General relativity and curvilinear coordinates

[George Jones]

But phi^(-1) won't map a "Cartesian" grid on V onto a "Cartesian" grid on U. Think of "grid" on a "rectangular" map mapped back onto the surface of the Earth.

 

[PeterDonis]

why is this an approximation?

Because if I set up local Cartesian coordinates centered on a point in a small patch of the manifold, the metric will only be exactly equal to the Euclidean metric at that point. In other words: I can specify the coordinates so that the metric is Euclidean, and the first derivatives of the metric coefficients are all zero, at a chosen point. But I cannot specify the second derivatives of the metric coefficients; those are given by the curvature of the manifold, and curvature is a geometric property, I can't make it vanish by choosing coordinates. So as I move away from the chosen point, the metric coefficients in my local "Cartesian" coordinate system will no longer be exactly equal to the Euclidean ones. They will only be approximately equal. In practice, since measurements always have some finite accuracy, the difference will be undetectable within some small but finite range of the chosen point, which is why, in practice, we can treat the local Cartesian coordinates, with their Euclidean metric, as valid on a small patch of the manifold centered on the chosen point.

 

["Don't panic!"]

Is the rest of the description I gave in that post ok? I'm I getting any closer?!

 

[PeterDonis]

when one transitions to a neighbouring coordinate patch on the manifold

How?

the coordinates describing this neighbouring patch will not be Cartesian

You can describe any local patch with Cartesian coordinates.

 

[stevendaryl]

You can describe any local patch with Cartesian coordinates.

Okay, I think we're telling him apparently contradictory things. A small patch can be approximately described with Cartesian coordinates, but it can't be exactly described with Cartesian coordinates. If the space is curved, then no matter what coordinate system you choose, there will be points within the patch where the metric will be different from the Cartesian case. If the patch is small enough, the difference might be negligible, but there will be a difference.

So I would say that when the curvatures is nonzero, then even in a small patch, you can't use Cartesian coordinates, except as an approximation.

 

[stevendaryl]

Ok, so is this anywhere near to being correct?
So as a manifold locally maps to a small patch of Euclidean space (why is this an approximation? I thought that a manifold was covered by a set of coordinate charts, each of which are homeomorphisms from small local patches on the manifold to small patches of Euclidean space, e.g. [itex]\phi :U\subset M\rightarrow V\subset\mathbb{R}^{n}[/itex]), a particular local subset of the manifold can be represented by local Cartesian coordinates.

Yes, every small patch can be mapped to Euclidean space, but not in a metric-preserving way. (metric-preserving just means distance-preserving)

Think about taking a globe, and cutting out a little square from it. Now, put that square onto a flat table top and press it flat against the table. You can't make it flat without stretching it in some parts. So after you've pressed it flat, the distances between points are not the same as they were on the globe.

 

[Cruz Martinez]

You seem to be confused about the meaning of the word "Euclidean". Usually, as I understand it, it means both that the manifold is ##\mathbb{R}^{n}## and that the metric is flat.

Also, the usual definition of "manifold" requires it to be locally Euclidean in the above sense; the usual term for this is "Riemannian". If we extend consideration to manifolds like spacetime, which have a metric that is not positive definite, those are referred to as "pseudo-Riemannian" manifolds, and they are locally Minkowskian (or Lorentzian), not Euclidean, since their metric is locally flat in the spacetime sense (i.e., locally the Minkowski metric).

What is the definition of manifold you refer to? The definition I know for a topological n-manifold is that it is a topological space that is second-countable and Hausdorff, and locally Euclidean of dimension n. A smooth manifold is an n-manifold with a smooth structure. In none of these definitions a metric comes into play.

 

[PeterDonis]

A small patch can be approximately described with Cartesian coordinates, but it can't be exactly described with Cartesian coordinates.

Yes. See my post #37.

 

[PeterDonis]

In none of these definitions a metric comes into play.

Yes, it does, in this one:

locally Euclidean

This means that you can always find a coordinate chart in which the metric is Euclidean (or Minkowskian, if you're in spacetime) at a chosen point. Which requires a metric.

