Gravity and the massless limit of linear massive

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: he gauge symmetry which kills the extra degrees of freedom appears only when the mass is strictly zero. The extra degrees of freedom are a massless vector and a massless scalar which couples to the trace of the energy momentum tensor. This extra scalar coupling is responsible for the vDVZ discontinuity. Taking m ! 0 straight away in the Lagrangian (3.1) does not yield a smooth limit, because degrees of freedom are lost. Rev. Mod. Phys., Vol. 84, No. 2, April–June 2012 To find the correct limit, the trick is to introduce new fields and gauge symmetries into the massive theory in a way that ¨ does not alter the theory. This is the Stuckelberg trick. Once this is done, a limit can be found in which no degrees of freedom are gained or lost. To introduce the idea, we consider a simpler case, the theory of a massive photon A coupled to a (not necessarily conserved) source J , Z 1 1 S ¼ dD x À F F À m2 A A þ A J ; (4.1) 4 2 where F  @ A À @ A . The mass term breaks the would-be gauge invariance A ¼ @ Ã, and for D ¼ 4 this theory describes the 3 degrees of freedom of a massive spin 1 particle. Recall that the propagator for a massive vector is ½Ài=ðp2 þ m2 ފð þ p p =m2 Þ, which is similar to $1=m2 for large momenta, invalidating the usual power counting arguments. As it stands, the limit m ! 0 of the Lagrangian (4.1) is not a smooth limit because we lose a degree of freedom; for m ¼ 0 we have Maxwell electromagnetism which in D ¼ 4 propagates only 2 degrees of freedom, the two polarizations of a massless helicity 1 particle. Also, the limit does not exist unless the source is conserved, as this is a consistency requirement in the massless case. ¨ The Stuckelberg trick consists of introducing a new scalar field , in such a way that the new action has gauge symmetry but is still dynamically equivalent to the original action. It will expose a different m ! 0 limit which is smooth, in that no degrees of freedom are gained or lost. We introduce a field  by making the replacement A ! A þ @ ; (4.2) following the pattern of the gauge symmetry we want to ¨ introduce (Stuckelberg, 1957). This is emphatically not a change of field variables. It is not a decomposition of A into transverse and longitudinal parts (A is not meant to identically satisfy @ A ¼ 0 after the replacement), and it is not a gauge transformation [the Lagrangian (4.1) is not gauge invariant]. Rather, this is creating a new Lagrangian from the old one, by the addition of a new field . F is invariant under this replacement, since the replacement looks similar to a gauge transformation and F is gauge invariant. The only thing that changes is the mass term and the coupling to the source, Z 1 1 S ¼ dD x À F F À m2 ðA þ @ Þ2 4 2  À @ J  : þ A J (4.3)  We have integrated by parts in the coupling to the source. The new action now has the gauge symmetry A ¼ @ Ã;  ¼ ÀÃ: (4.4) By fixing the gauge  ¼ 0, called...
View Full Document

This document was uploaded on 09/28/2013.

Ask a homework question - tutors are online