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}}})}
q
=
(
k
,
ω
c
)
{\displaystyle q=(\mathbf {k} ,{\frac {\omega }{c}})\quad }
◻
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')}}
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}}}}
G
(
x
,
x
′
)
=
−
1
(
2
π
)
4
∫
d
4
q
e
−
i
q
(
x
−
x
′
)
1
(
k
+
ω
c
)
(
k
−
ω
c
)
{\displaystyle G(x,x')={\frac {-1}{(2\pi )^{4}}}\int d^{4}qe^{-iq(x-x')}{\frac {1}{(k+{\frac {\omega }{c}})(k-{\frac {\omega }{c}})}}}
G
(
x
,
x
′
)
=
1
(
2
π
)
4
∫
d
3
k
e
−
i
k
(
x
−
x
′
)
∫
d
(
ω
c
)
e
i
ω
(
t
−
t
′
)
(
ω
c
−
k
)
(
ω
c
+
k
)
Θ
{\displaystyle G(x,x')={\frac {1}{(2\pi )^{4}}}\int d^{3}ke^{-i\mathbf {k} (x-x')}\int d({\frac {\omega }{c}}){\frac {e^{i\omega (t-t')}}{({\frac {\omega }{c}}-k)({\frac {\omega }{c}}+k)}}\Theta }
=
1
(
2
π
)
4
∫
d
3
k
e
−
i
k
(
x
−
x
′
)
(
2
π
i
e
i
c
k
(
t
−
t
′
)
−
e
−
i
c
k
(
t
−
t
′
)
2
k
)
Θ
{\displaystyle ={\frac {1}{(2\pi )^{4}}}\int d^{3}ke^{-i\mathbf {k} (x-x')}(2\pi i{\frac {e^{ick(t-t')}-e^{-ick(t-t')}}{2k}})\Theta }
=
−
2
π
(
2
π
)
4
∫
k
2
d
k
k
sin
(
c
k
(
t
−
t
′
)
)
2
π
∫
−
i
i
d
z
e
−
i
k
|
x
−
x
′
|
z
Θ
{\displaystyle ={\frac {-2\pi }{(2\pi )^{4}}}\int {\frac {k^{2}dk}{k}}\sin \left({ck(t-t')}\right)2\pi \int _{-i}^{i}dze^{-ik|\mathbf {x} -\mathbf {x'} |z}\Theta }
=
−
1
(
2
π
)
2
(
1
2
i
|
x
−
x
′
|
)
2
∫
d
k
s
i
n
(
c
k
(
t
−
t
′
)
)
s
i
n
(
k
|
x
−
x
′
|
)
Θ
{\displaystyle ={\frac {-1}{(2\pi )^{2}}}\left({\frac {1}{2i|\mathbf {x} -\mathbf {x'|} }}\right)2\int dksin(ck(t-t'))sin(k|\mathbf {x} -\mathbf {x'} |)\Theta }
=
1
(
2
π
)
2
2
|
x
−
x
′
|
2
π
4
[
2
δ
(
|
x
−
x
′
|
+
c
(
t
−
t
′
)
)
−
2
δ
(
|
x
−
x
′
|
−
c
(
t
−
t
′
)
)
]
Θ
{\displaystyle ={\frac {1}{(2\pi )^{2}}}{\frac {2}{|\mathbf {x} -\mathbf {x} '|}}{\frac {2\pi }{4}}\left[2\delta (|\mathbf {x} -\mathbf {x} '|+c(t-t'))-2\delta (|\mathbf {x} -\mathbf {x} '|-c(t-t'))\right]\Theta }
2
δ
(
|
x
−
x
′
|
+
c
(
t
−
t
′
)
)
→
0
∀
t
>
t
′
{\displaystyle 2\delta (|\mathbf {x} -\mathbf {x} '|+c(t-t'))\rightarrow 0\quad \forall \quad t>t'}
G
(
x
,
x
′
)
=
−
1
4
π
δ
(
|
x
−
x
′
|
−
c
(
t
−
t
′
)
)
|
x
−
x
′
|
{\displaystyle G(x,x')={\frac {-1}{4\pi }}\quad {\frac {\delta (|\mathbf {x} -\mathbf {x} '|-c(t-t'))}{|\mathbf {x} -\mathbf {x} '|}}}
1 (x,x') in the half-space with z>0 with the boundary condition G1 at x3 =z=0 we take the difference:
G
1
(
x
,
x
′
)
=
−
1
4
π
(
δ
(
|
x
−
x
′
|
−
c
(
t
−
t
′
)
)
|
x
−
x
′
|
−
δ
(
|
x
−
x
′
−
2
z
e
3
^
|
−
c
(
t
−
t
′
)
)
|
x
−
x
′
−
2
z
e
3
^
|
)
{\displaystyle 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} '-2z{\hat {e_{3}}}|-c(t-t'))}{|\mathbf {x} -\mathbf {x} '-2z{\hat {e_{3}}}|}}\right)}
F
=
A
(
x
)
∇
G
1
(
x
,
x
′
)
−
G
1
(
x
,
x
′
)
∇
A
(
x
)
{\displaystyle \mathbf {F} =A(x)\mathbf {\nabla } G_{1}(x,x')-G_{1}(x,x')\mathbf {\nabla } A(x)}
∫
∇
⋅
F
d
4
x
=
∫
c
d
t
∫
d
3
x
