引起時空彎曲 物理學 中，彎曲時空中的馬克士威方程組 （Maxwell's equations in curved spacetime）制約著彎曲時空 （其間的度規 可能不是閔考斯基性 的）中的電磁場 的動力學。它們可以被認為是真空中的馬克士威方程組 在廣義相對論 框架中的擴展，而真空中的馬克士威方程組只是一般化的馬克士威方程組在局部平直時空中的特殊形式。但由於在廣義相對論中電磁場本身的存在也會引起時空的彎曲，因此真空中的馬克士威方程組應被理解為一種出於方便的近似形式。
然而，這種形式的馬克士威方程組僅僅對真空情形下的馬克士威方程組有用，這也被稱作「微觀」馬克士威方程組。對於宏觀上與各向異性的物質相關的馬克士威方程組，物質的存在會建立一個參考系從而使方程組不再是協變 的。
閱讀本條目需要讀者了解平直時空中電磁理論的四維形式 。
電磁場本身要求其幾何描述與坐標選取無關，而馬克士威方程組在任何時空中的幾何描述都是一樣的，而不管這個時空是否是平直的。同時，當使用非笛卡爾 的局部坐標時平直閔考斯基空間中的方程組會做同樣的修改。例如本條目中方程組可以寫成球坐標 中的馬克士威方程組的形式。基於上述原因，更好的理解方法是將閔考斯基空間中的馬克士威方程組理解為一種特殊形式，而非將彎曲時空中的馬克士威方程組理解為一種相對論化的推廣。
廣義相對論 中真空中的電磁理論的方程式為
F
α
β
=
∂
α
A
β
−
∂
β
A
α
{\displaystyle F_{\alpha \beta }\,=\,\partial _{\alpha }A_{\beta }\,-\,\partial _{\beta }A_{\alpha }\,}
D
μ
ν
=
1
μ
0
g
μ
α
F
α
β
g
β
ν
−
g
{\displaystyle {\mathcal {D}}^{\mu \nu }\,=\,{\frac {1}{\mu _{0}}}\,g^{\mu \alpha }\,F_{\alpha \beta }\,g^{\beta \nu }\,{\sqrt {-g}}\,}
J
μ
=
∂
ν
D
μ
ν
{\displaystyle J^{\mu }\,=\,\partial _{\nu }{\mathcal {D}}^{\mu \nu }\,}
f
μ
=
F
μ
ν
J
ν
{\displaystyle f_{\mu }\,=\,F_{\mu \nu }\,J^{\nu }\,}
其中
g
α
β
{\displaystyle g^{\alpha \beta }\,}
g
α
β
{\displaystyle g_{\alpha \beta }\,}
g
{\displaystyle g\,}
度規張量 的行列式，
A
α
{\displaystyle A_{\alpha }\,}
四維勢 ，
F
α
β
{\displaystyle F^{\alpha \beta }\,}
電磁場的四維協變張量 ，
D
μ
ν
{\displaystyle D^{\mu \nu }\,}
位移電流 張量，
f
μ
{\displaystyle f_{\mu }\,}
勞侖茲力 的密度，
J
μ
{\displaystyle J_{\mu }\,}
四維電流密度 。儘管方程組中使用了偏導數 ，這些方程式仍然在任意曲面坐標變換下是協變的。也就是說如果將偏導數換成協變導數 ，引入的附加項會自動消去從而保持形式不變。
電磁場的四維勢
A
α
,
{\displaystyle A_{\alpha }\,,}
A
¯
β
=
∂
x
γ
∂
x
¯
β
A
γ
.
{\displaystyle {\bar {A}}_{\beta }={\frac {\partial x^{\gamma }}{\partial {\bar {x}}^{\beta }}}A_{\gamma }\,.}
電磁場 是一個協變的二階反對稱張量，它用電磁勢可以定義為
F
α
β
=
∂
α
A
β
−
∂
β
A
α
.
{\displaystyle F_{\alpha \beta }\,=\,\partial _{\alpha }A_{\beta }\,-\,\partial _{\beta }A_{\alpha }\,.}
為證明它的勞侖茲不變性 ，我們對其進行坐標變換
F
¯
α
β
=
∂
A
¯
β
∂
x
¯
α
−
∂
A
¯
α
∂
x
¯
β
{\displaystyle {\bar {F}}_{\alpha \beta }\,=\,{\frac {\partial {\bar {A}}_{\beta }}{\partial {\bar {x}}^{\alpha }}}\,-\,{\frac {\partial {\bar {A}}_{\alpha }}{\partial {\bar {x}}^{\beta }}}\,}
=
∂
∂
x
¯
α
(
∂
x
γ
∂
x
¯
β
A
γ
)
−
∂
∂
x
¯
β
(
∂
x
δ
∂
x
¯
α
A
δ
)
{\displaystyle =\,{\frac {\partial }{\partial {\bar {x}}^{\alpha }}}\left({\frac {\partial x^{\gamma }}{\partial {\bar {x}}^{\beta }}}A_{\gamma }\right)\,-\,{\frac {\partial }{\partial {\bar {x}}^{\beta }}}\left({\frac {\partial x^{\delta }}{\partial {\bar {x}}^{\alpha }}}A_{\delta }\right)\,}
=
∂
2
x
γ
∂
x
¯
α
∂
x
¯
β
A
γ
+
∂
x
γ
∂
x
¯
β
∂
A
γ
∂
x
¯
α
−
∂
2
x
δ
∂
x
¯
β
∂
x
¯
α
A
δ
−
∂
x
δ
∂
x
¯
α
∂
A
δ
∂
x
¯
β
{\displaystyle =\,{\frac {\partial ^{2}x^{\gamma }}{\partial {\bar {x}}^{\alpha }\,\partial {\bar {x}}^{\beta }}}A_{\gamma }\,+\,{\frac {\partial x^{\gamma }}{\partial {\bar {x}}^{\beta }}}{\frac {\partial A_{\gamma }}{\partial {\bar {x}}^{\alpha }}}\,-\,{\frac {\partial ^{2}x^{\delta }}{\partial {\bar {x}}^{\beta }\,\partial {\bar {x}}^{\alpha }}}A_{\delta }\,-\,{\frac {\partial x^{\delta }}{\partial {\bar {x}}^{\alpha }}}{\frac {\partial A_{\delta }}{\partial {\bar {x}}^{\beta }}}\,}
