Does spacetime curvature (for time dilation) cancel out at the point of center of mass (because curvature effects cancel out from all directions)?

Question

I have read this question and John Rennie's answer to it:

How does "curved space" explain gravitational attraction?

I am asking about the time dilation affects of it, I am not asking about the weight. Just the time dilation at the center, please reopen it or answer my question about the time dilation at the center.

Where he says:

"The general relativity equivalent to this is called the geodesic equation:

$$ {d^2 x^\mu \over d\tau^2} = - \Gamma^\mu_{\alpha\beta} u^\alpha u^\beta \tag{1} $$

This is a lot more complicated than Newton's equation, but the similarity should be obvious. On the left we have an acceleration, and on the right we have the GR equivalent of a force. The objects $\Gamma^\mu_{\alpha\beta}$ are the Christoffel symbols and these tell us how much spacetime is curved. The quantity $u$ is the four-velocity."

"and our equation for the radial acceleration becomes:

$$ {d^2 r \over d\tau^2} = - \frac{GM}{c^2r^2}\left(1 - \frac{2GM}{c^2r}\right) u^t u^t \tag{2} $$"

And this:

https://en.wikipedia.org/wiki/Gravitational_acceleration

where it says:

"General relativity

There is no gravitational acceleration, in that the proper acceleration and hence four-acceleration of objects in free fall are zero. Rather than undergoing an acceleration, objects in free fall travel along straight lines (geodesics) on the curved spacetime."

Now what I do not understand is, the gravitational acceleration here on earth is said to be constant G. But they mean that it is constant G on the surface right? Based on John Rennie's formulas it has to change if I move a little bit further from Earth or move closer to the center of the Earth (in a tunnel for example) ?

But if gravity is caused by spacetime-curvature, then the curvature itself has to be changing depending on the distance from the center of mass right (that's what the formula says too $$ {d^2 r \over d\tau^2} = - \frac{GM}{c^2r^2}\left(1 - \frac{2GM}{c^2r}\right) u^t u^t \tag{2} $$)?

Now things 'moving' on the geodesics in spacetime, so "objects in free fall travel along straight lines (geodesics) on the curved spacetime." There is a geodesic, specifically the one going through the center of mass.

Now here I came to a contradiction, since I understand that SR time dilation (and GR as well in a more curvature-depending way) says that clocks deeper inside the gravitational zone (closer to the center of mass) will seem to be ticking slower (compared to the clocks more outside the gravitational zone).

The contradiction is, that the center of mass point itself (which is on the geodesic going through it), has to have curvature zero, since there the gravitational curvature effects cancel out from all directions (Let's say mass-distribution is perfect spherical), but the formula for curvature does not give a value for r=0 ($$ \Gamma^r_{tt} = \frac{GM}{c^2r^2}\left(1 - \frac{2GM}{c^2r}\right) $$).

So if I would put a clock (let's say photon clock) into the point of center of mass (so i would put it onto a geodesic, and the clock would be in free fall in spacetime, so the clock would still 'move' in time with speed c, just in space dimensions it would be stationary), then it would seem to tick at the same rate as clocks in free space?

Question:

1. Does the gravitational acceleration (G) ($$ {d^2 r \over d\tau^2} = - \frac{GM}{c^2r^2}\left(1 - \frac{2GM}{c^2r}\right) u^t u^t \tag{2} $$) grow as we go closer to the center of mass ?

2. Does spacetime curvature ($\Gamma^\mu_{\alpha\beta}$) grow as we go closer to the center of mass?

3. If I would put a clock (let's say photon clock) into the point of center of mass, then it would seem to tick at the same rate as clocks in free space (because the center of mass point itself, has to have curvature zero, since there the gravitational curvature effects cancel out from all directions, so there curvature is 0)?

4. Is the reason for this the same as why time stops inside a black-hole (same reason, curvature but in a black-hole the curvature becomes infinity). But at the center of mass point of a black hole (r=0) clock should be running at a normal rate, because curvature effects cancel out from all directions? So curvature should be zero at the center of mass of a black hole?

Answer 1

A pure, completely formal answer will take more work than this, but the short resolution to this apparent contradiction is:

Gravitational acceleration depends on the gravitational force, which is encoded (within a reference frame) in the components of $\Gamma_{ab}{}^{c}$

Time dilation effects depend on the gravitational potential, which is encoded, within a reference frame, in the components of $g_{ab}$.

This isn't completely right, but at the level of this question, it's right enough. Note that this is consistent with the traditional idea, since the Christoffel symbols are derivatives of the metric components. To see this even more explicitly, I'll leave it as an excersise to go and calculate the components of the metric:

$$g_{ab}ds^{a}ds^{b} = -\left(1-\psi(r)\right)dt^{2} + \left(1 + \psi(r)\right)dr^{2} + r^{2} d\theta^{2} + r^{2}\sin^{2}\theta d\phi^{2}$$

Subject to the constraint $\psi \ll 1$, so which lets you ignore all terms of size $\psi^{2}$ and make assumptions like $\frac{\psi}{1 + \psi}\approx \psi\left(1-\psi\right)\approx \psi$.

What you'll find, when you substitute back into the geodesic equations, is that you'll recreate Newton's law: $\frac{d^{2}r}{d\tau^{2}} = - \frac{\partial \psi}{\partial r} = -\nabla \psi$

Finally, note that none of this directly talks about the curvature at the point, which depends on second derivatives of the potential.