Abstract
In this paper both the initial value problem and the conformal thin sandwich problem are written in a unified way.
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1 Some Geometry and Notation
The spacetime metric g μν will be written in a moving frame adapted to the slices of constant time as
where the spatial scalar N is the lapse function and βi is the (spatial) shift vector. In a natural (coordinate) basis, β i = g 0i (β i = g ij βj and g ij , g kl are taken as the 3 ×3 inverses of one another; they are riemannian). From this one can see that βi is a spatial vector and β i is a spatial one-form with respect to arbitrary spatial coordinate transformations provided these transformations are not time-dependent.
The spacetime cobasis is
and the dual vector basis is
We see that ∂ 0 is a Pfaffian derivative while ∂ t and ∂ i are natural derivatives. The basis vector ∂ 0 can be generalized to the operator on spatial tensors
which, it should be noted, commutes with∂ i and propagates orthogonally to t = const.slices. It is obvious that ∂ i = ∂∕∂x i and ∂ t = ∂∕∂t commute, because they are both natural derivatives. What is sometimes forgotten, but has been known for more than 50 years, is that (for example) the spatial Lie derivative £ β, for any vector field βj, commutes with ∂∕∂x i when they act on tensors and more general objects such as spatial connections. This result holds in an even more general form. See, for example, Schouten’s Ricci Calculus [20], p. 105, Eq.(10.17). It is noteworthy that \hat{∂ 0} acts orthogonally to t = constant slices and that it is actually the only time derivative that ever occurs in the 3+1 formulation of general relativity based on an exact or locally exact (integrable) “time” basis one-form θ0, such as dt.
The connection coefficients in our “Cauchy-adapted” frame are given by
where Γ denotes an ordinary Christoffel symbol, parentheses around indices denote the symmetric part (so A (βγ) = \frac{1} {2}(A βγ+A γβ)), and C denotes the coefficients of the commutator
Our spacetime covariant derivative convention associated with (5) is
The only non-vanishing C’s are
For the spatial covariant derivative we write
Because the shift βi is in the basis, the spacetime metric in the Cauchy basis that we use has no time–space components. A convenient consequence is that there is no ambiguity in writing g ij , g kl, or Γ jk i. For the spacetime metric determinant we will write detg μν = −N 2 g, g = detg ij ; and g ij is here and hereafter considered as a 3 ×3 symmetric tensor.
The ω’s are given next. Note that [MTW] write ωγβ α where we have ωβγ α for the same object. (We follow [6] here.) Of course, this convention does not matter in a coordinate basis, where all of the connection coefficients are symmetric. Here, note that only ω0j i and ω j0 i differ.
The Riemann tensor satisfies the commutator
where (Riem)αβ γ δ would be denoted (Riem)δαβ γ in [MTW]. Again, we are using here the conventions of [6].
There are a number of possible definitions for the second fundamental tensor or extrinsic curvature tensor K ij . This does not measure curvature in the sense of Gauss or Riemann, where curvature has dimensions of (length)−2. The extrinsic curvature is a measure, at a point on a spatial slice, of the curvature of a spacetime geodesic curve relative to a spatial geodesic curve to which it is tangent at the point. The dimensions are therefore (length)−1. (See, for example, [18]and the Appendix of[19] for a detailed discussion of extrinsic curvature tensors.) This has the same dimension as a connection symbol
Also
where∇ i is the spatial covariant derivative with connection γ jk i = ω jk i = Γ jk i; and the final term, apart the sign, is ℒ β g ij .
The Riemann tensor components are in accord with the Ricci identity (13)
The spatial Riemann tensor is denoted R ij k l . These curvatures are related by the Gauss–Codazzi–Mainardi equations for codimension one (see, for example, [20])
One can likewise form and decompose the Ricci tensor, which has the definition
Then we can construct
where ΔN denotes the spatial “rough” or scalar Laplacian acting on the lapse function: ΔN≡g ij∇ i ∇ j N≡∇2 N. The trace of K ij is K, the “mean curvature,” of the time slice.
