JPaquim
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I recently posted another thread on the General Physics sub forum, but didn't get as much feedback as I was hoping for, regarding this issue. Let's say I have two Lagrangians:
$$ \mathcal{L}_1 = -\frac{1}{2}(\partial_\mu A_\nu)(\partial^\mu A^\nu) + \frac{1}{2}(\partial_\mu A^\mu)^2 $$
$$ \mathcal{L}_2 = -\frac{1}{2}(\partial_\mu A_\nu)(\partial^\mu A^\nu) + \frac{1}{2}(\partial_\nu A_\mu)(\partial^\mu A^\nu) $$
They refer to the same Maxwell in a vacuum field theory, so they give rise to the same equations of motion. As such they should differ only by a total derivative. I can test if something is a total derivative by plugging it into the Euler-Lagrange equations, and see if I get something that can be reduced to [itex]0\,=\,0[/itex]. In this case, there should be some vector [itex]B^\mu[/itex] such that the difference between the two Lagrangians, is equal to [itex]\partial_\mu B^\mu[/itex], right? My question is, how can you find [itex]B^\mu[/itex] (up to the addition of a divergence-less term)?
$$ \mathcal{L}_1 = -\frac{1}{2}(\partial_\mu A_\nu)(\partial^\mu A^\nu) + \frac{1}{2}(\partial_\mu A^\mu)^2 $$
$$ \mathcal{L}_2 = -\frac{1}{2}(\partial_\mu A_\nu)(\partial^\mu A^\nu) + \frac{1}{2}(\partial_\nu A_\mu)(\partial^\mu A^\nu) $$
They refer to the same Maxwell in a vacuum field theory, so they give rise to the same equations of motion. As such they should differ only by a total derivative. I can test if something is a total derivative by plugging it into the Euler-Lagrange equations, and see if I get something that can be reduced to [itex]0\,=\,0[/itex]. In this case, there should be some vector [itex]B^\mu[/itex] such that the difference between the two Lagrangians, is equal to [itex]\partial_\mu B^\mu[/itex], right? My question is, how can you find [itex]B^\mu[/itex] (up to the addition of a divergence-less term)?