There is no such thing as 'the frame that is at rest'.What would you say any of this implies about how spacetime is structured and how I should know which frame is at rest? — JulianMau
The ones at the edge will have the least elapsed time because those are accelerating the most. — noAxioms
This experiment has already been done, using Earth as the disk. — noAxioms
Yes, but look at it from the perspective of the edge. According to SR, a clock on the edge, in its own frame, is stationary and thus runs faster, not slower, than the moving clock at the center. But that frame is not maintained for even a moment, so SR doesn't really apply. The change of reference frame, and thus the change of the instant with which the center-clock is simultaneous, is what makes the center clock steadily gain time, not lose it. The effect is a moment-of-acceleration thing: acceleration component multiplied by the distance of the reference object in the direction of said acceleration component.Not quite. From the point of view of special relativity, as analysed with respect to an inertial frame, only the scalar speed along the tangent of the trajectory, and not the acceleration, is relevant to the ratio of time dilation of a moving clock. — Pierre-Normand
Yes, linear speed (in the frame of center) is tangential speed and is constant, and the spatial distance from the center is fixed, since they're all welded to this disk and have really no choice about it. The radius should remain constant for the duration of the experiment.Isn't linear speed synonymous to tangential speed in this case? I'm not convinced that the spatial distance between the clocks located away from the centre of the disk would change considering that we don't even know whether the disk itself is Born rigid. — TimeLine
Yes, but look at it from the perspective of the edge. According to SR, a clock on the edge, in its own frame, is stationary and thus runs faster, not slower, than the moving clock at the center. — noAxioms
Yes, but look at it from the perspective of the edge. According to SR, a clock on the edge, in its own frame, is stationary and thus runs faster, not slower, than the moving clock at the center. But that frame is not maintained for even a moment, so SR doesn't really apply. The change of reference frame, and thus the change of the instant with which the center-clock is simultaneous, is what makes the center clock steadily gain time, not lose it. The effect is a moment-of-acceleration thing: acceleration component multiplied by the distance of the reference object in the direction of said acceleration component. — noAxioms
Isn't linear speed synonymous to tangential speed in this case? I'm not convinced that the spatial distance between the clocks located away from the centre of the disk would change considering that we don't even know whether the disk itself is Born rigid. — TimeLine
There is no issue with rigidity if the disk is assumed to be rotating at constant angular velocity. — Pierre-Normand
Is that taking into consideration the paradigm between local and global spatial geometry? — TimeLine
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