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Curvature is an almost directly measurable quantity:

  • I measure the surface area $S$ of a body (massive ball, e.g. Earth, how much color I need to paint it).

  • I measure its radius $r$ (by digging a hole into its center and applying a ruler).

I divide the numbers $S/r^2$ and if the result differs from $4 \pi$ than the curvature is non-zero. How can this result (presumably existing in the real world) be explained in the teleparallel gravity (TG)? The assumed flatness is not to be taken seriously? Is there some theoretical "academic" concept of curvature, which is flat in TP and a "physical curvature" which is nonzero and is a derived quantity in TG? Is it then correct to claim flatness? Because when I hear "curvature" I understand the number measured in the experiment I proposed.

Edit (on 29/04/2025 13:45)

My logical steps are as follows:

  1. There is only one flat space: the Minkowski space. If TG is flat than it has Minkowski metric.
  2. TP has much richer structure than just metrics, but I can forget it. I need to measure distances (areas) and for that metrics is enough.
  3. It is easy to measure in Minkowsky space.
  • By symmetry the choice of a path directly falling from surface of a spherical body to its center is unambiguous. I measure its length using metrics, get radius.
  • I build a triangular mesh on the surface, measure triangle's lengths, compute surface.
  1. Divide numbers.

What do I get? Is it $4 \pi$? I presume yes, in a flat and stationary spacetime. If I get $4 \pi$ then it is probably different from what one gets in GR, where the inner volumes decreases (see Sun volume). Or does the TG indeed reproduce the GR result?

It seems to me that GR and TG are equivalent when dynamics (movement of bodies) is concerned, but I believe the curvature is objective and cannot see how a flat space can mimic it.

PS: Maybe there is some issue concerning clocks (I am not sure). On a surface I can synchronize clocks by symmetry reasons: they should show the same time when a light signal originating from the center of the body crosses the surface. Then I measure (in the triangular mesh) distance between various spacetime points which show the same time. There should by some global minimum. But maybe it is an issue when measuring radius: Which (space)time points to take? I do not know. (I found this : GR radius)

F. Jatpil
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1 Answers1

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You seem to be talking about extrinsic curvature in your example, not intrinsic curvature (i.e., Riemann curvature). The latter is what we should be talking about with reference to affine connections and teleparallel gravity (TG).

Intrinsic curvature is by definition a property of the chosen of connection. The Levi-Civita Riemann tensor (which exists independently) will still be non-zero in TG even if the curvature of the chosen affine connection is flat: e.g., bodies of mass described by given metric tensor still have non-zero Levi-Civita Christoffel symbols and lead to a non-zero Levi-Civita curvature. Despite this, the TG curvature is identically zero. (Explicitly, the Riemann tensor calculated from the TG connection is identically zero). However, what really matters is things we can detect and test with experiments, such as the paths of physical objects. In TG, particles still follow Levi-Civita geodesics (i.e., minimising $S = \int ds $ leads to the standard geodesic equation with the Levi-Civita connection). With this in mind, I cannot conceive of any measurable way to differentiate between a flat, teleparallel connection versus the Levi-Civita connection if the dynamical field equations are also the same. And in the teleparallel equivalents of General Relativity, the field equations are indeed the same.

So, to answer some your question, yes it is correct to claim flatness, and it should be taken seriously. But this doesn't affect the measured quantity in your example.

Eletie
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