Black hole solutions in N>4 Gauss-Bonnet Gravity

1. 4th International Seminar on High Energy Physics QUARKS'2006Repino, St.Petersburg, Russia, May 19-25, 2006 Black hole solutions in N>4 Gauss-Bonnet Gravity S.Alexeyev*1, N.Popov2, T.Strunina3 1Sternberg Astronomical Institute, Moscow, Russia 2Computer Center of Russian Academy of Sciences, Moscow 3Ural State University, Ekaterinburg, Russia

2. Main publications • S.Alexeyev and M.Pomazanov,Phys.Rev.D55, 2110 (1997) • S.Alexeyev, A.Barrau, G.Boudoul, M.Sazhin, O.Khovanskaya,Astronomy Letters 28, 489 (2002) • S.Alexeyev, A.Barrau, G.Boudoul, O.Khovanskaya, M.Sazhin,Class.Quant.Grav. 19, 4431 (2002) • A.Barrau, J.Grain, S.Alexeyev, Phys.Lett.B584, 114 (2004) • S.Alexeyev, N.Popov, A.Barrau, J.Grain, Journal of Physics: Conference Series 33, 343 (2006) • S.Alexeyev, N.Popov, T.Strunina, A.Barrau, J.Grain, in preparation

3. Fundamental Planck scale shift • Large extra dimensions scenario (MD – D dimensional fundamental Planck mass, MPl – 4D Planck mass) MD = [MPl2/ VD-4]1/(D-2)

4. Planck energy in 4D representation ↓ 1019 GeV Fundamental Planck energy ↓ ≈ 1 TeV Planck Energy shift

5. Extended Schwarzschild solution in (4+n)D Tangherlini, ‘1963, Myers & Perry, ‘1986 Metric: ds2=-R(r)dt2+R(r)-1dr2+r2dΩn+22 Metric function: R(r) = 1 – [rs / r]n+1

6. (4+n)D Low Energy Effective String Gravity with higher order (second order in our consideration) curvature corrections S=(16πG)-1∫dDx(-g)½[R + Λ +λ SGB + …] Gauss-Bonnet term SGB = Rμναβ Rμναβ – 4Rαβ Rαβ + R2

7. Einstein-GB equations R.Cai, ‘2003 Rµν - ½ gµνR - Λgµν – α(½ gµνSGB – 2 RRµν + 4 RµγRγν + 4 RγδRγµδν– 2 RµγδλRνγδν) = 0 SGB = Rμναβ Rμναβ – 4Rαβ Rαβ + R2

8. (4+n)D Schwarzschild-Gauss-Bonnet black hole solution (Boulware, Dieser, ‘1986, R.Cai, ‘2003) • Metric representation: ds2 = - e2ν dt2 + e2α dr2 + r2 hij dxi dxj • Metric functions:

9. Mass and Temperature • Mass • Temperature

10. Hawking Temperature Twith GB/Twithout GB M/MPl Twith GB/Twithout GB M/MPl

11. “Toy model” (4+n)D Kerr-Gauss-Bonnet solution with one momentum (“degenerated solution”). Necessity: to compare with the usual Kerr one in the complete range of dimensions: N=5,…,11

12. “Degenerated” solution ds2 = - (du + dr)2 + dr2 + ρ2dθ2 + (r2 + a2) sin2θdφ2 + 2 a sin2θ dr dφ + β(r,θ) (du – a sin2θ dφ)2 +r2cos2θ (dx52 + sin2x5 (dx62 + sin2x6 (…dxN2)…) here β(r,θ) is the function to be found, ρ2 = r2 + a2 cos2θ N.Deruelle, Y.Morisawa, Class.Quant.Grav.22:933-938,2005, S.Alexeyev, N.Popov, A.Barrau, J.Grain, Journal of Physics: Conference Series 33, 343 (2006)

13. (UR) equation for β(r,θ) [h1(r) β + h0(r,θ)] (dβ/dr) + [g2(r,θ) β2 + g1(r,θ) β + g0 (r,θ)] = 0 For 6D case, for example h1 = 24 α r3 h0 = r ρ2 (r2 + ρ2) g2 = 4 α (3r4 + 6 r2 a2 cos2θ – a4 cos4θ) / ρ2 g1 = (r2 + ρ2) (2r2 + ρ2) g0 = Λ r2ρ4

