Why is gravity a fictitious force?

[Roberto Pavani]

Yes, "large enough" to "feel different gravity". But the second derivative is just enough to detect it.

 

[PeterDonis]

Yes, "large enough" to "feel different gravity". But the second derivative is just enough to detect it.

But, as has already been pointed out, none of this can be done within a single locally flat patch. You have to look at a region large enough for the effects of spacetime curvature (tidal gravity) to be observable. And such a region, by definition, is too large to be covered by a single locally flat patch.

 

[Dale]

Yes, "large enough" to "feel different gravity". But the second derivative is just enough to detect it.

This is tidal gravity. It is curved spacetime, not flat. It is physical gravitation and cannot be removed by a change of reference frame, nor described as a fictitious force.

 

[Matterwave]

Yes, "large enough" to "feel different gravity". But the second derivative is just enough to detect it.

If the "second derivative is (just) enough to detect it" then it means you are measuring curvature. :)

The geodesic deviation equation (in abstract index notation): $u^b\nabla_b (u^a\nabla_a \xi^c) = R^c_{\: abd}u^a u^b \xi^d$ explicitly includes the Riemann tensor in it. It shows, this is a coordinate independent "real" effect.