[
∇
A
⋅
∇
G
+
A
∇
2
G
1
−
∇
G
⋅
∇
A
−
G
1
∇
2
A
]
{\displaystyle \int \mathbf {\nabla } \cdot \mathbf {F} d^{4}x=\int cdt\int d^{3}x[\mathbf {\nabla } A\cdot \mathbf {\nabla } G+A\nabla ^{2}G_{1}-\mathbf {\nabla } G\cdot \mathbf {\nabla } A-G_{1}\nabla ^{2}A]}
∇
2
G
1
(
x
,
x
′
)
=
δ
4
(
x
−
x
′
)
+
1
c
2
∂
2
∂
t
2
G
1
(
x
,
x
′
)
{\displaystyle \nabla ^{2}G_{1}(x,x')=\delta ^{4}(x-x')+{\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}G_{1}(x,x')}
∇
2
A
(
x
)
=
μ
j
(
x
)
+
1
c
2
∂
2
∂
t
2
A
(
x
)
{\displaystyle \nabla ^{2}A(x)=\mu j(x)+{\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}A(x)}
j
(
x
)
=
0
{\displaystyle j(x)=0\quad }
∫
∇
⋅
F
d
4
x
=
A
(
x
′
)
+
1
c
2
∫
d
4
x
[
A
∂
2
∂
t
2
G
1
−
G
1
∂
2
∂
t
2
A
]
{\displaystyle \int \nabla \cdot \mathbf {F} d^{4}x=A(x')+{\frac {1}{c^{2}}}\int d^{4}x\left[A{\frac {\partial ^{2}}{\partial t^{2}}}G_{1}-G_{1}{\frac {\partial ^{2}}{\partial t^{2}}}A\right]}
1 (x,x')and A(x) fall off sufficiently fast at
t
→
∞
{\displaystyle t\rightarrow \infty }
∫
∇
⋅
F
d
4
x
=
A
(
x
′
)
{\displaystyle \int \nabla \cdot \mathbf {F} d^{4}x=A(x')}
z
>
0
{\displaystyle z>0\quad }
A
(
x
′
)
=
−
∫
d
2
x
∫
c
d
t
[
A
(
x
)
∂
∂
t
G
1
(
x
,
x
′
)
−
G
1
(
x
,
x
′
)
∂
∂
z
A
(
x
)
]
{\displaystyle A(x')=-\int d^{2}x\int cdt\left[A(x){\frac {\partial }{\partial t}}G_{1}(x,x')-G_{1}(x,x'){\frac {\partial }{\partial z}}A(x)\right]}
G
1
(
z
=
0
)
{\displaystyle G_{1}(z=0)\quad }
Failed to parse (syntax error): {\displaystyle 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> <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)}
, where
x
″
=
x
′
−
2
z
′
e
3
^
{\displaystyle \mathbf {x} ''=\mathbf {x} '-2z'{\hat {e_{3}}}}
![{\displaystyle ={\frac {1}{(2\pi )^{2}}}{\frac {2}{|\mathbf {x} -\mathbf {x} '|}}{\frac {2\pi }{4}}\left[2\delta (|\mathbf {x} -\mathbf {x} '|+c(t-t'))-2\delta (|\mathbf {x} -\mathbf {x} '|-c(t-t'))\right]\Theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/49a5589bd5af5cc12be1c4a9741ef26549f7b6c1)
![{\displaystyle \int \mathbf {\nabla } \cdot \mathbf {F} d^{4}x=\int cdt\int d^{3}x[\mathbf {\nabla } A\cdot \mathbf {\nabla } G+A\nabla ^{2}G_{1}-\mathbf {\nabla } G\cdot \mathbf {\nabla } A-G_{1}\nabla ^{2}A]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f5fe0f18d836935d1cc0d1e9a8b3ce7268539426)
![{\displaystyle \int \nabla \cdot \mathbf {F} d^{4}x=A(x')+{\frac {1}{c^{2}}}\int d^{4}x\left[A{\frac {\partial ^{2}}{\partial t^{2}}}G_{1}-G_{1}{\frac {\partial ^{2}}{\partial t^{2}}}A\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5fcea9b36150eabba8a50f39f374da0533b34e68)
![{\displaystyle A(x')=-\int d^{2}x\int cdt\left[A(x){\frac {\partial }{\partial t}}G_{1}(x,x')-G_{1}(x,x'){\frac {\partial }{\partial z}}A(x)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8bfdcf3960be5fabffc33340cfee023ee0608735)
