I was reading this question on the Physics Stack Exchange, and I'm still not quite sure how I can understand the relationship between locality and local gauge invariance using this example. Consider the Lagrangian density for the electromagnetic field $A_\mu$, coupled to a complex scalar field $\phi$: \begin{equation*} \mathcal{L}_{{\rm gauge}} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+(D_\mu\phi)^*D^\mu\phi, \end{equation*} where $D_\mu = \partial_\mu-iqA_\mu$, and $F_{\mu\nu} = \partial_\mu A_\nu-\partial_\nu A_\mu$. Is it right to say the local gauge invariance $\phi\rightarrow e^{i\theta(x)}\phi$ preserves the Lagrangian density $\mathcal{L}_{{\rm gauge}}$, which makes the corresponding action local, therefore the field theory is local?

I think I don't quite understand how we define 'locality' in the context of field theory. Also for this example, are we assuming local gauge invariance, instead of global symmetries?

  • 2
    $\begingroup$ The fact that the Lagrangian is a function of one position is what makes the action and therefore the field theory local. This is a necessary condition. You can also have local field theories for which no Lagrangian or action exists. "Local gauge invariance" is a bad term because it means the same thing as "gauge invariance". $\endgroup$ Nov 5, 2022 at 21:52
  • $\begingroup$ @Connor Behan Thanks for the comment! Sorry I think my understanding is still a bit vague. The Lagrangian is a function of $x$ here, is that because of the gauge invariance $πœ™β†’π‘’^{π‘–πœƒ(π‘₯)}πœ™$? Do we care about the time dependence of the lagrangian density when talking about the locality in field theory? $\endgroup$
    – IGY
    Nov 5, 2022 at 22:00
  • $\begingroup$ This might help $\endgroup$ Nov 5, 2022 at 22:54

1 Answer 1


"Local" gauge invariance has nothing, per se, to do with locality of the Lagrangian. Locality of the Lagrangian is about having all the products of fields and derivatives in the Lagrange density be evaluated at the same point. If you allow for combinations of fields at different spacetime points, like $\phi(x)\phi(y)$, you are almost certain to violate causality, with faster-than-light signaling from $x$ to $y$ and vice-versa.

On the other hand, what is "local" about a gauge transformation of the second kind* is that the parameter $\theta$ in the gauge transformation $\psi\rightarrow\phi'=e^{i\theta(x)}\phi$ can be chosen to be an arbitrary function of spacetime (subject to suitable differentiability conditions). The transformation is "local" in the sense that it is allowed to be completely different in different local neighborhoods. Concretely, you could choose $\theta(x)$ to be zero outside some small spacetime region, in which case the transformation would clearly be "local" in the sense that only in that local neighborhood is the field actually transformed in a nontrivial way.

*"Gauge transformation of the second kind" is probably a better name, since it does not introduce this confusion about with another meaning of "local."

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    $\begingroup$ It has to do with locality in the sense that whenever you impose local gauge invariance on matter fields, you end up with a local description of the matter's interaction with another field (EM or Weak or Strong). Other descriptions of the interaction are non-local (like Wilson loops). This is what the linked Andrea's answer said $\endgroup$
    – Ryder Rude
    Nov 6, 2022 at 4:10

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