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A double Myers-Perry black hole in five dimensions. Carlos A. R. Herdeiro, Carmen Rebelo, Miguel Zilhão and Miguel S. Costa. Centro de Física do Porto Faculdade de Ciência da Universidade do Porto.
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A double Myers-Perry black hole in five dimensions Carlos A. R. Herdeiro, Carmen Rebelo, Miguel Zilhão and Miguel S. Costa Centro de Física do Porto Faculdade de Ciência da Universidade do Porto Published in JHEP 0807:009,2008. (arXiv:0805.1206)
d = 4 • d = 5 Kerr BH Myers-Perry BH Black Ring Conical Sing. Multi-black hole solutions Emparan and Reall, arXiv:0801.3471 Double-Kerr BHs 0 – Motivation - The “phase” space of regular and asymptotically flat black objects is rather richer in five than in four dimensions. Stationary Vaccum Solutions (M, J) A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
Di-Rings Black Saturn Bicycling Rings 0 – Motivation - Why not study the Double Myers-Perry Solution? • We can generate it using the Inverse Scattering Method (ISM); • It might be of interest in studying spin-spin interactions in 5D general relativity. A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
Di-Rings Black Saturn Bicycling Rings - Overview: • How ISM works? • - Stationary and axisymmetric vacuum soluitons, • - One easy way to use it • The generation of a double Myers-Perry solution in 5D • Main Results • Conclusions A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
form a completely integrable system dressing matrix 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions - Stationary vaccum solutions with (D-3) angular killing vectors: The Einstein’s equations separate into two groups, one for the matrix Gab and the other for the metric factor f(ρ,z). λ=0 A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
form a completely integrable system 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions - Stationary vaccum solutions with (D-3) angular killing vectors: The Einstein’s equations separate into two groups, one for the matrix Gab and the other for the metric factor f(ρ,z). V.A. Belinsky and V.E.Zakharov; Sov.Phys.JETP 48(1978)985 and 50(1979)1 n-soliton dressing matrix → ISM • number of solitons we want to add, • associate (D-2)-component BZ vector λ=0 A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
4D Minkowski space: this potential is generate by an infinite rod of zero thickness and linear mass density ½ along ρ=0 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions - Stationary vaccum solutions with (D-3) angular killing vectors: can be characterized in terms of their rod structure.. (T.Harmark,Phys.Rev.D70:124002,2004) Rod Struture → “rod” sources of Ui along the axis. A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
soliton anti-soliton t φ ψ a0 t φ (ρ=0) z 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions - Stationary vaccum solutions with (D-3) angular killing vectors: can be characterized in terms of their rod structure.. (T.Harmark,Phys.Rev.D70:124002,2004) • 5D Minkowski space: • 4D Minkowski space: A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
f(ρ,z) • 5D Myesr-Perry BH: (1,Ωф,Ωψ) (0,1,0) Ut Uφ (0,0,1) • 5D Tangherlini BH: t t (1,0,0) φ φ ψ ψ (0,1,0) (0,0,1) a3 = ∞ a0 = -∞ a1 a2 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions • 4D Schwarzschild BH: t φ a1 a2 Rod Struture → rod intervals [ak-1,ak] → rod direction v(k) (T.Harmark,Phys.Rev.D70:124002,2004) A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
t G0 φ a1 a2 t G φ a1 a2 t GR φ a1 a2 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions n-soliton dressing matrix → ISM • number of solitons we want to add, • associate (D-2)-component BZ vector λ=0 • Adding: • anti-soliton, at z=a1 with BZ vector (1,0) • soliton, at z=a2 with BZ vector (1,0) det G ≠ - ρ2 But we could rescale G to have a physical solution! Schwarzschild BH A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
Removing: • anti-soliton, at z=a1 with BZ vector (1,0) • soliton, at z=a2 with BZ vector (1,0) GStatic adding... G0 • Adding: • anti-soliton, at z=a1 with BZ vector (1,b) • soliton, at z=a2 with BZ vector (1,c) t t t GStationary Kerr-NUT BH φ φ φ a1 a1 a1 a2 a2 a2 (1,Ωф) (2bNUT,1) (-2bNUT,1) One easy way → 1-GStatic 2-G0 (removing…) 3-Ψ0 4-GStationary(adding…) 5- f(ρ,z) A. A. Pomeransky; Phys.Rev.D73:044004,2006. 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions • - we have three degrees of freedom: a21,b,c • M, J, bNUT A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
1. Double Tangherlini BH a1 a2 a3 a4 a5 (1,0,0) (1,0,0) t (0,1,0) (0,1,0) φ (0,0,1) (0,0,1) ψ 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
3. 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1. Double Tangherlini BH (1,0,0) (1,0,0) t (0,1,0) (0,1,0) φ (0,0,1) (0,0,1) ψ a1 a2 a3 a4 a5 2. A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
Rod-Structure: (1,Ω1ф,0) (1,Ω2ф,0) t (h,1,0) (0,1,0) The solution has six parameters: φ (0,0,1) (0,0,1) - a21, a32 , a43, a54, b, c ψ a1 a2 a3 a4 a5 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 4. A Double Myers-Perry BH • We add: • anti-soliton, at z=a1 with BZ vector (1,b,0) • soliton, at z=a2 with BZ vector (1,0,0) • anti-soliton, at z=a4with BZ vector (1,c,0) • soliton, at z=a5 with BZ vector (1,0,0) two masses (M1,M2) two ang. momenta (J1ф,J2ф) distance ... A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
t φ DoubleTangherlini BH ψ a3 a5 a1 a1 a2 a3 a4 a5 Background geometry is asymptotically flat (1,0,0) (1,0,0) t (0,1,0) (0,1,0) φ (0,0,1) (0,0,1) ψ 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions - Conical Singularities • the background geometry is not flat space. A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
t φ ψ a3 a5 a1 a1 a2 a3 a4 a5 (1,0,0) (1,0,0) t (0,1,0) (0,1,0) φ (0,0,1) (0,0,1) ψ 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions - Conical Singularities • the background geometry is not flat space. DoubleTangherlini BH Conical escesses: A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
Background geometry: • 3 three fixed points • 2 conical singularities a1 a3 a5 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions - Conical Singularities A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
a1 a5 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions - Conical Singularities a2 a3 a4 A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions - Conical Singularities a1 a2 a3 a5 a4 …the introduction of rotaion could eliminate either of the conical singularities, but not both simultaneously! A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
- Torsion Singularity v1=(1,Ω1ф,0) v2=(1,Ω2ф,0) Spinning Rod v = (h,1,0) =Δaxis 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions - Conical Singularities a1 a2 a3 a4 a5 A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
Δaxis = 0 v1=(1,Ω1ф,0) v2=(1,Ω2ф,0) v = (0,1,0) 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions a1 a2 a3 a4 a5 …the requirement for either of the conical singularities to vanish is incompatiblewith the axis condition. A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
v1=(1,Ω1ф,0) v2=(1,Ω2ф,0) 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions - ADM mass a1 a2 a3 a4 a5 v = (h,1,0) A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
a1 a3 a5 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions • the solution is built upon a non-trivial background geometry with conical singularities; • the solution is built upon a non-trivial background geometry with conical singularities A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
a1 a2 a3 a3 a4 a5 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions • the solution is built upon a non-trivial background geometry with conical singularities; • the addition of rotation, to the black holes, brings a torsion singularity and changes both conical singularities; A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
a1 M1 Maxis a2 a3 a4 a5 M2 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions • the solution is built upon a non-trivial background geometry with conical singularities; • the addition of rotation, to the black holes, brings a torsion singularity and changes both conical singularities; • if Δaxis ≠ 0 then there is an extra contribution to the MADM; a3 A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
Double-Kerr BHs 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions • the solution is built upon a non-trivial background geometry with conical singularities; • the addition of rotation, to the black holes, brings a torsion singularity and changes both conical singularities; a1 • if Δaxis ≠ 0 then there is an extra contribution to the MADM; a2 • imposing Δaxis = 0, the torsion singularitydisappears but the conical singularities are still present. a3 a3 a4 a5 A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions • the solution is built upon a non-trivial background geometry with conical singularities; • the addition of rotation, to the black holes, brings a torsion singularity and changes both conical singularities; a1 • if Δaxis ≠ 0 then there is an extra contribution to the MADM; a2 • imposing Δaxis = 0, the torsion singularitydisappears but the conical singularities are still present. a3 a3 a4 a5 • It remains to be seen if, by including the second angular momentum parameter, such singularities can be removed. A double Myers-Perry black hole in 5D Salamanca, 15th September 2008
1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions • the solution is built upon a non-trivial background geometry with conical singularities; • the addition of rotation, to the black holes, brings a torsion singularity and changes both conical singularities; a1 • if Δaxis ≠ 0 then there is an extra contribution to the MADM; a2 • imposing Δaxis = 0, the torsion singularitydisappears but the conical singularities are still present. a3 a3 a4 a5 • It remains to be seen if, by including the second angular momentum parameter, such singularities can be removed. • Regarding spin-spin interactions, it would be interesting to have a physical interpretation of Δaxis, δф and δψin terms of the different forces and torques that play a role in this solution. A double Myers-Perry black hole in 5D Salamanca, 15th September 2008