# Determinant of the metric tensor

• LayMuon
In summary, the conversation discusses the equivalence principle and its application to non-inertial frames and gravitational fields. The concept of a "Galilean metric" is questioned and the possibility of a Riemannian metric in general relativity is explored. The conversation also touches on the determinant of the metric and how it can be negative in real spacetime. The Ashtekar formulation of GR is mentioned as a possible generalization.
LayMuon
We are stating with equivalence principle that passing locally to non inertial frame would be analogous to the presence of gravitational field at that point, so $g^'_{ij}=A g_{nm} A^{-1}$ where g' is the galilean metric and g is the metric in curved space, and A is the transformation which eliminates gravity at the point.

How do you prove that det[g] <0?

Doesn't it follow from the above realtion that det[g] is exactly -2 because det[g'] = -2, and not just simply negative? What did I miss?

Thanks!

Det(AB) = Det(A) Det(B)

LayMuon said:
We are stating with equivalence principle that passing locally to non inertial frame would be analogous to the presence of gravitational field at that point, so $g^'_{ij}=A g_{nm} A^{-1}$ where g' is the galilean metric and g is the metric in curved space, and A is the transformation which eliminates gravity at the point.

Curvature, unlike a gravitational field, is coordinate-independent, so you can't introduce or eliminate curvature by a change of coordinates. "Galilean metric" also doesn't make sense, because there is no spacetime metric in Galilean relativity. (If you try to define one, it ends up being degenerate.)

LayMuon said:
How do you prove that det[g] <0?

That would depend on what initial assumptions you were making. It could be taken as a postulate. There is nothing in GR, the field equations, etc., that prohibits a Riemannian rather than semi-Riemannian metric.

bcrowell said:
That would depend on what initial assumptions you were making. It could be taken as a postulate. There is nothing in GR, the field equations, etc., that prohibits a Riemannian rather than semi-Riemannian metric.

I would say the field equations without the assumption of locally SR = pseudo-Riemannian metric is a different theory, not GR.

PAllen said:
I would say the field equations without the assumption of locally SR = pseudo-Riemannian metric is a different theory, not GR.

Sure. However, it's not like anything in the formalism breaks down if you assume a different signature. I believe the Ashtekar formulation of GR can even handle geometries in which the signature differs from point to point.

bcrowell said:
Curvature, unlike a gravitational field, is coordinate-independent, so you can't introduce or eliminate curvature by a change of coordinates. "Galilean metric" also doesn't make sense, because there is no spacetime metric in Galilean relativity. (If you try to define one, it ends up being degenerate.)

That would depend on what initial assumptions you were making. It could be taken as a postulate. There is nothing in GR, the field equations, etc., that prohibits a Riemannian rather than semi-Riemannian metric.

Then how should I understand this paragraph from Landau Lifgarbagez "Classical Theory of Fields" (chapter 82):

"We note that, after reduction to diagonal form at a given point, the metric of the quantities g_ik has one positive and three negative principal values. From this it follows in particular, that the determinant g, formed from the quantitiers g_ik, is always negative for a real spacetime: g<0."

LayMuon said:
Then how should I understand this paragraph from Landau Lifgarbagez "Classical Theory of Fields" (chapter 82):

"We note that, after reduction to diagonal form at a given point, the metric of the quantities g_ik has one positive and three negative principal values. From this it follows in particular, that the determinant g, formed from the quantitiers g_ik, is always negative for a real spacetime: g<0."

Like I said, it depends on what initial assumptions you're making. The first sentence states an initial assumption or something that follows from some other initial assumptions.

But if signature changes from point to point there should be points where the metric is degenerate, no?

LayMuon said:
Then how should I understand this paragraph from Landau Lifgarbagez "Classical Theory of Fields" (chapter 82):

"We note that, after reduction to diagonal form at a given point, the metric of the quantities g_ik has one positive and three negative principal values. From this it follows in particular, that the determinant g, formed from the quantitiers g_ik, is always negative for a real spacetime: g<0."
At a given point you can diagonalize the metric in case of curved spacetime.Real spacetime refers to spacetime of general relativity which will of course have g<0.

martinbn said:
But if signature changes from point to point there should be points where the metric is degenerate, no?

Right. This is why the standard formalism of GR can't handle a change of signature; it's expressed using tensors with indices raised and lowered, which you can't do if you have a degenerate metric. In practice, if you get a degenerate metric in GR, it usually means you chose bad coordinates: http://lightandmatter.com/html_books/genrel/ch06/ch06.html#Section6.4

Last edited by a moderator:
bcrowell said:
Right. This is why the standard formalism of GR can't handle a change of signature; it's expressed using tensors with indices raised and lowered, which you can't do if you have a degenerate metric. In practice, if you get a degenerate metric in GR, it usually means you chose bad coordinates: http://lightandmatter.com/html_books/genrel/ch06/ch06.html#Section6.4

OK, then Ashtekar formulation of GR is actually a generalization, not just a different formulation, is that right? By the way, where can a look at it?

Last edited by a moderator:

## 1. What is the determinant of the metric tensor?

The determinant of the metric tensor is a mathematical quantity that describes the scaling factor of a coordinate system. It is denoted by g and is used in the calculation of distances and volumes in an n-dimensional space.

## 2. Why is the determinant of the metric tensor important?

The determinant of the metric tensor is important because it is used in the calculation of physical quantities such as distance, volume, and curvature in a given coordinate system. It also plays a crucial role in the formulation of Einstein's theory of general relativity.

## 3. How is the determinant of the metric tensor calculated?

The determinant of the metric tensor is calculated by taking the determinant of the matrix formed by the components of the metric tensor. This involves multiplying the diagonal elements of the matrix and subtracting the product of the off-diagonal elements.

## 4. What does a non-zero determinant of the metric tensor indicate?

A non-zero determinant of the metric tensor indicates that the coordinate system is not flat and has a non-trivial geometry. This means that distances and angles in this coordinate system are not preserved.

## 5. How does the determinant of the metric tensor change under coordinate transformations?

The determinant of the metric tensor changes under coordinate transformations according to a specific rule. It is multiplied by the square of the Jacobian of the transformation, which describes how the coordinates change from one system to another. This ensures that physical quantities, such as volumes, remain invariant under coordinate transformations.

Replies
4
Views
3K
Replies
1
Views
1K
Replies
11
Views
627
Replies
8
Views
2K
Replies
19
Views
750
Replies
10
Views
524
Replies
24
Views
905
Replies
8
Views
555
Replies
7
Views
2K
Replies
11
Views
2K