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Colliding Black Holes in AdS Hans Bantilan Queen Mary University of - PowerPoint PPT Presentation

Colliding Black Holes in AdS Hans Bantilan Queen Mary University of London July 1, 2016 Outline Motivation Setup Simulations Summary Motivation Heavy ion collisions Why collisions? to probe the quark and gluon constituents


  1. Colliding Black Holes in AdS Hans Bantilan Queen Mary University of London July 1, 2016

  2. Outline • Motivation • Setup • Simulations • Summary

  3. Motivation Heavy ion collisions ◦ Why collisions? to probe the quark and gluon constituents of nuclei ◦ Why heavy ions? to get as many p + and n 0 as possible to hit each other The STAR, PHENIX experiments at RHIC, the ALICE, ATLAS, CMS experiments at LHC ◦ Strip gold ( 197 79 Au) or lead ( 208 82 Pb) nuclei of electrons ◦ Accelerate to speeds close to c ◦ Arrange for a collision ◦ Collision energies of 200[GeV] per nucleon at RHIC, 2.76[TeV] per nucleon at LHC

  4. Motivation • A non-perturbative problem in QCD • Lattice QCD has no access to real-time dynamics • Experimental data are well described by relativistic viscous hydrodynamic simulations • But, several competing models for the pre-equilibrium stage that yield different initial energy density and flow velocity profiles for matching onto the hydrodynamic stage • Would be desirable to have a single model to describe both the pre-equilibrium stage and the hydrodynamic stage

  5. Motivation Pre-equilibrium stage ◦ Duration: 0.2-0.4 fm/c ◦ A model: classical Yang-Mills dynamics of gluons ◦ Resulting energy density and flow velocity profiles are used to match onto a hydrodynamic form of the stress tensor in subsequent hydrodynamic stage Hydrodynamic stage ◦ Duration: 5-10 fm/c ( ↑ for higher collision energies) ◦ A model: relativistic viscous hydrodynamics ◦ Resulting hydrodynamic output is used to match onto particle distributions in subsequent hadronic stage Hadronic stage ◦ Duration: remaining evolution time ◦ A model: microscopic kinetic description

  6. Motivation Figure: BH-BH collision in a Poincar´ e patch of AdS 5 , with black hole masses M 1 , M 2 , boosts γ 1 , γ 2 , and impact parameter b . Adapted from hep-th/0805.1551

  7. Motivation AdS/CFT correspondence between an asymptotically AdS spacetime in d + 1 dimensions and a CFT in d dimensions Proposed use to find a gravity description of non-perturbative problems in QCD Major obstacle is the current lack of a gravity dual for QCD Possible approach: try to capture some features of QCD with a CFT toy model for which there is a known gravity dual N = 4 SYM 4 at strong coupling ← → AdS 5 classical gravity

  8. Motivation AdS/CFT correspondence between an asymptotically AdS spacetime in d + 1 dimensions and a CFT in d dimensions Proposed use to find a gravity description of non-perturbative problems in QCD Major obstacle is the current lack of a gravity dual for QCD Possible approach: try to capture some features of QCD with a CFT toy model for which there is a known gravity dual N = 4 SYM 4 at strong coupling ← → AdS 5 classical gravity

  9. Setup Classical gravity in d + 1 dimensions with cosmological constant Λ = d ( d − 1) / (2 L 2 ), coupled to real scalar field matter 1 : � 1 dx d +1 √− g � 16 π ( R − 2Λ) − 1 � 2 g αβ ∂ α ϕ∂ β ϕ − V ( ϕ ) S = The corresponding field equations take the local form 2 : dV ✷ ϕ = dϕ 2Λ � 1 � d − 1 T αα g µν R µν = d − 1 g µν + 8 π T µν − 1 We will use scalar field collapse as a convenient mechanism to form BHs ` 1 2 Real scalar field: T µν = ∂ µ ϕ∂ ν ϕ − g µν 2 g αβ ∂ α ϕ∂ β ϕ + V ( ϕ ) ´

  10. Setup µ, ν = 1 , ..., d + 1 2Λ � 1 � d − 1 T αα g µν 0 = R µν − d − 1 g µν − 8 π T µν −

  11. Setup µ, ν = 1 , ..., d + 1 2Λ � 1 � d − 1 T αα g µν 0 = − d − 1 g µν − 8 π T µν − R µν

  12. Setup µ, ν = 1 , ..., d + 1 2Λ � 1 � d − 1 T αα g µν 0 = − d − 1 g µν − 8 π T µν − − 1 2 g αβ g µν,αβ + g αβ g β ( µ,ν ) α + 1 2 g αβ,α ( g αβ,ν − g νµ,β + g βν,µ ) log √− g log √− g ,β Γ βµν − Γ ανβ Γ βαν � � � � − ,µν +

