Maxwell方程组的洛伦兹协变¶
电磁场张量¶
\[
F^{\mu \gamma}=\begin{pmatrix}
0 & \frac{E_x}{c} & \frac{E_y}{c} & \frac{E_z}{c} \\
-\frac{E_x}{c} & 0 & B_z & -B_y \\
-\frac{E_y}{c} & -B_z & 0 & B_x \\
-\frac{E_z}{c} & B_y & -B_x & 0 \\
\end{pmatrix}
\]
同时存在与 \(F^{\mu \gamma}\)对偶的场张量 \(G^{\mu \gamma}\)
麦克斯韦方程的场张量形式¶
\[
\frac{\partial\boldsymbol{F}^{\mu \gamma}}{\partial{x^\gamma}}=\partial_\gamma \boldsymbol{F}^{\mu \gamma} = \mu_0J^\mu
\]
\[
\frac{\partial\boldsymbol{G}^{\mu \gamma}}{\partial{x^\gamma}} = 0 \Leftrightarrow \sum_{\lambda, \mu, \gamma}\partial_\lambda \boldsymbol{F}_{\mu \gamma}
\]
其中下标表示对\(\lambda, \mu, \gamma\)轮换求和；方程(2)和方程(3)分别对应两个 \(Maxwell\)方程
Lorentz协变性证明¶
\(\partial_\gamma \boldsymbol{F}^{\mu \gamma} = \mu_0J^\mu\) 协变证明¶
\[
\partial_\gamma \boldsymbol{F}^{\mu \gamma}=\partial_\gamma \Lambda_\lambda^\mu \Lambda_\sigma^\gamma\boldsymbol{F}^{\mu \gamma}
\]
同时，可以通过链式法则证明，对于四维矢量的梯度\(\partial_\mu\)也满足Lorentz变化关系式，也就是\(\frac{\partial}{\partial x^\mu}\)本身构成一个四维矢量，因此可以对(4)继续做Lorentz变换：
\[
\partial_\gamma \boldsymbol{F}^{\mu \gamma}=\partial_\gamma \Lambda_\lambda^\mu \Lambda_\sigma^\gamma\boldsymbol{\bar{F}}^{\lambda \sigma}=\Lambda_\lambda^\mu \Lambda_\sigma^\gamma \Lambda_{\gamma}^\sigma \bar{\partial}_\sigma \boldsymbol{\bar{F}}^{\lambda \sigma}=\Lambda_\lambda^\mu (\Lambda_\sigma^\gamma \Lambda_{\gamma}^\sigma) \bar{\partial}_\sigma \boldsymbol{\bar{F}}^{\lambda \sigma}
\]
括号中是一对正变换和逆变换：
\[
\Lambda_\sigma^\gamma \Lambda_{\gamma}^\sigma = 1
\]
因此：
\[
\partial_\gamma \boldsymbol{F}^{\mu \gamma}=\Lambda_\lambda^\mu \bar{\partial}_\sigma \boldsymbol{\bar{F}}^{\lambda \sigma}
\]
等式两边同时乘上Lorentz逆变换矩阵：
\[
\Lambda_\mu^\lambda \partial_\gamma \boldsymbol{F}^{\mu \gamma}=\bar{\partial}_\sigma \boldsymbol{\bar{F}}^{\lambda \sigma}
\]
利用(2)式：
\[
\Lambda_\mu^{\lambda}\mu_0J^\mu=\mu_0\bar{J}^\mu=\bar{\partial}_\sigma \boldsymbol{\bar{F}}^{\lambda \sigma}
\]
\(\sum_{\lambda, \mu, \gamma}\partial_\lambda \boldsymbol{F}_{\mu \gamma}\)协变证明¶
\[
\partial_\lambda \boldsymbol{F}_{\mu \gamma}=\Lambda^\mu_\sigma\Lambda^\gamma_\rho\Lambda_\lambda^\kappa\bar{\partial}_\kappa \boldsymbol{\bar{F}}_{\sigma \rho}
\]
对于剩下两个表达式，\(\Lambda^\mu_\sigma\Lambda^\gamma_\rho\Lambda_\lambda^\kappa\bar{\partial}_\kappa\)交换求和不变，因此：
$$ \sum_{\kappa, \sigma \rho}\bar{\partial}\kappa \boldsymbol{\bar{F}}=0 $$ 结合(9),(11)，证明了Maxwell方程组具有Lorentz协变.