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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.

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  • $\begingroup$ 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. $\endgroup$ – ACuriousMind Apr 1 '15 at 17:44
  • $\begingroup$ I've provided more information @ACuriousMind $\endgroup$ – Okazaki Apr 1 '15 at 18:02
  • $\begingroup$ Another question by OP about the same Lagrangian density: physics.stackexchange.com/q/175514/2451 $\endgroup$ – Qmechanic Apr 13 '15 at 10:55
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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).

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  • $\begingroup$ 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 $\endgroup$ – Okazaki Apr 2 '15 at 8:39
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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 $\mathcal{L}$.

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