 

[Cruz Martinez]

Yes, it does, in this one:

Locally euclidean means locally homeomorphic to R^n. It has nothing to do with a metric.

 

[pervect]

OK - let's try this example:

Draw a triangle on the surface of the Earth, using great circles - the shortest distance between two points that lies on the surface. The three vertices of the triangle are at the north pole, the other two are on the equator at 0 degrees longitude and 90 degrees longitude. Sum the angles - the triangle has three right angles, so the sum of the angles is 270 degrees.

Now, draw the same triangle on a flat map of the Earth. In contrast to the triangle on the Earth or globe, the sum of the angles on the map will be180 (as long as the lines are straight on the map, at least). Either way, the flat map does not accurately represent the geomtry of the sphere, it can't because the sphere is non-euclidean. Either the lines that are straight on the sphere are not straight on the map, or the sum of the angles is wrong.

You can also show that the angular excess is related to the area of the triangle IIRC, which explains why small triangles (low area) are approximately Euclidean, while large triangles are not. See for instance http://en.wikipedia.org/wiki/Spherical_trigonometry#Area_and_spherical_excess

 

[PAllen]

Locally euclidean means locally homeomorphic to R^n. It has nothing to do with a metric.

We're talking about Riemannian or pseudo-Riemannian manifolds. These are equipped with a metric. The same topology with different metric are considered two completely different Riemannian manifolds.

 

[Cruz Martinez]

We're talking about Riemannian or pseudo-Riemannian manifolds. These are equipped with a metric. The same topology with different metric are considered two completely different Riemannian manifolds.

This is why I asked for the definition Peter was referring to. I would be surprised if the same term "locally euclidean" was used to mean two such different things.

 

[PeterDonis]

Locally euclidean means locally homeomorphic to R^n.

There may well be different conventions for what these terms mean. The convention I'm using is my understanding of the usual one in GR, where, as PAllen says, we are always dealing with Riemannian or pseudo-Riemannian manifolds, and the metric is physically relevant--two different metrics on the same topological space are physically different solutions.

 

[Cruz Martinez]

There may well be different conventions for what these terms mean. The convention I'm using is my understanding of the usual one in GR, where, as PAllen says, we are always dealing with Riemannian or pseudo-Riemannian manifolds, and the metric is physically relevant--two different metrics on the same topological space are physically different solutions.

Oh, I see, so what you mean is that the Christoffel symbols can be made to vanish at any spacetime point by a suitable coordinate transformation, right?

 

[PAllen]

Oh, I see, so what you mean is that the Christoffel symbols can be made to vanish at any spacetime point by a suitable coordinate transformation, right?

Also, the metric can be made diag(1,1,1...) at any chosen point (for Riemannian), or diag(1,-1,-1,-1) [or diag(-1,1,1,1) depending on signature convention] for pseudo-Riemannan.

 

[Cruz Martinez]

Also, the metric can be made diag(1,1,1...) at any chosen point (for Riemannian), or diag(1,-1,-1,-1) [or diag(-1,1,1,1) depending on signature convention] for pseudo-Riemannan.

I understand. This was just a very unfortunate use of terminology, I guess.

 

["Don't panic!"]

So is the basic point that as the geometry of the manifold is generally non-Euclidean, even though we can locally map patches of the manifold to [itex]\mathbb{R}^{n}[/itex] we can not use Cartesian coordinates to describe the points in the manifold in every patch as this would imply that there is a global Cartesian coordinate system and hence the geometry of the manifold will be flat globally. Thus, in general, although it might be possible to use Cartesian coordinates for a single (small) patch of the manifold, this will not be true for every patch and hence we require more general curvilinear coordinate systems in order to be able to describe every patch on the manifold and construct an atlas?

 

["Don't panic!"]

Part of my issue is that I keep picturing the concept of "maps locally to Euclidean space" as embedded in Euclidean space and imagining a rectilinear (Cartesian) coordinate system with (for example) a sphere embedded inside such that a can relate the coordinates on the sphere (using spherical polar coordinates, for example) to Cartesian coordinates in the Euclidean space that it is embedded in.

