Changes

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>

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

Translational symmetry implies

Translational symmetry implies

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

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

<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>\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>