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Cosmology & CMB Set5:Initial Conditions and Inflation Davide - PowerPoint PPT Presentation

Cosmology & CMB Set5:Initial Conditions and Inflation Davide Maino Universit` a degli Studi di Milano ISAPP 2011 Ph.D. School Einstein Equations - I For perturbation we need ( 0 ) and similarly and and their time


  1. Cosmology & CMB Set5:Initial Conditions and Inflation Davide Maino Universit` a degli Studi di Milano ISAPP 2011 Ph.D. School

  2. Einstein Equations - I • For perturbation Θ we need Θ( 0 ) and similarly Φ and Ψ and their time derivatives • Einstein equations describe the evolution. Take the time-time component � � R 00 − 1 = ( − 1 + 2 Ψ) R 00 − R G 0 0 = g 00 2 g 00 R 2 and taking the perturbed part and working in Fourier space 0 = − 6 H ∂ t Ψ + 6 Ψ H 2 − 2 Φ k 2 δ G 0 a 2 = 8 π G δ T 0 0 where T 0 0 = − ρ γ ( 1 + 4 Θ 0 ) . The same is true also for other species with density contrast δ i , where i = b , cdm

  3. Einstein Equations - II • Moving to conformal time k 2 Φ+ 3 ˙ � Φ − ˙ � a a ˙ = 4 π Ga 2 ( ρ cdm δ cdm + ρ b δ b + 4 ρ γ Θ 0 + 4 ρ ν N 0 ) a Ψ a where N 0 is monopole term for neutrinos • This is Poisson equation taking into account expansion. • Taking the space-space equation k 2 (Φ + Ψ) = 32 ρπ a 2 G ( ρ γ Θ 2 + ρ ν N 2 ) where Θ 2 and N 2 are quadrupole moment for photons and neutrinos. • Scattering is effective, only monopole and dipole are important Φ = − Ψ

  4. Initial Conditions - I • BE for photons, baryons and dark matter ˙ − ˙ Φ − ik µ Ψ − ˙ τ [Θ 0 − Θ + µ v b ] Θ + ik µ Θ = ˙ − 3 ˙ δ b ( cdm ) + ikv b ( cdm ) = Φ • At early times all scales of interest are outside the horizon k η ≪ 1 ˙ Θ ≃ Θ Θ Θ ˙ η k Θ ≫ 1 → ˙ η → k Θ ≃ Θ ≫ Θ we neglect all Θ and terms with k ˙ Θ 0 + ˙ ˙ δ b ( cdm ) = − 3 ˙ Φ = 0 Φ

  5. Initial Conditions - II • With this approximation and taking contribution only from radiation, the time-time equation reads 3 ˙ � Φ − ˙ � a a ˙ = 16 π Ga 2 ρ γ Θ 0 a Ψ a • For RD a ∝ t 1 / 2 ∝ η and ˙ a / a ∝ 1 /η � ˙ � 2 ρ γ ˙ η 2 = 28 π Ga 2 Φ η − Ψ a ρ Θ 0 ≃ 2 ρ γ Θ 0 = 2 η 2 Θ 0 3 a using Friedmann equation in conformal time

  6. Initial Conditions - III • Take the conformal time derivative and using ˙ Θ 0 + ˙ Φ = 0 Φ + η ¨ ˙ Φ − ˙ − 2 ˙ Ψ = Φ η ¨ Φ + 4 ˙ Φ = 0 • Search solution of the form Φ = η p . Two solutions with p = 0 (time independent) and p = − 3 (rapidly decreasing). Φ = 2 Θ 0 • For density contrast ˙ δ = 3 ˙ Θ 0 → δ = 3 Θ + constant and perturbations are classified according to value for the constant. If zero perturbations are called adiabatic otherwise iso-curvature

  7. Inflation - I • Why Φ � = 0 and how is possible Φ = 2 Θ 0 ? • Conformal time η is maximum distance a photon can travels from the beginning: particle horizon • Scales λ ∝ k − 1 with k η ≪ 1 are larger than particle horizon hence no causal contact. Only when re-enter horizon physics (i.e. Compton Scattering) plays its game • However CMB is highly isotropic all over the sky: how is this possible?

