AlphaNumeric
Nov27-06, 11:41 AM
It's physics based but actually a maths question so I'm asking it here rather than the physics forums.
I = \int \mathcal{L}\; d^{4}x
I is invariant under some transformation \delta_{\epsilon} if \delta_{\epsilon}\mathcal{L} = \partial_{\mu}X^{\mu} for some function/tensor/field thingy X^{\mu}. This I've no problem with.
However, is the same true for a covariant derivative? If \delta_{\epsilon}\mathcal{L} = D_{\mu}X^{\mu} where D_{\mu}\varphi = \partial_{\mu}\varphi + g[A_{\mu},\varphi], as you get in nonabelian field theory. Is the action still invariant? Obviously the \partial_{\mu} part of D_{\mu} represents no problem but I don't know if the g[A_{\mu},\varphi] 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 :)
I = \int \mathcal{L}\; d^{4}x
I is invariant under some transformation \delta_{\epsilon} if \delta_{\epsilon}\mathcal{L} = \partial_{\mu}X^{\mu} for some function/tensor/field thingy X^{\mu}. This I've no problem with.
However, is the same true for a covariant derivative? If \delta_{\epsilon}\mathcal{L} = D_{\mu}X^{\mu} where D_{\mu}\varphi = \partial_{\mu}\varphi + g[A_{\mu},\varphi], as you get in nonabelian field theory. Is the action still invariant? Obviously the \partial_{\mu} part of D_{\mu} represents no problem but I don't know if the g[A_{\mu},\varphi] 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 :)