From Special Relativity to Feynman Diagrams.pdf

# 0 since the time of turnaround of s from the point of

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0 since the time of turnaround of S 0 , from the point of view of the inertial frame S , can be neglected compared with t 1 and t 3 : Note that the times t 0 i are proper times since B is at rest in S 0 : Setting t 1 ¼ t 3 the total duration of the journey of B from the point of view of A is: t 0 ¼ 2 t 1 1 ± 1 2 v 2 c 2 µ : ð C : 4 Þ Thus, if we take v ¼ 9 ² 10 7 m = s and t 1 ¼ 20 years , and if the two twins were, say, 22 years old when B departed, as they re-meet after the trip, their age difference will be t 1 v 2 c 2 % 2 years : A will be 62 and B 60. It is instructive to derive ( C.4 ) from geometric considerations, see Fig. C.1 . Let us plot on a space–time diagram, relative to S , the trajectories (world-lines) of the two twins. Let the points O and R in the diagram be the events in which they depart and meet again, respectively. The twin A is at rest in S and thus its world- line is vertical, directed along the time direction. Suppose, for the sake of simplicity that the twin B moves forth and back along the x -axis, so as to describe, Fig. C.1 World-lines of the two twins in a space–time diagram 542 Appendix C: The Twin Paradox

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in the diagram, two segments: One, OP , with positive slope D x c D t ¼ v c [ 0 during the forward journey, and an other, PR , with slope ± v c \ 0, during the backward journey. The lengths of the two world-lines, divided by c , measures the proper- time intervals relative to A and B (i.e. the times measured by A and B , respectively) between the two events O and R . Since A is at rest in S , its proper time interval is t ¼ j OR j c ¼ t 1 þ t 2 þ t 3 2 t 1 : As for B , its proper time interval is t 0 ¼ t 0 1 þ t 0 2 þ t 0 3 2 t 0 1 ¼ 2 c j OP j : ð C : 5 Þ From the diagram one would naively conclude that t 0 [ t since the length of the trajectory of B appears to be greater than that of A . Recall, however, that we are in Minkowski space and that lengths are measured with the Lorentzian signature for the metric. As a consequence, in contrast to the Pythagorean theorem which holds in Euclidean geometry, the squared length of the hypotenuse of the right triangle OPP 0 is given by the difference of the squared lengths of the catheti, instead of the sum (in other words the hypotenuse is shorter than each of the catheti): j OP j ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ j OP 0 j 2 ± j PP 0 j 2 q ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ c 2 t 2 1 ± D x 2 q ¼ ct 1 ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 1 ± v 2 c 2 r ; ð C : 6 Þ where we have used D x ¼ vt 1 : Substituting the above result in ( C.5 ), and expanding the square root to the first order in v 2 = c 2 , we find ( C.4 ). Let us now compute the duration of the journey from the point of view of B himself (frame of reference S 0 ).
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• Fall '17
• Chris Odonovan

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