I think what has been confusing me is that one can transform from curvilinear coordinates such as cylindrical, spherical polar (etc.) coordinates to Cartesian coordinates (at least locally), for example, one can cover a sphere with to coordinate charts using spherical polar coordinates (with suitable restrictions on [itex]\theta, \phi[/itex]). Is it the case, however, that when one talks of a manifold (in general, regardless of a metric being defined), the statement "maps locally to Euclidean space" means that we can locally represent points on the manifold as [itex]n[/itex]-tuples [itex](x^{1},\ldots ,x^{n})[/itex] in [itex]\mathbb{R}^{n}[/itex] and doesn't imply that the manifold is locally flat or that we can use Cartesian coordinates (that is, it maps locally to [itex]\mathbb{R}^{n}[/itex], but may be equipped with a different metric than Euclidean). Thus, for a general manifold, we can map a point locally to an [itex]n[/itex]-tuple of non-Cartesian coordinates and there will be no transformation that we can make to express these coordinates as Cartesian coordinates.

Is it that in certain cases, for example, where there is constant curvature, the manifold is locally flat (i.e. has a Euclidean metric) and as such it is always possible to map between non-Cartesian and Cartesian coordinates in these cases, but in general this will not be possible?

 

[PeterDonis]

in general, although it might be possible to use Cartesian coordinates for a single (small) patch of the manifold, this will not be true for every patch

Wrong. Every small patch can be described by a Cartesian coordinate system, with the Cartesian metric (strictly speaking, this is an approximation, as we've said before). We've said that before in this thread.

But there is no single global Cartesian coordinate system, with the Cartesian metric, that describes the entire manifold. That means the individual Cartesian coordinate systems on different patches cannot be combined into a single Cartesian coordinate system.

we require more general curvilinear coordinate systems in order to be able to describe every patch on the manifold and construct an atlas?

No, you don't. You can construct an atlas that only contains Cartesian coordinate systems, one for each small patch. But, as above, there will be no way to combine any of those individual systems into a larger Cartesian coordinate system.

Part of my issue is that I keep picturing the concept of "maps locally to Euclidean space" as embedded in Euclidean space

Well, then you should stop doing that. But for what it's worth, I think your issue is that you keep forgetting things that have already been said in this thread, so you go back to the same wrong statements, instead of realizing that you need to just discard all your intuitions and look at what we're actually telling you.

For example, even though we've told you, repeatedly, that every small patch of a manifold can be described by a Cartesian coordinate system, you keep asking if that isn't true. (See above, and below.) If you keep on circling back to that, you won't get anywhere.

for a general manifold, we can map a point locally to an

n-tuple of non-Cartesian coordinates and there will be no transformation that we can make to express these coordinates as Cartesian coordinates.

Wrong. As above, you can always define Cartesian coordinates on a single small patch, so for any other coordinates defined on that small patch, there will always be a transformation that expresses them in terms of Cartesian coordinates on that small patch.

Is it that in certain cases, for example, where there is constant curvature, the manifold is locally flat (i.e. has a Euclidean metric) and as such it is always possible to map between non-Cartesian and Cartesian coordinates in these cases, but in general this will not be possible?

No. See above.

 

["Don't panic!"]

But there is no single global Cartesian coordinate system, with the Cartesian metric, that describes the entire manifold. That means the individual Cartesian coordinate systems on different patches cannot be combined into a single Cartesian coordinate system.

Ok, I think that's starting to clear things up a bit.

You can construct an atlas that only contains Cartesian coordinate systems, one for each small patch. But, as above, there will be no way to combine any of those individual systems into a larger Cartesian coordinate system.

So in principle, given a manifold, it is possible to construct an atlas in which each individual coordinate chart uses Cartesian coordinates (with a Euclidean metric defined in each patch?), but there is no way to combine these patches such that they create a globally Cartesian coordinate system for the manifold. Would this have anything to do with the transition functions in the coordinate chart overlaps?

you can always define Cartesian coordinates on a single small patch, so for any other coordinates defined on that small patch, there will always be a transformation that expresses them in terms of Cartesian coordinates on that small patch.

Why would one ever choose to use another coordinate system then? Is it just that it makes the analysis easy and allows one to study properties on the manifold over several coordinate patches?

