Coupling constant in Yang-Mills Lagrangian

[spaghetti3451]

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?

 

[dextercioby]

The second term is not in the L_gauge, but it's the coupling of the gauge fields with matter. If you simply rescale the gauge potential, then j is unaffected.

 

[samalkhaiat]

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?

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

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

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