Relation between det(spacetime metric) and det(spatial metric)

  • #1

Main Question or Discussion Point

I have a metric g on spacetime and a spatial metric ##\gamma## such that the components of g can be written in matrix form as

$$ g_ {\alpha, \beta} = \begin{pmatrix} g_{00} & g_{0 j} \\
g_{i 0} & \gamma_{ij} \end{pmatrix} $$

where ##i,j = 1,2,3## and ##\alpha = 0,1,2,3##. Now I want to find a relation between the determinant of ##g## and the determinant of ##\gamma## expressed in terms of the components of ##g##. Using Cramer's rule I get

$$ det(g_{\alpha, \beta}) \equiv g = \frac{C_{00}}{g^{00}}$$

where ##C_{00}## is the cofactor of the (0,0)-element of the matrix above; i.e. it is ##det(\gamma_{ij}) = \gamma##. But in order to find the full relation I will still need to find the (0,0) element of the inverse. What is the best way to go about doing this? Do I really have to solve the full set of equations ##g^{\alpha \lambda}g_{\lambda \beta} = \delta^{\alpha}_{\ \ \ \beta}##? Or is there a better way?
 

Answers and Replies

  • #2
612
23
I have a metric g on spacetime and a spatial metric ##\gamma## such that the components of g can be written in matrix form as

$$ g_ {\alpha, \beta} = \begin{pmatrix} g_{00} & g_{0 j} \\
g_{i 0} & \gamma_{ij} \end{pmatrix} $$

where ##i,j = 1,2,3## and ##\alpha = 0,1,2,3##.
I'm assuming you mean ##\displaystyle \mathrm{g}_{\alpha,\beta}=\begin{bmatrix} g_{0,0} & g_{0,1} & g_{0,2} & g_{0,3} \\ g_{1,0} & \gamma_{1,1} & \gamma_{1,2} & \gamma_{1,3} \\ g_{2,0} & \gamma_{2,1} & \gamma_{2,2} & \gamma_{2,3} \\ g_{3,0} & \gamma_{3,1} & \gamma_{3,2} & \gamma_{3,3}\end{bmatrix}##, right?

Now I want to find a relation between the determinant of ##g## and the determinant of ##\gamma## expressed in terms of the components of ##g##. Using Cramer's rule I get

$$ det(g_{\alpha, \beta}) \equiv g = \frac{C_{00}}{g^{00}}$$

where ##C_{00}## is the cofactor of the (0,0)-element of the matrix above; i.e. it is ##det(\gamma_{ij}) = \gamma##.
I'm probably misinterpreting something here, but I'm not seeing how you're applying Cramer's Rule.

If we were talking about finding the determinant of the given matrix, you'd get $$\det(\mathrm{g}_{\alpha,\beta})=g_{0,0}~\det\begin{bmatrix}\gamma_{1,1} & \gamma_{1,2} & \gamma_{1,3} \\ \gamma_{2,1} & \gamma_{2,2} & \gamma_{2,3} \\ \gamma_{3,1} & \gamma_{3,2} & \gamma_{3,3}\end{bmatrix}-g_{1,0}~\det\begin{bmatrix} g_{0,1} & g_{0,2} & g_{0,3} \\ \gamma_{2,1} & \gamma_{2,2} & \gamma_{2,3} \\ \gamma_{3,1} & \gamma_{3,2} & \gamma_{3,3}\end{bmatrix}+g_{2,0}~\det\begin{bmatrix} g_{0,1} & g_{0,2} & g_{0,3} \\ \gamma_{1,1} & \gamma_{1,2} & \gamma_{1,3} \\ \gamma_{3,1} & \gamma_{3,2} & \gamma_{3,3}\end{bmatrix}-g_{3,0}~\det\begin{bmatrix}g_{0,1} & g_{0,2} & g_{0,3} \\ \gamma_{1,1} & \gamma_{1,2} & \gamma_{1,3} \\ \gamma_{2,1} & \gamma_{2,2} & \gamma_{2,3}\end{bmatrix}$$ by Laplace Expansion.

Given, I'm not really knowledgeable in general relativity (or differential geometry for that matter :tongue:) so I could be mistaking this determinant for something else, or misunderstanding your notation.
 

Related Threads on Relation between det(spacetime metric) and det(spatial metric)

  • Last Post
Replies
12
Views
5K
  • Last Post
Replies
7
Views
6K
  • Last Post
Replies
1
Views
5K
Replies
5
Views
2K
  • Last Post
Replies
3
Views
11K
Replies
3
Views
8K
  • Last Post
Replies
4
Views
990
  • Last Post
Replies
1
Views
2K
Replies
20
Views
3K
Top