In Weinberg's book, it is said that a given metric ##g_{\mu \nu}## could be describing a true gravitational field or can be just the metric ##\eta_{\alpha \beta}## of special relativity written in curvilinear coordinates. Then it is said that in the latter case, there will be a set of Minkowskian coordinates ##\xi^\alpha (x)## such that(adsbygoogle = window.adsbygoogle || []).push({});

$$\eta^{\alpha \beta} = g^{\mu \nu}\frac{\partial \xi^{\alpha}(x)}{\partial x^\mu}\frac{\partial \xi^{\beta}(x)}{\partial x^\nu} \ (1)$$everywhere, not just at every point ##X## and its infinitesimal neighborhood as the equivalence principle states.

Now what is the difference of everywhere to every point ##X##? Since if we pick up every point ##X## we will end up with all points that make the space we are considering?

Also, it is said in the book that for the metric which has coefficients

$$g_{rr} = 1; \ g_{\theta \theta} = r^2; \ g_{\varphi \varphi} = r^2 sin^2 \theta; \ g_{tt} = -1$$ it's possible to find a set of Minkowskian coordinates

$$\xi^1 = rsin \theta cos \varphi; \ \xi^2 = rsin \theta sin \varphi; \ \xi^3 = rcos \theta; \ \xi^4 = t$$ that satisfies (1) above.

I don't understand, since by the above reasoning a sphere would'nt be a curved space, for it's possible to find that set of Minkowskian coordinates above that satisfy (1) at every point on the sphere. But we know that the sphere is a intrinsically curved space. Could the Riemann curvature tensor vanish for the sphere when we use spherical coordinates to evaluate it?

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# I Gravitation vs Curvilinear Coordinates

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