No perfect circles in space...

[polymath257]

As I said, a geodesic is a "path"; it can, of course, be a circle.  It does not have to be, and I did not intend to imply such.

P.S.  An "arc"  can be a straight line, just with zero curvature.

Not quite. The definition of a geodesic on a surface is that it is a path that minimizes distances along the path. So, for a flat plane, a geodesic is a straight line. On a sphere, it is a great circle, not any other path or circle.

---

I'm not being intentionally difficult, I'm seriously trying to grasp this.

On a flat surface, the circumference of a circle with a diameter of 10 cm would be 31.4 (ish)cm.

Ok, now we have a flat surface with a dimple.  Let's say the dimple extends 1cm below (or above, doesn't really matter) the rest of the surface. We choose our centre point 5 cm from the dimple, giving us a radius of 5 and a diameter of 10.  But the line marking the circumference of the circle would, when passing through the dimple, go down 1 cm then up 1 cm, making the circumference of this circle 33.4 cm, so C = pi x d wouldn't apply, making this not a perfect circle. 

What am I missing?

Boru

A dimple is fine in 2D; but, how do we imagine such in 3D?

Through practice, mostly. Start by trying to imagine a sphere in 4 dimensions: it provides an example of a curved *three* dimensional manifold.

I might recommend the book 'The Shape of Space' as an introduction:

https://www.amazon.com/Shape-Space-Chapm...e+of+space

---

A dimple is fine in 2D; but, how do we imagine such in 3D?

I can't imagine a dimple on a flat surface in anything other than 3D - dimpling the a flat surface changes two dimensions to three, doesn't it?

Boru

No. It is still a 2 dimensional surface. Only two coordinates *on the surface* are required to determine a point.