It is important to know the spacetime scalar curvature, which we call (trRic):
in the form
where R is the spatial scalar curvature, because the spacetime scalar curvature density is the lagrangian density of the Hilbert action principle [14], explicitly modified in [24] to conform to the fixation of the metric on the boundary. The scalar curvature itself will be needed later. It is found from (24) and (21), (22), and (23) to be
2 Einstein’s Equations
Einstein used his insights about the principle of equivalence and his principle of general covariance (spacetime coordinate freedom plus a pseudo-riemannian metric not given a priori) in arriving at the final form of his field equations. As is now well known, his learning tensor analysis from Marcel Grossmann was an essential enabling step. The equations, using the Einstein tensor
are
where κ=8πG, c = 1, G = Newton’s constant, and the stress–energy–momentum tensor of fields other than gravity (the “source” tensor) T μν must satisfy, as Einstein reasoned, in analogy to the conservation laws of special relativity,
corresponding to
which is an identity, the “third Bianchi identity” or the “(twice) contracted Bianchi identity [23].” For purposes of the discussion below, besides c = 1 we also take Dirac’s form of Planck’s constant to be one: ℏ = 1. (Mass is now inverse length.)
Here we will consider only the vacuum theory. This is non-trivial because the equations are non-linear (gravity acts as a source of itself) and because the global topology and (or) boundary conditions are not prescribed by the equations. We assume that the object of solving the equations is to find the metric. In the 3+1 form of the equations, which is very close to a hamiltonian framework, the object is to obtain g ij and K ij , along with a workable specification of α=Ng −1∕2 and βi which, as we shall see shortly, are not determined by Einstein’s equations. For the vacuum case, one can use in four spacetime dimensions either of these two equations
or
For completeness, we note that the form of equation (32) with “sources” using the Ricci tensor is
where T = T α α, the trace of the stress–energy tensor.
We now remark on a couple of points that are sometimes useful to bear in mind. The curvature equations we have given in purely geometric form. They can be converted into physical gravity equations explicitly by assigning the physical dimensions length (L), mass (M), and time (T). We have already chosen c = 1, so T = L. We shall work in terms of L. Next we choose to make action dimensionless by setting ℏ = 1, which yields M = L −1. Then G (and κ) have dimension L 2. This is a handy viewpoint for the physicist and mathematician even if quantum effects are not considered. It enables us to display rather easily the nonlinear self-coupling that arises even in vacuum from the particular geometric nature of General Relativity.
We take the metric to be dimensionless while t and x i have dimension L. (Think of natural locally Riemannian normal coordinates to make this view palatable.) The canonical form of the action based on (2κ)−1 g 1∕2 N(trRic) is given by [1, 2, 24]. It yields in place of K ij the closely related field canonical momentum [2]
Inverting this expression in three dimensions yields
where π=π k k is the trace. We note that from (35) we can obtain for the mean curvature
the well-known, and vital, integrand of the boundary term of the Hilbert action [14] that can be seen [24] to convert it to the canonical action [1, 2].
Denoting the terms in the rectangular brackets in (35) by μ ij , which has dimension L, then
Now we could rewrite the curvature equations with the gravitational interaction explicit. For example, the Gauss–Codazzi–Mainardi equations become
3 The 3+1-Form of Einstein’s Equations
It is helpful to write out the ten vacuum equations using both (Ric) and (Ein):
This form was noted by Lichnerowicz [16] in the case of zero shift as being revealing. First, recall the geometric identity (15)
or
From the first equation in (42), one can obtain
The second and third equations in (42) contain no terms ∂ t K ij (i.e., no “accelerations” ∂ t ∂ t g ij ) and are, therefore, constraints on the initial values of g ij and K ij . As previously mentioned, in this “canonical”-like 3+1 form, there are no time derivatives of N = αg 1∕2 or of βi. In a second-order formalism, ∂ t N and ∂ t βi would appear, as we see from (43). To make second order wave operators on all components of the spacetime metric, the (original) harmonic coordinate conditions (−g)−1∕2 ∂ μ[(−g)1∕2 g μν] = 0 in natural coordinates were introduced and shown, along with the constraints, to be conserved by the resulting “reduced” equations if the constraints were assumed to hold at the “initial” time [4]. But no powers of \dot{N} and \dot{β}i appear in (25), the lagrangian of Hilbert’s action for the Einstein’s equations. We have thus an easy way of seeing that \dot{N} and \dot{β}i are dynamically irrelevant. We find from the second and third equations of (42), respectively,
These equations were derived in detail and displayed in [25], without a 3+1 splitting of the basis frames and coframes. An arbitrary spacetime basis was used there in order to remove what some people regarded as the “taint” of using particular coordinates. They are the standard 3+1 equations, wrongly called the standard ADM equations. In regard to evolving g ij and K ij , I do not claim that (43) and (44) are preferred for any other reason other than their maximum simplicity and absolute correctness when written in explicitly canonical form, using the variable πij, defined below, in place of K ij . I say nothing here about the numerical properties of (43) and(44).
