Determinant g = g_{00} det |g_{ij}|

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In summary: So, in summary, there is a relation between the determinant of the metric tensor and the determinant of its space-space part, given by g_{00} det | g_{ij} |, which can be derived from Cramer's rule.
  • #1
wandering.the.cosmos
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If we let latin alphabets {i,j,k,...} denote the spatial indices, and the greek ones run from 0 to 3, then I've seen the following

[tex]det| g_{\mu \nu} | = g_{00} det | g_{ij} |[/tex]

both in Landau's "Classical theory of fields", as well as ADM's paper on the initial value formulation of GR ("Dynamics of General Relativity", Arnowitt et al.) I don't quite understand it though, or am I reading it right?

Specifically, what about the terms that involve [itex]g_{01}, g_{02}, g_{03}[/itex] when we do the co-factor expansion? Do they somehow vanish by symmetry?

In (2+1) dimensions I got the determinant to be

[tex]det| g_{\mu \nu} | = g_{00} det | g_{ij} | - g_{10}(g_{01} g_{22} - g_{02} g_{21}) + g_{20} (g_{01} g_{12} - g_{11} g_{02})[/tex]

How could the last four terms cancel out?
 
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  • #2
[itex]g_{\mu\nu}[/itex] can be any symmetric matrix.

If we try for an example

[tex]
\left[ \begin {array}{cccc} m&1&2&3\\\noalign{\medskip}1&4&5&6\\\noalign{\medskip}2&5&7&8\\\noalign{\medskip}3&6&8&9\end {array}
\right]
[/tex]

we find the determinant is -m-2, which is not a multiple of m.

So I don't think your statement can be correct as written. Start reading the fine print :-).

(Is the ADM paper online anywhere?)
 
  • #3
It is equation 3.12 of "The Dynamics of General Relativity", Arnowitt, Deser and Misner:

http://arxiv.org/abs/gr-qc/0405109

Perhaps someone could explain to me if I am mis-reading it.
 
  • #4
That is correct assuming that the metric tensor is of the form
[tex]\left[ \begin {array}{cccc} g_{00}&0&0&0\\\noalign{\medskip}0&g_{11}&g_{12}&g_{13}\\\noalign{\medskip}0&g_{21}&g_{22}&g{23}\\\noalign{\medskip}0&g_{31}&g_{32}&g_{33}\end {array} \right] [/tex], the "three plus one dimensional decomposition of the Einstein field" referred to in the text.
 
  • #5
wandering.the.cosmos said:
If we let latin alphabets {i,j,k,...} denote the spatial indices, and the greek ones run from 0 to 3, then I've seen the following

[tex]det| g_{\mu \nu} | = g_{00} det | g_{ij} |[/tex]

both in Landau's "Classical theory of fields",

When a metric is static, coordinates can be chosen such that the components of the metric take the form given by HallsofIvy.

as well as ADM's paper on the initial value formulation of GR ("Dynamics of General Relativity", Arnowitt et al.)

I don't think equations (3.10) and (3.12) of the ADM paper give this.
 
  • #6
I have indeed mis-read the ADM paper. What it says is

[tex]g^{00} = \frac{\textrm{det} |g_{ij}|}{\textrm{det} |g_{\mu \nu}|}[/tex]

where [itex]\textrm{det} |g_{ij}|[/itex] is the determinant of the space-space part of the metric and [itex]\textrm{det} |g_{\mu \nu}|[/itex] is the determinant of the whole metric tensor.

This follows from Cramer's rule.
 

1. What is the determinant g?

The determinant g, also known as the metric determinant, is a mathematical concept used in the study of geometry and physics. It is a scalar quantity that represents the volume or size of a geometric object or space.

2. What does g_{00} represent in the formula for determinant g?

g_{00} is a component of the metric tensor, which is a mathematical object used to describe the geometry of a space. In the formula for determinant g, g_{00} represents the element in the first row and first column of the metric tensor.

3. How is the determinant g related to the metric tensor?

The determinant g is calculated using the metric tensor, specifically by taking the square root of the determinant of the metric tensor's spatial components (represented by g_{ij}). This relationship is described by the formula g = g_{00} det |g_{ij}|.

4. What is the significance of the determinant g in physics?

In physics, the determinant g is used to calculate various physical quantities, such as the length, volume, and area of objects or spaces. It is also used in the study of relativity, where it helps determine the curvature of space-time.

5. How is the determinant g calculated and what are its units?

The determinant g is calculated by taking the square root of the determinant of the metric tensor's spatial components. Its units depend on the units of the metric tensor's components and can vary depending on the specific application or context.

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