I was reading the last part this answer by John Rennie, and I am bit confused about magnitude of each component of 4-acceleration when standing on the surface of the Earth compared to standing in an elevator accelerating upwards in outer space by $g$. I want to check if I understood correctly.
In both situations, (assume the origin of the inertia reference frame is on the surface with positive $z$-axis oriented radially outward) the z-component of the 4-acceleration is
$$A^3=A_\text{SR}^3+A_\text{GR}^3=g$$
right?
When you're at rest on the Earth, $A_\text{SR}^3=0$ and $A_\text{GR}^3=g$. On the other hand, if you're at rest in an accelerating elevator, $A_\text{SR}^3=g$ and $A_\text{GR}^3=0$?
What about other components $A^0, A^1, A^2$? Are they all zero?
On a related note, the comments to the answer of the recent question of mine gave me some food for thought. In particular, I learned that the one reason why the components 4-acceleration vanish during free fall on Earth is because
$$A^\lambda=\frac{dU^\lambda }{d\tau } + \Gamma^\lambda {}_{\mu \nu}U^\mu U^\nu=-g+g=0$$
I wonder if it is true for $\lambda=3$ (z-component) only? For other components $\lambda=0,1,2$: $$\frac{dU^\lambda }{d\tau }=\Gamma^\lambda {}_{\mu \nu}U^\mu U^\nu=0$$ right?
