Electromagnetic Field Tensor

Let us now investigate whether we can write the components of the electric and magnetic fields as the components of some proper 4-tensor. There is an obvious problem here. How can we identify the components of the magnetic field, which is a pseudo-vector, with any of the components of a proper-4-tensor? The former components transform differently under parity inversion than the latter components. Consider a proper-3-tensor whose covariant components are written $B_{ik}$ , and which is antisymmetric:

$$B_{ij} = -B_{ji}.$$ (1762)

This immediately implies that all of the diagonal components of the tensor are zero. In fact, there are only three independent non-zero components of such a tensor. Could we, perhaps, use these components to represent the components of a pseudo-3-vector? Let us write

$$B^i = \frac{1}{2}\, \epsilon^{\,ijk} \,B_{jk}.$$ (1763)

It is clear that $B^i$ transforms as a contravariant pseudo-3-vector. It is easily seen that

$$B^{ij}=B_{ij} = \left(\begin{array}{ccc} 0& B_z & -B_y\\ [0.5ex] -B_z & 0 & B_x \\ [0.5ex] B_y & -B_x & 0 \end{array} \right),$$ (1764)

where $B^{\,1}=B_1 \equiv B_x$ , et cetera. In this manner, we can actually write the components of a pseudo-3-vector as the components of an antisymmetric proper-3-tensor. In particular, we can write the components of the magnetic field ${\bf B}$ in terms of an antisymmetric proper magnetic field 3-tensor which we shall denote $B_{ij}$ .

Let us now examine Equations (1762) and (1763) more carefully. Recall that ${\mit\Phi}_\mu = (-c\,{\bf A},\, \phi)$ and $\partial_\mu =(\nabla, \,c^{-1} \partial/\partial t)$ . It follows that we can write Equation (1762) in the form

$$E_i = -\partial_i {\mit \Phi}_4 +\partial_4 {\mit \Phi}_i.$$ (1765)

Likewise, Equation (1763) can be written

$$c\,B^{\,i} = \frac{1}{2}\,\epsilon^{\,ijk} \,c\,B_{jk} = -\epsilon^{\,ijk}\,\partial_j {\mit \Phi}_k.$$ (1766)

Let us multiply this expression by $\epsilon_{\,iab}$ , making use of the identity

$$\epsilon_{iab} \,\epsilon^{\,ijk} = \delta_a^{\,j} \,\delta_b^{\,k} - \delta_b^{\,j}\,\delta_a^{\,k}.$$ (1767)

We obtain

$$\frac{c}{2} \left(B_{ab} - B_{ba}\right) = - \partial_a {\mit \Phi}_b +\partial_b{\mit \Phi}_a,$$ (1768)

or

$$c\,B_{ij} = -\partial_i {\mit \Phi}_j + \partial_j {\mit \Phi}_i,$$ (1769)

because $B_{ij} = -B_{ji}$ .

Let us define a proper-4-tensor whose covariant components are given by

$$F_{\mu\nu} = \partial_\mu {\mit \Phi}_\nu -\partial_\nu {\mit \Phi}_\mu.$$ (1770)

It is clear that this tensor is antisymmetric:

$$F_{\mu\nu} = -F_{\nu\mu}.$$ (1771)

This implies that the tensor only possesses six independent non-zero components. Maybe it can be used to specify the components of ${\bf E}$ and ${\bf B}$ ?

Equations (1767) and (1772) yield

$$F_{4i} = \partial_4{\mit\Phi}_i -\partial_i{\mit\Phi}_4 = E_i.$$ (1772)

Likewise, Equations (1771) and (1772) imply that

$$F_{ij} = \partial_i{\mit\Phi}_j -\partial_j{\mit\Phi}_i= - c\, B_{ij}.$$ (1773)

Thus,

$$F_{i4} =-F_{4i} = -E_i,$$ (1774)

$$F_{ij} = -F_{ji} = -c\,B_{ij}.$$ (1775)

In other words, the completely space-like components of the tensor specify the components of the magnetic field, whereas the hybrid space and time-like components specify the components of the electric field. The covariant components of the tensor can be written

$$F_{\mu\nu} = \left\lgroup \begin{array}{cccc} 0 & -c\,B_z & +c\,B... ...B_y& +c\,B_x& 0& -E_z\\ [0.5ex] +E_x & +E_y &+E_z & 0\end{array}\right \rgroup.$$ (1776)

Not surprisingly, $F_{\mu\nu}$ is usually called the electromagnetic field tensor. The previous expression, which appears in all standard textbooks, is very misleading. Taken at face value, it is simply wrong. We cannot form a proper-4-tensor from the components of a proper-3-vector and a pseudo-3-vector. The expression only makes sense if we interpret $B_x$ (say) as representing the component $B_{23}$ of the proper magnetic field 3-tensor $B_{ij}$

The contravariant components of the electromagnetic field tensor are given by

$$F^{\,i4} =-F^{\,4i} = +E^{\,i},$$ (1777)

$$F^{\,ij} = -F^{\,ji} = -c\,B^{\,ij},$$ (1778)

or

$$F^{\mu\nu} = \left\lgroup \begin{array}{cccc} 0 & -c\,B_z & +c\,B... ...B_y& +c\,B_x& 0& +E_z\\ [0.5ex] -E_x & -E_y &-E_z & 0\end{array}\right \rgroup.$$ (1779)

Let us now consider two of Maxwell's equations:

$$\nabla\cdot {\bf E} =\frac{\rho}{\epsilon_0},$$ (1780)

$$\nabla\times {\bf B} = \mu_0 \left({\bf j} + \epsilon_0\, \frac{\partial {\bf E}} {\partial t}\right).$$ (1781)

Recall that the 4-current is defined $J^{\,\mu} = ({\bf j}, \,\rho \,c)$ . The first of these equations can be written

$$\partial_i E^{\,i} = \partial_i F^{\,i4} +\partial_4 F^{\,44} = \frac{J^{\,4}}{c\,\epsilon_0}.$$ (1782)

because $F^{\,44}= 0$ . The second of these equations takes the form

$$\epsilon^{\,ijk} \,\partial_j( c\,B_k) - \partial_4 E^{\,i} = \ep... ...n_{kab}\, c \,B^{\,ab} ) + \partial_4 F^{\,4i} = \frac{J^{\,i}}{c\,\epsilon_0}.$$ (1783)

Making use of Equation (1769), the previous expression reduces to

$$\frac{1}{2}\,\partial_j(c\, B^{\,ij} - c \,B^{\,ji}) +\partial_4 ... ...,4i} = \partial_j F^{\,ji} +\partial_4 F^{\,4i} =\frac{J^{\,i}}{c\,\epsilon_0}.$$ (1784)

Equations (1784) and (1786) can be combined to give

$$\partial_\mu F^{\,\mu\nu} = \frac{J^{\,\nu}}{c\,\epsilon_0}.$$ (1785)

This equation is consistent with the equation of charge continuity, $\partial_\mu J^{\,\mu} =0$ , because of the antisymmetry of the electromagnetic field tensor.