Baumgarte T., Shapiro S. Numerical Relativity. Solving Einstein’s Equations on the Computer

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3.2 Conformal transverse-traceless decomposition 71

Inserting the solution (3.65) into equation (3.50)wenowﬁnd

= (

LW)

¯

j)kl

. (3.66)

Since J

satisﬁes

= 0, it must be a vector with constant coefﬁcients when expressed

in Cartesian coordinates. Exercise 3.11 shows that for large r the extrinsic curvature

(3.66) agrees with that of a Kerr black hole (see exercise 2.34), suggesting that our

solution describes a rotating black hole with angular momentum J

. We will conﬁrm this

identiﬁcation in exercise 3.29.

Exercise 3.11 Show that in spherical polar coordinates the only nonvanishing com-

ponents of W

in equation (3.65) and the extrinsic curvature

in equation (3.66)

are

=−

(3.67)

and

rφ

sin

θ. (3.68)

Here J is the magnitude of the vector J

aligned with the polar axis. Further show

that, with suitable choices for

and ψ , this solution agrees asymptotically with

the Kerr solution of exercise 2.34.

In order to construct a complete solution to the constraint equations we would have

to insert the extrinsic curvature (3.66) back into the Hamiltonian constraint (3.37)and

solve for the conformal factor ψ. This equation is nonlinear and in general can only be

solved numerically. We will return to this problem in several places throughout this book;

in Chapter 12.2.2, for example, we will discuss two approaches for solving this equation in

the context of binary black holes. In the meantime, we can ﬁnd an approximate, analytical

solution by expanding around the nonrotating case J = 0, as outlined in exercise 3.12.

Exercise 3.12 Consider the solution (3.68) for a rotating black hole and assume

= 0 to show that the only nonvanishing source term in the Hamiltonian

constraint (3.37) arises from

18J

sin

θ. (3.69)

For J = 0 we recover the solution (3.18), which we will denote here as ψ

(0)

.The

leading-order correction to this solution must therefore scale with J

, which suggests

the ansatz

ψ = ψ

(0)

(2)

+ O(J

) = 1 +

(2)

+ O(J

). (3.70)

Show that inserting this ansatz into the Hamiltonian constraint (3.37) yields

(2)

=−

(r + M/2)

sin

θ. (3.71)

72 Chapter 3 Constructing initial data

Now express the angular dependence sin

θ in terms of the Legendre polynomials

(cos θ ) = 1andP

(cos θ ) = (3 cos

θ − 1)/2, split ψ

(2)

into

(2)

= ψ

(2)

(r)P

(cos θ ) + ψ

(2)

(r)P

(cos θ ), (3.72)

and show that the resulting equations are solved by

(2)

(r) =−



1 +



−5







+ 4











(3.73)

and

(2)

(r) =−



1 +



−5





. (3.74)

With the conformal factor ψ we now have a complete solution to the constraint equations

describing a rotating black hole. But we already know that the solution for a stationary,

rotating black hole is given by the Kerr solution of exercise 2.34. This raises an interesting

question, namely whether the above Bowen–York rotating black hole is identical to the Kerr

black hole. The answer is no, even though it is not as easy to see this as one might think.

One complication is that the Kerr solution describes a complete spacetime solution,

whereas the Bowen–York solution describes initial data only. These initial data represent

the solution on a certain time slice , but a priori it is not clear on which slice. For example,

there is no reason to expect that this slice corresponds to a slice of constant Boyer–Lindquist

time t (see Chapter 1.2 and exercise 2.34), making any direct comparison very difﬁcult. We

could evolve the Bowen–York initial data dynamically, thereby constructing a spacetime

solution. Again, we would have no reason to expect that this spacetime solution would be

represented in Boyer–Lindquist or any other coordinate system in which the Kerr solution

is known, and it would be hard to decide whether or not the two spacetimes are identical,

i.e., whether or not there exists a coordinate transformation that relates the two. One way

to distinguish the resulting spacetimes is to compare gauge-invariant quantities, and one

such example is the emitted gravitational radiation as measured by a distant observer. The

Kerr solution is stationary and does not emit any gravitational radiation, but a dynamical

evolution of the Bowen–York initial data does lead to a burst of gravitational radiation

before it settles down into a stationary solution.

This demonstrates that the Bowen–York

data do not represent a spatial slice of the Kerr solution. Instead, it may be considered a

Kerr solution plus an initial perturbation.

