Conformal Thin Sandwich Decomposition

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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
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