Emel Altas∗
Department of Physics, Karamanoglu Mehmetbey University,
70100, Karaman, Turkey
Bayram Tekin†
Department of Physics, Middle East Technical University,
06800, Ankara, Turkey (Dated: April 23, 2021)
We construct analytical initial data for a slowly moving and rotating black hole for generic orientations of the linear momentum and the spin. We solve the Hamiltonian constraint approximately and work out the properties of the apparent horizon and show the dependence of its shape on the angle between the spin and the linear momentum. In particular a dimple, whose location depends on the mentioned angle, arises on the 2-sphere geometry of the apparent horizon. We exclusively work in the case of conformally flat initial metrics.
I. INTRODUCTION
Since the first observation of black hole merger [1], there have been many observations of merger of compact objects via gravitational waves. The gravitational waves produced by these mergers are consistent with the numerical solutions of the field equations of General Relativity. Besides the highly accurate numerical results, it always pays to have approximate solutions of relativistic gravitating systems. Here we give an approximate analytical descrip-tion of a self-gravitating system that has a conserved total energy, total spin and a linear momentum in an asymptotically flat spacetime. The initial configuration is expected to evolve and settle to a single rotating black hole after emitting some gravitational radiation. The problem was studied in [2] in the case of vanishing linear momentum but with a nonzero spin; and in [3] in the case of vanishing spin with a nonzero linear momentum. See a remarkable exposition in [4]. Here we assume both of these quantities to be nonzero and pointing arbitrarily in three dimensional space. It will turn out that the shape of the apparent horizon depends on the angle between the linear momentum and the spin: even though at the next to leading order, the magnitude of the spin does not appear in the shape of the apparent horizon, its direction does. On the other hand, the shape of the apparent horizon depends on the magnitude of the linear momentum at the first order. The area of the apparent horizon does not depend on the angle between the spin and linear momentum. We also observe that a dimple arises on the 2-sphere geometry of the apparent horizon.
The layout of the paper is as follows. In the next section, we discuss briefly the con-∗Electronic address: emelaltas@kmu.edu.tr
†Electronic address: btekin@metu.edu.tr
straint equations in General Relativity and present the Bowen-York method [5] in finding solutions to the initial value problem. In section III, we give the approximate solution of the Hamiltonian constraint for a slowly rotating and moving black hole. In section IV, we compute the position of the apparent horizon as a function of the angle between the spin and the linear momentum.
II. INITIAL DATA FOR A BLACK HOLE WITH MOMENTUM AND SPIN
Assuming the usual ADM split of the metric [6]
ds2 = (NiNi− N2)dt2 + 2Nidtdxi+ γijdxidxj, i, j ∈ (1, 2, 3), (1)
the Einstein equations in vacuum without a cosmological constant split into constraints and the evolution equations. The constraint equations are given as
−ΣR − K2+ K
ijKij = 0,
−2DkKik+ 2DiK = 0, (2)
where Σ is the Cauchy surface; Kij = Kij(t, xk) is its extrinsic curvature defined as
Kij = 1 2N ˙γij − DiNj − DjNi , ˙γij = ∂ ∂tγij, (3)
with the trace K := γijK
ij; and Diγkl= 0. For further details of the construction, including
the evolution equations which we do not depict here, see the Appendix of [7]. Following Bowen-York [5], let us assume that Σ is conformally flat with the metric
γij = ψ4fij, ψ > 0, (4)
with fij denoting the flat metric in some generic coordinates. One also sets the extrinsic
curvature of the hypersurface to be given as Kij = ψ−2Kˆij. Furthermore, we assume that Σ
is a maximally embedded hypersurface in the spacetime such that the trace of the extrinsic curvature vanishes1
K = 0. (5)
Under these conditions the Hamiltonian constraint reduces to a nonlinear elliptic equation ˆ DiDˆiψ = − 1 8ψ −7ˆ Kij2, (6)
and the momentum constraint reduces to ˆ
DiKˆij = 0, (7)
with ˆDifjk = 0. The momentum constraint equations can be solved easily, following [5], let
us choose the 6 parameter solution ˆ Kij = 3 2r2 pinj + pjni+ (ninj− fij)p · n + 3 r3J lnkε kilnj+ εkjlni , (8)
1 For physically relevant decay conditions in the case of asymptotically flat initial data, we refer the reader to Section III C of [7] where a slightly extended discussion is compiled.
