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

In summary, the conversation discusses a metric g on spacetime and a spatial metric ##\gamma##, with components of g written in matrix form. The goal is to find a relation between the determinants of g and ##\gamma##, and Cramer's rule is used to find a partial solution. However, further work is needed to find the full relation.
  • #1
center o bass
560
2
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?
 
Physics news on Phys.org
  • #2
center o bass said:
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?

center o bass said:
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.
 

1. What is the difference between spacetime metric and spatial metric?

The spacetime metric is a mathematical representation of the geometry of spacetime, which includes both space and time dimensions. The spatial metric is a subset of the spacetime metric that only describes the geometry of space, without considering the time dimension.

2. How are the two metrics related?

The two metrics are related through the concept of covariance, which means that the equations that describe them are the same in all reference frames. This is a fundamental principle in the theory of relativity, and it allows for a unified understanding of the geometry of spacetime.

3. What does the determinant of the spacetime metric represent?

The determinant of the spacetime metric represents the volume scaling factor between different reference frames. In other words, it tells us how much space and time are distorted in different reference frames.

4. How does the determinant of the spatial metric affect the curvature of space?

The determinant of the spatial metric is related to the curvature of space through the Einstein field equations. This determinant, along with other components of the metric, determines the curvature of spacetime, which is a measure of how gravity affects the motion of objects.

5. Can the determinant of the spacetime metric be negative?

Yes, the determinant of the spacetime metric can be negative in certain reference frames, which indicates that space and time are "flipped" in these frames. This can happen in highly curved regions of spacetime, such as near massive objects like black holes.

Similar threads

Replies
1
Views
1K
  • Special and General Relativity
Replies
4
Views
2K
  • Differential Geometry
Replies
4
Views
3K
  • Differential Geometry
Replies
3
Views
1K
  • Special and General Relativity
Replies
1
Views
1K
  • Special and General Relativity
Replies
2
Views
812
  • Special and General Relativity
Replies
1
Views
650
  • Advanced Physics Homework Help
Replies
18
Views
1K
Replies
1
Views
2K
  • Differential Geometry
Replies
10
Views
4K
Back
Top