# Deriving EM energy and momentum with noether's theorem

1. May 16, 2007

### jostpuur

I think I know how to derive conserved energy and momentum currents of a free EM field. Lagrangian is

$$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$

I then substitute $$x^\mu\mapsto x^\mu + \lambda u^\mu$$, and take the derivative in respect to lambda. With some trickery I've got

$$\partial_\mu (\mathcal{L}\delta^\mu_\nu)u^\nu =-(\partial_\nu\partial_\mu A_\alpha)F^{\mu\alpha} u^\nu=\ldots$$

Now since $$\partial_\mu F^{\mu\alpha}=0$$ is the equation of motion, we can add $$(\partial_\nu A_\alpha)(\partial_\mu F^{\mu\alpha})$$ to the expression, and get

$$\ldots = -\partial_\mu((\partial_\nu A_\alpha)F^{\mu\alpha})u^\nu$$

and the conserving currents $$T^{\mu\nu}=-(\partial^\nu A_\alpha)F^{\mu\alpha} - \mathcal{L}\delta^{\mu\nu}$$ can be recognized.

$$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - j_\mu A^\mu$$

the similar trickery, now using $$\partial_\mu F^{\mu\nu}=j^\nu$$, leads into an equation

$$\partial_\mu (\mathcal{L}\delta^\mu_\nu)u^\nu = -\partial_\mu((\partial_\nu A_\alpha)F^{\mu\alpha})u^\nu -(\partial_\nu j_\alpha)A^\alpha u^\nu$$

and there's the problem. Only question I can now ask is, that could it be somehow acceptable to ignore $$\partial_\nu j_\alpha$$, or is there some reasons why the change of current density in translation should have been ignored in the beginning? I'll be glad to hear any other remarks as well, if they can show some light on the matter.

Last edited: May 16, 2007