  8. Inflation - I

  9. Inflation - II • The comoving horizon � dt � da 1 η = a = a aH logarithmic integral of the comoving Hubble radius ( aH ) − 1 , the maximum distance a particle can travel in a unit expansion time • Particles separated now by a distance > ( aH ) − 1 cannot talk now but if the Hubble radius were larger in the past they can talk each other • But H − 1 ∝ a 2 in RD or H − 1 ∝ a 3 / 2 in MD

  10. Inflation - II

  11. Inflation - III • ( aH ) − 1 has to decrease i.e. ( aH ) has to increase = d 2 a d � ada / dt � dt 2 > 0 dt a which is an accelerated expansion usually called inflation • With H almost constant during inflation da a = Hdt → a ( t ) = a e exp [ H ( t − t e )] t < t e

  12. Inflation - III

  13. Inflation - III • Accelerated expansion ¨ a a = − 4 π G 3 ( ρ + 3 p ) which is < 0 for both radiation and matter • ¨ a > 0 ↔ ρ + 3 p < 0 ↔ p < − ρ/ 3 • This new ingredient has to dominate the Universe energy content

  14. Inflation and horizon problem • During de Sitter phase horizon scale H − 1 is almost constant • If inflation lasts long enough, all interesting scales are outside the horizon • e-foldings N = ln [ H I ( t e − t i )] • To solve horizon we require that present horizon scale H − 1 was 0 reduced during inflation to a scale λ H 0 ( t i ) < H − 1 I � � a t i � a t e � � T 0 � e − N ≤ H − 1 λ H 0 ( t i ) = H − 1 = H − 1 0 0 I a t 0 a t e T f • In units of h = c = G = 1 Hubble constant H 0 = 9 . 6 × 10 − 62 and T 0 = 1 . 9 × 10 − 32 � T 0 � T e � � N ≥ ln ≃ 67 + ln H 0 H I

  15. Inflation and flatness problem • During de Sitter phase H − 1 is constant a 2 H 2 ∝ 1 k Ω − 1 = a 2 • The fine-tuning required to have Ω 0 ≃ 1 today is | Ω − 1 | ∼ 10 − 60 at the end of inflation � a i � 2 | Ω − 1 | t = t e = e − 2 N = | Ω − 1 | t = t i a e • Taking | Ω − 1 | t = t i ≃ 1 the e-foldings required are ≃ 70 • Inflation ⇒ Ω 0 = 1 but does not change geometry but magnify curvature radius so that locally the Universe appears flat

  16. Inflation with a scalar field - I • Scalar field: next step of complication after simple constant ( Φ , Ψ , g µν ) • In QFT each particle has a quantum field associated but none of the known particle has a field able to do inflation, not even Higgs boson • The field φ ( x , t ) = φ 0 ( t ) + δφ ( x , t ) φ 0 is classical part of the field, δφ are quantum fluctuations • Stress-energy tensor for a scalar field � 1 � T α β = g α ν ∂ ν φ∂ β φ − g α 2 g µν ∂ µ ∂ ν φ + V ( φ ) β

  17. Inflation with a scalar field - II • Take the classical part, depends only on t � 1 � φ 0 ) 2 + g α φ 0 ) 2 − V ( φ ) T α β = − g α β ( ˙ 2 ( ˙ 0 g 0 β • T 0 0 = − ρ and T i i = p 0 + 1 0 − V ( φ 0 ) ⇒ ρ = 1 − ˙ ˙ ˙ φ 2 φ 2 φ 2 − ρ = 0 + V ( φ 0 ) 2 2 1 ˙ φ 2 = 0 − V ( φ 0 ) p 2 • If potential energy dominates p = − ρ and inflation may occur

  18. Equation of motion � ˙ � ˙ � 2 � φ 2 = 8 π G 3 ρ = 8 π G a 0 2 + V ( φ 0 ) a 3 • Take time derivative � ˙ � � 2 � 2 da / dt ¨ 8 π G a a � φ 0 + V ′ ˙ � φ 0 ¨ ˙ a − = φ 0 a a 3 2 da / dt � � ρ − 8 π G � 8 π G � � φ 0 + V ′ ˙ � φ 0 ¨ ˙ − 4 π G 3 + p 3 ρ = φ 0 a 3 8 π G φ 0 + V ′ ˙ � � − 8 π GH ˙ φ 0 ¨ ˙ φ 2 = φ 0 0 3 φ 0 + V ′ = 0 φ 0 + 3 H ˙ ¨

