Solving the Euler-Lagrange equation with the Axion Lagrangian

I am trying to show that for a constant axion field $$\theta(\textbf{x},t)=const.$$ the axion Lagrangian $$\mathcal{L}_\theta=-\frac{\kappa\theta}{4\mu_0}F_{\mu\nu}\tilde{F}^{\mu\nu}$$ does not lead to a change in the equations of motion - I'm trying to solve the Euler-Lagrange equation for this, but I have little background in field theory.

So far, I have shown that $$\mathcal{L}_\theta=-2\partial_\mu(\varepsilon^{\mu\nu\rho\sigma}A_\nu \partial_\rho A_\sigma)$$

which I plug in into the Euler-Lagrange equation $$\partial_\mu\left(\frac{\partial\mathcal{L}_\theta}{\partial(\partial_\mu A_\nu)}\right)-\frac{\partial\mathcal{L}_\theta}{\partial A_\mu}=0$$

Since the Lagrangian does not depend on $$A_\mu$$, the latter term results equals zero. Hence I focus on the first term. Since my Lagrangian already contains $$\mu,\nu,\rho,\sigma$$, I use $$\alpha,\beta$$ to distinguish the indices of the derivatives from the indices of the Lagrangian - this is inspired by the procedure as shown here (https://quantummechanics.ucsd.edu/ph130a/130_notes/node452.html) for Maxwell's Lagrangian.

I get

$$\partial\alpha\left[-2\varepsilon^{\mu\nu\rho\sigma}\left(\frac{\partial(\partial_\mu A_\nu)}{\partial(\partial_\alpha A_\beta)}\partial_\rho A_\sigma + \partial_\mu A_\nu\frac{\partial(\partial_\rho A_\sigma)}{\partial(\partial_\alpha A_\beta)}\right)\right]$$

I have tried expanding this in cases for $$\alpha=\mu,\nu,\rho,\sigma$$ to solve them separately. For $$\alpha=\mu$$, I get

$$\partial\mu\left[-2\varepsilon^{\mu\nu\rho\sigma}\left(\frac{\partial(\partial_\mu A_\nu)}{\partial(\partial_\mu A_\nu)}\partial_\rho A_\sigma + \partial_\mu A_\nu\frac{\partial(\partial_\rho A_\sigma)}{\partial(\partial_\mu A_\nu)}\right)\right]$$

The first term $$\delta_\mu^\mu\delta_\nu^\nu=1$$, and the latter $$\delta_\mu^\rho\delta_\nu^\sigma$$ would result in zero in combination with $$\varepsilon^{\mu\nu\rho\sigma}$$. Why is this the case?

When solving for $$\alpha=\rho$$, I get (using the same reasoning)

$$-\partial_\rho2\varepsilon^{\mu\nu\rho\sigma}\partial_\mu A_\nu$$

With the same reasoning, for $$\alpha=\nu,\sigma$$ the Kroneckers would result into nothing. So finally I have

$$\partial_\mu\left(\frac{\partial\mathcal{L}_\theta}{\partial(\partial_\mu A_\nu)}\right)=-\partial_\mu2\varepsilon^{\mu\nu\rho\sigma}\partial_\rho A_\sigma -\partial_\rho2\varepsilon^{\mu\nu\rho\sigma}\partial_\mu A_\nu$$

How do I proceed from here?

• I have already solved the first subquestion, namely that $\varepsilon^\mu\nu\rho\sigma$ is zero when the indices $\mu,\nu,\rho,\sigma$ are not all unique, so that the combination $\delta_\mu^\rho\delta_\nu^\sigma$ can never yield anything nonzero. I am still working on the second question, so any help would be great! Jul 10, 2019 at 12:23
• HI did you ever figure this out? I have the same problem. May 10, 2023 at 1:45
• So for just calculating $\frac{\partial}{\partial (\partial_{\mu} A_{\nu})}$ I was thinking that the second term, $\partial_{\mu}A_{\nu}\frac{\partial (\partial_{\rho}A_{\sigma})}{\partial (\partial_{\mu} A_{\nu})}$ is zero. Then can write $2\epsilon^{\mu\nu\rho\sigma}\partial_{\rho}A_{\sigma} = \epsilon^{\mu\nu\rho\sigma} (\partial_{\rho}A_{\sigma} + \partial_{\rho}A_{\sigma}) = \epsilon^{\mu\nu\rho\sigma}(\partial_{\rho}A_{\sigma} -\partial_{\sigma}A_{\rho})$. But then I am missing a factor of two. I am not sure how to resolve this. May 10, 2023 at 5:06
• Hi @JosephSanders , I have solved it (with help of a local professor) a while ago. I'll put the answer (written in a LaTeX document with the Physics package installed) as images below, which'll save me the trouble of converting all equations. I hope that this can give you enough information to get through your calculation! Jul 5, 2023 at 13:24