Changes

The gauge condition for the Lorentz gauge is

The gauge condition for the Lorentz gauge is

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

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

Introduce the Green's function at <math> r=(\mathbf{r},t)</math> from some impulse source at <math> r'=(\mathbf{r}',t')</math>.

Introduce the Green's function at <math> r=(\mathbf{r},t)</math> from some impulse source at <math> r'=(\mathbf{r}',t')</math>

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

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

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

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

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

Translational symmetry implies:<br><br>

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

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

Translational symmetry implies<br><br>

:<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>

&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>