York–Lichnerowicz conformal decompositions

2.2 York–Lichnerowicz conformal decompositions

For general initial-data configurations, the most widely used class of constraint decompositions are the York–Lichnerowicz conformal decompositions. At their heart are a conformal decomposition of the metric and certain components of the extrinsic curvature, together with a transverse-traceless decomposition of the extrinsic curvature.

First, the metric is decomposed into a conformal factor $ψ$ multiplying an auxiliary 3-metric [69 , 107 , 108 ]:

The auxiliary 3-metric $&tidle;γij$ is often called the conformal or background 3-metric, and it carries five degrees of freedom. Its natural definition is given by

leaving $&tidle;γ = 1$ , but we are free to choose any normalization for $&tidle;γ$ . Using (16

), we can rewrite the Hamiltonian constraint (14

) as

where $&tidle;∇2 ≡ &tidle;∇i∇&tidle; i$ is the scalar Laplace operator, and $∇&tidle; i$ and $R&tidle;$ are the covariant derivative and Ricci scalar associated with $&tidle;γij$ . Equation (18

) is a quasilinear elliptic equation for the conformal factor $ψ$ , and we see that the Hamiltonian constraint naturally constrains the 3-metric.

The conformal decomposition of the Hamiltonian constraint was proposed by Lichnerowicz. But, the key to the full decomposition is the treatment of the extrinsic curvature introduced by York [109 , 110 ]. This begins by splitting the extrinsic curvature into its trace and tracefree parts,

The decomposition proceeds by using the fact that we can covariantly split any symmetric tracefree tensor as follows:

Here, $T ij$ is a symmetric, transverse-traceless tensor (i.e., $∇jT ij = 0$ and $T i= 0 i$ ) and

After separating out the transverse-traceless portion of $𝒮ij$ , what remains, $(𝕃X )ij$ , is referred to as its “longitudinal” part. We now want to apply this transverse-traceless decomposition to the tracefree part of the extrinsic curvature $Aij$ . However, the conformal decomposition of the metric leaves us with at least two ways to proceed.

The goal of the decomposition is to produce a coupled set of elliptic equations to be solved with some prescribed boundary conditions. We have already reduced the Hamiltonian constraint to an elliptic equation being solved on a background space in terms of differential operators that are compatible with the conformal 3-metric. In the end, we want to reduce the momentum constraints to a set of elliptic equations based on differential operators that are compatible with the same conformal 3-metric. But, the longitudinal operator (21 ) can be defined with respect to any metric space. In particular, it is natural to consider decompositions with respect to both the physical and conformal 3-metrics.

2.2.1 Conformal transverse-traceless decomposition

Let us first consider decomposing $Aij$ with respect to the conformal 3-metric [111 , 116 ]. As we will see, when certain assumptions are made, this decomposition has the advantage of producing a simpler set of elliptic equations that must be solved. The first step is to define the conformal tracefree extrinsic curvature $ij A&tidle;$ by

Next, the transverse-traceless decomposition is applied to the conformal extrinsic curvature,

Note that the longitudinal operator $&tidle;𝕃$ and the symmetric, transverse-tracefree tensor $Q&tidle;ij$ are both defined with respect to covariant derivatives compatible with $&tidle;γij$ .

Applying equations (16 ), (19 ), (21 ), (22 ), and (23 ) to the momentum constraints (15 ), we find that they simplify to

whereUpdate

and we have used the fact that

for any symmetric tracefree tensor $ij 𝒮$ .

In deriving equation (24 ), we have also used the fact that $Q&tidle;ij$ is transverse (i.e. $&tidle;∇j &tidle;Qij = 0$ ). However, in general, we will not know if a given symmetric tracefree tensor, say $M&tidle;ij$ , is transverse. By using (20 ) we can obtain its transverse-traceless part $&tidle;Qij$ via

and using the fact that if $&tidle;Qij$ is transverse, we find

Thus, Eqs. (27

) and (28

) give us a general way of constructing the required symmetric transverse-traceless tensor from a general symmetric traceless tensor.

Using the linearity of $&tidle;𝕃$ , we can rewrite (23 ) as

where

Similarly, using the linearity of $&tidle; Δ 𝕃$ , we can rewrite (24

) as

By solving directly for $V i$ , we can combine the steps of decomposing $M&tidle; ij$ with that of solving the momentum constraints.

