Is Non-Zero Metric Determinant Enough for a Global Coordinate System?

In summary, determining if a coordinate system is global on a manifold involves checking if the chart covers the entire maximal analytic extension of the manifold. Simply having a nonzero determinant of the metric is not enough to determine if a coordinate system is global, as shown by the example of Painleve coordinates on maximally extended Schwarzschild spacetime. Further analysis, such as computing geometric invariants and the Einstein tensor, is needed to fully determine the global nature of a coordinate system.
  • #1
spaghetti3451
1,344
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Is there a universal criteria to determine if a coordinate system is global?

I think that it is sufficient for the determinant of the metric to be non-zero in order for a coordinate system to be global. Is this so?

For example, take the metric

##ds^{2} = \ell^{2}(-\cosh^{2}\rho\ dt^{2} + d\rho^{2} + \sinh^{2}\rho\ d\phi^{2}).##

I'd say that this metric is not global because ##\sinh\rho = 0## for ##\rho = 0##. What do you think?
 
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  • #2
I think that you want to learn about geodesic completeness.
 
  • #3
Dale said:
I think that you want to learn about geodesic completeness.

I suppose you're talking about genuine singularities of the manifold. My question is more pedestrian. It's to check if the metric is global, or if it is degenerate at some point.
 
  • #4
Yes, I think that is geodesic completeness, but I could be wrong.
 
  • #5
spaghetti3451 said:
Is there a universal criteria to determine if a coordinate system is global?

By "global" I assume you mean "covers the entire manifold", correct?

In order to answer this question, you first have to know what "the entire manifold" means. The accepted meaning in GR is the maximal analytic extension.

spaghetti3451 said:
I think that it is sufficient for the determinant of the metric to be non-zero in order for a coordinate system to be global. Is this so?

No. For a counterexample, consider Painleve coordinates on maximally extended Schwarzschild spacetime. Their determinant is nonzero everywhere on the patch that they cover, but the patch that they cover is not the entire manifold.

I think that a nonzero determinant is a necessary condition for a coordinate chart to cover the entire maximal analytic extension (certainly it seems intuitively obvious that it ought to be), but as the above example shows, it is not sufficient. I am not sure if it is known what the sufficient conditions are.
 
  • #6
Dale said:
I think that is geodesic completeness

Not quite, because the maximal analytic extension of a manifold can still be geodesically incomplete (Schwarzschild spacetime is the obvious example). What I think you might mean here is that the chart covers all geodesics in the manifold as completely as possible, i.e., to all values of their affine parameters that are included in the maximal analytic extension. I think that works as a heuristic, but I don't know how you would formulate it mathematically in terms of properties of the line element expressed in the chart.
 
  • #7
PeterDonis said:
By "global" I assume you mean "covers the entire manifold", correct?

In order to answer this question, you first have to know what "the entire manifold" means. The accepted meaning in GR is the maximal analytic extension.
No. For a counterexample, consider Painleve coordinates on maximally extended Schwarzschild spacetime. Their determinant is nonzero everywhere on the patch that they cover, but the patch that they cover is not the entire manifold.

I think that a nonzero determinant is a necessary condition for a coordinate chart to cover the entire maximal analytic extension (certainly it seems intuitively obvious that it ought to be), but as the above example shows, it is not sufficient. I am not sure if it is known what the sufficient conditions are.

Ah! So does that mean that the metric

##ds^{2} = \ell^{2}(-\cosh^{2}\rho\ dt^{2} + d\rho^{2} + \sinh^{2}\rho\ d\phi^{2})##

is not global because ##\sinh^{2}\rho = 0## at ##\rho = 0##?
 
  • #8
spaghetti3451 said:
So does that mean that the metric

$$
ds^{2} = \ell^{2}(-\cosh^{2}\rho\ dt^{2} + d\rho^{2} + \sinh^{2}\rho\ d\phi^{2})
$$

is not global because ##\sinh^{2}\rho = 0## at ##\rho = 0##?

