General relativity and curvilinear coordinates

["Don't panic!"]

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.

Sorry, I worded this part badly, I was just specifically referring to the sphere example where one can embed it in 3 dimensional Euclidean space.

 

["Don't panic!"]

So is the general point then that we are free to choose any coordinate system we please to represent points in a given coordinate patch, as after all they are just labels and have no deeper meaning than that. So we can choose to use Cartesian coordinates if we consider small enough patches, or other coordinate systems that are able to cover much larger patches on the manifold?!

 

[Cruz Martinez]

So is the general point then that we are free to choose any coordinate system we please to represent points in a given coordinate patch, as after all they are just labels and have no deeper meaning than that. So we can choose to use Cartesian coordinates if we consider small enough patches, or other coordinate systems that are able to cover much larger patches on the manifold?!

I think this is a good way to think about it, yes.

 

[PeterDonis]

I was just specifically referring to the sphere example where one can embed it in 3 dimensional Euclidean space.

I know, but it's still a bad idea to think of it that way, because the local Cartesian coordinate chart that covers a small patch of the sphere centered on a chosen point is not the same as a Cartesian coordinate chart on the 3-d space in which the sphere is embedded.

 

["Don't panic!"]

the local Cartesian coordinate chart that covers a small patch of the sphere centered on a chosen point is not the same as a Cartesian coordinate chart on the 3-d space in which the sphere is embedded.

Ah ok. Sorry if this is another stupid question, but why is the local Cartesian coordinate system given in 3d when the sphere is a 2d manifold?

 

["Don't panic!"]

Apologies, I've just realized that the above question I posed is stupid as the sphere is the set of all points [itex]x^{2}+y^{2}+z^{2}=1[/itex] and so the problem is still two dimensional as the third coordinate is determined by the other two.

 

[PeterDonis]

why is the local Cartesian coordinate system given in 3d

It isn't. A local Cartesian coordinate system centered on a point on a 2-sphere has only two coordinates, ##x## and ##y##.

This is getting very frustrating; it's like you're not even reading what we write. You've been told repeatedly not to even think about the 3-d space in which the 2-sphere is embedded; you've even admitted yourself that it's a bad idea. Yet you continue to do it.

the sphere is the set of all points ##x^2+y^2+z^2=1##

And here you're doing it again.

 

["Don't panic!"]

the sphere is the set of all points x2+y2+z2=1x^{2}+y^{2}+z^{2}=1

I've seen it defined this way in a set of differential geometry notes, albeit [itex]S^{2}=\lbrace (x^{1},x^{2},x^{3})\in\mathbb{R}^{3}\;\vert\;(x^{1})^{2}+(x^{2})^{2}+(x^{3})^{2}=1\rbrace[/itex], I was just being lazy in the previous post and didn't write out what I'd read explicitly (apologies for that).

This is getting very frustrating; it's like you're not even reading what we write.

Rest assured, I am reading everything you write carefully. I have been wording the last few posts badly, as I wasn't thinking of the sphere embedded in 3-d space, I just was thinking about this early in the morning and made a stupid mistake, which I realized and put in my previous post (c.f. #76). Sorry, I'm not deliberately trying to annoy anyone, I guess coming from a physics background (where the level of mathematical rigour isn't great), I'm struggling to let go of the more elementary ways of studying geometry.

Let me try again. You know we were discussing larger coordinate patches, such that the manifold is non-trivially curved over this patch. Is it correct to say that we cannot use Cartesian coordinates as a coordinate chart for this patch? If so, is this so because the patch is large enough that the geometry of the patch is non-Euclidean?

 

[PeterDonis]

I've seen it defined this way in a set of differential geometry notes

Sure, if you are viewing the 2-sphere as being embedded in a 3-dimensional Euclidean space. But that's precisely the thing I thought we had agreed not to do. If you don't do that, then this definition makes no sense, because there are only two dimensions.

You know we were discussing larger coordinate patches, such that the manifold is non-trivially curved over this patch. Is it correct to say that we cannot use Cartesian coordinates as a coordinate chart for this patch?

First, let's be clear about exactly which "Cartesian coordinates" we are talking about. We are talking about two-dimensional Cartesian coordinates, ##x## and ##y##. Describing a patch of a 2-sphere using these coordinates means assigning a pair of numbers ##(x, y)## to each point in the patch, and computing distances between the points using the metric ##ds^2 = dx^2 + dy^2##. Note that these "distances" are distances along a geodesic (a great circle) of the 2-sphere.

Now, if the patch of the 2-sphere is small enough, we can assign 2-d Cartesian coordinates on the patch as described above, and the distances we computing using them as described above will be the same as the actual distances we measure (at least, to within the accuracy of our measurements). That is what we mean by saying that we can use Cartesian coordinates to describe the patch.

