局域化的時變電荷 和電流密度 在真空中是電磁波 的源。在有源的情形下，麥克斯韋方程組 可以寫成一個非齊次的電磁波方程 （英文：Inhomogeneous electromagnetic wave equation ）的形式，正是因為波源的存在使得偏微分方程 變為非齊次。
真空中的麥克斯韋方程組在含有電荷
ρ
{\displaystyle \rho }
J
{\displaystyle \mathbf {J} }
矢勢 和標勢 表示為
∇
2
φ
+
∂
∂
t
(
∇
⋅
A
)
=
−
ρ
ε
0
{\displaystyle \nabla ^{2}\varphi +{{\partial } \over \partial t}\left(\nabla \cdot \mathbf {A} \right)=-{\rho \over \varepsilon _{0}}}
∇
2
A
−
1
c
2
∂
2
A
∂
t
2
−
∇
(
1
c
2
∂
φ
∂
t
+
∇
⋅
A
)
=
−
μ
0
J
{\displaystyle \nabla ^{2}\mathbf {A} -{1 \over c^{2}}{\partial ^{2}\mathbf {A} \over \partial t^{2}}-\nabla \left({1 \over c^{2}}{{\partial \varphi } \over {\partial t}}+\nabla \cdot \mathbf {A} \right)=-\mu _{0}\mathbf {J} }
此時電場和磁場分別為
E
=
−
∇
φ
−
∂
A
∂
t
{\displaystyle \mathbf {E} =-\nabla \varphi -{\partial \mathbf {A} \over \partial t}}
以及
B
=
∇
×
A
{\displaystyle \mathbf {B} =\nabla \times \mathbf {A} }
如果加上洛倫茨規範條件
1
c
2
∂
φ
∂
t
+
∇
⋅
A
=
0
{\displaystyle {1 \over c^{2}}{{\partial \varphi } \over {\partial t}}+\nabla \cdot \mathbf {A} =0}
則非齊次的波動方程 為
∇
2
φ
−
1
c
2
∂
2
φ
∂
t
2
=
−
ρ
ε
0
{\displaystyle \nabla ^{2}\varphi -{1 \over c^{2}}{\partial ^{2}\varphi \over \partial t^{2}}=-{\rho \over \varepsilon _{0}}}
∇
2
A
−
1
c
2
∂
2
A
∂
t
2
=
−
μ
0
J
{\displaystyle \nabla ^{2}\mathbf {A} -{1 \over c^{2}}{\partial ^{2}\mathbf {A} \over \partial t^{2}}=-\mu _{0}\mathbf {J} }
在厘米-克-秒制 下，方程的形式為
∇
2
φ
−
1
c
2
∂
2
φ
∂
t
2
=
−
4
π
ρ
{\displaystyle \nabla ^{2}\varphi -{1 \over c^{2}}{\partial ^{2}\varphi \over \partial t^{2}}=-{4\pi \rho }}
∇
2
A
−
1
c
2
∂
2
A
∂
t
2
=
−
4
π
c
J
{\displaystyle \nabla ^{2}\mathbf {A} -{1 \over c^{2}}{\partial ^{2}\mathbf {A} \over \partial t^{2}}=-{4\pi \over c}\mathbf {J} }
電場和磁場的形式為
E
=
−
∇
φ
−
1
c
∂
A
∂
t
{\displaystyle \mathbf {E} =-\nabla \varphi -{1 \over c}{\partial \mathbf {A} \over \partial t}}
B
=
∇
×
A
{\displaystyle \mathbf {B} =\nabla \times \mathbf {A} }
洛倫茨規範條件為
1
c
∂
φ
∂
t
+
∇
⋅
A
=
0
{\displaystyle {1 \over c}{{\partial \varphi } \over {\partial t}}+\nabla \cdot \mathbf {A} =0}
如果採取有時在高維相對論場合計算中使用的洛倫茲-赫維賽德單位制 ，電荷和電流密度需要從厘米-克-秒制變換為
ρ
→
ρ
4
π
{\displaystyle \rho \rightarrow {\rho \over {4\pi }}}
J
→
1
4
π
J
{\displaystyle \mathbf {J} \rightarrow {1 \over {4\pi }}\mathbf {J} }
在狹義相對論 中，麥克斯韋方程組可以寫成協變 的形式：
◻
A
μ
=
d
e
f
∂
β
∂
β
A
μ
=
d
e
f
A
μ
,
β
β
=
−
μ
0
J
μ
{\displaystyle \Box A^{\mu }\ {\stackrel {\mathrm {def} }{=}}\ \partial _{\beta }\partial ^{\beta }A^{\mu }\ {\stackrel {\mathrm {def} }{=}}\ {A^{\mu ,\beta }}_{\beta }=-\mu _{0}J^{\mu }}
◻
A
μ
=
d
e
f
∂
β
∂
β
A
μ
=
d
e
f
A
μ
,
β
β
=
−
4
π
c
J
μ
{\displaystyle \Box A^{\mu }\ {\stackrel {\mathrm {def} }{=}}\ \partial _{\beta }\partial ^{\beta }A^{\mu }\ {\stackrel {\mathrm {def} }{=}}\ {A^{\mu ,\beta }}_{\beta }=-{\frac {4\pi }{c}}J^{\mu }}
（厘米-克-秒制）
其中
J
μ
{\displaystyle J^{\mu }\,}
四維電流密度 ：
J
μ
=
(
c
ρ
,
J
)
{\displaystyle J^{\mu }=\left(c\rho ,\mathbf {J} \right)}
∂
∂
x
a
=
d
e
f
∂
a
=
d
e
f
,
a
=
d
e
f
(
∂
/
