Inflation with Superstrings? Grant Mathews Univ. Notre Dame - PowerPoint PPT Presentation

Inflation with Superstrings? Grant Mathews – Univ. Notre Dame arXiv:1701.00577 Gravitation and Cosmology 2018 Yukawa Institute for Theoretical Physics Feb. 28, 2018 1

It is natural that the universe is born out of a landscape of superstrings 2

Is there new trans-Plankian physics imprinted on the Cosmic Microwave Background? 3

New physics always shows up as small deviations 4

WMAP 9yr 5

Some possible explanations for dip at l = 20 • Cosmic Variance: Planck XX arXiv:1502.02114 • Modified inflation effective potential – Harza, et al. arXiv:1405.2012, – Kitazawa and Sagnotti 1411.6396v2, – Yang and Ma arXiv:1501.00282 • …. • …. • Planck-mass particles coupled to inflation – GJM, M. R. Gangopadhyay, K. Ichiki, and T. Kajino, Phys. Rev. D92, 123519 (2015). arXiv: 1504.06913 6

How does this work? Classical and Quantum Gravity 31 053001 (2014). [35] D. J. H. Chung, E. W. Kolb, A. Riotto, and I. I. Tkachev, Phys. Rev. D 62 , 043508 (2000). [36] G. J. Mathews, D. Chung, K. Ichiki, T. Kajino, and M. Orito, Phys. Rev. D70 , 083505 (2004). Mathews et al. PRD (2015) • The total Lagrangian density is given as : L tot = 1 2 @ µ �@ µ � − V ( � ) 1 = 2 ∂ µ φ∂ µ φ − V ( φ ) L tot i ¯ ψγ µ ψ − m ¯ ψψ + N λφ ¯ + i ¯ @ µ − m ¯ + N �� ¯ + ψψ / • Then the fermion has the e ff ective mass : M ( φ ) = m − N λφ • This vanishes for a critical value of the inflaton field, φ ∗ = m / N λ 7

When φ = φ *, resonant particle production occurs Z 1 p | β k | 2 = N λ 3 / 2 n ⇤ = 2 2 π 3 | ˙ φ ⇤ | 3 / 2 dk p k 2 π 2 0 − π k 2 ✓ ◆ | β k | 2 = exp . ⇤ N λ | ˙ a 2 φ ⇤ | h ¯ i = n ⇤ Θ ( t � t ⇤ ) exp [ � 3 H ⇤ ( t � t ⇤ )] 8

How does this affect inflation? • Causes a jump in the evolution of the scalar field ˙ ˙ � ( t > t ⇤ ) = � ⇤ exp [ � 3 H ( t � t ⇤ )] � V 0 ( � ) ⇤ ⇥ ⇤ 1 � exp [ � 3 H ( t � t ⇤ )] 3 H ⇤ + N � n ⇤ ( t � t ⇤ ) exp [ � 3 H ⇤ ( t � t ⇤ )] 9

Alters the primordial power spectrum ∞ δ H ( a ) = H 2 ⎛ ⎞ dk 2 2 k 2 ( k ) ∫ C l = π j l ⎟ δ H ⎜ k H 0 ⎝ ⎠ 5 π ˙ φ 0 • In this case using the above equation for the fluctuation as it exists the horizon the perturbation in the primordial power spectrum is : [ δ H ( a )] N λ =0 δ H = (6) φ ∗ | H ∗ )( a ∗ / a ) 3 ln ( a / a ∗ ) 1 + Θ ( a − a ∗ )( N λ n ∗ / | ˙ Causes Dip 10

2 new parameters in fit to CMB ⇤ k ⇤ = ` ⇤ g k ⇤ /k = a ⇤ /a , multipole r lss A = | ˙ Amplitude φ ∗ | − 1 N λ n ∗ H − 1 ∗ [ � H ( a )] N λ =0 � H ( k ) = 1 + Θ ( k � k ⇤ ) A ( k ⇤ /k ) 3 ln ( k/k ⇤ ) 11

MCMC fit to CMB A Power l = 20 Spectrum k * 3000 PLANCK best fit 2500 power law 2000 1500 l = 2 1000 500 Mathews et al. PRD (2015) 0 12 5 10 15 20 25 30 35 40 45 50 1/k

