# Proof that the E.M Field is invariant under guage transformation.

1. Apr 16, 2009

### hob

To prove:

$$F$$ $$\overline{} \mu\nu$$ = $$\nabla$$$$\overline{} \mu$$$$A$$ $$\overline{} \nu$$ - $$\nabla$$$$\overline{} \nu$$$$A$$ $$\overline{} \mu$$

is invariant under the gauge transformation:

$$A$$ $$\overline{} \mu$$ $$\rightarrow$$ $$A$$ $$\overline{} \mu$$ + $$\nabla$$$$\overline{} \mu$$$$\Lambda$$

I end up with:

$$F$$ $$\overline{} \mu\nu$$ = $$F$$ $$\overline{} \mu\nu$$ + [$$\nabla$$$$\overline{} \mu$$,$$\nabla$$$$\overline{} \nu$$]$$\Lambda$$

Which I guess is invariant provided $$\nabla$$$$\overline{} \mu$$ & $$\nabla$$$$\overline{} \nu$$ commute?

Do they commute? and if so why?

Many thanks.

2. Apr 16, 2009

### xepma

Yes, they commute. In the case of normal minkowski space time and the Abelian gauge group U(1) the differential operators reduce to ordinary derivatives.