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}

 

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