I'm trying to solve the Euler-Lagrange equation for the MCS Lagrangian density as given by Kharzeev in this article (Eqn. 7):

$$ \mathcal{L}_{\textrm{MCS}} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}-A_\mu J^{\mu} - \frac{c}{4}\theta\tilde{F}^{\mu\nu}F_{\mu\nu} $$

I already know that the first two terms yield Maxwell's equations,

$$ \partial_\mu F^{\mu\nu} = \mu_0{J^\nu} $$

Now, I am trying to solve the Euler-Lagrange equation for the third term. In principle, $\theta = \theta(\textbf{x},t)$, but I first would like to focus on the case where $\theta$ is constant - for this, the equations of motion should not be altered by this term.

So far, for this third term, I've calculated

$$ \mathcal{L}_3 \propto \tilde{F}^{\mu\nu}F_{\mu\nu} = 2\varepsilon^ {\mu\nu\rho\sigma}\partial_\mu A_\nu \partial_\rho A_\sigma$$

This can be plugged in into the Euler-Lagrange equation,

$$ \partial_\mu\left(\frac{\partial \mathcal{L}_3}{\partial(\partial_\mu A_\nu)}\right)-\frac{\partial \mathcal{L}_3}{\partial A_\mu} = 0 $$

The latter term is zero, and when I focus on the part in brackets of the first term, I find

$$ \frac{\partial \mathcal{L}_3}{\partial(\partial_\alpha A_\beta)} = 2\varepsilon^{\alpha\beta\rho\sigma}\partial_\rho A_\sigma + 2\varepsilon^{\mu\nu\alpha\beta}\partial_\mu A_\nu$$ $$ = 2\varepsilon^{\alpha\beta\rho\sigma}\partial_\rho A_\sigma - 2\varepsilon^{\alpha\beta \nu\mu}\partial_\mu A_\nu $$ $$ = 2\varepsilon^{\alpha\beta\rho\sigma}(\partial_\rho A_\sigma - \partial_\sigma A_\rho) = 2\varepsilon^{\alpha\beta\rho\sigma}F_{\rho\sigma} = 4\tilde{F}^{\alpha\beta} $$

such that the result of the EL equation would be

$$ 4\partial_\alpha \tilde{F}^{\alpha\beta}=0 $$

Recently, I found a document of a colleague of mine that also did similar calculations and he ended up with

$$ -8\Box A^{\nu} = 0 $$

For what follows in his calculations on a varying $\theta(\textbf{x},t)$ this term is reoccurring, and because I did not get his result I was wondering:

Is my calculation correct, or is his - and if so, why?


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