Scale-invariance from spontaneously broken conformal invariance - PowerPoint PPT Presentation

Scale-invariance from spontaneously broken conformal invariance Austin Joyce Center for Particle Cosmology University of Pennsylvania Hinterbichler, Khoury arXiv:1106.1428 Hinterbichler, AJ, Khoury arXiv:1202.6056 March 17, 2012 Austin Joyce (UPenn) March 17, 2012 1 / 20

Introduction At early times, the universe is described by a (nearly) CFT on (nearly) flat space Universe is driven to be homogeneous and flat Explain scale invariance of primordial perturbations in terms of the symmetry breaking pattern so (4 , 2) − → so (4 , 1) 1 from some scalar operator in the CFT getting a VEV ∼ t ∆ Under general conditions, scale-invariance for spectator fields follows solely from symmetry breaking Specific examples: ◮ Quartic U(1), Negative φ 4 — Rubakov 0906.3693; Hinterbichler, Khoury 1106.1428 ◮ Galilean Genesis — Creminelli, Nicolis & Trincherini 1007.0027 Austin Joyce (UPenn) March 17, 2012 2 / 20

Simple example — negative quartic potential � � V � Φ � � − 1 2( ∂φ ) 2 + λ d 4 x 4 φ 4 S φ = Φ symmetry algebra so (4 , 2) δ P µ φ = − ∂ µ φ , δ J µν φ = ( x µ ∂ ν − x ν ∂ µ ) φ , δ D φ = − (∆ + x µ ∂ µ ) φ , − 2∆ x µ − 2 x µ x ν ∂ ν + x 2 ∂ µ � � δ K µ φ = φ . Consider coupling a weight-0 field χ in the CFT � � 2 φ 2 ( ∂χ ) 2 − m 2 � − 1 2 λφ 4 χ 2 + κ χ d 4 x 2 φ � φχ 2 S χ = . Austin Joyce (UPenn) March 17, 2012 3 / 20

Simple example — negative quartic potential � � V � Φ � � − 1 2( ∂φ ) 2 + λ d 4 x 4 φ 4 S φ = Φ symmetry algebra so (4 , 2) δ P µ φ = − ∂ µ φ , δ J µν φ = ( x µ ∂ ν − x ν ∂ µ ) φ , δ D φ = − (∆ + x µ ∂ µ ) φ , − 2∆ x µ − 2 x µ x ν ∂ ν + x 2 ∂ µ � � δ K µ φ = φ . Consider coupling a weight-0 field χ in the CFT � � 2 φ 2 ( ∂χ ) 2 − m 2 � − 1 2 λφ 4 χ 2 + κ χ d 4 x 2 φ � φχ 2 S χ = . µν = φ 2 η µν This field couples to the effective metric g eff Austin Joyce (UPenn) March 17, 2012 3 / 20

Negative quartic model cont. φ − λφ 3 = 0 Equation of motion: ¨ � Zero energy solution: ¯ 2 1 φ ( t ) = with −∞ < t < 0 attractor! λ ( − t ) � � This breaks some of the symmetries—the generators δ P i , δ D , δ J ij , δ K i still annihilate the background; they can be repackaged as δ J 5 i = 1 δ J 6 i = 1 δ J ij ; δ J 56 = δ D ; 2 ( δ P i + δ K i ) ; 2 ( δ P i − δ K i ) . which have the commutation relations of the so (4 , 1) algebra, [ δ J ab , δ J cd ] = η ac δ J bd − η bc δ J ad + η bd δ J ac − η ad δ J bc , where η ab = diag ( δ ij , 1 , − 1). Austin Joyce (UPenn) March 17, 2012 4 / 20

Negative quartic model cont. φ − λφ 3 = 0 Equation of motion: ¨ � Zero energy solution: ¯ 2 1 φ ( t ) = with −∞ < t < 0 attractor! λ ( − t ) � � This breaks some of the symmetries—the generators δ P i , δ D , δ J ij , δ K i still annihilate the background; they can be repackaged as δ J 5 i = 1 δ J 6 i = 1 δ J ij ; δ J 56 = δ D ; 2 ( δ P i + δ K i ) ; 2 ( δ P i − δ K i ) . which have the commutation relations of the so (4 , 1) algebra, [ δ J ab , δ J cd ] = η ac δ J bd − η bc δ J ad + η bd δ J ac − η ad δ J bc , where η ab = diag ( δ ij , 1 , − 1). symmetry breaking: so (4 , 2) → so (4 , 1). µν = ¯ φ 2 η µν ∼ 1 In the broken phase, χ couples to g eff t 2 η µν . Austin Joyce (UPenn) March 17, 2012 4 / 20

