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A Coupling constant in Yang-Mills Lagrangian

  1. Mar 23, 2017 #1
    The Yang-MIlls Lagrangian is given by ##\mathcal{L}_{\text{gauge}}
    = F_{\mu\nu}^{a}F^{\mu\nu a} + j_{\mu}^{a}A^{\mu a}.##

    We can rescale ##A_{\mu}^{a} \to \frac{1}{g}A_{\mu}^{a}## and then we have ##\frac{1}{g^{2}}F_{\mu\nu}^{a}F^{\mu\nu a}.##

    How does the second term change? Does the current have any effect on the pre-factor?
  2. jcsd
  3. Mar 24, 2017 #2


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    The second term is not in the Lgauge, but it's the coupling of the gauge fields with matter. If you simply rescale the gauge potential, then j is unaffected.
  4. Mar 25, 2017 #3


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    If the interaction Lagrangian is written as [itex]J^{\mu a}A^{a}_{\mu}[/itex], then the coupling is hidden in the matter field current. For example, look at the QCD Lagrangian

    [tex]\mathcal{L} = -\frac{1}{4} F^{a}_{\mu\nu}F^{\mu\nu a} + i \bar{\psi} \gamma^{\mu} D_{\mu}\psi .[/tex] Expand this using [itex]D_{\mu} = \partial_{\mu} - i g A^{a}_{\mu}T^{a}[/itex], you get

    [tex]\mathcal{L} = -\frac{1}{4} F^{a}_{\mu\nu}F^{\mu\nu a} + i \bar{\psi} \gamma^{\mu}\partial_{\mu}\psi + g A^{a}_{\mu} \left( \bar{\psi} \gamma^{\mu}T^{a}\psi \right) .[/tex]
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