Master's Thesis in Theoretical Physics - Universiteit Utrecht

such that the potential term is given by (4.72). As shown in the previous section, if we wouldhave an off-diagonal kinetic term, we can remove this by a redefinition of our fields. Theonly consequence is extra potential terms. We have numerically verified that these extrapotential terms have no significant influence on the dynamics of inflation. Therefore, wetake our action to be∫S =d 4 x −g(− ∑i=1,2g µν (∂ µ φ i ) ∗ (∂ ν φ i ) −∑i, j=1,2ξ i j Rφ ∗ i φ j − V (φ 1 ,φ 2 )), (4.90)where the potential V (φ 1 ,φ 2 ) is given in Eq. (4.72). By a variation of the action with respectto the metric g µν we find the Einstein tensorG µν =1M 2 p − 2 ∑ i, j=1,2 ξ i j φ ∗ i φ j×{2 ∑ [(δ i j − 2ξ i j )(∂ µ φ i ) ∗ (∂ ν φ j ) − ( 1 ]2 δ i j − 2ξ i j )g µν g ρσ (∂ ρ φ i ) ∗ (∂ σ φ j )i, j=1,2−g µν V (φ 1 ,φ 2 ) − 2∑i, j=1,2ξ i j[φ ∗ i (g µν g ρσ ∇ ρ ∇ σ − ∇ µ ∇ ν )φ j + φ j (g µν g ρσ ∇ ρ ∇ σ − ∇ µ ∇ ν )φ ∗ i]}.We can compare this to Eq. (4.3) and we see that this is indeed the extension from onereal field to two complex fields. Note the factors of 2, which would disappear if we wouldwrite the complex fields in terms of real fields as φ i = 1 (φ R + iφ I ). If we again assume a2 i ihomogeneous and isotropic background, we find H 2 from the 00 component of the Einsteintensor as{H 2 1∑=3(M 2 p − 2 ∑ i, j=1,2 ξ i j φ ∗ i φ | φ˙i | 2 + V (φ 1 ,φ 2 ) + 6H∑ξ i j [φ ∗ iφ˙j + φ j φ˙}. ∗ ij)] (4.91)i=1,2i, j=1,2Similarly we find the equation for Ḣ from the diagonal spatial components of G i j and byusing the equation for H 2 . This gives1Ḣ ={− ∑M 2 p − 2 ∑ i, j=1,2 ξ i j φ ∗ i φ | ˙φi | 2 − ∑ji=1,2Finally the equation of motion for φ i isi, j=1,2ξ i j [−(φ ∗ ¨iφ j + ¨φ∗i φ j)+H(φ ∗ i ˙φ}j + ˙φ∗i φ j)−2( ˙φ∗ ˙iφ j )] .(4.92)¨φ i + 3H ˙φi + R ∑ kξ ik φ k + ∂V (φ 1,φ 2 )∂φ ∗ i= 0. (4.93)The equation of motion for φ ∗ iis obtained by complex conjugation. We can again eliminate¨φ i from the expression for Ḣ by using the equation of motion. Then we find the Ricci scalarR = 6[Ḣ + H 2 ] in terms of φ i and ˙φi asR =1{M 2 p − 2 ∑ i, j,k=1,2 φ ∗ i ξ −2i j(δ jk − 6ξ jk )φ k−6∑i, j=1,2ξ i j [φ ∗ ∂V (φ 1 ,φ 2 )i∂φ ∗ j∑i, j=1,2(δ i j − 6ξ i j )( ˙φ∗ φ˙j )+ ∂V (φ 1,φ 2 )∂φ iφ j ] + 4V (φ 1 ,φ 2 )i}. (4.94)Note the conformal coupling value of the nonminimal matrix, i.e. ξ i j = 1 6 δ i j. If we takeour inflationary potential to be (4.72), we see that R vanishes for the conformal coupling