Can Metric Tensors Have Equal Determinants?

In summary, the conversation discusses the relationship between two points in a Lorentzian manifold and how the determinant of the metric at these points are related. It is stated that in general, there is no relation between the two determinants. However, if the points are actually two different coordinate representations of the same point, then the relationship is g' = J^2 g, where J is the Jacobian of the coordinate transformation. It is also noted that the metric should be free of intrinsic singularities, and in cases of singularities, they are defined not to be points in the manifold. The conversation also touches on the concept of "cheating" in mathematics, which is a well-respected method in mathematical physics.
  • #1
PLuz
64
0
Hello,

So, given two points, [itex]x[/itex] and [itex]x'[/itex], in a Lorentzian manifold (although I think it's the same for a Riemannian one). If in [itex]x[/itex] the determinant of the metric is [itex]g[/itex] and in the point [itex]x'[/itex] is [itex]g'[/itex]. How are [itex]g[/itex] and [itex]g'[/itex] related?By any means can [itex]g=g'[/itex]? In what conditions?

I'm sorry if this is a dumb question but when prooving an equation I found out that it holds only if and only if [itex]g=g'[/itex] and I don't think that this is always true.

Thank you.
 
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  • #2
These are two different points, right? If so, there is, in general, no relation between [itex]g[/itex] and [itex]g'[/itex].

If it is actually two different coordinate representations of the same point, then [itex]g' = J^2 g[/itex], where [itex]J[/itex] is the Jacobian of the coordinate transformation.
 
  • #3
Thank you for the prompt response.
The notation might have not been the best, sorry; it were two distinct points in a manifold.

What if the point [itex]x'[/itex] were in a sufficiently small neighborhood of [itex]x[/itex] and vice versa (the neighborhoods may not be the same). Aren't tensors defined in a small neighborhood of a point: since they represent a multilinear map?

Thank you, and sorry if I'm really confused about the concepts.
 
  • #4
Given a manifold, you can put any metric on it that you like. Just choose any smooth function that has the same signature everywhere.

It would be unusual if the signature of the metric varied from point to point. If this happens in actual applications, it's usually a sign that you chose a bad coordinate system, which can be related via a singular transformation to some other, better system of coordinates. (Differential geometry is agnostic about coordinates, but only up to a smooth, one-to-one mapping, not up to any mapping at all.)

I think the metric does have to be free of intrinsic singularities (as opposed to mere coordinate singularities). When a singularity turns up in general relativity, for example, we define the singularity not to be a point in the manifold at all. So for this reason, I think the answer to your #3 is that yes, g and g' are related if x and x' are close. For x and x' sufficiently close, g and g' can be made as close as desired.
 
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  • #5
George Jones said:
These are two different points, right? If so, there is, in general, no relation between [itex]g[/itex] and [itex]g'[/itex].

If it is actually two different coordinate representations of the same point, then [itex]g' = J^2 g[/itex], where [itex]J[/itex] is the Jacobian of the coordinate transformation.
I don't get this. The tensor transo. law of the metric can be considered to be active. Then I can use your formula. And relate g(x) and g'(x'), where x and x' are different points (different coordinates in the same chart).
 
  • #6
bcrowell said:
When a singularity turns up in general relativity, for example, we define the singularity not to be a point in the manifold at all.

How is this different from cheating (mathematically speaking)?
 
  • #7
TrickyDicky said:
How is this different from cheating (mathematically speaking)?
It isn't. But, after all, "cheating" is a well-respected mathematical method!
 
  • #8
HallsofIvy said:
It isn't. But, after all, "cheating" is a well-respected mathematical method!

:rofl:

Well, in this particular case I interpreted that "we" in "we define..." to be referring to physicists, so it would be more of a well-respected "mathematical physics" method. :tongue2:
 

1. What is the metric tensor determinant?

The metric tensor determinant is a mathematical quantity used in differential geometry to measure the infinitesimal distance between two points in a curved space. It is denoted by g, and is the determinant of the metric tensor, a matrix that describes the local geometry of a space.

2. How is the metric tensor determinant calculated?

The metric tensor determinant is calculated by taking the determinant of the metric tensor, which is a square matrix with elements that represent the inner products of basis vectors in a space. It can also be calculated by raising the components of the metric tensor to the power of the dimension of the space and then multiplying them together.

3. What is the significance of the metric tensor determinant?

The metric tensor determinant is significant because it provides information about the curvature of a space. In flat spaces, the determinant is equal to 1, while in curved spaces, it can vary from point to point, indicating the amount of curvature at that specific point.

4. How does the metric tensor determinant relate to the metric tensor?

The metric tensor determinant is a single value that is derived from the metric tensor. It can be seen as a scalar quantity that summarizes the information contained in the metric tensor, which is a higher-order object that describes the local geometry of a space.

5. Can the metric tensor determinant be negative?

Yes, the metric tensor determinant can be negative. This indicates a negative curvature in a space, such as in a saddle-shaped surface. In contrast, a positive determinant indicates a positive curvature, as in a sphere. A zero determinant indicates a flat space.

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