Straight is only defined the way you think it is in an orthonormal basis.

Straight is the shortest distance between two points, regardless of the coordinate system. Straight is a curve on a sphere. What would be the shortest distance between NY and Moscow on your model?

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After coord transform we're working in (lat,long) based coordinates. So haversine gives the shortest distance for 2 points on earth.

Not all transformations are isometric, which just means a transformation that it doesn’t distort angles or distances.

Here is the mathematical proof. (starts around slide 20) I can’t help you much working through it, but I suggest that unless you can contradict the math, you really don’t have any basis to say that your model doesn’t have any distortions.

The problem in the mathematical proof is on slide 5:

We imagine trying to draw a map of a region D ⊂ S2. The map is a smooth function F : D → E 2

to the

**Euclidean** plane that preserves as much information about the geometry of D as is possible.

I am not working in a euclidean plane. After coords transform my basis is not orthonormal.

I've told you this before: you're drawing an orthonormal basis next to mine and then start measuring things according to your base. That's now how coord transforms, maths or physics work.

Also i found this definition for isometry:

https://en.wikipedia.org/wiki/Isometry.

It's basically any transformation that preserves distances. And my transform preserves distances. It is isometric.

What you're doing is taking a coord transformed base, putting an orthonormal basis next to it, and then start measuring. That's not the definition of isometric. You have to use the distance metric of your coord transfromed metric space.

Since the curvature of the sphere does not vanish, it CANNOT BE LOCALLY ISOMETRICALLY MAPPED TO THE EUCLIDEAN PLANE

except my "disc" is not an euclidean plane. It does not have an orthnormal base.

- You've had me calculate curvature of my space using gauss and it came out indistinguishable from a sphere.

- You then had me do parallel transports, which were indistinguishable from a sphere.

- I've given you the calculation for riemann tensor and ricci scalar, and it was comparable with a sphere.

You never found fault in any of the math, and yet you now call it a euclidean plane which violates all of the above.

But here's a very simple visual example: Take a compass and draw a circle anywhere on the AE map.

Circles represent all points at the same distance from the center. Does this circle represent all points at equal distance?

Obviously not, you said yourself there's distortion.

So my "disc" is demonstrably not an euclidean geometry.

At what point have i tried to "flatten" the sphere?

-> By trying to project a sphere onto a flat surface without any distortion.

Not treating it as a euclidean geometry.

If it has intrinsic curvature, it doesn't matter how you view it. It will always have intrinsic curvature and you can always detect it with some simple tests.

Agreed. Mathematically it behaves like a euclidean globe. But i can represent it whatever way i like.

I can represent it in an orthonormal base and say i live on a sphere with straight light.

Or i can represent it in my way, and look at reality as a flat earth with curvy light.

I'm able to construct a consistent world view either way and i'm able to make the same predictions of reality in either representation.

However when i switch to FE representation, i should be aware not to use any properties exclusive to euclidean geometries.