If, OTOH, the patch is larger, then the distances we compute using the Cartesian coordinates, as above, will be detectably different from the actual distances we measure, because of the curvature of the 2-sphere. This is what we mean by saying that we cannot use Cartesian coordinates to describe larger patches of the 2-sphere.

Let's illustrate all this by a concrete example. Suppose we are trying to describe distances on the Earth, which we'll idealize as perfectly spherical. We pick a point on the Earth's surface: say the intersection of the prime meridian with the equator. This point has ##(\theta, \phi)## coordinates (i.e., latitude, longitude) of ##(\pi / 2, 0)## (the usual convention for ##\theta## in spherical coordinates is that it is zero at the south pole and ##\pi## at the north pole, and we are measuring angles in radians).

We now want to describe a patch of the Earth's surface that is 2 meters across, centered on our chosen point, using Cartesian coordinates. We put the origin at our chosen point, so it has ##(x, y)## coordinates of ##(0, 0)##. The coordinate patch then extends to ##x## and ##y## values of ##+1## and ##-1##; the ##x## direction is along the equator, and the ##y## direction is along the prime meridian.

How do these coordinates correspond to the ##(\theta, \phi)## coordinates? Well, we want the distances computed in both charts to be the same, so we need the change in ##\theta## or ##\phi## corresponding to a distance of 1 meter. To obtain this, we must specify a number ##R## which is usually called the "radius of curvature" of the 2-sphere (but it's important to not think of this as the radius of the 2-sphere embedded in 3-d space, since we're explicitly not visualizing the 2-sphere that way--the number ##R## is just an intrinsic property of the 2-sphere). The metric in spherical coordinates is then ##ds^2 = R^2 \left( d \theta^2 + \sin^2 \theta d \phi^2 \right)##. For our idealized Earth, we will use ##R = 6.378 \times 10^6## meters.

To compare distances computed using the two charts, we then simply equate the formulas for ##ds^2## for the same distance. For example, consider the distance from the origin to the point 1 meter east along the equator. This point has Cartesian coordinates ##(1, 0)## and spherical coordinates ##(\pi / 2, \Phi)##, where ##\Phi## is the unknown we want to solve for. We have ##ds^2 = dx^2 + dy^2 = R^2 \left( d \theta^2 + \sin^2 \theta d \phi^2 \right)##. The differences in coordinate values are ##dx = 1##, ##dy = 0##, ##d\theta = 0##, ##d\phi = \Phi##. So we have ##ds^2 = 1 + 0 = \left( 6.378 \times 10^6 \right)^2 \left( 0 + \Phi^2 \right)##. This obviously gives ##\Phi = 1 / 6.378 \times 10^6 \approx 1.568 \times 10^{-7}##. A similar computation shows us that the point ##(x, y) = (0, 1)## in Cartesian coordinates has spherical coordinates ##(\theta, \phi) = (\pi / 2 + \Phi, 0)##.

To check whether our Cartesian description is accurate enough, however, we need to look at a distance that is not along one of our coordinate axes. For example, consider the point ##(x, y) = (1, 1)##, i.e., our Cartesian chart says it is exactly ##\sqrt{2}## meters northeast of the origin. What do we get when we compute this distance in spherical coordinates? The key is the ##\sin^2 \theta## factor in the ##d\phi^2## term in the metric. It didn't come into play before, because we were only considering line segments where either ##\theta## was constant or ##\phi## was constant. Now we have to consider a segment where both ##\theta## and ##\phi## are changing; the spherical coordinates of our point now are ##(\theta, \phi) = (\pi / 2 + \Phi, \Phi)##.

The rigorous way to compute the distance in spherical coordinates along this line segment would be to integrate our formula for ##ds^2##; but since ##\sin \theta## is linear at ##\theta = \pi / 2##, we can get a very good approximation by just averaging the two limiting values. So we have

$$
ds^2 = R^2 \Phi^2 \left( 1 + \frac{\sin \pi / 2 + \sin \left( \pi / 2 + \Phi \right)}{2} \right)
$$

This gives ##\sqrt{2}## meters to at least ten decimal places (at least according to my calculator), because ##\Phi## is so small that ##\sin \left( \pi / 2 + \Phi \right)## is 1 to many decimal places. So on this small patch of the Earth's surface, distances computed using Cartesian coordinates match up with actual distances to a very good approximation, certainly good enough for even very accurate surveying equipment to be unable to detect any difference.

Now, suppose we try to extend our Cartesian coordinates to a patch of the Earth's surface extending 100 kilometers from the origin. We still use meters as our units, so now we are using Cartesian coordinates extending to values of ##x = \pm 100000## and ##y = \pm 100000##. This corresponds to differences in the angular coordinates of ##\theta = \pi / 2 \pm \Theta## and ##\phi = \pm \Theta##, where ##\Theta = 100000 / R \approx 1.568 \times 10^2##. So we can still match distances exactly along the coordinate axes.

