# Lagrangian Gauge Transformation Q

• div curl F= 0

#### div curl F= 0

Dear All,

I'd be grateful for a bit of help with the following problems:

Consider the Lagrangian:
$$\displaystyle \mathcal{L} = (\partial_{\mu} \phi) (\partial^{\mu} \phi^{\dagger}) - m^2 \phi^{\dagger} \phi$$
where $$\phi = \phi(x^{\mu})$$

Now making a U(1) gauge transformation:
$$\displaystyle \phi \longmapsto e^{i \Lambda(x^{\mu})} \phi$$

does the Lagrangian become:

$$\displaystyle \mathcal{L} = (\partial_{\mu} \phi) \cdot (\partial^{\mu} \phi^{\dagger}) - m^2 \phi^{\dagger} \phi + \phi \phi^{\dagger} (\partial_{\mu} \Lambda) \cdot (\partial^{\mu} \Lambda) + i \partial_{\mu} \Lambda \cdot (\phi \partial^{\mu} \phi^{\dagger} - \phi^{\dagger} \partial^{\mu} \phi)$$ ?

I realize you can add in another field to counteract the gauge transformation so the Lagrangian becomes gauge invariant, but how exactly would you determine the field to "add in" by inspection?

Thanks for any replies

Uhm, looks right. The point of gauge invariance is that you want a gauge transformation that commutes with the derivative. In other words, if
$$\phi \to g \phi$$,
then
$$\partial_\mu \phi \to g \partial_\mu \phi + (\partial_\mu g) \phi$$
whereas we would like covariance:
$$D_\mu \phi \to g D_\mu \phi$$
which implies that
$$D_\mu \to g D_\mu g^{-1}$$ (the derivative now acts on everything to its right).

That transformation: $$D_{\mu} \to g D_{\mu} g^{-1}$$
Yes, $$g X g^{-1}$$ is group action in the adjoint representation.