Conformal Thin-Sandwich Data

The conformal thin-sandwich method chooses data that describe an instantaneous spatial geometry and its intended rate of change. Instead of prescribing a transverse-traceless momentum directly, it solves for the shift and conformal factor—and, in the extended system, the lapse—from a trace-free trial velocity. This makes quasi-equilibrium choices intuitive, but it does not remove elliptic kernels or guarantee uniqueness.

Required background. Analysis for Mathematical Relativity supplies elliptic solvability, The Einstein Equation supplies the constraints, Hypersurface Normals and Tangential Projectors supplies lapse, shift, and extrinsic curvature, and Differential Equations for Relativity supplies coupled boundary-value systems.

Helpful background. Causality and Global Structure explains Cauchy data, while PDEs and Green Functions for Gravitation reviews elliptic operators and kernels.

Velocity data and extrinsic curvature

Write the spacetime metric in lapse–shift form and use

$$K_{ij}=-\frac{1}{2N} \left(\partial_t h_{ij}-\mathcal L_\beta h_{ij}\right).$$

Taking the trace-free part gives

$$A^{ij}=\frac{1}{2N} \left((L\beta)^{ij}-u^{ij}\right),$$

where $u_{ij}$ is the trace-free part of $\partial_th_{ij}$ and $L$ is the conformal Killing operator. The signs follow from the volume convention $K_{ij}=-h_i{}^\mu h_j{}^\nu\nabla_\mu n_\nu$; changing the sign of $K$ changes both this relation and the momentum constraint.

Choose conformal free data

$$(\widetilde h_{ij},\widetilde u_{ij},\tau,\widetilde N), \qquad \widetilde h^{ij}\widetilde u_{ij}=0,$$

and seek $\phi>0$ and $\beta^i$. With

$$h_{ij}=\phi^4\widetilde h_{ij}, \qquad N=\phi^6\widetilde N,$$

the trace-free curvature is

$$A^{ij}=\phi^{-10}\widetilde A^{ij}, \qquad \widetilde A^{ij} =\frac{1}{2\widetilde N} \left((\widetilde L\beta)^{ij}-\widetilde u^{ij}\right).$$

Here $\widetilde N\gt0$, $\widetilde\Delta=\widetilde D^i\widetilde D_i$, and $(\widetilde L\beta)_{ij}=2\widetilde D_{(i}\beta_{j)}-\tfrac23\widetilde h_{ij}\widetilde D_k\beta^k$. In vacuum with $\Lambda=0$, the actual coupled equations are

$$\widetilde D_j\!\left[\frac{1}{2\widetilde N} (\widetilde L\beta)^{ij}\right] =\widetilde D_j\!\left[\frac{\widetilde u^{ij}}{2\widetilde N}\right] +\frac23\phi^6\widetilde D^i\tau,$$ $$-8\widetilde\Delta\phi+\widetilde R\phi -|\widetilde A|_{\widetilde h}^2\phi^{-7} +\frac23\tau^2\phi^5=0.$$

The lapse belongs inside the vector divergence. If $\widetilde N$ varies spatially, replacing the first operator by $(2\widetilde N)^{-1}\widetilde D_j(\widetilde L\beta)^{ij}$ loses first-derivative terms. The construction and its conformal covariance are developed in Bartnik and Isenberg 2004, §4.3, equations (71)–(75).

What is freely specified

The intended interpretation of each datum is:

• $\widetilde h_{ij}$ fixes the conformal spatial geometry;
• $\widetilde u_{ij}$ fixes a conformal trace-free trial velocity;
• $\tau=K$ fixes the mean curvature, related to local volume expansion;
• $\widetilde N$ fixes a densitized-lapse choice;
• boundary and asymptotic conditions select the physical solution class.

The unknown shift supplies a longitudinal correction so the reconstructed curvature satisfies the momentum constraint; for nonconstant $\tau$ this is not simply an orthogonal projection of the trial velocity. The conformal factor enforces the Hamiltonian constraint.