After applying (19 ) and (22 ) to the Hamiltonian constraint (18 ), we obtain the following full decomposition, which I will list together here for convenience:

γij = ψ4 &tidle;γij, ij −10 ij 1 −4 ij K = ψ A&tidle; + 3ψ &tidle;γ K, A&tidle;ij = (&tidle;𝕃V )ij + &tidle;M ij, (32 ) i 2 6 i ij 10 i &tidle;Δ 𝕃V − 3ψ &tidle;∇ K = − ∇&tidle;j M&tidle; + 8 πG ψ j , &tidle;∇2ψ − 1ψ &tidle;R − -1ψ5K2 + 1ψ−7A&tidle; A&tidle;ij = − 2πG ψ5 ρ. 8 12 8 ij

In the decomposition given by (32 ), we are free to specify a symmetric tensor $&tidle;γ ij$ as the conformal 3-metric, a symmetric tracefree tensor $&tidle; ij M$ , and a scalar function $K$ . Then, with given matter energy and momentum densities, $ρ$ and $i j$ , and appropriate boundary conditions, the coupled set of constraint equations for $ψ$ and $V i$ are solved. Finally, given the solutions, we can construct the physical initial data, $γij$ and $Kij$ .

The decomposition outlined above has the interesting property that if we choose $K$ to be constant and if the momentum density vanishes⁴ , then the momentum constraint equations fully decouple from the Hamiltonian constraint. As we will see later, this simplification has proven to be useful.

2.2.2 Physical transverse-traceless decomposition

Alternatively, we can decompose $Aij$ with respect to the physical 3-metric [82 , 83 , 84 ]. We decompose the extrinsic curvature as

In this case, the longitudinal operator $¯𝕃$ and the symmetric transverse-tracefree tensor $ij Q$ are both defined with respect to covariant derivatives compatible with $γij$ .

Applying equations (16 ), (19 ), (21 ), (33 ), and (26 ) to the momentum constraint (15 ), we find that it simplifies to

where we have used the fact that

As in the previous section, we will obtain the symmetric transverse-traceless tensor $Qij$ from a general symmetric tracefree tensor $&tidle;ij M$ by using (20 ). In this case, we take

and use the fact that $Qij$ is transverse, to obtain

Again, we can define

and use the linearity of $&tidle;𝕃$ and $&tidle;Δ 𝕃$ to combine the process of obtaining the transverse-traceless part of $M ij$ and solving the momentum constraints. We obtain the following full decomposition, which I will list together here for convenience⁵ :

4 γij = ψ &tidle;γij(, ) Kij = ψ− 4 A&tidle;ij + 1γ&tidle;ijK , 3 &tidle;Aij = (&tidle;𝕃V )ij + ψ −6M&tidle;ij, (39 ) i ij 2 i −6 ij 4 i Δ&tidle;𝕃V + 6 (𝕃&tidle;V ) &tidle;∇j lnψ = 3∇&tidle; K − ψ ∇&tidle;j M&tidle; + 8πG ψ j , ∇&tidle;2 ψ − 1ψR&tidle; − 1-ψ5K2 + 1ψ5 &tidle;A &tidle;Aij = − 2πG ψ5ρ. 8 12 8 ij

In the decomposition given by (39 ), we are again free to specify a symmetric tensor $&tidle;γij$ as the conformal 3-metric, a symmetric tracefree tensor $&tidle;M ij$ , and a scalar function $K$ . Then, with given matter energy and momentum densities, $ρ$ and $i j$ , and appropriate boundary conditions, the coupled set of constraint equations for $ψ$ and $i V$ are solved. Finally, given the solutions, we can construct the physical initial data, $γij$ and $Kij$ .

Notice that, while very similar to the decomposition from Section 2.2.1, the sets of equations are distinctly different. In general, if we make the same choices for the freely specifiable data in both decompositions (i.e., we choose $&tidle;γij$ , $&tidle; ij M$ , and $K$ the same), we will produce two different sets of initial data. Both will be equally valid solutions of the constraint equations, but they will have distinct physical properties.

There is at least one exception to this. Assume we have a valid set of initial data $γij$ and $Kij$ , which satisfies the constraint equations (14 ) and (15 ). For any everywhere-positive function $Ψ$ , we define our freely specifiable data as follows:

−4 &tidle;γij ≡ Ψ γij, M&tidle; ij ≡ Ψ10 (Kij − 1γijK ) , (40 ) i 3 K ≡ K i.

Then the solution to both sets of equations, assuming we use correct boundary conditions, will be $ψ = Ψ$ and $i V = 0$ , which yields the original data as the solution for each decomposition.