 

[PeterDonis]

Would this have anything to do with the transition functions in the coordinate chart overlaps?

Yes.

Why would one ever choose to use another coordinate system then?

Because, as we've said before in this thread, there are features that we want to describe that do not fit in a single small patch of the manifold.

 

["Don't panic!"]

Referring to the transition functions - is this because it won't be possible to construct such maps such that a Cartesian coordinate system in one patch will be mapped to a Cartesian coordinate system in the overlap with its neighbouring patch and hence an inconsistency arises as the neighbouring patch will have a region that both can be described by Cartesian coordinates and not at the same time?

there are features that we want to describe that do not fit in a single small patch of the manifold

So by using other coordinate systems we can describe larger patches of the manifold (i.e. we can construct a coordinate chart that covers a larger region of the manifold) allowing us to analyse features that extend over a larger region of the manifold. And the point is that when we analyse small enough regions within one of these coordinate patches we can construct a locally invertible mapping between these coordinates and Cartesian coordinates?

 

[Nugatory]

Why would one ever choose to use another coordinate system then? Is it just that it makes the analysis easy?

That's one reason, but this is neither surprising (consider that we routinely use polar coordinates instead of cartesian coordinates to solve central force problems in classical physics, just because it's easier) nor unimportant (it can be the difference between a successful analysis and total failure).

There's also the need to avoid coordinate singularities. Polar coordinates are a poor choice for problems in classical mechanics that involve calculating values at the origin; Schwarzschild coordinates are a poor choice for calculating values at the event horizon of a black hole;

 

[PeterDonis]

is this because it won't be possible to construct such maps such that a Cartesian coordinate system in one patch will be mapped to a Cartesian coordinate system in the overlap with its neighbouring patch

No; this can be done. But the transition functions will not be trivial. In a flat manifold, the transformation between any two Cartesian coordinate charts is trivial, in the sense that in general it's just a translation of the origin plus a rotation of the axes. So the transition functions are just the trivial ones corresponding to those. In a curved manifold, the transition functions will not take those simple forms; they will be more complicated.

when we analyse small enough regions within one of these coordinate patches we can construct a locally invertible mapping between these coordinates and Cartesian coordinates?

Yes.

 

["Don't panic!"]

But the transition functions will not be trivial.

Does this imply that the metric will change in this overlap (or at least that its second-order derivative is non-trivial such that one can deduce that the manifold you are on is curved)?

 

[Cruz Martinez]

Is it the case, however, that when one talks of a manifold (in general, regardless of a metric being defined), the statement "maps locally to Euclidean space" means that we can locally represent points on the manifold as [itex]n[/itex]-tuples [itex](x^{1},\ldots ,x^{n})[/itex] in [itex]\mathbb{R}^{n}[/itex] and doesn't imply that the manifold is locally flat or that we can use Cartesian coordinates (that is, it maps locally to [itex]\mathbb{R}^{n}[/itex], but may be equipped with a different metric than Euclidean).

This part is very correct. The statement "maps locally to Euclidean space" simply means that you can assign an n-tuple of real numbers to every one of the points in a neighborhood of each point (hence locally), this neighborhood might be big or small. It simply means "you can use elements of R^n as suitable labels for the points of the manifold". There are some finer points to address but you get the idea.

 

[PeterDonis]

Does this imply that the metric will change in this overlap

No. The metric of each local Cartesian coordinate system is the same, in terms of the coordinates it corresponds to. But the transition functions, that tell you how to convert coordinates in one local Cartesian coordinate system to those of a neighboring one that partly overlaps the first one, will not be simple (i.e., not simple translations and rotations) if the manifold is curved.

 

["Don't panic!"]

There are some finer points to address but you get the idea.

Could you elaborate on this?

Pictorially, when we use a coordinate chart to represent points in a patch on a manifold (supposing that the patch is large enough that the manifold is curved within the patch), would it be correct to visualise the chart as a coordinate grid being "laid down" on the manifold over the patch we are considering. As the manifold is curved in this patch, the coordinate lines in this patch will be curved. We can map these coordinate lines locally into [itex]\mathbb{R}^{n}[/itex] along with the coordinates of the points in this patch. We can then map the coordinate description (in the curvilinear coordinate system) to Cartesian coordinates in a pointwise fashion?