But when we look at distances along line segments that aren't along the coordinate axes, we run into trouble. Consider the segment running northeast, as before, but now going out to ##(x, y) = (100000, 100000)##, which corresponds to ##(\theta, \phi) = (\pi / 2 + \Theta, \Theta)##. The Cartesian distance is ##100000 * \sqrt{2} \approx 141421## meters. The actual distance, computed using the metric in spherical coordinates and using the averaging approximation we used above, is

$$
ds^2 = R^2 \Theta^2 \left( 1 + \frac{\sin \pi / 2 + \sin \left( \pi / 2 + \Theta \right)}{2} \right)
$$

This works out to about ##141419## meters, which means the Cartesian distance is about 2 meters too long. This is still a fairly small fraction of the total distance, but an error of this size is easily detectable using modern equipment. So for this larger patch, Cartesian coordinates no longer accurately represent actual distances on the Earth's surface.

 

["Don't panic!"]

Thanks for the detailed example, I think the penny is finally starting to drop, thanks for bearing with me!

A couple of questions though.

The metric in spherical coordinates is then ds2=R2(dθ2+sin2θdϕ2)

In practice, how does one determine the metric (at each point) on the manifold intrinsically (i.e. without resorting to any embedding)?
Do we need to know the form of the metric before we can determine whether the coordinates we have chosen correctly/accurately describe the patch of the manifold that we are considering?

Another question, although slightly aside, when one talks of a topology defining the geometry of a manifold, is it meant that from a topology we can determine (in a sense) how the points are related to one another ("nearness" to one another in the set, etc.) and thus the manifold has a geometrical structure even before introducing a metric?

 

[PeterDonis]

In practice, how does one determine the metric (at each point) on the manifold intrinsically (i.e. without resorting to any embedding)?

You assign coordinates to points, measure distances between the points, and use that information to determine what the metric is, i.e., what the form of the line element ##ds^2## must be in terms of the coordinates in order to give the correct distances.

when one talks of a topology defining the geometry of a manifold, is it meant that from a topology we can determine (in a sense) how the points are related to one another ("nearness" to one another in the set, etc.)

Yes, but the notion of "nearness" here does not have to be the same as the one implied by the geometry. See below.

and thus the manifold has a geometrical structure even before introducing a metric?

The manifold does have a structure independent of the metric, but that structure might not match up with the geometry given by the metric. For example, consider points lying along a null worldline in spacetime. These points can be ordered, and a topology can be defined on them that gives a notion of "nearness"--for example, a light ray traveling from the Earth to the Moon passes through intermediate points, and it is meaningful to say which points are closer to Earth vs. closer to the Moon. But geometrically, according to the metric of spacetime, the distance along the null worldline is zero.

 

["Don't panic!"]

the form of the line element ds2ds^2 must be in terms of the coordinates in order to give the correct distances.

So in the sphere example, in order for the line element to give the correct distance, it must be in terms of [itex](\theta , \phi)\in\mathbb{R}^{2}[/itex] only? Sorry to be a pain, but would you mind showing me how one can determine the metric of a sphere without embedding?

The manifold does have a structure independent of the metric, but that structure might not match up with the geometry given by the metric

Would it be correct to say that the topology is required so that one has a set of open subsets such that one can "stitch together" neighbouring open subsets to reconstruct the manifold (such that the manifold can be constructed from local patches of [itex]\mathbb{R}^{n}[/itex]?

Would it also be correct to say that the geometry of the manifold is determined by the metric - without a metric defined on it there is no notion of geometry on the manifold?

 

[PeterDonis]

in order for the line element to give the correct distance, it must be in terms of ##(\theta , \phi)\in\mathbb{R}^{2}## only?

If you're using those coordinates, yes. (Note that the ranges of the coordinates are restricted; ##\theta## goes from ##0## to ##\pi## and ##\phi## goes from ##0## to ##2 \pi##.) Remember that you can choose any coordinates you like, but the line element will be different depending on which ones you choose. Think, for example, of a Mercator projection or a stereographic projection of the Earth's surface. Those give different coordinates, and a different form of the line element, but the actual distance between two given points--say New York and London--will be the same no matter which coordinates/line element you use to calculate it. The only thing all the coordinate charts have to have in common is that they assign 2-tuples of numbers to points (not 3-tuples or 4-tuples or single numbers, etc.), because the manifold is two-dimensional.

would you mind showing me how one can determine the metric of a sphere without embedding?

As I said: you assign coordinates to points, measure the distances between the points, and figure out what the line element has to be as a function of the coordinates to give you the correct distances.

I understand that this is hard for us moderns to imagine, because we all know the Earth is round and is embedded in a 3-dimensional space, and the usual coordinates we use globally on the Earth are defined using global features (the equator, the poles, etc.), not things we can measure locally. In fact, even the ancients didn't use purely local, embedded measurements to learn that the Earth was round; Eratosthenes, the Greek geometer who first estimated the size of the Earth, used the difference in sun angle at noon on the summer solstice between Syene and Alexandria in Egypt. So what I've been describing has not really ever been done from scratch as I've described it, at least not for the Earth.

