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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

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Triangular solution to the general relativistic three-body problem

Kei Yamada Hirosaki University

with Ichita-san & Asada-san

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Contents

  • Introduction
  • Equilateral triangular solution in GR
  • Triangular solution in GR: general masses
  • Summary
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SLIDE 3
  • Introduction
  • Equilateral triangular solution in GR
  • Triangular solution in GR: general masses
  • Summary

Contents

  • Introduction
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Three-body problem

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

  • J. L. Lagrange
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Equilateral triangular solution

C

C : the center of mass

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60 60 L1 L2 L3 L4 L5

Lagrange points

Jupiter Sun

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60 60 L1 L2 L3 L4 L5

Lagrange points

Trojan asteroids Jupiter Sun

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60 60 L1 L2 L3 L4 L5

Lagrange points

Trojan asteroids Jupiter Sun

What happens in the general relativity (GR)?

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GR effects of Solar system

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

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EIH equation of motion

Einstein-Infeld-Hoffman (EIH) equation of motion for N bodies We look for an equilibrium solution in a circular motion

Newtonian term GR correction by mass GR correction by velocity

mK d2rK dt2 = X

A6=K

rAK GmAmK r3

AK

" 1 − 4 X

B6=K

GmB c2rBK − X

C6=A

GmC c2rCA ✓ 1 − rAK · rCA 2r2

CA

◆ + ⇣vK c ⌘2 + 2 ⇣vA c ⌘2 − 4 ⇣vA c ⌘ · ⇣vK c ⌘ − 3 2 vA

c

  • · rAK

rAK !2 # − X

A6=K

h (vA c ) − (vK c ) i GmAmK r3

AK

rAK · h 3 ⇣vA c ⌘ − 4 ⇣vK c ⌘i + 7 2 X

A6=K

X

C6=A

rCA GmCmK r3

CA

GmA c2rAK

Triple product

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

Contents

  • Equilateral triangular solution in GR
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Equilateral triangular configuration

Equilateral triangle

rIJ = a (I, J = 1, 2, 3)

a a a r1 r2 r3

m2 m3 m1

v3 v2 v1

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Center of mass at 1PN

In general, this is different from the Newtonian one

rG = P

A

νArA " 1 + 1 2 @ ⇣vA c ⌘2 − X

B6=A

GmB c2rAB 1 Aλ # P

C

νC " 1 + 1 2 @ ⇣vC c ⌘2 − X

D6=C

GmD c2rCD 1 Aλ # ,

λ ⌘ GM c2a ⌧ 1

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Center of mass at 1PN

In general, this is different from the Newtonian one

rG = P

A

νArA " 1 + 1 2 @ ⇣vA c ⌘2 − X

B6=A

GmB c2rAB 1 Aλ # P

C

νC " 1 + 1 2 @ ⇣vC c ⌘2 − X

D6=C

GmD c2rCD 1 Aλ #

In this case, this agrees with the Newtonian case But

, λ ⌘ GM c2a ⌧ 1

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Equilateral triangular solution at the 1PN

At 1PN order, EOM for becomes m1 n1 n⊥1 is normal to n1 ≡ r1/|r1|, n⊥1 ≡ v1/|v1|,

ω : angular velocity, νI ≡ mI/M, M =

  • I

mI (I = 1, 2, 3)

gP N1 = λ 16(ν2

2 + ν2ν3 + ν2 3)

M a3 × h 48(ν2

2 + ν2ν3 + ν2 3) − 2(8ν3 2 + 7ν2 2ν3 + 7ν2ν2 3 + 8ν3 3)

+ (16ν4

2 + 41ν3 2ν3 + 84ν2 2ν2 3 + 41ν2ν3 3 + 16ν4 3)

i

−ω2n1 = −M a3 n1 + gP N1n1 + √ 3 16 M a3 ν2ν3(ν2 − ν3) ν2

2 + ν2ν3 + ν2 3

⇥ 6 + 9

  • ν2 + ν3

⇤ λn⊥1

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Equilateral triangular solution at the 1PN

At 1PN order, EOM for becomes m1 n1 n⊥1 is normal to n1 ≡ r1/|r1|, n⊥1 ≡ v1/|v1|,

ω : angular velocity, νI ≡ mI/M, M =

  • I

mI (I = 1, 2, 3)

gP N1 = λ 16(ν2

2 + ν2ν3 + ν2 3)

M a3 × h 48(ν2

2 + ν2ν3 + ν2 3) − 2(8ν3 2 + 7ν2 2ν3 + 7ν2ν2 3 + 8ν3 3)

+ (16ν4

2 + 41ν3 2ν3 + 84ν2 2ν2 3 + 41ν2ν3 3 + 16ν4 3)

i

−ω2n1 = −M a3 n1 + gP N1n1 + √ 3 16 M a3 ν2ν3(ν2 − ν3) ν2

2 + ν2ν3 + ν2 3

⇥ 6 + 9

  • ν2 + ν3

⇤ λn⊥1

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Equilateral triangular solution at the 1PN

At 1PN order, EOM for becomes m1 n1 n⊥1 is normal to n1 ≡ r1/|r1|, n⊥1 ≡ v1/|v1|,

ω : angular velocity, νI ≡ mI/M, M =

  • I

mI (I = 1, 2, 3)

gP N1 = λ 16(ν2

2 + ν2ν3 + ν2 3)

M a3 × h 48(ν2

2 + ν2ν3 + ν2 3) − 2(8ν3 2 + 7ν2 2ν3 + 7ν2ν2 3 + 8ν3 3)

+ (16ν4

2 + 41ν3 2ν3 + 84ν2 2ν2 3 + 41ν2ν3 3 + 16ν4 3)

i

−ω2n1 = −M a3 n1 + gP N1n1 + √ 3 16 M a3 ν2ν3(ν2 − ν3) ν2

2 + ν2ν3 + ν2 3

⇥ 6 + 9

  • ν2 + ν3

⇤ λn⊥1

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In only 2 cases, bodies satisfy EOM;

  • mass ratio 1 : 1 : 1 ● mass ratio 0 : 0 : 1

Equilateral triangular solution at the 1PN

−ω2n1 = −M a3 n1 + gP N1n1 + √ 3 16 M a3 ν2ν3(ν2 − ν3) ν2

2 + ν2ν3 + ν2 3

⇥ 6 + 9

  • ν2 + ν3

⇤ λn⊥1

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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

Equilateral triangular solution at the 1PN

[Ichita, KY & Asada, PRD 83, 084026 (2011)] −ω2n1 = −M a3 n1 + gP N1n1 + √ 3 16 M a3 ν2ν3(ν2 − ν3) ν2

2 + ν2ν3 + ν2 3

⇥ 6 + 9

  • ν2 + ν3

⇤ λn⊥1

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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)]

?

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SLIDE 21
  • Introduction
  • Equilateral triangular solution in GR
  • Triangular solution in GR: general masses
  • Summary

Contents

  • Triangular solution in GR: general masses
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m3

Corrections of distance

a(1 + ε31) a(1 + ε12) a(1 + ε23) m2 m1

PN inequilateral triangle

rIJ = a(1 + εIJ), εIJ = O(1PN)

ω = ωN

We can ignore the 1PN correction to the center of mass

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Triangular solution at the 1PN

EOM for becomes m1 ω = ωN

−ω2r1 = − ω2

Nr1

+ ν2 ✓ −3 + ν1ν2 + ν2ν3 + ν3ν1 − 3 8ν3[5 − 3(ν1 + ν2)] ◆ λr21 + ν3 ✓ −3 + ν1ν2 + ν2ν3 + ν3ν1 − 3 8ν2[5 − 3(ν3 + ν1)] ◆ λr31 − 3(ν2ε12r21 + ν3ε31r31)

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Triangular solution at the 1PN

EOM for becomes m1 ω = ωN = 0

−ω2r1 = − ω2

Nr1

+ ν2 ✓ −3 + ν1ν2 + ν2ν3 + ν3ν1 − 3 8ν3[5 − 3(ν1 + ν2)] ◆ λr21 + ν3 ✓ −3 + ν1ν2 + ν2ν3 + ν3ν1 − 3 8ν2[5 − 3(ν3 + ν1)] ◆ λr31 − 3(ν2ε12r21 + ν3ε31r31)

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Triangular solution at the 1PN

EOM for becomes m1 ω = ωN = 0

−ω2r1 = − ω2

Nr1

+ ν2 ✓ −3 + ν1ν2 + ν2ν3 + ν3ν1 − 3 8ν3[5 − 3(ν1 + ν2)] ◆ λr21 + ν3 ✓ −3 + ν1ν2 + ν2ν3 + ν3ν1 − 3 8ν2[5 − 3(ν3 + ν1)] ◆ λr31 − 3(ν2ε12r21 + ν3ε31r31)

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Triangular solution at the 1PN

Triangular solution for the arbitrary mass ratio at 1PN [KY & Asada, submitted] As a result, we could uniquely express εIJ

ε12 = −  1 − 1 3(ν1ν2 + ν2ν3 + ν3ν1) + 1 8ν3[5 − 3(ν1 + ν2)]

  • λ,

ε23 = −  1 − 1 3(ν1ν2 + ν2ν3 + ν3ν1) + 1 8ν1[5 − 3(ν2 + ν3)]

  • λ,

ε31 = −  1 − 1 3(ν1ν2 + ν2ν3 + ν3ν1) + 1 8ν2[5 − 3(ν1 + ν3)]

  • λ.
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Application for Solar system

Planet Sun-Planet Sun-L4 (L5) Planet-L4 (L5) Earth

  • 1477
  • 1477
  • 1477 -923

Jupiter

  • 1477
  • 1477
  • 1477 -922

Corrections for L4 (L5) of Solar system [m] The sign + denotes increase of distance

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

Contents

  • Summary
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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
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Ongoing & Future works

  • The Stability
  • The Gravitational wave
  • Higher order PN approximation
  • An elliptical motion
  • Four (or more) body systems
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Thank you for your attention