I Riemann tensor in 3d Cartesian coordinates

[davidge]

Suppose we wish to use Cartesian coordinates for points on the surface of a sphere. Then all derivatives of the metric would vanish and so the Riemann curvature tensor would vanish. But it would give us a wrong result, namely that the space is not curved. So it means that if we want to get correct results we must necessarily use spherical coordinates in this case?

 

[Orodruin]

You are using Cartesian coordinates in an embedding space. This is not the same thing as describing the intrinsic properties of the surface and even if the metric in the embedding space is flat, the induced metric on the submanifold can be curved.

 

[Nugatory]

First you have to precise about what you mean by "using Cartesian coordinates on the surface of a sphere". If you're just attaching ordinary x,y and z coordinates to points on the surface of a soccer ball on the table in front of you, then you're working in a three-dimensional flat space (the room you're in) that happens to contain a soccer ball. A straight line between any two points on the surface of the ball will pass through the interior of the ball; a two dimensional creature living on the two dimensional surface of the ball would interpret that straight line as a path through the mystical third dimension that connects the two points without going through any of the points in between.

The two-dimensional surface of the ball is curved, and it will be identically curved no matter what two-dimensional coordinates you use to label the points on its surface. Latitude and longitude are familiar and easy to work with, but there are others. Note that in this two dimensional world, the paths through the interior of the ball do not exist; a straight line between any two points is a great circle along the surface.

 

[davidge]

Thanks Nugatory and Orodruin for your reply.

You are using Cartesian coordinates in an embedding space

If you're just attaching ordinary x,y and z coordinates to points on the surface of a soccer ball on the table in front of you, then you're working in a three-dimensional flat space (the room you're in) that happens to contain a soccer ball.

a two dimensional creature living on the two dimensional surface of the ball would interpret that straight line as a path through the mystical third dimension that connects the two points without going through any of the points in between.

Then two-dimensional surface of the ball is curved, and it will be identically curved no matter what two-dimensional coordinates you use to label the points on its surface

I see I think. Good example Nugatory.

So to get non zero components of the Riemann tensor we would have to make a displacement on the surface of the sphere? Using spherical coordinates it's easy, because one just have to set the radius constant. Using Cartesian coordinates I think it is more difficult to keep on a "straight line", because we would have to change three parametres, namely x, y and z. In this case, we would get non-zero derivatives of the metric. Is this right?

 

[Nugatory]

Using Cartesian coordinates I think it is more difficult to keep on a "straight line", because we would have to change three parametres, namely x, y and z. In this case, we would get non-zero derivatives of the metric. Is this right?

The surface of the sphere is a two-dimensional surface, so you only get two coordinates (if you introduce a third, you are working with a three-dimensional space of which the points making up the surface of your sphere is an arbitrary subset). No matter what coordinates you use, you will find that there is no two-dimensional coordinate system in which the metric components are $g_{ij}=\delta_{ij}$ everywhere on the surface of the sphere - which is to say that surface is not flat and it cannot be properly described with two-dimensional Cartesian coordinates.

Of course "flat" can be a pretty good approximation, and then we can use Cartesian coordinates locally. For example, if you ask a Manhattanite for directions, you're likely to hear something like "three blocks uptown, two blocks crosstown" - that's Cartesian coordinates with axes aligned along the street grid. But formally what we're doing is defining a completely different two-dimensional surface, namely a plane that happens to be tangent to the surface of the Earth at the point that we're standing and using Cartesian coordinates to give directions for motion in that plane, not on the surface of the earth.

 

[davidge]

No matter what coordinates you use, you will find that there is no two-dimensional coordinate system in which the metric components are gij=δijgij=δijg_{ij}=\delta_{ij} everywhere on the surface of the sphere - which is to say that surface is not flat and it cannot be properly described with two-dimensional Cartesian coordinates.

Could it be accepted as a definition of Cartesian coordinates? That is, Cartesian coordinates are the ones which describe flat spaces, no matter what their dimension is?

formally what we're doing is defining a completely different two-dimensional surface, namely a plane that happens to be tangent to the surface of the Earth at the point that we're standing and using Cartesian coordinates to give directions for motion in that plane, not on the surface of the earth.

I see.

 

[Orodruin]

Could it be accepted as a definition of Cartesian coordinates? That is, Cartesian coordinates are the ones which describe flat spaces, no matter what their dimension is?

