cianfa72 said:
According to that post
https://www.physicsforums.com/threads/simple-reasoning-that-the-equivalence-principle-suggests-curvature.1002557/#post-6487001 there is a physical procedure involving Alice, Bob and Carol to ascertain the curvature of the underlying spacetime (i.e. tidal gravity) by mean of timelike geodesics.
What about if at least one of the 3 timelike geodesics is replaced by a
spacelike geodesic, is there still a similar physical interpretation ?
The technique as stated in that post only applies to timelike geodesics. It's closely related to a technique that works for three space-like geodesics to detect and determined the curvature of what one might term the 3d subspace spanned by the three space-like geodesics. To be as rigorous as I can be, I suppose I need to note that what we do is take a tangent space at some point P on a 4d space-time manifold, then three space-like vectors at that point P generate a 3d subspace of the 4 dimensional manifold at that point.
By using the exponential map, we essentially conflate two slightly different 3 dimensonal spaces, one is a 3d space which exists in the tangent space, and one of which exists on the manifold. These two different spaces are connected via a mapping, the exponential map.
The 3d surface in the tangent space is flat, by definitiotion, but the 3d surface defined by mapping the tangent space into the manifold via the exponential map has the possibility of being curved.
The easiest thing to measure is to consider only two space-like vectors at some point P on the manifold rather than 3. Then it genrates a 2 dimensional surface, which has a curvature that can be represented by a single number, due to the constraints on the Riemann curvature tensor.
So we draw a plane surface in the tangent space, defined by two basis vectors in the tangent space. This generates a two dimensonal surface on the manifold, and we can talk about whether or not this 2 dimensonal surface is flat or curved. The experimental procedure here would be to draw a "triangle" in the manifold whose "sides" are actually geodesics. Then we can measure the angles between the geodesics where they meet, and if the space of the 2d submanifold is flat, the sum of the interior angles will be 180 degrees. If the space of the submanifold is not flat, that sum of the interior angles will be different than 180 degrees.
Dr. Greg's post that you mentioned essentially reformulates this idea to time-like geodesics rather than spatial geodesics.
A procedure I have suggested in the past for understanding what spatial curvature means in three spatial dimensons (note that this is different from what space-time curvature means in 4 dimensions) is that we consider three orthonormal basis vectors at some point P, and consider each of the three 2 dimensonal submanifolds genrated by pairs of basis vectors.
This technique can prove that a 3d space is not flat, as if any of the 2d subspaces is not flat, the 3d space is not flat. To actually prove it is flat requires it to be refined a bit. I think that this post would get to be too long if I went into the necessary refinements, but I hope this helps clarify things somewhat.
Essentially there are 6 numbers (degree3s of freedom) of the Rieman curvature tensor, by itself the above technique only makes three of them zero, and we need all six to be zero to have no curvature.
If you're interested in more ramblings about this or other aspects of inter preting the Riemann curvature tensor accor ding to my part iclar preference, just ask.