r t rt trs - - PowerPoint PPT Presentation
r t rt trs history, and rsts The example of binary black-holes in Einstein-Maxwell-Dilaton (EMD) theory Nathalie
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5 10 15
z1/M
5 10 15
z2/M
2750 2800 2850 2900 2950 3000
t / M
0.1 0.2
h
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µ Fνλ − 1 4gµνF 2
2e−2aϕF 2
r
r
r
2 eϕ∞ r
2 ln
r
EX – 9
1 4πr+
2 eϕ∞ r+
+
r+
4π )
2
2r+ − 1 2
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1 16π
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√−g
ν
2gµνF 2
2gµνT
√−g
2e−2aϕF 2+4π
√−g dm dϕ
µν
r
t
r
r dm dϕ|∞
µν
r
t
2 eϕ∞ r , ϕasym = ϕ∞ − r− 2r
dϕ, q =
2 e−ϕ∞|∞ so that q2 = 2mdm dϕe2ϕ|∞
2
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2 e−ϕ∞, r+ = 2m∞, r− = 2dm dϕ|∞, and m(ϕ) =
2
2
2r+ − 1 2
+
r+
dϕδϕ|∞ = 0
4π
4π + e2ϕ 2 Q2
4 and M 2 irr = S 4π
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1 16π
A
A qA
A
4π + e2ϕ 2 Q2 A
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A =
√−g
Auν A)
ν
2gµνF 2
A
µν − 1 2gµνT A
δ(4)(x−zA) √−g
A
2e−2aϕF 2 + 4π A
δ(4)(x−zA) √−g dmA dϕ
A m∞
A vi A
rA
A m
′∞ A
rA + · · · , etc
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dt + qAAµ dxµ
A
dt
1P N = −(mA + mB) +
2(mAv2 A + mBv2 B) + GAB mAmB R
8(mAv4 A + mBv4 B)
R
2(v2 A + v2 B) − 7 2(vA.vB) − 1 2(N.vA)(N.vB) + ¯
2R2
A/mA)|∞, βA = α′ A|ϕ∞
−4αAαB+3eAeB 2(1+αAαB−eAeB)
2 βAαB2−2eAeB(aαB−αAαB)+e2
B(1+aαA−e2 A)
1+αAαB−eAeB
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A/mA)|∞, βA = α′ A|ϕ∞
−4αAαB+3eAeB 2(1+αAαB−eAeB)
2 βAαB2−2eAeB(aαB−αAαB)+e2
B(1+aαA−e2 A)
1+αAαB−eAeB
A, α2 B or αAαB. Now,
4π + e2ϕ 2 Q2 A
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A/2
A/mA)|∞ = 1 1+exp 2
√ 2 qA
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1P N at 1PN and the state-of-the-art
1P N )
1P N )
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1P N = −(mA + mB) +
2(mAv2 A + mBv2 B) + GAB mAmB R
8(mAv4 A + mBv4 B)
R
2(v2 A + v2 B) − 7 2(vA.vB) − 1 2(N.vA)(N.vB) + ¯
2R2
2P N is given by LST 2P N with some replacements and modulo 3 coeffi-
2P N depends on the positions, velocities and accelerations of A and B
B/R2
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2µ − GAB µM R
H1PN µ
1
2
R + h1PK 3
R) + (h1PK
4
ˆ P 2+h1PK
5
ˆ P 2
R)
ˆ R
6
ˆ R2 H2PN µ
1
2
R + h2PK 3
R + h2PK 4
R) (h2PK
5
ˆ P 4+h2PK
6
ˆ P 2
R ˆ
P 2+h2PK
7
ˆ P 4
R)
ˆ R
8
ˆ P 2+h2PK
9
ˆ P 2
R)
ˆ R2
10
ˆ R3
A/mA)|∞, βA = α′ A|ϕ∞ and β′ A
a
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2µ − GAB µM R
H1PN µ
1
2
R + h1PK 3
R) + (h1PK
4
ˆ P 2+h1PK
5
ˆ P 2
R)
ˆ R
6
ˆ R2 H2PN µ
1
2
R + h2PK 3
R + h2PK 4
R) (h2PK
5
ˆ P 4+h2PK
6
ˆ P 2
R ˆ
P 2+h2PK
7
ˆ P 4
R)
ˆ R
8
ˆ P 2+h2PK
9
ˆ P 2
R)
ˆ R2
10
ˆ R3
G(Q,p) R pr
r + γ1 ˆ R
r + γ2ˆ
r + δ2P2 ˆ R + ǫ2 ˆ p2
r
ˆ R + η2 ˆ R2
He(q,p) µ
µ
2
µ
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e = −A(r)dt2+B(r)dr2+r2dφ2
r
B + p2
φ
ˆ r2
r + a2 r2 + a3 r3 + · · ·
r + b2 r2 + · · ·
r
GABM r
3 GABM r
GABM r
2 GABM r
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He(q,p) µ
µ
2
µ
r
B + p2
φ
ˆ r2
µ − 1
r
B + p2
φ
r2
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e = −A(r)dt2+B(r)dr2+r2dφ2
r
B + p2
φ
ˆ r2
A′′ A′ = (Au2)′′ (Au2)′
ju2A GABME√ 1+2ν(E−1)
r
A′ (Au2)′ ,
(Au2)′
EOBNR(u ; ν) + 2ǫ1PKu2 + (ǫ0 2PK + νǫν 2PK)u3
4π + Q2 A e2ϕ 2
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(Ω)
2PK = ǫν 2PK]
5[AGR EOBNR(u ; ν) + 2ǫ1PKu2 + (ǫ0 2PK + νǫν 2PK)u3]
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