Physical Transverse-Traceless Decomposition

From GRwiki

Jump to: navigation, search

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

Freely specify the following:

   {\tilde g}_{ab} &  & {\rm conformal\ metric} \\
    {\tilde M}_{ab} & & {\rm symmetric\  trace\  free\ tensor} \\
   K & & {\rm trace\ of\ the\ extrinsic\ curvature}

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:

   {\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 


   {\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

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

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

   (\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

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:

   (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

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

Personal tools