advanced_tools:gauge_symmetry:stueckelberg_trick
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advanced_tools:gauge_symmetry:stueckelberg_trick [2017/11/08 14:37] jakobadmin [Student] |
advanced_tools:gauge_symmetry:stueckelberg_trick [2022/06/07 10:04] (current) 192.84.145.254 fixed aligns |
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| This lagrangian is not gauge invariant, but gauge invariance can be restored by coupling in new fields, $\pi^{a}(x)$, with $a\in\{1,\ldots,N^2-1\}$, 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, | This lagrangian is not gauge invariant, but gauge invariance can be restored by coupling in new fields, $\pi^{a}(x)$, with $a\in\{1,\ldots,N^2-1\}$, 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} | \begin{align} | ||
| - | A_{\mu}&\longmapsto U^\dagger(\pi)(A_{\mu}+\partial_{\mu})U(\pi)\equiv A'_{\mu} | + | 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} | F_{\mu\nu}&\longmapsto U^\dagger(\pi) F_{\mu\nu}U(\pi)\equiv F'_{\mu\nu}\label{StuckelbergTransformationExample} | ||
| \end{align} | \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, | 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} | \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} | + | \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} | &=\!-\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} | \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 | 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} | \begin{align} | ||
| - | A_{\mu}&\longmapsto V^\dagger(x)(A_{\mu}+ {1}\partial_{\mu})V(x) | + | A_{\mu}&\longmapsto V^\dagger(x)(A_{\mu}+ {1}\partial_{\mu})V(x) \\ |
| U(\pi)&\longmapsto V^\dagger(x)U(\pi)~, | U(\pi)&\longmapsto V^\dagger(x)U(\pi)~, | ||
| \end{align} | \end{align} | ||
advanced_tools/gauge_symmetry/stueckelberg_trick.txt ยท Last modified: 2022/06/07 10:04 by 192.84.145.254
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