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

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 $G$?

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).
• You mean $\partial_\mu G^\mu = \partial_{[\mu} \epsilon^{\mu\nu\alpha\beta} A_{\nu]} \partial_{[\alpha} A_{\beta]}$ and that is it? But then that $\partial_\mu$ acts onto the $A_\beta$ as well. Don't I need to write $\epsilon^{\mu\nu\alpha\beta} (\partial_{[\mu} A_{\nu]}) (\partial_{[\alpha}A_{\beta]})$ which then prevents this pulling out? – Martin Ueding Jan 20 '13 at 12:44
• If you write $G^{\mu} = \epsilon^{\mu\nu\alpha\beta}A_{\nu}\partial_{[\alpha}A_{\beta]}$ then $\partial_{\mu}G^{\mu}$ is what you want. The epsilon automatically antisymmetrizes $\mu$ and $\nu$ for you. – twistor59 Jan 20 '13 at 12:52
• In your definition of $F$, the $\partial$ only acts on $A$. But with that $G$, the $\partial_\mu$ will act on both $A_\nu$ and $A\beta$ with the product rule. Or does some (anti-)symmetry remove that part? – Martin Ueding Jan 20 '13 at 12:55
• Right, so you will get some terms like $\partial_{[\mu}F_{\alpha]\beta}$, but the epsilon will fully antisymmetrize over $\mu$, $\alpha$ and $\beta$, and this vanishes by Maxwell equations. – twistor59 Jan 20 '13 at 13:03