Master's Thesis in Theoretical Physics - Universiteit Utrecht

where in all the equations z = k∆η. Using these relations we can take the derivatives, andwe find|f |20 = i✁∂χ(x) +2 5 π 2 i ✁∂(∂ 2 0 + k2 ) ei⃗ k·⃗x ∫ ηdη ′{ ln(µ 2 ∆η 2 )sin(k∆η) − cos(k∆η)Si(2k∆η)k[( )] k∆η }+sin(k∆η) Ci(2k∆η) − γ E − ln χ(η ′ )2|f |2 [+ ln(a)i2 5 π 2 ✁∂χ(x) + i✁∂ ( ln(a)χ(x) )]−|f | 2 (1 − ɛ)2 HH ′ a(ν 22 4 π 2 − 1 )4×{i✁∂ ei⃗ k·⃗x ∫ ηdη ′ a ′{ [ln(α∆η 2 ) + 1 ] sin(k∆η) − cos(k∆η)Si(2k∆η)k[( )] k∆η }+sin(k∆η) Ci(2k∆η) − γ E − ln χ(η ′ )2+ln(a)i✁∂ ei⃗ k·⃗x ∫ ηdη ′ a ′ sin(k∆η)χ(η ′ ) + i✁∂ ei⃗ k·⃗xkk∫ ηdη ′ a ′ ln(a ′ )sin(k∆η)χ(η ′ )}.(6.96)We now make a distinction between the terms. During inflation the scale factor a growsexponentially, and therefore the second term (the conformal anomaly) grows as ln(a) andthe third term as aa ′ and aa ′ ln(aa ′ ). The first term is the conformal vacuum contributionand does not depend on the scale factor. Thus during inflation, the first term does not growand is not relevant. Note that there are also higher order terms but these scale as s ≪ 1and are therefore suppressed (see the discussion below Eq. (5.104). To summarize, we keeponly the conformal anomaly term and the third term. Now we writei✁∂χ(x) = (iγ 0 ∂ 0 + iγ i ∂ i )e i⃗ k·⃗x χ(k,η) = (iγ 0 ∂ 0 −⃗γ ·⃗k)e i⃗ k·⃗x χ(k,η),such that the modified Dirac equation simplifies to0 = (iγ 0 ∂ 0 −⃗γ ·⃗k)χ(k,η)|f[|2+2 5 π 2 ln(a)(iγ 0 ∂ 0 −⃗γ ·⃗k)χ(k,η) + (iγ 0 ∂ 0 −⃗γ ·⃗k) ( ln(a)χ(k,η) )]−|f | 2 (1 − ɛ)2 HH ′ a(ν 22 4 π 2 − 1 )4{× (iγ 0 ∂ 0 −⃗γ ·⃗k) 1 ∫ ηdη ′ a ′{ [ln(α∆η 2 ) + 1 ] sin(k∆η) − cos(k∆η)Si(2k∆η)k[( )] k∆η }+sin(k∆η) Ci(2k∆η) − γ E − ln χ(η ′ )2+ln(a)(iγ 0 ∂ 0 −⃗γ ·⃗k) 1 ∫ ηdη ′ a ′ sin(k∆η)χ(η ′ )k+(iγ 0 ∂ 0 −⃗γ ·⃗k) 1 ∫ η}dη ′ a ′ ln(a ′ )sin(k∆η)χ(η ′ ) . (6.97)kWe can simplify the above equation by noticing that−(iγ 0 −⃗γ ·⃗k) 1 k Θ(∆η)sin(k∆η)