No. Cartesian coordinates are coordinates on a Euclidean space. There is no necessity for a flat space to be Euclidean.

 

[davidge]

No. Cartesian coordinates are coordinates on a Euclidean space. There is no necessity for a flat space to be Euclidean.

Can you give me an example of a flat, non-Euclidean space in which we cannot use Cartesian coordinates?

 

[Orodruin]

Can you give me an example of a flat, non-Euclidean space in which we cannot use Cartesian coordinates?

A flat torus, a flat Möbius strip, a cylinder, a flat Klein bottle, a cone with the apex removed, etc.

 

[davidge]

a cylinder

In what sense a cylinder is flat?

 

[Orodruin]

In what sense a cylinder is flat?

The usual sense. A cylinder embedded in R3 has zero intrinsic curvature. You can cut it open and lay it flat on a table.

 

[Ibix]

A cylinder has no intrinsic curvature. Draw a triangle on a piece of paper and roll it into a cylinder. The angles still add to 180. Circles still have circumference $2\pi r$ if the radius is measured in the surface of the cylinder. Etcetera. It's pretty much the canonical simple example of a space with extrinsic curvature but no intrinsic curvature.

 

[Orodruin]

Also, in general, a metric need not be induced by an embedding.

 

[DrGreg]

In what sense a cylinder is flat?

You can tightly wrap a flat piece of paper round it without stretching, tearing or crumpling. So its "intrinsic" local structure (as a manifold) is no different than a flat piece of paper's. Of course they have different "extrinsic" properties (which aren't part of the manifold, but of the embedding in 3D Euclidean space) and different global properties, but flatness is a local intrinsic property.

 

[Nugatory]

In which sense a cylinder is flat?

Take a sheet of ordinary paper. It will pass any test for local flatness you try: the interior angles of triangles will add to 180 degrees, the pythagorean theorem works, there is a coordinate system in which the metric components are $g_{ij}=\delta_{ij}$. Roll the sheet up into tube and it will form the surface of a cylinder, but it will still have all those flatness properties. The easiest way to see this is to imagine that you drew some geometric shape on the piece of paper before you rolled it up - the marks on the sheet of paper don't move around on the surface so their geometric relationships don't change.

The surface of a sphere is fundamentally different because there's no way of forming the sheet of paper into a sphere without stretching the paper (which doesn't work because paper doesn't stretch - better to use a rubber sheet instead) and altering the geometric relationships between nearby points.

This would be a good time to google search for the difference between "intrinsic curvature" and "extrinsic curvature" if you are not already familiar with those terms.

 

[Orodruin]

It is official. This thread is now a test of how many different ways one can say "make a paper roll".

 

[davidge]

@Orodruin , @DrGreg , @Nugatory and @Ibix, thanks for explaining what is the meaning of intrinsic and extrinsic curvature.

No matter what coordinates you use, you will find that there is no two-dimensional coordinate system in which the metric components are $g_{ij} = g_{ji} = \delta_{ij}$ everywhere on the surface of the sphere

Trying to understand the meaning of this, I have tried the following:

Suppose we want to calculate the distance between two points on a sphere, where one point is on the pole and the other is on the equator of the sphere. We can set our coordinate axes such that one coordinate, say z, is always zero.

The infinitesimal Euclidean distance is $ds^2 = dx^2 + dy^2$, but $x^2 + y^2 = r^2$ in this case, and so $ds^2 = \frac{x^2}{y^2}dx^2 + \frac{y^2}{x^2}dy^2$. (Is this right?) It seems clear that the metric will have first, second, etc... derivatives. Now it seems that that space (surface of the sphere) has a curvature, although I have not worked on the components of the Riemann tensor. (There will be any non-zero component?)

 

[Nugatory]

Suppose we want to calculate the distance between two points on a sphere, where one point is on the pole and the other is on the equator of the sphere. We can set our coordinate axes such that one coordinate, say z, is always zero.

This picture makes no sense as a way to assign coordinates to the two-dimensional surface of the sphere because there are points on all three coordinate axes that do not lie on the surface of the sphere - so whatever set of points are being labeled by those coordinates, it's not the set of points that corresponds to the surface of the sphere. Furthermore, you have three coordinates not two - taking $z=0$ reduces the number of coordinates by one, but the surface that is being described by the two remaining coordinates is not the surface of the sphere, it's a plane that intersects the sphere.

