Ngô Quốc Anh

December 8, 2012

Derivation of the Einstein constraint equations: The momentum constraint

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 3:28

This entry is a continuation of the previous entry where we showed in detail the derivation of the Hamiltonian constrain in general relativity. Today, we derive the so-called momentum constraint equation.

First, let us recall the Codazzi equation which is given by the following identity

\displaystyle {R^ \bot }(X,Y)Z = ({\nabla _X}\mathrm{I\!I})(Y,Z) - ({\nabla _Y}\mathrm{I\!I})(X,Z).

Sometimes, we call it the Codazzi–Mainardi equation or the Ricci identity, which expresses the curvature of the normal bundle in terms of the second fundamental form. Using this, we obtain

\displaystyle\sum\limits_{i = 1}^n {{R^ \bot }({e_i},Y){e_i}} = \sum\limits_{i = 1}^n {({\nabla _{{e_i}}}\mathrm{I\!I})(Y,{e_i})} - \sum\limits_{i = 1}^n {({\nabla _Y}\mathrm{I\!I})({e_i},{e_i})}

for any tangent vector Y. Since R(e_i,Y)e_i, for all i \geqslant 1, belongs to the tangent space, there hold

\displaystyle\left\langle {R({e_i},Y){e_i},{e_0}} \right\rangle = 0 \quad \forall i = \overline{1,n}

and thus we can write

\displaystyle\left\langle {{R^ \bot }({e_i},Y){e_i},{e_0}} \right\rangle = \left\langle {R({e_i},Y){e_i},{e_0}} \right\rangle\quad \forall i = \overline{1,n}.

Moreover, by the antisymmetric property of the Riemann curvature tensor, i.e., R_{ijkl}=-R_{ijlk}, we know that

\displaystyle\left\langle {R({e_0},{e_i}){e_0},{e_0}} \right\rangle = 0\quad \forall i = \overline{1,n}

which immediately implies


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