Let's say we are working in QED. The lagrangian is

\begin{equation} \mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\bar{\psi}\big(i\not{D}-m\big)\psi \end{equation}

where $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ and $\not{D}=\gamma^\mu\big(\partial_\mu+iqA_\mu)$. Here $A_\mu$ is the EM gauge field and $q$ is the charge of the electron. This lagrangian has a $U(1)$ gauge symmetry

\begin{equation} \begin{cases} \psi\rightarrow\psi'=e^{iq\alpha(x)}\psi \\ \bar{\psi}\rightarrow\bar{\psi}'=e^{-iq\alpha(x)}\bar{\psi} \\ A_\mu\rightarrow A_\mu'=A_\mu-\partial_\mu\alpha(x) \end{cases} \end{equation}

where $\alpha(x)$ is a smooth arbitrary function of spacetime coordinate $x^\mu$. In many textbooks (for example Tong's lectures on QFT on p.125 eq. 6.11) it is required that $\alpha(x)$ falls off at infinity. Now, I understand this requirement for physical fields like $F_{\mu\nu}$ since we want finite energy, charge, etc. and if the fields are non-zero at infinity then by integration we would find an infinite total charge. However, I don't understand why we also require that $\alpha(x)$ dies as $\vec{r}\rightarrow\infty$. Regardless of the behaviour of $\alpha(x)$ and wether its behaviour makes $A_\mu$ take a constant value at infinity (or even blow up), the physical fields $\vec{E}$ and $\vec{B}$ will still be zero at the boundary as long as the gauge-independent part of $A_\mu$ falls off nicely.

Comment: Just for context, I'm rethinking the foundamentals of gauge theory because I'm studying Strominger's lectures on the infrared structure of gravity [1] where he relies heavily on gauge transformations at infinity.

[1] https://arxiv.org/abs/1703.05448

  • $\begingroup$ I think Tong imposes the condition to have a "real" gauge symmetry, i.e., symmetries that do not change the physics. However, if you want to have nontrivial symmetries (which change the physics) you allow for more general fall-off (as Strominger does.). $\endgroup$
    – ungerade
    Dec 7 '19 at 20:44
  • $\begingroup$ Btw. Tong clarifies this on page 138. $\endgroup$
    – ungerade
    Dec 7 '19 at 20:47

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