General relativity and curvilinear coordinates

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Curvilinear coordinate systems are essential in general relativity because they accommodate the curvature of spacetime caused by gravity, which cannot be accurately described using global Cartesian coordinates. While local Cartesian coordinates can describe small patches of a manifold, they fail to capture global features like event horizons or paths across curved surfaces. Curvilinear coordinates allow for a more accurate representation of geometric objects over larger areas of the manifold, as they can adapt to the local curvature. The discussion emphasizes that curvature exists even locally, necessitating the use of curvilinear coordinates to maintain accuracy in physical descriptions. Overall, curvilinear coordinates are crucial for understanding the complex geometry of spacetime in the presence of matter.
  • #121
micromass said:
This induces a second topology on the spacetime which is not the same as the a priori topology.

Is what would be induced by the Minkowski inner product even a valid topology? It gives zero distance between distinct points.
 
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  • #122
Yes, it's a valid topology. But it will be non-Hausdorff and very ugly.
 
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  • #123
Well, there is the Hawking-King-McCarthy topology,

http://authors.library.caltech.edu/11027/

Fifteen years ago, I wrote "If anybody knows of applications of the Hawking-King-McCarthy topology to anything, I would be most interested."

I am still interested.
 
  • #124
micromass said:
induces the topology. Many other distance functions also induces that same topology.

So by defining the topology this way we don't determine the geometry a priori then, as we haven't specified the form of the distance function?
Is this why we are able to map patches of a manifold, in which the geometry is non-Euclidean, into open sets of \mathbb{R}^{n} (and thus have to use non-Cartesian coordinates)?
 
  • #125
"Don't panic!" said:
So by defining the topology this way we don't determine the geometry a priori then, as we haven't specified the form of the distance function?
Is this why we are able to map patches of a manifold, in which the geometry is non-Euclidean, into open sets of \mathbb{R}^{n} (and thus have to use non-Cartesian coordinates)?
Without the metric tensor and the Levi-Civita connection the notion of cartesian or non-cartesian isnt'ty even meaninful.
Think about it as layers of structure. First we have a set which has no structure, on that set define a locally euclidean (locally homeomorphicto R^n) topology, then a smooth structure. Up to this point you can have coordinate charts, they are just the homeomorphisms from open sets of the manifold to R^n, but these coordinates are not cartesian or non-cartesian, it doesn't even make sense to ask that question yet.
Now once you define the metric tensor and the Levi-Civita connection you can ask wether a particular choice of coordinates is cartesian or not. If the Levi-Civita connection induced by the metric is curved then you can't have cartesian coordinates except for a very small patch on the manifold.
 
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  • #126
Cruz Martinez said:
Now once you define the metric tensor and the Levi-Civita connection you can ask wether a particular choice of coordinates is cartesian or not. If the Levi-Civita connection induced by the metric is curved then you can't have cartesian coordinates except for a very small patch on the manifold.

Ok, I think it's all starting to make sense a bit more now.
Just to clarify though (and then I'll stop bugging everyone), if we consider a manifold with a metric that induces a non-Euclidean geometry, then if we consider a patch on such a manifold that is large enough that the local geometry cannot be considered as Euclidean, the coordinate maps that we use to map points in such a patch will be non-Cartesian as it will not be possible to construct such coordinate maps (unless we consider smaller patches around each point in the patch)?
 
  • #127
"Don't panic!" said:
Ok, I think it's all starting to make sense a bit more now.
Just to clarify though (and then I'll stop bugging everyone), if we consider a manifold with a metric that induces a non-Euclidean geometry, then if we consider a patch on such a manifold that is large enough that the local geometry cannot be considered as Euclidean, the coordinate maps that we use to map points in such a patch will be non-Cartesian as it will not be possible to construct such coordinate maps (unless we consider smaller patches around each point in the patch)?

Correct.
 
  • #128
So am i correct in saying that the Cartesian coordinate system is a special kind of mapping which directly relates the intrinsic distance between two points on a manifold to the 'numerical' distance between their coordinates in \mathbb{R}^{n}. As, in general, a coordinate patch on a manifold will have a non-Euclidean geometry, although it will be possible to construct a one-to-one mapping such that these points can be labeled by coordinates in \mathbb{R}^{n}, it will not be possible to construct a map that preserves the intrinsic distance between two points in this patch such that it corresponds to the 'coordinate distance' between their corresponding coordinates in \mathbb{R}^{n}. In other words, although we will be able to construct a coordinate map, it will be impossible to construct a Cartesian coordinate map for this patch (apart from within a small neighbourhood around each point in this patch)?
 

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