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Can I use a coordinate system that has an intrinsic curvature to describe a flat space (such as Minkowski space)?

Say, for example, I wanted to provide coordinates for an infinite flat plane (${E}^{2}$) it seems apparent that I couldn't use spherical coordinates (i.e. longitude and latitude) globally to do this because they have an intrinsic constant positive curvature (curve back on themselves), whereas the flat plane has zero curvature.

So, in general relativity, when I have a flat metric, and do a coordinate transformation to another coordinate system, must that coordinate system also be "flat"? Or, is it true that all reasonable coordinate systems are "flat" in an infinitesimal neighbourhood of a given point and that consequently, as the tensor change of coordinate transformations involve only derivatives, the intrinsic curvature of the new coordinate system is irrelevant?

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    $\begingroup$ Hmmm... Any of them? Isn't it what general principle of relativity is about? $\endgroup$ – Prof. Legolasov Nov 12 '16 at 12:20
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This characteristic of flat is actually a statement about curvature; it is intuitively clear that the Minkowski (Euclidean) metric has no curvature, but the mathematical statement of flatness is that the Riemann tensor $R_{ijkl} = 0$ everywhere. From this, it should be clear that any coordinate transformation leaves the Riemann tensor invariant if the space is flat, which shows that you can never transform from a flat space to a curved space.

On the other hand, if we start with a curved space we can always choose coordinates such that at a given point $x$, $g_{\mu\nu} (x) = \eta_{\mu\nu} \Rightarrow \Gamma_{\nu\sigma}^{\mu} (x) = 0$. Hence, we can always choose coordinates such that the space looks flat locally. As you note, with such a procedure we should not expect it to work globally in general.

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  • $\begingroup$ So @Aaron , If I'm in Minkowski space and choose to transform to a uniformly rotating frame of reference this is: a) permitted and b) the underlying space still has the same curvature (i.e. 0). Therefore (and this is the thing I really wish to understand), I can state the Faraday EM two-form F in an inertial frame, move to a uniformly rotating frame apply the tensor coordinate transformations to F and obtain the correct form of F despite being in a rotating frame? $\endgroup$ – Geoffrey Bryan Nov 10 '16 at 4:01
  • $\begingroup$ @GeoffreyBryan Yes, that is correct. You can think of this even in the non-relativistic case. One can move from a inertial frame with a particle at rest to a rotating frame of reference by introducing centrifugal forces, which is essentially what the christoffel symbols do in the covariant derivative under a generic coordinate transformation. $\endgroup$ – Aaron Nov 10 '16 at 4:34
  • $\begingroup$ @AccidentalFourierTransform Thanks for the catch, I've changed it. $\endgroup$ – Aaron Jan 21 '18 at 22:41

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