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?


2 Answers 2


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.

This remains true when our gauge theory becomes non-abelian, and the corresponding term in the action is known as a topological term or theta-term.


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