Global on what manifold? That's the question. Just writing down a line element doesn't tell you what the manifold is.

What ##\sinh^2 \rho = 0## at ##\rho = 0## tells you is that you have a coordinate singularity at ##\rho = 0##. But that's not enough information to know whether the manifold extends beyond that point or not; just the fact that ##\rho < 0## is mathematically possible doesn't necessarily mean that coordinate range maps to anything in the manifold (##\rho## could be like ##r## in Schwarzschild spacetime, where only the range ##r > 0## maps to points in the manifold). In other words, it's not enough information to tell whether, for example, ##\rho = 0## is like the singularity at ##r = 0## in Schwarzschild spacetime, which really does mark an "end" of the manifold, or like the singularity at ##r = 2M## in Schwarzschild spacetime, which doesn't.

As I said before, I don't know exactly what the right conditions are, but computing geometric invariants at ##\rho = 0## (such as the Ricci scalar or the Kretschmann scalar) would help to determine what is going on there. Also, computing the Einstein tensor for your metric would be useful since it would tell you what kind of stress-energy would be needed to produce such a metric.
 
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  • #9
PeterDonis said:
Global on what manifold? That's the question. Just writing down a line element doesn't tell you what the manifold is.

What ##\sinh^2 \rho = 0## at ##\rho = 0## tells you is that you have a coordinate singularity at ##\rho = 0##. But that's not enough information to know whether the manifold extends beyond that point or not; just the fact that ##\rho < 0## is mathematically possible doesn't necessarily mean that coordinate range maps to anything in the manifold (##\rho## could be like ##r## in Schwarzschild spacetime, where only the range ##r > 0## maps to points in the manifold). In other words, it's not enough information to tell whether, for example, ##\rho = 0## is like the singularity at ##r = 0## in Schwarzschild spacetime, which really does mark an "end" of the manifold, or like the singularity at ##r = 2M## in Schwarzschild spacetime, which doesn't.

As I said before, I don't know exactly what the right conditions are, but computing geometric invariants at ##\rho = 0## (such as the Ricci scalar or the Kretschmann scalar) would help to determine what is going on there. Also, computing the Einstein tensor for your metric would be useful since it would tell you what kind of stress-energy would be needed to produce such a metric.

Global on AdS##_{3}##. See what's written below equation (9.4) in page 97 of these lecture notes http://www.hartmanhep.net/topics2015/gravity-lectures.pdf.
 
  • #10
PeterDonis said:
I think you might mean here is that the chart covers all geodesics in the manifold as completely as possible, i.e., to all values of their affine parameters that are included in the maximal analytic extension.
Yes, this is what I was thinking about. I didn't mention the maximal analytic extension, but that is what I was thinking about.
 

1. What is a global coordinate system?

A global coordinate system is a mathematical framework used to describe the physical location of objects in three-dimensional space. It is used to determine the position, orientation, and velocity of objects relative to a fixed reference point, such as the Earth's center.

2. How is a global coordinate system different from a local coordinate system?

A global coordinate system is fixed and does not change, while a local coordinate system is relative and can change depending on the reference point. Global coordinates are also used to describe positions on a large scale, while local coordinates are used for smaller, more specific areas.

3. What are the main components of a global coordinate system?

The main components of a global coordinate system are a set of axes (usually x, y, and z), a reference point (such as the Earth's center), and a unit of measurement (such as meters or degrees). These components work together to determine the position of objects in three-dimensional space.

4. How is a global coordinate system used in navigation?

A global coordinate system is used in navigation to determine the position of objects, such as ships or aircraft, in relation to other objects or fixed points on the Earth's surface. This allows for accurate navigation and mapping of locations on a global scale.

5. Can a global coordinate system be used on other planets?

Yes, a global coordinate system can be used on other planets, as long as the reference point and unit of measurement are adjusted to fit the specific planet. For example, on Mars, the reference point may be the planet's center and the unit of measurement may be kilometers instead of meters.

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