Triangular solution to the general relativistic three-body problem - PowerPoint PPT Presentation

Triangular solution to the general relativistic three-body problem Kei Yamada Hirosaki University with Ichita-san & Asada-san

Contents • Introduction • Equilateral triangular solution in GR • Triangular solution in GR: general masses • Summary

Contents • Introduction • Introduction • Equilateral triangular solution in GR • Triangular solution in GR: general masses • Summary

Three-body problem Particular solutions to the three-body problem Euler’s collinear solution (1765) & J. L. Lagrange Lagrange’s equilateral triangular solution (1772)

Equilateral triangular solution C : the center of mass C

Lagrange points L 4 60 � Jupiter L 3 L 1 L 2 Sun 60 � L 5

Lagrange points L 4 60 � Jupiter Trojan asteroids L 3 L 1 L 2 Sun 60 � L 5

Lagrange points L 4 60 � Jupiter Trojan asteroids L 3 L 1 L 2 Sun 60 � L 5 What happens in the general relativity (GR)?

GR effects of Solar system Two-body systems: (e.g. the perihelion precession of Mercury) Three-body systems: ? It is interesting as a new test of GR

EIH equation of motion Einstein-Infeld-Hoffman (EIH) equation of motion for N bodies Newtonian term GR correction by mass " d 2 r K ✓ ◆ Gm A m K Gm B Gm C 1 − r AK · r CA X X X = 1 − 4 m K r AK − r 3 2 r 2 dt 2 c 2 r BK c 2 r CA AK CA A 6 = K B 6 = K C 6 = A ! 2 # � v A � − 3 ⌘ 2 ⌘ 2 · r AK ⇣ v K ⇣ v A ⇣ v A ⌘ ⇣ v K ⌘ c + + 2 − 4 · 2 c c c c r AK GR correction by velocity i Gm A m K h ( v A c ) − ( v K h ⇣ v A ⌘ ⇣ v K ⌘i X c ) 3 − 4 r AK · − r 3 c c AK A 6 = K + 7 Gm C m K Gm A Triple product X X r CA r 3 2 c 2 r AK CA A 6 = K C 6 = A We look for an equilibrium solution in a circular motion

Contents • Introduction • Equilateral triangular solution in GR • Equilateral triangular solution in GR • Triangular solution in GR: general masses • Summary

Equilateral triangular configuration v 3 m 3 Equilateral triangle r IJ = a ( I, J = 1 , 2 , 3) r 3 a a v 2 r 2 r 1 m 1 m 2 a v 1

Center of mass at 1PN 0 1 " # ⇣ v A ⌘ 2 Gm B 1 + 1 X P ν A r A A λ @ − c 2 r AB c 2 A B 6 = A λ ⌘ GM r G = # , c 2 a ⌧ 1 0 1 " ⇣ v C Gm D 1 + 1 ⌘ 2 X P ν C A λ − @ c 2 r CD c 2 C D 6 = C In general, this is different from the Newtonian one

Center of mass at 1PN 0 1 " # ⇣ v A ⌘ 2 Gm B 1 + 1 X P ν A r A A λ @ − c 2 r AB c 2 A B 6 = A λ ⌘ GM r G = , c 2 a ⌧ 1 0 1 " # ⇣ v C Gm D 1 + 1 ⌘ 2 X P ν C A λ − @ c 2 r CD c 2 C D 6 = C In general, this is different from the Newtonian one But In this case, this agrees with the Newtonian case

Equilateral triangular solution at the 1PN At 1PN order, EOM for becomes m 1 − ω 2 n 1 = − M a 3 n 1 + g P N 1 n 1 √ 3 ν 2 ν 3 ( ν 2 − ν 3 ) M ⇥ � �⇤ + 6 + 9 ν 2 + ν 3 λ n ⊥ 1 ν 2 2 + ν 2 ν 3 + ν 2 16 a 3 3 λ M g P N 1 = 16( ν 2 2 + ν 2 ν 3 + ν 2 3 ) a 3 h 48( ν 2 2 + ν 2 ν 3 + ν 2 3 ) − 2(8 ν 3 2 + 7 ν 2 2 ν 3 + 7 ν 2 ν 2 3 + 8 ν 3 3 ) × i + (16 ν 4 2 + 41 ν 3 2 ν 3 + 84 ν 2 2 ν 2 3 + 41 ν 2 ν 3 3 + 16 ν 4 3 ) � ω : angular velocity, ν I ≡ m I /M, M = m I ( I = 1 , 2 , 3) I n ⊥ 1 is normal to n 1 ≡ r 1 / | r 1 | , n ⊥ 1 ≡ v 1 / | v 1 | , n 1

Equilateral triangular solution at the 1PN At 1PN order, EOM for becomes m 1 − ω 2 n 1 = − M a 3 n 1 + g P N 1 n 1 √ 3 ν 2 ν 3 ( ν 2 − ν 3 ) M ⇥ � �⇤ + 6 + 9 ν 2 + ν 3 λ n ⊥ 1 ν 2 2 + ν 2 ν 3 + ν 2 16 a 3 3 λ M g P N 1 = 16( ν 2 2 + ν 2 ν 3 + ν 2 3 ) a 3 h 48( ν 2 2 + ν 2 ν 3 + ν 2 3 ) − 2(8 ν 3 2 + 7 ν 2 2 ν 3 + 7 ν 2 ν 2 3 + 8 ν 3 3 ) × i + (16 ν 4 2 + 41 ν 3 2 ν 3 + 84 ν 2 2 ν 2 3 + 41 ν 2 ν 3 3 + 16 ν 4 3 ) � ω : angular velocity, ν I ≡ m I /M, M = m I ( I = 1 , 2 , 3) I n ⊥ 1 is normal to n 1 ≡ r 1 / | r 1 | , n ⊥ 1 ≡ v 1 / | v 1 | , n 1

Equilateral triangular solution at the 1PN At 1PN order, EOM for becomes m 1 − ω 2 n 1 = − M a 3 n 1 + g P N 1 n 1 √ 3 ν 2 ν 3 ( ν 2 − ν 3 ) M ⇥ � �⇤ + 6 + 9 ν 2 + ν 3 λ n ⊥ 1 ν 2 2 + ν 2 ν 3 + ν 2 16 a 3 3 λ M g P N 1 = 16( ν 2 2 + ν 2 ν 3 + ν 2 3 ) a 3 h 48( ν 2 2 + ν 2 ν 3 + ν 2 3 ) − 2(8 ν 3 2 + 7 ν 2 2 ν 3 + 7 ν 2 ν 2 3 + 8 ν 3 3 ) × i + (16 ν 4 2 + 41 ν 3 2 ν 3 + 84 ν 2 2 ν 2 3 + 41 ν 2 ν 3 3 + 16 ν 4 3 ) � ω : angular velocity, ν I ≡ m I /M, M = m I ( I = 1 , 2 , 3) I n ⊥ 1 is normal to n 1 ≡ r 1 / | r 1 | , n ⊥ 1 ≡ v 1 / | v 1 | , n 1

Equilateral triangular solution at the 1PN − ω 2 n 1 = − M a 3 n 1 + g P N 1 n 1 √ 3 ν 2 ν 3 ( ν 2 − ν 3 ) M ⇥ � �⇤ + 6 + 9 ν 2 + ν 3 λ n ⊥ 1 ν 2 2 + ν 2 ν 3 + ν 2 16 a 3 3 In only 2 cases, bodies satisfy EOM; ● mass ratio 1 : 1 : 1 ● mass ratio 0 : 0 : 1

Equilateral triangular solution at the 1PN − ω 2 n 1 = − M a 3 n 1 + g P N 1 n 1 √ 3 ν 2 ν 3 ( ν 2 − ν 3 ) M ⇥ � �⇤ + 6 + 9 ν 2 + ν 3 λ n ⊥ 1 ν 2 2 + ν 2 ν 3 + ν 2 16 a 3 3 In only 2 cases, bodies satisfy EOM; ● mass ratio 1 : 1 : 1 ● mass ratio 0 : 0 : 1 This solution does not always exist in GR [Ichita, KY & Asada, PRD 83 , 084026 (2011)]

Equilateral triangular solution at the 1PN For the arbitrary mass ratio, a solution exists? ? cf. [Krefetz, Astron. J. 72 , 471 (1967)] for restricted 3-body problem, used by [Seto & Muto, PRD 81 , 103004 (2010)]

Contents • Introduction • Equilateral triangular solution in GR • Triangular solution in GR: general masses • Triangular solution in GR: general masses • Summary

Corrections of distance m 3 PN inequilateral triangle r IJ = a (1 + ε IJ ) , ε IJ = O(1PN) ω = ω N a (1 + ε 23 ) a (1 + ε 31 ) m 2 m 1 a (1 + ε 12 ) We can ignore the 1PN correction to the center of mass

Triangular solution at the 1PN EOM for becomes m 1 − ω 2 r 1 = − ω 2 N r 1 ✓ ◆ − 3 + ν 1 ν 2 + ν 2 ν 3 + ν 3 ν 1 − 3 + ν 2 8 ν 3 [5 − 3( ν 1 + ν 2 )] λ r 21 ✓ ◆ − 3 + ν 1 ν 2 + ν 2 ν 3 + ν 3 ν 1 − 3 + ν 3 8 ν 2 [5 − 3( ν 3 + ν 1 )] λ r 31 − 3( ν 2 ε 12 r 21 + ν 3 ε 31 r 31 ) ω = ω N

Triangular solution at the 1PN EOM for becomes m 1 − ω 2 r 1 = − ω 2 N r 1 ✓ ◆ − 3 + ν 1 ν 2 + ν 2 ν 3 + ν 3 ν 1 − 3 + ν 2 8 ν 3 [5 − 3( ν 1 + ν 2 )] λ r 21 ✓ ◆ − 3 + ν 1 ν 2 + ν 2 ν 3 + ν 3 ν 1 − 3 + ν 3 8 ν 2 [5 − 3( ν 3 + ν 1 )] λ r 31 − 3( ν 2 ε 12 r 21 + ν 3 ε 31 r 31 ) = 0 ω = ω N

Triangular solution at the 1PN EOM for becomes m 1 − ω 2 r 1 = − ω 2 N r 1 ✓ ◆ − 3 + ν 1 ν 2 + ν 2 ν 3 + ν 3 ν 1 − 3 + ν 2 8 ν 3 [5 − 3( ν 1 + ν 2 )] λ r 21 ✓ ◆ − 3 + ν 1 ν 2 + ν 2 ν 3 + ν 3 ν 1 − 3 + ν 3 8 ν 2 [5 − 3( ν 3 + ν 1 )] λ r 31 − 3( ν 2 ε 12 r 21 + ν 3 ε 31 r 31 ) = 0 ω = ω N

Triangular solution at the 1PN As a result, we could uniquely express ε IJ  1 − 1 3( ν 1 ν 2 + ν 2 ν 3 + ν 3 ν 1 ) + 1 � ε 12 = − 8 ν 3 [5 − 3( ν 1 + ν 2 )] λ ,  � 1 − 1 3( ν 1 ν 2 + ν 2 ν 3 + ν 3 ν 1 ) + 1 ε 23 = − 8 ν 1 [5 − 3( ν 2 + ν 3 )] λ ,  � 1 − 1 3( ν 1 ν 2 + ν 2 ν 3 + ν 3 ν 1 ) + 1 ε 31 = − 8 ν 2 [5 − 3( ν 1 + ν 3 )] λ . Triangular solution for the arbitrary mass ratio at 1PN [KY & Asada, submitted]

Application for Solar system Corrections for L4 (L5) of Solar system [m] Planet Sun-Planet Sun-L4 (L5) Planet-L4 (L5) Earth -1477 -1477 -1477 -923 Jupiter -1477 -1477 -1477 -922 The sign + denotes increase of distance

Contents • Introduction • Equilateral triangular solution in GR • Triangular solution in GR: general masses • Summary • Summary

Summary • We found a triangular solution at the 1PN order • The PN triangle is smaller than the Newtonian one (for same mass ratio), and changed from an equilateral triangle • This solution may also be applied to near SMBHs and compact binaries • Future observations are needed

Ongoing & Future works • The Stability • The Gravitational wave • Higher order PN approximation • An elliptical motion • Four (or more) body systems

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