- #1

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Let's say we have a multicomponent matter field ##\{\phi^m(x)\}## which transforms according to some Lie group ##G## of internal symmetry:

$$\phi'(x) = \phi(x) + \delta_0\phi(x)$$

$$\delta_0\phi(x) = \theta^aT_a\phi(x) \equiv \theta\phi(x)$$

where ##\theta^a## are parameters, ##T_a## are generators, and ##a## is multiplet index.

We localize this internal symmetries by ##\theta^a \rightarrow \theta^a(x)##. In order to maintain invariance of the Lagrangian, we introduce the covariant derivative:

$$\nabla_\mu \phi(x) = (\partial_\mu + A_\mu)\phi(x) \qquad A_\mu \equiv A^a_\mu T_a$$

where ##A_\mu## is the introduced gauge field.

Now when we localize the symmetry and introduce the covariant derivative in such a way that keeps Lagrangian invariant, we have that the covariant derivative of the field transforms according to the rule(by definition):

$$\delta_0\nabla_\mu\phi(x) = \theta\nabla_\mu\phi(x)$$

And we have that equations of motion can be written in covariant form:

$$\frac{\partial \mathcal{L}}{\partial \phi} - \nabla_\mu\frac{\partial\mathcal{L}}{\partial\nabla_\mu\phi} =0$$

We define the following quantity:

$$K^\mu = \frac{\partial\mathcal{L}}{\partial\nabla_\mu\phi}$$

and we're looking for the transformation properties of this quantity. I have solved it by saying that the quantity ##K^\mu\nabla_\mu\phi## must be gauge invariant, that is:

$$\delta_0(K^\mu\nabla_\mu\phi) = 0$$

from which it is easy to find that ##\delta_0K^\mu = -K^\mu\theta##. This is the correct solution, however I'm not sure whether my argument is correct. I have a gut feeling that it must be correct, but in that case I don't see why it would rigorously be true, although it seems trivially true for quadratic Lagrangian.

So it would be good if someone would look at this and point it out if I did this correctly. Thanks.

Antarres