# General relativity and curvilinear coordinates

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1. Jun 8, 2015

### "Don't panic!"

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.

Last edited: Jun 8, 2015
2. Jun 8, 2015

### Staff: Mentor

This looks ok, heuristically, but I think there is a simpler way to put it: Cartesian (non-curvilinear) coordinate systems only exist globally in a flat manifold, and spacetime in the presence of gravity is not flat. So we have to use curvilinear coordinate systems for spacetime in the presence of gravity because they're the only ones that exist.

3. Jun 8, 2015

### "Don't panic!"

Thanks for taking a look. Is it also fair to say that, in order to describe a curved space one requires a set of curvilinear coordinates defined locally. We can find a mapping from such a curvilinear coordinate system to local Cartesian coordinates. For example, if the manifold is a 2-sphere it can be described by a curvilinear coordinate system $(r,\theta)$, that can be locally mapped to a Cartesian coordinate system, such that we can describe the whole surface through a collection of coordinate patches from open sets of $(r,\theta)$ to Cartesian coordinates?!

4. Jun 8, 2015

### Staff: Mentor

Any coordinates will be "defined locally"; there are no other kinds of coordinates.

5. Jun 8, 2015

### "Don't panic!"

And is this why we consider curvilinear coordinates, as locally implies that that coordinates change from point to point?! (Sorry, I think I've gone and confused myself now)

6. Jun 8, 2015

### Staff: Mentor

No, it doesn't. "Locally" just means that coordinates are assigned to points in the manifold, and nearby points get assigned nearby coordinate values, and within a small enough patch around a given point we can use flat coordinates, at least to a good approximation. But that will be true whether the manifold is globally flat or globally curved. Curvature is something that doesn't show up locally.

7. Jun 8, 2015

### "Don't panic!"

I get that, but then why do we use curvilinear coordinates at all, couldn't we just use Cartesian coordinates locally? Or is the point that points on the manifold exist independently of the coordinates that we assign them, and so we are free to use whaf ever coordinates we like?

8. Jun 8, 2015

### Staff: Mentor

Because many of the questions we would like to answer are not local questions. For example, if we want to describe the event horizon of a black hole, we can't do it using local coordinates; the horizon is a global feature of the spacetime geometry, not a local one. (It's the same problem we would have if we tried, for example, to describe the route an airliner takes from Los Angeles to Sydney; there's no way to do it using local Cartesian coordinates.)

9. Jun 8, 2015

### "Don't panic!"

So, for example, if we wanted to describe a path across the surface of a sphere we would need to do so using curvilinear coordinates (in this case spherical polar). I think what confuses me, is that coordinate systems on a manifold are usually only valid over a particular patch. Is the point that the patch that a particular coordinate system is valid over may be large enough that curvature needs to be taken into account? (Intuitively I can picture a coordinate grid "wrapped" around a sphere, for example)

10. Jun 8, 2015

### Staff: Mentor

Yes.

Or Mercator, or stereographic, or...

The point being that "curvilinear coordinates" is a very general term; for any manifold, there will be lots of different possible curvilinear coordinates that could be chosen.

Yes. For example, spherical polar coordinates on a sphere are valid everywhere except the poles.

11. Jun 8, 2015

### "Don't panic!"

Great, I think that's started to clear things up for me.
Is it correct to say that, in general, curvilinear coordinates are required to describe points, curves etc on a manifold, as if one used Cartesian only one would only ever be able to describe very small patches on the manifold and would be unable to describe paths across the manifold (which would be pretty useless).

12. Jun 8, 2015

### Staff: Mentor

To describe curves and other geometric objects that cover more than a small local patch of the manifold, yes. Points, strictly speaking, do not fall in this category; you can always describe a single point and a small local patch around it with local Cartesian coordinates.

13. Jun 8, 2015

### "Don't panic!"

Ah, ok. So we could choose to use local Cartesian coordinates to describe small patches, but in general this wouldn't be useful as we would be unable to describe curves, vector fields etc. that are defined over several coordinate patches.

14. Jun 8, 2015

### PAllen

Well, even here trying to use lots of small Cartesian patches, you run into the fact that none of them can be precisely Cartesian if the manifold is curved. Curvature is non-zero at a single point. In rough analogy with simple calculus, the deviation between a tangent and a curve gets smaller the closer you get to the point of tangency; yet the second derivative does not approach zero, and has a specific value at every point.

15. Jun 9, 2015

### "Don't panic!"

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?

Last edited: Jun 9, 2015
16. Jun 9, 2015

### martinbn

This depends on what one calls Cartesian coordinates. For example if the manifold is $\mathbb R^3$, then are the usual coordinates $(x, y, z)$ Cartesian? I'd guess that most people would say yes, and that they are global. Is this manifold flat? Well, that is not a meaningfull question until a metric is specified (or just a connection). It can be flat (with the usual structure) or it could be non-flat.

17. Jun 9, 2015

### stevendaryl

Staff Emeritus
I would say that coordinates are Cartesian only if the metric (or line element) is $ds^2 = dx^2 + dy^2 + dz^2$. So you can't have Cartesian coordinates on a curved space, even on a tiny patch.

18. Jun 9, 2015

### Staff: Mentor

Just to clarify, when I said "flat manifold", I meant a manifold with a flat metric (zero Riemann curvature tensor) defined on it.

19. Jun 9, 2015

### Staff: Mentor

Not really. The problem is not that you can't assign Cartesian coordinates to points on the manifold. The problem is that the actual metric on the manifold is not the flat Cartesian metric. So if you use the Cartesian coordinates, with their metric, to compute physical quantities, you will get incorrect answers.

Not exactly. The coordinate lines seem "curved" only because you are visualizing them in a higher dimensional flat space--for example, great circles on the Earth seem curved to you because you are visualizing them as being embedded in flat 3-dimensional space. But with respect to the intrinsic metric of the Earth's surface, great circles are not curved; they are straight lines (geodesics). So coordinate lines do not have to be curved in any intrinsic sense in order to cover a curved manifold. (Note that they can be curved--lines of latitude on the Earth, except for the equator, are not geodesics, even though lines of longitude are. But they do not have to be curved.)
It is true that any coordinates in which gravity causes coordinate acceleration will be curvilinear, yes. But the main thing the term "coordinate acceleration" is emphasizing is the physical fact that objects moving solely under gravity are in free fall; they feel zero proper acceleration. So any talk of "acceleration due to gravity" can only refer to coordinate acceleration. This is important because coordinate acceleration is (obviously) dependent on the coordinates you choose, whereas proper acceleration is an invariant.

Last edited: Jun 9, 2015
20. Jun 9, 2015

### "Don't panic!"

I think part of my confusion arises because we tend to relate Cartesian coordinates to more general coordinates. For example, in a polar coordinate system $(r,\theta)$ we can map to Cartesian coordinates via $$x=r\cos (\theta),\qquad y=r\sin (\theta)$$ Sorry, I've got myself into a real muddle here, I thought I understood this all before, but now having started to try and learn differential geometry I'm having severe doubts about my understanding!

Is the point that if we use a curvilinear coordinate system then we can treat it as Cartesian (in it's own system, i.e. set up a coordinate grid, for example, in polar coordinates there would be an $r$ -coordinate line and a $\theta$ - coordinate line) if we then map to an actual Cartesian coordinate system though, and represent the curvilinear coordinate system in a Cartesian coordinate system the coordinates lines of the curvilinear coordinate system will appear curved?!

Last edited: Jun 9, 2015