The canonical equations derived by Arnowitt, Deser, and Misner [2] and by Dirac [12, 13], are not equivalent to (44) given above, even when written in the same formalism, that is, with g ij and K ij . This is because their equation of motion is G ij = 0 rather than R ij = 0. Although G μν = 0 and R μν = 0 are equivalent, this is not true of spatial components. Instead, one has the key identity [1]
or
Therefore, G ij = 0 is not the correct equation of motion unless the constraint C = 0 also holds.
For the interested reader, I remark that this means the hamiltonian vector field of the ADM and Dirac canonical formalisms is not well-defined throughout the phase space. There is an easy cure in the ADM approach, which is based on a canonical action principle. When the metric g ij is varied, one must hold fixed the “weighted” or “densitized” lapse function α=g −1∕2 N, instead of just the scalar lapse N. Thus, one carries out independent variations of g ij , α, βi, and πij = (2κ)−1 g 1∕2 (Kg ij−K ij)[1, 3].
4 Conformal Transformations
A very useful technique for transforming the constraints C i and C ((45) and (46)) into a well posed problem involving elliptic partial differential equations is to use conformal transformations of the essential spatial objects g ij , K ij , N, and βi. Along the way we will again encounter the densitized lapse
which has coordinate weight (−1): N is a scalar with respect to spatial time-independent coordinate transformations (weight zero by definition).
The conformal factor will be denoted by φ. It will be assumed that φ>0 throughout. The conformal transformation is defined by its action on the metric
It is called “conformal” because it preserves angles between vectors intersecting at a given point, whether one constructs the scalar product and vector magnitudes with g ij or \bar{g} ij . The power “4” in (50) is convenient for three dimensions. For dimension n≥3, the “convenient power” is 4(n−2)−1 for the metric conformal deformation. The neatness of this choice comes out most clearly in the relation of the scalar curvatures \bar{R}=R(\bar{g}) and R = R(g) below. From (50) and the fact that the spatial connection is simply the “Christoffel symbol of the second kind { i\abovewithdelims\lbrace\rbrace0.0pt _ j k} ” which we denote by Γ jk i,
we find that
From \bar{Γ} jk i we can find the relationship between \bar{R} ij and R ij .
There is no need to derive the transformation of the Riemann tensor, for in three dimensions R ijkl can be expressed in terms of g ij and R ij . One can see that this must be so, for both the Ricci and Riemann tensors have six algebraically independent components. The formula relating them has long been known. It is displayed for example in [27]. This formula can be obtained from the identical vanishing of the Weyl conformal curvature tensor in three dimensions. But riemannian three-spaces are not conformally flat, in general. The Weyl tensor in three dimensions is replaced by the Cotton tensor [10], which is conformally invariant and vanishes if the three-space is conformally flat
where
Its dual [28], found by using the inverse volume form εmjk on the skew pair [jk] and by raising the index i to l is a symmetric tensor∗C lm with trace identically zero and covariant divergence identically zero. The Cotton tensor has third derivatives of the metric and is therefore not a curvature tensor, but rather is a differential curvature tensor with dimensions (length)−3. Therefore, the dual Cotton tensor is
Under conformal transformations, we have that \bar{g}ij = φ−4 g ij, \bar{ε} ijk = φ6ε ijk (the volume three-form), \bar{C} ijk = C ijk , and \bar{ε}ijk = φ−6εijk. Therefore,
The properties of∗C ij hold for an entire conformal equivalence class, that is, for all sets of conformally related riemannian metrics as in (50) for all 0<φ<∞. The divergence of∗C ij is identically zero, so it need not be surprising that
using the barred and unbarred connections to form the covariant divergence operators \bar{∇ i } and∇ j . We see that (57) is the natural conformal transformation law for symmetric, traceless type \binom{2}\binom{{0} tensors in three dimensions, whose divergence may or may not vanish. Note that in obtaining (58), the scaling (57) was used. But no properties of∗C ij were employed except the symmetry type of the tensor representation: symmetric with zero trace.
