The 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:

$$B_{ij} = -B_{ji}.$$ (1467)

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

$$B^i = \frac{1}{2} \epsilon^{ijk} B_{jk}.$$ (1468)

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

$$B^{ij}=B_{ij} = \left(\begin{array}{ccc} 0& B_z & -B_y\ [0.... ... -B_z & 0 & B_x \ [0.5ex] B_y & -B_x & 0 \end{array} \right),$$ (1469)

where $B^1=B_1 \equiv B_x$, etc. 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 Eqs. (1465) and (1466) 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 Eq. (1465) in the form

$$E_i = -\partial_i {\mit \Phi}_4 +\partial_4 {\mit \Phi}_i.$$ (1470)

Likewise, Eq. (1466) can be written

$$c B^i = \frac{1}{2} \epsilon^{ijk} c B_{jk} = -\epsilon^{ijk} \partial_j {\mit \Phi}_k.$$ (1471)

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

$$\epsilon_{iab} \epsilon^{ijk} = \delta_a^j \delta_b^k - \delta_b^j \delta_a^k.$$ (1472)

We obtain

$$\frac{c}{2} \left(B_{ab} - B_{ba}\right) = - \partial_a {\mit \Phi}_b +\partial_b{\mit \Phi}_a,$$ (1473)

or

$$c B_{ij} = -\partial_i {\mit \Phi}_j + \partial_j {\mit \Phi}_i,$$ (1474)

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

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

$$F_{\mu\nu} = \partial_\mu {\mit \Phi}_\nu -\partial_\nu {\mit \Phi}_\mu.$$ (1475)

It is clear that this tensor is antisymmetric:

$$F_{\mu\nu} = -F_{\nu\mu}.$$ (1476)

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 (1470) and (1475) yield

$$F_{4i} = \partial_4{\mit\Phi}_i -\partial_i{\mit\Phi}_4 = E_i.$$ (1477)

Likewise, Eqs. (1474) and (1475) imply that

$$F_{ij} = \partial_i{\mit\Phi}_j -\partial_j{\mit\Phi}_i= - c B_{ij}.$$ (1478)

Thus,

$$F_{i4} = -F_{4i} = -E_i,$$ (1479)

$$F_{ij} = -F_{ji} = -c B_{ij}.$$ (1480)

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

$$F_{\mu\nu} = \left\lgroup \begin{array}{cccc} 0 & -c B_z & ... ...-E_z\ [0.5ex] +E_x & +E_y &+E_z & 0\end{array}\right \rgroup.$$ (1481)

Not surprisingly, $F_{\mu\nu}$ is usually called the electromagnetic field tensor. The above 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

$$F^{i4} = -F^{4i} = +E^i,$$ (1482)

$$F^{ij} = -F^{ji} = -c B^{ij},$$ (1483)

or

$$F^{\mu\nu} = \left\lgroup \begin{array}{cccc} 0 & -c B_z & ... ...+E_z\ [0.5ex] -E_x & -E_y &-E_z & 0\end{array}\right \rgroup.$$ (1484)

Let us now consider two of Maxwell's equations:

$$\nabla\!\cdot\!{\bf E} = \frac{\rho}{\epsilon_0},$$ (1485)

$$\nabla\times{\bf B} = \mu_0 \left({\bf j} + \epsilon_0 \frac{\partial {\bf E}} {\partial t}\right).$$ (1486)

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

$$\partial_i E^i = \partial_i F^{i4} +\partial_4 F^{44} = \frac{J^4}{c \epsilon_0}.$$ (1487)

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

$$\epsilon^{ijk} \partial_j( c B_k) - \partial_4 E^i = \ep... ... c B^{ab} ) + \partial_4 F^{4i} = \frac{J^i}{c \epsilon_0}.$$ (1488)

Making use of Eq. (1472), the above expression reduces to

$$\frac{1}{2} \partial_j(c B^{ij} - c B^{ji}) +\partial_4... ...rtial_j F^{ji} +\partial_4 F^{4i} =\frac{J^i}{c \epsilon_0}.$$ (1489)

Equations (1487) and (1489) can be combined to give

$$\partial_\mu F^{\mu\nu} = \frac{J^\nu}{c \epsilon_0}.$$ (1490)

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