Write $\epsilon_{\mu\nu\alpha\beta} F^{\mu\nu} F^{\alpha\beta}$ as a total divergence $\partial_\mu G^\mu$

Question

I have the following homework problem in theoretical electrodynamics:

Show that the gauge invariant Lagrange density $\epsilon_{\mu\nu\alpha\beta} F^{\mu\nu} F^{\alpha\beta}$ can be written as a total divergence of a four-vector.

This total divergence should be something like $\partial_\mu G^\mu$, right?

With $$ \hat F^{\mu\nu} = \frac 12 \epsilon^{\mu\nu\alpha\beta} F_{\alpha\beta} $$ I figured that I write this Lagrange density (I'll refer to it as $L_2$) could be written as: $$ L_2 = 2 \hat F^{\mu\nu} F_{\mu\nu} $$

I simplified this by using the matrix representations of each $F$ and got it down to: $$ L_2 = -\frac 4c B_i E^i $$

$B$ and $E$ can be expressed in Terms of $A$ like so: $$ B_i = \epsilon_{ijk} \partial_j A^j ,\quad E_i = -\partial_i A^0 -\partial_0 A^i $$

How would I continue to find that four-vector $G$?

Answer 1

If you write $$F_{\mu\nu} = \partial_{[\mu}A_{\nu ]} $$ Then your Lagrange density is $${ \mathcal{L}} = \epsilon^{\mu\nu\alpha\beta}\partial_{[\mu}A_{\nu ]}\partial_{[\alpha}A_{\beta ]}$$

Now we want a vector whose divergence is ${ \mathcal{L}}$. The $\partial_{\mu}$ looks promising, so we ask if we can bring that outside so it acts on the remaining vector, i.e. $${ \mathcal{L}} = \partial_{\mu}(\epsilon^{\mu\nu\alpha\beta}A_{\nu }\partial_{[\alpha}A_{\beta ]}) \ \ (1)$$ We don't need to worry that we've lost the antisymmetrization brackets on $\mu$ and $\nu$ because the epsilon symbol forces this.

As you pointed out, this isn't quite what we started with because we have an additional term of the form $$(\epsilon^{\mu\nu\alpha\beta}A_{\nu }\partial_{\mu}\partial_{[\alpha}A_{\beta ]}) $$ However the total antisymmetry of the epsilon symbol means we can treat the $\mu$ $\alpha$ $\beta$ contribution as $$\partial_{[\mu}F_{\alpha\beta]} $$ which vanishes due to the Maxwell equations. Hence the anzatz (1) holds. (The vector whose divergence we take looks like the abelian Chern Simons current).