引起时空弯曲 物理学 中，弯曲时空中的麦克斯韦方程组 （Maxwell's equations in curved spacetime）制约着弯曲时空 （其间的度规 可能不是闵可夫斯基性 的）中的电磁场 的动力学。它们可以被认为是真空中的麦克斯韦方程组 在广义相对论 框架中的扩展，而真空中的麦克斯韦方程组只是一般化的麦克斯韦方程组在局部平直时空中的特殊形式。但由于在广义相对论中电磁场本身的存在也会引起时空的弯曲，因此真空中的麦克斯韦方程组应被理解为一种出于方便的近似形式。
然而，这种形式的麦克斯韦方程组仅仅对真空情形下的麦克斯韦方程组有用，这也被称作“微观”麦克斯韦方程组。对于宏观上与各向异性的物质相关的麦克斯韦方程组，物质的存在会建立一个参考系从而使方程组不再是协变 的。
阅读本条目需要读者了解平直时空中电磁理论的四维形式 。
电磁场本身要求其几何描述与坐标选取无关，而麦克斯韦方程组在任何时空中的几何描述都是一样的，而不管这个时空是否是平直的。同时，当使用非笛卡尔 的局部坐标时平直闵可夫斯基空间中的方程组会做同样的修改。例如本条目中方程组可以写成球坐标 中的麦克斯韦方程组的形式。基于上述原因，更好的理解方法是将闵可夫斯基空间中的麦克斯韦方程组理解为一种特殊形式，而非将弯曲时空中的麦克斯韦方程组理解为一种相对论化的推广。
广义相对论 中真空中的电磁理论的方程为
F
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{\displaystyle F_{\alpha \beta }\,=\,\partial _{\alpha }A_{\beta }\,-\,\partial _{\beta }A_{\alpha }\,}
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{\displaystyle {\mathcal {D}^{\mu \nu }\,=\,{\frac {1}{\mu _{0}\,g^{\mu \alpha }\,F_{\alpha \beta }\,g^{\beta \nu }\,{\sqrt {-g}\,}
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{\displaystyle J^{\mu }\,=\,\partial _{\nu }{\mathcal {D}^{\mu \nu }\,}
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{\displaystyle f_{\mu }\,=\,F_{\mu \nu }\,J^{\nu }\,}
其中
g
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{\displaystyle g^{\alpha \beta }\,}
g
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β
{\displaystyle g_{\alpha \beta }\,}
g
{\displaystyle g\,}
度规张量 的行列式，
A
α
{\displaystyle A_{\alpha }\,}
四维势 ，
F
α
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{\displaystyle F^{\alpha \beta }\,}
电磁场的四维协变张量 ，
D
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ν
{\displaystyle D^{\mu \nu }\,}
位移电流 张量，
f
μ
{\displaystyle f_{\mu }\,}
洛伦兹力 的密度，
J
μ
{\displaystyle J_{\mu }\,}
四维电流密度 。尽管方程组中使用了偏导数 ，这些方程仍然在任意曲面坐标变换下是协变的。也就是说如果将偏导数换成协变导数 ，引入的附加项会自动消去从而保持形式不变。
电磁场的四维势
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{\displaystyle A_{\alpha }\,,}
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{\displaystyle {\bar {A}_{\beta }={\frac {\partial x^{\gamma }{\partial {\bar {x}^{\beta }A_{\gamma }\,.}
电磁场 是一个协变的二阶反对称张量，它用电磁势可以定义为
F
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{\displaystyle F_{\alpha \beta }\,=\,\partial _{\alpha }A_{\beta }\,-\,\partial _{\beta }A_{\alpha }\,.}
为证明它的洛伦兹不变性 ，我们对其进行坐标变换
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{\displaystyle {\bar {F}_{\alpha \beta }\,=\,{\frac {\partial {\bar {A}_{\beta }{\partial {\bar {x}^{\alpha }\,-\,{\frac {\partial {\bar {A}_{\alpha }{\partial {\bar {x}^{\beta }\,}
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{\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)\,}
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{\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 }\,}
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{\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 }\,}
=
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{\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)\,}
=
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F
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{\displaystyle =\,{\frac {\partial x^{\delta }{\partial {\bar {x}^{\alpha }{\frac {\partial x^{\gamma }{\partial {\bar {x}^{\beta }\,F_{\delta \gamma }\,.}
这一定义暗示了电磁场张量满足关系
∂
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{\displaystyle \partial _{\lambda }F_{\mu \nu }+\partial _{\mu }F_{\nu \lambda }+\partial _{\nu }F_{\lambda \mu }=0\,}
这一关系包含了法拉第电磁感应定律 和磁场的高斯定理 ，因此也叫做法拉第-高斯方程。具体而言，
∂
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{\displaystyle \partial _{\lambda }F_{\mu \nu }+\partial _{\mu }F_{\nu \lambda }+\partial _{\nu }F_{\lambda \mu }\,}
=
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{\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\,.}
虽然这个关系包含了64个分量方程，但只有四个是独立的。借助电磁场张量的反对称性，可知只有
λ
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{\displaystyle \lambda ,\mu ,\nu \,}
法拉第-高斯方程有时也写作下面的形式
F
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{\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)\,}
=
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3
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{\displaystyle =\,{\frac {1}{3}\left(\partial _{\lambda }F_{\mu \nu }+\partial _{\mu }F_{\nu \lambda }+\partial _{\nu }F_{\lambda \mu }\right)=0\,}
其中按照惯例用分号表示协变导数 ，逗号表示偏导数，方括号表示反对称形式。电磁场张量的协变导数为
F
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μ
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{\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 }\,}
克里斯托费尔符号 ，两个下标是对称的。
位移电场
D
,
{\displaystyle \mathbf {D} \,,}
附屬磁场
H
,
{\displaystyle \mathbf {H} \,,}
D
μ
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g
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F
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−
g
.
