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 ${\bf E}$ 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

$$E^i =\frac{1}{2} \epsilon^{ijk} E_{jk}.$$ (1491)

It is easily demonstrated that

$$E^{ij}=E_{ij} = \left(\begin{array}{ccc} 0& E_z & -E_y\ [0.... ... -E_z & 0 & E_x \ [0.5ex] E_y & -E_x & 0 \end{array} \right),$$ (1492)

in a right-handed coordinate system.

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

$$G^{\mu\nu} = \frac{1}{2} \epsilon^{\mu\nu\alpha\beta} F_{\alpha\beta}.$$ (1493)

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

$$G^{4i} = \frac{1}{2} \epsilon^{4ijk} F_{jk} = -\frac{1}{2... ...4} F_{jk} = \frac{1}{2} \epsilon^{ijk} c B_{jk} = c B^i,$$ (1494)

plus

$$G^{ij} = \frac{1}{2} (\epsilon^{ijk4} F_{k4} + \epsilon^{ij4k} F_{4k}) = \epsilon^{ijk} F_{k4},$$ (1495)

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

$$G^{ij} = -\epsilon^{ijk} E_k =-\frac{1}{2} \epsilon^{ijk} \epsilon_{kab} E^{ab} = - E^{ij}.$$ (1496)

It follows that

$$G^{i4} = - G^{4i} = - c B^i,$$ (1497)

$$G^{ij} = -G^{ji} = -E^{ij},$$ (1498)

or

$$G^{\mu\nu} = \left\lgroup \begin{array}{cccc} 0 & -E_z & +E_... ....5ex] +c B_x &+ c B_y &+c B_z & 0\end{array}\right \rgroup.$$ (1499)

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 ${\bf E}$. 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

$$G_{i4} = - G_{4i} = + cB_i,$$ (1500)

$$G_{ij} = -G_{ji} = -E_{ij},$$ (1501)

or

$$G_{\mu\nu} = \left\lgroup \begin{array}{cccc} 0 & -E_z & +E_... ....5ex] -c B_x & -c B_y &-c B_z & 0\end{array}\right \rgroup.$$ (1502)

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

$$\nabla\!\cdot\!{\bf B} = 0,$$ (1503)

$$\nabla\times{\bf E} = -\frac{\partial{\bf B}}{\partial t}.$$ (1504)

The first of these equations can be written

$$-\partial_i (c B^i) = \partial_i G^{i4} +\partial_4 G^{44} = 0,$$ (1505)

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

$$\epsilon^{ijk}\partial_j E_k = \epsilon^{ijk}\partial_j (1/2... ...on_{kab} E^{ab}) = \partial_j E^{ij} = -\partial_4 (c B^i),$$ (1506)

or

$$\partial_j G^{ji} + \partial_4 G^{4i} = 0.$$ (1507)

Equations (1505) and (1507) can be combined to give

$$\partial_\mu G^{\mu\nu} = 0.$$ (1508)

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

$$\partial_\mu F^{\mu\nu} = \frac{J^\nu}{c \epsilon_0},$$ (1509)

$$\partial_\mu G^{\mu\nu} = 0.$$ (1510)

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.