[Physics] the value of the Riemann tensor at the center of Earth and why is it not 0

Question

I have read this:

What is the general relativity explanation for why objects at the center of the Earth are weightless?

And this:

Gravitational time dilation at the earth's center

And this:

http://en.wikipedia.org/wiki/Riemann_curvature_tensor

Where it says:

"
Converting to the [[tensor index notation]], the Riemann curvature tensor is given by:
$$R^\rho{}_{\sigma\mu\nu} = dx^\rho(R(\partial_{\mu},\partial_{\nu})\partial_{\sigma})$$
where
$$\partial_{\mu} = \partial/\partial x^{\mu}$$
are the coordinate vector fields. The above expression can be written using [[Christoffel symbols]]:
$$R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho{}_{\nu\sigma}
– \partial_\nu\Gamma^\rho{}_{\mu\sigma}
+ \Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma}
– \Gamma^\rho{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\sigma}$$ "

And this:

Time dilation without curvature at the center of mass, how is that possible?

Where John Rennie says in the comments:

"a stationary particle is weightless if all the Christoffel symbols ΓαttΓαtt are zero. However this does not mean the Riemann tensor has to be zero as well. It's possible for the Riemann tensor to be non-zero, i.e. there is curvature, and the Christoffel symbols to be zero. This is what happens at the centre of the Earth."

" the curvature is described by the Riemann tensor not by the Christoffel symbols (though they can be derived from each other). The Riemann tensor is not zero at the centre of the sphere."

What I do not understand is, he says

"the Riemann tensor can be non-zero, even if the Christoffel symbols are all zero."

But this seems to me to contradict the formula:

"$$R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho{}_{\nu\sigma}
– \partial_\nu\Gamma^\rho{}_{\mu\sigma}
+ \Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma}
– \Gamma^\rho{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\sigma}$$ "

Because if all the Christoffel symbols are zero, then the Riemann tensor seems to be zero here.

The curvature if I understand correctly is just calculated with the formula of the Riemann tensor as I understand.

Question:

1. Can somebody please help me understand the math formula of the curvature (which is just the Riemann tensor as I understand) at the center, so that I see it is not zero when r=0?

Answer 1

The notation:

$$ \partial_\mu\Gamma^\rho{}_{\nu\sigma} $$

means:

$$ \frac{\partial}{\partial x^\mu}\left(\Gamma^\rho{}_{\nu\sigma}\right) $$

so we are taking the derivative of $\Gamma^\rho{}_{\nu\sigma}$ with respect to the coordinates $t$, $r$, $\theta$ and $\phi$. The fact a function has the value zero at some point does not necessarily mean its derivative is zero at that point, and indeed this is the case at the centre of a sphere. That's why the Riemann tensor can be non-zero at $r=0$.