# Spherical coordinates metric

Mr-R
Dear all,

As I was reading my book. It said that the line element of a particular coordinate system (spherical) in $R^{3}$ is so and so. Then it said that the metric is flat. I don't get how the metric is flat in spherical coordinate. Could someone shed some light on this please?

Thanks

Mentor
I don't get how the metric is flat in spherical coordinate.

The metric is flat if the Riemann curvature tensor is zero. That's true regardless of what coordinates you use. Spherical coordinates can be used in a flat space, just as polar coordinates can be used on a flat plane.

Mr-R
The metric is flat if the Riemann curvature tensor is zero. That's true regardless of what coordinates you use. Spherical coordinates can be used in a flat space, just as polar coordinates can be used on a flat plane.

Hmmm I think I kind of get it. Polar coor. is a curved system which can be used on a flat Euclidean space. And spherical coor. is a curved system which can be used on a 3D Euclidean space. Right?

I am not sure what is confusing me. Let's say that I live on a sphere. So it has a curved surface. I want to describe some points on it. I choose to use spherical coordinates. So did I just use a flat metric to describe a curved surface?

Really sorry if I am mixing up concepts

Thanks

Gold Member
Polar coor. is a curved system which can be used on a flat Euclidean space. And spherical coor. is a curved system which can be used on a 3D Euclidean space. Right?

The coordinate systems are not curved. They use angles and a distance from the origin to specify points, but are just another way of doing the same thing as Cartesian coordinates. Choosing a coordinate system is often done to simplify calculations based on the problem at hand.

• 1 person
Mentor
Polar coor. is a curved system which can be used on a flat Euclidean space. And spherical coor. is a curved system which can be used on a 3D Euclidean space. Right?

For the appropriate definition of "curved", this is right, yes. But note that that definition doesn't generalize: it only works if the underlying space is Euclidean. If you are trying to use these coordinates on a non-Euclidean underlying space, they may not be curved in terms of that underlying space. See below.

Lets say that I live on a sphere. So it has a curved surface. I want to describe some points on it. I choose to use spherical coordinates. So did I just use a flat metric to describe a curved surface?

No. You used coordinates that, on that particular surface, happen to be "flat"--more precisely, partially flat (see below)--but the coordinates are not the same as the metric. The metric is still curved.

I said "partially flat" above because if we use, say, latitude and longitude as coordinates on the Earth's surface, lines of constant longitude are geodesics (the closest thing to "straight lines" on the sphere's surface, i.e., the equivalent of "flat", non-curved lines on a plane), lines of constant latitude are not (except for the equator). So we can't say that the coordinates are entirely flat or entirely curved. (Note, btw, that the same is true for polar coordinates on a plane or spherical coordinates in 3-D Euclidean space: some coordinate lines are straight and some are curved.)

Also, it's worth nothing that the term "spherical coordinates" on the surface of a 2-sphere is actually ambiguous: it can mean something like latitude and longitude, which is what I assumed you meant above, but it can also mean a system of coordinates something like polar coordinates, where we pick a particular point as the "origin" (say the North Pole), and use "radius" from that point (i.e., distance along a line of longitude from the North Pole) and "angle" (which line of longitude) as our coordinates. The coordinate lines will be the same as for latitude and longitude, but the actual coordinate values will be somewhat different. This kind of system, with one dimension added (so that we have a radius and two angular coordinates, like spherical coordinates), is commonly used in astronomy to describe the universe as a whole.

• 1 person
Mr-R
PeterDonis and TumblingDice,

Thank you both for your valuable replies. I think I should read more about coordinate systems and their relation with the metric and space.

Staff Emeritus
I'd add that while I agree that a flat piece of paper is indeed still flat even if you use polar coordinates, the Christoffel symbols are nonzero in polar coordinates. If you want a formal way to say that polar coordinates are "curvy", that way would be to say that they have nonzero Christoffel symbols in a coordinate basis.

The Christoffel symbols are one way (fairly advanced) of computing the derivative operators ##\nabla## or their 4-d equivalent (covariant derivatives). They can be thought of as describing how the basis vectors change as you move from point to point.

For the case of a flat piece of paper, introductory texts will use a way other than Christoffel symbols to compute the derivative operators. Christoffel symbols are (as far as I know) pretty essential for computing derivative operators on non-flat manifolds.

