The Cosmological Moduli Problem and Non-thermal Histories of the Universe

Transcription

1 The Cosmological Moduli Problem and Non-thermal Histories of the Universe Kuver Sinha Mitchell Institute for Fundamental Physics Texas A M University LEPP Seminar Cornell University

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3 Introduction Affleck-Dine Baryogenesis Late-time Baryogenesis? Radiation dominated? Coincidence Problem Conclusion

4 Outline Introduction: some interesting problems in non-thermal cosmological histories Aspects of baryon asymmetry

5 Introduction The Cosmological Moduli Problem Scalar field decaying gravitationally Γ τ m3 τ M 2 p Light moduli disallowed from Big Bang Nucleosynthesis constraints m τ > 20TeV Banks, Kaplan, Nelson (1994), Coughlan et al (1983), Hall, Lykken, Weinberg (1983) Corresponds to a temperature 1 MeV

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7 Mass: O(20) TeV - O(1000) TeV The most well-studied moduli stabilization models have such moduli... In KKLT, m 3/2 W flux 30TeV, (2 ReT ) 3/2 m σ F T,T a ReT m 3/2 1000TeV, m soft F T ReT m 3/2 a ReT,

8 Late-decaying moduli with T r in the interesting range occur in KKLT Large Volume Models M theory and type IIA flux vacua Fluxless M theory compactifications on G 2 manifolds Generically have string moduli m 3/2? Acharya et al arxiv: , Covi et al arxiv: , arxiv:

9 T r : Highest temperature of the most recent radiation dominated epoch of the Universe Standard cosmology: T r large Dark matter produced thermally, reaches equilibrium before freeze-out Radiation dominated at freeze-out Baryogenesis/Leptogenesis or above Electroweak scale No entropy production after DM freeze-out/baryogenesis

10 T r = ( m ) 3/2 ( ) τ Mp 5 MeV 100 TeV Λ Entropy produced Thermal abundances ( ) 3 suppressed: Tr T EW DM produced from direct decay Protecting baryon asymmetry: interesting challenge. Interesting axion physics

11 If enough DM produced, annihilation cross-section enhanced Ω DM m χ T H T 2 σv 1 T σv But T = T r, not T f DM candidates with larger cross-sections start fitting Ω... PAMELA arxiv: [PRD] (B. Dutta, L. Leblond, KS)

12 String moduli can affect early cosmology in dramatic ways Role through decay (i) Matter - anti matter asymmetry (main focus of this talk) arxiv: [PRD] (R. Allahverdi, B. Dutta, KS), arxiv: [PRD] (B. Dutta, KS) (ii) Remarkably, string moduli offer ways to connect dark matter physics and baryogenesis Work near completion (iii) Dark matter physics arxiv: [PRD] (B. Dutta, L. Leblond, KS) Low-scale inflation: Effect on supersymmetry breaking? arxiv: [PRD] (R. Allahverdi, B. Dutta, KS)

13 Thoughts on Inflation Can you get a model of inflation to leave footprints at colliders?

14 The Kallosh-Linde problem V Σ Prevent decompactification H Supersymmetry breaking scale General question: what effect would a low-scale inflation model have on supersymmetry breaking?

15 Matter anti-matter asymmetry

16 Baryogenesis: A Classic Problem We see matter around us and not antimatter Cosmology teaches us that their abundances are equal in the early universe Intervening phase of Baryogenesis n b n b n γ Sakharov (1968): (i) Violate baryon number B symmetry, (ii) Violate C and CP, (iii) Depart from thermal equilibrium Standard Model has all three but not enough

17 Lots of ways to do this Electroweak Baryogenesis Leptogenesis Affleck-Dine Baryogenesis Need a mechanism that survives dilution from late-time modulus decay...

