Varying the action with respect to A_\mu

[PhyAmateur]

Was reading a paper and trying to work out the author's calculations: I am trying to vary the action,
$$S_A= \int d^3x \left[ \, -\frac{1}{4}F_{\mu\nu}(A)F^{\mu\nu}(A) +\frac{1}{2}m\, \epsilon^{\mu\nu\rho}A_\mu F_{\nu\rho}(A) \right] $$

with respect to $$A_\mu$$. I am finding difficulty deriving this because $$A_\mu$$ is embedded in $$F_{\mu \nu}$$. So my attempt was writing this all in terms of $$A_\mu$$:

$$S_A= \int d^3x \, \left[ -\frac{1}{4}(\partial_\mu A_\nu - \partial_\nu A_\mu) (\partial^\mu A^\nu - \partial^\nu A^\mu) +\frac{1}{2}m\, \epsilon^{\mu\nu\rho}A_\mu (\partial_\nu A_\rho - \partial_\rho A_\nu) \right] $$ and then I got stuck. If you could please lead me from here.

Reference, section 2: http://arxiv.org/pdf/hep-th/9705122.pdf

 

[jvicens]

Thank you for your question. It seems like you are trying to vary the action with respect to the vector field $$A_\mu$$. This can be done by using the Euler-Lagrange equations, which state that the variation of the action with respect to a field is equal to zero. In this case, we can write:

$$\delta S_A = \int d^3x \left[-\frac{1}{4}\delta(F_{\mu\nu}F^{\mu\nu}) + \frac{1}{2}m\epsilon^{\mu\nu\rho}\delta(A_\mu F_{\nu\rho})\right] = 0$$

Using the product rule and the fact that $$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$$, we can expand the first term as:

$$\delta(F_{\mu\nu}F^{\mu\nu}) = \delta(\partial_\mu A_\nu - \partial_\nu A_\mu)(\partial^\mu A^\nu - \partial^\nu A^\mu) = \partial_\mu(\delta A_\nu)(\partial^\mu A^\nu - \partial^\nu A^\mu) - \partial_\nu(\delta A_\mu)(\partial^\mu A^\nu - \partial^\nu A^\mu)$$

The second term can be expanded similarly as:

$$\delta(A_\mu F_{\nu\rho}) = \delta A_\mu (\partial_\nu A_\rho - \partial_\rho A_\nu) + A_\mu \delta(\partial_\nu A_\rho - \partial_\rho A_\nu)$$

Substituting these expressions into the variation of the action and using integration by parts, we can write:

$$\int d^3x \left[\partial_\mu(\delta A_\nu)(\partial^\mu A^\nu - \partial^\nu A^\mu) - \partial_\nu(\delta A_\mu)(\partial^\mu A^\nu - \partial^\nu A^\mu) + \frac{1}{2}m\epsilon^{\mu\nu\rho}(\delta A_\mu)(\partial_\nu A_\rho - \partial_\rho A_\nu) \right] = 0$$

Since the variation