# Conformal Thin Sandwich Decomposition

#### Notation

Let Newton's constant $G=1$. The matter energy density is $\rho$, the momentum density is $j^a$, and $s$ is the trace of the spatial stress.

#### Original Thin Sandwich Decomposition

Freely specify the following:

$\begin{array}{rcl} {\tilde g}_{ab} & & {\rm conformal\ metric} \\ {\tilde u}_{ab} & & {\rm symmetric\ trace\ free\ tensor} \\ K & & {\rm trace\ of\ the\ extrinsic\ curvature} \\ {\tilde\alpha} & & {\rm scalar\ density\ of\ weight\ }{-1} \end{array}$

The variable $\tilde u_{ab}$ is the trace-free part of the time derivative of the conformal metric $\tilde g_{ab}$. Also, $\tilde\alpha$ is related to the scalar lapse function $\alpha$ by $\alpha = \tilde\alpha\sqrt{g}$, where $g$ is the determinant of the physical spatial metric $g_{ab}$.

Let ${\tilde D}_a$ and ${\tilde R}_{ab}$ denote the covariant derivative and Ricci tensor built from the conformal metric. Indices on ${\tilde D}_a$, ${\tilde u}_{ab}$ and ${\tilde R}_{ab}$ are raised with the inverse conformal metric. Define the operators

$\begin{array}{rcl} (\tilde L \beta)^{ab} & \equiv & {\tilde D}^a \beta^b + {\tilde D}^b \beta^a - \frac{2}{3} {\tilde g}^{ab} {\tilde D}_c \beta^c \\ {\tilde \Delta}_{L} \beta^a & \equiv & {\tilde D}_b (\tilde L \beta)^{ab} = {\tilde D}_b {\tilde D}^b \beta^a + \frac{1}{3} {\tilde D}^a {\tilde D}_b \beta^b + {\tilde R}^a_b \beta^b \end{array}$

Solve the following equations for a scalar $\psi$ and a vector $\beta^a$:

$\begin{array}{rcl} {\tilde D}^a {\tilde D}_a \psi & = & \frac{1}{8} \psi {\tilde R} + \frac{1}{12} \psi^5 (K)^2 - \frac{1}{8} \psi^{-7} {\tilde A}_{ab}{\tilde A}^{ab} -2\pi \psi^5 \rho \\ {\tilde \Delta}_L \beta^a & = & {\tilde D}_b {\tilde u}^{ab} + 2\sqrt{\tilde g} {\tilde A}^{ab}{\tilde D}_b \tilde\alpha + \frac{4}{3} \tilde\alpha \sqrt{\tilde g} \psi^6 {\tilde D}^a K + 16\pi \tilde\alpha \sqrt{\tilde g} \psi^{6} j^a \end{array}$

where

${\tilde A}^{ab} \equiv \frac{1}{2\tilde\alpha\sqrt{\tilde g}} \left[ (\tilde L\beta)^{ab} - {\tilde u}^{ab} \right]$

The physical metric and extrinsic curvature are

$\begin{array}{rcl} g_{ab} & = & \psi^4 {\tilde g}_{ab} \\ K^{ab} & = & \psi^{-10} {\tilde A}^{ab} + \frac{1}{3} \psi^{-4} {\tilde g}^{ab} K \end{array}$

If the initial data $g_{ab}$, $K_{ab}$ are evolved using $\alpha = \sqrt{ g}\tilde\alpha$ for the initial scalar lapse and $\beta^a$ for the initial shift vector, then:

• the initial value of $\partial g_{ab}/\partial t$ will be given by
$\partial g_{ab}/\partial t = \psi^4 {\tilde u}_{ab} - \frac{2}{3} \psi^{10} \tilde\alpha K \tilde g_{ab} + 4 \psi^3 {\tilde g}_{ab} \beta^c {\tilde D}_c \psi + \frac{2}{3} \psi^4 {\tilde g}_{ab} {\tilde D}_c\beta^c$

#### Extended Thin Sandwich Decomposition

Freely specify the following:

