(adsbygoogle = window.adsbygoogle || []).push({}); No, it isn't. You are failing to take into account that the geometry of spacetime, locally, is not Euclidean--it's Lorentzian. Lines of simultaneity are not perpendicular to the observer's worldline in the Euclidean sense, which is how you have drawn them. They are "perpendicular" (orthogonal) in the Lorentzian sense. sergiokapone said: ↑The lines of simultaneity was drawn according to Einstein's synchronization procedure.

Furthermore, since you have adopted coordinates in which the metric is not Minkowski, even the Lorentzian concept of orthogonality will not be represented in your diagram the way it would be represented in a standard spacetime diagram of the kind used in special relativity. In fact, without some specific coordinate chart and metric in mind, there is no way to even know how orthogonality would be represented in your diagram.

The way I would proceed would be to draw an SR-style spacetime diagram with ##\tilde{t}## as the vertical (time) axis and ##\ell## as the horizontal (space) axis--since we know that, locally, the metric takes the Minkowski form in terms of those quantities, i.e., ##d\tau^2 = d\tilde{t}^2 - d\ell^2## (in units where c = 1). In that diagram, lines of simultaneity in the sense you mean (i.e., lines of constant ##\tilde{t}##) are horizontal, so it's easy to draw them, and it's also easy to see how they relate to the round-trip light paths, since those are 45 degree lines in terms of ##\tilde{t}## and ##\ell## in units where c = 1, by definition (note that in your diagram, you can't even assume that--it's quite possible that in the general coordinates of your diagram, light rays travel on curves that are not 45 degree lines).

Once you have draw that diagram, you can then try to draw what the coordinate "grid lines" in some general chart ##x^0, x^i## would look like. But again, that will depend, in general, on the coordinates.

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# I Velocity measurement by a stationary observer in GR

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