(ed.). Gravitational waves (IOP, 2001)(422s).

298 Elementary introduction to pre-big bang cosmologyto introduce the more conventional E-frame (16.21) in which the dilaton isminimally coupled to the metric, with a canonical kinetic term.In the two frames the field equations are different, and the perturbationequations are also different. This seems to raise a potential problem: which frameis to be used to evaluate the physical effects of the cosmological perturbations?The problem is only apparent, however, because physical observables (likethe perturbation spectrum) are the same in both frames. The reason is that thereis a compensation between the different perturbation equations and the differentbackground solution around which we expand. A general proof of this result canbe given by using the notion of canonical variable (see subsection 16.5.3). HereI will give only an explicit example for tensor perturbations in a d = 3, isotropicand spatially flat background.Let us start in the E-frame, with the background equations:R µν = 1 2 ∂ µφ∂ ν φ, (16.43)referring to the ‘tilded’ variables of (16.42) (we will omit the ‘tilde’, forsimplicity, and will explicitly reinsert it at the end of the computation).Considering the transverse-traceless part of metric perturbations:δ (1) φ = 0, δ (1) g µν = h µν , δ (1) g µν =−h µν , ∇ ν h µ ν = 0 = h µ ν (16.44)(∇ µ denotes covariant differentiation with respect to the unperturbed metric g,and the indices of h are also raised and lowered with g). The perturbation of thebackground equations gives:δ (1) R ν µ = 0. (16.45)We can then work in the synchronous gauge, whereTo first order in h we getg 00 = 1, g 0i = 0, g ij =−a 2 δ ij ,h 00 = 0, h 0i = 0, g ij h ij = 0, ∂ j h j i = 0. (16.46)δ (1) Ɣ j 0i = 1 j 2ḣi , δ (1) Ɣ 0 ij =− 1 2 ḣij,δ (1) Ɣ k ij = 1 2 (∂ i h k j + ∂ j h k i − ∂ k h ij ). (16.47)The (0, 0) component of equation (16.45) is trivially satisfied (as well as theperturbation of the scalar field equation); the (i, j) components, by using theidentities (see for instance [54])g jk ḣ ik = ḣ j i + 2Hh j i ,g jk ḧ ik = ḧ j i + 2Ḣh j i + 4H ḣ j i + 4H 2 h j i , (16.48)giveδ (1) R j i =− 1 ()ḧ j i + 3H ḣ j i − ∇22a 2 h i j ≡− 1 2 £h i j = 0. (16.49)