Black-Hole Binary Initial Data: Getting the Spin Right Gregory B. Cook Wake Forest University October 5, 2005 Collaborators: Harald Pfeiffer[7] (Caltech), Jason D. Grigsby (WFU), & Matthew Caudill (WFU)
Motivation Black hole binaries are among the most likely sources for early detection with LIGO, VIRGO, GEO, TAMA,... Available computed waveforms should increase chance of detecting collision events. Quasiequilibrium Binary Data General Relativity doesn t permit true equilibrium for astrophysical binary systems. When the bodies are sufficiently far apart, the timescale for orbital decay is much larger than the orbital period. If the orbit is nearly circular (quasi-circular) then there is a corotating reference frame in which the binary appears to be at rest. Quasiequilibrium gives us a physical condition to guide us in fixing boundary conditions and data that is not otherwise constrained.
The 3 + 1 Decomposition Lapse : α Spatial metric : γ ij αn µ δt t + δt Shift vector : β i Extrinsic Curvature : K ij n µ t µ δt t ds 2 Time vector : t µ = αn µ + β µ = α 2 dt 2 + γ ij (dx i + β i dt)(dx j + β j dt) β µ β µ δt γ µν = g µν + n µ n ν K µν = 1 2 γα µ γβ ν L ng αβ Constraint equations R + K 2 K ij K ij = 16πρ j K ij γ ij K = 8πj i S µν j µ T µν γ α µ γβ ν T αβ γ ν µ nα T να ρ n µ n ν T µν = S µν + 2n (µ j ν) + n µ n ν ρ Evolution equations t γ ij = 2αK ij + i β j + j β i t K ij = i j α + αh Rij 2K il K l j + KK ij i 8πS ij + 4πγ ij (S ρ) + β l l K ij + K il j β l + K jl i β l
Conformal Thin-Sandwich Decomposition αn µ δt n µ β µ t µ δt β µ δt t + δt γ ij = ψ 4 γ ij t K ij = ψ 10 2 α h ( Lβ) ij ũ iji + 1 3 ψ 4 γ ij K Hamiltonian Const. 2 ψ 1 8 ψ R 1 12 ψ5 K 2 + 1 8 ψ 7 à ij à ij = 2πψ 5 ρ Momentum Const. j ( Lβ) ij ( Lβ) ij j α = 4 3 αψ6 i K + α 1 j + 16π αψ 10 j i Const. Tr(K) eqn. h 2 (ψ 7 α) (ψ 7 1 α) 8 ψ R i + 5 12 ψ5 K 2 + 7 8 ψ 7 à ij à ij ψ 5 β i i K à ij 1 2 α h( Lβ) ij ũ iji Constrained vars : ψ, β i, and α ψ 6 α Freely specified : γ ij K ũ ij t γ ij and t K = 2πψ 5 K(ρ + 2S) ψ 5 t K 8 < t γ ij = 0 Quasiequilibrium : t K = 0
Ham. & Mom. const. eqns., & Const Tr(K) eqn. from Conf. TS + ũ ij = t K = 0 Equations of Quasiequilibrium 9 >= >; Eqns. of Quasiequilibrium With γ ij = f ij and K = 0, these equations have been widely used to construct binary neutron star initial data[3, 12, 4, 15]. Binary neutron star initial data require: boundary conditions at infinity compatible with asymptotic flatness and corotation. «i ψ r = 1 β i r = Ω 0 α r = 1 φ compatible solution of the equations of hydrostatic equilibrium. ( Ω 0 )
Ham. & Mom. const. eqns., & Const Tr(K) eqn. from Conf. TS + ũ ij = t K = 0 Equations of Quasiequilibrium 9 >= >; Eqns. of Quasiequilibrium With γ ij = f ij and K = 0, these equations have been widely used to construct binary neutron star initial data[3, 12, 4, 15]. Binary neutron star initial data require: boundary conditions at infinity compatible with asymptotic flatness and corotation. «i ψ r = 1 β i r = Ω 0 α r = 1 φ compatible solution of the equations of hydrostatic equilibrium. ( Ω 0 ) Binary black hole initial data require: a means for choosing the angular velocity of the orbit Ω 0. with excision, inner boundary conditions are needed for ψ, β i, and α. Gourgoulhon, Grandclément, & Bonazzola[10, 11]: Black-hole binaries with γ ij = f ij & K = 0, inversion-symmetry, and Killing-horizon conditions on the excision boundaries. Solutions require constraint violating regularity condition imposed on inner boundaries!
