On the stability of a superspinar Ken-ichi Nakao (OCU, Japan) In - PowerPoint PPT Presentation

On the stability of a superspinar Ken-ichi Nakao (OCU, Japan) In collaboration with Hideyuki Tagoshi (ICCR, Japan) Tomohiro Harada (Rikkyo University, Japan) Pankaj S. Joshi (TIFR, India) Prashant Kocharakota (TIFR, India) Jun-qi Guo (University of Jinan, China) Mandar Patil (Polish Academy of Science, Poland) Andrzej Krolak (Polish Academy of Science, Poland) arXiv:1707.07242 Based on

� § 1 Introduction Kerr spacetime: vacuum, stationary, axi-symmetric, asymptotically flat In Boyer-Lindquist coordinates 2 $ ' A dt 2 + A Σ sin 2 θ d ϕ − 2 aMr ds 2 = − ΣΔ + Σ Δ dr 2 + Σ d θ 2 dt & ) % A ( ( ) = r 2 − 2 Mr + a 2 where Δ r 2 − Δ r ( ) a 2 sin 2 θ ) = r 2 + a 2 ) = r 2 + a 2 cos 2 θ ( ) ( A r , θ ( Σ r , θ −∞ < t < + ∞ , −∞ < r < + ∞ , 0 ≤ θ ≤ π , 0 ≤ ϕ < 2 π M : ADM mass a : Kerr parameter (angular momentum L = aM ) 𝑁 % − 𝑏 % a 2 ≤ M 2 : larger root of D ( r ) = 0 𝑠 = 𝑁 + ≥ 𝑁 is the event horizon a 2 > M 2 : always D ( r ) > 0 holds. No event horizon. Ring singularity at r = 0 , q = p /2 is naked.

� .. Reissner-Nordstrom spacetime Maxwell field, static, spherically symmetric, asymptotically flat 𝑒𝑡 % = −𝑔 𝑠 𝑒𝑢 % + 𝑒𝑠 % Metric 𝑔 𝑠 + 𝑠 % 𝑒𝜄 % + 𝑠 % sin % 𝜄𝑒𝜒 % where 𝑠 + 𝑅 % 𝑔 𝑠 = 1 − 2𝑁 𝑠 % M : ADM mass Q : Charge parameter Gauge one-form 𝐵 9 = − 𝑅 𝑠 , 0,0,0 𝑁 % − 𝑅 % Q 2 ≤ M 2 : larger root of f ( r ) = 0 𝑠 = 𝑁 + ≥ 𝑁 is the event horizon Q 2 > M 2 : always f ( r ) > 0 holds. No event horizon. Singularity at r = 0 is naked.

From a point of view of SUSY, 𝑅 % ≤ 𝑁 % if the system is supersymmetric, BPS bound holds. 𝑏 % ≤ 𝑁 % Kerr bound does not have a special meaning. In the framework of superstring theory, over-spinning very compact entity named the superspinar may exist. Stringy effects will make any singularities harmless. Gimon and Horava (2007) The over-spinning Kerr geometry around the naked singularity is very interesting.

� � In Boyer-Lindquist coordinates 2 $ ' A dt 2 + A Σ sin 2 θ d ϕ − 2 aMr ds 2 = − ΣΔ + Σ Δ dr 2 + Σ d θ 2 dt & ) A % ( ( ) = r 2 − 2 Mr + a 2 where Δ r 2 − Δ r ( ) a 2 sin 2 θ ) = r 2 + a 2 ) = r 2 + a 2 cos 2 θ ( ) ( A r , θ ( Σ r , θ 𝜖 𝜖𝑢 : time coordinate basis=Killing vector field 𝜖 𝜖𝑢 , 𝜖 𝜖 𝜖𝑢 = − 1 − 2𝑁𝑠 Ergo-region (ergo-sphere): is spacelike 𝑕 > 0 𝜖𝑢 Σ 𝑁 % − 𝑏 % cos % 𝜄 𝑁 % − 𝑏 % cos % 𝜄 𝑁 − < 𝑠 < 𝑁 + Killing “energy” of a particle: 𝐹 = −𝒗 F 𝜖 𝜖𝑢 can be negative in the ergo-region, even if 𝒗 is future directed timelike vector.

