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Take a Lagrangian $L \rightarrow L+\partial_{\mu}F^{\mu}$.

If we can show that the total derivative $\partial_{\mu}F^{\mu}$ identically satisfies the Euler-Largrange equation, then we have shown that the equations of motion will be unchanged.

How can we show this, even for the simple example of one space coordinate (and its time derivative)?

aRockStr
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  • This happens because of an application of Stokes' theorem. Check the first paragraph of this answer: https://physics.stackexchange.com/a/391538/133418 – Avantgarde Aug 21 '18 at 21:22
  • @Avantgarde Thank you for your reply! In reference to said paragraph, if the fields are only assumed to vanish at infinity, and we integrate over the boundary $\partial\Sigma$, does that mean the boundary is taken to be at infinity, i.e. all of spacetime? – aRockStr Aug 21 '18 at 23:09
  • You're welcome. Yes, a volume integral is converted into a surface integral. The surface is the edge of spacetime, where we assume the fields to vanish. – Avantgarde Aug 21 '18 at 23:48
  • Possible duplicates: https://physics.stackexchange.com/q/87628/2451 and links therein. – Qmechanic Aug 22 '18 at 05:02

1 Answers1

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A total derivative $\partial_\mu F^\mu$ does not, necessarily, identically satisfies the Euler-Lagrange equation, as you are thinking. The point is that the total derivative in $\mathcal{L}^\prime=\mathcal{L}+\partial_\mu F^\mu$, assuming that $F^\mu$ vanishes at the boundaries, will not contribute to the action due to Stokes theorem, and therefore you will derive the same equations of motion as you would if you have just $\mathcal{L}$. In other words, if you have the Lagrangian $\mathcal{L}^\prime=\mathcal{L}+\partial_\mu F^\mu$, the equations of motion that you will get varying the action will be the Euler-Lagrange equations for $\mathcal{L}$ and not $\mathcal{L}^\prime$.

Davi Bastos
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