The Inner Boundary s i i τ τ h ij γ ij s i s j k µ 1 2 (n µ + s µ ) Σ ḱ µ n µ k µ ḱ µ 1 2 (n µ s µ ) Extrinsic curvature of S embedded in s i spacetime S Σ µν 1 2 hα µ hβ ν L kg αβ Σ µν 1 2 hα µ hβ ν L ḱ g αβ Extrinsic curvature of S embedded in Σ H ij 1 2 hk i hl j L sγ kl Projections of K ij onto S Σ ij = 1 2 (H ij J ij ) Σ ij = 1 2 (H ij + J ij ) J ij J i h k i hl j K kl h k i sl K kl J h ij J ij = h ij K ij Expansion of null rays θ h ij Σ ij = 1 2 (H J) θ h ij Σij = 1 2 (H + J) σ ij Shear of null rays Σ ij 1 2 h ijθ σ ij Σ ij 1 2 h ij θ
AH and QE Conditions on the Inner Boundary The quasiequilibrium inner boundary conditions start with the following assumptions: 1. The inner boundary S is a (MOTS): marginally outer-trapped surface θ = 0 2. The horizons are in quasiequilibrium: σ ij = 0 and no matter is on S Raychaudhuri s equation implies that MOTS initially evolves along k µ. L k θ = 1 2 θ2 σ µν σ µν ω µν ω µν R µν k µ k ν = 0 3. The time evolution vector lies in outgoing null surface through S: t µ k µ S = 0 k µ 1 2 (n µ + s µ ) t µ = αn µ + β µ 9 = ; = α S = β i s i S β S
AH/Quasiequilibrium Boundary Conditions θ = ψ 2 h hij i i s j + 4 s k k ln ψ ψ 2 J 2 σ ij = 1 (H ij 1 2 h ijh) 1 β «2 α 1 2 ψ 4 α n o D(i β j) 1 2 h ij Dk β k 1 2 [ h ik hjl ũ kl 1 2 h ij hkl ũ kl ]
AH/Quasiequilibrium Boundary Conditions θ = ψ 2 h hij i i s j + 4 s k k ln ψ ψ 2 J 2 σ ij = 1 (H ij 1 2 h ijh) 1 β «2 α 1 2 ψ 4 α n o D(i β j) 1 2 h ij Dk β k 1 2 [ h ik hjl ũ kl 1 2 h ij hkl ũ kl ] s k k ln ψ = 1 4 ( h ij i s j ψ 2 J) β i = αψ 2 s i + β i 0 = D (i β j) 1 2 h ij Dk β k
AH/Quasiequilibrium Boundary Conditions θ = ψ 2 h hij i i s j + 4 s k k ln ψ ψ 2 J 2 σ ij = 1 (H ij 1 2 h ijh) 1 β «2 α 1 2 ψ 4 α n o D(i β j) 1 2 h ij Dk β k 1 2 [ h ik hjl ũ kl 1 2 h ij hkl ũ kl ] s k k ln ψ = 1 4 ( h ij i s j ψ 2 J) β i = αψ 2 s i + β i 0 = D (i β j) 1 2 h ij Dk β k The boundary condition on the lapse is not fixed by QE assumptions. It has been shown to be associated with the initial temporal gauge choice and is taken to be freely specifiable.
