Changes

Where:

Where:

<math>A^\mu = (\mathbf{A},\frac{\Phi} {c}), \square^2=\part_\mu \part^\mu = \nabla^2 - \frac{1}{c^2} \frac{\part}{\part t^2}</math>

<math>A^\mu = (\mathbf{A},\frac{\Phi} {c}), \square^2=\part_\mu \part^\mu = \nabla^2 - \frac{1}{c^2} \frac{\part}{\part t^2}</math>

<math>j^\mu = (\mathbf{j},c\rho), \part_\mu= (\mathbf{\nabla}, \frac{1}{c} \frac{\part}{\part t})</math>

<math>j^\mu = (\mathbf{j},c\rho), \part_\mu= (\mathbf{\nabla}, \frac{1}{c} \frac{\part}{\part t})</math><br><br>

 

Lorentz Gauge: <math>A^\mu = 0 \rArr \mathbf{\nabla} \cdot \mathbf{A}+\frac{1}{c^2} \frac{\part\Phi}{\part t}=0</math><br><br>

Lorentz Gauge: <math>A^\mu = 0 \rArr \mathbf{\nabla} \cdot \mathbf{A}+\frac{1}{c^2} \frac{\part\Phi}{\part t}=0</math>

Introduce Green's function at (x=t) from some impulse source at x'=(<b>x'</b>,t')<br><br>

 

<math>\square^2_xG(x,x')=\delta^4(x-x')</math><br><br>

Introduce Green's function at (x=t) from some impulse source at x'=(<b>x'</b>,t')<br>

Let <math> \tilde{G} (q) = \frac{1}{(2\pi)^2} \int d^4x e^{iqx} G(x,0)</math><br><br>

<math>\square^2_xG(x,x')=\delta^4(x-x')</math>

Then <math> G(q)=\frac{1}{(2\pi)^2} \int d^4qe^{iqx} \tilde{G}(x,0)</math><br><br>

 

In free space, translational symmetry implies:<br><br>

Let <math> \tilde{G} (q) = \frac{1}{(2\pi)^2} \int d^4x e^{iqx} G(x,0)</math>

<math>G(x-x',0)=G(x,x') \quad </math><br><br>

 

Then <math> G(q)=\frac{1}{(2\pi)^2} \int d^4qe^{iqx} \tilde{G}(x,0)</math>

 

In free space, translational symmetry implies:<br>

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

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

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

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

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

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>

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

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