One way or another you need two coordinates, not three, that cover the surface of the sphere without picking up any points outside the sphere (informally, this means that the coordinate axes lie on the surface of the sphere). Latitude and longitude would work, as would stereographic coordinates. You can calculate the value of the components of the metric tensor for the surface of the sphere in your chosen coordinate system by using the concept that @Orodruin mentioned above: it's an induced metric once you've decide to think of the two-dimensional sphere as something embedded in Euclidean three-dimension space.

 

[davidge]

Ok

You can calculate the value of the components of the metric tensor for the surface of the sphere in your chosen coordinate system by using the concept that @Orodruin mentioned above: it's an induced metric once you've decide to think of the two-dimensional sphere as something embedded in Euclidean three-dimension space.

How would this induced metric look like in terms of Cartesian coordinates?

 

[Nugatory]

How would this induced metric look like in terms of Cartesian coordinates?

You mean the Cartesian coordinates of the three-dimensional Euclidean space in which the sphere is embedded?
Suppose you go with latitude $\phi$ and longitude $\theta$ as your coordinates on the surface of the sphere. We have (although wise people will check my algebra):
$x=R\cos\phi\sin\theta$
$y=R\cos\phi\cos\theta$
$z=R\sin\phi$
relating the $\theta$ and $\phi$ coordinates of points on the two-dimensional surface of the sphere to the x,y, and z coordinates of points in the three-dimensional Euclidean space in which the sphere is embedded. Some algebra and the formula in the wikipedia article I linked will see you home from there.

 

[davidge]

You mean the Cartesian coordinates of the three-dimensional Euclidean space in which the sphere is embedded?

Yes

Suppose you go with latitude ϕϕ\phi and longitude θθ\theta as your coordinates on the surface of the sphere. We have (although wise people will check my algebra):
x=Rcosϕsinθx=Rcos⁡ϕsin⁡θx=R\cos\phi\sin\theta
y=Rcosϕcosθy=Rcos⁡ϕcos⁡θy=R\cos\phi\cos\theta
z=Rsinϕz=Rsin⁡ϕz=R\sin\phi
relating the θθ\theta and ϕϕ\phi coordinates of points on the two-dimensional surface of the sphere to the x,y, and z coordinates of points in the three-dimensional Euclidean space in which the sphere is embedded. Some algebra and the formula in the wikipedia article I linked will see you home from there.

Yea, in this case we would obtain the components in terms of $r, \theta$ and $\phi$. I would like to get them in terms of $x,y$ and $z$ instead.

 

[Orodruin]

YesYea, in this case we would obtain the components in terms of $r, \theta$ and $\phi$. I would like to get them in terms of $x,y$ and $z$ instead.

You are missing the point. The radius $r$ is not a coordinate on the sphere. The sphere is two-dimensional and is described with only two coordinates.

 

[davidge]

The radius rrr is not a coordinate on the sphere

I'm sorry for including the $r$ into above. So how can we obtain the components in terms of $x,y$ instead of $\theta, \phi$?

 

[Orodruin]

I'm sorry for including the $r$ into above. So how can we obtain the components in terms of $x,y$ instead of $\theta, \phi$?

Start from the line element and the relation $ds^2 = dx^2 + dy^2 + dz^2$ and the expression of z in terms of x and y $z = \sqrt{R^2 -x^2-y^2}$ to find $dz$.

 

[pervect]

@Orodruin , @DrGreg , @Nugatory and @Ibix, thanks for explaining what is the meaning of intrinsic and extrinsic curvature.Trying to understand the meaning of this, I have tried the following:

Suppose we want to calculate the distance between two points on a sphere, where one point is on the pole and the other is on the equator of the sphere. We can set our coordinate axes such that one coordinate, say z, is always zero.

The infinitesimal Euclidean distance is $ds^2 = dx^2 + dy^2$, but $x^2 + y^2 = r^2$ in this case, and so $ds^2 = \frac{x^2}{y^2}dx^2 + \frac{y^2}{x^2}dy^2$. (Is this right?) It seems clear that the metric will have first, second, etc... derivatives. Now it seems that that space (surface of the sphere) has a curvature, although I have not worked on the components of the Riemann tensor. (There will be any non-zero component?)