=
∂
x
γ
∂
x
¯
β
∂
x
δ
∂
x
¯
α
∂
A
γ
∂
x
δ
−
∂
x
δ
∂
x
¯
α
∂
x
γ
∂
x
¯
β
∂
A
δ
∂
x
γ
{\displaystyle =\,{\frac {\partial x^{\gamma }}{\partial {\bar {x}}^{\beta }}}{\frac {\partial x^{\delta }}{\partial {\bar {x}}^{\alpha }}}{\frac {\partial A_{\gamma }}{\partial x^{\delta }}}\,-\,{\frac {\partial x^{\delta }}{\partial {\bar {x}}^{\alpha }}}{\frac {\partial x^{\gamma }}{\partial {\bar {x}}^{\beta }}}{\frac {\partial A_{\delta }}{\partial x^{\gamma }}}\,}
=
∂
x
δ
∂
x
¯
α
∂
x
γ
∂
x
¯
β
(
∂
A
γ
∂
x
δ
−
∂
A
δ
∂
x
γ
)
{\displaystyle =\,{\frac {\partial x^{\delta }}{\partial {\bar {x}}^{\alpha }}}{\frac {\partial x^{\gamma }}{\partial {\bar {x}}^{\beta }}}\,\left({\frac {\partial A_{\gamma }}{\partial x^{\delta }}}\,-\,{\frac {\partial A_{\delta }}{\partial x^{\gamma }}}\right)\,}
=
∂
x
δ
∂
x
¯
α
∂
x
γ
∂
x
¯
β
F
δ
γ
.
{\displaystyle =\,{\frac {\partial x^{\delta }}{\partial {\bar {x}}^{\alpha }}}{\frac {\partial x^{\gamma }}{\partial {\bar {x}}^{\beta }}}\,F_{\delta \gamma }\,.}
這一定義暗示了電磁場張量滿足關係
∂
λ
F
μ
ν
+
∂
μ
F
ν
λ
+
∂
ν
F
λ
μ
=
0
{\displaystyle \partial _{\lambda }F_{\mu \nu }+\partial _{\mu }F_{\nu \lambda }+\partial _{\nu }F_{\lambda \mu }=0\,}
這一關係包含了法拉第電磁感應定律 和磁場的高斯定理 ，因此也叫做法拉第-高斯方程式。具體而言，
∂
λ
F
μ
ν
+
∂
μ
F
ν
λ
+
∂
ν
F
λ
μ
{\displaystyle \partial _{\lambda }F_{\mu \nu }+\partial _{\mu }F_{\nu \lambda }+\partial _{\nu }F_{\lambda \mu }\,}
=
∂
λ
∂
μ
A
ν
−
∂
λ
∂
ν
A
μ
+
∂
μ
∂
ν
A
λ
−
∂
μ
∂
λ
A
ν
+
∂
ν
∂
λ
A
μ
−
∂
ν
∂
μ
A
λ
=
0
.
{\displaystyle =\,\partial _{\lambda }\partial _{\mu }A_{\nu }-\partial _{\lambda }\partial _{\nu }A_{\mu }+\partial _{\mu }\partial _{\nu }A_{\lambda }-\partial _{\mu }\partial _{\lambda }A_{\nu }+\partial _{\nu }\partial _{\lambda }A_{\mu }-\partial _{\nu }\partial _{\mu }A_{\lambda }\,=0\,.}
雖然這個關係包含了64個分量方程式，但只有四個是獨立的。藉助電磁場張量的反對稱性，可知只有
λ
,
μ
,
ν
{\displaystyle \lambda ,\mu ,\nu \,}
法拉第-高斯方程式有時也寫作下面的形式
F
[
μ
ν
;
λ
]
=
F
[
μ
ν
,
λ
]
=
1
6
(
∂
λ
F
μ
ν
+
∂
μ
F
ν
λ
+
∂
ν
F
λ
μ
−
∂
λ
F
ν
μ
−
∂
μ
F
λ
ν
−
∂
ν
F
μ
λ
)
{\displaystyle F_{[\mu \nu ;\lambda ]}\,=\,F_{[\mu \nu ,\lambda ]}\,=\,{\frac {1}{6}}\left(\partial _{\lambda }F_{\mu \nu }+\partial _{\mu }F_{\nu \lambda }+\partial _{\nu }F_{\lambda \mu }-\partial _{\lambda }F_{\nu \mu }-\partial _{\mu }F_{\lambda \nu }-\partial _{\nu }F_{\mu \lambda }\right)\,}
=
1
3
(
∂
λ
F
μ
ν
+
∂
μ
F
ν
λ
+
∂
ν
F
λ
μ
)
=
0
{\displaystyle =\,{\frac {1}{3}}\left(\partial _{\lambda }F_{\mu \nu }+\partial _{\mu }F_{\nu \lambda }+\partial _{\nu }F_{\lambda \mu }\right)=0\,}
其中按照慣例用分號表示協變導數 ，逗號表示偏導數，方括號表示反對稱形式。電磁場張量的協變導數為
F
α
β
;
γ
=
F
α
β
,
γ
−
Γ
μ
α
γ
F
μ
β
−
Γ
μ
β
γ
F
α
μ
{\displaystyle F_{\alpha \beta ;\gamma }\,=\,F_{\alpha \beta ,\gamma }-{\Gamma ^{\mu }}_{\alpha \gamma }F_{\mu \beta }-{\Gamma ^{\mu }}_{\beta \gamma }F_{\alpha \mu }\,}
其中
Γ
β
γ
α
{\displaystyle \Gamma _{\beta \gamma }^{\alpha }\,}
克里斯托費爾符號 ，兩個下標是對稱的。
位移電場
D
,
{\displaystyle \mathbf {D} \,,}
附屬磁場
H
,
{\displaystyle \mathbf {H} \,,}
D
μ
ν
=
1
μ
0
g
μ
α
F
α
β
g
β
ν
−
g
.
