the Jacobian of Lorentz Transform is:

$\partial / \partial x_i = L_{ij} \partial / \partial y_j$

The Maxwell’s equations can be viewed like this for matching the order and sign:

$-\nabla \cdot \vec{D} = -\rho$ ……………..(1)

$\partial \vec{D} / \partial x_0 - \nabla \times \vec{H} = - \vec{j}$……..(2)

and

$\nabla \cdot \vec{B} = 0$ ……………….(3)

$\partial \vec{B} / \partial x_0 + \nabla \times \vec{E} = 0$ ………..(4)

Lets simplify (2). the cross product can be changed to Einstein notation.

$\nabla \times \vec{H} = \frac {\partial}{\partial x_i} \epsilon_{ i j k }H_{j k }$

where $i , j , k = 1,2,3$ . Set a matrix F be :

$F_{ij} = \epsilon_{i j k} H_{j k } = \begin {pmatrix} 0 & H_3 & -H_2 \\ - H_3 & 0 & H_1 \\ H_2 & -H_1 & 0 \end {pmatrix}$

If we let the index i be zero. $F_{0 j } = D_j$,

$F_{ij} =\begin {pmatrix} D_1 & D_2 & D_ 3\\ 0& H_3 & -H_2 \\ - H_3 & 0 & H_1 \\ H_2 & -H_1 & 0 \end {pmatrix}$

the equation (2) can be written like :

$\partial /\partial x_i F_{i j } = - j_{i}$

where $i = 0, 1, 2, 3$ and $j = 1,2,3$. each column represent 1 equation. notices that the minus sign before the rot in equation (2) is automatically included.

If we extend the j to be zero, define $F_{i 0} = -D_i$, $D_0 = 0$. the minus sign perseveres the matrix F to be anti-symmetric. Then, the equation 1 will be absorbed. if we define a 4-current.

$\vec{J}= ( \rho , j_1, j_2, j_3 )$

the equation (1) and (2) can be combined into:

$\partial /\partial x_i F_{i j } = - J_i$

where

$F_{ij} =\begin {pmatrix} 0 & D_1 & D_2 & D_ 3\\ -D_1 & 0& H_3 & -H_2 \\ -D_2 & - H_3 & 0 & H_1 \\ -D_3 & H_2 & -H_1 & 0 \end {pmatrix}$

Using similar method, by defining a matrix G, such that:

$G_{ij} =\begin {pmatrix} 0 & B_1 & B_2 & B_ 3\\ -B_1 & 0& -E_3 & E_2 \\ -B_2 & E_3 & 0 & -E_1 \\ -B_3 & -E_2 & E_1 & 0 \end {pmatrix}$

$\partial /\partial x_i G_{i j } =0$

The transformation of field is simply use the Jacobian. Since the 4-current also need to transform, and the Zero 4-vector in equation G also. Thus

$L_{\alpha i} F_{i j } L_{j \beta} = F_{\alpha \beta}$