You might find articles on the Mercator projection helpful. If you look at a map of the globe, the most common map is the Mercator projection, though there are others. If you read about about cartography, which is less advanced than GR but seems applicable to what you're struggling with, you can get a better idea of how we create a 2d map of the 2d surface of the Earth - and the formulas we use to get distances from the map coordinates, which would basically be the metric.

Using x,y, and z as coordinates doesn't really work, because not all coordinate values are constrained to lie on the sphere. You could conceivably use x and y as coordinates on a hemisphere, but you won't have the amount of reference material available to you that you will have if you go with lattitude, longitude, and the mercator projection technique.

If you did use x and y as coordinates, I can quickly sketch out what I think you should get - which doesn't look particularly close to what you wrote earlier. You want to compute $ds^2 = dx^2 + dy^2 + dz^2$ with the constraint equation that defines z(x,y) = $\sqrt{R^2 - x^2 - y^2}$

Using the chain rule, we can write

$$dz = \frac{\partial z(x,y)}{\partial x}dx + \frac{\partial z(x,y)}{\partial y}dy$$

So we would get

$$
ds^2 = dx^2 + dy^2 + dz^2 = dx^2 + dy^2 + \left( \frac{\partial z(x,y)}{\partial x}dx + \frac{\partial z(x,y)}{\partial y}dy \right)^2
$$

And if you expand it out you'll get a line element in terms of dx and dy. Conceptually, you'd be mapping the Earth by a projection process, but instead of the mercator projection process I suggested earlier, you'd map points on half the sphere to a unit circle on the x,y plane. So you'd really need two maps to cover an entire sphere.

The mercator projection technique isn't that much more invovled, is a bit more standard, and maps all points on the globe except for the north and south poles on one map, so it's probably a preferable technique. I don't have any good references for it, alas - you can try the wiki, it does have an article, but I'm not sure how indepth it really is.

In any event, you'll wind up with 2 coordinates for every point on the 2d manifold that's the surface of the globe, which is the correct noumber of coordinates. And you'll find that there is some formula that will give you the distance between two points , but it won't be the pythagorean formula that you seem to want it to be, it will be some other more complex formula..

 

[davidge]

@pervect , thank you, it's interesting to know how this kind of mapping works. I will google it!

These are my conclusions
- Cartesian coordinates are any kind of coordinates we use to map points on a Euclidean space.
- For general non-Euclidean spaces we must find appropriate coordinates that describes properly that space, and we must relate it in some way with the Cartesian ones.*

Are these conclusions correct?

*This raise two questions:
1 - If we always define the another set of coordinates in terms of the Cartesian set, then the latter can be regarded as a fundamental or "natural" set of coordinates? Although it would be the same as treating Euclidean spaces as the fundamental ones, and I think you will say it is not correct to assume it.

2 - There is no way of finding such a coordinate system without knowing the shape of the manifold. For example, one cannot define spherical coordinates without learning before the structure of a sphere and what isometries it have? Likewise, one cannot correctly map points on a cylinder, without knowing about cylindrical coordinates(?)Now, @Orodruin and others mentioned that a cylinder has zero intrinsic curvature. Would it mean that we can use Cartesian coordinates on it and also on any shape which has zero intrinsic curvature?

 

[stevendaryl]

Now, @Orodruin and others mentioned that a cylinder has zero intrinsic curvature. Would it mean that we can use Cartesian coordinates on it and also on any shape which has zero intrinsic curvature?

@Orodruin has already joked about the fact that this question (why does a cylinder have zero intrinsic curvature) has been answered 4 times: posts #11, #12, #14 and #15.

If a manifold is flat (has zero intrinsic curvature), then that means that you can use Cartesian coordinates for a section of it. You can't necessarily use Cartesian coordinates for the whole thing, because that might lead to multiple coordinates for the same points. For example, on a cylinder with coordinate x representing the distance around the cylinder, and y representing the distance up and down the cylinder, then the points (x,y) and (x+L, y) are the same point. That can't happen with Cartesian coordinates (where L is the circumference of the cylinder). If you stick to a region with 0 < x < L, the coordinates work just like Cartesian coordinates.

 

[davidge]

the points (x,y)(x,y)(x,y) and (x+L,y)(x+L,y)(x+L, y) are the same point

Since their coordinates have a difference of L, I don't see why these are the same point.

 

[Orodruin]

Since their coordinates have a difference of L, I don't see why these are the same point.

Because if you go around the cylinder once you end up at the same point.