We now pass to the famous formula for the conformal transformation of the scalar curvature R. It was introduced in connection with an early treatment of the constraints in [16]. From (53) it follows that
So far every transformation has followed from the defining relation \bar{g} ij = φ4 g ij . A glance at both the constraints (45) and (46) shows that we must deal with K ij . The method here can be deduced by writing K ij as
where A ij is the traceless part of K ij. We treat A ij and K differently because A ij and Kg ij can be regarded as different irreducible types of symmetric two-index tensors. They also have different conformal transformations. Lichnerowicz took K = 0 [16]. But this is too restrictive, and even then the simplified “momentum constraint” (45) was not solved. Mme. Choquet-Bruhat first argued that one has to solve the momentum constraint with a second-order operator on a vector potential [5]. A very useful result was given in [26] and it was used for many years to solve the momentum constraint. Its imperfections were noted first by O’Murchadha [17], Isenberg [15], and the author. The inference was that the early method was only an ansatz. A better method yet, with no ambiguities, was displayed in [25] and [19]. It is given below.
Suppose an overbar denotes a solution of the constraints and the corresponding object without an overbar denotes a “trial function.” The strategy is to “deform” the trial objects conformally into barred quantities, that is, into solutions. Every object we deal with has, in effect, a “conformal dimension,” which is not given by its physical dimension or by its tensorial character but by how it transforms under a change of the basis or of the natural coordinates.
5 An Elliptic System
We write (45) and (46) in the barred variables as
Conformal transformations for objects that are purely concomitants of \bar{g} ij (or g ij ) are derived as above in a straightforward manner. But the extrinsic curvature variables have to be handled with a modicum of care. The transformations obtained by extending \bar{g} ij = φ4 g ij to all of the spacetime metric variables is not appropriate because the view that spacetime structure is primary is not helpful in a situation, as here, where there is as yet no spacetime.
We begin with \bar{K}. We hold it fixed because its inverse in the simpler cosmological models is the “Hubble time,” without a knowledge of which the epoch is not known. Data astronomers obtain from different directions in the sky, or at different “depths” back in time are basically correlated and they fix \bar{K} implicitly. Therefore, I long ago adopted the rule [24] of fixing the “mean curvature” \bar{K}
under conformal transformations. Thus it is specified a priori.
What to do about the symmetric tracefree tensor \bar{A}ij? The prior discussion of∗C ij indicates the transformation
But symmetric tensors “\bar{T}ij” have, in a curved space, three irreducible types that are formally L 2-orthogonal [11]. One is the trace (\bar{g}ij\bar{T} k k), another is like∗C ij, that is, a part with vanishing covariant divergence. Finally, a symmetric tracefree tensor can be constructed from a vector
the “conformal Killing form” of X i. (I have not found other constructions that are sufficiently well-behaved under conformal transformations to be useful in this problem.) This expression (65) vanishes if X i is a conformal killing vector of \bar{g} ij . Then, X i would be a conformal killing vector of every metric conformal to \bar{g} ij . Therefore,
and
which misses obeying our “rule” (57) or (64). For a long time, the mismatched powers required a work-around to obtain an ansatz for solving the constraints [27], but there is a simple and elegant solution [28]. The vectorial part (65) needs a weight factor and a corresponding change in the measure of orthogonality.
