Changes

The last term vanishes if G₁(x,x')and A(x) fall off sufficiently fast at <math>t\rightarrow\infin</math>. They do.  So:<br>

The last term vanishes if G₁(x,x')and A(x) fall off sufficiently fast at <math>t\rightarrow\infin</math>. They do.  So:<br>

<math>\int \nabla \cdot \mathbf{F} d^4x=A(x')</math><br>

<math>\int \nabla \cdot \mathbf{F} d^4x=A(x')</math><br><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><br>

<math>A(x')=-\int d^2x\int cdt\left[A(x)\frac{\part}{\part t}G_1(x,x')-G_1(x,x')\frac{\part}{\part z}A(x)\right]</math>, where the last term is zero by the constriction of<math>G_1(z=0) \quad</math><br><br>

<math>A(x')=-\int d^2x\int cdt\left[A(x)\frac{\part}{\part t}G_1(x,x')-G_1(x,x')\frac{\part}{\part z}A(x)\right]</math>, where the last term is zero by the constriction of<math>G_1(z=0) \quad</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><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><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>

<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>