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Next: The dual electromagnetic field Up: Relativity and electromagnetism Previous: Tensors and pseudo-tensors

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}.
\end{displaymath} (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}.
\end{displaymath} (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),
\end{displaymath} (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.
\end{displaymath} (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.
\end{displaymath} (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.
\end{displaymath} (1472)

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

c B_{ij} = -\partial_i {\mit \Phi}_j + \partial_j {\mit \Phi}_i,
\end{displaymath} (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.
\end{displaymath} (1475)

It is clear that this tensor is antisymmetric:
F_{\mu\nu} = -F_{\nu\mu}.
\end{displaymath} (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.
\end{displaymath} (1477)

Likewise, Eqs. (1474) and (1475) imply that
F_{ij} = \partial_i{\mit\Phi}_j -\partial_j{\mit\Phi}_i= - c  B_{ij}.
\end{displaymath} (1478)

$\displaystyle F_{i4}$ $\textstyle =$ $\displaystyle -F_{4i} = -E_i,$ (1479)
$\displaystyle F_{ij}$ $\textstyle =$ $\displaystyle -F_{ji} = -c B_{ij}.$ (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
\end{displaymath} (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

$\displaystyle F^{i4}$ $\textstyle =$ $\displaystyle -F^{4i} = +E^i,$ (1482)
$\displaystyle F^{ij}$ $\textstyle =$ $\displaystyle -F^{ji} = -c B^{ij},$ (1483)

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
\end{displaymath} (1484)

Let us now consider two of Maxwell's equations:

$\displaystyle \nabla\!\cdot\!{\bf E}$ $\textstyle =$ $\displaystyle \frac{\rho}{\epsilon_0},$ (1485)
$\displaystyle \nabla\times{\bf B}$ $\textstyle =$ $\displaystyle \mu_0 \left({\bf j} + \epsilon_0  \frac{\partial
{\bf E}}
{\partial t}\right).$ (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}.
\end{displaymath} (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}.
\end{displaymath} (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}.
\end{displaymath} (1489)

Equations (1487) and (1489) can be combined to give
\partial_\mu F^{\mu\nu} = \frac{J^\nu}{c \epsilon_0}.
\end{displaymath} (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.

next up previous
Next: The dual electromagnetic field Up: Relativity and electromagnetism Previous: Tensors and pseudo-tensors
Richard Fitzpatrick 2006-02-02