- #106

LikenTs

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PeterDonis said:The blue line in your chart, the one titled "Universal Time Simultaneity", is not a 45 degree line, which is what ##t = t'## describes.

When the ship stops at turnaround point the world line is vertical. That means rate is the same for ship's clock and earth's clock, so a 45 degree line in this graph. It is exaggerated on purpose to reflect this fact, but instead of a stoppage of a few seconds it seems like a stoppage of several months.

Regarding the rest of the questions, could you read this document? I think it's from a physics professor at Princeton, and maybe he explains it better than I can.

http://kirkmcd.princeton.edu/examples/clock.pdfKirk T. McDonald said:Moments When vB = 0

Thus, twin B can compute the “age” of twin A (presuming that twin A has remained at rest) at those moments when he (twin B) is at rest with respect to twin A, whether or not the two twins are in the same place.

Twin B’s Trip Includes Frequent Stops

The discussion in sec. 2.2.1 above offers a kind of solution to the issue of distant simultaneity(“age” of a distant clock) for an accelerated observer such as twin B, provided the distant clock A is somehow known to remain in a single inertial frame at all times. Namely, the accelerated observer B brings himself to rest with respect to the distant clock in inertial frame A whenever he wants to know its “age”, which is then the value, eq. (7), of the time on the clock frame of A that is next to the accelerated observer B.In principle, such “stops” could be very brief for the accelerated observerB, such that the result of the computation (7) is little different from that made just before a “stop”. Hence,the accelerated observer Bcand simply assign the “age” of the distant clock (i.e., of twin A in the present example) as the value tB(τ B) he computes from his accelerometer data.ouldreasonably omit the (disruptive) “stops”,