Defining the Spin of the Black Hole The spin parameters β i can be defined by demanding that the time vector associated with quasiequilibrium in the corotating frame must be null, forming the null generators of the horizon. k µ (n µ + s µ ) = k µ = h1, αs i β ii This vector k µ is null for any choice of α & β i. In the frame where a black hole is not spinning, the null time vector has components t µ = [1, 0]. Corotating Holes Corotating holes are at rest in the corotating frame, where we must pose boundary conditions. So, h k µ = 1, αs i β ii = [1, 0] Thus we find β i = αs i β i = 0 Irrotational Holes Irrotational holes are at rest in the inertial frame. With the time vectors in the inertial and corotating frames related by k µ = = + Ω 0 t t φ h 1, αs i β ii = [1, Ω 0 ( / φ) i ] Thus we find «i β i = αs i + Ω 0 β i = Ω 0 ξ i φ ξ i ( φ )i & D (i ξ j) 1 2 h ij Dk ξ k = 0
The Lapse BC & QE So far, nothing has fixed a boundary condition on the lapse α. One possibility[9] is to recall that θ θ is a Lorentz invariant and so to consider L ζ θ = 0 as a quasiequilibrium condition. L ζ θ = 0 J s i i α = ψ 2 (J 2 JK + D)α D ψ 4 [ h ij ( D i J i )( D j J j ) 1 22 R + 2 D2 ln ψ]
The Lapse BC & QE So far, nothing has fixed a boundary condition on the lapse α. One possibility[9] is to recall that θ θ is a Lorentz invariant and so to consider L ζ θ = 0 as a quasiequilibrium condition. L ζ θ = 0 J s i i α = ψ 2 (J 2 JK + D)α D ψ 4 [ h ij ( D i J i )( D j J j ) 1 22 R + 2 D2 ln ψ] This conditions is satisfied for stationary solutions, but seems to be degenerate with the other QE boundary conditions. To see this, note that the stationary maximal slicings of Schwarzschild form a 1-parameter family: ds 2 = α = β R = C R 2 s dr 2 1 2M R + + R 2 d 2 Ω C2 R s 4 1 2M R + C2 R 4 1 2M R + C2 R 4 K i j = C R 3 2 6 4 2 0 0 0 1 0 0 0 1 α S = C 4M 2 3 7 5
The Orbital Angular Velocity For a given choice of the Lapse BC, γ ij and K, we are still left with a family of solutions parameterized by the orbital angular velocity Ω 0. Except for the case of a single spinning black hole, it is not reasonable to expect more than one value of Ω 0 to correspond to a system in quasiequilibrium. GGB[10, 11] have suggested a way to pick the quasiequilibrium value of Ω 0 : Ω 0 is chosen as the value for which the ADM energy E ADM equals the Komar mass M K. Komar mass M K = 1 4π I γ ij ( i α β k K ik )d 2 S j Acceptable definition of the mass only for stationary spacetimes. ADM energy E ADM = 1 16π I γ ij k (G k i δk i G)d2 S j Acceptable definition of the mass for arbitrary spacetimes. G ij γ ij f ij
Summary of QE Formalism γ ij = ψ 4 γ ij K ij = ψ 10 à ij + 1 3 γij K à ij = ψ6 2α ( Lβ) ij t γ ij = 0 2 ψ 1 8 ψ R 1 12 ψ5 K 2 + 1 8 ψ 7 à ij à ij = 0 j ( Lβ) ij ( Lβ) ij j ln αψ 6 = 4 3 α i K h 2 1 (αψ) (αψ) R 8 + 5 12 ψ4 K 2 + 7 8 ψ 8 A ij A iji = ψ 5 β i i K t K = 0 s k k ln ψ S = 1 4 ( h ij i s j ψ 2 J) S θ = 0 β i S = 8 < : αψ 2 s i S αψ 2 s i S + Ω 0 ξ i S corotation irrotation σ ij = 0 t µ k µ = 0 α S = unspecified by QE ψ r = 1 β i r = Ω 0 φ α r = 1 «i The only remaining freedom in the system is the choice of the lapse boundary condition, the initial spatial and temporal gauge, and the initial dynamical( wave ) content found in α S, γ ij and K.
Corotating; Maximal Slice; Comparison; E b /µ vs J/µm -0.02-0.03 CO: MS - d(αψ)/dr = (αψ)/2r CO: HKV - GGB CO: EOB - 3PN CO: EOB - 2PN CO: EOB - 1PN E b /µ -0.04-0.05-0.05-0.06-0.06 3.4 3.6 3.6 4 4.4 4.8 J/µm
Irrotational; Maximal Slice; Comparison; E b /µ vs J/µm -0.02 IR: MS - d(αψ)/dr = (αψ)/2r IR: IVP Conf.Imaging/Eff.Pot. IR: EOB - 3PN IR: EOB - 2PN IR: EOB - 1PN -0.04 E b /µ -0.05-0.06-0.06-0.07-0.08-0.08 3 3.2 3.4 3.6 3.2 3.6 4.0 4.4 4.8 J/µm
Are We Finding Quasicircular orbits? The Komar mass criteria is used to choose the quasiequilibrium model which implies quasicircular orbits. Can we find an independent way of verifying this? Comparison with post-newtonian: 1. Compare plots as we have done already. 2. Look at post-newtonian definitions of ellipticity (eg. Mora & Will[14] or Memmesheimer, Gopakumar, & Schäfer[13]). An effective potential approach[6, 8].
