A double Myers-Perry black hole in five dimensions Published in JHEP 0807:009,2008. (arXiv:0805.1206) Carlos A. R. Herdeiro, Carmen Rebelo, Miguel Zilhão.

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Presentation transcript:

A double Myers-Perry black hole in five dimensions Published in JHEP 0807:009,2008. (arXiv: ) Carlos A. R. Herdeiro, Carmen Rebelo, Miguel Zilhão and Miguel S. Costa Centro de Física do Porto Faculdade de Ciência da Universidade do Porto

A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September – Motivation 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) d = 4 Kerr BH d = 5 Myers-Perry BH Black Ring Double-Kerr BHs Conical Sing. Multi-black hole solutions Emparan and Reall, arXiv:

Bicycling Rings 0 – Motivation 0 – Motivation Di-Rings Black Saturn A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 Double Myers-Perry - Why not study the Double Myers-Perry Solution? 1.We can generate it using the Inverse Scattering Method (ISM); 2.It might be of interest in studying spin-spin interactions in 5D general relativity.

Bicycling Rings Di-Rings Black Saturn A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September Overview: 1.How ISM works? - Stationary and axisymmetric vacuum soluitons, - One easy way to use it 2.The generation of a double Myers-Perry solution in 5D 3.Main Results 4.Conclusions

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions G ab f(ρ,z). The Einstein’s equations separate into two groups, one for the matrix G ab and the other for the metric factor f(ρ,z). form a completely integrable system A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September Stationary vaccum solutions with (D-3) angular killing vectors: λ=0 dressing matrix

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions G ab f(ρ,z). The Einstein’s equations separate into two groups, one for the matrix G ab and the other for the metric factor f(ρ,z). form a completely integrable system A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September Stationary vaccum solutions with (D-3) angular killing vectors: λ=0 n-soliton dressing matrix → ISM i.number of solitons we want to add, ii.associate (D-2)-component BZ vector V.A. Belinsky and V.E.Zakharov; Sov.Phys.JETP 48(1978)985 and 50(1979)1

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September 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) 4D Minkowski space: rod this potential is generate by an infinite rod of zero thickness and linear mass density ½ along ρ=0 Rod Struture → “rod” sources of U i along the axis.

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September 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) 4D Minkowski space: 5D Minkowski space:tφ z (ρ=0) tφ ψ a0a0 soliton anti-soliton

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 t φ 4D Schwarzschild BH: a1a1 a2a2 f(ρ,z) Rod Struture → rod intervals [ a k-1,a k ] → rod direction v (k) (T.Harmark,Phys.Rev.D70:124002,2004) a 0 = -∞ t φ ψ 5D Tangherlini BH: a1a1 a2a2 a 3 = ∞ (0,0,1) (1,0,0) (0,1,0) UφUφUφUφ UtUtUtUt 5D Myesr-Perry BH:t φ ψ (0,0,1) (1,Ω ф, Ω ψ ) (0,1,0)

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 λ=0 n-soliton dressing matrix → ISM i.number of solitons we want to add, ii.associate (D-2)-component BZ vector Adding: anti-soliton, at z=a 1 with BZ vector (1,0) soliton, at z=a 2 with BZ vector (1,0) G0G0tφ a1a1 a2a2 t φ a1a1 a2a2 G det G ≠ - ρ 2 G But we could rescale G to have a physical solution!tφ a1a1 a2a2 GRGR Schwarzschild BH

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September we have three degrees of freedom: a 21,b,c - M, J, b NUT G Statictφ a1a1 a2a2 Removing: anti-soliton, at z=a 1 with BZ vector (1,0) soliton, at z=a 2 with BZ vector (1,0) G0G0 t φ a1a1 a2a2 adding... Adding: anti-soliton, at z=a 1 with BZ vector (1,b) soliton, at z=a 2 with BZ vector (1,c) (1,Ω ф ) t φ a1a1 a2a2 (2b NUT,1) (-2b NUT,1) Kerr-NUT BH G Stationary One easy way → 1-G Static 2-G 0 (removing…) 3-Ψ 0 4-G Stationary (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 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September Double Tangherlini BH a1a1 a2a2 a3a3 a4a4 a5a5 t φ ψ (0,0,1) (1,0,0) (0,1,0) (0,0,1) (0,1,0) (1,0,0)

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 (1,0,0) t φ ψ (0,0,1) (0,1,0) (0,0,1) (0,1,0) a3a Double Tangherlini BH a1a1 a2a2 a4a4 a5a53.

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September A Double Myers-Perry BH We add: anti-soliton, at z=a 1 with BZ vector (1,b,0) soliton, at z=a 2 with BZ vector (1,0,0) anti-soliton, at z=a 4 with BZ vector (1,c,0) soliton, at z=a 5 with BZ vector (1,0,0) a1a1 a2a2 a3a3 a4a4 a5a5 (1,Ω 1 ф,0) (1,Ω 2 ф,0) t φ ψ (0,0,1) (0,1,0) (0,0,1) (h,1,0) Rod-Structure: The solution has six parameters: - a 21, a 32, a 43, a 54, b, c two masses (M 1,M 2 ) two ang. momenta (J 1 ф,J 2 ф ) distance...

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September Conical Singularities the background geometry is not flat space. a3a3 a5a5 t φ ψ a1a1 a1a1 a2a2 a3a3 a4a4 a5a5 t φ ψ (0,0,1) (1,0,0) (0,1,0) (0,0,1) (0,1,0) (1,0,0) DoubleTangherlini BH Background geometry is asymptotically flat

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September Conical Singularities the background geometry is not flat space. a1a1 a2a2 a3a3 a4a4 a5a5 t φ ψ (0,0,1) (1,0,0) (0,1,0) (0,0,1) (0,1,0) (1,0,0) a3a3 a5a5 t φ ψ a1a1 DoubleTangherlini BH Conical escesses:

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 Background geometry: - Conical Singularities a1a1 a3a3 a5a5 3 three fixed points 2 conical singularities

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September Conical Singularities a2a2 a4a4 a1a1 a5a5 a3a3

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September Conical Singularities a2a2 a3a3 a4a4 a1a1 a5a5 …the introduction of rotaion could eliminate either of the conical singularities, but not both simultaneously!

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 v 1 =(1,Ω 1 ф,0) v 2 =(1,Ω 2 ф,0) - Conical Singularities a2a2 a3a3 a4a4 a1a1 a5a5 = Δ axis - Torsion Singularity Spinning Rod Spinning Rod v = (h,1,0)

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 v 1 =(1,Ω 1 ф,0) v 2 =(1,Ω 2 ф,0) a2a2 a3a3 a4a4 a1a1 a5a5 Δ axis = 0Δ axis = 0 v = (0,1,0) incompatible …the requirement for either of the conical singularities to vanish is incompatible with the axis condition.

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 v 1 =(1,Ω 1 ф,0) v 2 =(1,Ω 2 ф,0) - ADM mass a2a2 a3a3 a4a4 a1a1 a5a5 v = (h,1,0)

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 a1a1 a3a3 a5a5 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

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 a3a3 a2a2 a3a3 a4a4 a1a1 a5a5 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;

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 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; a3a3 a2a2 a3a3 a4a4 a1a1 a5a5 if Δ axis ≠ 0 then there is an extra contribution to the M ADM; M1M1 M2M2 M axis

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 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; a3a3 a2a2 a3a3 a4a4 a1a1 a5a5 if Δ axis ≠ 0 then there is an extra contribution to the M ADM; Double-Kerr BHs imposing Δ axis = 0, the torsion singularity disappears but the conical singularities are still present.

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 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; a3a3 a2a2 a3a3 a4a4 a1a1 a5a5 if Δ axis ≠ 0 then there is an extra contribution to the M ADM; imposing Δ axis = 0, the torsion singularity disappears but the conical singularities are still present. It remains to be seen if, by including the second angular momentum parameter, such singularities can be removed.

1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions 1 – How ISM works? 2 -Double Myers-Perry 3 - Results 4 - Conclusions A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 A double Myers-Perry black hole in 5D Salamanca, 15 th September 2008 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; a3a3 a2a2 a3a3 a4a4 a1a1 a5a5 if Δ axis ≠ 0 then there is an extra contribution to the M ADM; imposing Δ axis = 0, the torsion singularity disappears but the conical singularities are still present. 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 Δ axis, δ ф and δ ψ in terms of the different forces and torques that play a role in this solution.