� � Collisional Penrose process M. Patil, T. Harada, KN, P.S. Joshi and M. Kimura (2015) when 𝑏 = 1 + 𝜁 𝑁 𝐹 I = 𝑛 2𝑁 − 𝑀 L 2𝑁 − 𝑀 % 0 < 𝜁 ≪ 1 2 𝑁𝜁 𝐹 N = 2𝑛 − 𝐹 I 𝐹 % = 𝑛, 𝑀 % 𝑠 = 0 𝜃 = 𝐹 I Singularity 2𝑛 → ∞ for 𝜁 → 0 𝑠 = 𝑁 No upper bound on efficiency! Ultra-high-energy cosmic ray! 𝐹 L = 𝑛, 𝑀 L

Efficiency of energy extraction from accreting matter ≈ 42%, 𝑏 % = 𝑁 % when Singularity ISCO: 𝑠 = 𝑁 + 0 in Boyer-Lindquist Accretion Disk

Efficiency of energy extraction from accreting matter=100%, 𝑏 % = 32 when 27 𝑁 % Gimon & Horava (2007) 𝑠 = 0 𝑠 = 2 Singularity ISCO: 3 𝑁 in Boyer-Lindquist Accretion Disk

Is a superspinar stable? How does a superspinar form?

Is a superspinar stable? How does a superspinar form?

§ 2 Stability of Superspinar Teukolsky equations determine perturbations in Kerr spacetime 𝜔 = 𝑓 WXYZ[X\] 𝑆 𝑠 𝑇 𝜄 Master variable of the perturbations: + 𝐿 % − 2𝑗𝑡 𝑠 − 𝑁 𝐿 ∆ Wa 𝑒 𝑒𝑠 ∆ a[L 𝑒𝑆 + 4𝑗𝑡𝜕𝑠 − 𝜇 𝑆 = 0 𝑒𝑠 ∆ % 1 𝑒𝜄 sin 𝜄 𝑒𝑇 𝑒 𝑏𝜕 cos 𝜄 + 𝑡 % − 𝑛 + 𝑡 cos 𝜄 𝑒𝜄 + − 𝑡 𝑡 − 1 + 𝐵 𝑇 = 0 sin 𝜄 sin 𝜄 where 𝐿 ≔ 𝑠 % + 𝑏 % 𝜕 − 𝑏𝑛 𝜇 ≔ 𝐵 + 𝑏 % 𝜕 % − 2𝑏𝑛𝜕 𝐵 = 𝐵 Yh\a : Eigen value of the equation for S 𝑚 ≥ 𝑛𝑏𝑦 𝑛 , |𝑡| 𝑡 = 0 : scalar 𝑡 = 1 : EM 𝑡 = 2 : GW

§ 2 Stability of Superspinar Teukolsky equations determine perturbations in Kerr spacetime 𝜔 = 𝑓 WXYZ[X\] 𝑆 𝑠 𝑇 𝜄 Master variable of the perturbations: + 𝐿 % − 2𝑗𝑡 𝑠 − 𝑁 𝐿 ∆ Wa 𝑒 𝑒𝑠 ∆ a[L 𝑒𝑆 + 4𝑗𝑡𝜕𝑠 − 𝜇 𝑆 = 0 𝑒𝑠 ∆ % 1 𝑒𝜄 sin 𝜄 𝑒𝑇 𝑒 𝑏𝜕 cos 𝜄 + 𝑡 % − 𝑛 + 𝑡 cos 𝜄 𝑒𝜄 + − 𝑡 𝑡 − 1 + 𝐵 𝑇 = 0 sin 𝜄 sin 𝜄 Ψ N = 𝑠 − 𝑗𝑏 cos 𝜄 WN 𝜔 with 𝑡 = −2 Outgoing GW

𝜔 = 𝑓 WXYZ[X\] 𝑆 𝑠 𝑇 𝜄 Master variable of the perturbations: + 𝐿 % − 2𝑗𝑡 𝑠 − 𝑁 𝐿 ∆ Wa 𝑒 𝑒𝑠 ∆ a[L 𝑒𝑆 + 4𝑗𝑡𝜕𝑠 − 𝜇 𝑆 = 0 𝑒𝑠 ∆ % 1 𝑒𝜄 sin 𝜄 𝑒𝑇 𝑒 𝑏𝜕 cos 𝜄 + 𝑡 % − 𝑛 + 𝑡 cos 𝜄 𝑒𝜄 + − 𝑡 𝑡 − 1 + 𝐵 𝑇 = 0 sin 𝜄 sin 𝜄 Ψ N = 𝑠 − 𝑗𝑏 cos 𝜄 WN 𝜔 with 𝑡 = −2 Outgoing GW 𝐷 𝑠 𝑓 XYn for Quasi-normal mode (QNM) for 𝑠 → ∞ 𝑆 → In black hole case Horizon ∆ 𝑠 = 0 : singular point of the radial Teukolsky equation Regularity at horizon Unique boundary condition 𝑆 → 0 for 𝑠 → 𝑠 [