That being the case, why am I emphasizing this? Because when it comes to spacetime, we have no choice; the only measurements we can make are local, embedded measurements. We can't look at the universe "from the outside" to see how it is embedded in some higher-dimensional space. We have no evidence that it even is so embedded. So the only way we can measure the geometry of the universe is intrinsically, by assigning coordinates to events and making measurements of distances and times between events, and figuring out what the line element has to be as a function of the coordinates to give the correct distances and times.

For more details about how one might go about this, you might try Taylor & Wheeler's introductory texts on relativity; I believe that both Spacetime Physics and Exploring Black Holes have some discussion of it. Misner, Thorne, and Wheeler does as well, but that tome is a hefty investment.

Would it be correct to say that the topology is required so that one has a set of open subsets such that one can "stitch together" neighbouring open subsets to reconstruct the manifold (such that the manifold can be constructed from local patches of ##\mathbb{R}^{n}##?

This is really a definition of a topology (or part of the definition).

Would it also be correct to say that the geometry of the manifold is determined by the metric - without a metric defined on it there is no notion of geometry on the manifold?

For the usual definition of "geometry", yes, a metric is required. Some mathematics texts distinguish between "metrical geometry", which is what you've defined here, and "affine geometry", which doesn't require a metric but does require more structure than just a topology. For purposes of physics, the distinction isn't important because we need to have the metric anyway in order to match the theory to experiments, since experiments give us distances (and times).

 

[victorneto]

I have just been asked why we use curvilinear coordinate systems in general relativity. I replied that, from a heuristic point of view, space and time are relative, such that the way in which you measure them is dependent on the reference frame that you observe them in. This implies that coordinate systems change from point to point in spacetime, i.e. they are, in general, curvilinear coordinate systems. From a more mathematical point of view, spacetime is represented by a 4 dimensional manifold which is, in general, curved (physically this is caused by the presence of matter, although it is also possible for the spacetime to be intrinsically curved, even in vacuum). We wish to be able to describe such a manifold without embedding in some higher dimensional space (as after all, we have no a priori reason to believe that our universe is embedded in a higher dimensional space), and therefore we can only describe it in terms of local coordinate maps, in which we can construct locally invertible maps between Euclidean space [itex]\mathbb{R}^{n}[/itex] (which itself is most straightforwardly described by Cartesian coordinates) and the manifold (generally non-Euclidean). Thus the coordinate systems are necessarily curvilinear, as they can only locally describe the manifold, and will change as we move across the manifold.

Would this be an acceptable answer? Any feedback, improvements would be much appreciated.

Olá,
Very good your answer. And add the following relevant data: one can not draw straight lines in curved spaces as required by Cartesian rectilinear systems!
On the surface of a sphere, it is still a shortest distance curve. Hence the convenience of curvilinear coordinate systems.

 

[rrogers]

Would it be correct to say that as Cartesian coordinates describe a (hyper) plane, then one could run into the situation where the manifold has intrinsic curvature at each particular point and so one could not construct a Cartesian coordinate patch around any point as one would "move off" the manifold Such that the coordinate system doesn't accurately describe the points on the manifold at that point (as per your analogy, the points in the Cartesian coordinate system would be in a hyperplane tangent to a point on the manifold and would deviate from the actual description of points in the patch it's trying to describe)?

I (maybe incorrectly) visualise it as follows. Locally, a manifold can be mapped to Euclidean space. Such a manifold will be curved in general (even locally) and so in order to accurately describe points on the manifold in a given coordinate patch in terms of coordinates in Euclidean space we require a coordinate system whose coordinate lines curve (such that they recreate the curvature of the manifold in that patch). For example, with a sphere we can choose a hemisphere as a coordinate patch and "wrap" our coordinate system over the hemisphere such that it recreates the curvature as we move between different points on the hemisphere. This can be achieved by using, for example, spherical polar coordinates which describe the coordinates on a spherically curved surface in Euclidean space.

As an aside, when people talk of gravity as causing coordinate acceleration is it meant that because gravity is the manifestation of curved spacetime (due to matter content) in a region, when we describe this region in a coordinate system we will need to use curvilinear coordinates which will change as we move in that region, hence give the illusion that objects are accelerating relative to this coordinate system?

Perhaps it should be mentioned that the small neighborhood and local euclidean coordinates are typically shorthand for a particular vector space that is well defined. Using a set of corrdinates on a space that has continuity; at each point we consider all possible paths through it. The vector space defined by all of the tangents to the curve at one point is used. By continuity this can typically be considered embedded in the originating manifold; but that is not necessary and is only a crutch. I would suggest section 2.2 of Hawkins and Ellis "The large scale structure of space-time". Believe it or not the whole book is quite readable and clear (with patience and thought); with some isolated parts.