{\displaystyle {\mathcal {D}^{\mu \nu }\,=\,{\frac {1}{\mu _{0}\,g^{\mu \alpha }\,F_{\alpha \beta }\,g^{\beta \nu }\,{\sqrt {-g}\,.}
注意这个方程是电磁理论中唯一有度规（即引力）存在的方程。并且这一方程具有尺度不变性，即将度规乘以一个常数不会改变方程的形式。也就是说，引力只能通过改变全局坐标中的光速 来影响相应的电磁场，由于光在引力场中会发生偏折，其效应等同于质量的引力场增加了周围时空的折射率 ，从而影响了对应的位移电场和附屬磁场。
更一般地，当物质中的磁化-极化张量不为零时，我们有
D
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{\displaystyle {\mathcal {D}^{\mu \nu }\,=\,{\frac {1}{\mu _{0}\,g^{\mu \alpha }\,F_{\alpha \beta }\,g^{\beta \nu }\,{\sqrt {-g}\,-\,{\mathcal {M}^{\mu \nu }\,.}
电磁位移张量的坐标变换规则为
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D
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{\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-极化张量存在，它和电磁位移张量具有同样的变换规则。
电磁位移张量的散度 被定义为电流 ，在真空中
J
μ
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.
{\displaystyle J^{\mu }\,=\,\partial _{\nu }{\mathcal {D}^{\mu \nu }\,.}
如果磁化-极化张量存在，则上式替换为
J
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.
{\displaystyle J_{\text{free}^{\mu }\,=\,\partial _{\nu }{\mathcal {D}^{\mu \nu }\,.}
这一方程包含了高斯定理 和安培环路定理 。
无论磁化-极化张量是否存在，电磁位移张量是反对称的这一事实暗示了电流是一个守恒量：
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{\displaystyle \partial _{\mu }J^{\mu }\,=\,\partial _{\mu }\partial _{\nu }{\mathcal {D}^{\mu \nu }=0\,}
这个二阶偏导数为零是由于偏导是对易 的。
电流的安培-高斯定义并不能决定它的大小，因为电磁四维势的大小并未确定。相反地，通常的步骤是使电流等于一个用其他场表示的表达式（主要是电荷），然后解得电磁位移、电磁场和电磁势。
电流是一个逆变 的矢量，因而其变换规则是
J
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{\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]\,.}
变换规则的验证如下：
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{\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)\,}
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{\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]\,+\,}
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{\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]\,}
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{\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]\,+\,}
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{\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 }\,}
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{\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]\,+\,}
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{\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 }\,}
=
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)
.
{\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)\,.}
下面只需证明
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ν
∂
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∂
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ρ
∂
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∂
2
x
σ
∂
x
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∂
x
¯
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=
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\,}
这是微分学 中一条已知定理的应用：
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∂
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∂
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=
∂
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∂
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∂
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∂
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+
∂
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∂
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2
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∂
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{\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
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ν
∂
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∂
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+
∂
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∂
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∂
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∂
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∂
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=
∂
∂
x
β
(
∂
x
σ
∂
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∂
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ν
∂
x
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)
{\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
(
∂
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¯
ν
∂
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¯
ν
)
=
∂
∂
x
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(
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\,.}
洛伦兹力 的密度是一个协变矢量：
f
μ
=
F
μ
ν
J
ν
.
