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, Moscow3Ural State University, Ekaterinburg, Russia

4th International Seminar on High Energy Physics QUARKS'2006Repino, St.Petersburg, Russia, May 19-25, 2006

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

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)

Planck Energy shift

Planck energy in 4D representation

↓

1019 GeV

Fundamental Planck energy

↓

≈ 1 TeV

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

(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

Einstein-GB equations

R.Cai, ‘2003

Rµν - ½ gµνR - Λgµν

– α (½ gµνSGB – 2 RRµν + 4 RµγRγν

+ 4 RγδRγµ

δν – 2 RµγδλRν

γδν) = 0

SGB = Rμναβ Rμναβ – 4Rαβ Rαβ + R2

(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:

Mass and Temperature

Mass

Temperature

Hawking Temperature

M/MPl

M/MPl

Twith GB/Twithout GB

Twith GB/Twithout GB

“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

“Degenerated” solution

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)

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 (dx6

2 + sin2x6 (…dxN2)…)

(UR) equation for β(r,θ)

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 r2 ρ4

[h1(r) β + h0(r,θ)] (dβ/dr)

+ [g2(r,θ) β2 + g1(r,θ) β + g0 (r,θ)] = 0

Λ = 0

β(r,θ) μ /[rN-5 (r2 + a2 cos2θ)] + …

Λ ≠ 0

β(r,θ) C(N) Λ r4 / [r2 + a2 cos2θ] + …

Behavior at the infinity

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 rh

2 a2 cos2θ – a4 cos4θ) (rh2 + a2 cos2θ)-1

+ (2 rh2 + a2 cos2θ) (3 rh

2 + a2 cos2θ)

+ Λ rh2 (rh

2 + a2 cos2θ)2] / [24 α rh3 + rh (2 rh

2 + a2 cos2θ)]

Usual form of metricds2 = - 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

Mass & angular momentum

Mass M = µ (N-2) AN-2/16πG

where AN-1 = 2 πN/2/Γ(N/2)

Angular momentum Jyix

i = 2 M ai/N

the same as in pure Kerr case

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

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

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

Real angular momentum tensor

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

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)Whereρ2 = r2 + a2 cos2θ + b2 sin2θ,β = β(r, θ) is unknown functiona, b - moments

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 ρ4 Λ

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)

5D solution

6D metric

ds2 = dt2 - dr2 – sin2ψ [(r2 + a2) sin2θ dφ12

+ (r2 + b2) cos2θ dφ22]

- (r2 + a2 cos2θ + b2 sin2θ) sin2ψ dθ2 - [r2 + (a2 sin2θ + b2 cos2θ) cos2ψ] dψ2 - 2 dr sin2ψ (a sin2θ dφ1 + b cos2θ dφ2) + 2 (b2 - a2) sinθ cosθ sinψ cosψ dθ dψ - β(r,θ,ψ) [dt – dr –sin2ψ (a sin2ψ dφ1 + b cos2θ dφ2)]2

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

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