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 ${\bf E}$ with any of the components of a pseudo-tensor. However, we can represent the components of ${\bf E}$ in terms of those of an antisymmetric pseudo-3-tensor $E_{ij}$ by writing

$$E^{\,i} =\frac{1}{2}\,\epsilon^{\,ijk}\, E_{jk}.$$ (1786)

It is easily demonstrated that

$$E^{\,ij}=E_{ij} = \left(\begin{array}{ccc} 0& E_z & -E_y\\ [0.5ex] -E_z & 0 & E_x \\ [0.5ex] E_y & -E_x & 0 \end{array} \right),$$ (1787)

in a right-handed coordinate system.

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

$$G^{\,\mu\nu} = \frac{1}{2}\, \epsilon^{\,\mu\nu\alpha\beta}\, F_{\alpha\beta}.$$ (1788)

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

$$G^{\,4i} = \frac{1}{2}\,\epsilon^{\,4ijk} \,F_{jk} = -\frac{1}{2}... ...ilon^{\,ijk4}\,F_{jk} = \frac{1}{2}\,\epsilon^{\,ijk} \,c\,B_{jk} = c\,B^{\,i},$$ (1789)

plus

$$G^{\,ij} = \frac{1}{2}\,(\epsilon^{\,ijk4}\, F_{k4} + \epsilon^{\,ij4k} \,F_{4k}) = \epsilon^{\,ijk}\, F_{k4},$$ (1790)

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

$$G^{\,ij} = -\epsilon^{\,ijk} \,E_k =-\frac{1}{2}\,\epsilon^{\,ijk} \epsilon_{kab} \,E^{\,ab} = - E^{\,ij}.$$ (1791)

It follows that

$$G^{i4} = - G^{4i} = - c\,B^i,$$ (1792)

$$G^{ij} = -G^{ji} = -E^{ij},$$ (1793)

or

$$G^{\mu\nu} = \left\lgroup \begin{array}{cccc} 0 & -E_z & +E_y & -... ...& 0& -c\,B_z\\ [0.5ex] +c\,B_x &+ c\,B_y &+c\,B_z & 0\end{array}\right \rgroup.$$ (1794)

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

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

$$G_{i4} = - G_{4i} = + cB_i,$$ (1795)

$$G_{ij} = -G_{ji} = -E_{ij},$$ (1796)

or

$$G_{\mu\nu} = \left\lgroup \begin{array}{cccc} 0 & -E_z & +E_y & +... ...& 0& +c\,B_z\\ [0.5ex] -c\,B_x & -c\,B_y &-c\,B_z & 0\end{array}\right \rgroup.$$ (1797)

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

Let us now consider the two Maxwell equations

$$\nabla\cdot{\bf B} =0,$$ (1798)

$$\nabla\times {\bf E} = -\frac{\partial{\bf B}}{\partial t}.$$ (1799)

The first of these equations can be written

$$-\partial_i\,(c\,B^{\,i}) = \partial_i G^{\,i4} +\partial_4 G^{\,44} = 0,$$ (1800)

because $G^{\,44}=0$ . The second equation takes the form

$$\epsilon^{\,ijk}\partial_j E_k = \epsilon^{\,ijk}\partial_j (1/2\,\epsilon_{kab} E^{\,ab}) = \partial_j E^{\,ij} = -\partial_4\, (c\,B^{\,i}),$$ (1801)

or

$$\partial_j G^{\,ji} + \partial_4 G^{\,4i} = 0.$$ (1802)

Equations (1802) and (1804) can be combined to give

$$\partial_\mu G^{\,\mu\nu} = 0.$$ (1803)

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

$$\partial_\mu F^{\,\mu\nu} = \frac{J^{\,\nu}}{c\,\epsilon_0},$$ (1804)

$$\partial_\mu G^{\,\mu\nu} =0.$$ (1805)

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.