This observation raises the next question: how does this perturbation arise in the CTT

decomposition? All evidence points towards our assumption of conformal ﬂatness, ¯γ

. Given this assumption, the resulting data can represent a Kerr black hole only if the

Kerr spacetime admits a slicing on which the spatial metric is conformally ﬂat. Whether

or not the Kerr spacetime admits such a slicing is again difﬁcult to decide, since we have

Gleiser et al. (1998).

See Gleiser et al. (1998) as well as the discussion in Burko et al. (2006).

3.2 Conformal transverse-traceless decomposition 73

no a priori knowledge of which slicing this might be. We can nevertheless consider certain

families of slices  and evaluate the Bach tensor (3.15). Slices of constant Boyer–Lindquist

time t, for example, are not conformally ﬂat, nor are axisymmetric foliations that smoothly

reduce to slices of constant Schwarzschild time in the Schwarzschild limit.

This suggests

very strongly (but does not prove) that, for nonzero angular momentum, the Kerr spacetime

does not admit spatial slices that are conformally ﬂat.

A boosted black hole

In an alternative approach to solving (3.60) we decompose W

−



∂

U + x

∂



(3.75)

(see equation B.7 in Appendix B), in which case (3.60) becomes equivalent to the set

∂

U = 0 (3.76)

∂

= 0. (3.77)

Note that both equations are now homogeneous. We can construct a simple solution by

assuming U = 0 and writing the solution for V

=−

, (3.78)

where P

is an arbitrary vector with constant coefﬁcients when expressed in Cartesian

coordinates (the general solution also allows for an arbitrary constant term, which would

drop out, however, when we compute

below). Inserting V

into the decomposition

(3.75) yields

=−

(7P

), (3.79)

where again l

= x

/r, and, from equation (3.50),

= (

LW)



+ P

− (η

−l



. (3.80)

In exercise 3.32 we will see that the linear momentum of this solution is P

. By virtue of

the linearity of the momentum constraint (3.60) we can add several terms of this form to

obtain a solution describing multiple, boosted black holes. If we would like these black

holes to spin, we could further add terms of the form (3.66). In fact, these solutions

form the basis of the binary black hole initial data that we discuss in Chapter 12.2.To

complete these data, of course, we still need to solve the Hamiltonian constraint (3.37),

as described in Chapter 12.2.2. As for the spinning black holes, this can in general only

See Garat and Price (2000).

74 Chapter 3 Constructing initial data

be done numerically, but we can again compute approximate solutions as an expansion

around the known Schwarzschild solution for P

= 0.

Exercise 3.13 Follow the approach outlined in exercise 3.12 to construct an approx-

imate, analytical solution to the Hamiltonian constraint (3.37) for the extrinsic

curvature (3.80). First show that



1 + 2cos



, (3.81)

where P

= P

is the square of the momentum’s magnitude. Then make an ansatz

ψ = ψ

(0)

(2)

+ O(P

) = 1 +

(2)

+ O(P

) (3.82)

and show that the Hamilton constraint is solved, to ﬁrst order in P

,by

(2)

= ψ

(2)

(r)P

(cos θ ) + ψ

(2)

(r)P

(cos θ ) (3.83)

where

(2)



1 +



−5

16r







+ 5





+ 10





+ 10





+ 5



(3.84)

and

(2)



1 +



−5











+ 378





+ 658





+539





+ 192





+ 15









/(2r)

1 + M/(2r)



(3.85)

Before closing this section it may be useful to discuss one more technical aspect.

We have seen that the longitudinal part

is constructed from the vector potential W

According to the decomposition (3.48) we can add to this longitudinal part a transverse part

to construct a general solution to the momentum constraint. According to equation

(3.49) the transverse part

has to be divergence-free, and we want to discuss brieﬂy

how a divergence-free tensor can be constructed from a symmetric tracefree tensor

Consider a solution Y

to the equation



. (3.86)

If we deﬁne

− (

LY)

, (3.87)

Gleiser et al. (2002).

Cook (2000).

