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Inflationary Cosmology Drew Jamieson Stony Brook University March - PowerPoint PPT Presentation

Inflationary Cosmology Drew Jamieson Stony Brook University March 23, 2016 Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 1 / 28 Overview Cosmic Expansion 1 Three Cosmological Problems 2 Mechanism for Inflationary Expansion


  1. Inflationary Cosmology Drew Jamieson Stony Brook University March 23, 2016 Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 1 / 28

  2. Overview Cosmic Expansion 1 Three Cosmological Problems 2 Mechanism for Inflationary Expansion 3 Exciting Quantum Oscillators According to Heisenberg 4 Generation of Cosmic Anisotropies 5 Comparison with CMB Data 6 Conclusions 7 Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 2 / 28

  3. The Metric ds 2 = − dt 2 + a ( t ) 2 γ ij dx i dx j 1 γ ij dx i dx j = 1 − Kr 2 dr 2 + r 2 d Ω ( spherical coordinates ) ➔ K = +1: Spherical ➔ K = 0: Flat ➔ K = − 1: Hyperbolic Einstein field equations: R µν − 1 2 g µν R = − 8 π GT µν � ˙ � 2 a = 4 π G ( ρ + Λ) − K a 2 a 3 Scale factor evolves according to energy density, cosmological constant, and curvature Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 3 / 28

  4. Matter d 4 x √− g L matter � S matter = 1 δ S matter √− g T µν = δ g µν 1 3 T i T 00 = ρ i = p Continuity equation: ∇ µ T µ ν = 0 ρ = − 3( ρ + p ) ˙ a ˙ a For p = w ρ : d ρ da = − 3(1 + w ) ρ a ρ ∼ a − 3(1+ w ) Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 4 / 28

  5. Λ Expansion Regimes a ( t ) K w a ( t ) ∼ √ t NR 1 Radiation: 3 2 Nonrelativistic matter: 0 ∼ t 3 R − 1 Curvature: ∼ t 3 ∼ e α t Λ: − 1 t Today: a = 1, ˙ a = H 0 Percentages of total energy density: Ω R + Ω NR + Ω K + Ω Λ = 1 � a ˙ Ω R a 4 + Ω NR + Ω K a = H ( a ) = H 0 a 2 + Ω Λ a 3 1 0.500 Ω NR = 0 . 308 ± 0 . 012 0.100 a ( t ) 0.050 Ω Λ = 0 . 692 ± 0 . 012 Ω R ≈ Ω K ≈ 0 0.010 0.005 0.01 0.05 0.10 0.50 1 5 10 Time ( Billion years ) Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 5 / 28

  6. Three problems Flatness: ρ T ≪ 10 − 16 during e + e − annihilation ρ K Ω K ≪ 1 = ⇒ Horizons: CMB temperature fluctuations: δ T T ∼ 10 − 5 even though sky is mostly causally disconnected Monopoles: If you like GUTs, you get roughly 1 magnetic monopole per nucleon Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 6 / 28

  7. Solution: Inflation: Early on, the universe underwent a period of temporary but rapid expansion ➔ a ( t ) ∼ e α t ➔ α → 0 eventually Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 7 / 28

  8. The Inflaton and Slow Roll Inflation d 4 x √− g � 1 � � 2 ∂ µ φ∂ µ φ − V ( φ ) S KG = ρ = 1 p = 1 φ 2 + V ( φ ) φ 2 − V ( φ ) ˙ ˙ 2 2 φ 2 − V ( φ ) 1 2 ˙ Slow Roll Inflaton Potential w = 2 ˙ φ 2 + V ( φ ) 1 If V ( φ ) ≫ ˙ φ 2 w ≈ − 1 Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 8 / 28

  9. Intermission Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 9 / 28

  10. Exciting an Oscillator Two mechanisms: a) A driving force shakes things up: q ( t ) = − ω 2 q ( t ) + j ( t ) ¨ b) A time-dependent frequency: q ( t ) = − ω ( t ) 2 q ( t ) ¨ ➔ Both involve resonance if periodic ➔ Classically ω ( t ) won’t work in the ground state ➔ In quantum mechanics, even ground state has fluctuations Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 10 / 28

  11. The Driven Oscillator In the Heisenberg picture: [ q ( t ) , p ( t )] = i 1 q † = q , p † = p 2 p ( t ) 2 + ω 2 H = 1 2 q ( t ) 2 − j ( t ) q ( t ) Heisenberg equation of motion: i ˙ f ( t ) = [ f ( t ) , H ( t )] q ( t ) = p ( t ) ˙ p ( t ) = − ω 2 q ( t ) + j ( t ) ˙ Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 11 / 28

  12. The Driven Oscillator Usual trick: 1 � a + a † � √ q ( t ) = 2 ω � ω � a − a † � p ( t ) = − i 2 The Hamiltonian simplifies: a † ( t ) a ( t ) + 1 − j ( t ) � � � � a † ( t ) + a ( t ) H ( t ) = ω √ 2 2 ω EOM for a ( t ): 1 √ i ˙ a ( t ) = ω a ( t ) − 2 ω j ( t ) Solution: � t i a ( t ) = a in e − i ω t + 1 dt ′ j ( t ′ ) e i ω ( t ′ − t ) √ 2 ω 0 Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 12 / 28

