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}=\begin{pmatrix}0 & iE_x & iE_y & ...
-iE_y & -B_z & 0 & B_x \cr
-iE_z & B_y & -B_x & 0 \cr \end{pmatrix} $\egroup

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

$\displaystyle \vec{B}$ $\displaystyle =\vec{\nabla}\times \vec{A}$    
$\displaystyle \vec{E}$ $\displaystyle =-\vec{\nabla}\phi-{1\over c}{\partial \vec{A}\over\partial t}$    
$\displaystyle F_{12}$ $\displaystyle ={\partial A_2\over\partial x_1}-{\partial A_1\over\partial x_2}=(\vec{\nabla}\times\vec{A})_z=B_z$    
$\displaystyle F_{13}$ $\displaystyle ={\partial A_3\over\partial x_1}-{\partial A_1\over\partial x_3}=-(\vec{\nabla}\times\vec{A})_y=-B_y$    
$\displaystyle F_{0i}$ $\displaystyle ={\partial A_i\over\partial x_0}-{\partial A_0\over\partial x_i} ...
...ial \phi\over\partial x_i}+{1\over c}{\partial A_i\over\partial t}\right) =iE_i$    

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

$\displaystyle {\partial F_{0\nu}\over\partial x_\nu}$ $\displaystyle ={j_0\over c}$    
$\displaystyle {\partial F_{0i}\over\partial x_i}$ $\displaystyle ={ic\rho\over c}$    
$\displaystyle {\partial (iE_i)\over\partial x_i}$ $\displaystyle =i\rho$    
$\displaystyle {\partial E_i\over\partial x_i}$ $\displaystyle =\rho$    
$\displaystyle \vec{\nabla}\cdot\vec{E}$ $\displaystyle =\rho$    

Jim Branson 2012-10-21