How to evaluate the Euler-Lagrange equation for the electromagnetic Lagrangian?

Question

I'm fascinated with field theories, but have little knowledge about them, so excuse Me if this is a dumb question.

We all know, that if we have a Lagrangian in terms of a field $\Phi $, we can just pop that into the Euler-Lagrange equation $$ {\partial{\mathcal L} \over \partial \Phi} - \partial_{\mu}\left( {\partial{\mathcal L} \over \partial (\partial_{\mu} \Phi)}\right) = 0$$ to get the equations of motion for that field.

I also know, that the electromagnetic Lagrangian takes the form $ \mathcal{L}_{EM}=-\frac{1}{4\mu_0}F_{\mu \nu}F^{\mu\nu}$ where $ F_{\mu \nu}$ is an antisymmetric tensor with zeros on the diagonal and various components of the electric and magnetic fields everywhere else.

My question is, how would we find the equations of motion? Should we just apply the E-L equation, but with E or B instead of $ \Phi$ ?

I suppose we should arrive at the result $ c^2\cdot E_{xx}=E_{tt}$ , the equation for the electric field of an EM wave.

Answer 1

In spite of what gauge theory may suggest, the field variable is the potential $A^\mu$, so not the force field tensor $F^{\mu\nu}$. In terms of the potential the Lagrangian is $$ \mathcal{L}_{EM}=\frac{\epsilon_0}{2} \left( \partial_\mu A_\nu \partial^\mu A^\nu - \partial_\nu A_\mu \partial^\mu A^\nu\right)~.$$ If you insert this in the EL equation you find $$\partial_\mu \partial^\mu A^\nu = 0$$ which in the Lorentz gauge becomes the wave equation for the potential. From this you will find the wave equations for E and B.