Riemann Curvature Tensor – Riemann Curvature Tensor in an Inertial Frame

Question

My understanding is that the mathematical definition of an inertial frame at $x_0$ is a choice of coordinates s.t:

1. $g_{\mu\nu}(x_0) = \eta_{\mu\nu}(x_0)$

2. $\partial_\rho g_{\mu\nu}(x_0) = 0$

I've also been told that the Riemann curvature tensor evaluates to:

$$R^\beta_{\rho\nu\mu} = \frac{\eta^{\beta\alpha}}{2}(\partial_\nu\partial_\rho g_{\alpha\mu}(x_0) + \partial_\mu\partial_\alpha g_{\nu\rho}(x_0)-\partial_\nu\partial_\alpha g_{\mu\rho}(x_0)-\partial_\mu\partial_\rho g_{\alpha\nu}(x_0))$$

Why is it not the case that, by (2), all these derivatives of the gravitational tensor evaluate to zero so the Riemann curvature tensor is 0 overall?

Probably fundamentally not understood what one of these results actually means so any help would be amazing, thank you.

Answer 1

The fact that a function's first derivative vanishes at a point does not mean that its second derivative vanishes at that point. Note that for $f(x)=x^2$, $f'(0)=0$ but $f''(0)=2$.