Measuring the Spin of a Black Hole Spin is only rigorously defined at spatial/null infinity. Must use quasi-local definition: e.g. Brown & York[5] or Ashtekar & Krishnan[2] S = 1 K ij ξ i s j hd 2 x 8π BH = 1 Ã ij ξ i s hd j 2 x ξ 8π i s i = 0 BH ξ i = 8 < : ξ i CK : Killing vector of h ij conformal Killing vector of h ij ξ i KV : Killing vector of h ij
Corotation: Effective Potential; E b /µ vs l/m -0.02 E b /µ -0.03-0.04-0.05-0.06-0.07-0.08 J=3.30 J=3.35 J=3.38 J=3.40 J=3.45 J=3.50 J=3.55 J=3.60 J=3.65 J=3.70 J=3.75 J=3.80 J=3.90 J=4.00 J=4.10 J=4.20 J=4.30 J=4.40 J=4.50 5 10 15 Komar 20 l/m
Corotation: Flat & True KV Spin; S/M 2 irr vs mω 0 0.2 0.15 S/M 2 0.1 0.05 CO: CK - d(αψ)/dr = (αψ)/2r CO: CK - d(αψ)/dr=0 CO: CK - αψ = 1/2 CO: KV - d(αψ)/dr = αψ CO: KV - d(αψ)/dr=0 CO: KV - αψ = 1/2 0 0 0.025 0.05 0.075 0.1 0.125 mω 0
Corotation: Flat & True KV Spin Error; (S S Kerr )/S Kerr vs mω 0 (S - S Kerr )/S Kerr 0-0.025-0.05-0.075 CO: CK - d(αψ)/dr = αψ CO: CK - d(αψ)/dr=0 CO: CK - αψ = 1/2 CO: KV - d(αψ)/dr = αψ CO: KV - d(αψ)/dr=0 CO: KV - αψ = 1/2-0.1 0 0.025 0.05 0.075 0.1 0.125 mω 0
Irrotation: Effective Potential; E b /µ vs l/m -0.02 E b /µ -0.03-0.04-0.05-0.06-0.07-0.08 J=3.90 J=3.80 J=3.75 J=3.70 J=3.65 J=3.60 J=3.55 J=3.50 J=3.45 J=3.30 J=3.20 J=3.10 J=3.05 J=3.00 Komar 5 10 15 20 l/m
Irrotation: Flat & True KV Spin; S/M 2 irr vs mω 0 0-0.025 S/M 2-0.05-0.075 IR: CK - d(αψ)/dr = (αψ)/2r IR: CK - d(αψ)/dr=0 IR: CK - αψ = 1/2 IR: KV - d(αψ)/dr = αψ IR: KV - d(αψ)/dr=0 IR: KV - αψ = 1/2-0.1 0 0.025 0.05 0.075 0.1 0.125 mω 0
Irrotation: Flat & True KV Spin Error; S/S Kerr vs mω 0 0-0.1 IR: CK - d(αψ)/dr = αψ IR: CK - d(αψ)/dr=0 IR: CK - αψ = 1/2 IR: KV - d(αψ)/dr = αψ IR: KV - d(αψ)/dr=0 IR: KV - αψ = 1/2 S/S Kerr -0.2-0.3-0.4 0 0.05 0.1 mω 0
True Irrotation Clearly the 0 th -order notion of defining irrotation is not adequate β i S = αψ 2 s i S + Ω 0 ξ i S. Instead, choose β i S = αψ 2 s i S + Ω BH ξ i S and choose Ω BH so that the quasilocal spin vanishes.