𝜔 = 𝑓 WXYZ[X\] 𝑆 𝑠 𝑇 𝜄 Master variable of the perturbations: + 𝐿 % − 2𝑗𝑡 𝑠 − 𝑁 𝐿 ∆ Wa 𝑒 𝑒𝑠 ∆ a[L 𝑒𝑆 + 4𝑗𝑡𝜕𝑠 − 𝜇 𝑆 = 0 𝑒𝑠 ∆ % 1 𝑒𝜄 sin 𝜄 𝑒𝑇 𝑒 𝑏𝜕 cos 𝜄 + 𝑡 % − 𝑛 + 𝑡 cos 𝜄 𝑒𝜄 + − 𝑡 𝑡 − 1 + 𝐵 𝑇 = 0 sin 𝜄 sin 𝜄 Ψ N = 𝑠 − 𝑗𝑏 cos 𝜄 WN 𝜔 with 𝑡 = −2 Outgoing GW 𝑆 → 𝐷 𝑠 𝑓 XYn for Quasi-normal mode (QNM) for 𝑠 → ∞ In superspinar (over-spinning Kerr) case ∆ 𝑠 > 0 : no singular point of the radial Teukolsky equation on the real axis of 𝑠. No horizon No unique boundary condition unless we know what is a superspinar.

� § 3 Stability of Superspinar (1) Cardoso, Pani, Cadoni, Cavaglia (2008); Pani, Barausse, Berti and Cardoso (2010) Quasi-normal modes of perturbations (no incoming wave at infinity) 𝑍 r = −𝑗𝑙𝑍 • Absorbing BC: 𝜔 = 𝑓 WXYZ[X\] 𝑆 𝑠 𝑇 𝜄 𝑠 % + 𝑏 % L % ⁄ 𝑆 where 𝑍 = ∆ a % ⁄ Boundary condition at 𝑠 = 𝑠 p = constant 𝑍” + 𝑊 𝑠 𝑍 = 0 Reflecting BC: 𝑆 = 0 • 𝑙 = 𝑊 𝑠 p Singularity 𝑠 = 0 Imaginary part of 𝜕 is positive → unstable 𝑠 = 𝑠 p Reflecting BC Absorbing BC

(Real Part of Angular Frequency) (Real Part of Angular Frequency) Reflection Boundary Condition (Radius of Boundary) (angular momentum) (Imaginary Part of Angular Frequency) (Imaginary Part of Angular Frequency) 𝜕 w > 0: unstable (Radius of Boundary) (angular momentum)

Absorbing Boundary Condition 𝜕 w > 0: unstable

Reflecting Boundary Condition

Pani et al state that the superspinar is, in general, unstable, since perturbations grow exponentially under both reflecting and absorbing boundary conditions. However, there are boundary conditions under which the superspinar is stable. Is their conclusion right?

Pani et al state that the superspinar is, in general, unstable, since perturbations grow exponentially under both reflecting and absorbing boundary conditions. However, there are boundary conditions under which the superspinar is stable. Is their conclusion right? Let’s consider!

In superspinar case, replace the the stability problem with “Does there exist the boundary condition under which the superspinar is stable?”

In superspinar case, replace the the stability problem with “Does there exist the boundary condition under which the superspinar is stable?” Answer: Yes! We assume stable angular frequency, i.e., 𝜕 = 𝜕 • + 𝑗𝜕 w with negative 𝜕 w Input Parameter

Solve the angular Teukolsky equation by imposing regularities at 𝜄 = 0, 𝜌 % 1 𝑒𝜄 sin 𝜄 𝑒𝑇 𝑒 𝑏𝜕 cos 𝜄 + 𝑡 % − 𝑛 + 𝑡 cos 𝜄 𝑒𝜄 + − 𝑡 𝑡 − 1 + 𝐵 𝑇 = 0 sin 𝜄 sin 𝜄 𝐵 is determined. 𝜇 ≔ 𝐵 + 𝑏 % 𝜕 % − 2𝑏𝑛𝜕

Solve the angular Teukolsky equation by imposing regularities at 𝜄 = 0, 𝜌 % 1 𝑒𝜄 sin 𝜄 𝑒𝑇 𝑒 𝑏𝜕 cos 𝜄 + 𝑡 % − 𝑛 + 𝑡 cos 𝜄 𝑒𝜄 + − 𝑡 𝑡 − 1 + 𝐵 𝑇 = 0 sin 𝜄 sin 𝜄 𝐵 is determined. 𝜇 ≔ 𝐵 + 𝑏 % 𝜕 % − 2𝑏𝑛𝜕 Solve the radial Teukolsky equation with QNM boundary condition + 𝐿 % − 2𝑗𝑡 𝑠 − 𝑁 𝐿 ∆ Wa 𝑒 𝑒𝑠 ∆ a[L 𝑒𝑆 + 4𝑗𝑡𝜕𝑠 − 𝜇 𝑆 = 0 𝑒𝑠 ∆ No singular point on real axis of 𝑠 obtained 𝑆 𝑠 is regular everywhere.