Spacetime coordinate smoothness requirement

[cianfa72]

TL;DR

About smoothness requirements for "good/smooth" coordinate systems in spacetime

Hi, I was keep reading the interesting book Exploring Black Holes - second edition from Taylor, Wheeler, Bertschinger. I'd like to better understand some points they made.

In Box 3 section 3-6 an example of coordinate singularity at point O in Euclidean plane in polar coordinates centered there is shown.

They claim at point O the angle $\phi$ is undefined since there are an infinite number of $\phi=const$ coordinate curves passing there. Therefore I believe both basis vectors $\frac {\partial} {\partial_r}$ and $\frac {\partial} {\partial_\phi}$ actually do not exist at O (basically there is no definite way to "move" in $r$ direction at O and there are infinite ways to "move" in $\phi$ direction at O). So far so good.

Next in section 5-9 they define the features of a "good" coordinate system/chart, namely:

FIRST REQUIREMENT: UNIQUENESS
The global coordinate system must provide a unique set of coordinates for each separate event in the spacetime region under consideration.


SECOND REQUIREMENT: SMOOTHNESS
The coordinates must vary smoothly from event to neighboring event. In practice, this means there must be a differentiable coordinate transformation that takes the global metric to a local inertial metric (except on a physical singularity).

Figure 8 (right) shows an example of a global coordinate system that fails to satisfy the uniqueness requirement.

About the second requirement (smoothness) in case of Euclidean plane the polar coordinates fulfill the requirement everywhere except that at point O: namely there is a differentiable coordinate transformation that brings the global metric (expressed in polar coordinates) into local inertial metric form (i.e. standard Euclidean form) everywhere except at point O.

Does the above coordinate transformation actually doesn't "count" as differentiable, therefore does not "qualify" polar coordinates as smooth ?

 

[PAllen]

In general, there may be no global ‘good’ coordinates. The existence of such coordinates is the exception for smooth manifolds. Instead, you can expect to cover the manifold with overlapping ‘good’ coordinate patches. Polar coordinates on a plane are good except for one point. On a 2 sphere, there are no global good coordinates, but it can be covered with as few as two patches.

 

[cianfa72]

In general, there may be no global ‘good’ coordinates. The existence of such coordinates is the exception for smooth manifolds. Instead, you can expect to cover the manifold with overlapping ‘good’ coordinate patches. Polar coordinates on a plane are good except for one point.

Strictly speaking the first requirement (uniqueness), as stated in the book, applies to just the "spacetime region under consideration" hence it might not actually be global.

Following their definitions, polar coordinates are indeed good/smooth since there is a differentiable transformation that brings the metric in standard form at any point in the region/patch over which the coordinates are defined.

Edit: polar coordinates lack both of uniqueness and smoothness at point O. Is there a coordinate system that fullfills only one of these at point O ?

 

[jbergman]

TL;DR Summary: About smoothness requirements for "good/smooth" coordinate systems in spacetime

Hi, I was keep reading the interesting book Exploring Black Holes - second edition from Taylor, Wheeler, Bertschinger. I'd like to better understand some points they made.

In Box 3 section 3-6 an example of coordinate singularity at point O in Euclidean plane in polar coordinates centered there is shown.
View attachment 346618

They claim at point O the angle $\phi$ is undefined since there are an infinite number of $\phi=const$ coordinate curves passing there. Therefore I believe both basis vectors $\frac {\partial} {\partial_r}$ and $\frac {\partial} {\partial_\phi}$ actually do not exist at O (basically there is no definite way to "move" in $r$ direction at O and there are infinite ways to "move" in $\phi$ direction at O). So far so good.

Next in section 5-9 they define the features of a "good" coordinate system/chart, namely:


Figure 8 (right) shows an example of a global coordinate system that fails to satisfy the uniqueness requirement.
View attachment 346619

About the second requirement (smoothness) in case of Euclidean plane the polar coordinates fulfill the requirement everywhere except that at point O: namely there is a differentiable coordinate transformation that brings the global metric (expressed in polar coordinates) into local inertial metric form (i.e. standard Euclidean form) everywhere except at point O.

Does the above coordinate transformation actually doesn't "count" as differentiable, therefore does not "qualify" polar coordinates as smooth ?

Polar coordinates are not a valid chart for an open set $U$ containing the origin since the transition function with the identity function isn't a local diffeomorphism. The point is that a real singularity at a point can only be detected in a valid chart for an open set containing that point.

 

[Dale]

Strictly speaking the first requirement (uniqueness), as stated in the book, applies to just the "spacetime region under consideration" hence it might not actually be global.

I think one other subtlety that is often overlooked is that the “region under consideration” has to be an open subset. So I think it is not just the center point that is problematic. I think you cannot define your coordinate patch from $-\pi \le \theta < \pi$ but rather $-\pi < \theta < \pi$ so the entire half-line $\theta=-\pi$ has to be covered with a second patch.

 

[cianfa72]

Polar coordinates are not a valid chart for an open set $U$ containing the origin since the transition function with the identity function isn't a local diffeomorphism. The point is that a real singularity at a point can only be detected in a valid chart for an open set containing that point.

You mean the composition of polar coordinates map with the identity map in the open region $r > 0, -\pi < \theta < \pi$. It is a differentiable map, however it is not a local diffeomorphism at the origin O.

Coming back to the definition given in the book, the transformation $x=rcos(\phi), y=rsin(\phi)$ is differentiable everywhere including the origin O, even though the Jacobian is singular at that point. Therefore according book's definition polar coordinates (in their domain of definition) are actually smooth.

 

[jbergman]

You mean the composition of polar coordinates map with the identity map in the open region $r > 0, -\pi < \theta < \pi$. It is a differentiable map, however it is not a local diffeomorphism at the origin O.

Coming back to the definition given in the book, the transformation $x=rcos(\phi), y=rsin(\phi)$ is differentiable everywhere including the origin O, even though the Jacobian is singular at that point. Therefore according book's definition polar coordinates (in their domain of definition) are actually smooth.

No, it isn't because the inverse map $$(x,y)\rightarrow (\sqrt{x^2 + y^2}, atan(y/x))$$ is not well defined at the origin. A diffeomorphism requires both the map and its inverse to be smooth.

 

[cianfa72]

No, it isn't because the inverse map $$(x,y)\rightarrow (\sqrt{x^2 + y^2}, atan(y/x))$$ is not well defined at the origin. A diffeomorphism requires both the map and its inverse to be smooth.

As far as I can understand, your point is that what must be differentiabile in the OP definition of smoothness (in order to bring the metric in standard form) is actually the inverse map. And as you highlighted for polar coordinates it is not at the origin O since it is not even defined there.

 

[jbergman]

As far as I can understand, your point is that what must be differentiabile in the OP definition of smoothness (in order to bring the metric in standard form) is actually the inverse map. And as you highlighted for polar coordinates it is not at the origin O since it is not even defined there.

I didn't realize that the source claimed that. Theirs is an imprecise definition, IMO, but I would have to look at the original resource more thoroughly to understand the meaning of that statement the authors intended.