Gauge transformation of Lagrangian

Suppose I have a Lagrangian density $\mathcal{L}(\phi^\mu,\sigma)$ depending on vector fields $\phi^\mu$ and their derivatives and a scalar field $\sigma$ and its derivatives. If I make a gauge transformation $\phi^\mu\rightarrow \phi^\mu+\partial^\mu\alpha$ does the field $\sigma$ transform? I've seen notes claiming $\sigma \rightarrow \sigma + \alpha$ but just wanted to make sure.

Edit: For more detail - I was looking at a problem that said the Lagrangian $\mathcal{L}(\phi^\mu)=-\frac{1}{2}(\partial_\mu\phi^\nu)^2+\frac{1}{2}(\partial_\mu\phi^\mu)^2+\frac{m^2}{2}(\phi_\mu\phi^\mu)^2$ is not invariant with respect to the gauge transformation $\phi^\mu\rightarrow \phi^\mu+\partial^\mu\alpha$ and asked to introduce a new 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 $\mathcal{L}'(\phi^\mu,0)=\mathcal{L}(\phi^\mu)$.

I've found a Lagrangian with these properties but it assumes $\sigma \rightarrow \sigma + \alpha$ and I'm not sure if this is correct.

• The question is far too unspecific. What kind of gauge transformation (it looks as if $\alpha$ is meant to be an infinitesimal $\mathrm{U}(1)$ transformation of a gauge field $\phi$)? Also, the way fields transform under internal gauge transformations is not fixed by their scalar/spinor/vector behaviour under Lorentz transformations, it has to be additionally specified, so there is no way to tell how anything transforms from the information you have given. Apr 1, 2015 at 17:44
• Another question by OP about the same Lagrangian density: physics.stackexchange.com/q/175514/2451 Apr 13, 2015 at 10:55

When you introduce a new field to make the Lagrangian gauge invariant, then you are at liberty to choose the transformation behaviour of the new field such that the Lagrangian becomes gauge invariant.

If $\sigma\mapsto\sigma+\alpha$ leads to an invariant Lagrangian, then you are free to choose $\sigma$ as a field transforming such. In the situation you presented, there's really not more to it.

Note though, that you have still not specified what the gauge group is, and if $\alpha$ is gauge Lie algebra valued rather than a real number, then you will have to let $\sigma$ take values in the Lie algebra as well rather than being just real valued (although it will still transform as a scalar under the Lorentz group).

• I saw a similar question that asked if it was possible to find the new Lagrangian with a "canonical kinetic term" $-\frac{1}{2}(\partial_\mu\sigma)^2$. I think the answer is no if we want the sigma to transform in the way I specified but I'm not sure if there's an intrinsic reason for this. @ACuriousMind Apr 2, 2015 at 8:39

Suppose I have a Lagrangian density $\mathcal{L}(\phi^\mu,\sigma)$ depending on vector fields $\phi^\mu$ and their derivatives and a scalar field $\sigma$ and its derivatives. If I make a gauge transformation $\phi^\mu\rightarrow \phi^\mu+\partial^\mu\alpha$ does the field $\sigma$ transform? I've seen notes claiming $\sigma\rightarrow \sigma + \alpha$ but just wanted to make sure.

If I interpret what you wrote in the usual sense, then $\phi^\mu$ and $\sigma$ are independent field variables. Thus changing the $\phi^\mu$ in independent of changing the $\sigma$. What it looks like you are trying to say/understand is that the change $\phi^\mu\to\phi^\mu+\partial^\mu\alpha$ is equivalent to $\sigma\to\sigma+\alpha$.

Let me give you an example. Suppose that $$\mathcal{L}=x-y$$ then the replacement $$x\to x+z$$ is equivalent to the replacement $$y\to y-z$$ because both have the same overall effect on $\mathcal{L}$.