 

["Don't panic!"]

Am I correct in thinking that Cartesian coordinates are only applicable where Euclidean geometry holds (i.e. where the parallel postulate, distance is determined via Pythagoras theorem etc. hold), thus if the patch on the manifold that we are considering is large enough that the geometry is non-Euclidean within this patch, then we will have to consider more general coordinates?

I know in previous posts we've discussed how, in practice, we can determine the metric in a region on a manifold by performing measurements, but is there a way to derive it from a purely mathematical approach? For example, how does one determine that the line element on a sphere is given by [tex]ds^{2}=R^{2}(d\theta^{2}+\sin^{2}(\theta)d\phi^{2})[/tex] from a purely mathematical point of view?

 

[PeterDonis]

Am I correct in thinking that Cartesian coordinates are only applicable where Euclidean geometry holds

Yes. The metric that goes with Cartesian coordinates is the metric of a Euclidean geometry, so obviously it won't work for a geometry that isn't Euclidean.

in practice, we can determine the metric in a region on a manifold by performing measurements, but is there a way to derive it from a purely mathematical approach?

If you don't know what the manifold is, how can you derive anything mathematically? If you already know what the manifold is, then you already know the metric, since that's part of the definition of the manifold, so there's nothing to derive.

The practical problem, when we're trying to determine the metric of a real thing like the Earth's surface or the spacetime of the universe, is that we don't know in advance what the exact manifold is. We find that out by making measurements.

 

[PeterDonis]

For example, how does one determine that the line element on a sphere is given by

$$ds^{2}=R^{2}(d\theta^{2}+\sin^{2}(\theta)d\phi^{2})$$

from a purely mathematical point of view?

You don't "determine" that, in the sense of deriving it from something else. That line element defines what a "2-sphere" is, as a manifold with metric.

Again, the practical problem, if we're dealing with something real like the Earth's surface, is that we don't know, a priori, whether it's an exact 2-sphere or some other manifold. (We know it has the topology of a 2-sphere, but we don't know the exact metric a priori.) We have to determine exactly which manifold it is (what the exact line element is) by measurements.

 

["Don't panic!"]

The metric that goes with Cartesian coordinates is the metric of a Euclidean geometry, so obviously it won't work for a geometry that isn't Euclidean.

So is this what is meant in texts that I've read that say one cannot construct a Cartesian coordinate system on a manifold that is non-Euclidean at every point on the manifold?

Is the approach then (at least from a purely mathematical perspective) that once a manifold is constructed one defines a metric on this manifold and this then provides information about it's geometry and this defines what object we are considering (for example, a 2-sphere)?

To go back to an earlier point then, in the case where the geometry is non-Euclidean around every point on the manifold, when we introduce a coordinate chart the coordinate map to [itex]\mathbb{R}^{n}[/itex] will be of a more general form (spherical polar or stereographic for example), such that we can assign coordinates to points over a finite patch of the manifold instead of just infinitesimally close to a given point . Is this why one can locally use Cartesian coordinates on a 2-sphere, as the geometry is Euclidean in the local neighbourhood around each point?

 

[PeterDonis]

So is this what is meant in texts that I've read that say one cannot construct a Cartesian coordinate system on a manifold that is non-Euclidean at every point on the manifold?

What do you mean by "non-Euclidean at every point on the manifold"? Does this apply to a 2-sphere? See below.

Is the approach then (at least from a purely mathematical perspective) that once a manifold is constructed one defines a metric on this manifold and this then provides information about it's geometry and this defines what object we are considering (for example, a 2-sphere)?

The metric is part of the definition of the manifold (at least, with the definition of "manifold" that is used in physics), so yes, it defines what object is being considered.

Is this why one can locally use Cartesian coordinates on a 2-sphere, as the geometry is Euclidean in the local neighbourhood around each point?

I'm confused about your terminology. You can use Cartesian coordinates locally (meaning, on a sufficiently small patch around a given point--how small is "sufficiently small" depends on how accurate your measurements are) on any manifold, since that's part of the definition of a manifold. But when you say "the geometry is Euclidean in the local neighborhood around each point", that sounds like it's different from "a manifold that is non-Euclidean at every point". So I'm not sure what you're trying to say.

 

["Don't panic!"]

But when you say "the geometry is Euclidean in the local neighborhood around each point", that sounds like it's different from "a manifold that is non-Euclidean at every point". So I'm not sure what you're trying to say.

Sorry, I didn't word this very well. What I meant really was that on a curved manifold, such as a 2-sphere, in a text that I've read it said that Cartesian coordinates can only be constructed in an infinitesimal neighbourhood of each point on the manifold. Is it possible then with more general coordinates to describe larger (finite) neighbourhoods around each point on the manifold? (using spherical polar coordinates one can cover a 2-sphere with two coordinate charts, right?)

 

[PeterDonis]

in a text that I've read it said that Cartesian coordinates can only be constructed in an infinitesimal neighbourhood of each point on the manifold.