18 Two ways... 1 Affleck-Dine Baryogenesis 2 Late-time Barogenesis

19 bary

20 Outline 1. Examine initial conditions for Affleck-Dine baryogenesis 2. Propose late-time baryogenesis mechanism 3. Address baryon-dark matter coincidence problem

21 Affleck-Dine Baryogenesis Flat directions common in supersymmetric theories Finite energy density breaks SUSY induces a SUSY breaking mass along φ. If this Hubble-induced mass is tachyonic, the field acquires a large vev during inflation. It starts to oscillate when the Hubble constant becomes smaller than the effective mass V (φ) m 3/2. The energy of the oscillations corresponds to a condensate of non-relativistic particles. Store baryon number in a condensate. After oscillations set in, a net baryon asymmetry may be produced depending on the magnitude of baryon number-violating terms in V (φ) Affleck, Dine ( 80s), Dine, Randall, Thomas ( 97)

22 H u L : flat direction φ is given by H u = 1 ( ) 0, L = 1 ( ) φ 2 φ 2 0 MSSM flat directions are lifted by non-renormalizable terms in the superpotential W = λ nm n 3 P φ n V (φ) = (c H H 2 + m 2 soft) φ λ 2 φ 2n 2 M 2n 6 P ( (A + a H H)λφ n nm n 3 P + h.c. )

23 For H m soft, c H < 0, φ ( ) 1 chm n 3 n 2 P (n 1)λ n discrete vacua in the phase of φ, field settles into one of them. When H m soft, φ oscillates around the new minimum φ = 0 Soft A-term becomes important and the field obtains a motion in the angular direction to settle into a new phase Baryon asymmetry n B n γ ( ) ( ) n Tr,inflaton MP n GeV m 3/2

24 Initial condition problem: is c H < 0? Strategy: Induce tachyonic masses along chiral superfields during inflation. Avoid tachyons in the final spectrum. arxiv: [PRD] (B. Dutta, KS) D = 4, N = 1 supergravity

25 D = 4, N = 1 supergravity ( V = e K K ij D i WD j W 3 W 2) Kahler potential and superpotential K = K (T i, T i ) + K αβ (T i, T i )φ α φ β +... W = Ŵ Y αβγφ αβγ. Normalize φ: φ normalized = K 1/2 φ

26 Scalar masses m 2 soft = m 2 3/2 + V 0 F i F j i j ln K. V 0 is the potential along the modulus, given by V 0 = F i F j Kij 3m 2 3/2 + V D where F i = e K /2 D j K ij and m 2 3/2 = e K W 2. hep-ph/ , hep-th/

27 Inflation due to modulus σ c H = m2 H 2 1 K σσ σ σ ln K + V D V 0 K σσ σ σ ln K. c H 1 K σσ σ σ ln K. K = 2 ln V What about K? Say something based on holomorphy of W hep-th/ , hep-th/

28 W = Ŵ Y αβγφ αβγ. Ŷ αβγ (τ s, U) = e K /2 Y αβγ (U) ( ) 1 2 Kα Kβ Kγ Should only depend on local geometric data τ s and complex structure moduli U, but not the overall volume ln K = 1 3 K + ln k(τ s, U),

29 c H 1 K σσ σ σ ln K. ln K = 1 3 K + ln k(τ s, U), Energy density is dominated by a modulus that is not a local modulus of the visible sector c H = 2 3. Energy density is dominated by a local modulus c H = 2 3 K TsTs Ts Ts ln k(τ s, U).

30 Extricate local condition from global: Ts Ts ln k(τ s, U) > 0. What can you say about k(τ s, U)? Not much... Kahler potential data hard to calculate... k(τ) = k 0 + k 1 τ p +....

31 Summary: Energy density during inflation dominated by non-local geometric modulus: AD baryogenesis doesn t work Energy density during inflation dominated by local modulus: Difficult to gain control over initial conditions

32 Inflation due to hidden sector scalar field ξ Planck suppressed operators mixing the visible and inflationary sectors in the Kahler potential induce negative masses by gravity mediation along flat directions if the dimensionless coupling is chosen appropriately. The contribution to soft masses from the hidden matter sector in the final stabilized vacuum at the end of inflation should be negligible. The inflationary dynamics should be compatible with moduli stabilization.

33 K = K (T i ) + K hidden (ξ) + K (T i, ξ)φφ K (T i, ξ) = 1 (1 + γ ξξ) V2/3 W = Ŵ (T i) + W hidden (ξ). If the F-term for ξ dominates during inflation, we obtain for ξ 1 c H 1 γ.

34 c H 1 γ For γ > 1, it is possible to obtain a negative induced mass during inflation. m 2 m3/2 2 F γ ξ 2 m3/2 2 (1 3γ) This leads to tachyons for γ > 1. To avoid tachyons and couplings that give rise to the flavor problem, the final supersymmetry breaking should not be matter dominated but sourced by another (sequestered) sector.

35 W (T ) = W flux + Ae at + Be bt V = e K V (ξ) + V (T ) + O(ξ/M p ) Take ξ as a pseudo-modulus. Supersymmetry preserving vacuum at ξ = ξ susy. For ξ ξ susy, flat potential inflation rolls out to ξ susy. Try SQCD W (ξ) = hq i ξ i j qj hµ 2 ξ i i W np = N(h N f Λ (N f 3N) m detξ) 1/N K (ξ) = ξ ξ + q q + q q

36 V 1 (T + T ) 3 µ4 ln( ξ 2 ) + V (T ) H F ξ µ 2, F ξ F T H µ 2 m 3/2 For γ > 1, the field ξ induces tachyonic masses along visible sector flat directions. Inflation ends with ξ rolling out to ξ susy, restoring supersymmetry.

37 Baryogenesis after modulus decay

38 Late-time Baryogenesis Challenges: Sphaleron transitions are exponentially suppressed, thus rendering scenarios like electroweak baryogenesis and leptogenesis inapplicable. Baryon asymmetry by the direct decay of moduli, through CP and baryon number violating couplings to baryons. W λtu c d c d c /M P Since u c d c d c is odd under R-parity, the Lightest Supersymmetric Particle (LSP) will be unstable unless T = 0 at the minimum.

39 Strategy 1. Modulus decays, produces MSSM + extra matter (X) non-thermally 2. Extra matter X has baryon violating couplings to MSSM 3. Decay violates CP Sakharov conditions are satisfied.

40 Cosmological history of modulus Equation of motion of a scalar field with gravitational strength decay rate in a FRW background: τ + (3H + Γ τ ) τ + V = 0 Γ τ = c mτ 3 2π Λ 2 After inflation, the initial field vev is τ = τ in For H > m τ, the friction term dominates, field frozen at τ = τ in. Universe radiation dominated Oscillations start at temperature T Γ τ. Matter dominated universe. Energy of τ dominates until it decays

41 Calculate reheat temperature At reheat, the lifetime of the modulus (Γ 1 τ ) is equal to the expansion rate at the time of reheating t = 2 3H. Right after reheating the universe becomes radiation dominated with H = π 2 g 90 T 2 r M p Combine, get T r ( g = c 1/2 ( g ) 1/4 Γ τ M p ) 1/4 ( m τ 100 TeV ) 3/2 ( ) Mp 5 MeV. Λ

42 Important quantity: Y τ = n τ s 3 4 ( = T r m τ m τ 100 TeV ) 1/2 ( ) Mp Λ 5c 1/2 10 8

43 Estimates for baryogenesis Net baryon asymmetry η = n B n B s = ɛ n X s n X s T r = 2 n τ s (Br) X = 3 (Br) 2 m X τ T r = ɛ 3 (Br) 2 m X τ (Br) X 0.1 Typically, ɛ 1 8π λ 4 Trλ 2 T r m τ Need ɛ 10 1

44 The Model Two flavors of X = (3, 1, 4/3), X = (3, 1, 4/3) Singlet N arxiv: [PRD] (R. Allahverdi, B. Dutta, KS) W extra = λ iα Nu c i X α + λ ijα d c i d c j X α (1) + M N 2 NN + M X,(α)X α X α. M 500 GeV. Can be obtained by the Giudice-Masiero mechanism if the modulus has non-zero F -term. R parity conserving model X = (X +1, ψ 1 ) N = (Ñ 1, N +1 )

45 d c ψ 1 d c N d c ψ 1 ψ1 ψ2 ψ 2 u c d c B = +2/3

46 N ψ 1 ψ1 u c d c N ψ 1 ψ 2 ψ2 d c u c B = 1/3

47 ψ 1 di c d j c and ψ1 Ñuc k, Nũc i Total asymmetry from ψ 1 and ψ 1 decays: ɛ 1 = 1 8π ) i,j,k (λ Im k1 λ k2λ ij1 λ ij2 i,j λ ij1 λ ij1 + k λ k1 λ k1 where, for M 2 M 1 > Γ ψ1, we have F S (x) = 2 x x 1. F S (M 2 2 M 2 1 )

48 Same asymmetry from ψ 1 and ψ 1 decays since ψ 1 and ψ c 1 form a four-component fermion with hypercharge quantum number 4/3. In the limit of unbroken supersymmetry, we get exactly the same asymmetry from the decay of scalars X 1, X1 and their antiparticles X 1, X 1. In the presence of supersymmetry breaking the asymmetries from fermion and scalar decays will be similar provided that m 1,2 M 1,2 Similarly, the decay of the scalar and fermionic components of X 2, X 2 will result in an asymmetry ɛ 2, with 1 2.

49 η = M 1 M 2 8π M2 2 M2 1 [ Br 1 i,j,k i,j λ ij1 λ ij1 + k λ k1 λ k1 ( ) Im λ k1 λ k2λ ij1 λ ij2 Br 2 + i,j λ ij2 λ ij2 + k λ k2 λ k2 ]. For λ i1 λ i2 λ ij1 λ ij2 and CP violating phases of O(1) in λ and λ, we need couplings O(0.1 1) to generate the correct asymmetry.

50 Variations Single flavor of X, two flavors of singlets N W extra = λ iα N α u c i X + λ ij d c i d c j X (2) + M N αβ 2 N αn β + M X XX

51 u c Nα X u c X Nβ Nα u c X X Nα Nβ u c u c X

52 ɛ α = ( ) i,j,β Im λ iα λ iβ λ jβ λ jα 24π i λ iα λ iα [ where F S ( M 2 β M 2 α ) + F V ( M 2 β M 2 α F S (x) = 2 x x 1, F V = ( x ln ) x Choose λs, can get required BAU )]

53 Other variations: singlets replaced by iso-doublet color triplet fields Y, Y with charges 5/3. W extra = λ iα YQ i X α + λ ijα d c i d c j X α (3) + M Y Y Y + M X,(α) X α X α

54 Comments on Phenomenology Stable LSP dark matter (X, X) pair produced at LHC cascade decays into LSP neutralino via squarks, heavier neutralinos, charginos and sleptons final states contain multi jets plus multi leptons and missing energy M eff of four highest E T jets and missing energy gives mass scale of X.

55 n n oscillations G = λ2 1 λ 2 12 MX 4 M (u c d c s c ) 2 N Oscillation time t = 1/( G) < sec G < GeV 5 Using this bound, for M X M N 1 TeV, we find (λ 1 λ 12 ) < 10 4

56 Non-thermal Approach to the Dark Matter - Baryon Coincidence Problem

57 Why is Ω baryon Ω darkmatter? Common origin from modulus decay near 1 MeV gives a natural answer...

58 Y τ naturally small Y τ = n τ s 3 T r 4 m τ n b n b 10 10, s n b n b = Y τ ɛ (Br) s N Count degrees of freedom (Br) N 1% 10% ɛ is loop-suppressed Baryogenesis occurs naturally in non-thermal scenarios...

59 Strategy LSP and N both produced from modulus τ decay Physics of annihilation is particular to the dark sector. Makes sense to render annihilation irrelevant use Y τ Number densties n χ and n b related by simple branching fractions In the absence of symmetries, branching fractions similar

60 Cladogenesis: The evolutionary change and diversification resulting from the branching off of new taxa from common ancestral lineages.

61 A collective apology: Darkogenesis Hylogenesis Xogenesis Baryomorphosis Aidnogenesis

62 dρ τ dt dρ R dt dn χ dt = 3Hρ τ Γ τ ρ τ, [ = 4Hρ R + (m τ N LSP m χ )Γ τ n τ + σv 2m χ nχ 2 ( nχ eq ) ] 2, [ = 3Hn χ + N LSP Γ τ n χ σv nχ 2 ( nχ eq ) ] 2 Modulus decays when H Γ τ. Initial condition: modulus dominates energy density at the freeze-out of χ. χ is non-relativistic at the time of freeze-out (with ( ) nχ eq mχt 3/2 = g 2π e m χ/t ) and reaches equilibrium before reheating occurs.

63 Dark matter freeze-out occurs when the annihilation rate is equal to the rate of expansion T f m χ /20 1 GeV Γ χ = n eq χ (T f ) σv = H(T f ). Entropy production during reheat with T r 1 MeV dilutes the initial density of dark matter by a factor of Tr 3 /Tf Number density of non-thermally produced dark matter if dark matter overproduced by modulus it annihilates back into radiation Maximal density of dark matter is n eq χ (T r ) 3H(T r ) 2 σv 3Γ τ 2 σv

64 Y χ (T ) For small N LSP, n χ s(t ) = Y χ (T ) π 2 g M p T r σv n χ s(t ) = B τ χy τ Therefore, Y χ (T r ) = min ( Br χ Y τ, ) π 2 g M p T r σv. Moroi, Randall ( 99) Consider Br χ 10 3, Y τ for annihilation to be important, need very large annihilation cross-section

65 Ω b Ω DM 1 5 = m b m χ Y b Y χ = m b m χ ɛ Br N Br χ We know that for Y τ 10 8 ɛ 0.1, Br N 0.1 Therefore m b 50GeV for Br χ 10 3

66 Decay modes of the modulus Consider Kahler moduli in type IIB string theory The gauginos and gauge bosons couple through the gauge kinetic function. We consider a scenario in which the visible sector is constructed on D7 branes wrapping a cycle Σ, with gauge coupling given by 1/g 2 = V (Σ), where V (Σ) = ReT = τ is the volume of Σ in string units. Visible sector fermions and scalars couple to the modulus through the Kahler potential and soft terms.

67 Normalize moduli: (τ) i = j C ij (τ n ) j where the C ij are eigenvectors of the matrix K 1 2 V.

68 Decays to Gauge Bosons L τgg = (Ref ) ( 14 F µνf µν ) = 1 4M P Ref j τi Ref C ij (τ n ) j F µν F µν For simplicity, assume that τ i is predominantly aligned along a single normalized eigenstate τ n, with a coefficient C i. Γ Ti gauge = N g 128π 1 τ 2 C2 i m 3 T i M 2 P where N g = 12 is the number of gauge bosons.

69 L τi λλ = Ref Decays to Gauginos ( 1 ) 2 λdλ F i i f λr λ R + h.c. L Ti λλ 1 ( p F i T p + p F i ) Tp λ R λ R + h.c. 4M P p T p = C p (T n ) p Γ Ti g g = p Γ T i g g = p N g 128π C2 p N g 128π C2 p p F i 2 mt 3 i MP 2 p F i 2 mt 3 i MP 2

70 Decay to Visible Sector Fermions and Scalars K K (τ) φφ (4) Γ K 1 τ K 2 m 2 soft m 2 τ m2 soft m 2 τ (5) Suppressed

71 Typically unsuppressed two-body decays of τ to Higgs K K (τ) φφ + Z (τ) H u H d Γ 1 8π C2 i 1 K ( i Z ) 2 m3 T i M 2 P

72 Decay to Gravitino L = 1 4 ɛklmn ( G,Ti k T G,T i k T ) ψ l σ m ψ n 1 2 eg/2 ( G,Ti T + G,T i T i ) [ψm σ mn ψ n + ψ m σ mn ψ n ], where G = K + log W 2 is the Kahler function. The decay width to helicity ±1/2 components is given by Γ Ti gravitino 1 288π ( ) m G Ti 2 K 1 2 T mt 3 i T i Ti m3/2 2 MP 2

73 Branching Conditions Acceptable reheat and suppression of decay to R parity odd particles p( p) 1 τ C i 1 C p( p) ( p( p) F T i ) m Ti m Ti m 3/2 G Ti K 1/2 T i Ti (6)

74 Conditions on branching to LSP can be made even milder for Y τ 10 9 Can happen if the overall constant c in the decay width Γ τ = c 2π of the modulus takes on extremely small values m 3 τ M 2 p Physically: a compactification with large volume V, and a number of local geometric moduli that are decoupled in the Kahler metric by powers of 1/V. In that case, going to the eigenbasis φ i of K 1 2 V, one obtains τ i = O(V p i ) φ i + j i O(V p j ) φ j, (7) with p j < p i.

75 Conclusion Modulus decay near the MeV scale leads to interesting physics Baryogenesis is a challenge, but very natural methods exist Coincidence problem becomes simpler to address. Reason: Y τ is small, and remarkably close to the observed abundance of baryons and dark matter. Y τ is small because moduli interact gravitationally (that s also why they decay late). Don t throw away a small number you got for free!