Adding a total derivative to the Lagrangian

  • Thread starter JPaquim
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  • #1
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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)?
 

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  • #2
dextercioby
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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
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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.

I'm sorry, but I'm just not finding it trivial. Could you write it out for me, please?
 
  • #4
ChrisVer
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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
 
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  • #5
Bill_K
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or you can just try the Lorentz gauge...
No, you cannot impose gauge conditions on the Lagrangian!

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

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