where ni is the unit normal on a sphere of radius r. For other solutions, see [8]. Assuming
the following asymptotic behavior for the conformal factor
ψ(r) = 1 + E
2r + O(1/r
2), (9)
one can easily show that (see [9]) the pi in the solution (8) corresponds to the total conserved
linear momentum via
Pi = 1 8π ˆ S2 ∞ dS njKij = 1 8π ˆ S2 ∞ dS njKˆij. (10)
Similarly one can show that Ji corresponds to the total conserved angular momentum
ex-pressed in terms of the coordinates and the extrinsic curvature as
Ji = 1 16πεijk ˆ S2 ∞ dS nl xjKkl− xkKjl= 1 16πεijk ˆ S2 ∞ dS nl xjKˆkl− xkKˆjl. (11)
Finally the ADM energy
EADM = 1 16π ˆ S2 ∞ dS ni ∂jhij − ∂ihjj , (12) becomes EADM = − 1 2π ˆ S2 ∞ dS ni∂iψ, (13)
and so using the asymptotic form (9) one finds EADM = E. This has been a brief description
of the solution of the momentum constraints. Now, the important task is to solve the Hamil-tonian constraint, which as we noted, is a nonlinear elliptic equation and thus, generically, it can only be solved numerically. But in the next section, we shall give an approximate solution for small momentum and small rotation.
III. INITIAL DATA WITH SMALL MOMENTUM AND SMALL SPIN
Computation of ˆKijKˆij (from equation (8)) yields
ˆ KijKˆij = 9 2r4 p2+ 2(~p · ~n)2+ 18 r5 ~ J × ~n· ~p + 18 r6 ~ J × ~n·J × ~~ n. (14)
Without loss of generality, let us assume that the direction of the spin is the ˆk direction,
namely
~
J = J ˆk, (15)
and ~p is lying in the xz plane and given as ~
p = p sin θ0ˆi + p cos θ0kˆ (16)
with θ0 a fixed angle. To simplify the notation of the following discussion, let us denote
The Hamiltonian constraint, after these conventions becomes ˆ DiDˆiψ = ψ−7 9J p 4r5 c1sin θ sin φ − 9J2 4r6 sin 2θ − 9p2
16r4(1 + 2(c1sin θ cos φ + c2cos θ) 2)
!
.
(18) As it clear from the right-hand side, the correct perturbative expansion in terms of the momentum and spin reads
ψ(r, θ, φ) := ψ(0)+ J2ψ(J )+ p2ψ(p)+ J pψ(J p)+ O(p4, J4, p2J2), (19)
where the functions on the right-hand side depend on all coordinates (r, θ, φ). At the lowest order, one has
ˆ
DiDˆiψ(0) = 0. (20)
To proceed, let us discuss the boundary conditions that we shall employ. Following [10] and [3], we chose the following boundary conditions
lim r→∞ψ(r) = 1, ψ(r) > 0, (21) and lim r→ 0ψ(r) = ψ (0). (22)
At the lowest order, the solution satisfying these boundary conditions reads
ψ(0) = 1 + a
r. (23)
Inserting (19) into (18), one arrives at three linear partial differential equations to be solved: ˆ DiDˆiψ(J ) = − 9 4sin 2θ r (r + a)7, DˆiDˆ iψ(J p) = 9 4c1sin θ sin φ r2 (r + a)7, (24) and also ˆ DiDˆiψ(p) = − 9 16
1 + 2(c1sin θ cos φ + c2cos θ)2
r3
(r + a)7. (25)
