Adding a total derivative to the Lagrangian

1. Dec 16, 2013

JPaquim

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)?

2. Dec 16, 2013

dextercioby

Since the first 2 terms are identical, you need to show that the second term of L1 = L2 + 4 div, thing which I find trivial.

3. Dec 16, 2013

JPaquim

I'm sorry, but I'm just not finding it trivial. Could you write it out for me, please?

4. Dec 16, 2013

ChrisVer

or you can just try the Lorentz gauge... for the 1st
For the second you just need to do a partial derivative, and use again the Lorentz gauge...
You'll have the same lagrangians, differing by a total derivative term

Last edited: Dec 16, 2013
5. Dec 17, 2013

Bill_K

No, you cannot impose gauge conditions on the Lagrangian!

Aμ,νAν,μ = (Aμ,νAν) - Aμ,νμAν = (Aν,μAμ) - Aμ,νμAν = (Aν,μAμ - Aμ,μAν) + Aμ,μAν,ν

Last edited: Dec 17, 2013