General relativity and curvilinear coordinates

["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 n-tuples (x^{1},\ldots ,x^{n}) in \mathbb{R}^{n} and doesn't imply that the manifold is locally flat or that we can use Cartesian coordinates (that is, it maps locally to \mathbb{R}^{n}, 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 \mathbb{R}^{n} 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 \mathbb{R}^{n} 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 (\theta , \phi). 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. (\theta , \phi)\mapsto (\sin (\theta)\cos (\phi), \sin (\theta)\sin (\phi), \cos (\theta)) =(x,y,z)\in\mathbb{R}^{3}?

 

[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 \mathbb{R}^{n} 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. (\theta , \phi)\mapsto (\sin (\theta)\cos (\phi), \sin (\theta)\sin (\phi), \cos (\theta)) =(x,y,z)\in\mathbb{R}^{3}?

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.

 

["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 x^{2}+y^{2}+z^{2}=1 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 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, 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 (\theta , \phi)\in\mathbb{R}^{2} 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 \mathbb{R}^{n}?

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 \mathbb{R}^{n} (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 ds^{2}=R^{2}(d\theta^{2}+\sin^{2}(\theta)d\phi^{2}) 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 \mathbb{R}^{n} 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.