From Special Relativity to Feynman Diagrams.pdf

Let a and b be two twins which are initially both at

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Let A and B be two twins which are initially both at rest on earth. Suppose the twin B makes a journey on a high speed spaceship with constant velocity v and then comes back to earth meeting again the twin A . Let S be the frame of reference of earth and S 0 the one attached the spaceship. If D t is the time duration, relative to the earth’s system S , of the total journey of B , if we were to naively apply the special relativity formulas given in Chap. 1 , and since the events A , B occur in the same place relative to S 0 , the corresponding time D t 0 elapsed in the spaceship frame S 0 is related to D t by the time dilation relation D t ¼ D t 0 c ð v Þ : It follows that the twin B must be younger than the twin A when they meet again. This result appears to be paradoxical, since from the principle of relativity it follows that it is the same thing to consider B traveling with velocity v with respect to A or A traveling with velocity ± v with respect to B . Since time dilation depends on v 2 , considering B at rest and A traveling, it should be also possible to argue that A is younger than B . This puzzling result can be easily seen not to be correct if we recall that the special relativity effects can be applied only to frames of reference in relative uniform motion . If the two twins are to meet again to find out who is the younger, the spaceship system S 0 must invert its motion in order to come back to earth and therefore there is a part of its motion which is accelerated with respect to S . The situation is therefore not symmetrical since the S frame always remains inertial , while the frame S 0 is non-inertial during the inversion of its motion. There is thus no logical contradiction in saying that B is younger than A . Even if the analysis of the twin paradox can be made entirely within the framework of special relativity we shall give its solution by applying the principle of equivalence discussed in this Chapter and showing that in both reference systems S and S 0 the twin B is younger than the twin A . We shall perform the computation at the first order in v 2 c 2 and we shall denote by t 1 ; t 3 ; t 2 the time R. D’Auria and M. Trigiante, From Special Relativity to Feynman Diagrams , UNITEXT, DOI: 10.1007/978-88-470-1504-3, Ó Springer-Verlag Italia 2012 541
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durations of the forth and back journeys and the inversion of motion, respectively. In the frame of reference S 0 the corresponding times lapses will be denoted by t 0 1 ; t 0 2 ; t 0 3 . Let us first compute the total time duration of the journey from the point of view of the twin A , that is relative to the frame of reference S . The B twin in the frame S 0 , measures a total duration of the journey t 0 ¼ t 0 1 þ t 0 2 þ t 0 3 , while A measures t ¼ t 1 þ t 2 þ t 3 where: t 0 1 ¼ t 1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ± v 2 = c 2 p t 1 1 ± 1 2 v 2 = c 2 µ ; ð C : 1 Þ t 0 3 ¼ t 3 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ± v 2 = c 2 p t 1 1 ± 1 2 v 2 = c 2 µ ; ð C : 2 Þ t 0 2 0 t 2 0 ; ð C : 3 Þ where we have set t 0 2 ¼ t 2
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