True Irrotation: Effective Potential; E b /µ vs l/m -0.04 E b /µ -0.05-0.06-0.07 J=3.7 J=3.5 J=3.4 J=3.3 J=3.25 J=3.2 J=3.15 J=3.1 Komar -0.08-0.09 5 10 15 l/m
True Irrotation: Flat & True KV Spin; S/M 2 irr vs mω 0 0-0.0025 S/M 2-0.005-0.0075-0.01 TI: CK - d(αψ)/dr = (αψ)/2r TI: CK - d(αψ)/dr=0 TI: CK - αψ = 1/2 TI: KV - d(αψ)/dr = αψ TI: KV - d(αψ)/dr=0 TI: KV - αψ = 1/2 0 0.05 0.1 0.15 mω 0
True Irrotation: Flat & True KV Spin Error; S/S Kerr vs mω 0 0-0.01 S/S Kerr -0.02-0.03 TI: CK - d(αψ)/dr = αψ TI: CK - d(αψ)/dr=0 TI: CK - αψ = 1/2 TI: KV - d(αψ)/dr = αψ TI: KV - d(αψ)/dr=0 TI: KV - αψ = 1/2 0 0.05 0.1 0.15 mω 0
True Irrotation; Maximal Slice; Comparison; E b /µ vs J/µm -0.02 IR: MS - d(αψ)/dr = (αψ)/2r TI: MS - d(αψ)/dr = (αψ)/2r IR: IVP Conf.Imaging/Eff.Pot. IR: EOB - 3PN IR: EOB - 2PN IR: EOB - 1PN -0.04 E b /µ -0.05-0.06-0.06-0.07-0.08-0.08 3 3.2 3.4 3.6 3.2 3.6 4.0 4.4 4.8 J/µm
Maximal Slice; Comparison of ISCO; E b /M irr vs ΩM irr E b / m -0.014-0.016-0.018 CO: QE CO: HKV-GGB CO: PN EOB CO: PN standard IR: QE IR: IVP conf IR: PN EOB IR: PN standard TI: QE -0.02-0.022 0.08 0.12 0.16 mω 0
Maximal Slice; Comparison of ISCO; E b /M irr vs J/M 2 irr -0.014-0.016 E b / m -0.018-0.02-0.022 CO: QE CO: HKV-GGB CO: PN EOB CO: PN standard IR: QE IR: IVP conf IR: PN EOB IR: PN standard TI: QE 0.75 0.8 0.85 0.9 J / m 2
Maximal Slice; Comparison of ISCO; J/M 2 irr vs ΩM irr J / m 2 0.9 0.85 0.8 CO: QE CO: HKV-GGB CO: PN EOB CO: PN standard IR: QE IR: IVP conf IR: PN EOB IR: PN standard TI: QE 0.75 0.08 0.12 0.16 mω 0
Understanding the Higher Order Effects Irrotation Ω BH is the amount by which we must counter-spin the hole to achieve no angular momentum: Ω BH Ω 0 Corotation S We find that M S. In this context, if we let 2 QL M 2 Kerr(Ω 0 ) Ω BH represent the angular velocity of a Kerr BH with spin S QL : Ω BH Ω 0 Ω BH Ω 0 = 1 mω 0 S M 2 QL ( 4 + S M 2 QL ) 2
Tidal Field Rotation Rate From the local inertial frame of one black hole, the tidal field of the second black hole is rotating with an angular velocity ω[1] ( m ) ( m ) ] 3/2 ω = Ω 0 [1 η + O b b To 1PN order m b = (mω 0) 2/3 [ 1 + ( 1 1 3 η) (mω 0 ) 2/3 + O (mω 0 ) 4/3] Ω BH Ω 0 = ω Ω 0 = 1 η (mω 0 ) 2/3 η ( 1 1 3 η) (mω 0 ) 4/3 +...
Ratio of Spin and Orbital Angular Velocities; Ω BH /Ω 0 vs mω 0 1 0.95 0.9 Ω BH /Ω 0 0.85 0.8 0.75 0.7 0.65 TI: d(αψ)/dr = αψ TI: d(αψ)/dr=0 TI: αψ = 1/2 CO: d(αψ)/dr = αψ CO: d(αψ)/dr=0 CO: αψ = 1/2 1PN 1PN+ 0 0.05 0.1 0.15 mω 0
Spin Error Using 1PN Kerr Spin; (S S 1PN+ )/S 1PN+ vs mω 0 0-0.005 (S - S 1PN+ )/S 1PN+ -0.01-0.015 CO: CK - d(αψ)/dr = αψ CO: CK - d(αψ)/dr=0 CO: CK - αψ = 1/2 CO: KV - d(αψ)/dr = αψ CO: KV - d(αψ)/dr=0 CO: KV - αψ = 1/2-0.02 0 0.025 0.05 0.075 0.1 0.125 mω 0
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