# Topological term in Lagrangian of Maxwell's theory

I should derive the equation of motion for the case, when the Lagrangian of Maxwell's theory has the additional term

$$\Delta \mathcal{L}=cF_{\mu\nu}\widetilde{F}^{\mu\nu}$$

with the dual tensor $$\widetilde{F}^{\mu\nu}:=\frac{1}{2}\varepsilon^{\mu\nu\sigma\rho}F_{\sigma\rho}$$.

First of all, I rewrite $$\Delta \mathcal{L}$$:

$$\Delta \mathcal{L}=...=2c\varepsilon^{\alpha\beta\gamma\delta}\partial_{\alpha}A_{\beta}\partial_{\gamma}A_{\delta}$$

I am pretty sure that this is right, because I checked it very often.

Now, the only additional term in the equation of motion is the expression:

$$\partial_{\mu}\frac{\partial\Delta \mathcal{L}}{\partial(\partial_{\mu}A_{\nu})} =...=2c\varepsilon^{\mu\nu\alpha\beta}\partial_{\mu}\partial_{\alpha}A_{\beta}$$

This should also be right....But according to the solution, the equations doesn`t change...How can I see that my additional term $$\varepsilon^{\mu\nu\alpha\beta}\partial_{\mu}\partial_{\alpha}A_{\beta}$$ is equal to zero?

By definition of the Levi-Civita symbol it is antisymmetric under permutations of its indices. So under permutation of $$\alpha$$ and $$\mu$$ it would obtain a minus sign. On the other hand the double derivative $$\partial_\alpha\partial_\mu$$ is symmetric under this permutation. The contraction of a symmetric and an antisymmetric object is 0.
The term you computed is identically zero, since partial derivatives commute, and the $$\epsilon$$ symbol is fully antisymmetric.
In the language of differential forms, the extra term in the Lagrangian is $$\mathrm{d}A\wedge\mathrm{d}A$$, and the contribution to the equations of motion is simply $$\mathrm{d}(\mathrm{d}A)=0$$, since the exterior derivative is nilpotent ($$\mathrm{d}^2=0$$) by construction. We can also note that $$\mathrm{d}A\wedge\mathrm{d}A=\mathrm{d}(A\wedge\mathrm{d}A)$$, which is a total derivative.