{\displaystyle f_{\mu }\,=\,F_{\mu \nu }\,J^{\nu }\,.}
如果在一个测试粒子上的作用力只有引力和电磁力，则有
d
p
α
d
t
=
Γ
α
γ
β
p
β
d
x
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d
t
+
q
F
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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 }\,}
四维动量 ，
t
{\displaystyle t\,}
世界线 参数化的任意时间坐标。式中第一项含有克里斯托费尔符号，表示引力项，第二项含有电荷，表示电磁力项。
方程具有洛伦兹不变性，对时间坐标变换的验证方法为将方程乘以
d
t
d
t
¯
{\displaystyle {\frac {dt}{d{\bar {t}\,}
链式法则 。
对空间坐标变换，通过对克里斯托费尔符号进行坐标变换
Γ
¯
α
γ
β
=
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x
¯
β
∂
x
ϵ
∂
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δ
∂
x
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α
∂
x
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∂
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γ
Γ
δ
ζ
ϵ
+
∂
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∂
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 }\,}
我们得到
d
p
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α
d
t
−
Γ
¯
α
γ
β
p
¯
β
d
x
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γ
d
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−
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
∂
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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
(
∂
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∂
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α
)
p
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−
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∂
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β
∂
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η
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2
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)
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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
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t
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)
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d
x
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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\,}
在真空中经典电磁场的拉格朗日量 （在密度意义下的单位为焦耳/米3 ）是一个标量：
L
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4
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0
F
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F
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g
+
A
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{\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 }\,.}
如果我们将自由电流与束缚电流分离开来，拉格朗日量则变为
L
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−
1
4
μ
0
F
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β
F
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−
g
+
A
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J
free
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+
1
2
F
α
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M
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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 }\,.}
作为爱因斯坦引力场方程 的源，电磁场的应力-能量张量 是一个协变的对称张量
T
μ
ν
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1
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μ
ν
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g
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g
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)
{\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 })\,}
并且它是无迹的
T
μ
ν
g
μ
ν
=
0
{\displaystyle T_{\mu \nu }g^{\mu \nu }\,=\,0\,}
这是由于电磁场的传播速度在不同参考系下是不变的。
为了表达能量和动量的守恒律，电磁应力-能量张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}.}
对于上式可以证明
T
μ
ν
;
ν
+
f
μ
=
0
{\displaystyle {\mathfrak {T}_{\mu }^{\nu }_{;\nu }\,+\,f_{\mu }\,=\,0\,}
并可重写为
−
T
μ
ν
,
ν
=
−
Γ
μ
ν
σ
T
σ
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+
f
μ
{\displaystyle -{\mathfrak {T}_{\mu }^{\nu }_{,\nu }\,=\,-\Gamma _{\mu \nu }^{\sigma }{\mathfrak {T}_{\sigma }^{\nu }\,+\,f_{\mu }\,}
这个方程的意义是，电磁场能量的减少相当于电磁场对引力场以及通过洛伦兹力对物质所做的功，类似地，电磁场动量的减少率相当于作用在引力场上的电磁力以及作用在物质上的洛伦兹力。
守恒律的推导过程为
T
μ
ν
;
ν
+
f
μ
=
−
1
μ
0
(
F
μ
α
;
ν
g
α
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F
β
γ
g
γ
ν
+
F
μ
α
g
α
β
F
β
γ
;
ν
g
γ
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−
1
2
δ
μ
ν
F
σ
α
;
ν
g
α
β
F
β
ρ
g
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σ
)
−
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+
{\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
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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
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)
−
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}\,}
所得的结果是为零的，因为它等于自身的负值（对照推导的第二步）。
从电磁理论的狭义相对论形式 可以借助电磁场张量修改得到非齐次的电磁波方程
◻
F
a
b
=
d
e
f
F
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b
;
d
d
=
−
2
R
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c
b
d
F
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+
R
a
e
F
e
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−
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}\,}
是黎曼张量 的协变形式，而
◻
{\displaystyle \Box \,}
达朗贝尔算符 在协变导数情形下的推广。
波方程可以写成四维势的形式（参见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}
是里奇张量 。而这里假设了洛伦茨规范 在弯曲时空中的推广具有下面的形式
A
a
;
a
=
0
{\displaystyle {A^{a}_{;a}=0}
这个波方程与平直时空中的波方程具有十分类似的形式，除了导数被替换为协变导数，以及在表示源的项中多了一项描述时空曲率的项。这个方程和弯曲时空中的洛伦兹力也有相似之处，这里的四维势
A
a
{\displaystyle {A^{a}\,}
在考虑麦克斯韦方程组本身与背景时空无关时，时空度规在广义相对论中被认为是受电磁场影响的动态变化的变量，这导致了电磁波方程和麦克斯韦方程组是非线性的。这一点可以从爱因斯坦场方程 中的曲率张量依赖于应力-能量张量 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}
是爱因斯坦张量 ，
G
{\displaystyle G\,}
万有引力常数 ，
R
{\displaystyle R\,}
标量曲率 ）是里奇张量的迹。应力-能量张量来源于粒子的应力 和能量，但同时也来源于电磁场，这造就了非线性。
![{\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)
