Partial and Covariant derivatives in invarint actions

  • #1
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Main Question or Discussion Point

It's physics based but actually a maths question so I'm asking it here rather than the physics forums.

[tex]I = \int \mathcal{L}\; d^{4}x[/tex]

I is invariant under some transformation [tex]\delta_{\epsilon}[/tex] if [tex]\delta_{\epsilon}\mathcal{L} = \partial_{\mu}X^{\mu}[/tex] for some function/tensor/field thingy [tex]X^{\mu}[/tex]. This I've no problem with.

However, is the same true for a covariant derivative? If [tex]\delta_{\epsilon}\mathcal{L} = D_{\mu}X^{\mu}[/tex] where [tex]D_{\mu}\varphi = \partial_{\mu}\varphi + g[A_{\mu},\varphi][/tex], as you get in nonabelian field theory. Is the action still invariant? Obviously the [tex]\partial_{\mu}[/tex] part of [tex]D_{\mu}[/tex] represents no problem but I don't know if the [tex]g[A_{\mu},\varphi][/tex] term vanishes or not within the integral.

I've been doing some supersymmetry and a number of times I've got the answer the question has asked to find plus a covariant derivative of something. :cry: If I just got a mess of terms I'd know I'm way off, but the fact everything collects nicely into a covariant derivative makes me feel I'm at least on the right track.

Thanks for any help :)
 

Answers and Replies

  • #2
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If ##[A_\mu,X^\mu]=0## we are done, otherwise we just do not have enough information to draw a conclusion. The reason is, that ##\partial_\mu## is a local property, and ##D_\mu## connects this local property with another one ##[A_\mu,X^\mu]## at another location. So ##\mathcal{L}## could still be invariant or equally not.
 

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