  13. Finite Duration Driving Force The force acts only for 0 < t < T � T i 0 dt ′ j ( t ′ ) e i ω ( t ′ − T ) , then: Define j 0 = √ 2 ω     a in e − i ω t for t < 0           � t  a in e − i ω t + 1 0 dt ′ j ( t ′ ) e i ω ( t ′ − t ) for 0 < t < T i a ( t ) = √ 2 ω       ( a in + j 0 ) e − i ω t for t > T         Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 13 / 28

  14. Finite Duration Driving Force Before After a in , a † a out , a † in out in a in + 1 � out a out + 1 � � � a † a † H = ω H = ω 2 2 � n | Ω in � � n | Ω out � 1 1 � � a † a † √ | n in � = √ | n out � = out in n ! n ! {| n out �} form complete basis on {| n in �} form complete basis on the Hilbert space the Hilbert space � � n + 1 Both | n in � and | n out � have energy ω , but a out = a in + j 0 2 | n in � � = | n out � , Heisenberg picture | Ω in � � = | Ω out � Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 14 / 28

  15. Finite Duration Driving Force In the ground state: | ψ � = | Ω in � : After Before Fluctuations: Fluctuations: � T dt ′ sin (( t − t ′ ) ω ) j ( t ′ ) � Ω in | q ( t ) | Ω in � = � Ω in | q ( t ) | Ω in � = 0 ω 0 Energy: Energy: � Ω in | H | Ω in � = ω � 1 � 2 + | j o | 2 � Ω in | H | Ω in � = ω 2 What is | Ω in � at late times? a out | Ω in � = ( a in + j 0 ) | Ω in � = j 0 | Ω in � A coherent state! Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 15 / 28

  16. Finite Duration Driving Force Early vacuum state is superposition of late time excited states: 2 | j 0 | 2 j n e − 1 0 � √ | Ω in � = | n out � , ( Bogoliubov transformation ) n ! n Particle production: 0.08 | j 0 |= 5 0.06 P ( n ) = | � n out | Ω in � | 2 = | j 0 | 2 n n ! e −| j 0 | 2 P ( n ) 0.04 0.02 0.00 0 10 20 30 40 50 n Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 16 / 28

  17. Quantum Fluctuations of the Inflaton Consider the a Klein-Gordon field φ I = ¯ φ ( t ) + φ ( t ,� x ) in flat FRW: d 4 x √− g � 1 2 g µν ∂ µ φ∂ ν φ − 1 � � 2 m 2 φ 2 S KG = ds 2 = a ( η ) 2 ( − d η 2 + d � Let dt = a ( η ) d η : x · d � x ) Let χ ( η,� x ) = a ( η ) φ ( η,� x ) � 2 − m 2 a 2 − a ′′ � � � � � χ ′ 2 − � � d 3 xd η χ 2 S KG = ∇ χ a Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 17 / 28

  18. Quantum Fluctuations of the Inflaton Equation of motion: m 2 a 2 − a ′′ � � χ ′′ − ∇ 2 χ + χ = 0 a Fourier Transform: d 3 x � x ) e − i � k · � x χ k ( η ) = χ ( η,� 3 (2 π ) 2 k 2 + m 2 a 2 − a ′′ � � χ ′′ k + χ k = 0 a Each k-mode is a Harmonic oscillator with time dependent frequency: � k 2 + m 2 a ( η ) 2 − a ′′ ( η ) ω k ( η ) = a ( η ) Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 18 / 28

  19. η Quantum Fluctuations of the Inflaton � k 2 + m 2 a ( η ) 2 − a ′′ ( η ) ω k ( η ) = a ( η ) For exponential expansion: a = e Ht = − 1 H η , assuming m ≪ H : 2.0 � k 2 − 2 ω k ( η ) ≈ 1.5 k = 2 η 2 1.0 k = 1 Becomes imaginary for: ω 2 0.5 √ 2 H < a ( η ) 2 π 1 0.0 | η | < or k k - 1 - 0.5 k =  2  - 1.0 λ ( η ) ≈ H − 1 = ⇒ - 3.5 - 3.0 - 2.5 - 2.0 - 1.5 - 1.0 A mode oscillates until its wavelength is stretched larger than the Hubble radius, when it crosses this horizon the mode grows This is an effective amplification of vacuum fluctuations! Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 19 / 28

  20. Quantum Fluctuations of the Inflaton Three cases: small wavelength Start small, never cross the horizon, ( k >> η − 1 ) these quantum fluctuations are not amplified medium wavelength Start small but eventually cross the horizon, ( k ≈ η − 1 ) freezing out, these are amplified by inflation large wavelength Are always larger than the horizon, ( k << η − 1 ) physically unclear what these are (Λ)? Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 20 / 28

  21. Quantum Fluctuations of the Inflaton χ ′′ k + ω k ( η ) 2 χ k = 0 Solve Hermiticity, commutation relations, and equation of motion with: 1 � k ( η ) a k + v k ( η ) a † � v ∗ χ k ( η ) = √ k 2 where v ∗ k ( η ) v ′ k ( η ) − v ′∗ k ( η ) v k ( η ) = − 2 i (Wronskian for ODE) If mode functions v k ( η ) solve KG equation, so does χ k ( η ) k ( η ) = k 2 + m 2 H 2 η 2 − 2 Solution: ω 2 η 2 � π | η | � � v k ( η ) = J n ( k | η | ) − iY n ( k | η | ) 2 � 4 − m 2 9 H 2 ≈ 3 where n = 2 Drew Jamieson (SBU) Inflationary Cosmology March 23, 2016 21 / 28

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