Sometimes in physics the idea of "curvature" gets overloaded and confused in popular presentations. For instance, an accelerating elevator is commonly held up as an instance of gravity, and gravity is described as the curvature of space-time. But a flat space-time remains flat even if you are in an accelerating elevator. This leads to the obvious question - where did the gravity come from? This is an example of how the simple popular explanations about gravity can be ambiguous and seemingly contradictory, resulting in confusion about the intended message is.

The only real thing I can say at the moment is that sometimes "gravity" is described by Christoffel symbols, and that other times it's described by the Riemann curvature tensor. The later (the Riemann) is more directly tied to tidal gravity than gravity, but the statement is coordinate independent. The former description (Christoffel symbols) is more directly tied to gravity in the usual coordinate systems, but because it's a coordinate dependent statement its applicability depends on what coordinates are used.

I'm not aware of any really clean way out of the confusion, one just has to interpret what "gravity" and "curvature' mean in context when reading popular literature:(. In formal papers, people will usually specify exactly what tensor they're talking about, but this may not help the lay reader.

• 1 person
Mr-R
I'd add that while I agree that a flat piece of paper is indeed still flat even if you use polar coordinates, the Christoffel symbols are nonzero in polar coordinates. If you want a formal way to say that polar coordinates are "curvy", that way would be to say that they have nonzero Christoffel symbols in a coordinate basis.

The Christoffel symbols are one way (fairly advanced) of computing the derivative operators ##\nabla## or their 4-d equivalent (covariant derivatives). They can be thought of as describing how the basis vectors change as you move from point to point.

For the case of a flat piece of paper, introductory texts will use a way other than Christoffel symbols to compute the derivative operators. Christoffel symbols are (as far as I know) pretty essential for computing derivative operators on non-flat manifolds.

Sometimes in physics the idea of "curvature" gets overloaded and confused in popular presentations. For instance, an accelerating elevator is commonly held up as an instance of gravity, and gravity is described as the curvature of space-time. But a flat space-time remains flat even if you are in an accelerating elevator. This leads to the obvious question - where did the gravity come from? This is an example of how the simple popular explanations about gravity can be ambiguous and seemingly contradictory, resulting in confusion about the intended message is.

The only real thing I can say at the moment is that sometimes "gravity" is described by Christoffel symbols, and that other times it's described by the Riemann curvature tensor. The later (the Riemann) is more directly tied to tidal gravity than gravity, but the statement is coordinate independent. The former description (Christoffel symbols) is more directly tied to gravity in the usual coordinate systems, but because it's a coordinate dependent statement its applicability depends on what coordinates are used.

I'm not aware of any really clean way out of the confusion, one just has to interpret what "gravity" and "curvature' mean in context when reading popular literature:(. In formal papers, people will usually specify exactly what tensor they're talking about, but this may not help the lay reader.

Thanks pervect. I am going through D'Inverno's book. Currently doing the Tensor calculus excersices. I still need to understand the physics or geometrical meaning of the symbols and tensors. D'Inverno explains the topic in a fairly abstract way. For me atleast.

Staff Emeritus
Dear all,

As I was reading my book. It said that the line element of a particular coordinate system (spherical) in $R^{3}$ is so and so. Then it said that the metric is flat. I don't get how the metric is flat in spherical coordinate. Could someone shed some light on this please?

Thanks

Roughly speaking, a coordinate system is a way to slice up space (or spacetime). In 3D rectangular coordinates, you slice up space into planes of constant x. Then each plane can be sliced into lines of constant y. Then each line can be sliced into points of constant z.

In the case of spherical coordinates, you slice space into spheres of constant r (or altitude, if you are describing things relative to the Earth). Then you slice each sphere into circles of constant $\theta$ (or latitude), and then slice each circle into points of constant $\phi$ (or longitude).

The slices of constant r are spheres, and they have a curved geometry. But the curvature of space itself doesn't depend on how you slice it up. You can slice flat space into curved pieces (and the other way around, I suppose, but that's harder).

• 1 person
Mr-R
Roughly speaking, a coordinate system is a way to slice up space (or spacetime). In 3D rectangular coordinates, you slice up space into planes of constant x. Then each plane can be sliced into lines of constant y. Then each line can be sliced into points of constant z.

In the case of spherical coordinates, you slice space into spheres of constant r (or altitude, if you are describing things relative to the Earth). Then you slice each sphere into circles of constant $\theta$ (or latitude), and then slice each circle into points of constant $\phi$ (or longitude).

The slices of constant r are spheres, and they have a curved geometry. But the curvature of space itself doesn't depend on how you slice it up. You can slice flat space into curved pieces (and the other way around, I suppose, but that's harder).

Nice simple explanation stevendaryl.
Thanks