Correct. How small "infinitesimal" is, in practice, depends on how accurate your measurements are.

Is it possible then with more general coordinates to describe larger (finite) neighbourhoods around each point on the manifold

Of course. Isn't that what we've been saying all along?

(using spherical polar coordinates one can cover a 2-sphere with two coordinate charts, right?)

Yes.

 

["Don't panic!"]

Of course. Isn't that what we've been saying all along?

I think what has confused me earlier is that we discussed how points on a manifold could be locally described by [itex]n[/itex]-tuples in [itex]\mathbb{R}^{n}[/itex] and we can do this because the manifold is locally homeomorphic to Euclidean space, but to me this implies that the manifold is locally flat and so Cartesian coordinates are applicable. However, if we do consider larger patches that are curved (in the sense that the geometry over them is non-Euclidean) how can we still describe the points in terms of [itex]n[/itex]-tuples of coordinates in [itex]\mathbb{R}^{n}[/itex]?

 

[stevendaryl]

I think what has confused me earlier is that we discussed how points on a manifold could be locally described by [itex]n[/itex]-tuples in [itex]\mathbb{R}^{n}[/itex] and we can do this because the manifold is locally homeomorphic to Euclidean space, but to me this implies that the manifold is locally flat and so Cartesian coordinates are applicable. However, if we do consider larger patches that are curved (in the sense that the geometry over them is non-Euclidean) how can we still describe the points in terms of [itex]n[/itex]-tuples of coordinates in [itex]\mathbb{R}^{n}[/itex]?

Well, take the example of the surface of the Earth. The "patch" that is the entire Earth except for the north and south poles can be described by the pair of coordinates:

(latitude, longitude).​

So it's a 2-D manifold described by two coordinates. It's certainly non-Euclidean, though.

Here is perhaps a way to think about it: You have a two-dimensional space. That means that the points on that space can be described by two real numbers: [itex](x,y)[/itex]. For simplicity, let me assume that it is Riemannian, which just means that the relevant notion of "length" of a curve through space is always positive. Let [itex]\mathcal{P}_0[/itex] and [itex]\mathcal{P}_1[/itex] be two points that are close enough together that there is a unique minimal-distance path connecting the two. Let the coordinate of the two point be written as:

[itex]\mathcal{P}_0 = (x, y)[/itex]
[itex]\mathcal{P}_1 = (x+\delta x, y + \delta y)[/itex]

Let [itex]\delta s[/itex] be the distance between the points (as measured along the unique minimal-distance path).

[itex]\delta s^2[/itex] can be expressed as a double power series in [itex]\delta x[/itex] and [itex]\delta y[/itex] as follows:

[itex]\delta s^2 = A_{xx} \delta x^2 + A_{xy} \delta x \delta y + A_{yx} \delta y \delta x + A_{yy} \delta y^2[/itex]
[itex] + B_{xxx} \delta x^3 + B_{xxy} \delta x^2 \delta y + ... [/itex]
[itex] + C_{xxxx} \delta x^4 + C_{xxxy} \delta x^3 \delta y + ...[/itex]
[itex] + ...[/itex]

The first line groups together the terms that are second-order in the differences [itex]\delta x, \delta y[/itex]. The second line are the terms of 3rd order, the third line are the terms of 4th order, etc.

So far, that's true for any 2D patch, and for any coordinate system on that patch. We can define the coordinate system to be "locally cartesian at point [itex]\mathcal{P}_0[/itex]" if

1. [itex]A_{xx} = A_{yy} = 1[/itex]
2. [itex]A_{xy} = A_{yx} = 0[/itex]
3. For all [itex]i, j, k[/itex], [itex]B_{ijk} = 0[/itex]

For every 2-D patch, and for every point on that patch, there is a coordinate system that is locally cartesian at that point.

For the coordinate system to be globally cartesian within the patch, it would have to be the case that all the terms are zero except for the first line (only 2nd order terms).

 

["Don't panic!"]

Well, take the example of the surface of the Earth. The "patch" that is the entire Earth except for the north and south poles can be described by the pair of coordinates:
(latitude, longitude).So it's a 2-D manifold described by two coordinates. It's certainly non-Euclidean, though.

Thanks for your detailed explanation.
I think my confusion is, if the manifold is non-Euclidean, and we consider a patch that is large enough that it's geometry is non-Euclidean, when talks of assigning coordinates to points in this patch, is it that we can map the points to [itex]n[/itex]-tuples in [itex]\mathbb{R}^{n}[/itex] with respect to some coordinate map (defining a coordinate system) and not necessarily that the geometry has to Euclidean in order to do this?

 

[PeterDonis]

if the manifold is non-Euclidean, and we consider a patch that is large enough that it's geometry is non-Euclidean, when talks of assigning coordinates to points in this patch, is it that we can map the points to ##n##-tuples in ##\mathbb{R}^{n}## with respect to some coordinate map (defining a coordinate system) and not necessarily that the geometry has to Euclidean in order to do this?