Perturbations — φ Writing φ = ¯ φ + ϕ , the Fourier modes of the perturbations ϕ satisfy � k 2 − 6 � ϕ k + ¨ ϕ k = 0 t 2 As k → 0, this is solved by ϕ k ∼ 1 t 2 and ϕ k ∼ ( − t ) 3 . The growing mode is just a constant time shift of the background solution � 2 1 φ ( t ) + ε ˙ φ ( t + ε ) = ¯ ¯ φ ( t ) = ¯ ¯ φ ( t ) + ε t 2 , λ hence the φ ∼ 1 / t solution is an attractor. Austin Joyce (UPenn) March 17, 2012 5 / 20

Perturbations — φ Writing φ = ¯ φ + ϕ , the Fourier modes of the perturbations ϕ satisfy � k 2 − 6 � ϕ k + ¨ ϕ k = 0 t 2 As k → 0, this is solved by ϕ k ∼ 1 t 2 and ϕ k ∼ ( − t ) 3 . The growing mode is just a constant time shift of the background solution � 2 1 φ ( t ) + ε ˙ φ ( t + ε ) = ¯ ¯ φ ( t ) = ¯ ¯ φ ( t ) + ε t 2 , λ hence the φ ∼ 1 / t solution is an attractor. Quantum fluctuations — | ϕ k | 2 ∼ 1 / k 5 t 4 Red spectrum � blue spectrum for ζ Austin Joyce (UPenn) March 17, 2012 5 / 20

Perturbations — χ Expanding around χ = 0, the quadratic χ lagrangian is λ t 2 ( ∂χ ) 2 − 2 m 2 χ + 2 κ L χ = − 1 χ 2 λ t 4 � 2 1 Defining ˆ χ = ( − t ) · χ , its mode functions satisfy λ � k 2 − 2(1 − m 2 � χ − κ ) ¨ χ k + ˆ χ k = 0 ˆ t 2 Austin Joyce (UPenn) March 17, 2012 6 / 20

Perturbations — χ Expanding around χ = 0, the quadratic χ lagrangian is λ t 2 ( ∂χ ) 2 − 2 m 2 χ + 2 κ L χ = − 1 χ 2 λ t 4 � 2 1 Defining ˆ χ = ( − t ) · χ , its mode functions satisfy λ � k 2 − 2(1 − m 2 � χ − κ ) ¨ χ k + ˆ χ k = 0 ˆ t 2 If | m 2 χ | , | κ | ≪ 1, then this is solved by (assuming adiabatic vacuum initial conditions) χ k = e − ikt � 1 − i � ˆ √ kt 2 k In the long-wavelength limit, χ is scale-invariant 1 λ 2 π 2 k 3 | χ k | 2 ≃ P χ = 2(2 π ) 2 Austin Joyce (UPenn) March 17, 2012 6 / 20

Coupling to gravity — cosmology Einstein Frame We consider minimal coupling to gravity � M 2 d 4 x √− g � � Pl S = 2 R + L CFT [ g µν ] . 1 Conformal symmetry is broken at the M Pl level. Austin Joyce (UPenn) March 17, 2012 7 / 20

Coupling to gravity — cosmology Einstein Frame We consider minimal coupling to gravity � M 2 d 4 x √− g � � Pl S = 2 R + L CFT [ g µν ] . 1 Conformal symmetry is broken at the M Pl level. At early times, gravity is negligible, ¯ 1 φ ∼ ( − t ) is an approximate solution Austin Joyce (UPenn) March 17, 2012 7 / 20

