How do I verify the invariance on Yang-Mills' Lagrangian:

$$L = -\frac{1}{4} \sum_{a} \left(\partial_\mu A_\nu^a - \partial_\nu A_\mu^a + gf^{abc}A_\mu^bA_\nu^c \right)^2$$

under the transformation:

$$A_\mu^a (x) \rightarrow A_\mu^a (x) + \frac{1}{g} \partial_\mu \alpha^a(x) -f^{abc}\alpha^b(x)A_\mu^c(x)$$

The first term on the gauge transformation obviously recovers the old lagrangian, but the second and third terms are getting very problematic. Can someone help me with those? I'm note sure about my argument on terms $\partial_\nu \partial_\mu \alpha^a$ and $\partial_\mu \partial_\nu \alpha^a$, and the third one just get worse. I'm pretty sure that I must use the fact that the structure constant is totally antisymmetric, but it's not getting better.

  • $\begingroup$ Not totally sure what the issue is, but $\alpha$ is infinitesimal, so $\alpha^2$ terms should be dropped $\endgroup$ – user1379857 May 21 at 18:55
  • $\begingroup$ Look here physics.stackexchange.com/q/456875 $\endgroup$ – Vicky May 21 at 19:00
  • $\begingroup$ The problem is entirely mathematical. The $\alpha^2$ terms will sure drop, but there are many other terms from the gauge transformation, mainly at $A_\mu^b A_\nu^c$, that i couldn't make it cancel with each other to show the lagrangian invariance. $\endgroup$ – otto May 21 at 19:26
  • $\begingroup$ @otto First, if you want to write an answer to a comment use '@nickname'. Otherwise the person you are answering will see nothing. Second, in the link I provided your question is solved in an 'exact' way, I mean without Taylor expansions. Make a full reading of it $\endgroup$ – Vicky May 24 at 11:17

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