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.

2 thoughts on “Electromagnetic field tensor: a couple of Maxwell’s equations

  1. Pingback: Electromagnetic field tensor: cyclic derivative relation | Physics tutorials

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