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.

See also:

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Blog at WordPress.com.

%d bloggers like this: