On solving post-Newtonian accurate Kepler Equation

On solving post-Newtonian accurate KeplerEquation

Yannick Boetzel

Physik-Institut der Universitat Zurich

In collaboration with: A. Susobhanan, A. Gopakumar, A. Klein, P. Jetzer

La Thuile, 30.03.2017

Yannick Boetzel Rencontres de Moriond On solving PN accurate Kepler Equation

Eccentric GW templates

We work on providing accurate & efficient prescriptions for h+,×(t)associated with compact binaries merging along eccentric orbits



We work on providing accurate & efficient prescriptions for h+,×(t)associated with compact binaries merging along eccentric orbits

h+,×(t) = −G m η

c2 R ′x∞∑p=0

[Cp+,× cos(pl) + Sp+,× sin(pl)

]

C1+ = s2i

(−e +

e3

8− e5

192+

e7

9216

)+ c2β(1 + c2i )

(−3e

2+

2e3

3− 37e5

768+

11e7

7680

)S1+ = s2β(1 + c2i )

(−3e

2+

23e3

24+

19e5

256+

371e7

5120

)



h+,×(r , r , φ, φ)

Invoke quasi-Keplerian solutionto the conservative dynamics ⇓

(r , r , φ, φ) ⇒ κi (x , et , u(t),m, η)

Explicit expression for u(t) bysolving Kepler equation ⇓

h+,× (κi (x , et , u(t),m, η)) ⇒ h+,×(t) for conservative dynamics

Impose GW emission inducedvariations in x and et

⇓h+,× (κi (x(t), et(t), u(t),m, η)) ⇒ h+,×(t) with radiation reaction

⇓h+,×(t) for binaries inspiralling in PN-accurate eccentric orbits


Classical Kepler problem

Equation of motion

d2~r

dt2= −Gm

r2r



Keplerian parametrization

r = a (1− e cos u)

φ− φ0 = v(u)

tanv

2=

√1 + e

1− etan

u

2

where v is called true anomaly and u iscalled eccentric anomaly



Keplerian parametrization

r = a (1− e cos u)

φ− φ0 = v(u)

Classical Kepler Equation

l = u − e sin u

where l = n (t − t0) is called mean anomaly


Solving the classical Kepler Equation

Fourier series

u(l)− l =∞∑s=1

As sin(sl)

As =2

π

∫ π

0

(u(l)− l) sin(sl)dl

=2

sπ

∫ π

0

cos(sl)du

=2

s

{1

π

∫ π

0

cos(su − se sin u)du

}=

2

sJs(se)


Solving the classical Kepler Equation

Solution to the classical Kepler Equation

u = l +∞∑s=1

2

sJs(se) sin(sl)


Relativistic Kepler problem

Equation of motion

d2~r

dt2= −Gm

r2r + A r + B v

A = A1 + A2 + A2.5 + A3 + A3.5 + ...

B = B1 + B2 + B2.5 + B3 + B3.5 + ...

A’s & B’s are functions of r , r , φ, φ, m and η

Radiation reaction terms (2.5PN & 3.5PN) cause orbit to shrink

There exists a quasi-Keplerian parametrization to the 3PNconservative dynamics



Quasi-Keplerian parametrization

r = ar (1− er cos u)

φ− φ0 = (1 + k)v + (f4φ + f6φ) sin(2v) + (g4φ + g6φ) sin(3v)

+ i6φ sin(4v) + h6φ sin(5v)

tanv

2=

√1 + eφ1− eφ

tanu

2

The coefficients a, er , eφ, k , f , g , i , h are PN-accurate functions of x ,et and η = µ/m

k represents precession of periastron



3PN-accurate Kepler Equation

l = u − et sin u + (g4t + g6t) (v − u)

+ (f4t + f6t) sin(v) + i6t sin(2v) + h6t sin(3v)



3PN-accurate Kepler Equation

l = u − et sin u + (g4t + g6t) (v − u)

+ (f4t + f6t) sin(v) + i6t sin(2v) + h6t sin(3v)

Objective

Invert the 3PN-accurate Kepler Equation to get u(l) accurate to 3PNorder


Solving the 3PN-accurate Kepler Equation

v − u = 2∞∑j=1

βj

jsin(ju)

sin(v) =2√

1− e2φ

eφ

∞∑j=1

βj sin(ju)

sin(2v) =4√

1− e2φ

e2φ

∞∑j=1

βj(j√

1− e2φ − 1)

sin(ju)

sin(3v) =2√

1− e2φ

e3φ

∞∑j=1

βj(

4− e2φ − 6j√

1− e2φ + 2j2(1− e2φ))

sin(ju)

where β =1−√

1−e2φ

eφ



l = u − et sin u +∞∑j=1

αj(x , et , η) sin(ju)

αj(x , et , η) =2βj

√1− e2φ

e3φ

((f4t + f6t)e

2φ +

(g4t + g6t)e3φ

j√

1− e2φ

+ 2i6teφ[j√

1− e2φ − 1]

+ h6t[4− e2φ − 6j

√1− e2φ + 2j2(1− e2φ)

])



Fourier series

u − l =∞∑s=1

As sin(sl)

As =2

π

∫ π

0

(u − l) sin(sl)dl

=2

sπ

∫ π

0

cos(sl)du

=2

s

{1

π

∫ π

0

cos(su − se sin u + s

∞∑j=1

αj sin(ju))du

}



As =2

sπ

∫ π

0

cos (su − set sin u) du −2

π

∞∑j=1

αj

∫ π

0

sin (su − set sin u) sin(ju)du

=2

sJs (set) +

1

π

∞∑j=1

αj

∫ π

0

{cos ((s + j)u − set sin u)− cos ((s − j)u − set sin u)

}du

=2

sJs (set) +

∞∑j=1

αj {Js+j (set)− Js−j (set)}



Solution to the 3PN-accurate Kepler Equation

u = l +∞∑s=1

As sin(sl)

As =2

sJs(set) +

∞∑j=1

αj {Js+j(set)− Js−j(set)}


Numerical comparison

Numerical solution using Newton’s method

We plot integrated error 12π

(∫ 2π

0

(unum−uanlunum−l

)2dl

)1/2


Numerical comparison

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9et

10−7 10−7

10−6 10−6

10−5 10−5

10−4 10−4

10−3 10−3

10−2 10−2

10−1 10−1

100 100

Inte

grat

eder

ror

x = 0.01

x = 0.02

x = 0.03

x = 0.05

x = 0.07

x = 0.10


Brief summary

We derived an elegant series solution to the 3PN accurate Keplerequation

We use this solution to calculate eccentric GW templates


On solving post-Newtonian accurate Kepler Equation

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