# Making Lagrangian gauge invariant

1. Apr 3, 2015

### Maybe_Memorie

1. The problem statement, all variables and given/known data

The Lagrangian $\mathcal{L}\frac{1}{2}(\partial_\mu\phi^\nu)^2+\frac{1}{2}(\partial_\mu\phi^\mu)^2+\frac{m^2}{2}(\phi_\mu\phi^\mu)^2$ for the vector field $\phi^\mu$ is not invariant with respect to the gauge transformation $\phi^\mu\rightarrow \phi^\mu+\partial^\mu\alpha$. Introduce a new real scalar field $\sigma$ and find a new interacting Lagrangian $\mathcal{L}'(\phi^\mu,\sigma)=\mathcal{L}(\phi^\mu)+\tilde{\mathcal{L}}(\phi^\mu,\sigma)$ which is gauge invariant under the given transformation and satisfies $\mathcal{L}'(\phi^\mu,0)=\mathcal{L}(\phi^\mu)$.

Can we solve this with a $\sigma$ that has a canonical kinetic term $-\frac{1}{2}(\partial_\mu\sigma)^2$?

2. Relevant equations

3. The attempt at a solution

It's the part about the canonical kinetic term I don't understand. On the requirement that $\sigma \rightarrow \sigma + \alpha$ I found a $\tilde{\mathcal{L}}(\phi^\mu,\sigma)$ that satisfies the required properties. I won't write it because it's rather long. I was told the answer to whether or not there can be a canonical kinetic term is "no", but I don't know why.

Is there a particular reason why this should be? Something about having unbounded negative energy perhaps?

2. Apr 8, 2015