Intuition regarding Riemann curvature tensor

[Hill]

TL;DR Summary

"Why" there are terms quadratic in first derivatives of metric?

The Riemann curvature tensor contains second derivatives of metric and squares of the first derivatives. The second derivatives have to be there because they are the ones that cannot be eliminated locally by a choice of coordinates. But other than being a mathematical artifact, is there a meaning for the squares of the first derivatives to be there?

 

[Dale]

The first derivatives tell you directions. Curvature can be different in different directions. And the curvature in one direction can change as you move in a different direction

 

[PeterDonis]

The second derivatives have to be there because they are the ones that cannot be eliminated locally by a choice of coordinates.

If you look at how this actually works, you find that in Riemann normal coordinates, which are the ones that eliminate everything that can be eliminated by a choice of coordinates, the first derivatives are zero at the origin of coordinates, i.e., at the point the coordinates are "centered" on. So at that point, the first derivatives indeed do not appear at all in the Riemann tensor.

But as soon as you move away from the origin, the first derivatives are no longer zero (because the second derivatives weren't zero at the origin), and they will be different depending on which direction you move. So you do need them to capture the complete behavior once you are away from the origin.

 

[cianfa72]

But as soon as you move away from the origin, the first derivatives are no longer zero (because the second derivatives weren't zero at the origin)

lndeed, a zero second derivatives w.r.t. a coordinate basis at the origin, would mean constant first derivatives in a neighborhood of the origin, but first derivatives by definition vanish at the origin in Riemann normal coordinates centered on it.