{\displaystyle {\mathcal {D}}^{\mu \nu }\,=\,{\frac {1}{\mu _{0}}}\,g^{\mu \alpha }\,F_{\alpha \beta }\,g^{\beta \nu }\,{\sqrt {-g}}\,.}
注意這個方程式是電磁理論中唯一有度規（即重力）存在的方程式。並且這一方程式具有尺度不變性 ，即將度規乘以一個常數不會改變方程式的形式。也就是說，重力只能通過改變全局坐標中的光速 來影響相應的電磁場，由於光在重力場中會發生偏折，其效應等同於質量的重力場增加了周圍時空的折射率 ，從而影響了對應的位移電場和附屬磁場。
更一般地，當物質中的磁化-極化張量不為零時，我們有
D
μ
ν
=
1
μ
0
g
μ
α
F
α
β
g
β
ν
−
g
−
M
μ
ν
.
{\displaystyle {\mathcal {D}}^{\mu \nu }\,=\,{\frac {1}{\mu _{0}}}\,g^{\mu \alpha }\,F_{\alpha \beta }\,g^{\beta \nu }\,{\sqrt {-g}}\,-\,{\mathcal {M}}^{\mu \nu }\,.}
電磁位移張量的坐標變換規則為
D
¯
μ
ν
=
∂
x
¯
μ
∂
x
α
∂
x
¯
ν
∂
x
β
D
α
β
det
[
∂
x
σ
∂
x
¯
ρ
]
{\displaystyle {\bar {\mathcal {D}}}^{\mu \nu }\,=\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial {\bar {x}}^{\nu }}{\partial x^{\beta }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,}
其中使用了雅可比行列式 。如果磁化-極化張量存在，它和電磁位移張量具有同樣的變換規則。
電磁位移張量的散度 被定義為電流 ，在真空中
J
μ
=
∂
ν
D
μ
ν
.
{\displaystyle J^{\mu }\,=\,\partial _{\nu }{\mathcal {D}}^{\mu \nu }\,.}
如果磁化-極化張量存在，則上式替換為
J
free
μ
=
∂
ν
D
μ
ν
.
{\displaystyle J_{\text{free}}^{\mu }\,=\,\partial _{\nu }{\mathcal {D}}^{\mu \nu }\,.}
這一方程式包含了高斯定理 和安培環路定理 。
無論磁化-極化張量是否存在，電磁位移張量是反對稱的這一事實暗示了電流是一個守恆量：
∂
μ
J
μ
=
∂
μ
∂
ν
D
μ
ν
=
0
{\displaystyle \partial _{\mu }J^{\mu }\,=\,\partial _{\mu }\partial _{\nu }{\mathcal {D}}^{\mu \nu }=0\,}
這個二階偏導數為零是由於偏導是對易 的。
電流的安培-高斯定義並不能決定它的大小，因為電磁四維勢的大小並未確定。相反地，通常的步驟是使電流等於一個用其他場表示的表達式（主要是電荷），然後解得電磁位移、電磁場和電磁勢。
電流是一個逆變 的向量，因而其變換規則是
J
¯
μ
=
∂
x
¯
μ
∂
x
α
J
α
det
[
∂
x
σ
∂
x
¯
ρ
]
.
{\displaystyle {\bar {J}}^{\mu }\,=\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,J^{\alpha }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,.}
變換規則的驗證如下：
J
¯
μ
=
∂
∂
x
¯
ν
(
D
¯
μ
ν
)
=
∂
∂
x
¯
ν
(
∂
x
¯
μ
∂
x
α
∂
x
¯
ν
∂
x
β
D
α
β
det
[
∂
x
σ
∂
x
¯
ρ
]
)
{\displaystyle {\bar {J}}^{\mu }\,=\,{\frac {\partial }{\partial {\bar {x}}^{\nu }}}\left({\bar {\mathcal {D}}}^{\mu \nu }\right)\,=\,{\frac {\partial }{\partial {\bar {x}}^{\nu }}}\left({\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial {\bar {x}}^{\nu }}{\partial x^{\beta }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\right)\,}
=
∂
2
x
¯
μ
∂
x
¯
ν
∂
x
α
∂
x
¯
ν
∂
x
β
D
α
β
det
[
∂
x
σ
∂
x
¯
ρ
]
+
∂
x
¯
μ
∂
x
α
∂
2
x
¯
ν
∂
x
¯
ν
∂
x
β
D
α
β
det
[
∂
x
σ
∂
x
¯
ρ
]
+
{\displaystyle =\,{\frac {\partial ^{2}{\bar {x}}^{\mu }}{\partial {\bar {x}}^{\nu }\partial x^{\alpha }}}\,{\frac {\partial {\bar {x}}^{\nu }}{\partial x^{\beta }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial ^{2}{\bar {x}}^{\nu }}{\partial {\bar {x}}^{\nu }\partial x^{\beta }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,}
∂
x
¯
μ
∂
x
α
∂
x
¯
ν
∂
x
β
∂
D
α
β
∂
x
¯
ν
det
[
∂
x
σ
∂
x
¯
ρ
]
+
∂
x
¯
μ
∂
x
α
∂
x
¯
ν
∂
x
β
D
α
β
∂
∂
x
¯
ν
det
[
∂
x
σ
∂
x
¯
ρ
]
{\displaystyle {\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial {\bar {x}}^{\nu }}{\partial x^{\beta }}}\,{\frac {\partial {\mathcal {D}}^{\alpha \beta }}{\partial {\bar {x}}^{\nu }}}\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial {\bar {x}}^{\nu }}{\partial x^{\beta }}}\,{\mathcal {D}}^{\alpha \beta }\,{\frac {\partial }{\partial {\bar {x}}^{\nu }}}\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,}
=
∂
2
x
¯
μ
∂
x
β
∂
x
α
D
α
β
det
[
∂
x
σ
∂
x
¯
ρ
]
+
∂
x
¯
μ
∂
x
α
∂
2
x
¯
ν
∂
x
¯
ν
∂
x
β
D
α
β
det
[
∂
x
σ
∂
x
¯
ρ
]
+
{\displaystyle =\,{\frac {\partial ^{2}{\bar {x}}^{\mu }}{\partial x^{\beta }\partial x^{\alpha }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial ^{2}{\bar {x}}^{\nu }}{\partial {\bar {x}}^{\nu }\partial x^{\beta }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,}