Correct. Euclidean geometry means the coordinate n-tuples plus a particular metric, the Euclidean one. Non-Euclidean geometry means the coordinate n-tuples plus a different metric, a non-Euclidean one.

 

["Don't panic!"]

Correct. Euclidean geometry means the coordinate n-tuples plus a particular metric, the Euclidean one. Non-Euclidean geometry means the coordinate n-tuples plus a different metric, a non-Euclidean one.

So in saying that a manifold is locally homeomorphic to [itex]\mathbb{R}^{n}[/itex] is this the statement that each point on a manifold can be represented as an [itex]n[/itex]-tuple in [itex]\mathbb{R}^{n}[/itex], however in general it is not possible to construct coordinate maps in such a way that they cover the whole of a manifold, but can cover a patch of it (hence are local). The coordinate maps that map each point within a given patch on the manifold to an [itex]n[/itex]-tuple in [itex]\mathbb{R}^{n}[/itex] will in general be non-Cartesian as the geometry within the patch that they are describing will be generally non-Euclidean. Despite this, the geometry in a sufficiently small neighbourhood around each point in a given patch will be Euclidean and thus one can construct local Cartesian coordinates within such a neighbourhood.

 

[PeterDonis]

So in saying that a manifold is locally homeomorphic to ##\mathbb{R}^{n}##

"Homeomorphic" is a topological term, not a geometric term; it doesn't include the metric. A manifold meets a stronger condition: the metric is locally Euclidean.

is this the statement that each point on a manifold can be represented as an ##n##-tuple in ##\mathbb{R}^{n}##

Not exactly. "Locally homeomorphic" means that any open neighborhood of any point can be assigned coordinates in this way, i.e., that we can set up a one-to-one mapping between the points in any open neighborhood and ##n##-tuples of real numbers. But this in itself says nothing about the metric; a metric doesn't even have to exist. Manifolds in physics always have metrics, so they meet stronger conditions than just this one, as above.

however in general it is not possible to construct coordinate maps in such a way that they cover the whole of a manifold, but can cover a patch of it (hence are local).

This is because, in general, there might not be an open neighborhood of every point that covers the entire manifold. For example, there is no open neighborhood of any point on a 2-sphere that covers the entire 2-sphere (there can't be, since the 2-sphere as a whole is closed).

The coordinate maps that map each point within a given patch on the manifold to an ##n##-tuple in ##\mathbb{R}^{n}## will in general be non-Cartesian as the geometry within the patch that they are describing will be generally non-Euclidean.

Yes, but, as above, this is a stronger condition than just homeomorphism, because it involves the metric. A 2-sphere and an irregular blob with a closed 2-dimensional surface are homeomorphic (and both are locally homeomorphic to ##\mathbb{R}^n## ), but they do not have the same metric (i.e., geometry).

 

["Don't panic!"]

Ah ok, I think I'm starting to get it a bit more now. So the homeomorphism guarantees that we can assign n-tuples of real numbers to each point and then the metric on a given patch determines the type of coordinate maps we can construct from that patch to [itex]\mathbb{R}^{n}[/itex] (by this I mean that if the patch is small enough that the metric is Euclidean then we can use Cartesian coordinate maps, as well as others. However, if the patch is such that the metric is non-Euclidean within it then it will not be able to construct Cartesian coordinate maps (apart from infinitesimally close to each point), however other (more general) coordinate maps will be possible).

 

[Ibix]

I'm not sure if this will help (maybe I'm teaching my grandmother this), but...

One can cover (nearly) the whole of the surface of a cylinder with coordinates (φ,z). One can then simply perform x=φ, y=z and draw a chart on the x-y plane. There is a one-to-one correspondence between the points (φ,z) on the cylinder and the points (x,y) on the chart (there's a bit of a problem over how to represent φ=0 or 2π, but we'll ignore that for the time being). Furthermore, distances are preserved by the map - if I move in a straight line from (φ₀,z₀) to (φ₀+dφ,z₀+dz) then (assuming I was careful about the choice of scale for z) the distance ##\sqrt{d\phi^2+dz^2}## on the cylinder and the distance ##\sqrt{dx^2+dy^2}## on the chart are equal (and that holds as d → Δ).

One can also cover (nearly) the whole surface of a 2-sphere with coordinates (φ,z). Once again one can simply perform x=φ, y=z and draw a chart on the x-y plane (this is not quite a Mercator projection - not sure if it has a proper name). There is a one-to-one correspondence between the points (φ,z) on the sphere and the points (x,y) on the chart (although there are now problems at the poles as well as φ=0 or 2π, but again we'll ignore that). However, in this case distances are not preserved. In general, ##\sqrt{d\phi^2+dz^2} \neq \sqrt{dx^2+dy^2}## (however careful I was about the choice of z). You can easily see this by considering a half circle centered on the north pole (a full circle isn't legit since we'd have to cross the coordinate singularity in our naive choice of map). The path length on the sphere gets arbitrarily small as z tends towards 1; on the map it is π whatever the value of y.

