# Electromagnetic field tensor: a couple of Maxwell’s equations

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 6; Problem 6.10, 7.2.

The electromagnetic field tensor ${F^{ij}}$ is

$\displaystyle F^{ij}=\left[\begin{array}{cccc} 0 & E_{x} & E_{y} & E_{z}\\ -E_{x} & 0 & B_{z} & -B_{y}\\ -E_{y} & -B_{z} & 0 & B_{x}\\ -E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (1)$

The various electromagnetic laws can be expressed by the equation

$\displaystyle \partial_{i}F_{jk}+\partial_{k}F_{ij}+\partial_{j}F_{ki}=0 \ \ \ \ \ (2)$

The lowered version ${F_{ij}}$ of ${F^{ij}}$ in the flat metric of special relativity is

$\displaystyle F_{ij}=\left[\begin{array}{cccc} 0 & -E_{x} & -E_{y} & -E_{z}\\ E_{x} & 0 & B_{z} & -B_{y}\\ E_{y} & -B_{z} & 0 & B_{x}\\ E_{z} & B_{y} & -B_{x} & 0 \end{array}\right] \ \ \ \ \ (3)$

Because ${F_{ij}=-F_{ji}}$, if we set two of the indices equal in this equation, the LHS is identically zero. For example if ${i=j}$:

 $\displaystyle \partial_{i}F_{ik}+\partial_{k}F_{ii}+\partial_{i}F_{ki}$ $\displaystyle =$ $\displaystyle -\partial_{i}F_{ki}+0+\partial_{i}F_{ki}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (5)$

Since the three terms are cyclic permutations of each other, setting any other pair of indices equal gives the same result.

If we choose ${i=y}$, ${j=x}$, ${k=z}$, we get

$\displaystyle -\partial_{y}B_{y}-\partial_{z}B_{z}-\partial_{x}B_{x}=0 \ \ \ \ \ (6)$

This is the law ${\nabla\cdot\mathbf{B}=0}$.

If we now choose ${i=y}$, ${j=x}$, ${k=t}$, we get

$\displaystyle -\partial_{y}E_{x}+\partial_{t}B_{z}+\partial_{x}E_{y}=0 \ \ \ \ \ (7)$

This is the ${z}$ component of Faraday’s law ${\nabla\times\mathbf{E}+\frac{\partial\mathbf{B}}{\partial t}=0}$.

Other choices give the remaining components of Faraday’s law.