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

[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.

 

[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.

 

[sir-pinski]

Yes, \nabla \overline{} \mu and \nabla \overline{} \nu commute. This is because the commutator of two vector fields is given by:

[\nabla \overline{} \mu, \nabla \overline{} \nu] = \nabla_{\overline{} \mu} \nabla_{\overline{} \nu} - \nabla_{\overline{} \nu} \nabla_{\overline{} \mu}

where \nabla_{\overline{} \mu} and \nabla_{\overline{} \nu} are the covariant derivatives with respect to the vector fields \overline{} \mu and \overline{} \nu, respectively.

Now, the covariant derivative of a vector field with respect to another vector field is given by:

\nabla_{\overline{} \mu} \overline{} \nu = \partial_{\overline{} \mu} \overline{} \nu + \Gamma^{\lambda}_{\mu \nu} \overline{} \lambda

where \partial_{\overline{} \mu} is the partial derivative with respect to the coordinate \overline{} \mu and \Gamma^{\lambda}_{\mu \nu} are the Christoffel symbols of the second kind. These symbols represent the connection coefficients of the metric tensor and are symmetric in their lower indices.

Since the Christoffel symbols are symmetric in their lower indices, we have:

\Gamma^{\lambda}_{\mu \nu} = \Gamma^{\lambda}_{\nu \mu}

This means that the covariant derivative of a vector field with respect to another vector field is symmetric, i.e.:

\nabla_{\overline{} \mu} \overline{} \nu = \nabla_{\overline{} \nu} \overline{} \mu

Therefore, the commutator of two covariant derivatives, and thus two vector fields, is zero, i.e.:

[\nabla \overline{} \mu, \nabla \overline{} \nu] = 0

This shows that \nabla \overline{} \mu and \nabla \overline{} \nu commute, and thus the E.M. field is invariant under gauge transformation.