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

1. Dec 23, 2014

### 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:

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

2. Dec 23, 2014

### 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.

3. Dec 23, 2014

### Staff: Mentor

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, $g_{\mu \nu} \neq \eta_{\mu \nu}$.

4. Dec 23, 2014

### Einj

It makes perfect sense! Thanks a lot!

5. Dec 23, 2014

### 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.