 

[PeterDonis]

As the manifold is curved in this patch, the coordinate lines in this patch will be curved.

Once again, this is not necessarily true. You appear to be again visualizing the manifold as embedded in a higher dimensional Euclidean space, which you have already admitted is a bad idea. Even in a curved manifold, coordinate grid lines can be geodesics, which means they are not curved in any intrinsic sense.

 

["Don't panic!"]

Once again, this is not necessarily true. You appear to be again visualizing the manifold as embedded in a higher dimensional Euclidean space, which you have already admitted is a bad idea.

Yes sorry, I got this idea from reading a section of John Lee's book but I guess I misinterpreted it (just seemed like a nice heuristic way to understand why one would use curvilinear coordinates to parametrise a coordinate patch.

 

[Cruz Martinez]

Could you elaborate on this?

Well, the finer points I meant are topological concepts. The maps from the set M to R^n should be homeomorphisms, this gives M a topology, which is locally identical to the topology of R^n. (Note I use locally in the sense of neighborhoods).
Btw, which book by Lee are you reading?

 

["Don't panic!"]

which book by Lee are you reading?

"Introduction to Smooth Manifolds".

With regards to my earlier post that you said was correct:
So is the statement "locally Euclidean" specifically saying that locally points on the manifold can be represent by n-tuples of real numbers in [itex]\mathbb{R}^{n}[/itex] and says nothing about whether the coordinate space is flat or that the coordinates are Cartesian?! Is the point that coordinates are just labels to keep track of where all the points on the manifold are, so within a given patch we are free to choose any coordinate system we like (although in practice we would choose one that suited the problem at hand), not just Cartesian or spherical polar etc.?Also, for the example of the sphere, is it that we represent points in a given cooordinate patch by spherical polar coordinates (for example), such that a given point has a coordinate representation [itex](\theta , \phi)[/itex]. As the manifold is locally flat within a sufficiently small neighbourhood of each point we can then map these coordinate representations to coordinates in the ambient 3-dimensional Euclidean space, whose coordinates are Cartesian, i.e. [itex](\theta , \phi)\mapsto (\sin (\theta)\cos (\phi), \sin (\theta)\sin (\phi), \cos (\theta)) =(x,y,z)\in\mathbb{R}^{3}[/itex]?

 

[PeterDonis]

As the manifold is locally flat within a sufficiently small neighbourhood of each point we can then map these coordinate representations to coordinates in the ambient 3-dimensional Euclidean space

Please, please, please stop visualizing the manifold as being embedded in a higher-dimensional space! You've admitted it's a bad idea; we've agreed it's a bad idea; don't do it. That is not what the manifold being locally flat means.

 

[Cruz Martinez]

"Introduction to Smooth Manifolds".

With regards to my earlier post that you said was correct:
So is the statement "locally Euclidean" specifically saying that locally points on the manifold can be represent by n-tuples of real numbers in [itex]\mathbb{R}^{n}[/itex] and says nothing about whether the coordinate space is flat or that the coordinates are Cartesian?! Is the point that coordinates are just labels to keep track of where all the points on the manifold are, so within a given patch we are free to choose any coordinate system we like (although in practice we would choose one that suited the problem at hand), not just Cartesian or spherical polar etc.?

I think this is correct, though locally euclidean can mean different things in different contexts, as I have just learnt, so be aware of that.

As the manifold is locally flat within a sufficiently small neighbourhood of each point we can then map these coordinate representations to coordinates in the ambient 3-dimensional Euclidean space, whose coordinates are Cartesian, i.e. [itex](\theta , \phi)\mapsto (\sin (\theta)\cos (\phi), \sin (\theta)\sin (\phi), \cos (\theta)) =(x,y,z)\in\mathbb{R}^{3}[/itex]?

This on the other hand doesn't sound correct to me. The locally cartesian coordinates have nothing to do with the ambient space. They are just different labels, you can keep the sphere embedded in R^3 for visualization purposes, but it's better if you don't rely on that picture too much.