Constant-roll inflation

Martin, HM, Suyama, PRD 87, 023514 (2013) , arXiv:1211.0083HM, Starobinsky, Yokoyama, JCAP 1509, 018 (2015), arXiv:1411.5021HM, Starobinsky, EPL 117, 39001 (2017), arXiv:1702.05847HM, Starobinsky, arXiv:1704.08188HM, Hu, arXiv:1706.06784

2017.06.22 PASCOS at IFT, Madrid

Hayato MotohashiIFIC / CSIC-University of Valencia

Canonical single field Inflation

3𝐻# =�̇�#

2 + 𝑉

−2�̇� = �̇�#

�̈� + 3𝐻�̇� +𝜕𝑉𝜕𝜙 = 0

Slow-roll approximation �̈� ≃ 0Ultra slow-roll : constant potential �̈� = −3𝐻�̇�Constant-roll generalization �̈� = 𝛽𝐻�̇� (𝛽: constant)

𝛽0−3

USR SR

Ultra slow-roll inflation𝑉(𝜙)

𝜙Constant potential 𝑉 ≃ 𝑉2

�̈� = −3𝐻�̇� ⇒ �̇� ∝ 𝑎78

3𝐻# =�̇�#

2 + 𝑉 ≃ 𝑉2Slow-roll parameter

𝜖: ≡ −�̇�𝐻# =

�̇�#

2𝐻# ∝ 𝑎7<

Kinney, gr-qc/0503017

Ultra slow-roll inflation Ultra slow roll �̈� = −3𝐻�̇� ⇒ 𝜖: ∝ 𝑎7<

Curvature perturbation on superhorizon scales

𝜁> = 𝐴> + 𝐵> A𝑑𝑡𝑎8𝜖:

�

�

Constant mode

Slow-roll 𝜖: ≃ const ≪ 1 : decaying mode Ultra slow-roll 𝜖: ∝ 𝑎7< : growing mode

While 𝜖: itself is small, GHIJKGL

= −6 is not small.⇒ Violation of slow roll.

Generalization to constant roll Constant roll condition �̈� = 𝛽𝐻�̇� ⇒ 𝜖: ∝ 𝑎#N

Curvature perturbation on superhorizon scales

𝜁> = 𝐴> + 𝐵> A𝑑𝑡𝑎8𝜖:

�

�

Constant mode

Martin, HM, Suyama, 1211.0083

𝑑ln𝜖:𝑑𝑁 = 2𝛽 − 3><

Growing mode Decaying mode𝛽

−8#

0−3USR SR

PBH & slow-roll violationClosely related notion is enhancement of ΔU#(𝑘) for PBH (subhorizon evolution). Leading order slow roll

ΔU# 𝑘 ≈𝐻#

8𝜋#𝜖:No go for slow roll: GHIJK

GL< −0.38 is necessary for PBH

as DM from single field inflation. Since𝜖[𝜖:

= 1 +1

6 − 2𝜖:𝑑ln𝜖:𝑑𝑁

#

it is not appropriate to use

𝜖: ≈ 𝜖[, ΔU# 𝑘 ≈𝑉

24𝜋#𝜖[

Blue tilt

⇔GHIJKGL

< 0

𝜖: ≡ −�̇�/𝐻#𝜖[ ≡ 𝑉`/𝑉 #/2

HM, Hu, 1706.06784

Case study: Inflection model HM, Hu, 1706.06784

Garcia-Bellido, Morales, 1702.03901Ezquiaga, Garcia-Bellido, Morales, 1705.04861

Bezrukov, Pauly, Rubio, 1706.05007

Case study: Inflection model

Nonnegligible

HM, Hu, 1706.06784

Case study: Inflection model

Exact

ΔU# 𝑘 ≈𝑉

24𝜋#𝜖[

HM, Hu, 1706.06784

Constant-roll potentialWith constant-roll condition

�̈� = 𝛽𝐻�̇�

the evolution equation −2�̇� = �̇�# or �̇� = −2 G:Ga

implies𝑑#𝐻𝑑𝜙# = −

𝛽2 𝐻

Thus, 𝐻 is given by linear combination of 𝑒± 7N/#� a.The potential is then given exactly by

𝑉 = 3𝐻# − 2𝑑𝐻𝑑𝜙

#

which is linear combination of 𝑒± 7#N� a.For each 𝑉(𝜙), we can get 𝜙 𝑡 , 𝐻 𝑡 , 𝑎 𝑡 analytically.

HM, Starobinsky, Yokoyama, 1411.5021

Constant-roll potentialThe potential includes

a) 𝑉 ∝ 𝑒 7#N� a with 𝛽 < 0: Power-law inflationbut 𝑟 = 8 1 − 𝑛f ≈ 0.28 is too large.

b) 𝑉 ∝ cosh( −2𝛽� 𝜙) + const with 𝛽 < 0: Known

c) 𝑉 ∝ cos( 2𝛽� 𝜙) + const with 𝛽 > 0: New potential

Barrow, 1994

Abbott, Wise, 1984Lucchin, Matarese, 1985

Growing mode Decaying mode𝛽

−8#

cosh pot. cos pot.0−3USR SR

cosh potential (𝛽 < 0)𝑉(𝜙)𝑀#𝑀lm

# = 3 1 −3 + 𝛽6 1 − cosh −2𝛽� 𝜙

𝑀lm

Assume inflation ends before 𝜙 = 0.