Amplitude, A, relates to the inflaton coupling λ and number N of degenerate Fermions ⇤ N λ 5 / 2 1 A = | ˙ φ ∗ | − 1 N λ n ∗ H − 1 √ ≈ ∗ p 5 π 7 / 2 2 δ H ( k ⇤ ) | λ =0 A ∼ 1 . 3 N λ 5 / 2 COBE Normalization λ ≈ (1 . 0 ± 0 . 5) N 2 / 5

k ∗ relates to the fermion mass m for a given inflation model: The fermi be deduced from e m ≈ φ ⇤ / λ 3 / 2 . m = N λφ ⇤ . => For this pur ✓ φ ◆ α V ( φ ) = Λ φ m 4 p 2 α N ∗ m pl φ ∗ = pl m pl √ p m = N λ 2 α N − ln ( k ∗ /k H ) m ⇠ (8 � 11) m pl α = 2 / 3 => � 3 / 2

Suppose this particle is a Superstring: How could you know? • There should be similar resonant couplings corresponding to different excitations of the same string. • Could this be the l =2 suppression or more? Resonant Superstring Excitations during Inflation G. J. Mathews 1 , 2 , M. R. Gangopadhyay 1 , K. Ichiki 3 , T. Kajino 2 , 4 , 5 15 1 Center for Astrophysics, Department of Physics, arXiv:1701.00577

TT EE 2500 1 2000 0.5 π π 1500 TT /2 EE /2 l(l+1)C l l(l+1)C l 1000 0 500 0 -0.5 0 10 20 30 40 50 60 70 0 10 20 30 40 50 60 70 80 90 l l ` ≈ 2 , A = 1 . 7 ± 1 . 5 , k ⇤ ( n + 1) = 0 . 0004 ± 0 . 0003 h Mpc � 1 ` ≈ 20 , A = 1 . 7 ± 1 . 5 , k ⇤ ( n ) = 0 . 0015 ± 0 . 0005 h Mpc � 1 ` ≈ 60 , A = 1 . 7 ± 1 . 5 , k ∗ ( n − 1) = 0 . 005 ± 0 . 004 h Mpc − 1 MCMC fit to multiple dips in the CMB power spectrum Gangopadhyay, Mathews, Ichiki, Kajino arXiv:1701.00577 16

Do these states look like Excitation modes of a superstring? 1. Momentum states in the compact dimension 2. Oscillations 3. Winding around the compact dimensions 17

A simple example: D=26 Bosonic closed string with 1 dimension compacted in a circle of radius R M 2 = n 2 R 2 + w 2 R 2 + 2 α 0 ( N + ˜ N + 2) α 0 2 Oscillations Winding Momentum X ( α µ N = − n α nµ + α − n α n ) Potential States ˜ X α µ N = ( − ˜ − n ˜ α nµ + ˜ α − n ˜ α n ) Energy N − ˜ N + nw = 0

Special Cases: only oscillations ✓ N osc + ξ ◆ M 2 ≈ , Case I . α 0 ✓ n ◆ 2 ξ ≡ α 0 . R Only momentum ✓ n 2 + ξ ◆ M 2 ≈ , Case II . R 2 or winding states ξ = 2 R 2 α 0 ( N + ˜ N − 2) 19

Can fix the ratio of mass states ≈ M 2 ( ` ⇤ = 2) M 2 ( ` ⇤ = 20) ≡ R +1 ≈ N − ln ( k ⇤ ( n + 1) /k H ) N − ln ( k ⇤ ( n ) /k H ) wer spectrum. That is, we infer from es: M 2 ( ` ∗ =2) M 2 ( ` =20) ≡ R +1 = 1 . 024 ± 0 . 050 . M 2 ( ` ⇤ = 20) N − ln ( k ⇤ ( n ) /k H ) M 2 ( ` ⇤ = 60) ≡ R � 1 ≈ N − ln ( k ⇤ ( n − 1) /k H ) M ( ain M 2 ( ` ∗ =20) M 2 ( ` =60) ≡ R − 1 = 1 . 024 ± 0 . 030 . of how this might relate to string par

Case of simple oscillations ≈ M 2 ( ` ⇤ = 2) M 2 ( ` ⇤ = 20) ≡ R +1 ≈ N − ln ( k ⇤ ( n + 1) /k H ) N − ln ( k ⇤ ( n ) /k H ) R +1 = ( N osc + 1) N osc 1 N osc = R +1 − 1 − => ce N osc = 42 + ∞ − 28 ly, the uncertaint 21

Physical properties of the Superstring Coupling Constant is � ⇡ (1 . 0 ± 0 . 5) . Small because N is large N 2 / 5 m ⇠ (8 � 11) m pl Very large � 3 / 2 − Very uncertain ce N osc = 42 + ∞ − 28 ly, the uncertaint 22

Conclusions • Marginal evidence of sequential dips in the CMB power spectrum • These could be caused by resonant coupling to successive excitations of a superstring during inflation. • The regular spacing and constant amplitude of the dips is consistent with mass eigenstates corresponding to successive oscillations or momentum states of a single closed superstring. • Uncertainties are too large to make definitive conclusion 23