In finding the solutions to these equations, we will need the following spherical harmonics :
Y00(θ, φ) = √1 4π, Y 0 1(θ, φ) = s 3 4πcos θ, Y 0 2(θ, φ) = s 5 16π(3 cos 2θ − 1), (26) Y1−1(θ, φ) = s 3 4πsin θ sin φ, Y 1 2(θ, φ) = s 15
4πsin θ cos θ cos φ, Y 1
1(θ, φ) = s
3
4πsin θ cos φ. Then ansatz for ψ(J ) can be taken as
ψ(J )(r, θ, φ) = ψ0(J )(r)Y00(θ, φ) + ψ(J )1 (r)Y20(θ, φ), (27) which upon insertion to the first equation of (24) yields two ordinary differential equations equations d dr r2dψ (J ) 0 (r) dr = −3√π r 3 (r + a)7, d dr r2dψ (J ) 1 (r) dr − 6ψ(J )1 (r) = 3 rπ 5 r3 (r + a)7. (28)
The solution obeying the boundary conditions (21) reads ψ(J )(r, θ, φ) = (a 4+ 5a3r + 10a2r2 + 5ar3+ r4) 40a3(a + r)5 − r2 40a(a + r)5(3 cos 2θ − 1). (29) Similarly setting ψ(J p)(r, θ, φ) = ψ(J p)0 (r)Y00(θ, φ) + ψ1(J p)(r)Y1−1(θ, φ) (30) in the second equation of (24), one finds that ψ0(J p)(r) = 0 satisfies the boundary conditions; and the ψ(J p)1 (r) piece satisfies
d dr r2dψ (J p) 1 (r) dr − 2ψ(J p)1 (r) = 3 √ 3π 2 c1 r4 (r + a)7 (31)
of which the solution can be found and one has
ψ(J p)(r, θ, φ) = −c1r (a
2+ 5ar + 10r2)
80a(a + r)5 sin θ sin φ. (32)
Finally, let us do the ψ(p)(r, θ, φ) part which is slightly more complicated. One sets
ψ(p) = ψ0(p)(r)Y00(θ, φ) + ψ1(p)(r)Y11(θ, φ)2+ ψ(p)2 (r)Y21(θ, φ) + ψ3(p)(r)Y10(θ, φ)2 (33) to arrive at four equations, two of which are
d dr r2dψ (p) 0 dr +√3 π(ψ (p) 1 + ψ (p) 3 ) = − 9 8 √ π r 5 (r + a)7, (34) and d dr r2dψ (p) 1 dr − 6ψ1(p) = −3 2πc 2 1 r5 (r + a)7. (35)
ψ2(p)(r) equation can be obtained from (35) with the replacement c21 →q 3
5πc1c2 and ψ (p) 3 (r) equation can be obtained from (35) via c2
1 → c22. The solutions read, respectively, as follows
ψ0(p)(r) = − √
π (84a6+ 378a5r + 653a4r2+ 514a3r3+ 142a2r4− 35ar5− 25r6)
80ar2(a + r)5 −21 √ πa 20r3 log a a + r, (36) and ψ1(p)(r) = πc 2
1(84a5+ 378a4r + 658a3r2+ 539a2r3 + 192ar4+ 15r5) 40r2(a + r)5 +21πac 2 1 10r3 log a r + a, (37)
Recall that for the ADM energy computation, we need the dominant terms up to and including O(1
r) in ψ(r, θ, φ). Collecting these parts in the above solutions, one gets
ψ(r) = 1 + a r + J2 40a3r + 5p2 32ar + O( 1 r2). (38)
Therefore from (9), the ADM energy of the solution reads
EADM= 2a +
J2
20a3 + 5p2
16a. (39)
Observe that the J p term does not contribute to the energy since it is of O(r12).
Next, as in [3], let us express the ADM energy in terms of the irreducible mass Mirr which is defined[11] as
Mirr :=
s
A
16π (40)
with A being the area of a section of the event horizon. But as the event horizon is a 4 dimensional concept, which cannot be derived from the 3 dimensional initial data, we will approximate this with the area of the apparent horizon, AAH, following [3].