 

[davidge]

Because if you go around the cylinder once you end up at the same point.

but one faces this same issue if one uses cylindrical coordinates

 

[Orodruin]

but one faces this same issue if one uses cylindrical coordinates

So what? Steven just told you that you cannot have global Cartesian coordinates on the cylinder. In fact, I would not call them Cartesian as that usually refers to coordinates in a Euclidean space and a cylinder is not Euclidean.

 

[davidge]

at? Steven just told you that you cannot have global Cartesian coordinates on the cylinder

yea, but others have said in previous posts that it's possible to correctly describe points on such surfaces if we use the correct coordinate system. For example, we could correctly map points on the surface of a sphere using spherical coordinates. Now it seems that this issue of going to the same point still persists even if we use the "correct" coordinate system.

 

[Orodruin]

yea, but others have said in previous posts that it's possible to correctly describe points on such surfaces if we use the correct coordinate system. For example, we could correctly map points on the surface of a sphere using spherical coordinates. Now it seems that this issue of going to the same point still persists even if we use the "correct" coordinate system.

I still don't see your point. The problem you have is that you use too many coordinates. A two dimensional manifold will be described with two coordinates - not three.

 

[davidge]

The problem you have is that you use too many coordinates

This is not the only problem. The thing (that motivated me to start the thread) is that it appears we don't need just to use the correct number of coordinates, but the correct type of coordinate system.

A two dimensional manifold will be described with two coordinates - not three

yes, but one has to account for what coordinate system is being using, not just the number of coordinates, is it not so?

 

[stevendaryl]

but one faces this same issue if one uses cylindrical coordinates

Right. That's why the cylinder cannot be described by Cartesian coordinates.

 

[Orodruin]

but the correct type of coordinate system

You inherently seem to think of a manifold in terms of its embedding in a higher dimensional space. You need to get rid of this prejudice and start considering them from their intrinsic properties only. From the manifold point of view, there are not different types of coordinate systems - a coordinate system is just a continuous bijective map from a part of the manifold to a subset of R^n. You can describe the manifold using whatever coordinates you like that satisfy this. But again - get rid of the embedding thinking.

 

[PeterDonis]

it appears we don't need just to use the correct number of coordinates, but the correct type of coordinate system.

This is not correct. You can use any coordinates you want. You just need to use the correct expression for the metric for the coordinates you are using.

Let's consider a very simple example: the 2-d Euclidean plane. You can use Cartesian coordinates $x, y$ on this plane; if you do, the metric is $ds^2 = dx^2 + dy^2$. Or you can use polar coordinates $r, \theta$; if you do that, the metric is $ds^2 = dr^2 + r^2 d\theta^2$. Both of these $ds^2$ expressions represent e, the same line element, just in different coordinates.

To illustrate with a concrete example, suppose we are considering the line element $ds^2$ between the origin and the point $0.1, 0.1$ in Cartesian coordinates. The squared length of this line element is $ds^2 = dx^2 + dy^2 = (0.1)^2 + (0.1)^2 = 0.02$. But if we are using polar coordinates, the same point is described by coordinates $0.1414, \pi/4$, and the squared length is $ds^2 = dr^2 + r^2 d\theta^2 = (0.1414)^2 + (0.1414)^2 * (0)^2 = 0.02$--the same squared length, because it's the same line element, just represented in different coordinates.

It would of course be quite wrong to take the Cartesian coordinate values and plug them into the metric in polar coordinates, or vice versa. You have to use the correct metric for the coordinates you pick; but as long as you do that, you can pick any coordinates you like.

 

[PeterDonis]

That's why the cylinder cannot be described by Cartesian coordinates.

There seems to be a terminology issue here. By "Cartesian coordinates" I think you mean coordinates in which the metric is $ds^2 = dx^2 + dy^2$ and the ranges of the coordinates are $- \infty < x < \infty$ and $- \infty < y < \infty$, with no periodicity in either coordinate (i.e., distinct values of each coordinate map to distinct points). But one can drop the second requirement without dropping the first; i.e., one can describe a cylinder using coordinates in which the metric is $ds^2 = dx^2 + dy^2$ but the $y$ coordinate, for example, is periodic, so $y = - L/2$ and $y = L/2$ describe the same point (if we fix the value of $x$).

 

[stevendaryl]

There seems to be a terminology issue here. By "Cartesian coordinates" I think you mean coordinates in which the metric is $ds^2 = dx^2 + dy^2$ and the ranges of the coordinates are $- \infty < x < \infty$ and $- \infty < y < \infty$, with no periodicity in either coordinate (i.e., distinct values of each coordinate map to distinct points). But one can drop the second requirement without dropping the first; i.e., one can describe a cylinder using coordinates in which the metric is $ds^2 = dx^2 + dy^2$ but the $y$ coordinate, for example, is periodic, so $y = - L/2$ and $y = L/2$ describe the same point (if we fix the value of $x$).

Yes, but some people require that a coordinate system must be a one-to-one map from an open set of R^N to an open set of the manifold. If (x,y) = (x,y+L), then it's not one-to-one. It's a picky point, but in the book that I read on differential geometry, there was that requirement.

 

[PeterDonis]

Since their coordinates have a difference of L, I don't see why these are the same point.

Because the $x$ coordinate in this chart is periodic; it acts like an angular coordinate (for example, the $\theta$ coordinate in cylindrical coordinates) in this respect.

one faces this same issue if one uses cylindrical coordinates

Yes, and that's fine. Coordinate charts in which a coordinate is periodic are perfectly fine.

we could correctly map points on the surface of a sphere using spherical coordinates.

Yes, you can. And you can also do so in "rectangular" coordinates (I'll use that term instead of "Cartesian" to avoid any confusion); but the metric will not be the same as the metric of a flat plane even if you call the coordinates $x$ and $y$. (For example, pervect mentioned Mercator coordinates on the 2-sphere; they are "rectangular" but the metric is not the same as that of a flat plane.)

Also, a 2-sphere has a property that a flat plane or a cylinder do not: a 2-sphere cannot be entirely covered by a single coordinate chart. Standard spherical coordinates do not cover the "poles", since there are coordinate singularities there. You can cover a 2-sphere with two charts (the simplest choices are stereographic projections around two antipodal points), but not with a single chart.

 

[PeterDonis]

If $(x,y)=(x,y+L)$, then it's not one-to-one.

Wouldn't that invalidate any chart with angular coordinates? For example, polar coordinates on a plane? I have never seen a chart like that labeled as invalid (but of course I have not delved very deeply into differential geometry texts written from a mathematician's rather than a physicist's point of view).

 

[stevendaryl]

Wouldn't that invalidate any chart with angular coordinates? For example, polar coordinates on a plane? I have never seen a chart like that labeled as invalid (but of course I have not delved very deeply into differential geometry texts written from a mathematician's rather than a physicist's point of view).

It's a picky point that nobody cares about, but I would say that polar coordinates aren't really valid for the entire plane. The most egregious problem is that the metric components are singular at the origin. The fact that (r, \theta) and (r, \theta + 2\pi) are identified is not very serious. Basically, using \theta modulo 2\pi IS using multiple charts: You have one chart where \theta goes from 0 to 2\pi, and another where \theta goes from -\pi to \pi, and the mapping is just the modulo mapping.

 

[PeterDonis]

I would say that polar coordinates aren't really valid for the entire plane. The most egregious problem is that the metric components are singular at the origin

I agree there is a coordinate singularity at the origin, but I'm not asking about that.

Basically, using $\theta$ modulo $2\pi$ IS using multiple charts

Hm, ok.

 

[davidge]

You inherently seem to think of a manifold in terms of its embedding in a higher dimensional space

a coordinate system is just a continuous bijective map from a part of the manifold to a subset of R^n. You can describe the manifold using whatever coordinates you like that satisfy this. But again - get rid of the embedding thinking.

Oh ok.

Because the xxx coordinate in this chart is periodic; it acts like an angular coordinate (for example, the θθ\theta coordinate in cylindrical coordinates) in this respect.

Yes, and that's fine. Coordinate charts in which a coordinate is periodic are perfectly fine.

Yes, you can. And you can also do so in "rectangular" coordinates (I'll use that term instead of "Cartesian" to avoid any confusion); but the metric will not be the same as the metric of a flat plane even if you call the coordinates xxx and yyy. (For example, pervect mentioned Mercator coordinates on the 2-sphere; they are "rectangular" but the metric is not the same as that of a flat plane.)

Thanks for the answers to these questions

You can use any coordinates you want. You just need to use the correct expression for the metric for the coordinates you are using.