  13. Setup µ, ν = 1 , ..., d + 1 2Λ � 1 � d − 1 T αα g µν 0 = − d − 1 g µν − 8 π T µν − −∇ ( µ C ν ) − 1 2 g αβ g µν,αβ + g αβ g β ( µ,ν ) α + 1 2 g αβ,α ( g αβ,ν − g νµ,β + g βν,µ ) log √− g log √− g ,β Γ βµν − Γ ανβ Γ βαν � � � � − ,µν + C µ ≡ H µ − � x µ (physical solutions satisfy C µ = 0)

  14. Setup µ, ν = 1 , ..., d + 1 2Λ � 1 � d − 1 T αα g µν 0 = − d − 1 g µν − 8 π T µν − −∇ ( µ H ν ) + ∇ ( µ ✷ x ν ) − 1 2 g αβ g µν,αβ + g αβ g β ( µ,ν ) α + 1 2 g αβ,α ( g αβ,ν − g νµ,β + g βν,µ ) log √− g log √− g ,β Γ βµν − Γ ανβ Γ βαν � � � � − ,µν + C µ ≡ H µ − � x µ (physical solutions satisfy C µ = 0)

  15. Setup µ, ν = 1 , ..., d + 1 2Λ � 1 � d − 1 T αα g µν 0 = − d − 1 g µν − 8 π T µν − −∇ ( µ H ν ) + ✘✘✘✘ ∇ ( µ ✷ x ν ) ✘ − 1 ✭✭✭✭✭✭✭✭✭✭✭✭✭✭ 1 ✭ 2 g αβ g µν,αβ + ✘✘✘✘✘ g αβ g β ( µ,ν ) α + ✘ 2 g αβ,α ( g αβ,ν − g νµ,β + g βν,µ ) log √− g log √− g ✘ ,β Γ βµν − Γ ανβ Γ βαν − g αβ ✭ − ✘✘✘✘✘✘ � � ,µν + ✭✭✭✭✭✭✭✭ � � , ( µ g ν ) α,β C µ ≡ H µ − � x µ (physical solutions satisfy C µ = 0)

  16. Setup µ, ν = 1 , ..., d + 1 2Λ � 1 � d − 1 T αα g µν 0 = − d − 1 g µν − 8 π T µν − −∇ ( µ H ν ) + ✘✘✘✘ ∇ ( µ ✷ x ν ) ✘ − 1 ✭✭✭✭✭✭✭✭✭✭✭✭✭✭ 1 ✭ 2 g αβ g µν,αβ + ✘✘✘✘✘ g αβ g β ( µ,ν ) α + ✘ 2 g αβ,α ( g αβ,ν − g νµ,β + g βν,µ ) log √− g log √− g ✘ ,β Γ βµν − Γ ανβ Γ βαν − g αβ ✭ − ✘✘✘✘✘✘ � � ,µν + ✭✭✭✭✭✭✭✭ � � , ( µ g ν ) α,β C µ ≡ H µ − � x µ (physical solutions satisfy C µ = 0) choose some H µ = f µ ( g ) (this sets � x µ = f µ ( g ) as long as C µ = 0)

  17. Setup µ, ν = 1 , ..., d + 1 2Λ � 1 � d − 1 T αα g µν 0 = − d − 1 g µν − 8 π T µν − 2 n ( µ C ν ) − (1 + κ 2 ) g µν n α C α � � −∇ ( µ H ν ) + ✘✘✘✘ ∇ ( µ ✷ x ν ) − κ 1 ✘ ✭ − 1 ✭✭✭✭✭✭✭✭✭✭✭✭✭✭ 1 2 g αβ g µν,αβ + ✘✘✘✘✘ g αβ g β ( µ,ν ) α + 2 g αβ,α ( g αβ,ν − g νµ,β + g βν,µ ) ✘ log √− g log √− g ✘ ,β Γ βµν − Γ ανβ Γ βαν − g αβ ✭ − ✘✘✘✘✘✘ � � ,µν + ✭✭✭✭✭✭✭✭ � � , ( µ g ν ) α,β C µ ≡ H µ − � x µ (physical solutions satisfy C µ = 0) choose some H µ = f µ ( g ) (this sets � x µ = f µ ( g ) as long as C µ = 0)

