Conformal thin-sandwich decomposition

2.3 Conformal thin-sandwich decomposition

The two general initial-value decompositions outlined in Section 2.2.1 and Section 2.2.2 require identical freely specified data ( $&tidle;γij$ , $M&tidle; ij$ , and $K$ ), yet they usually produce different physical initial data. One shortcoming of these approaches is that they provide no direct insight into how to choose the freely specifiable data. All of the data are determined by schemes that involve only a single spacelike hypersurface. The resulting constraint equations are independent of the kinematical variables $α$ and $i β$ that govern how the coordinates move through spacetime, and thus there is no connection to dynamics. York’s conformal thin-sandwich decomposition [115 ] takes a different approach by considering the evolution of the metric between two neighboring hypersurfaces (the thin sandwich). This decomposition is very similar to an approach originated by Wilson [104

, 46

], but is somewhat more general. Perhaps the most attractive feature of this decomposition is the insight it yields into the choice of the freely specifiable data.

The decomposition begins with the standard conformal decomposition of the 3-metric (16 ). However, we next make use of the evolution equation for the metric (13 ) in order to connect the 3-metrics on the two neighboring hypersurfaces. Label the two slices by $t$ and $′ t$ , with $′ t = t + δt$ , then $′ γij = γij + (∂tγij) δt$ . We would like to specify how the 3-metric evolves, but we do not have full freedom to do this. We know we can freely specify only the conformal 3-metric, and similarly, we are free to specify only the evolution of the conformal 3-metric. We make the following definitions:

and

The latter definition is made for convenience, so that we can treat $ψ$ , $&tidle;γij$ , and $u&tidle;ij$ as regular scalars and tensors instead of as scalar- and tensor-densities within this thin-sandwich formalism.

The conformal scaling of $uij$ follows directly from (16 ), (41 ), (42 ), (43 ), and the identity that, for any small perturbation, $δln γ = γijδγij$ . The result is

which relies on the useful intermediate result that⁶

Equation (41 ) represents the tracefree part of the evolution of the 3-metric, so (13 ) becomes

Using the conformal scalings (22

), (35

), and (44

), we obtain

York has pointed out that it is natural to use the following conformal rescaling of the lapse:

This rescaling follows naturally from the “slicing function” that replaces the usual lapse ( $√ -- α = γ&tidle;α$ ) which has been critical in solving several problems [4 ]. It also results in the natural conformal scaling (22

) postulated for the tracefree part of the extrinsic curvature. Substituting (48

) into (47

) yields what is taken as the definition of the tracefree part of the conformal extrinsic curvature,

Because the tracefree extrinsic curvature satisfies the normal conformal scaling, the Hamiltonian constraint will take on the same form as in (32 ). However, the momentum constraint will have a very different form. Combining equations (16 ), (19 ), (21 ), (22 ), and (49 ) with the momentum constraint (15 ), we find that it simplifies toUpdate

Let us, for convenience, group together all the equations that constitute the conformal thin-sandwich decomposition:

γij = ψ4&tidle;γij, ij − 10 &tidle;ij 1 −4 ij K = ψ A + 3ψ &tidle;γ K, &tidle; ij -1-( &tidle; ij ij) A = 2&tidle;α (𝕃β) − u&tidle; , (51 ) &tidle; i &tidle; ij &tidle; 4 6 &tidle;i &tidle; (1--ij) 10 i Δ 𝕃β − (𝕃 β) ∇j ln &tidle;α − 3α&tidle;ψ ∇ K = &tidle;α∇j &tidle;α&tidle;u + 16 πG &tidle;αψ j, &tidle;∇2 ψ − 1ψ R&tidle;− 1-ψ5K2 + 1ψ −7 &tidle;Aij &tidle;Aij = − 2πG ψ5ρ. 8 12 8

In this decomposition (51

)Update , we are free to specify a symmetric tensor $&tidle;γij$ as the conformal 3-metric, a symmetric tracefree tensor $&tidle;uij$ , a scalar function $K$ , and the scalar function $&tidle;α$ . Solving this set of equations with appropriate boundary conditions yields initial data $γij$ and $Kij$ on a single hypersurface. However, we also know the following: If we chose to use the shift vector obtained from solving (50

) and the lapse from (48

) via our choice of $&tidle;α$ and our solution to the Hamiltonian constraint, then the rate of change of the physical 3-metric is given by

( ) ∂tγij = uij + 2γij ¯∇kβk − αK 3 [ ( )] (52 ) = ψ4 &tidle;uij + 23 &tidle;γij &tidle;∇k βk + 6βk∇&tidle;k ln ψ − ψ6 &tidle;αK .

This direct information about the consequences of our choices for the freely specifiable data is something not present in the previous decompositions. As we will see later, this framework has been used to construct initial data that are in quasiequilibrium.