Changes

:<math>\square^2_rG(r,r')=\delta^4(r-r')</math><br>

:<math>\square^2_rG(r,r')=\delta^4(r-r')</math><br>

and its Fourier transform

and its Fourier transform

:<math> \tilde{G} (q) = \frac{1}{(2\pi)^2} \int d^4r e^{-iq\cdot r} G(r,0)</math>

:<math> \tilde{G} (q) = \frac{1}{(2\pi)^2} \int d^4r\, e^{-iq\cdot r} G(r,0)</math>

:<math> G(r,0)=\frac{1}{(2\pi)^2} \int d^4qe^{iq\cdot r} \tilde{G}(q)</math><br>

:<math> G(r,0)=\frac{1}{(2\pi)^2} \int d^4q\, e^{iq\cdot r} \tilde{G}(q)</math><br>

Translational symmetry implies

Translational symmetry implies

:<math>G(r-r',0)=G(r,r')</math>

:<math>G(r-r',0)=G(r,r')\quad</math>

&there4;<math> G(r,r')=\frac{1}{(2\pi)^2}\int d^4q e^{iq\cdot (r-r')} \tilde{G} (q)</math><br> <math>\square^2_rG(r,r')=\frac{1}{(2\pi)^2}\int d^4qe^{iq\cdot (r-r')}\tilde{G}(q)</math><br><br>

<math>\square^2_rG(r,r')=\frac{1}{(2\pi)^2}\int d^4qe^{iq\cdot (r-r')}(-k^2+\frac{\omega^2}{c^2})</math>, where <math>q=(\mathbf{k},\frac{\omega}{c}) \quad</math><br><br>

:<math> G(r,r')=\frac{1}{(2\pi)^2}\int d^4q\, e^{iq\cdot (r-r')} \tilde{G} (q)</math><br>

But, <math>\square^2_rG(r,r')=\delta^4(r-r')=\frac{1}{(2\pi)^4}\int d^4q e^{iq\cdot (r-r')}</math><br><br>

:<math>\square^2_rG(r,r')=\frac{1}{(2\pi)^2}\int d^4q\,(-q^2)e^{iq\cdot (r-r')}\tilde{G}(q)</math><br>

&there4;<math>\tilde{G}(q)=\frac{(2\pi)^2}{(2\pi)^4}\frac{1}{-q^2}= \frac{-1}{(2\pi)^2q^2}</math><br><br>

:<math>\square^2_rG(r,r')=\frac{1}{(2\pi)^2}\int d^4q\, e^{iq\cdot (r-r')}(-k^2+\frac{\omega^2}{c^2})</math>

<math>G(r,r')=\frac{-1}{(2\pi)^4} \int d^4qe^{iq\cdot (r-r')} \frac{1}{(k+\frac{\omega}{c})(k-\frac{\omega}{c})}</math><br><br>

where <math>q=(\mathbf{k},\frac{\omega}{c})</math>.  But

Chose the "retarded" solution, such that the function is zero unless t>t'<br><br>

:<math>\square^2_rG(r,r')=\delta^4(r-r')=\frac{1}{(2\pi)^4}\int d^4q\, e^{iq\cdot (r-r')}</math><br>

<math>G(r,r')=\frac{1}{(2\pi)^4}\int d^3ke^{i\mathbf{k}\cdot (r-r')}\int d(\frac{\omega}{c}) \frac{e^{-i\omega(t-t')}}{(\frac{\omega}{c}-k)(\frac{\omega}{c}+k)}\Theta(t-t')</math><br><br>

:<math>\tilde{G}(q)=\frac{(2\pi)^2}{(2\pi)^4}\frac{1}{-q^2}= \frac{-1}{(2\pi)^2q^2}</math><br>

<math>=\frac{1}{(2\pi)^4}\int d^3ke^{i\mathbf{k}\cdot (r-r')}(2\pi i  \frac{e^{ick(t-t')}-e^{-ick(t-t')}}{2k})\Theta</math><br><br>

:<math>G(r,r')=\frac{-1}{(2\pi)^4} \int d^4q\, e^{iq\cdot (r-r')} \frac{1}{(k+\frac{\omega}{c})(k-\frac{\omega}{c})}</math><br>

<math>=\frac{-2\pi}{(2\pi)^4}\int_0 \frac{k^2dk}{k} \sin\left({ck(t-t')}\right) 2\pi\int_{-1}^1 dze^{ik|\mathbf{r}-\mathbf{r'}|z}\Theta</math><br><br>

Chose the "retarded" solution, such that the function is zero unless t>t'.<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>G(r,r')=\frac{1}{(2\pi)^4}\int d^3k\, e^{i\mathbf{k}\cdot (r-r')}\int d(\frac{\omega}{c}) \frac{e^{-i\omega(t-t')}}{(\frac{\omega}{c}-k)(\frac{\omega}{c}+k)}\Theta(t-t')</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><br>

::<math>=\frac{1}{(2\pi)^4}\int d^3k\, e^{i\mathbf{k}\cdot (r-r')}(2\pi i  \frac{e^{ick(t-t')}-e^{-ick(t-t')}}{2k})\Theta</math><br><br>

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

::<math>=\frac{-2\pi}{(2\pi)^4}\int_0 \frac{k^2dk}{k} \sin\left({ck(t-t')}\right) 2\pi\int_{-1}^1 dze^{ik|\mathbf{r}-\mathbf{r'}|z}\Theta</math><br><br>

&there4;<math>  G(r,r')=\frac{-1}{4\pi}\quad \frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}</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>

Now to get the <math>G_1(r,r')\quad </math> in the half-space with z>0 with the boundary condition <math>G_1\quad </math> at<math> r_3=z=0  \quad</math> we take the difference:<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>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}'+2z'\hat{e_3}|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'+2z'\hat{e_3}|}\right)</math><br><br>

 

Now use Green's theorem:<br><br>

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

Let  <math>F^\mu=A(r)\part_\mu G_1(r,r')-G_1(r,r')\part_\mu A(r)</math><br><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 to get the <math>G_1(r,r')\quad </math> in the half-space with z>0 with the boundary condition <math>G_1\quad </math> at<math> r_3=z=0  \quad</math> &nbsp; we take the difference:<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}'+2z'\hat{e_3}|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'+2z'\hat{e_3}|}\right)</math><br><br>

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>\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>

But  <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>