# Ngô Quốc Anh

## December 31, 2012

### n+1 decomposition of the spacetime metric

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 17:29

Let us introduce the components $\gamma_{ij}$ of the $n$-metric $\gamma$ of the hypersurface $M$ with respect to the coordinates $(x^i)$

$\displaystyle \gamma=\gamma_{ij}dx^i\otimes dx^j.$

From the definition of the shift vector $\beta$, we can write

$\displaystyle \beta_i=\gamma_{ij}\beta^j.$

For the spacetime metric $g$ of $V$, with respect to the coordinates $(x^\alpha)$ we can also write

$\displaystyle g=g_{\alpha\beta}dx^\alpha\otimes dx^\beta.$

Keep in mind that $\gamma=g|_M$. Thanks to $\partial_t = Nn+\beta$, we obtain

$\displaystyle \begin{array} {lcl}{g_{00}} &=&\displaystyle g({\partial _t},{\partial _t}) \hfill \\&=&\displaystyle g(Nn + \beta ,Nn + \beta ) \hfill \\ &=&\displaystyle {N^2}g(n,n) + g(\beta ,\beta ) \hfill \\ &=&\displaystyle - {N^2} + \gamma (\beta ,\beta ) \hfill \\ &=&\displaystyle - {N^2} + {\beta _i}{\beta ^i}. \hfill \\ \end{array}$

Similarly, one can also obtain

$\displaystyle\begin{array}{lcl} {g_{0i}} &=&\displaystyle g({\partial _t},{\partial _i}) \hfill \\&=&\displaystyle g(Nn + \beta ,{\partial _i}) \hfill \\&=&\displaystyle g({\beta _j}d{x^j},{\partial _i}) \hfill \\&=&\displaystyle {\beta _i}. \hfill \\ \end{array}$

Finally, it is easy to see that

$\displaystyle {g_{ij}} = g({\partial _i},{\partial _j}) = \gamma ({\partial _i},{\partial _j}) = {\gamma _{ij}}$

for all $i,j \geqslant 1$.