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Imagine that somebody wants to generalize special relativity to non-inertial frames of reference. For example I am going around a point and the metrics of space is non-Euclidean from my point of view. Free bodies move along geodesics which have very complicated equations. But from the point of view of an inertial frame of reference the geodesics allow simple linear parameterization.

So what are the conditions for this metrics if there's a coordinate system where geodesics allow linear parameterization? Are they just Einstein's equations without mass-energy? Or not necessarily?

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For any affine connection, in geodesic normal coordinates at a point $p$, the geodesics passing through $p$ are linear. But if we want all geodesics passing through all points in some open set to become lines in some coordinates, this is precisely that the connection must be projectively flat, i.e. its induced projective connection has zero curvature. See:

Molzon, Robert(1-KY); Mortensen, Karen Pinney(1-KY) The Schwarzian derivative for maps between manifolds with complex projective connections.(English summary) Trans. Amer. Math. Soc. 348(1996), no.8, 3015–3036.

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  • $\begingroup$ But can I derive Einstein's equations for zero stress–energy tensor only using this condition? Or they will be more general? $\endgroup$
    – Марат Рамазанов
    Commented Jul 13 at 8:50
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    $\begingroup$ It is stronger than vanishing of the stress energy tensor. In a spacetime with no other energy sources, just gravity, the stress energy tensor is the traceless Ricci. But in fact the traceless Weyl projective curvature also vanishes, so the only curvature left is the scalar curvature, which is forced to be constant. The spacetime has constant curvature, and is a space form. $\endgroup$
    – Ben McKay
    Commented Jul 13 at 9:07
  • $\begingroup$ Yes, there are non-Euclidean metrics that allow straight geodesic world lines. For example Klein model. But does this have a phisical meaning? Can a frame of reference moving with zero acceleration have a non-Euclidean metrics, have a non-zero curvature tensor? $\endgroup$
    – Марат Рамазанов
    Commented Jul 14 at 11:59
  • $\begingroup$ @МаратРамазанов: I am not a physicist, so I am not the best person to ask about what has a physical meaning. There are physicists who take an interest in locally de Sitter and also in locally anti de Sitter spacetimes, and even in de Sitter and anti de Sitter globally. If you have a frame of reference moving with zero acceleration in all directions, i.e. forming a parallel framing, then clearly the curvature vanishes, so the metric is flat, i.e. locally isometric to Minkowski space. $\endgroup$
    – Ben McKay
    Commented Jul 14 at 12:36
  • $\begingroup$ Maybe empty space has a non-zero curvature by default. It makes sence, because there must be something in space, at least gravity. Otherwise it'd be space with nothing in it, but space is expected to contain something $\endgroup$
    – Марат Рамазанов
    Commented Jul 14 at 12:43

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