# What is the action for E-M in terms of E & B?

Ankerbrau
Typically the action for E-M is

$$F_{\mu\nu}F^{\mu \nu}$$
where
$$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$$
since the equations of motion for
$$A_{\mu}$$
are the inhomogenous Maxwell equations.

However, here comes my problem:
If one expresses this action in terms of the electric and magnetic
field E and B
$$F_{\mu\nu}F^{\mu \nu}=B^2-E^2$$
the equations of motion for those fields
would be
E=0
and
B=0.

So, where is the trick and what is the correct action
for the fields E and B?

Homework Helper
Gold Member
The correct canonical variables for electrodynamics are the components of 4-vector potential and their derivatives. Perhaps the best way to see this is to include the Lagrangian for a test particle, which requires coupling the momentum of the particle to the 4-vector potential. It's therefore obvious that the E and B fields cannot be canonical variables. Landau and Lifschitz, The Classical Theory of Fields, is a good reference for this.

Homework Helper
$\mathcal{L} = E^2 - B^2$ is not a valid Lagrangian density, because it contains no derivatives.

Ankerbrau
Thanks, obviously

$$E^2-B^2$$ is not the right Lagrangian.
My question was more:
Is there any Lagrangian at all in terms of E and B,
no matter how awkward it looks?

Today a colleague explained me that
experimentally the Aharonov-Bohm effect showed
that the real physical can only be $$A_\mu$$,
theory wise it is connected to gauge invariant quantities
that do not depend on E and B like
$$\int A_\mu dx^\mu$$

I would say that $$E^2-B^2$$ is the right Lagrangian (density), but the variational principle must involve the potentials $$A_\mu$$ as canonical variables. As I said, if you couple electrodynamics to matter, this is more obviously forced on you.