# Gluon Propagator

The photon propagator is obtained from the Yang-Mills Lagrangian applying the Faddeev-Popv procedure:

$$D_{\mu\nu}(k) ~=~ \frac{i}{k^2}[-g_{\mu\nu} + (1-\zeta)\frac{k_\mu k_\nu}{k^2}]$$

To get to the gluon propagator from here, we just multiply an extra color factor of $\delta_{ab}$. It is not obvious to me why. What is the reasoning behind this?

• Did you expect a different result? Commented Nov 14, 2016 at 19:42
• I did not expect the kronecker delta. Commented Nov 14, 2016 at 19:44
• That just says the color stays the same, though. Commented Nov 14, 2016 at 19:48
• The gluon field carries color, so there must be color indices in the gluon propagator. Commented Nov 14, 2016 at 19:48
• okay, so its something we put in by hand then? to say that the color stays the same? Commented Nov 14, 2016 at 19:49

The quadratic part of the Lagrangian of the gluon is exactly the same as the photon Lagrangian, except that the fields have color indices: $$(\partial_\mu A_\nu^a - \partial_\nu A_\mu^a)(\partial_\mu A_\nu^a - \partial_\nu A_\mu^a)$$ which is the same as $$\delta^{ab} (\partial_\mu A_\nu^a - \partial_\nu A_\mu^a)(\partial_\mu A_\nu^b - \partial_\nu A_\mu^b)$$ whence the $\delta$.