∂
x
¯
μ
∂
x
α
∂
D
α
β
∂
x
β
det
[
∂
x
σ
∂
x
¯
ρ
]
+
∂
x
¯
μ
∂
x
α
∂
x
¯
ν
∂
x
β
D
α
β
det
[
∂
x
σ
∂
x
¯
ρ
]
∂
x
¯
ρ
∂
x
σ
∂
2
x
σ
∂
x
¯
ν
∂
x
¯
ρ
{\displaystyle {\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial {\mathcal {D}}^{\alpha \beta }}{\partial x^{\beta }}}\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial {\bar {x}}^{\nu }}{\partial x^{\beta }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]{\frac {\partial {\bar {x}}^{\rho }}{\partial x^{\sigma }}}{\frac {\partial ^{2}x^{\sigma }}{\partial {\bar {x}}^{\nu }\partial {\bar {x}}^{\rho }}}\,}
=
0
+
∂
x
¯
μ
∂
x
α
∂
2
x
¯
ν
∂
x
¯
ν
∂
x
β
D
α
β
det
[
∂
x
σ
∂
x
¯
ρ
]
+
{\displaystyle =\,0\,+\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial ^{2}{\bar {x}}^{\nu }}{\partial {\bar {x}}^{\nu }\partial x^{\beta }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,}
∂
x
¯
μ
∂
x
α
J
α
det
[
∂
x
σ
∂
x
¯
ρ
]
+
∂
x
¯
μ
∂
x
α
D
α
β
det
[
∂
x
σ
∂
x
¯
ρ
]
∂
x
¯
ρ
∂
x
σ
∂
2
x
σ
∂
x
β
∂
x
¯
ρ
{\displaystyle {\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,J^{\alpha }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]{\frac {\partial {\bar {x}}^{\rho }}{\partial x^{\sigma }}}{\frac {\partial ^{2}x^{\sigma }}{\partial x^{\beta }\partial {\bar {x}}^{\rho }}}\,}
=
∂
x
¯
μ
∂
x
α
J
α
det
[
∂
x
σ
∂
x
¯
ρ
]
+
∂
x
¯
μ
∂
x
α
D
α
β
det
[
∂
x
σ
∂
x
¯
ρ
]
(
∂
2
x
¯
ν
∂
x
¯
ν
∂
x
β
+
∂
x
¯
ρ
∂
x
σ
∂
2
x
σ
∂
x
β
∂
x
¯
ρ
)
.
{\displaystyle =\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,J^{\alpha }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\left({\frac {\partial ^{2}{\bar {x}}^{\nu }}{\partial {\bar {x}}^{\nu }\partial x^{\beta }}}\,+\,{\frac {\partial {\bar {x}}^{\rho }}{\partial x^{\sigma }}}{\frac {\partial ^{2}x^{\sigma }}{\partial x^{\beta }\partial {\bar {x}}^{\rho }}}\right)\,.}
下面只需證明
∂
2
x
¯
ν
∂
x
¯
ν
∂
x
β
+
∂
x
¯
ρ
∂
x
σ
∂
2
x
σ
∂
x
β
∂
x
¯
ρ
=
0
{\displaystyle {\frac {\partial ^{2}{\bar {x}}^{\nu }}{\partial {\bar {x}}^{\nu }\partial x^{\beta }}}\,+\,{\frac {\partial {\bar {x}}^{\rho }}{\partial x^{\sigma }}}{\frac {\partial ^{2}x^{\sigma }}{\partial x^{\beta }\partial {\bar {x}}^{\rho }}}\,=\,0\,}
這是微分學 中一條已知定理的應用：
∂
2
x
¯
ν
∂
x
¯
ν
∂
x
β
+
∂
x
¯
ρ
∂
x
σ
∂
2
x
σ
∂
x
β
∂
x
¯
ρ
=
∂
x
σ
∂
x
¯
ν
∂
2
x
¯
ν
∂
x
σ
∂
x
β
+
∂
x
¯
ν
∂
x
σ
∂
2
x
σ
∂
x
β
∂
x
¯
ν
{\displaystyle {\frac {\partial ^{2}{\bar {x}}^{\nu }}{\partial {\bar {x}}^{\nu }\partial x^{\beta }}}\,+\,{\frac {\partial {\bar {x}}^{\rho }}{\partial x^{\sigma }}}{\frac {\partial ^{2}x^{\sigma }}{\partial x^{\beta }\partial {\bar {x}}^{\rho }}}\,=\,{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\nu }}}{\frac {\partial ^{2}{\bar {x}}^{\nu }}{\partial x^{\sigma }\partial x^{\beta }}}\,+\,{\frac {\partial {\bar {x}}^{\nu }}{\partial x^{\sigma }}}{\frac {\partial ^{2}x^{\sigma }}{\partial x^{\beta }\partial {\bar {x}}^{\nu }}}\,}
=
∂
x
σ
∂
x
¯
ν
∂
2
x
¯
ν
∂
x
β
∂
x
σ
+
∂
2
x
σ
∂
x
β
∂
x
¯
ν
∂
x
¯
ν
∂
x
σ
=
∂
∂
x
β
(
∂
x
σ
∂
x
¯
ν
∂
x
¯
ν
∂
x
σ
)
{\displaystyle =\,{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\nu }}}{\frac {\partial ^{2}{\bar {x}}^{\nu }}{\partial x^{\beta }\partial x^{\sigma }}}\,+\,{\frac {\partial ^{2}x^{\sigma }}{\partial x^{\beta }\partial {\bar {x}}^{\nu }}}{\frac {\partial {\bar {x}}^{\nu }}{\partial x^{\sigma }}}\,=\,{\frac {\partial }{\partial x^{\beta }}}\left({\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\nu }}}\,{\frac {\partial {\bar {x}}^{\nu }}{\partial x^{\sigma }}}\right)\,}
=
∂
∂
x
β
(
∂
x
¯
ν
∂
x
¯
ν
)
=
∂
∂
x
β
(
4
)
=
0
.
