Additional Term added to Electromagnetic Lagrangian

[n1person]

I am currently reading and trying to solve most of the problems in Carroll's Geometry and Spacetime. I am generally okay at the math (I've done some mathy Riemannian Geometry type stuff), but am not overly good at some of the higher-level physics.

Homework Statement

(Chapter 1, Question 13)
Consider adding to the Lagrangian for electromagnetism the additional term:

$$\mathcal{L}'=\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}$$

a) Express this in terms of E and B
b) Show that including this term doesn't effect Maxwell's equations. Can you think of a deep reason for this?

Homework Equations

Using -1,+1,+1,+1 flat metric

E&M Lagrangian:
$$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + A_{\mu} J^{\mu}$$

Levi-Civita Symbol:
$$\tilde{\epsilon}_{\mu\nu\rho\sigma}$$
= -1 for odd permutations of 0123, +1 for even permutations, 0 otherwise

Electromagnetic Field Strength Tensor:
$$F_{\alpha\beta}=\partial_\alpha A_\beta - \partial_\beta A_\alpha$$
(too lazy to type the whole matrix :P)Euler Lagrange Equations for a flat space-time field theory:

$$\frac{\partial\mathcal{L}}{\partial\Phi^i}- \partial_{\mu}(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi^i)} ) = 0$$

The Attempt at a Solution

a)
Just using some casework, I was able to get that it was equal to:
$$\mathcal{L}'=-\textbf{E} \cdot \textbf{B}$$

b)
This one is less clear to me, following the procedure in the book:
$$\mathcal{L}_N=-\frac{1}{4}F_{\mu \nu}F^{\mu \nu} + A_{\mu} J^{\mu}+\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu \nu}F^{\rho\sigma}= \mathcal{L} + \mathcal{L}'$$
$$\Phi^i=A_\nu$$
$$\frac{\partial\mathcal{L}'}{\partial A_\mu}= 0$$
Index Lower Fun!
$$\mathcal{L}'=\tilde{\epsilon}_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma} = \eta^{\alpha\mu}\eta^{\beta\nu}\eta^{\gamma\rho} \eta^{\delta \sigma} \tilde{\epsilon}_{\mu\nu\rho\sigma}F_{\alpha\beta}F_{\gamma\delta}$$
$$\frac{\partial}{\partial_\mu A_\nu} (\eta^{\alpha\mu}\eta^{\beta\nu}\eta^{\gamma\rho} \eta^{\delta \sigma} \tilde{\epsilon}_{\mu\nu\rho\sigma}F_{\alpha\beta}F_{\gamma\delta}) = \eta^{\alpha\mu}\eta^{\beta\nu}\eta^{\gamma\rho} \eta^{\delta \sigma} \tilde{\epsilon}_{\mu\nu\rho\sigma} (\frac{\partial}{\partial_\mu A_\nu} ( F_{\alpha\beta}) F_{\gamma\delta} + \frac{\partial}{\partial_\mu A_\nu} (F_{\gamma\delta}) F_{\alpha\beta})$$
$$\frac{\partial}{\partial_\mu A_\nu} (F_{\gamma\delta}) = \delta^\mu_\gamma \delta^\nu_\delta - \delta^\mu_\delta \delta^\mu_\gamma$$

So when we plug this in we get:

$$\eta^{\alpha\mu}\eta^{\beta\nu}\eta^{\gamma\rho} \eta^{\delta \sigma} \tilde{\epsilon}_{\mu\nu\rho\sigma} ((\delta^\mu_\gamma \delta^\nu_\delta - \delta^\mu_\delta \delta^\mu_\gamma) F_{\gamma\delta} + (\delta^\mu_\alpha \delta^\nu_\beta - \delta^\mu_\beta \delta^\mu_\alpha) F_{\alpha\beta})$$

At this point I think, well if whenever there are duplicates in $$\mu\nu\rho\sigma$$, the Levi-Civita Symbol is zero, and whenever there are aren't duplicates all the kroncher-deltas are zero, so this entire term is just zero.

Is this reasoning correct? I feel I am missing something cause I cannot think of a deep reason for this... I cannot think of an interesting physical interpretation...

Thanks for following along and reading till here :)

 

[n1person]

Potential answers might be daughted/distracted by everything else. I suppose a lot of this question boils down to the following: does
$$\tilde{\epsilon}_{\mu\nu\rho\sigma} (\delta^\mu_\gamma \delta^\nu_\delta - \delta^\mu_\delta \delta^\nu_\gamma)$$ =0 ?