# I Locally Cartesian coordinates on the sphere

#### kent davidge

I was trying to construct locally Euclidean metrics. Consider the sphere with the usual coordinate system induced from spherical coordinates in $\mathbb R^3$. Consider a point $p$ in the Equator having coordinates $(\theta_0, \phi_0) = (\pi/2, 0)$. If you make the coordinate change $\xi^1 = \theta - \pi/2$ and $\xi^2 = \phi$, you find that the metric will be $(d \xi^1)^2 + \sin^2 (\xi^1 + \pi/2) (d\xi^2)^2$ in these coordinates. Close enough to $p$, $\xi^1 \approx 0$ and the metric is flat.

But if I try to employ the same procedure for the north pole $(\theta_0, \phi_0) = (0,0)$ this doesnt work. Why?

Related Special and General Relativity News on Phys.org

#### Ibix

Because the coordinate system is singular there. Your longitude is not well defined. It should work in other coordinate systems.

#### kent davidge

Because the coordinate system is singular there.
It should work in other coordinate systems.
No, I believe its something else, because I have seen that its possible to get the flat metric around the north pole in this usual coordinate system, but using geodesics... It must be something wrong with "my" method.

#### PeterDonis

Mentor
I was trying to construct locally Euclidean metrics.
What do you mean by "locally Euclidean"? And what do you mean by "metric"?

Every Riemannian manifold with metric is locally Euclidean in the sense that a small enough piece of it can be approximated by a piece of Euclidean space of the same dimension. This is a coordinate-independent statement; "metric" here means the geometry of the manifold, independent of what coordinates you choose.

Every Riemannian manifold with metric can also have "locally Euclidean" coordinates chosen on a small enough piece of it. But those coordinates will be different for different small pieces. For example, you basically figured out how to construct a locally Euclidean chart on a small piece of a 2-sphere (which is a Riemannian manifold with metric) centered on the point $\theta = \pi /2$, $\phi = 0$ in the usual spherical coordinates (i.e., the point where the prime meridian intersects the equator). But the specific transformation to a locally Euclidean chart that you wrote down only works for that particular small piece of the sphere; it doesn't work anywhere else. (Nor do you have to go to the north pole of the sphere to see this; even if all you do is go around the equator to, say, $\phi = \pi$, you will have to change your transformation equation to be $\xi^2 = \phi - \pi$ to make the line element approximately like that of a Euclidean 2-plane on the small patch.)

In any curved Riemannian manifold, there will not exist any global coordinate chart that is "locally Euclidean" everywhere in the sense of having the line element in those coordinates be sufficiently close to the Euclidean one everywhere. So if that's what you are looking for, stop looking; it doesn't exist.

#### kent davidge

Thanks for the reply Peter. I was not clear enough on my previous posts. I did not try to go to the north pole using that same transformation equations, rather I tried to use the same procedure for the north pole: there seems to be no linear coordinate transformations that achieve the flat metric around the north pole if your original coordinate system is the usual spherical coordinate system.

On the other hand, the coordinate transformation $\xi^1 = \theta \cos \phi$ and $\xi^2 = \theta \sin \phi$ will do the job there.

#### Nugatory

Mentor
On the other hand, the coordinate transformation $\xi^1 = \theta \cos \phi$ and $\xi^2 = \theta \sin \phi$ will do the job there.
What is the value of the $\phi$ coordinate at the pole?

#### kent davidge

What is the value of the $\phi$ coordinate at the pole?
I would say it's zero, but I'm not sure.

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
ξ1≈0\xi^1 \approx 0 and the metric is flat
No it isn’t. Do not confuse locally orthonormal with flat. The sphere has non-zero Ricci scalar everywhere. This implies non-zero curvature. The sphere is not flat anywhere.

Also, do not confuse locally orthonormal with locally flat. Locally flat is a property of an embedding of a manifold into a higher-dimensional space.

#### Nugatory

Mentor
I would say it's zero, but I'm not sure.
is there a reason why zero is a better answer than $\pi/2$, or $\pi/4$, or for that matter $\pi/437.8$? @Ibix gave you a big hint in post #2 of this thread.

#### PeterDonis

Mentor
Close enough to $p$, $\xi^1 \approx 0$ and the metric is flat.
No, it isn't, as @Orodruin pointed out. But it is approximately flat in a small enough patch of spacetime around point $p$, and the approximation gets better the closer you get to point $p$.

#### pervect

Staff Emeritus
I was trying to construct locally Euclidean metrics.
"Locally Euclidean" metric isn't standard terminology.

Consider the sphere with the usual coordinate system induced from spherical coordinates in $\mathbb R^3$. Consider a point $p$ in the Equator having coordinates $(\theta_0, \phi_0) = (\pi/2, 0)$. If you make the coordinate change $\xi^1 = \theta - \pi/2$ and $\xi^2 = \phi$, you find that the metric will be $(d \xi^1)^2 + \sin^2 (\xi^1 + \pi/2) (d\xi^2)^2$ in these coordinates. Close enough to $p$, $\xi^1 \approx 0$ and the metric is flat.
Every line element is a quadratic form, and what you're apparently looking for is to make the line element locally $\delta_{ij} dx^i dx^j$ at a specific point. Basically, you are trying to find coordinates that diagonalize the quadratic form of the line element (near a point), and also make these diagonal elements unity.

This doesnt' make the original metric flat - because the Riemann curvature tensor is still nonzero, as Orodruin and others have pointed out.

But if I try to employ the same procedure for the north pole $(\theta_0, \phi_0) = (0,0)$ this doesnt work. Why?
The difficulty you are running into here is that the spherical coordinates are singular near $\theta=0$. $\delta$ here is the kronecker delta function, which is 1 if i=j and 0 otherwise. Look at the induced metric. I may be using different conventions than you are, but I believe the line element would be <<link>>

$$d\theta^2 + \sin^2 \theta \, d\phi^2$$

So, what happens near $\theta = 0$? The line element because approximately $d\theta^2 + 0 \, d\phi^2$

The reason this happens isn't due to anything odd of the geometry of the sphere at $\theta$=0, it has to do with the coordinate singularity there.https://en.wikipedia.org/w/index.php?title=Metric_tensor&oldid=907252658#The_round_metric_on_a_sphere

"Locally Cartesian coordinates on the sphere"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving