The Worldvolume Action of Kink Solitons Kurt Hinterbichler, Justin - PowerPoint PPT Presentation

The Worldvolume Action of Kink Solitons Kurt Hinterbichler, Justin Khoury, Burt A. Ovrut, James Stokes The University of Pennsylvania March 18, 2012

Outline Motivation Scalar Kinks in d = 5 Flat Spacetime Scalar Kinks in d = 5 AdS Spacetime Scalar Kinks in d = 5 Heterotic M-theory Spacetime 2 of 17

The Universe on a Domain Wall? Motivation from String Theory � E 8 × E 8 Heterotic string E 2 ≪ α ′ ⇒ d = 11 SUGRA on R 1 , 9 × S 1 = ℓ s g s / Z 2 with E 8 SYM on the orbifold fixed planes � Compactify 6 of the 9 non-compact spacelike directions on a Calabi-Yau 3-fold = ⇒ spacetime is a warped product of R 1 , 3 × CY 3 and S 1 / Z 2 � There exists a regime in which the universe appears 5-dimensional (Heterotic M-theory) � 5 + 1-dimensional “5-branes” in the 11D bulk = ⇒ 3 + 1-dimensional “3-branes” in 5D effective theory � The physics of the brane-bending mode π of the 3-brane has implications for 4D cosmology (e.g. Galileons) 3 of 17

Problem Compute the effective action for the brane bending mode of a probe domain-wall kink soliton in maximally symmetric spacetime as a precursor for the Heterotic M-theory 5-brane. UV Toy Model � 1 � d 5 x √− g ( R − 2Λ) − 1 � 2 g mn ∂ m Φ ∂ n Φ − V (Φ) S 5 D = , (1) 2 κ 2 M 5 5 � For Λ = 0 (flat spacetime) choose 1 V (Φ) = λ (Φ 2 − η 2 ) 2 (probe brane approximation) � “Kink” solution describes a static Φ 1 domain wall of width: ℓ = √ 2 λ η √ � 1 Φ = ηφ (0) , φ (0) = tanh ( η 2 λ y ) � l l y (2) 4 of 17

Procedure � Consider classical fluctuations about the classical background configuration which depend on both the internal space y and the tangential 3-brane coordinates σ µ � The fluctuation of the position of the zero-locus of Φ is described by a field π = π ( σ µ ). What is its effective action? � Expect on general grounds that operators in the effective action organize into geometric invariants √ √ √ √ − hK 2 − h , − hK , − hR , , . . . (3) � �� � � �� � � �� � dim = 0 dim = 1 dim = 2 � Define the effective Lagrangian by integrating out the internal space � Compute Lagrangian by expanding in powers of the wall thickness ℓ 5 of 17

Gauss Codazzi formalism (Carter and Gregory hep-th/9410095) � The dynamics of the wall is fully described by the scalar wave equation ( g mn = (1 , η µν ) flat) g mn ∇ m ∇ n Φ − 4 λ Φ(Φ 2 − η 2 ) = 0 (4) � This is equivalent to n Φ + K L n Φ + g mn D m D n Φ − 4 λ Φ(Φ 2 − η 2 ) = 0 L 2 (5) where h mn = g mn − n m n n , K mn = ∇ m n n (6) are subject to the constraints L n K mn = K mp K p L n h mn = 2 K mn , (7) n � Work in gaussian normal coordinates adapted to the worldvolume ( σ µ , y ) so L n = ∂/∂ y 6 of 17

Rescalings � Let L = characteristic length of wall fluctuation, ℓ = wall width, ǫ = ℓ/ L � Rescale worldvolume quantities (e.g., K , σ µ ) by L and quantities transverse to the brane (i.e, y ) by ℓ � EOM in dimensionless variables h ′ mn = 2 ǫκ mn (8) κ ′ mn = ǫκ mp κ p (9) n 0 = φ ′′ + ǫκφ ′ − 2 φ ( φ 2 − 1) + ǫ 2 ˆ D m ˆ D m φ (10) � Expand φ, h mn , κ mn in dimensionless small parameter ǫ , φ = φ (0) + ǫφ (1) + ǫ 2 2 φ (2) + O ( ǫ 3 ) , (11) h mn = h (0) mn + ǫ h (1) mn + ǫ 2 2 h (2) mn + O ( ǫ 3 ) , (12) 2 κ (2) mn + ǫ 2 κ mn = 1 ǫ κ (0) mn + κ (1) mn + ǫ 6 κ (3) mn + O ( ǫ 3 ) (13) 7 of 17

ǫ -expansion � At zeroth the Gauss-Codazzi equations are trivially solved by h (0) mn = ˆ h (0) mn ( σ ) , κ (0) mn = 0 , φ (0) = tanh( u ) (14) � At first order in ǫ the induced metric and extrinsic curvature are simply h (1) mn = 2 u ˆ κ (1) mn , κ (1) mn = ˆ κ (1) mn ( σ ) (15) � The first-order scalar equation φ ′′ (1) − 2(3 φ 2 κ (1) ( σ ) φ ′ (0) − 1) φ (1) + ˆ (0) = 0 can be solved with the ansatz φ (1) ( σ, u ) = ˆ κ (1) ( σ ) f ( u ) u →± 0 u →±∞ � Taking boundary conditions to be Φ − → 0, Φ − → ± η gives � y � � y � + O ( ℓ 2 ) Φ = η tanh + ηℓ K ( σ ) f (16) ℓ ℓ where f ( u ) = ± 1 2 tanh( u ) − 1 � 2 3 ± u cosh 2 ( u ) − 1 1 6 e ∓ 2 u . � 2 + (17) 2 8 of 17

Effective action in R 1 , 4 √ � d 4 σ S 4 D = − h L 4 D ( σ ) (18) M 4 where √− g � L 4 D ( σ ) ≡ dy J L 5 D ( σ, y ) , J = √ (19) − h In Gaussian normal coordinates ( σ µ , y ), J = 1 + y K + 1 2 y 2 � R (4) − R (5) + R (5) mn n m n n � + · · · (20) where K , R (4) , R (5) , . . . are evaluated on the brane world volume ( y = 0) and R (5) mn = 0 in flat spacetime. It follows that to order ℓ 2 L 4 D = − 4 η 2 1 + C I R (4) + C II K 2 + · · · � � (21) 3 ℓ where 2 = π 2 − 6 ℓ 2 ℓ 2 C I = I II C II = − I III 2 = − 1 3 ℓ 2 . ℓ 2 , (22) 24 I I I I � + ∞ � + ∞ � + ∞ (0) = π 2 − 6 (0) = 4 (0) = 8 du φ ′ 2 du u 2 φ ′ 2 du f φ ′ I I = 3 , I II = , I III = 9 . 9 −∞ −∞ −∞ (23) 9 of 17

Gauss Codazzi formalism (Hinterbichler, Khoury, Ovrut, JS, To Appear) � The dynamics of the wall is fully described by the scalar wave equation ( g mn = ( e 2 y / R , η µν ) AdS) √ g mn ∇ m ∇ n Φ − 4 λ Φ(Φ 2 − η 2 ) − 4 2 λ � η 2 − Φ 2 � = 0 . (24) R � This is equivalent to √ n Φ + K L n Φ + D m D m Φ − 4 λ Φ(Φ 2 − η 2 ) − 4 2 λ η 2 − Φ 2 � � L 2 = 0 , (25) R where h mn = g mn − n m n n , K mn = ∇ m n n (26) are subject to the constraints n + 1 n − R (5) L n K mn = K mp K p rspq n s n q h r m h p n = K mp K p L n h mn = 2 K mn , R 2 h mn � Work in gaussian normal coordinates adapted to the worldvolume ( σ µ , y ) so L n = ∂/∂ y 10 of 17

ǫ -expansion � At zeroth the Gauss-Codazzi equations are solved by ( δ = ℓ/ R ) h (0) mn = e 2 δ u ˆ h (0) mn ( σ ) , κ (0) mn = δ h (0) mn , φ (0) = tanh( u ) (27) � At first order in ǫ the induced metric and extrinsic curvature are simply h (1) mn = 1 � � κ (1) mn = e 2 δ u ˆ κ (1) mn − ˆ κ (1) mn ( σ ) , κ (1) mn ( σ ) (28) δ � The first-order scalar equation � � φ ′′ (1) + κ (0) φ ′ (1) + κ (1) φ ′ 3 φ 2 (0) − 2 (0) − 1 φ (1) + 8 δφ (0) φ (1) = 0 can be solved with the ansatz φ (1) ( σ, u ) = ˆ κ (1) ( σ ) F ( u ) u →± 0 u →±∞ � Taking boundary conditions to be Φ − → 0, Φ − → ± η gives � y � � y � K ( σ ) = K ( σ ) − 4 + ηℓ ˆ + O ( ℓ 2 ) , ˆ Φ = η tanh K ( σ ) F ℓ ℓ R (29) where F ( u ) can be determined numerically 11 of 17

Effective action in AdS 5 In Gaussian normal coordinates ( σ µ , y ), � � J = 1 + y K + 1 R (4) + 16 2 y 2 + · · · (30) R 2 where K and R (4) are evaluated on the brane world volume ( y = 0). It follows that to order ℓ 2 L 4 D = − 4 η 2 1 + C 0 K + C I R (4) + C II K 2 � 1 − 6 δ 2 ( I III − I II ) � �� (31) 3 ℓ where � � � 3 ℓ 2 � 3 ℓ 2 I ǫδ + I III I II I III C 0 = 3 � ℓδ, C I = 8 , C II = − 8 . � 1 − 6 δ 2 ( I III − I II ) � 1 − 6 δ 2 ( I III − I II ) � 1 − 6 δ 2 ( I III − I II ) (32) � + ∞ � + ∞ � + ∞ (0) = π 2 − 6 (0) = 4 φ (0) − 1 du φ ′ 2 3 φ 3 du u 2 φ ′ 2 � � I I = 3 , I ǫδ = du u , I II = , (0) 9 −∞ −∞ −∞ � + ∞ du F φ ′ I III = (0) −∞ 12 of 17

Evaluation of I ǫδ � + ∞ � � φ (0) − 1 � Note that I ǫδ = 3 φ 3 −∞ du u is naively quadratically (0) divergent since the argument of the integrand is even and unbounded � The Gaussian normal coordinate patch is only defined up to the point where the geodesics converge, which is determined by the minimum of L or R � After introducing the appropriate cut-off, the integral is finite and can be estimated to be of order  1 /ǫ 2 if δ < ǫ  I ǫδ ∝ (33)  1 /δ 2 if ǫ < δ 13 of 17

C 0 � l 1 � 6 ∆ 2 � I III � I II � 10 1.2 8 1.0 6 0.8 0.6 4 0.4 2 0.2 ∆ 0.6 ∆ 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.1 0.2 0.3 0.4 0.5 C I � l 2 C II � l 2 0.6 ∆ 0.15 0.1 0.2 0.3 0.4 0.5 � 0.02 0.10 � 0.04 � 0.06 0.05 � 0.08 � 0.10 0.6 ∆ 0.0 0.1 0.2 0.3 0.4 0.5 � 0.12 � � , C 0 / l , C I / l 2 and C II / l 2 1 − 6 δ 2 ( I III − I II ) Figure: Numerical calculation of as functions of δ . Of the four coefficients, only C 0 / l depends on I ǫδ and, hence, on the value of the cut-off ratio R L = ǫ δ . Therefore, to evaluate C 0 / l we must specify a value for ǫ . In Figure 2(B), we choose ǫ = 0 . 2. Note that C 0 / l is defined piecewise and changes behavior at δ ∼ ǫ 14 of 17 .

Heterotic M-theory (Lukas, Ovrut, Stelle, Waldram, hep-th/9806051) In the absence of any 3-branes, the bosonic part of the action for d = 5, N = 1 supersymmetric heterotic M-theory � d 4 xdy √− g S = − 1 � 1 2 R + 1 4 g mn ∂ m φ∂ n φ + 1 3 α 2 e − 2 φ � { 2 κ 2 5 M 5 � � d 4 x √− g 2 α e − φ − d 4 x √− g 2 α e − φ } , + (34) M 4 M 4 ds 2 = e 2 A ( y ) dx µ dx ν η µν + e 2 B ( y ) dy 2 , φ = φ ( y ) (35) For SUSY preserving backgrounds A and B satisfy the BPS equations e − B A ′ = − α e − B φ ′ = − 2 α e − φ 3 e − φ , (36) 15 of 17