{\displaystyle =\,{\frac {\partial }{\partial x^{\beta }}}\left(\,{\frac {\partial {\bar {x}}^{\nu }}{\partial {\bar {x}}^{\nu }}}\right)\,=\,{\frac {\partial }{\partial x^{\beta }}}\left(\mathbf {4} \right)\,=\,0\,.}
勞侖茲力 的密度是一個協變向量：
f
μ
=
F
μ
ν
J
ν
.
{\displaystyle f_{\mu }\,=\,F_{\mu \nu }\,J^{\nu }\,.}
如果在一個測試粒子上的作用力只有重力和電磁力，則有
d
p
α
d
t
=
Γ
α
γ
β
p
β
d
x
γ
d
t
+
q
F
α
γ
d
x
γ
d
t
{\displaystyle {\frac {dp_{\alpha }}{dt}}\,=\,\Gamma _{\alpha \gamma }^{\beta }\,p_{\beta }\,{\frac {dx^{\gamma }}{dt}}\,+\,q\,F_{\alpha \gamma }\,{\frac {dx^{\gamma }}{dt}}\,}
其中
p
α
{\displaystyle p^{\alpha }\,}
四維動量 ，
t
{\displaystyle t\,}
世界線 參數化的任意時間坐標。式中第一項含有克里斯托費爾符號，表示重力項，第二項含有電荷，表示電磁力項。
方程式具有勞侖茲不變性，對時間坐標變換的驗證方法為將方程式乘以
d
t
d
t
¯
{\displaystyle {\frac {dt}{d{\bar {t}}}}\,}
鏈式法則 。
對空間坐標變換，通過對克里斯托費爾符號進行坐標變換
Γ
¯
α
γ
β
=
∂
x
¯
β
∂
x
ϵ
∂
x
δ
∂
x
¯
α
∂
x
ζ
∂
x
¯
γ
Γ
δ
ζ
ϵ
+
∂
x
¯
β
∂
x
η
∂
2
x
η
∂
x
¯
α
∂
x
¯
γ
{\displaystyle {\bar {\Gamma }}_{\alpha \gamma }^{\beta }\,=\,{\frac {\partial {\bar {x}}^{\beta }}{\partial x^{\epsilon }}}\,{\frac {\partial x^{\delta }}{\partial {\bar {x}}^{\alpha }}}\,{\frac {\partial x^{\zeta }}{\partial {\bar {x}}^{\gamma }}}\,\Gamma _{\delta \zeta }^{\epsilon }\,+{\frac {\partial {\bar {x}}^{\beta }}{\partial x^{\eta }}}\,{\frac {\partial ^{2}x^{\eta }}{\partial {\bar {x}}^{\alpha }\partial {\bar {x}}^{\gamma }}}\,}
我們得到
d
p
¯
α
d
t
−
Γ
¯
α
γ
β
p
¯
β
d
x
¯
γ
d
t
−
q
F
¯
α
γ
d
x
¯
γ
d
t
{\displaystyle {\frac {d{\bar {p}}_{\alpha }}{dt}}\,-\,{\bar {\Gamma }}_{\alpha \gamma }^{\beta }\,{\bar {p}}_{\beta }\,{\frac {d{\bar {x}}^{\gamma }}{dt}}\,-\,q\,{\bar {F}}_{\alpha \gamma }\,{\frac {d{\bar {x}}^{\gamma }}{dt}}\,}
=
d
d
t
(
∂
x
δ
∂
x
¯
α
p
δ
)
−
(
∂
x
¯
β
∂
x
θ
∂
x
δ
∂
x
¯
α
∂
x
ι
∂
x
¯
γ
Γ
δ
ι
θ
+
∂
x
¯
β
∂
x
η
∂
2
x
η
∂
x
¯
α
∂
x
¯
γ
)
∂
x
ϵ
∂
x
¯
β
p
ϵ
∂
x
¯
γ
∂
x
ζ
d
x
ζ
d
t
−
q
∂
x
δ
∂
x
¯
α
F
δ
ζ
d
x
ζ
d
t
{\displaystyle =\,{\frac {d}{dt}}\left({\frac {\partial x^{\delta }}{\partial {\bar {x}}^{\alpha }}}\,p_{\delta }\right)\,-\,\left({\frac {\partial {\bar {x}}^{\beta }}{\partial x^{\theta }}}\,{\frac {\partial x^{\delta }}{\partial {\bar {x}}^{\alpha }}}\,{\frac {\partial x^{\iota }}{\partial {\bar {x}}^{\gamma }}}\,\Gamma _{\delta \iota }^{\theta }\,+\,{\frac {\partial {\bar {x}}^{\beta }}{\partial x^{\eta }}}\,{\frac {\partial ^{2}x^{\eta }}{\partial {\bar {x}}^{\alpha }\partial {\bar {x}}^{\gamma }}}\right)\,{\frac {\partial x^{\epsilon }}{\partial {\bar {x}}^{\beta }}}\,p_{\epsilon }\,{\frac {\partial {\bar {x}}^{\gamma }}{\partial x^{\zeta }}}\,{\frac {dx^{\zeta }}{dt}}\,-\,q\,{\frac {\partial x^{\delta }}{\partial {\bar {x}}^{\alpha }}}\,F_{\delta \zeta }\,{\frac {dx^{\zeta }}{dt}}\,}