3.3 Conformal thin-sandwich decomposition 75

then

is clearly divergence-free,

−

(

LY)

−



= 0. (3.88)

In fact, this construction can be conveniently embedded in the above solution of the

constraint equations. Given that

L is linear, equation (3.48) can be written

− (

LY)

+ (

LW)

+ (

LV)

, (3.89)

wherewehavedeﬁned

= W

− Y

. (3.90)

Given that the vector Laplacian is linear, we also have



−



¯γ

K −

+ 8πψ

. (3.91)

Solving this equation for V

instead of equation (3.53)forW

allows us to construct a

general solution

from an arbitrary symmetric tracefree tensor

(equation 3.89)

directly, rather than having to ﬁrst ﬁnd a transverse part

3.3 Conformal thin-sandwich decomposition

Solving the conformal transverse-traceless decomposition yields data γ

and K

intrinsic

to one spatial slice , but this solution does not tell us anything about how it will evolve in

time away from , nor does the formalism allow us to determine any such time evolution.

In some circumstances, for example when we are interested in constructing equilibrium

or quasiequilibrium solutions, we would like to construct data in such a way that they do

have a certain time evolution. The conformal thin-sandwich, or “CTS” approach

offers

an alternative to the transverse-traceless decomposition that does allow us to determine

the evolution of the spatial metric. Instead of providing data for γ

and K

on one time

slice, it provides data for γ

on two time slices, or, in the limit of inﬁnitesimal separation

of the two slices, data for γ

and its time derivative.

We start by deﬁning u

as the traceless part of the time derivative of the spatial metric,

≡ γ

1/3

∂

(γ

−1/3

), (3.92)

in terms of which the evolution equation (2.134) becomes

=−2α A

+ (Lβ)

. (3.93)

Here L is the vector gradient deﬁned in equation (3.50), except that the “unbarred” L is

deﬁned in terms of the physical metric γ

.Wealsodeﬁne

≡ ∂

¯γ

(3.94)

See Yo r k , J r . (1999), as well as Isenberg (1978)andWilson and Mathews (1995) for related earlier approaches.

76 Chapter 3 Constructing initial data

and

¯γ

≡ 0. (3.95)

It may seem odd that we need both of these deﬁnitions to specify

. The reason for this

is the arbitrariness in choosing the determinant of the conformal metric ¯γ .Sofarwehave

only chosen ¯γ

and hence ¯γ on the slice , but we have not yet speciﬁed a normalization

¯γ away from . If deﬁned from equation (3.94) alone, this would leave

deﬁned only

up to some overall factor. This factor is ﬁxed with the help of equation (3.95), which in

effect determines the time derivative of ¯γ :

0 = ¯γ

= ¯γ

∂

¯γ

= ∂

ln ¯γ. (3.96)

With this result we can now show that

= ψ

. (3.97)

Exercise 3.14 Derive equation (3.97).

Exercise 3.15 Show that

(Lβ)

= ψ

−4

(

Lβ)

. (3.98)

From equations (3.97) and (3.98), together with the scaling relation (3.35), we can

conformally transform equation (3.93) to obtain

2α



(

Lβ)

−



. (3.99)

At this point it seems natural to introduce a conformally rescaled or “densitized” lapse,

α = ψ

¯α, (3.100)

in terms of which (3.99) becomes

2¯α



(

Lβ)

−



. (3.101)

This equation relates

to the shift vector β

. Inserting this equation into the momentum

constraint (3.38), yields an equation for the shift,

(



β)

− (

Lβ)

ln( ¯α) = ¯α

(¯α

−1

) +

¯αψ

K + 16π ¯αψ

. (3.102)

We can now construct a solution of the thin-sandwich formulation as summarized in

Box 3.2. We ﬁrst choose the background metric ¯γ

as well as its time derivative

Given choices for the densitized lapse ¯α and the trace of the extrinsic curvature K ,we

can then solve the Hamiltonian constraint (3.37) and the momentum constraint (3.102)for

the conformal factor ψ and the shift β

. With these solutions, we can then construct

from equation (3.101) and ﬁnally the physical quantities γ

and K

. This version of the

3.3 Conformal thin-sandwich decomposition 77

conformal thin-sandwich formalism is sometimes called the “original” version, in contrast

to the “extended” version that we will discuss below.

Box 3.2 The original conformal thin-sandwich (CTS) decomposition

Freely speciﬁable variables are ¯γ

, K and ¯α. Given these, the momentum constraint

(



β)

− (

Lβ)

ln( ¯α) = ¯α

(¯α

−1

) +

¯αψ

K + 16π ¯αψ

(3.103)

is solved for β

, and the Hamiltonian constraint

ψ −

R −

−7

=−2πψ

ρ, (3.104)

where

2¯α



(

Lβ)

−



, (3.105)

is solved for ψ. The physical solution is then constructed from

= ψ

¯γ

= A

K = ψ

−2

α = ψ

¯α.

