The dual electromagnetic field tensor

We have seen that it is possible to write the components of the electric and magnetic fields as the components of a proper-4-tensor. Is it also possible to write the components of these fields as the components of some pseudo-4-tensor? It is obvious that we cannot identify the components of the proper-3-vector ${\bfm E}$ with any of the components of a pseudo-tensor. However, we can represent the components of ${\bfm E}$ in terms of those of an antisymmetric pseudo-3-tensor $E_{ij}$ by writing

$$E^i =\frac{1}{2}\,\epsilon^{ijk} E_{jk}.$$ (218)

It is easily demonstrated that

$$E^{ij}=E_{ij} = \left(\begin{array}{ccc} 0& E_z & -E_y\\ [0.... ... -E_z & 0 & E_x \\ [0.5ex] E_y & -E_x & 0 \end{array} \right),$$ (219)

in a right-handed coordinate system.

Consider the dual electromagnetic field tensor $G^{\mu\nu}$, which is defined

$$G^{\mu\nu} = \frac{1}{2}\, \epsilon^{\mu\nu\alpha\beta} F_{\alpha\beta}.$$ (220)

This tensor is clearly an antisymmetric pseudo-4-tensor. We have

$$G^{4i} = \frac{1}{2}\,\epsilon^{4ijk} \,F_{jk} = -\frac{1}{2... ...{ijk4}\,F_{jk} = \frac{1}{2}\,\epsilon^{ijk} \,cB_{jk} = cB^i,$$ (221)

plus

$$G^{ij} = \frac{1}{2}\,(\epsilon^{ijk4} F_{k4} + \epsilon^{ij4k} F_{4k}) = \epsilon^{ijk} F_{k4},$$ (222)

where use has been made of $F_{\mu\nu}=-F_{\nu\mu}$. The above expression yields

$$G^{ij} = -\epsilon^{ijk} \,E_k =-\frac{1}{2}\,\epsilon^{ijk} \epsilon_{kab} \,E^{ab} = - E^{ij}.$$ (223)

It follows that

$$G^{i4} = - G^{4i} = - cB^i,$$ (224)

$$G^{ij} = -G^{ji} = -E^{ij},$$ (225)

or

$$G^{\mu\nu} = \left\lgroup \begin{array}{cccc} 0 & -E_z & +E_... ...z\\ [0.5ex] +cB_x &+ cB_y &+cB_z & 0\end{array}\right \rgroup.$$ (226)

The above expression is, again, slightly misleading, since $E_x$ stands for the component $E^{23}$ of the pseudo-3-tensor $E^{ij}$ and not for an element of the proper-3-vector ${\bfm E}$. Of course, in this case $B_x$ really does represent the first element of the pseudo-3-vector ${\bfm B}$. Note that the elements of $G^{\mu\nu}$ are obtained from those of $F^{\mu\nu}$ by making the transformation $cB^{ij} \rightarrow E^{ij}$ and $E^i\rightarrow -cB^i$.

The covariant elements of the dual electromagnetic field tensor are given by

$$G_{i4} = - G_{4i} = + cB_i,$$ (227)

$$G_{ij} = -G_{ji} = -E_{ij},$$ (228)

or

$$G_{\mu\nu} = \left\lgroup \begin{array}{cccc} 0 & -E_z & +E_... ...z\\ [0.5ex] -cB_x & -cB_y &-cB_z & 0\end{array}\right \rgroup.$$ (229)

The elements of $G_{\mu\nu}$ are obtained from those of $F_{\mu\nu}$ by making the transformation $cB_{ij} \rightarrow E_{ij}$ and $E_i\rightarrow -cB_i$.

Let us now consider the two Maxwell equations

$$\nabla\!\cdot\!{\bfm B} = 0,$$ (230)

$$\nabla\wedge{\bfm E} = -\frac{\partial{\bfm B}}{\partial t}.$$ (231)

The first of these equations can be written

$$-\partial_i\,cB^i = \partial_i G^{i4} +\partial_4 G^{44} = 0,$$ (232)

since $G^{44}=0$. The second equation takes the form

$$\epsilon^{ijk}\partial_j E_k = \epsilon^{ijk}\partial_j (1/2... ...psilon_{kab} E^{ab}) = \partial_j E^{ij} = -\partial_4\, cB^i,$$ (233)

or

$$\partial_j G^{ji} + \partial_4 G^{4i} = 0.$$ (234)

Equations (2.186) and (2.188) can be combined to give

$$\partial_\mu G^{\mu\nu} = 0.$$ (235)

Thus, we conclude that Maxwell's equations for the electromagnetic fields are equivalent to the following pair of 4-tensor equations:

$$\partial_\mu F^{\mu\nu} = \frac{J^\nu}{c\,\epsilon_0},$$ (236)

$$\partial_\mu G^{\mu\nu} = 0.$$ (237)

It is obvious from the form of these equations that the laws of electromagnetism are invariant under translations, rotations, special Lorentz transformations, parity inversions, or any combination of these transformations.