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All models of Euclidean geometry are isomorphic as models to $R^2$ for some real closed field $R$ (see this previous question, or here), or, if we additionally assume that the model has to be a complete metric space, then isomorphic as models to $\mathbb{R}^2$ (by this result).

For the purposes of this question only, I will (unconventionally) define "Euclidean geometry" to be: $$Euclidean\ geometry = (model\ of\,Tarski'\!s\,axioms)+(complete\,metric\,space) $$

Question: Is equivalence as models a stricter or looser notion than equivalence as metric spaces or Riemannian manifolds?

In other words, do there exist non-linear coordinate systems which are also models for Euclidean geometry, and how does Tarski's representation theorem relate to the answer of that question?

Note: technically polar coordinates are only models for Euclidean geometry minus a point (I think), so that would suggest one would have to look further.

Background: Wikipedia states the following about models for hyperbolic geometry:

All models essentially describe the same structure. The difference between them is that they represent different coordinate charts laid down on the same metric space [emphasis mine], namely the hyperbolic space. The characteristic feature of the hyperbolic space itself is that it has a constant negative Gaussian curvature, which is indifferent to the coordinate chart used. The geodesics are similarly invariant: that is, geodesics map to geodesics under coordinate transformation. Hyperbolic geometry generally is introduced in terms of the geodesics and their intersections on the hyperbolic space.

From this, it is clear that all models of hyperbolic geometry are equivalent (isomorphic) in the category of metric spaces. It also seems implied from the above that all models of hyperbolic geometry are equivalent (isomorphic) in the category of Riemannian manifolds.

Are the different models of hyperbolic geometry also isomorphic as models in model theory?

I assume the answer is yes. My motivation is to compare this with the Euclidean case. In particular, if isomorphic models of hyperbolic geometry represent different coordinate charts on the same complete metric space, then this should imply that there could exist non-linear coordinate charts for Euclidean geometry which are still isomorphic as models to $\mathbb{R}^2$.

Chill2Macht
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    Can you state exactly what you mean by "Euclidean geometry"? And it seems that every time you say "elementarily equivalent", you actually just mean "isomorphic"... – Eric Wofsey Jun 29 '17 at 07:27
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    In any case, it is unclear what your question is supposed to mean...if you are not requiring completeness, then most models of Euclidean geometry are not metric spaces or Riemannian manifolds at all... – Eric Wofsey Jun 29 '17 at 07:31
  • @EricWofsey doesn't elementarily equivalent mean "isomorphic as models"? By Euclidean geometry, I either mean "a model of Tarski's axioms" or "a model of Tarski's axioms + complete metric space", depending upon the context. If it's still unclear, then the default assumption should be "a model of Tarski's axioms + complete metric space". – Chill2Macht Jun 29 '17 at 07:34
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    @Chill2Macht: "Elementarily equivalent" means "satisfies the same logical formulas" (usually we talk about first-order logic). As an example, every formula in the language of ordered sets (i.e. using just logic and $\leq$; things like $\forall x,y: x < y \implies \exists z: x<z<y$) is satisfied by the rationals if and only if it is satisfied by the reals –  Jun 29 '17 at 08:19
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    @Hurkyl You're right -- I misinterpreted this page https://plato.stanford.edu/entries/modeltheory-fo/ : it says only that "isomorphic as models" $\implies$ "elementarily equivalent"; it doesn't say anything about the converse, which I assume in general is false https://math.stackexchange.com/questions/1329960/simple-examples-of-non-isomorphic-but-elementarily-equivalent-structures. I will fix the post. Tarski's representation theorem only says isomorphic as models. – Chill2Macht Jun 29 '17 at 10:39
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    While it is the case that all 2D models of hyperbolic geometry are representing the same hyperbolic plane (an hyperbolic plane with curvature - 1) it is not the case that all hyperbolic planes have curvature -1 , but all hyperbolic planes are scalable to a plane with curvature -1 , hope this helps but if it only confuses ignore it – Willemien Jun 29 '17 at 12:46
  • @Willemien The important thing, is that (1) the curvature is constant and (2) the curvature is negative, right? – Chill2Macht Jun 29 '17 at 12:51
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    @chill2macht Yes very correct the curvature is constant (the plane is isotropic ) and the curvature is negative – Willemien Jun 29 '17 at 19:05

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