# Physical Transverse-Traceless Decomposition

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 $D_a$, ${\tilde M}_{ab}$ and ${\tilde R}_{ab}$ are raised with the inverse conformal metric. Solve the following equations for $\psi$ and $V^a$:

$\begin{array}{rcl} {\tilde D}^a {\tilde D}_a \psi - \frac{1}{8} \psi {\tilde R} - \frac{1}{12} \psi^5 (K)^2 + \frac{1}{8} \psi^{5} {\tilde A}_{ab}{\tilde A}^{ab} & = & -2\pi \psi^5 \rho \\ {\tilde D}^b {\tilde D}_b V^a + \frac{1}{3} {\tilde D}^a {\tilde D}_b V^b + {\tilde R}_b^a V^b + 6\psi^{-1} ({\tilde D}_b\psi) (\tilde L V)^{ab} - \frac{2}{3} {\tilde D}^a K + \psi^{-6} {\tilde D}_b {\tilde M}^{ab} & = & 8\pi \psi^4 j^a \end{array}$

where

${\tilde A}^{ab} \equiv {\tilde D}^a V^b + {\tilde D}^b V^a - \frac{2}{3} {\tilde g}^{ab} {\tilde D}_c V^c + \psi^{-6} {\tilde M}^{ab}$

Indices on ${\tilde A}^{ab}$ are lowered with the conformal metric. The physical metric and extrinsic curvature are

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

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 $A^{ab}$.