In the extended system one replaces the prescribed $\widetilde N$ by a prescribed $\partial_t\tau$ and solves for the lapse as an additional unknown. One must not prescribe both independently. The extra elliptic equation comes from the traced vacuum evolution equation,

$$(\partial_t-\beta^iD_i)\tau =-D^iD_iN+N\left(A_{ij}A^{ij}+\frac13\tau^2\right).$$

Setting $\widetilde u=0$ and $\partial_t\tau=0$ is often called quasi-equilibrium data, but it is a statement about the chosen coordinates and conformal representative. It does not prove the existence of a spacetime Killing field: the full spatial and normal Killing equations impose further conditions.

Kernels and nonuniqueness

If $Y^i$ is a conformal Killing field of $\widetilde h$, then $\widetilde L Y=0$. Consequently $\beta$ and $\beta+Y$ produce the same longitudinal tensor unless boundary data remove the kernel. The vector equation is solvable only when its source is orthogonal to this adjoint kernel.

Nonuniqueness can also be nonlinear. In the extended system, the lapse equation couples to powers of $\phi$ and may admit folds or multiple branches. A numerical elliptic solver converging to one branch is therefore not evidence of uniqueness. One must state the free-data regime, function spaces, positivity of lapse and conformal factor, kernel conditions, and the theorem that selects a branch. Carlotto discusses these structural issues in Carlotto 2021, §2.

The method is especially useful for compact-binary constructions because boundary conditions can encode asymptotic inertial behavior, approximate helical motion, or excised horizons. Those applications introduce additional physical assumptions and belong to numerical initial-data treatments. Here the mathematical point is narrower: the trial velocity enters only through its longitudinally corrected combination with the shift.

Conformal covariance provides another check. Under $\widetilde h'_{ij}=\vartheta^4\widetilde h_{ij}$, use $\phi'=\phi/\vartheta$, $\widetilde u'_{ij}=\vartheta^4\widetilde u_{ij}$, and $\widetilde N'=\vartheta^6\widetilde N$, with the same $\beta^i$ and $\tau$. Then $h$, $N$, and $A^{ij}$ are unchanged. In particular, $\widetilde A'^{ij}=\vartheta^{-10}\widetilde A^{ij}$ compensates exactly for $(\phi')^{-10}$. If a proposed formula changes the physical data under this representative change, at least one conformal weight is inconsistent.

Independent reconstruction check

Once $(\phi,\beta)$ is found, reconstruct $h_{ij}$ and $K_{ij}$ and check both constraints directly. This catches conformal-weight and sign errors. One should also recompute the trace-free velocity:

$$(\partial_t h_{ij})^{\mathrm{TF}} =-2N A_{ij}+(L\beta)_{ij}.$$

It must reproduce the physical tensor obtained from $\widetilde u_{ij}$ after the declared conformal rescaling. This check is stronger than verifying only the scalar Lichnerowicz equation.

Check your understanding

Let $Y$ be a conformal Killing field of $\widetilde h$. Why can $\beta$ and $\beta+Y$ represent the same conformal trace-free extrinsic curvature?

Solution

Because $(\widetilde L Y)^{ij}=0$, replacing $\beta$ by $\beta+Y$ leaves $\widetilde A^{ij}=[(\widetilde L\beta)^{ij}-\widetilde u^{ij}]/(2\widetilde N)$ unchanged. Boundary conditions or a normalization must fix this kernel before the shift itself is unique.

References

• Bartnik, Robert, and Jim Isenberg. “The Constraint Equations.” In The Einstein Equations and the Large Scale Behavior of Gravitational Fields, 1–38. Birkhäuser, 2004. arXiv:gr-qc/0405092.
• Carlotto, Alessandro. “The General Relativistic Constraint Equations.” Living Reviews in Relativity 24 (2021): 2. DOI.