- #1
JPaquim
- 33
- 0
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 0\,=\,0. In this case, there should be some vector B^\mu such that the difference between the two Lagrangians, is equal to \partial_\mu B^\mu, right? My question is, how can you find B^\mu (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 0\,=\,0. In this case, there should be some vector B^\mu such that the difference between the two Lagrangians, is equal to \partial_\mu B^\mu, right? My question is, how can you find B^\mu (up to the addition of a divergence-less term)?
