# 经典电动力学中电磁张量

\begin{align}
\nabla\cdot \bf{E}=\rho \\
\nabla \times \bf B – \frac 1 c \frac{\partial \bf E}{\partial t}=j \\
\nabla \cdot \bf B = 0 \\
\nabla\times \bf E + \frac 1 c \frac{\partial \bf B}{\partial t}=0
\end{align}

We would introduce a vector potential $$\bf A$$ and scalar potential $$\phi$$, which are defined as $$\bf B=\nabla\times\bf A$$ and $$\bf E=-\nabla\phi – \frac{\partial \bf A}{\partial t}$$. More generally, it can be extended to a rank four vector $$A^a=(\phi,\bf A)$$.

This would remind us of a EM tensor $$F^{ab}=\partial^a A^b$$. However, it is really annoying that F should have only six components.

Then we would like to transform it into a symmetric and traceless tensor $$F^{ab}=\partial^aA^b+\partial^bA^a-\frac 1 2\delta^{ab}F^{a}_b$$. After a simple and straight forward calculation, I find the final expression of Maxwell Eqns are not beautiful enough because a lot of quantities can not be merged.

Another method is to make F a antisymmetric one, i.e., $$F^{ab}=\partial^aA^b-\partial^bA^a$$. And this leads to good results. Here we reaches the standard definition of $$F^{ab}$$.

A more advanced understanding is , it is better to define a quantity with a certain meaning. Here we would like to define F as a flux through a volume element $$\bf{\epsilon}$$, which is in correspondence with the old quantities we use. That is to say, F is a 2-form. So it is definitely anti-symmetric tensor.