The difference between the two cases is because the cylinder comes equipped with a Euclidean metric while the sphere does not. The cylinder has no intrinsic curvature while the sphere does.

However - look again at the sphere on the equator. Distances up the lines of constant longitude are pretty much correct close to the equator (they'd be perfect if we'd picked a Mercator projection) and distances along the equator are spot on, since the total length of the equator is 2πR in both the map and the sphere. So locally we have a very close approximation to a Euclidean metric. As long as we don't stray too far from the equator the map we've drawn is fine. It's only if we go too far away that we start to notice distortion.

For the manifolds we work with in GR we can always find a way to project the manifold onto ℝⁿ so that there is a "small" region like the equatorial belt in the sphere example around the point of interest. In the sphere example, all we need to do is move the pole to 90° away from our point of interest and hey presto - this is effectively what a street atlas of your town has done. How small is small depends on how strongly curved the manifold is and how precise we need to be. But if there is curvature, inevitably there comes a point where "let's pretend it's Euclidean" will have your GPS driving you into walls...

 

[PeterDonis]

So the homeomorphism guarantees that we can assign n-tuples of real numbers to each point and then the metric on a given patch determines the type of coordinate maps we can construct from that patch to ##\mathbb{R}^{n}##

Not quite. An assignment of n-tuples of real numbers to each point is a coordinate map from the patch to ##\mathbb{R}^{n}##. The geometry of the manifold determines the form of the line element ##ds^2## when expressed in terms of that coordinate map.

 

[Cruz Martinez]

So in saying that a manifold is locally homeomorphic to [itex]\mathbb{R}^{n}[/itex] is this the statement that each point on a manifold can be represented as an [itex]n[/itex]-tuple in [itex]\mathbb{R}^{n}[/itex], however in general it is not possible to construct coordinate maps in such a way that they cover the whole of a manifold, but can cover a patch of it (hence are local). The coordinate maps that map each point within a given patch on the manifold to an [itex]n[/itex]-tuple in [itex]\mathbb{R}^{n}[/itex] will in general be non-Cartesian as the geometry within the patch that they are describing will be generally non-Euclidean. Despite this, the geometry in a sufficiently small neighbourhood around each point in a given patch will be Euclidean and thus one can construct local Cartesian coordinates within such a neighbourhood.

This is entirely correct, except I wouldn't say "represented", I would say "mapped" or "labeled". Also the points in the manifold are not mapped to n-tuples of R^n, but to single elements of R^n, which are n-tuples of numbers, this is just a small correction to what you said.

 

[Cruz Martinez]

Not exactly. "Locally homeomorphic" means that any open neighborhood of any point can be assigned coordinates in this way, i.e., that we can set up a one-to-one mapping between the points in any open neighborhood and ##n##-tuples of real numbers. But this in itself says nothing about the metric; a metric doesn't even have to exist. Manifolds in physics always have metrics, so they meet stronger conditions than just this one, as above.

Locally homeomorphic to R^n means that around each point we can find a neighborhood of the point and a homeomorphism from the neighborhood to R^n. This doesn't imply we can do this for any neighborhood in the manifold, only that we can always find one such neighborhood around each point.
Also, manifolds in physics don't always have a metric, what about phase-space in classical mechanics?

This is because, in general, there might not be an open neighborhood of every point that covers the entire manifold. For example, there is no open neighborhood of any point on a 2-sphere that covers the entire 2-sphere (there can't be, since the 2-sphere as a whole is closed).

The whole 2-sphere is also open in the subspace topology, in fact it is clopen. The fact that a set is closed in some topology doesn't mean it isn't open as well.

 

[PeterDonis]

This doesn't imply we can do this for any neighborhood in the manifold

Can you give an example of a manifold that has an open neighborhoods which doesn't have a homeomorphism to ##\mathbb{R}^n##?

manifolds in physics don't always have a metric, what about phase-space in classical mechanics?

Hm, good point. You can define a metric on phase space, but it won't be unique, and AFAIK it isn't really used for anything. There is, however, a unique volume form on phase space, so the notion of "phase space volume" has a well-defined meaning, even though "phase space distance" doesn't.

The whole 2-sphere is also open in the subspace topology, in fact it is clopen.

The subspace topology on a 2-sphere isn't the relevant topology for this discussion.

 

[micromass]

Can you give an example of a manifold that has an open neighborhoods which doesn't have a homeomorphism to ##\mathbb{R}^n##?

Manifold: the circle ##S^1##. The open neighborhood: the entire circle ##S^1##.

The subspace topology on a 2-sphere isn't the relevant topology for this discussion.

Then which one is, because your statement confused me.