IV. COMPUTATION OF THE APPARENT HORIZON FOR THE
BOOSTED, ROTATING SOLUTIONS
Let S be a 2 dimensional subspace of Σ and si be the normalized unit vector of S, i.e.
sis
i = 1. Then the metric on S is the pull-back metric from Σ given as
mij := γij − sisj. (41)
The expansion of the null geodesic congruence vanishes at the apparent horizon by definition,
i.e. it is a marginally trapped surface and the defining equation becomes
γij − sisj(D
isj − Kij) = 0. (42)
Assuming the surface to be defined as a level set of a function
Φ := r − h(θ, φ) = 0, (43)
then the normal one-form reads
si := λmi = λ∂iΦ, (44)
which explicitly becomes
si = λ (1, −∂θh, −∂φh) . (45)
Recall that the metric on Σ is
γij = ψ4 1 0 0 0 r2 0 0 0 r2sin2θ , (46)
si = λγrr, −γθθ∂θh, −γφφ∂φh , (47) and λ = γrr+ γθθ(∂θh)2+ γφφ(∂φh)2 −1/2 . (48)
Equation (42) reads more explicitly as
γij∂imj − γijΓkijmk− λ2mimj∂imj+ λ2mimjmkΓkij + λm
imjK
ij = 0, (49)
where we have used γijK
ij = K = 0. After working out each piece, one arrives at
−γθθ∂2 θh − γ φφ∂2 φh − 1 2 (γrr)2∂rγrr− γθθγrr∂rγθθ− γφφγrr∂rγφφ+ ∂θhγφφγθθ∂θγφφ +λ2 (γθθ)2(∂θh)2∂θ2h + (γ φφ)2(∂ φh)2∂2φh + 2γ φφγθθ∂ φh∂θh∂θ∂φh +λ 2 2 (γrr)3∂rγrr+ (γθθ)2γrr(∂θh)2∂rγθθ + (γφφ)2γrr(∂φh)2∂rγφφ −(∂φh)2∂θh(γφφ)2γθθ∂θγφφ +λ (γrr)2Krr+ (γθθ)2(∂θh)2Kθθ+ (γφφ)2(∂φh)2Kφφ− 2γrrγθθ∂θhKrθ −2γrrγφφ∂ φhKrφ+ 2γθθγφφ∂θh∂φhKθφ = 0. (50)
An exact solution to this equation is beyond reach and we do not really need it. All we need is an approximate solution of the form
h(θ, φ) = h0+ php+ J hJ + O(p2, J2, J p), (51)
where
∂rh = 0, ∂rh0 = 0 = ∂θh0 = ∂φh0. (52)
Note that to compute the area of the apparent horizon and the irreducible mass up to and including the O(p2, J2, J p) terms, one only needs the shape of the horizon up to and
including the O(p, J ) terms which becomes clear when one studies the area integral. [See also[3].] Ignoring the higher order terms such as (∂θh)2, (∂φh)2 and ∂θh∂φh, the apparent
horizon equation becomes
− γθθ∂2 θh − γ φφ∂2 φh − 1 2 (γrr)2∂rγrr− γθθγrr∂rγθθ− γφφγrr∂rγφφ+ ∂θhγφφγθθ∂θγφφ + λ 2 2 (γ rr)3∂ rγrr+ λγrr γrrKrr− 2γθθ∂θhKrθ− 2γφφ∂φhKrφ = 0. (53) To proceed, we need the components of the extrinsic curvature in the (r, θ, φ) coordinates. After coordinate transformations, one finds
ˆ
Krr =
3p
r2
c1sin θ cos φ + c2cos θ
, Kˆrθ =
3p 2r
c1cos θ cos φ − c2sin θ
and ˆ Krφ = − 3p 2rc1sin θ sin φ + 3J r2 sin 2θ. (55)
Therefore the resulting equation is
∂θ2h + 1 sin2θ∂ 2 φh + cot θ∂θh − 2r − 4r2 ∂rψ ψ + 6J ψ4r2∂φh − 3p ψ4
c1sin θ cos φ + c2cos θ
= 0. (56) At order, O(p0, J0), this equation yields
1 + 2r∂rψ
ψ = 0, (57)
where ψ = 1 + ar . And setting r = h, one finds
h0 = a. (58)
This explains the physical meaning of the parameter a: it is the location of the apparent horizon at the lowest order. The next order contribution, which we shall find below, will be perturbations to this location. At O(p) and O(J ) we have the following equations, respectively ∂θ2hp+ 1 sin2θ∂ 2 φh p+ cot θ∂ θhp− hp− 3 16
c1sin θ cos φ + c2cos θ = 0, (59) and ∂θ2hJ + 1 sin2θ∂ 2 φh J + cot θ∂θhJ − hJ = 0. (60)
These are linear PDE’s and a close scrunity shows that hJ equation is the homogenous
Helmholtz equation on a sphere (S2), while hP equation is the inhomogeneous Helmholtz
equation with a non-trivial source. So the next task is to find everywhere finite solutions of the following equation