14. Behavior at the infinity Λ = 0 β(r,θ)  μ/[rN-5 (r2 + a2 cos2θ)] + … Λ ≠ 0 β(r,θ)  C(N) Λ r4/ [r2 + a2 cos2θ] + …

15. Behavior at the horizon β(r,θ) = 1 + b1(θ) (r - rh) + b2(θ) (r – rh)2 + … For 6D case b1 = [4 α (3 rh4 + 6 rh2 a2 cos2θ – a4 cos4θ) (rh2 + a2 cos2θ)-1 + (2 rh2 + a2 cos2θ) (3 rh2 + a2 cos2θ) + Λ rh2 (rh2 + a2 cos2θ)2]/[24 α rh3 + rh (2 rh2 + a2 cos2θ)]

16. Usual form of metric ds2 = - dt2 (1 – β2) + dr2 [(r4(1 – β2) + a2 (r2 + β2a2cos4θ) / Δ2] + ρ2dθ2 – 2aβ2sin2θ dtdφ + dφ2 sin2θ [r2 + a2 + a2β2 sin2θ] + r2 cos2θ (dx5 + …) Δ = r2 + a2 - ρ2 β2 ρ2 =r2 + a2 cos2θ

17. Mass & angular momentum • Mass M = µ (N-2) AN-2/16πG where AN-1 = 2 πN/2/Γ(N/2) • Angular momentum Jyixi = 2 M ai/N  the same as in pure Kerr case

18. 6D plot of β=β(r,a ∙ cosθ) in asymptotically flat case (Λ=0), λ=1

19. 6D plot of β=β(r,a ∙ cosθ) when Λ≠ 0, λ=1

20. While considering “degenerated solution” there are no any new types of particular points, so, there is no principal difference from pure Kerr case all the difference will occur only in temperature and its consequences

21. Real angular momentum tensor

22. Number of angular momentums According to the existence of [Ns/2] Casimirs of SO(N) (Ns is the number of space dimensions) • For N=4 (Ns=3) there is 1 moment • For N=5 (Ns=4) there are 2 moments • For N=6 (Ns=5) there are 2 moments • For N=7 (Ns=6) there are 3 moments • For N=8 (Ns=7) there are 3 moments • For N=9 (Ns=8) there are 4 moments • For N=10 (Ns=9) there are 4 moments • For N=11 (Ns=10) there are 5 moments

23. 5D Kerr metric (complete version) ds2 = dt2 - dr2 - (r2+a2) sin2θ dφ1 - (r2+b2) cos2θ dφ2 – ρ2 dθ2 - 2 dr (a sin2θ dφ1 + b cos2θ dφ2) - β (dt – dr – a sin2θ dφ1 - b cos2θ dφ2) Where ρ2 = r2 + a2 cos2θ + b2 sin2θ, β = β(r, θ) is unknown function a, b - moments

24. θθ component A β’’ + B β’2 + C β’ + D β + E = 0 Where A = r ρ2 (4 αβ – ρ2) B = 4 α r ρ2 C = 2 [ 4 αβ (ρ2 - r2) – ρ2 (ρ2 + r2) ] D = 2 r (2 r2 – 3 ρ2) E = 2 r ρ4Λ

25. Solution manipulations This equation could be divided into 2 parts • A(r,ρ)β’’+B(r,ρ)β’+C(r,ρ)β+D(r,ρ,Λ)=Z(r,ρ,β) • E(r,ρ)(ββ’)’+F(r,ρ)(ββ’) =Z(r,ρ,β)

26. 5D solution

27. 6D metric ds2 = dt2 -dr2 –sin2ψ [(r2+a2)sin2θdφ12 +(r2+b2)cos2θdφ22] -(r2+a2cos2θ+b2sin2θ)sin2ψdθ2 - [r2+(a2sin2θ+b2cos2θ)cos2ψ]dψ2 -2drsin2ψ(asin2θdφ1+bcos2θdφ2) +2(b2 - a2)sinθcosθsinψcosψdθdψ -β(r,θ,ψ) [dt–dr–sin2ψ(asin2ψdφ1 +bcos2θdφ2)]2

28. Conclusions Taking into account 5D case one can see that in the general form of Kerr-Gauss-Bonnet solution there are no any new types of particular points, so, there is no principal difference from pure Kerr case all the difference will occur only in temperature and its consequences

29. Thank you for your kind attention! And for your questions!