=
∂
x
δ
∂
x
¯
α
(
d
p
δ
d
t
−
Γ
δ
ζ
ϵ
p
ϵ
d
x
ζ
d
t
−
q
F
δ
ζ
d
x
ζ
d
t
)
+
{\displaystyle =\,{\frac {\partial x^{\delta }}{\partial {\bar {x}}^{\alpha }}}\,\left({\frac {dp_{\delta }}{dt}}\,-\,\Gamma _{\delta \zeta }^{\epsilon }\,p_{\epsilon }\,{\frac {dx^{\zeta }}{dt}}\,-\,q\,F_{\delta \zeta }\,{\frac {dx^{\zeta }}{dt}}\right)\,+\,}
d
d
t
(
∂
x
δ
∂
x
¯
α
)
p
δ
−
(
∂
x
¯
β
∂
x
η
∂
2
x
η
∂
x
¯
α
∂
x
¯
γ
)
∂
x
ϵ
∂
x
¯
β
p
ϵ
∂
x
¯
γ
∂
x
ζ
d
x
ζ
d
t
{\displaystyle {\frac {d}{dt}}\left({\frac {\partial x^{\delta }}{\partial {\bar {x}}^{\alpha }}}\right)\,p_{\delta }\,-\,\left({\frac {\partial {\bar {x}}^{\beta }}{\partial x^{\eta }}}\,{\frac {\partial ^{2}x^{\eta }}{\partial {\bar {x}}^{\alpha }\partial {\bar {x}}^{\gamma }}}\right)\,{\frac {\partial x^{\epsilon }}{\partial {\bar {x}}^{\beta }}}\,p_{\epsilon }\,{\frac {\partial {\bar {x}}^{\gamma }}{\partial x^{\zeta }}}\,{\frac {dx^{\zeta }}{dt}}\,}
=
0
+
d
d
t
(
∂
x
δ
∂
x
¯
α
)
p
δ
−
∂
2
x
ϵ
∂
x
¯
α
∂
x
¯
γ
p
ϵ
d
x
¯
γ
d
t
=
0
{\displaystyle =\,0\,+\,{\frac {d}{dt}}\left({\frac {\partial x^{\delta }}{\partial {\bar {x}}^{\alpha }}}\right)\,p_{\delta }\,-\,{\frac {\partial ^{2}x^{\epsilon }}{\partial {\bar {x}}^{\alpha }\partial {\bar {x}}^{\gamma }}}p_{\epsilon }\,{\frac {d{\bar {x}}^{\gamma }}{dt}}\,=\,0\,}
在真空中古典電磁場的拉格朗日量 （在密度意義下的單位為焦耳/米3 ）是一個純量：
L
=
−
1
4
μ
0
F
α
β
F
α
β
−
g
+
A
α
J
α
{\displaystyle {\mathcal {L}}\,=\,-{\frac {1}{4\mu _{0}}}\,F_{\alpha \beta }\,F^{\alpha \beta }\,{\sqrt {-g}}\,+\,A_{\alpha }\,J^{\alpha }\,}
其中
F
α
β
=
g
α
γ
F
γ
δ
g
δ
β
.
{\displaystyle F^{\alpha \beta }=g^{\alpha \gamma }F_{\gamma \delta }g^{\delta \beta }\,.}
如果我們將自由電流與束縛電流分離開來，拉格朗日量則變為
L
=
−
1
4
μ
0
F
α
β
F
α
β
−
g
+
A
α
J
free
α
+
1
2
F
α
β
M
α
β
.
{\displaystyle {\mathcal {L}}\,=\,-{\frac {1}{4\mu _{0}}}\,F_{\alpha \beta }\,F^{\alpha \beta }\,{\sqrt {-g}}\,+\,A_{\alpha }\,J_{\text{free}}^{\alpha }\,+\,{\frac {1}{2}}\,F_{\alpha \beta }\,{\mathcal {M}}^{\alpha \beta }\,.}
作為愛因斯坦重力場方程式 的源，電磁場的應力-能量張量 是一個協變的對稱張量
T
μ
ν
=
−
1
μ
0
(
F
μ
α
g
α
β
F
β
ν
−
1
4
g
μ
ν
F
σ
α
g
α
β
F
β
ρ
g
ρ
σ
)
{\displaystyle T_{\mu \nu }\,=\,-{\frac {1}{\mu _{0}}}(F_{\mu \alpha }g^{\alpha \beta }F_{\beta \nu }\,-\,{\frac {1}{4}}g_{\mu \nu }\,F_{\sigma \alpha }g^{\alpha \beta }F_{\beta \rho }g^{\rho \sigma })\,}
並且它是無跡的
T
μ
ν
g
μ
ν
=
0
{\displaystyle T_{\mu \nu }g^{\mu \nu }\,=\,0\,}
這是由於電磁場的傳播速度在不同參考系下是不變的。
為了表達能量和動量的守恆律，電磁應力-能量張量的最佳表示方法是
T
μ
ν
=
T
μ
γ
g
γ
ν
−
g
.
