
#19
Mar2912, 02:35 PM

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The principle of relativity states that the mathematical form of the laws of physics are the same in all inertial frames, it does not state that astronauts who make mistakes in calculations should nevertheless get the right answer. The laws of physics say that his clock ran slow while he was travelling and so his clock is not so simply related to the size of the light sphere in the earth frame. 



#20
Mar2912, 04:51 PM

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#21
Mar2912, 07:11 PM

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#22
Mar2912, 07:14 PM

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#23
Mar2912, 07:20 PM

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#24
Mar2912, 08:25 PM

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#25
Mar2912, 08:52 PM

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If you're thinking that Dale's obviously wrong and your'e obviously right  I'd think again. If you're thinking that you're getting frustrated and need to take a break from talking about this thread for a while, that's a bit more reasonable. 



#26
Mar3012, 03:14 PM

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The answer is rather more simple. The time t' was accumulated in two other inertial frames and bears no relevance to the propagation of light as measured in the Earth frame. In that frame a time of t has elapsed and the radius is ct. All observers who are at rest in that frame will agree on the radius regardless of their past history. If you are thinking you can somehow glue together all of the inertial frames in which the astronaut has been at rest, and all the noninertial frames when he accelerated, to produce a "composite" single coordinate system, in which the astronaut is always at rest, well, yes, it is technically possible to do this (and, as others have pointed out, there is more than one way of doing so). But this frame is not an inertial frame and so the rules of inertial frames don't apply. In particular there is no requirement for the speed of light to be constant in such a frame. It can speed up, slow down, stop*, or even go backwards* in an arbitrary coordinate system. This is because in a noninertial coordinate system, the "time" coordinate need not tick at the same rate as a propertime clock at rest. (This may be described as "gravitational"  or "pseudogravitational"  time dilation.) ___________ *To be mathematically precise, if the time coordinate stops or goes backwards, then it's technically incorrect to describe it as a "coordinate". 



#27
Apr212, 11:23 PM

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As natural as FermiNormal coordinates (and Rindler coordinates for uniform acceleration) are, they have a key limitation  they don't generally cover all of spacetime. In the case of Rindler, they cover just one section of spacetime. The reason for this is that they would multiply map these regions, which is not allowed for a coordinate patch. This happens because the spacelike geodesics 4orthogonal to different points on the 'time axis' world line intersect. Wherever this occurs, you have a region that cannot be covered with these coordinates. In the case of FermiNormal coordinates for the twin situation with rapid or instant acceleration then coasting away, then rapid or instant turnaround, then rapid or instant deceleration to a stop, there are three patches of spacetime not covered by the coordinates (actually six for instant accelerations, so lets assume nearly instant instead). If we assume the traveler goes to the right and back, then the uncovered patches are: 1) To the left of the initial acceleration 2) To the right of the turnaround 3) To the left of the stopping acceleration As a result, for a signal emitted by the stay at home when the traveler starts, both the left going light path and the right going light path would pass out of the coordinate patch. Thus, any concept of computing overall coordinate speed of this signal is impossible in these coordinates. It is true that using these coordinates (but not any computation of the whole path). once the travel stops, the coordinate position of the right going ray will match the stay at home twin. The left going ray will remain outside of the coordinate patch for the traveler for a long time after they have stopped, but eventually will pass to a covered region, at which point the traveler will agree with stay at home on its distance. Radar coordinates are able to cover the whole spacetime, even for sudden turnarounds. In an earlier post, Dalespam explained you would see speed of c for the signals (part of the definition of radar coordinates  any light ray to or from the time axis world line has speed c; rays between other points, not generally); and you would determine smaller distance than stay at home. They would never come to agreement on coordinate features of these light rays, but would for later ones, after they are back together. The upshot is that Dr. Greg's final answer (also said in several earlier posts in this thread) is the only way the traveler agrees with the stay at home: once back, they simply use stay at home measurements. 



#28
Apr312, 01:45 AM

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Colocation of points with different time coordinates, , temporal disordering etc.?? Am I right that this is essentially the same as Minkowski hypersurfaces of simultaneity wrt an accelerating frame??? [with the added complication of an internal time differential.] If this is correct would it also be true that planes of simultaneity in Minkowski space are equally false for those regions in the case of acceleration?? Do you think that the regions of invalidity only start at the point of intersection? Or is it more likely that the divergence from correspondence with reality simply diminishes from that point as it approaches the bounds of the physical system?? Thanks 


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