Why is it interesting?

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[W]e review the implementation of the Stückelberg trick in the theory of massive $SU(N)$ YM gauge bosons which, for simplicity, all have the same mass $m$, \begin{align} \mathcal{L}&=-\frac{1}{4g^{2}}{\rm tr}F_{\mu\nu}F^{\mu\nu}-\frac{m^{2}}{2g^{2}}{\rm tr} A_{\mu}A^{\mu}\ .\label{MassiveYMforStuckelbergSection} \tag{ 3.22} \end{align}

This lagrangian is not gauge invariant, but gauge invariance can be restored by coupling in new fields, $\pi^{a}(x)$, with $a\in\{1,\ldots,N^2-1\}$, {\it i.e.}, one field for each generator of $SU(N)$. In order to insert the $\pi^{a}(x)$'s appropriately, one first performs a gauge transformation with $\pi^{a}(x)$ as the gauge parameter, \begin{align} A_{\mu}&\longmapsto U^\dagger(\pi)(A_{\mu}+\partial_{\mu})U(\pi)\equiv A'_{\mu} F_{\mu\nu}&\longmapsto U^\dagger(\pi) F_{\mu\nu}U(\pi)\equiv F'_{\mu\nu}\label{StuckelbergTransformationExample} \end{align} where $U(\pi)=e^{\pi^{a}(x)T_{a}}$ is an element of $SU(N)$. We then define a new lagrangian $\mathcal{L}'$ by taking \eqref{MassiveYMforStuckelbergSection} and replacing $A_{\mu}\mapsto A_{\mu}'$ and $F_{\mu\nu}\mapsto F_{\mu\nu}'$. That is, \begin{align} \mathcal{L}'&=-\frac{1}{4g^{2}}{\rm tr}F'_{\mu\nu}F'^{\mu\nu}-\frac{m^{2}}{2g^{2}}{\rm tr}A'_{\mu}A'^{\nu} &=\!-\frac{1}{4g^{2}}\!{\rm tr}F_{\mu\nu}F^{\mu\nu}\!-\!\frac{m^{2}}{2g^{2}} {\rm tr}D_{\mu}U(\pi)D^{\mu}U^\dagger(\pi)~, \label{StuckelbergedLagrangianEx} \end{align} where $D_{\mu}U(\pi)= \partial_{\mu}U(\pi)+A_{\mu}U(\pi)$ is the gauge covariant derivative of $U(\pi)$. The lagrangian $\mathcal{L}'$ then enjoys a gauge symmetry under which we simultaneously change \begin{align} A_{\mu}&\longmapsto V^\dagger(x)(A_{\mu}+ {1}\partial_{\mu})V(x) U(\pi)&\longmapsto V^\dagger(x)U(\pi)~, \end{align} where $V(x)\in SU(N)$.

The physics of the $\mathcal{L}$ and $\mathcal{L}'$ lagrangians is the same, we have just made the degrees of freedom in Eq. (3.22) manifest. In $\mathcal{L}'$, we introduced $N^{2}-1$ new fields, but also restored $N^{2}-1$ gauge symmetries and hence degree of freedom counting is the same for both cases. We can demonstrate the equivalence explicitly by using the gauge symmetry of $\mathcal{L}'$ to go ``unitary gauge" in which we set $U(\pi)\to {1}$, where the two lagrangians coincide.

The above process and its generalizations are known collectively as the Stückelberg trick, which is often a useful tool for elucidating the physics in certain regimes of theories, especially at high energies.

https://arxiv.org/abs/1405.5532

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