Conformal Thin Sandwich Decomposition

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[edit] 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.

[edit] 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} \\ {\alpha} & & {\rm scalar\ density\ of\ weight\ }{-1} \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 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 \alpha + \frac{4}{3} \alpha \sqrt{\tilde g} \psi^6 {\tilde D}^a K + 16\pi \alpha \sqrt{\tilde g} \psi^{6} j^a \end{array}

where

{\tilde A}^{ab} \equiv \frac{1}{2\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}\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} \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

[edit] 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 \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 \alpha + \frac{4}{3} \alpha \sqrt{\tilde g} \psi^6 {\tilde D}^a K + 16\pi \alpha \sqrt{\tilde g} \psi^{6} j^a \\ {\tilde D}^a{\tilde D}_a \alpha & = & - \psi^{-2} ({\dot K} - \beta^a\partial_a K) /\sqrt{\tilde g} - \frac{1}{6} \alpha \psi^4 (K)^2 - \frac{3}{4}\alpha \tilde R + \frac{7}{4} \alpha \psi^{-8} {\tilde A}^{ab}{\tilde A}_{ab} \\ & & - 14\psi^{-1} {\tilde D}^a\psi {\tilde D}_a\alpha - 42\alpha \psi^{-2} {\tilde D}^a\psi {\tilde D}_a\psi + \frac{1}{2} \alpha \psi^4 (s + 4\rho) \end{array}

where

{\tilde A}^{ab} \equiv \frac{1}{2\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}\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} \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
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