∂
c
t
,
∇
)
{\displaystyle {\partial \over {\partial x^{a}}}\ {\stackrel {\mathrm {def} }{=}}\ \partial _{a}\ {\stackrel {\mathrm {def} }{=}}\ {}_{,a}\ {\stackrel {\mathrm {def} }{=}}\ (\partial /\partial ct,\nabla )}
是四維梯度 ，而電磁四維勢 為
A
μ
=
(
φ
,
A
c
)
{\displaystyle A^{\mu }=(\varphi ,\mathbf {A} c)}
A
μ
=
(
φ
,
A
)
{\displaystyle A^{\mu }=(\varphi ,\mathbf {A} )}
洛倫茨規範為
∂
μ
A
μ
=
0
{\displaystyle \partial _{\mu }A^{\mu }=0}
這裡
◻
=
∂
β
∂
β
=
∇
2
−
1
c
2
∂
2
∂
t
2
{\displaystyle \Box =\partial _{\beta }\partial ^{\beta }=\nabla ^{2}-{1 \over c^{2}}{\frac {\partial ^{2}}{\partial t^{2}}}}
達朗貝爾算符 。電磁波方程在彎曲時空中需要做兩處修正，分別是偏導數被替換為協變導數 ，以及增加了一項有關時空曲率的項。在國際單位制下
−
A
α
;
β
β
+
R
α
β
A
β
=
μ
0
J
α
{\displaystyle -{A^{\alpha ;\beta }}_{\beta }+{R^{\alpha }}_{\beta }A^{\beta }=\mu _{0}J^{\alpha }}
其中
R
α
β
{\displaystyle {R^{\alpha }}_{\beta }}
是里奇曲率張量 。 這裡分號表示對角標求協變導數。對於厘米-克-秒制下的方程，需要用
4
π
/
c
{\displaystyle 4\pi /c}
真空磁導率 。
這裡假設洛倫茨規範在彎曲時空中的推廣為
A
μ
;
μ
=
0
{\displaystyle {A^{\mu }}_{;\mu }=0}
在波源周圍沒有邊界條件 的情形下，非齊次波方程在厘米-克-秒制下的解為
φ
(
r
,
t
)
=
∫
δ
(
t
′
+
|
r
−
r
′
|
c
−
t
)
|
r
−
r
′
|
ρ
(
r
′
,
t
′
)
d
3
r
′
d
t
′
{\displaystyle \varphi (\mathbf {r} ,t)=\int {{\delta \left(t'+{{\left|\mathbf {r} -\mathbf {r} '\right|} \over c}-t\right)} \over {\left|\mathbf {r} -\mathbf {r} '\right|}}\rho (\mathbf {r} ',t')d^{3}r'dt'}
以及
A
(
r
,
t
)
=
∫
δ
(
t
′
+
|
r
−
r
′
|
c
−
t
)
|
r
−
r
′
|
J
(
r
′
,
t
′
)
c
d
3
r
′
d
t
′
{\displaystyle \mathbf {A} (\mathbf {r} ,t)=\int {{\delta \left(t'+{{\left|\mathbf {r} -\mathbf {r} '\right|} \over c}-t\right)} \over {\left|\mathbf {r} -\mathbf {r} '\right|}}{\mathbf {J} (\mathbf {r} ',t') \over c}d^{3}r'dt'}
其中
δ
(
t
′
+
|
r
−
r
′
|
c
−
t
)
{\displaystyle {\delta \left(t'+{{\left|\mathbf {r} -\mathbf {r} '\right|} \over c}-t\right)}}
是狄拉克δ函數 。
對於國際單位制，
ρ
→
ρ
4
π
ε
0
{\displaystyle \rho \rightarrow {\rho \over {4\pi \varepsilon _{0}}}}
J
→
μ
0
4
π
J
{\displaystyle \mathbf {J} \rightarrow {\mu _{0} \over {4\pi }}\mathbf {J} }
對於洛倫茲-赫維賽德單位制，
ρ
→
ρ
4
π
{\displaystyle \rho \rightarrow {\rho \over {4\pi }}}
J
→
1
4
π
J
{\displaystyle \mathbf {J} \rightarrow {1 \over {4\pi }}\mathbf {J} }
這些解被稱作推遲解，它們表示的是一族由波源向外發出的並從現在向未來傳播的球面電磁波的線性疊加 。
此外還有所謂超前解，表示為
φ
(
r
,
t
)
=
∫
δ
(
t
′
−
|
r
−
r
′
|
c
−
t
)
|
r
−
r
′
|
ρ
(
r
′
,
t
′
)
d
3
r
′
d
t
′
{\displaystyle \varphi (\mathbf {r} ,t)=\int {{\delta \left(t'-{{\left|\mathbf {r} -\mathbf {r} '\right|} \over c}-t\right)} \over {\left|\mathbf {r} -\mathbf {r} '\right|}}\rho (\mathbf {r} ',t')d^{3}r'dt'}
以及
A
(
r
,
t
)
=
∫
δ
(
t
′
−
|
r
−
r
′
|
c
−
t
)
|
r
−
r
′
|
J
(
r
′
,
t
′
)
c
d
3
r
′
d
t
′
{\displaystyle \mathbf {A} (\mathbf {r} ,t)=\int {{\delta \left(t'-{{\left|\mathbf {r} -\mathbf {r} '\right|} \over c}-t\right)} \over {\left|\mathbf {r} -\mathbf {r} '\right|}}{\mathbf {J} (\mathbf {r} ',t') \over c}d^{3}r'dt'}
它們表示的是一族由波源向外發出的並從未來向現在傳播的球面電磁波的線性疊加。
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