  18. Setup Ingredients Evolution Equations Initial Data Boundary Conditions Gauge Choice

  19. Setup Ingredients • Evolution Equations Initial Data Boundary Conditions Gauge Choice

  20. Setup Evolution Equations − 1 2 g αβ g µν,αβ − g αβ 0 = , ( µ g ν ) α,β − H ( µ,ν ) + H α Γ αµν − Γ αβµ Γ βαν 2 n ( µ C ν ) − (1 + κ 2 ) g µν n α C α � � − κ 1 − 2Λ � 1 � d − 1 T αα g µν d − 1 g µν − 8 π T µν − ↓ 0 = E ( g µν ) ( d + 2)( d + 1) / 2 such equations, one for each g µν H µ = f µ ( g ) constraint damping terms ∼ κ 1 , designed to damp towards C µ = 0

  21. Setup g tt dt 2 + 2 g tz dtdz + 2 g tx 1 dtdx 1 + 2 g tx 2 dtdx 2 + g µν dx µ dx ν = g zz dz 2 + 2 g zx 1 dzdx 1 + 2 g zx 2 dzdx 2 + g x 1 x 1 dx 2 1 + 2 g x 1 x 2 dx 1 dx 2 + g x 2 x 2 dx 2 2 g µν = g µν ( t, z, x 1 , x 2 )

  22. Setup g tt dt 2 + 2 g tz dtdz + 2 g tx 1 dtdx 1 + 2 g tx 2 dtdx 2 + 2 g tx 3 dtdx 3 + g µν dx µ dx ν = g zz dz 2 + 2 g zx 1 dzdx 1 + 2 g zx 2 dzdx 2 + 2 g zx 3 dzdx 3 + g x 1 x 1 dx 2 1 + 2 g x 1 x 2 dx 1 dx 2 + 2 g x 1 x 3 dx 1 dx 3 + g x 2 x 2 dx 2 2 + g x 2 x 3 dx 2 dx 3 + g x 3 x 3 dx 2 3 + g µν = g µν ( t, z, x 1 , x 2 , x 3 )

  23. Setup g tt dt 2 + 2 g tz dtdz + 2 g tx 1 dtdx 1 + 2 g tx 2 dtdx 2 + 2 g tx 3 dtdx 3 + g µν dx µ dx ν = g zz dz 2 + 2 g zx 1 dzdx 1 + 2 g zx 2 dzdx 2 + 2 g zx 3 dzdx 3 + g x 1 x 1 dx 2 1 + 2 g x 1 x 2 dx 1 dx 2 + 2 g x 1 x 3 dx 1 dx 3 + g x 2 x 2 dx 2 2 + g x 2 x 3 dx 2 dx 3 + g x 3 x 3 dx 2 3 + g µν = g µν ( t, z, x 1 = 0 , x 2 , x 3 ) with SO (2) in x 1 , x 2

  24. Setup g tt dt 2 + 2 g tz dtdz + 2 g tx 1 dtdx 1 + 2 g tx 2 dtdx 2 + 2 g tx 3 dtdx 3 + g µν dx µ dx ν = g zz dz 2 + 2 g zx 1 dzdx 1 + 2 g zx 2 dzdx 2 + 2 g zx 3 dzdx 3 + g x 1 x 1 dx 2 1 + 2 g x 1 x 2 dx 1 dx 2 + 2 g x 1 x 3 dx 1 dx 3 + g x 2 x 2 dx 2 2 + g x 2 x 3 dx 2 dx 3 + g x 3 x 3 dx 2 3 + g µν = g µν ( t, z, x 1 = 0 , x 2 , x 3 ) with SO (2) in x 1 , x 2 L ξ g µν = 0 ∂ ∂ L ξ H µ = 0 ξ = x 2 − x 1 ∂x 1 ∂x 2 L ξ ϕ = 0

  25. Setup Ingredients • Evolution Equations Initial Data Boundary Conditions Gauge Choice

  26. Setup Ingredients Evolution Equations • Initial Data Boundary Conditions Gauge Choice

  27. Setup Initial Data ( d ) R + K 2 − K ij K ij − 2Λ − 16 πρ 0 = D j K ji − D i K − 8 πj i 0 = ↓ 0 = E ( ζ µ ) ( d + 1) such equations, one for each ζ µ where 1 n µ = − α∂ µ t, ρ = n µ n ν T µν , j i = − g µi n ν T µν , K ij = − 1 2 L n g ij = − 1 2 α ( − ∂ t g ij + D i β j + D j β i ) 1 Here, α is the lapse function and β i is the shift vector

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