}'|=\sqrt{r^2+z'^2}=S, dS=\frac{rdr}{\sqrt{r^2+z'^2}}</math><br><br>