Conformal Transverse-Traceless Decomposition

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Notation: Let Newton's constant G=1. The energy density is \rho and the momentum density is j^a.

Freely specify the following:

\begin{array}{rcl} {\tilde g}_{ab} & & {\rm conformal\ metric} \\ {\tilde M}_{ab} & & {\rm symmetric\ trace\ free\ tensor} \\ K & & {\rm trace\ of\ the\ extrinsic\ curvature} \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 M}_{ab} and {\tilde R}_{ab} are raised with the inverse conformal metric. Solve the following equations for \psi and V^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 D}^b {\tilde D}_b V^a + \frac{1}{3} {\tilde D}^a {\tilde D}_b V^b + {\tilde R}_b^a V^b - \frac{2}{3} {\tilde D}^a K + {\tilde D}_b {\tilde M}^{ab} & = & 8\pi \psi^4 j^a \end{array}

where

{\tilde A}^{ab} \equiv {\tilde D}^a V^b + {\tilde D}^b V^a - \frac{2}{3} {\tilde g}^{ab} {\tilde D}_c V^c + {\tilde M}^{ab}

Indices on {\tilde A}^{ab} are lowered with the conformal metric. 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} {\tilde g}^{ab} K \end{array}

Useful relations: The equations above are written compactly with the operators

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

Also define L by (LV)^{ab} \equiv D^a V^b + D^b V^a - (2/3)g^{ab} D_c V^c where D_a is the covariant derivative built from the physical metric. Let R denote the physical curvature scalar. The following relations hold:

\begin{array}{rcl} (LV)^{ab} & = & \psi^{-4} (\tilde L V)^{ab} \\ D_b A^{ab} & = & \psi^{-10} {\tilde D}_b (\psi^{10} A^{ab} ) \\ R & = & \psi^{-4}{\tilde R} - 8 \psi^{-5} {\tilde D}_a {\tilde D}^a \psi \end{array}

for any vector V^a and any symmetric trace-free tensor A^{ab}.

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