We start off with Maxwell's Equation in the Lorentz gauge: ◻ 2 A μ ( x , t ) = ◻ 2 A μ ( x ) = ( − μ 1 j μ ( x ) ) {\displaystyle \square ^{2}A^{\mu }(\mathbf {x} ,t)=\square ^{2}A^{\mu }(x)=(-\mu _{1}j^{\mu }(x))}
Where: A μ = ( A , Φ c ) , ◻ 2 = ∂ μ ∂ μ = ∇ 2 − 1 c 2 ∂ ∂ t 2 {\displaystyle A^{\mu }=(\mathbf {A} ,{\frac {\Phi }{c}}),\square ^{2}=\partial _{\mu }\partial ^{\mu }=\nabla ^{2}-{\frac {1}{c^{2}}}{\frac {\partial }{\partial t^{2}}}} j μ = ( j , c ρ ) , ∂ μ = ( ∇ , 1 c ∂ ∂ t ) {\displaystyle j^{\mu }=(\mathbf {j} ,c\rho ),\partial _{\mu }=(\mathbf {\nabla } ,{\frac {1}{c}}{\frac {\partial }{\partial t}})}
Lorentz Gauge: A μ = 0 ⇒ ∇ ⋅ A + 1 c 2 ∂ Φ ∂ t = 0 {\displaystyle A^{\mu }=0\Rightarrow \mathbf {\nabla } \cdot \mathbf {A} +{\frac {1}{c^{2}}}{\frac {\partial \Phi }{\partial t}}=0}
Introduce Green's function at (x=t) from some impulse source at x'=(x',t') ◻ x 2 G ( x , x ′ ) = δ 4 ( x − x ′ ) {\displaystyle \square _{x}^{2}G(x,x')=\delta ^{4}(x-x')}
Let G ~ ( q ) = 1 ( 2 π ) 2 ∫ d 4 x e i q x G ( x , 0 ) {\displaystyle {\tilde {G}}(q)={\frac {1}{(2\pi )^{2}}}\int d^{4}xe^{iqx}G(x,0)}
Then G ( q ) = 1 ( 2 π ) 2 ∫ d 4 q e i q x G ~ ( x , 0 ) {\displaystyle G(q)={\frac {1}{(2\pi )^{2}}}\int d^{4}qe^{iqx}{\tilde {G}}(x,0)}
In free space, translational symmetry implies: G ( x − x ′ , 0 ) = G ( x , x ′ ) {\displaystyle G(x-x',0)=G(x,x')\quad }
∴ G ( x , x ′ ) = 1 ( 2 π ) 2 ∫ d 4 q e − i q ( x − x ′ ) G ~ ( q ) {\displaystyle G(x,x')={\frac {1}{(2\pi )^{2}}}\int d^{4}qe^{-iq(x-x')}{\tilde {G}}(q)} ◻ x 2 G ( x , x ′ ) = 1 ( 2 π ) 2 | i n t d 4 q e − i q ( x − x ′ ) G ~ ( q ) {\displaystyle \square _{x}^{2}G(x,x')={\frac {1}{(2\pi )^{2}}}|intd^{4}qe^{-iq(x-x')}{\tilde {G}}(q)} ◻ x 2 G ( x , x ′ ) = 1 ( 2 π ) 2 ∫ d 4 q e − i q ( x − x ′ ) ( − k 2 + ω 2 c 2 ) {\displaystyle \square _{x}^{2}G(x,x')={\frac {1}{(2\pi )^{2}}}\int d^{4}qe^{-iq(x-x')}(-k^{2}+{\frac {\omega ^{2}}{c^{2}}})} , where q = ( k , ω c ) {\displaystyle q=(\mathbf {k} ,{\frac {\omega }{c}})\quad } But, ◻ x 2 G ( x , x ′ ) = δ 4 ( x − x ′ ) = 1 ( 2 π ) 4 ∫ d 4 q e − i q ( x − x ′ ) {\displaystyle \square _{x}^{2}G(x,x')=\delta ^{4}(x-x')={\frac {1}{(2\pi )^{4}}}\int d^{4}qe^{-iq(x-x')}} &there4 G ~ ( q ) = ( 2 π ) 2 ( 2 π ) 4 1 − q 2 = − 1 ( 2 π ) 2 q 2 {\displaystyle {\tilde {G}}(q)={\frac {(2\pi )^{2}}{(2\pi )^{4}}}{\frac {1}{-q^{2}}}={\frac {-1}{(2\pi )^{2}q^{2}}}}
, where 