𝛽 = −3

𝛽 = −3.5

𝛽 = −2.5

𝑡 → ∞Attractor

Non-attractor

cosh potential (𝛽 < 0)𝑉(𝜙)𝑀#𝑀lm

# = 3 1 −3 + 𝛽6 1 − cosh −2𝛽� 𝜙

𝑀lm

Assume inflation ends before 𝜙 = 0.Analytic solution

𝜙 𝑡𝑀lm

=2−𝛽

�ln coth

−𝛽2 𝑀𝑡 → 0

𝐻 𝑡𝑀 = coth −𝛽𝑀𝑡 → 1

𝑎 𝑡 ∝ sinh7s/N −𝛽𝑀𝑡 → 𝑒tu

Slow-roll parameters (𝜖s ≡ −�̇�/𝐻#,𝜖vws ≡ 𝜖v̇/𝐻𝜖v)2𝜖s = 𝜖#vws = −𝛽/ cosh# −𝛽𝑀𝑡 → 0

𝜖#v = 2𝛽 tanh# −𝛽𝑀𝑡 → 2𝛽

(𝑡 → ∞)

New cos potential (𝛽 > 0)𝑉(𝜙)𝑀#𝑀lm

# = 3 1 −3 + 𝛽6 1 − cos 2𝛽� 𝜙

𝑀lm

Assume inflation ends before 𝜙 = 𝜙y.

𝑡 → −∞

𝛽 = 0.2

𝛽 = 0.5

𝜙y 𝜙y

Natural inflation + negative CC

New cos potential (𝛽 > 0)𝑉(𝜙)𝑀#𝑀lm

# = 3 1 −3 + 𝛽6 1 − cos 2𝛽� 𝜙

𝑀lm

Assume inflation ends before 𝜙 = 𝜙y. Analytic solution

𝜙 𝑡𝑀lm

= 22𝛽

�arctan(𝑒Ntu) → 0

𝐻 𝑡𝑀 = − tanh 𝛽𝑀𝑡 → 1

𝑎 𝑡 ∝ cosh7s/N 𝛽𝑀𝑡 → 𝑒tu

Slow-roll parameters2𝜖s = 𝜖#vws = 2𝛽/ sinh# 𝛽𝑀𝑡 → 0

𝜖#v = 2𝛽 tanh# 𝛽𝑀𝑡 → 2𝛽

(𝑡 → −∞)

Same asymptotic values

Curvature perturbationMukhanov-Sasaki equation

𝑣>`` + 𝑘# −𝑧``

𝑧 𝑣> = 0

where 𝑣> = 2� 𝑧𝜁> with 𝑧 = 𝑎 𝜖s� . Without approximation,

𝑧``

𝑧 = 𝑎#𝐻# 2 − 𝜖s +32 𝜖# +

14 𝜖#

# −12 𝜖s𝜖# +

12 𝜖#𝜖8

For both cosh potential and cos potential,𝑧``

𝑧 →𝛽 + 2 𝛽 + 1

𝜏# =𝜈# − 1/4

𝜏#where

𝜈 ≡ 𝛽 + 2 𝛽 + 1 + 1/4� = 𝛽 + 3/2

𝜖s ≡ −�̇�/𝐻#

𝜖vws ≡ 𝜖v̇/𝐻𝜖v

Curvature perturbationSince spectral index is given by

𝑛f − 1 = 3 − 2𝜈 = 3 − 2𝛽 + 3we obtain

𝛽 =𝑛f − 72 or

1 − 𝑛f2

For 𝑛f = 0.96, 𝛽 = −3.02or0.02

Growing mode Decaying mode𝛽

−8#

cosh pot. cos pot.0−3USR SRNon-

attractor

Curvature perturbation 𝛽 = −3.02

ln 𝑎

Growing

Subhorizon Superhorizon

Curvature perturbation 𝛽 = 0.02

ln 𝑎

Conserved

Subhorizon Superhorizon

Sufficient number of efolds

𝑡 → −∞

𝜙y(𝛽)𝜙<2(𝛽)

𝑁 = 60

∃𝜙�I�

Observational constraint Ade et al., 1510.09217

Observational constraintHM, Starobinsky, 1702.05847

𝑓(𝑅) constant-roll inflation

𝑆 = A𝑑�𝑥 −𝑔� 𝑓 𝑅2

�

�EOM

3𝐹𝐻# =12 𝑅𝐹 − 𝑓 − 3𝐻�̇�

2𝐹�̇� = −�̈� + 𝐻�̇�Constant-roll condition

�̈� = 𝛽𝐻�̇�allows one to find analytic solution for 𝑓(𝑅) (parametric form) and evolution or equivalently Einstein frame potential 𝑉(𝜙) and 𝜙(𝑡�).

HM, Starobinsky, 1704.08188

𝐹 ≡ 𝑑𝑓/𝑑𝑅

𝑓(𝑅)HM, Starobinsky, 1704.08188

Einstein frame potentialHM, Starobinsky, 1704.08188

Observational constraintHM, Starobinsky, 1704.08188

Summary• Even if 𝜖: itself is small, slow roll approximation

breaks down by GHIJKGL

. If GHIJKGL

< −3, 𝜁> grows on superhorizon scales.

• Constant-roll condition �̈� = 𝛽𝐻�̇� or �̈� = 𝛽𝐻�̇� allows one to reconstruct potential and solve inflationary dynamics fully analytically.

• Two-parametric phenomenological constant-roll potentials have parameter regions that satisfy observational constraint.