3
$\begingroup$

While in QED correlation functions including the photon field vanish when the Lorentz index of the gauge field is contracted with its own momentum (this fact is usually referred to as Ward Identity, as far as I know), in non-abelian gauge theories correlation functions with analogous contracted are related to correlation functions including ghosts on external lines (at least in covariant gauges, such as Lorenz), via identities known as Slavnov-Taylor. My question is: is it correct to state that when considering pure gluons on-shell scattering amplitudes (not off-shell correlation functions) the amplitude vanish whenever ANY number of spin-one polarization vector is replaced with its own momentum?

I guess it is correct because, regardless of the fact that the theory is non-abelian, massless spin-one asymptotic states are always defined modulo a factor proportional to the particle's momentum, which can't contribute to scattering amplitudes (which are physical).

Is this true also for perturbative Einstein Gravity?

$\endgroup$
  • $\begingroup$ One consequence of the Slaavnov-Taylor identity is that $p_\mu p_\nu G^{\mu\nu}\sim \xi$ where $\xi$ depends on the gauge fixing term and $G^{\mu\nu}$ is the gluon propagator. Can this be helpful? $\endgroup$ – apt45 Jan 8 '18 at 11:37
  • $\begingroup$ Mmm nope, I'm interested in scattering amplitudes sorry $\endgroup$ – user180825 Jan 8 '18 at 12:40
  • $\begingroup$ Is not an s-channel $2\rightarrow 2$ scattering of gluons an amplitude of interest? $\endgroup$ – apt45 Jan 8 '18 at 13:15
  • $\begingroup$ Of course but 1) I'm interested in the proof of my statement for any number of external gluons and 2) to get the amplitude 2->2 the gluon propagator has to be contracted with gluon vertices; moreover the s-channel Is only a Feynman diagram contribution to the 2->2 amplitude, so it doesn't have to be gauge invariant (indeed it isn't) and I don't expect It ti have the mentioned property $\endgroup$ – user180825 Jan 8 '18 at 17:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.