Coupling to gravity — cosmology Einstein Frame We consider minimal coupling to gravity � M 2 d 4 x √− g � � Pl S = 2 R + L CFT [ g µν ] . 1 Conformal symmetry is broken at the M Pl level. At early times, gravity is negligible, ¯ 1 φ ∼ ( − t ) is an approximate solution Dilatation symmetry then implies ρ CFT ∼ P CFT ∼ 1 t 4 — however, we 1 know ρ = const . to lowest order in M Pl so P CFT ≃ β ρ CFT ≃ 0 , t 4 . Quartic model, β > 0, Galilean genesis β < 0. Austin Joyce (UPenn) March 17, 2012 7 / 20

Cosmology cont. Pl ˙ We integrate M 2 H = − 1 2 ( ρ CFT + P CFT ) to find β β H ( t ) ≃ − , a ( t ) ≃ 1 − . 6( − t ) 3 M 2 12 t 2 M 2 Pl Pl The universe is therefore contracting (expanding) for β > 0 ( β < 0) Austin Joyce (UPenn) March 17, 2012 8 / 20

Cosmology cont. Pl ˙ We integrate M 2 H = − 1 2 ( ρ CFT + P CFT ) to find β β H ( t ) ≃ − , a ( t ) ≃ 1 − . 6( − t ) 3 M 2 12 t 2 M 2 Pl Pl The universe is therefore contracting (expanding) for β > 0 ( β < 0) √ β M Pl ( φ ∼ M Pl in φ 4 model) The universe is nearly static until t end = − Austin Joyce (UPenn) March 17, 2012 8 / 20

Cosmology cont. Pl ˙ We integrate M 2 H = − 1 2 ( ρ CFT + P CFT ) to find β β H ( t ) ≃ − , a ( t ) ≃ 1 − . 6( − t ) 3 M 2 12 t 2 M 2 Pl Pl The universe is therefore contracting (expanding) for β > 0 ( β < 0) √ β M Pl ( φ ∼ M Pl in φ 4 model) The universe is nearly static until t end = − The CFT equation of state w CFT ≃ P CFT = 12 β t 2 M 2 Pl . ρ CFT decreases from + ∞ to O (1). Austin Joyce (UPenn) March 17, 2012 8 / 20

Cosmology cont. Pl ˙ We integrate M 2 H = − 1 2 ( ρ CFT + P CFT ) to find β β H ( t ) ≃ − , a ( t ) ≃ 1 − . 6( − t ) 3 M 2 12 t 2 M 2 Pl Pl The universe is therefore contracting (expanding) for β > 0 ( β < 0) √ β M Pl ( φ ∼ M Pl in φ 4 model) The universe is nearly static until t end = − The CFT equation of state w CFT ≃ P CFT = 12 β t 2 M 2 Pl . ρ CFT decreases from + ∞ to O (1). | w | ≫ 1 drives the background to be flat, homogeneous and isotropic Gratton, Khoury, Steinhardt, Turok astro-ph/0301395 Pl = k a 2 + C matter + C radiation + C anisotropy C 3 H 2 M 2 + . . . + a 3 a 4 a 6 a 3(1+ w ) Austin Joyce (UPenn) March 17, 2012 8 / 20

Other examples Quartic U(1) model — Rubakov 0906.3693 L U ( 1 ) = − 1 ψ∂ψ + λ 2 ∂ ¯ 4 | ψ | 4 In polar coordinates, ψ = φ e i χ , this is a special case of the quartic model L U ( 1 ) = − 1 2( ∂φ ) + λ 4 φ 4 − 1 2 φ 2 ( ∂χ ) 2 Austin Joyce (UPenn) March 17, 2012 9 / 20

Other examples Quartic U(1) model — Rubakov 0906.3693 L U ( 1 ) = − 1 ψ∂ψ + λ 2 ∂ ¯ 4 | ψ | 4 In polar coordinates, ψ = φ e i χ , this is a special case of the quartic model L U ( 1 ) = − 1 2( ∂φ ) + λ 4 φ 4 − 1 2 φ 2 ( ∂χ ) 2 Galilean genesis — Creminelli, Nicolis & Trincherini 1007.0027 L Gal = 1 1 1 2 e 2 φ ( ∂φ ) 2 + 3 H 2 � φ ( ∂φ ) 2 + 6 H 2 ( ∂φ ) 4 This has a solution e φ = 1 H ( − t ) , which also breaks so (4 , 2) → so (4 , 1). This solution violates the NEC Perturbations can propagate superluminally Austin Joyce (UPenn) March 17, 2012 9 / 20