(3.106)

It is again instructive to count the degrees of freedom, and to compare with the transverse-

traceless decomposition of Section 3.2. There, we found that of the 12 independent variables

in γ

and K

, four were determined by the constraint equations, four were related to the

coordinate freedom, and four represented the dynamical degrees of freedom of general

relativity. The latter eight conditions can be chosen freely. In the thin-sandwich formalism

we count a total of 16 independent variables, of which we can freely choose 12: ﬁve

each in ¯γ

and

, and one each for ¯α and K . The four remaining variables, ψ and β

are then determined by the constraint equations. The four new independent variables are

accounted for by the lapse ¯α and the shift β

, which are absent in the transverse-traceless

decomposition. The CTT approach only deals with quantities intrinsic to one spatial slice

, and hence only requires coordinates on . The thin-sandwich approach, on the other

hand, also takes into account the evolution of the metric off the slice, and therefore requires

coordinates in a neighborhood of . As a consequence, the lapse α and the shift β

,which

describe the evolution of the coordinates away from , appear in the CTS approach, but

not in the CTT decomposition. The four new degrees of freedom hence reﬂect the time

derivatives of the coordinates.

Instead of ﬁxing the densitized lapse ¯α, we can specify alternatively the time derivative

of the mean curvature ∂

K together with K . From equation (2.137) we can then solve

α =−∂

K + α



+ 4π(ρ + S)



+ β

K (3.107)

78 Chapter 3 Constructing initial data

for the lapse together with equations (3.37) and (3.102). Condition (3.107) involves the

physical Laplace operator D

, but in this context the conformal Laplace operator

would

be more handy to evaluate. As it turns out, we can express equation (3.107)intermsof

by combining it with the Hamiltonian constraint (3.37) to yield

(αψ) = αψ



−8

R + 2πψ

(ρ + 2S)



−ψ

∂

K + ψ

K. (3.108)

The freely speciﬁable quantities are now the mean curvature K and the conformal metric

¯γ

, together with their time derivatives ∂

K and

. We can then solve the Hamiltonian

constraint (3.37) for the conformal factor ψ, the momentum constraint (3.102) for the shift

, and equation (3.108)forthelapseα. We summarize this “extended” version of the

conformal thin-sandwich decomposition in Box 3.3.

Box 3.3 The extended conformal thin-sandwich (CTS) decomposition

Freely speciﬁable variables are ¯γ

, K and ∂

K . Given these, the momentum constraint

(



β)

− (

Lβ)

ln( ¯α) = ¯α

(¯α

−1

) +

¯αψ

K + 16π ¯αψ

(3.109)

is solved for β

, the Hamiltonian constraint

ψ −

R −

−7

=−2πψ

ρ (3.110)

is solved for ψ, and a combination of the Hamiltonian constraint and the trace of the evolution

equation for K

(αψ) = αψ



−8

R + 2πψ

(ρ + 2S)



− ψ

∂

K + ψ

(3.111)

is solved for the product αψ = ¯αψ

. In the above equations

is given by

2¯α



(

Lβ)

−



. (3.112)

The physical solution is then constructed from

= ψ

¯γ

= A

K = ψ

−2

(3.113)

Exercise 3.16 Derive equation (3.108) from equations (3.107)and(3.37).

See Pfeiffer and York, Jr. (2003a). For some results on the uniqueness and nonuniqueness of solutions to these equations

see Pfeiffer and York, Jr. (2005); Baumgarte et al. (2007); Walsh (2007); and references therein.

3.3 Conformal thin-sandwich decomposition 79

The extended version of the thin-sandwich formalism seems particularly useful for the

construction of equilibrium or quasiequilibrium data, since it allows us to set the time

derivatives of the conformal metric and the mean curvature to zero,

= 0and∂

K = 0. (3.114)

In this case equation (3.99) reduces to

2α

(

Lβ)

, (3.115)

which looks very similar to expression (3.50)for

in the conformal transverse-traceless

decomposition. Here, however,

is not longitudinal because of the extra factor of ψ

/α.