  19. Slow-roll condition • To have inflation we require ˙ φ 2 0 ≪ V ( φ ) so potential φ 0 is slowly rolling down its potential i.e. potential is very flat • We can neglect ¨ φ 0 and Friedmann becomes H 2 ≃ 8 π G 3 V ( φ 0 ) dynamic is dominated by potential energy of the scalar field • Slow-roll requires ( V ′ ) 2 2 ≪ V ( φ 0 ) ˙ ≪ H 2 φ 0 ⇒ V V ′′ ≪ H 2 φ 0 ≪ 3 H ˙ ¨ φ 0 ⇒

  20. Slow-roll parameters • It is useful to define ˙ ˙ � V ′ � 2 φ 2 H 1 0 ǫ = − H 2 = 4 π G H 2 = 16 π G V � V ′′ V ′′ � 1 = 1 η = 8 π G H 2 V 3 ¨ φ 0 η − ǫ = − δ = H ˙ φ 0 • ǫ quantifies how much H changes during inflation ¨ a H + H 2 = ( 1 − ǫ ) H 2 a = ˙ inflation is possible only if ǫ < 1. • In general slow-roll is attained when ǫ ≪ 1 and | η | ≪ 1

  21. Slow-roll parameters • Take Friedmann equations ˙ φ 2 H 2 0 = 4 π G ǫ ¨ a ( 1 − ǫ ) H 2 = a a / a ) 2 take time derivative of the first and divide by (˙ ¨ φ 0 + ˙ ǫ − 2 H ǫ = 2 ˙ ǫ φ 0 • Replace ¨ φ 0 with δ ǫ = − 2 H ǫ ( δ + ǫ ) ˙ • Evolution of ǫ is second-order in parameters i.e. almost constant during inflation

  22. Choose a Frame • Gravity acts on all components: small fluctuations in φ are related to metric perturbation giving rise to perturbations of the curvature Ψ • Comiving slicing: orthogonal to comoving observers world-line. Free-falling observed for which expansion is isotropic δφ com = 0 • Scalar field transforms δφ = δφ − ˙ ˜ φ T and ˜ δφ = 0 with T = δφ/ ˙ φ Ψ → Ψ com = ψ + HT = Ψ + H δφ ˙ φ • Comoving curvature perturbation R = Ψ + H δφ ≃ H δφ ˙ ˙ φ φ

  23. Perturbations • Consider now also the perturbation term δφ and linearize the perturbation δφ + 2 ˙ a ¨ ˙ δφ + k 2 δφ + a 2 V ′′ δφ = 0 a and in slow-roll inflation V ′′ is negligible • For scales λ ≪ H − 1 we have k ≫ aH and the second term can be neglected ¨ δφ + k 2 δφ = 0 harmonic oscillator with frequency k 2 . Fluctuations do not grow • For scales λ ≫ H − 1 above the horizon, k ≪ aH neglect the last term δφ + 2 ˙ a ¨ ˙ δφ = 0 a fluctuations are constant

  24. Slow-Roll Evolution • Rewrite u ≡ a δφ to remove expansion damping � ˙ � � 2 � a k 2 − 2 ¨ u + u = 0 a • for conformal time measured from the end of inflation η ˜ = η − η end � a Ha 2 ≈ − 1 da η ˜ = aH a end • More compact slow-roll equation � k 2 − 2 � ¨ u + u = 0 η ˜

  25. Slow-Roll limit • Slow-roll equation has exact solution � � k ± i e ∓ ik ˜ η u = A η ˜ • For | k ˜ η | ≫ 1 (early times, inside the Hubble length) behaves as free oscillator η |→∞ u = Ake ∓ ik ˜ η lim | k ˜ • Normalization A is set by the origin of quantum fluctuations of φ

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