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    $\begingroup$ This is not what a "rest frame" is. You can perfectly define a chart $(U, \varphi)$ for an accelerated particle such that the local coordinates of the particle are $\varphi(p) = (t, x = 0)$. $\endgroup$ Commented Mar 18 at 23:44
  • $\begingroup$ @MarkusKlyver as accelerated particle you mean a (proper) accelerating one, i.e. a particle traveling along a non geodesic timelike curve. $\endgroup$ Commented Mar 20 at 6:44
  • $\begingroup$ @CarloC The answer didn't specify the nature of motion, but even if the curve is a geodesic (and you want an inertial chart) you still have to say that all geodesics ought be lines, it isn't enough to preserve only the time axis. Anyway, you do not work with inertial charts in general relativity. You work with inertial frames, and frames are sections of the frame bundle. $\endgroup$ Commented Mar 20 at 9:54
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    $\begingroup$ @CarloC If you want an inertial frame you have to use something like Fermi-Walker transport to prevent the frame from "rotating". An inertial frame would also require $\gamma$ (in my definition) to be a timelike geodesic as opposed to any timelike curve $\gamma$. The definition I gave you gives you a definition of an observer and not an inertial and non-rotating observer. You are correct that I did not include this in my definition. $\endgroup$ Commented Mar 21 at 18:42
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    $\begingroup$ @MarkusKlyver Ah ok, you mean to get an inertial frame (intended as a section of the frame bundle, not necessarily a chart) such a frame field is required to be Fermi-Walker transported along the timelike curves of the congruence (i.e. zero Fermi derivative of the frame's vector fields along the congruence). $\endgroup$ Commented Mar 21 at 20:56