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 $i,j \geqslant 1$.

By collecting the above calculation, we arrive at $\displaystyle g_{\alpha\beta}=\begin{pmatrix} g_{00} & g_{0j} \\ g_{i0} & g_{ij} \end{pmatrix}=\begin{pmatrix} -N^2+\beta_k\beta^k & \beta_j \\ \beta_i & \gamma_{ij} \end{pmatrix}.$

Equivalently, we can write $\displaystyle \begin{array}{lcl} g &=&\displaystyle {g_{\alpha \beta }}d{x^\alpha } \otimes d{x^\beta } \hfill \\ &=&\displaystyle ( - {N^2} + \underbrace {{\beta _i}}_{{\beta ^j}{\gamma _{ij}}}{\beta ^i})d{t^2} + \underbrace {{\beta _j}}_{{\beta ^i}{\gamma _{ij}}}dt \otimes d{x^j} + \underbrace {{\beta _i}}_{{\beta ^j}{\gamma _{ij}}}d{x^i} \otimes dt + {\gamma _{ij}}d{x^i} \otimes d{x^j} \hfill \\&=&\displaystyle - {N^2}d{t^2} + {\gamma _{ij}}(d{x^i} + {\beta ^i}dt) \otimes (d{x^j} + {\beta ^j}dt). \hfill \\ \end{array}$

The components of the inverse matrix $g^{\alpha\beta}$ can be calculated as given below $\displaystyle g^{\alpha\beta}=\begin{pmatrix} g^{00} & g^{0j} \\ g^{i0} & g^{ij} \end{pmatrix}=\begin{pmatrix} -\frac{1}{N^2} & \frac{\beta^j}{N^2} \\ \frac{\beta^i}{N^2} & \gamma^{ij}-\frac{\beta^i\beta^j}{N^2} \end{pmatrix}.$

Using the Cramer rule for expressing the inverse $(g^{\alpha\beta})$ of the matric $(g_{\alpha\beta})$, we have $\displaystyle g^{00}=\frac{C_{00}}{\det g}$

where $C_{00}$ is the element $(0,0)$ of the cofactor matrix associated with $g_{\alpha\beta}$. Clearly, $C_{00}=(-1)^0M_{00}=M_{00}$ where $M_{00}$ is the minor $(0,0)$ of the matrix $g_{\alpha\beta}$. Hence $M_{00}=\det \gamma$. In other words, there holds $\displaystyle g^{00}=\frac{\det \gamma}{\det g}.$

Thus, we have shown that $\displaystyle\sqrt { - \det g} = N\sqrt {\det \gamma }.$

Since $g^{00}<0$, our calculation above also shows that $\det g<0$.