$\begin{array}{rcl} {\tilde g}_{ab} & & {\rm conformal\ metric} \\ {\tilde u}_{ab} & & {\rm symmetric\ trace\ free\ tensor} \\ K & & {\rm trace\ of\ the\ extrinsic\ curvature} \\ {\dot K} & & {\rm scalar\ field} \end{array}$

Let ${\tilde D}_a$ and ${\tilde R}_{ab}$ denote the covariant derivative and Ricci tensor built from the conformal metric. Indices on ${\tilde D}_a$, ${\tilde u}_{ab}$ and ${\tilde R}_{ab}$ are raised with the inverse conformal metric. Define the operators

$\begin{array}{rcl} (\tilde L \beta)^{ab} & \equiv & {\tilde D}^a \beta^b + {\tilde D}^b \beta^a - \frac{2}{3} {\tilde g}^{ab} {\tilde D}_c \beta^c \\ {\tilde \Delta}_{L} \beta^a & \equiv & {\tilde D}_b (\tilde L \beta)^{ab} = {\tilde D}_b {\tilde D}^b \beta^a + \frac{1}{3} {\tilde D}^a {\tilde D}_b \beta^b + {\tilde R}^a_b \beta^b \end{array}$

Solve the following equations for a scalar $\psi$, vector $\beta^a$, and weight -1 scalar density $\tilde\alpha$:

$\begin{array}{rcl} {\tilde D}^a {\tilde D}_a \psi & = & \frac{1}{8} \psi {\tilde R} + \frac{1}{12} \psi^5 (K)^2 - \frac{1}{8} \psi^{-7} {\tilde A}_{ab}{\tilde A}^{ab} -2\pi \psi^5 \rho \\ {\tilde \Delta}_L \beta^a & = & {\tilde D}_b {\tilde u}^{ab} + 2\sqrt{\tilde g} A^{ab}{\tilde D}_b \tilde \alpha + \frac{4}{3} \tilde\alpha \sqrt{\tilde g} \psi^6 {\tilde D}^a K + 16\pi \tilde\alpha \sqrt{\tilde g} \psi^{6} j^a \\ {\tilde D}^a{\tilde D}_a \tilde \alpha & = & - \psi^{-2} ({\dot K} - \beta^a\partial_a K) /\sqrt{\tilde g} - \frac{1}{6} \tilde\alpha \psi^4 (K)^2 - \frac{3}{4}\tilde\alpha \tilde R + \frac{7}{4} \tilde\alpha \psi^{-8} {\tilde A}^{ab}{\tilde A}_{ab} \\ & & - 14\psi^{-1} {\tilde D}^a\psi {\tilde D}_a\tilde\alpha - 42\tilde\alpha \psi^{-2} {\tilde D}^a\psi {\tilde D}_a\psi + \frac{1}{2}\tilde \alpha \psi^4 (s + 4\rho) \end{array}$

where

${\tilde A}^{ab} \equiv \frac{1}{2\tilde\alpha\sqrt{\tilde g}} \left[ (\tilde L\beta)^{ab} - {\tilde u}^{ab} \right]$

The physical metric and extrinsic curvature are

$\begin{array}{rcl} g_{ab} & = & \psi^4 {\tilde g}_{ab} \\ K^{ab} & = & \psi^{-10} {\tilde A}^{ab} + \frac{1}{3} \psi^{-4} {\tilde g}^{ab} K \end{array}$

If the initial data $g_{ab}$, $K_{ab}$ are evolved using $\sqrt{ g}\tilde\alpha$ for the initial scalar lapse and $\beta^a$ for the initial shift vector, then:

• the initial value of $\partial K/\partial t$ will coincide with the scalar field $\dot K$.
• the initial value of $\partial g_{ab}/\partial t$ will be given by
$\partial g_{ab}/\partial t = \psi^4 {\tilde u}_{ab} - \frac{2}{3} \psi^{10} \tilde\alpha K \tilde g_{ab} + 4 \psi^4 {\tilde g}_{ab} \beta^c {\tilde D}_c \ln \psi + \frac{2}{3} \psi^4 {\tilde g}_{ab} {\tilde D}_c\beta^c$