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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# I Gravitation vs Curvilinear Coordinates

Have something to add?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**