Recall our statement that the densitized lapse α is the preferred undetermined multiplier (rather than the lapse N) in the action principle leading to 3+1 (or canonical) equations of motion. See [1] where this is made perfectly clear. This is not to say anything about the “best” form of the \hat{∂ 0}K ij (or \hat{∂ 0}πij) equation of motion for calculational purposes. In fact, the system for \hat{∂ 0}g ij and \hat{∂ 0}K ij is not hyperbolic. But only this form gives a hamiltonian vector field well-defined in the entire momentum phase space.
To proceed, we note that one does not scale undetermined multipliers. Therefore
But because \bar{α}=\bar{g}−1∕2\bar{N} and \bar{g}1∕2 = φ6 g 1∕2, then \bar{N}=\bar{g}1∕2α=g 1∕2α implies
I have known that the transformation (69) was useful since 1971. But, thinking that N was an undetermined multiplier – a lowly “C-number,” independent of the dynamical variables, in Dirac’s well-known parlance – I did not use (69). Then I learned about the densitized lapse and saw its role in the action principle. It then dawned on me that (69) was correct all along. It made its first appearance in the conformal thin sandwich problem [25].
The lapse becomes, thus, a dynamical variable [1, 3, 19]. A look at (69) and at the relation between ∂ t \bar{g} ij and \bar{K} ij gives us the scalar weight factor (−2\bar{N})−1 in the decomposition of \bar{A}ij
The subscript (δ) indicates that the covariant divergence of \bar{A}(δ) ij vanishes. Note that (70) does not mean that the extrinsic curvature is sensitive to N. It is not. What it does mean is that the identification of the divergence-free and trace-free part of the extrinsic curvature is, in part, dependent on N. Also note that the two parts of \bar{A}ij are formally L 2-orthogonal both before and after a conformal transformation, with the geometrical spacetime measure
instead of the spatial measure g 1∕2. Therefore, we have
Upon integration by parts, with suitable boundary conditions, or no boundary, each of the integrals (72) vanishes.
We construct \bar{A}(δ) ij or A (δ) ij by extracting from a freely given symmetric tracefree tensor \bar{F}ij = φ−10 F ij its transverse-tracefree part, which will be our A (δ) ij
with
The momentum constraints
become, with Z i = X i−Y i,
The solution for Z i, with given N, will give the parts of \bar{K} ij ,
However(75) contains φ and is coupled to the “hamiltonian constraint” (62) unless the “constant mean curvature” (CMC) condition K = constant (in space, ∂ j K = 0) can be employed, as introduced in [24]. This includes maximal slicing K = 0 [16]. (Lichnerowicz did not propose the CMC condition as claimed in [22].)
Gathering the transformations for \bar{R} and \bar{K}ij enables us to write the hamiltonian constraint as the general relativity version of the Laplace–Poisson equation
Suppose, for example, we choose N = 1. Then
We are certainly entitled to have chosen g ij such that g 1∕2 = 1, without loss of generality. Thus we recover Teitelboim’s gauge for the lapse equation
in his noted paper on the canonical path integral in general relativity [21].
A bit more generally, if we choose
we see that \bar{N} automatically satisfies the time gauge equation used by Choquet-Bruhat and Ruggeri [8].
The constraints have the same form as they do in the thin sandwich formulation: see [25]. Therefore, the space of solutions has the properties obtained in [7]; see also [9].
Write out from the formula for ∂ t \bar{g} ij its tracefree part \bar{u} ij , which we can use as the velocity of the conformal metric
with (\bar{Z}+\bar{β}) j ≡\bar{g} ij (Z i+βi). This has the form of the solution in the conformal thin sandwich problem. The choice of shift βi = −Z i is possible here and renders a simple final form.
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Acknowledgement
The author thanks Michael Brown for his invaluable assistance in preparing the manuscript.
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I dedicate this paper to John Archibald Wheeler, a wonderful mentor and colleague.
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York, J.W. (2010). Unified Form of the Initial Value Conditions. In: Ciufolini, I., Matzner, R. (eds) General Relativity and John Archibald Wheeler. Astrophysics and Space Science Library, vol 367. Springer, Dordrecht. https://doi.org/10.1007/978-90-481-3735-0_5
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