- #1

RedX

- 970

- 3

In general, for a gauge field with Yang-Mills Lagrangian

[tex]\mathcal L=-\frac{1}{4}F^{c}_{\mu \nu}F^{c \mu \nu} [/tex]

for each c it is impossible to find the resulting free Green's function G(k) in momentum space:

[tex](g^{\mu \nu}k^2-k^{\mu}k^{\nu})G_{\nu \eta}(k)=\delta^{\mu}_{\eta} [/tex]

since [tex](g^{\mu \nu}k^2-k^{\mu}k^{\nu}) [/tex] has an eigenvalue of zero along the direction of k, which means the determinant of the operator is zero and uninvertible.

However, it seems you are allowed to insert a delta function into the path integral and use the FP-method, and for a [tex]R_\xi [/tex] gauge this results in trying to invert:

[tex][g^{\mu \nu}k^2-(1-\frac{1}{\xi})k^{\mu}k^{\nu}]G_{\nu \eta}(k)=\delta^{\mu}_{\eta} [/tex]

which can readily be checked has an inverse:

[tex]G_{\nu \eta}(k)=\frac{1}{k^2}\left(g_{\nu \eta}-(1-\xi)\frac{k_\nu k_\eta}{k^2} \right) [/tex]

My question is with regards to gauge symmetry of the action. By inserting a delta function (that contains the information on the choice of gauge) into the path integral, it somehow manages to find its way into the exponential and alters the action itself. Is this new action in general gauge invariant?

Also, what if the choice of gauge is not Lorentz invariant? Will your scattering amplitudes still be Lorentz invariant? That is, if you boost the initial and final states to larger momentum, will the amplitude be the same as the amplitude for the slower initial and final states pre-boost?

Also, what if your choice of gauge is not linear? Can you just choose any equation involving the fields to be a gauge condition? Also, in all the books I've seen, the gauge condition doesn't mix up group indices. Often the gauge condition is one of spacetime, and each gluon field is given this same spacetime restriction. Can your gauge condition mix the different gluons for example?