Further assuming maximal time-slicing, K = 0, equations (3.37), (3.102) and (3.108)now

form the coupled system

ψ =

R −

−7

− 2πψ

ρ, (3.116)

(



β)

= 2

(αψ

−6

) + 16παψ

, (3.117)

(αψ) = αψ



−8

R + 2πψ

(ρ + 2S)



, (3.118)

for ψ, β

and α. With

= 0, K = 0and∂

K = 0 the only remaining freely speciﬁable

quantity is the conformal metric ¯γ

. The equations further simplify under the assumption

of conformal ﬂatness ¯γ

= η

, in which case

R = 0 and the differential operators become

much easier to invert.

Exercise 3.17 Show that under the additional assumption of conformal ﬂatness,

the CTS equations (3.116)–(3.118) reduce to

∂

ψ =−

−7

− 2πψ

ρ, (3.119)

∂

= 2

∂

(αψ

−6

) + 16παψ

, (3.120)

∂

(αψ) = αψ



−8

+ 2πψ

(ρ + 2S)



, (3.121)

in Cartesian coordinates.

The equations for the shift (3.117) and (3.120) again involve the vector Laplacian, which

we previously encountered in the conformal transverse-traceless decomposition of Section

3.2.

We discuss strategies for solving this operator in Appendix B. Interestingly, we will

rediscover the shift condition (3.117) in Chapter 4.5, where we will ﬁnd that it is identical

to the “minimal distortion” condition. The conformal thin-sandwich formalism therefore

Recall equation (3.60) in Cartesian coordinates and the discussion of Bowen–York solutions that follow.

80 Chapter 3 Constructing initial data

reduces to the Hamiltonian constraint for the conformal factor, the minimal distortion

condition for the shift, and the maximal slicing condition for the lapse.

If initial data for a time evolution calculation are constructed from the CTT decomposi-

tion, then the lapse and shift have to be chosen independently of the construction of initial

data. The CTS formalism, on the other hand, provides a lapse and a shift together with the

initial data γ

and K

. Obviously, once the initial data are determined, the lapse and shift

can always be chosen freely in performing subsequent evolution calculations. However,

the original relation between the time derivative of γ

and

only applies when the lapse

and shift as obtained from the CTS solution are employed in the dynamical simulation.

Before closing this section it may be of interest to consider the circumstances under

which the CTS formalism can reproduce a complete spacetime solution, in the sense that

a dynamical evolution of the initial data, using the lapse and the shift obtained from the

CTS solution, would lead to a time-independent solution. We would expect this to be

possible only if the spacetime is stationary, i.e., possesses a timelike Killing vector ﬁeld

. For the metric coefﬁcients to be independent of time during an evolution, our time-

vector t

,deﬁnedin(2.98), then has to be be aligned with ξ

. This is the case if the lapse

and the shift are the Killing lapse and Killing shift, as we discussed in the context of

equation (2.162).

To obtain a Killing lapse and shift we need to choose

= 0and∂

K = 0inthe

extended CTS formalism, but these conditions are not sufﬁcient. The point is that we also

have to choose the conformally related metric ¯γ

,aswellasK , and these choices may or

may not represent the stationary slices of the spacetime that we are seeking. For simple

spherical spacetimes, and suitable boundary conditions, the choice K = 0alwaysgives

rise to a static solution of Einstein’s equations. For example, in vacuum with ¯γ

chosen

to be ﬂat and K = 0, there is a particular choice of boundary conditions that yields the

familiar isotropic form of the Schwarzschild metric (2.35); other choices of boundary

conditions yield different static solutions, also with maximal slicing.

In the presence of

matter, static solutions to the equations of motion ∇

must be solved simultaneously

to obtain static spacetimes.

By contrast, as we discussed in Section 3.2, there exists strong evidence that rotating

Kerr black holes do not admit spatial slices that are conformally ﬂat. Adopting confor-

mal ﬂatness, then, would preclude our obtaining a spatial slice of a rotating Kerr black

hole. The dynamical evolution of any CTS initial data that we might construct for a

rotating black hole in vacuum thus must display some time dependence that we can

interpret as a gravitational wave perturbation of our rotating hole. This issue serves as

a motivation for the “waveless” approximation, yet another decomposition of the con-

straint equations, which we discuss brieﬂy in the following section. Before discussing

this approach, we note that numerical CTS solutions for a rotating black hole, adopting

See equations (4.23)–(4.25). Given that these solutions are spherically symmetric they are automatically conformally

ﬂat. A coordinate transformation from the areal radius R in equations (4.23)–(4.25) to an isotropic coordinate r (see

exercise H.1) brings the spatial metric into the form ψ