{\displaystyle {\mathfrak {T}}_{\mu }^{\nu }=T_{\mu \gamma }\,g^{\gamma \nu }\,{\sqrt {-g}}.}
對於上式可以證明
T
μ
ν
;
ν
+
f
μ
=
0
{\displaystyle {{\mathfrak {T}}_{\mu }^{\nu }}_{;\nu }\,+\,f_{\mu }\,=\,0\,}
並可重寫為
−
T
μ
ν
,
ν
=
−
Γ
μ
ν
σ
T
σ
ν
+
f
μ
{\displaystyle -{{\mathfrak {T}}_{\mu }^{\nu }}_{,\nu }\,=\,-\Gamma _{\mu \nu }^{\sigma }{\mathfrak {T}}_{\sigma }^{\nu }\,+\,f_{\mu }\,}
這個方程式的意義是，電磁場能量的減少相當於電磁場對重力場以及通過勞侖茲力對物質所作的功，類似地，電磁場動量的減少率相當於作用在重力場上的電磁力以及作用在物質上的勞侖茲力。
守恆律的推導過程為
T
μ
ν
;
ν
+
f
μ
=
−
1
μ
0
(
F
μ
α
;
ν
g
α
β
F
β
γ
g
γ
ν
+
F
μ
α
g
α
β
F
β
γ
;
ν
g
γ
ν
−
1
2
δ
μ
ν
F
σ
α
;
ν
g
α
β
F
β
ρ
g
ρ
σ
)
−
g
+
{\displaystyle {{\mathfrak {T}}_{\mu }^{\nu }}_{;\nu }\,+\,f_{\mu }\,=\,-{\frac {1}{\mu _{0}}}(F_{\mu \alpha ;\nu }g^{\alpha \beta }F_{\beta \gamma }g^{\gamma \nu }\,+\,F_{\mu \alpha }g^{\alpha \beta }F_{\beta \gamma ;\nu }g^{\gamma \nu }\,-\,{\frac {1}{2}}\delta _{\mu }^{\nu }\,F_{\sigma \alpha ;\nu }g^{\alpha \beta }F_{\beta \rho }g^{\rho \sigma }){\sqrt {-g}}\,+\,}
1
μ
0
F
μ
α
g
α
β
F
β
γ
;
ν
g
γ
ν
−
g
{\displaystyle {\frac {1}{\mu _{0}}}\,F_{\mu \alpha }\,g^{\alpha \beta }\,F_{\beta \gamma ;\nu }\,g^{\gamma \nu }\,{\sqrt {-g}}\,}
=
−
1
μ
0
(
F
μ
α
;
ν
F
α
ν
−
1
2
F
σ
α
;
μ
F
α
σ
)
−
g
{\displaystyle =\,-{\frac {1}{\mu _{0}}}(F_{\mu \alpha ;\nu }F^{\alpha \nu }\,-\,{\frac {1}{2}}F_{\sigma \alpha ;\mu }F^{\alpha \sigma }){\sqrt {-g}}\,}
=
−
1
μ
0
(
(
−
F
ν
μ
;
α
−
F
α
ν
;
μ
)
F
α
ν
−
1
2
F
σ
α
;
μ
F
α
σ
)
−
g
{\displaystyle =\,-{\frac {1}{\mu _{0}}}((-F_{\nu \mu ;\alpha }-F_{\alpha \nu ;\mu })F^{\alpha \nu }\,-\,{\frac {1}{2}}F_{\sigma \alpha ;\mu }F^{\alpha \sigma }){\sqrt {-g}}\,}
=
−
1
μ
0
(
F
μ
ν
;
α
F
α
ν
−
F
α
ν
;
μ
F
α
ν
+
1
2
F
σ
α
;
μ
F
σ
α
)
−
g
{\displaystyle =\,-{\frac {1}{\mu _{0}}}(F_{\mu \nu ;\alpha }F^{\alpha \nu }-F_{\alpha \nu ;\mu }F^{\alpha \nu }\,+\,{\frac {1}{2}}F_{\sigma \alpha ;\mu }F^{\sigma \alpha }){\sqrt {-g}}\,}
=
−
1
μ
0
(
F
μ
α
;
ν
F
ν
α
−
1
2
F
α
ν
;
μ
F
α
ν
)
−
g
{\displaystyle =\,-{\frac {1}{\mu _{0}}}(F_{\mu \alpha ;\nu }F^{\nu \alpha }-{\frac {1}{2}}F_{\alpha \nu ;\mu }F^{\alpha \nu }){\sqrt {-g}}\,}
=
−
1
μ
0
(
−
F
μ
α
;
ν
F
α
ν
+
1
2
F
σ
α
;
μ
F
α
σ
)
−
g
{\displaystyle =\,-{\frac {1}{\mu _{0}}}(-F_{\mu \alpha ;\nu }F^{\alpha \nu }\,+\,{\frac {1}{2}}F_{\sigma \alpha ;\mu }F^{\alpha \sigma }){\sqrt {-g}}\,}
所得的結果是為零的，因為它等於自身的負值（對照推導的第二步）。
從電磁理論的狹義相對論形式 可以藉助電磁場張量修改得到非齊次的電磁波方程式
◻
F
a
b
=
d
e
f
F
a
b
;
d
d
=
−
2
R
a
c
b
d
F
c
d
+
R
a
e
F
e
b
−
R
b
e
F
e
a
+
J
a
;
b
−
J
b
;
a
{\displaystyle \Box F_{ab}\ {\stackrel {\mathrm {def} }{=}}\ F_{ab;}{}^{d}{}_{d}=\,-2R_{acbd}F^{cd}+R_{ae}F^{e}{}_{b}-R_{be}F^{e}{}_{a}+J_{a;b}-J_{b;a}}
其中
R
a
c
b
d
{\displaystyle R_{acbd}\,}
是黎曼張量 的協變形式，而
◻
{\displaystyle \Box \,}
達朗貝爾算符 在協變導數情形下的推廣。
波方程式可以寫成四維勢的形式（參見ref 2, p. 569）
◻
A
a
=
A
a
;
b
b
=
−
μ
0
J
a
+
R
a
b
A
b
{\displaystyle \Box A^{a}={{A^{a;}}^{b}}_{b}=-\mu _{0}J^{a}+{R^{a}}_{b}A^{b}}
其中
R
a
b
=
d
e
f
R
s
a
s
b
{\displaystyle {R^{a}}_{b}\ {\stackrel {\mathrm {def} }{=}}\ {R^{s}}_{asb}}
是里奇張量 。而這裡假設了勞侖次規範 在彎曲時空中的推廣具有下面的形式
A
a
;
a
=
0
{\displaystyle {A^{a}}_{;a}=0}
這個波方程式與平直時空中的波方程式具有十分類似的形式，除了導數被替換為協變導數，以及在表示源的項中多了一項描述時空曲率的項。這個方程式和彎曲時空中的勞侖茲力也有相似之處，這裡的四維勢
A
a
{\displaystyle {A^{a}}\,}
在考慮馬克士威方程組本身與背景時空無關時，時空度規在廣義相對論中被認為是受電磁場影響的動態變化的變量，這導致了電磁波方程式和馬克士威方程組是非線性的。這一點可以從愛因斯坦場方程式 中的曲率張量依賴於應力-能量張量 看出。
G
a
b
=
8
π
G
c
4
T
a
b
{\displaystyle G_{ab}={\frac {8\pi G}{c^{4}}}T_{ab}}
其中
G
a
b
=
d
e
f
R
a
b
−
1
2
R
g
a
b
{\displaystyle {G}_{ab}\ {\stackrel {\mathrm {def} }{=}}\ {R}_{ab}-{1 \over 2}{R}g_{ab}}
是愛因斯坦張量 ，
G
{\displaystyle G\,}
萬有引力常數 ，
R
{\displaystyle R\,}
純量曲率 ）是里奇張量的跡。應力-能量張量來源於粒子的應力 和能量，但同時也來源於電磁場，這造就了非線性。
![{\displaystyle F_{[\mu \nu ;\lambda ]}\,=\,F_{[\mu \nu ,\lambda ]}\,=\,{\frac {1}{6}}\left(\partial _{\lambda }F_{\mu \nu }+\partial _{\mu }F_{\nu \lambda }+\partial _{\nu }F_{\lambda \mu }-\partial _{\lambda }F_{\nu \mu }-\partial _{\mu }F_{\lambda \nu }-\partial _{\nu }F_{\mu \lambda }\right)\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d8e2cede3dfb7a42657649d9d67336241e51e8a)
