The Electromagnetic Field Tensor

The transformation of electric and magnetic fields under a Lorentz boost we established even before Einstein developed the theory of relativity. We know that E-fields can transform into B-fields and vice versa. For example, a point charge at rest gives an Electric field. If we boost to a frame in which the charge is moving, there is an Electric and a Magnetic field. This means that the E-field cannot be a Lorentz vector. We need to put the Electric and Magnetic fields together into one (tensor) object to properly handle Lorentz transformations and to write our equations in a covariant way.

The simplest way and the correct way to do this is to make the Electric and Magnetic fields components of a rank 2 (antisymmetric) tensor.

\bgroup\color{black}$\displaystyle F_{\mu\nu}=\pmatrix{ 0 & B_z & -B_y & -iE_x \...
... & B_x & -iE_y \cr
B_y & -B_x & 0 & -iE_z \cr
iE_x & iE_y & iE_z & 0} $\egroup

The fields can simply be written in terms of the vector potential, (which is a Lorentz vector) \bgroup\color{black}$A_\mu=(\vec{A},i\phi)$\egroup.

\bgroup\color{black}$\displaystyle F_{\mu\nu}={\partial A_\nu\over\partial x_\mu}-{\partial A_\mu\over\partial x_\nu} $\egroup

Note that this is automatically antisymmetric under the interchange of the indices. As before, the first two (sourceless) Maxwell equations are automatically satisfied for fields derived from a vector potential. We may write the other two Maxwell equations in terms of the 4-vector \bgroup\color{black}$j_\mu=(\vec{j},ic\rho)$\egroup.

\bgroup\color{black}$\displaystyle {\partial F_{\mu\nu}\over\partial x_\nu}={j_\mu\over c} $\egroup

Which is why the T-shirt given to every MIT freshman when they take Electricity and Magnetism should say

``... and God said \bgroup\color{black}${\partial\over\partial x_\nu}\left({\partial A_\nu\over\partial x_\mu}-{\partial A_\mu\over\partial x_\nu}\right)
={j_\mu\over c}$\egroup and there was light.''

Of course he or she hadn't yet quantized the theory in that statement.

For some peace of mind, lets verify a few terms in the equations. Clearly all the diagonal terms in the field tensor are zero by antisymmetry. Lets take some example off-diagonal terms in the field tensor, checking the (old) definition of the fields in terms of the potential.

\vec{B}&=&\vec{\nabla}\times \vec{A} \\
...artial x_i}+{1\over c}{\partial A_i\over\partial t}\right)

Lets also check what the Maxwell equation says for the last row in the tensor.

{\partial F_{4\nu}\over\partial x_\nu}&=&{j_4\over c} \\
...over\partial x_i}&=&\rho \\
\vec{\nabla}\cdot\vec{E}&=&\rho \\

We will not bother to check the Lorentz transformation of the fields here. Its right.

Jim Branson 2013-04-22