Question

Lagrangian density of em solution in terms of electric and magnetic field

Answer 1

The Lagrangian for Electromagnetic Fields

There are not many ways to make a scalar Lagrangian from the field tensor. We already know that

\begin{displaymath}\bgroup\color{black} {\partial F_{\mu\nu}\over\partial x_\nu}={j_\mu\over c} \egroup\end{displaymath}

and we need to make our Lagrangian out of the fields, not just the current. Again, \bgroup\color{black}$x_\mu$\egroup cannot appear explicitly because that violates symmetries of nature. Also we want a linear equation and so higher powers of the field should not occur. A term of the form \bgroup\color{black}$mA_\mu A_\mu$\egroup is a mass term and would cause fields to fall off faster than \bgroup\color{black}${1\over r}$\egroup. So, the only reasonable choice is

\begin{displaymath}\bgroup\color{black} F_{\mu\nu}F_{\mu\nu}= 2(B^2-E^2). \egroup\end{displaymath}

One might consider

\begin{displaymath}\bgroup\color{black} e_{\mu\nu\lambda\sigma}F_{\mu\nu}F_{\lambda\sigma}=\vec{B}\cdot\vec{E} \egroup\end{displaymath}

but that is a pseudo-scalar, not a scalar. That is, it changes sign under a parity transformation. The EM interaction is known to conserve parity so this is not a real option. As with the scalar field, we need to add an interaction with a source term. Of course, we know electromagnetism well, so finding the right Lagrangian is not really guess work. The source of the field is the vector \bgroup\color{black}$j_\mu$\egroup, so the simple scalar we can write is \bgroup\color{black}$j_\mu A_\mu$\egroup.