I see

Let's consider a very simple example: the 2-d Euclidean plane

To illustrate with a concrete example, suppose we are considering the line element ds2ds2ds^2 between the origin and the point 0.1,0.10.1,0.10.1, 0.1 in Cartesian coordinates

Thanks Peter for introducing this example. If the two points were $(1 , 1)$ and $(2 , 1)$ instead of $(0 , 0)$ and $(0.1 , 0.1)$, what value would we put for $r$ in the $r^{2}(\Delta \theta)^2$ term? Or more generally, what value do we put for $r$ when the two points are very close together, but the $r^{2}(d\theta)^2$ term don't vanish?

 

[PeterDonis]

If the two points were $(1 , 1)$ and $(2 , 1)$ instead of $(0 , 0)$ and $(0.1 , 0.1)$, what value would we put for $r$ in the $r^{2}(\Delta \theta)^2$ term?

You would have to solve an integral, since (assuming the coordinates you give are Cartesian), both $r$ and $\theta$ are changing. The general form of the integral would be

$$
s = \int ds = \int \sqrt{ds^2} = \int \sqrt{dr^2 + r^2 d\theta^2}
$$

which is not solvable as it stands. To solve it, you would need either $\theta$ as a function of $r$, or $r$ as a function of $\theta$ (i.e., you need an equation of the curve along which you are integrating), so that you can reduce the integral to an integral over one variable.

 

[davidge]

Ah ok

Is it possible to calculate $s$ on the surface of a sphere using $\int_{A}^{B}dx^2 + dy^2$ with $dx$ and $dy$ as in the image below (where I've set $z = 0$ along the path)?

 

[Nugatory]

Ah ok

Is it possible to calculate $s$ on the surface of a sphere using $\int_{A}^{B}dx^2 + dy^2$ with $dx$ and $dy$ as in the image below (where I've set $z = 0$ along the path)?

Yes, although what you're calculating is the length of the curve in three-dimensional space using the metric for that three-dimensional space written in Cartesian coordinates ($dx^2+dy^2$ is just another way of writing $g_{xx}dx^2+g_{yy}dy^2+g_{zz}dz^2=g_{ij}x^ix^j$ when $dz$ is zero). That's a different calculation than the one you'd make for the surface of a two dimensional sphere that is not embedded in a three-dimensional euclidean space.

BTW, we're trying to get you to abandon the embedding because the four-dimensional spacetime of general relativity cannot be usefully embedded in a five-dimensional Euclidean space. Thus, you're going to have to learn the techniques of differential geometry without relying on embedding before you can move forward with GR.

 

[davidge]

Yes, although what you're calculating is the length of the curve in three-dimensional space using the metric for that three-dimensional space written in Cartesian coordinates

BTW, we're trying to get you to abandon the embedding because the four-dimensional spacetime of general relativity cannot be usefully embedded in a five-dimensional Euclidean space. Thus, you're going to have to learn the techniques of differential geometry without relying on embedding before you can move forward with GR.

Oh yea, now I understand that. I asked about the integral $\int_{A}^{B} dx^2 + dy^2$ just for curiousity. Yet, how could we evaluate it? I know we cannot just replace the $d's$ with $\Delta 's$ in the integral, because in this case we would have a straight line through the interior of the sphere, not the path through the surface of it.

Edit: oops, I know that we cannot just replace the $d's$ with $\Delta's$ in any integral (it is meaningless). What I meant was that the result would not be as just $\Delta x^2 + \Delta y^2$.

 

[Nugatory]

Yet, how could we evaluate it?

By far the easiest way to evaluate that integral is to transform into polar coordinates. Choose the origin and axes so that the curve lies in the equatorial plane so that $d\phi$ is zero, $dr$ is already zero, and then $ds^2=g_{ij}x^ix^j$ will simplify down to $ds=\sqrt{g_{\theta\theta}d\theta^2}=Rd\theta$ and the integral is trivial. (The integral also looks formally similar to the integral you'd do without relying on the embedding but using latitude and longitude coordinates on the surface of the sphere. That's a consequence of choosing those particular coordinate systems, not any fundamental truth about the geometry).

 

[davidge]

By far the easiest way to evaluate that integral is to transform into polar coordinates

Yea. The thing is that I would like to evaluate that integral without any dependence with another coordinate system. If I were to use polar coordinates on the integral, I would not even write down the metric in Cartesian coordinates.