n+1 decomposition of the spacetime metric

December 31, 2012

n+1 decomposition of the spacetime metric

Filed under: Uncategorized — Tags: General Relativity — 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 $\det g<0$ .

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

M	T	W	T	F	S	S
					1	2
3	4	5	6	7	8	9
10	11	12	13	14	15	16
17	18	19	20	21	22	23
24	25	26	27	28	29	30
31

	Han on The implicit function theorem:…
	Alan de Gois on Variation of the Riemann tenso…
	Jet Foncannon on A generalization of the Morera…
	ibrahim tuet (@yiuyu… on In a normed space, finite line…
	Tran Minh Nguyet on Uniformly boundedness implies…
	Ngô Quốc Anh on Uniformly boundedness implies…
	Tran Minh Nguyet on Uniformly boundedness implies…
	An on Tính tích phân suy rộng loại…
	An on Tính tích phân suy rộng loại…
	BỔ ĐỀ SCHWARZ VÀ ĐỊN… on Schwarz’s Lemma, Schwarz…

Ngô Quốc Anh

December 31, 2012