Riemann tensor and derivatives of ##g_{\mu\nu}##

[Einj]

Hello everyone,
I'm studying Weinberg's 'Gravitation and Cosmology'. In particular, in the 'Curvature' chapter it says that the Riemann tensor cannot depend on $g_{\mu\nu}$ and its first derivatives only since:

... at any point we can find a coordinate system in which the first derivatives of the metric tensor vanish, so in this coordinate system the desired tensor must be equal to one of those that can be constructed out of the metric tensor alone, ..., and since this is an equality between tensors it must be true in all coordinate systems.

What I don't understand is how introducing the second derivatives should change this situation. The point is that (and I'm not sure about that...) we can always find a locally inertial frame. In this frame $g_{\mu\nu}=\eta_{\mu\nu}$, which is constant, and hence all its derivatives should vanish.
What am I doing wrong?

Thanks a lot

 

[ShayanJ]

The wrong part of your reasoning is that only if spacetime is flat, you can find a coordinate system in which $g_{\mu\nu}=\eta_{\mu\nu}$ globally. In a curved spacetime, its only that there is always a coordinate system that at a given point P, $g_{\mu \nu}(x_p)=\eta_{\mu \nu}$. The first derivatives at P can be made to vanish too but not the second derivatives.

 

[PeterDonis]

In this frame $g_{\mu\nu}=\eta_{\mu\nu}$, which is constant, and hence all its derivatives should vanish.

You left out a key qualifier: in the local inertial frame, $g_{\mu \nu} = \eta_{\mu \nu}$ at a single point (the origin of the frame). At other points, $g_{\mu \nu} \neq \eta_{\mu \nu}$.

 

[Einj]

It makes perfect sense! Thanks a lot!

 

[Matterwave]

If you ever get confused on this again, just think about the function $f(x)=x^2$. The first derivative of this function $f'(x)=2x$ vanishes at $x=0$ but the second derivative $f''(x)=2$ does not. The first derivative measures a slope, while the second derivative measures concavity - which is roughly analogous to curvature.