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kent davidge

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But if I try to employ the same procedure for the north pole ##(\theta_0, \phi_0) = (0,0)## this doesn't work. Why?

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In summary, locally Euclidean metrics can be constructed on a small enough piece of a Riemannian manifold, but they will be different for different small pieces. There is no global coordinate chart that is locally Euclidean everywhere.

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kent davidge

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But if I try to employ the same procedure for the north pole ##(\theta_0, \phi_0) = (0,0)## this doesn't work. Why?

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Ibix

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kent davidge

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Ibix said:Because the coordinate system is singular there.

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.It should work in other coordinate systems.

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PeterDonis

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kent davidge said: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"

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.

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kent davidge

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

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Nugatory

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

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kent davidge

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

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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.kent davidge said:ξ1≈0\xi^1 \approx 0 and the metric is flat

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.

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PeterDonis

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kent davidge said:Close enough to ##p##, ##\xi^1 \approx 0## and the metric is flat.

No, it isn't, as @Orodruin pointed out. But it is

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kent davidge said: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 doesn't 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.ph...&oldid=907252658#The_round_metric_on_a_sphere

Locally Cartesian coordinates on the sphere are a coordinate system used to represent points on a spherical surface. They are similar to the traditional x, y, and z coordinates used in Cartesian coordinate systems, but are adapted to the curved surface of a sphere.

Locally Cartesian coordinates on the sphere are defined by choosing a central point on the sphere, known as the origin, and two perpendicular lines passing through the origin, known as the axes. The coordinates of a point on the sphere are then determined by the distance along each axis from the origin.

One advantage of using locally Cartesian coordinates on the sphere is that they allow for easier calculations and measurements on a curved surface. They also provide a more intuitive way of visualizing and understanding the geometry of the sphere.

Yes, locally Cartesian coordinates can be used for any point on the sphere. However, they are most commonly used for points that are close to the origin, as the accuracy of the coordinates decreases as the distance from the origin increases.

No, there are other coordinate systems that can be used to represent points on the sphere, such as spherical coordinates or latitude and longitude. The choice of coordinate system depends on the specific application and the desired level of accuracy.

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