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

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- Thread starter kent davidge
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- #1

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

- #2

Ibix

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- #3

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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.

- #4

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

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.

- #5

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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.

- #6

Nugatory

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

- #7

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

- #8

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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.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.

- #9

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No, it isn't, as @Orodruin pointed out. But it isClose enough to ##p##, ##\xi^1 \approx 0## and the metric is flat.

- #11

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

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.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.

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

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

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

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

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