~ ∇2 S2 + k f (θ, φ) = g (θ, φ) , (61) where ~∇2 S2 is the Laplacian on S2: ~ ∇2 S2 = ∂θ2+ cot θ∂θ+ 1 sin2θ∂ 2 φ. (62)
It is clear that the Green’s function technique is the most suitable approach to this problem. For the Helmholtz operator on the sphere, the Green function G(ˆx, ˆx0) is defined as
~ ∇2 S2 + λ(λ + 1) G(ˆx, ˆx0) = δ(2)(ˆx − ˆx0), (63) which can be found to be (for example, see [12])
G(ˆx, ˆx0) = 1 4 sin πλ ∞ X n=0 1 (n!)2 Γ(n − λ) Γ(−λ) Γ(n + λ + 1) Γ(λ + 1) 1 + ˆx · ˆx0 2 !n , (64)
where ˆx = sin θ cos φˆi + sin θ sin φˆj + cos θˆk and ˆx0 is a similar expression with some other θ and φ. Employing this Green’s function with λ = −1+i
√ 3
Figure 1: The shape of the apparent horizon when the angle between ~p and ~J is 45 degrees; and to be able to see the dimple, we have chosen p/a = 8√2 which is outside the validity of the approximation we have worked with. But the dimple exists for even small p.
hp = − 1
16(c1sin θ cos φ + c2cos θ) (65)
and hJ = 0. Therefore the apparent horizon is located at
r = h(θ, φ) = a − 1
16
~
p · ˆJ cos θ + |~p ∧ ˆJ | sin θ cos φ
, (66)
where ˆJ = JJ~. In the limit θ0 = 0, h reduces to the form given in [3], that is h(θ) = a−16p cos θ; and the apparent horizon in this axially symmetric case is a squashed sphere from the North pole. Note that the shape of the apparent horizon 66 at this order does not depend on the magnitude of the spin but it does depend on its orientation with respect to the linear momentum. In figure 1, we plot the apparent horizon. To be able to see the dimple clearly in the whole figure, we have chosen a high momentum value.
Let us now evaluate the area of the apparent horizon from the formula
AAH = 2π ˆ 0 dφ π ˆ 0 dθ√det m, (67)
which at the order we are working yields
AAH = 2π ˆ 0 dφ π ˆ 0 dθ sin θ ψ4h2 1 + 1 h2(∂θh) 2 + 1 h2sin2θ(∂φh) 21/2 . (68)
This is a pretty long computation since the conformal factor is quite complicated. But at the end, one finds
AAH= 64πa2+ 4πp2+
11πJ2
Note that the angle between the spin and the linear momentum does not appear in the area. Then the irreducible mass Mirr reads
Mirr = 2a +
p2
16a + 11J2
320a3. (70)
Comparing with EADM we have
EADM= Mirr+ p2 2Mirr + J 2 8M3 irr (71) which matches the slow momentum and spin limit of the result in [11].
V. CONCLUSIONS
Momentum constraints in General Relativity are easily solved with the method of Bowen-York while the Hamiltonian constraint is a nontrivial elliptic equation. Here, extending earlier works [2], [3] we gave an approximate analytical solution that describes a spinning and moving system with a conserved spin and linear momentum pointing in arbitrary directions. We computed the properties of the apparent horizon, such as its shape and surface area and showed the dependence of the shape on the angle between the spin and the linear momentum. We calculated the relation between the conserved quantities such as the ADM mass, the spin, the linear momentum and the irreducible mass. The area of the apparent horizon does not depend on the angle between the spin and the linear momentum, but a dimple arises in the apparent horizon whose location depends on this angle.
Acknowledgments
We would like to thank Fethi Ramazanoğlu for useful discussions.
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