![{\displaystyle {\bar {\mathcal {D}}}^{\mu \nu }\,=\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial {\bar {x}}^{\nu }}{\partial x^{\beta }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4c5bd1af20b02ce57b4f9538cdd596e171ca96a)
![{\displaystyle {\bar {J}}^{\mu }\,=\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,J^{\alpha }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0639f83cc7b02d4d546f786a651079a4ab59de83)
![{\displaystyle {\bar {J}}^{\mu }\,=\,{\frac {\partial }{\partial {\bar {x}}^{\nu }}}\left({\bar {\mathcal {D}}}^{\mu \nu }\right)\,=\,{\frac {\partial }{\partial {\bar {x}}^{\nu }}}\left({\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial {\bar {x}}^{\nu }}{\partial x^{\beta }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\right)\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/19ed9ccb7c46e15aa8718636417c009bbdde8551)
![{\displaystyle =\,{\frac {\partial ^{2}{\bar {x}}^{\mu }}{\partial {\bar {x}}^{\nu }\partial x^{\alpha }}}\,{\frac {\partial {\bar {x}}^{\nu }}{\partial x^{\beta }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial ^{2}{\bar {x}}^{\nu }}{\partial {\bar {x}}^{\nu }\partial x^{\beta }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79db9659ef320850fecda40c5c20b8bf5e040151)
![{\displaystyle {\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial {\bar {x}}^{\nu }}{\partial x^{\beta }}}\,{\frac {\partial {\mathcal {D}}^{\alpha \beta }}{\partial {\bar {x}}^{\nu }}}\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial {\bar {x}}^{\nu }}{\partial x^{\beta }}}\,{\mathcal {D}}^{\alpha \beta }\,{\frac {\partial }{\partial {\bar {x}}^{\nu }}}\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc708d6a890aaef612bde28b40cb4b4d41bff61b)
![{\displaystyle =\,{\frac {\partial ^{2}{\bar {x}}^{\mu }}{\partial x^{\beta }\partial x^{\alpha }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial ^{2}{\bar {x}}^{\nu }}{\partial {\bar {x}}^{\nu }\partial x^{\beta }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ad38011349456602c0723033ad27a81cbed67438)
![{\displaystyle {\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial {\mathcal {D}}^{\alpha \beta }}{\partial x^{\beta }}}\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial {\bar {x}}^{\nu }}{\partial x^{\beta }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]{\frac {\partial {\bar {x}}^{\rho }}{\partial x^{\sigma }}}{\frac {\partial ^{2}x^{\sigma }}{\partial {\bar {x}}^{\nu }\partial {\bar {x}}^{\rho }}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b6ad2cb30ba59b95234a39403ae0a02cac8f8ce8)
![{\displaystyle =\,0\,+\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\frac {\partial ^{2}{\bar {x}}^{\nu }}{\partial {\bar {x}}^{\nu }\partial x^{\beta }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7cec56ce4063cb77af4bc97d7edb2aff46830b8)
![{\displaystyle {\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,J^{\alpha }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]{\frac {\partial {\bar {x}}^{\rho }}{\partial x^{\sigma }}}{\frac {\partial ^{2}x^{\sigma }}{\partial x^{\beta }\partial {\bar {x}}^{\rho }}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48b1341f4d0de76fabafb3e709b38d1e4eb09daa)
![{\displaystyle =\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,J^{\alpha }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\,+\,{\frac {\partial {\bar {x}}^{\mu }}{\partial x^{\alpha }}}\,{\mathcal {D}}^{\alpha \beta }\,\det \left[{\frac {\partial x^{\sigma }}{\partial {\bar {x}}^{\rho }}}\right]\left({\frac {\partial ^{2}{\bar {x}}^{\nu }}{\partial {\bar {x}}^{\nu }\partial x^{\beta }}}\,+\,{\frac {\partial {\bar {x}}^{\rho }}{\partial x^{\sigma }}}{\frac {\partial ^{2}x^{\sigma }}{\partial x^{\beta }\partial {\bar {x}}^{\rho }}}\right)\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3ef199bdac0c3bf785194955a3825f8fb370373)
