On Einstein original paper "On the electrodynamics of moving bodies", on section 2 of the first part (Kinematics), the following thought experiment is described: a rod is imparted constant speed v in the direction of the xx axis (growing x). At both ends of the rod, A and B, there is a clock, both clocks are synchronous in the stationary system. Then, a ray of light is sent from A at time tA, which is reflected at B at time tB, and arrives at A at time t′A. It is then stated that "taking into consideration the principle of the constancy of the velocity of light"

zabuheljz 2022-08-12 Answered
On Einstein original paper "On the electrodynamics of moving bodies", on section 2 of the first part (Kinematics), the following thought experiment is described: a rod is imparted constant speed v in the direction of the x x axis (growing x). At both ends of the rod, A and B, there is a clock, both clocks are synchronous in the stationary system. Then, a ray of light is sent from A at time t A , which is reflected at B at time t B , and arrives at A at time t A . It is then stated that "taking into consideration the principle of the constancy of the velocity of light", we get:
t B t A = r A B c v and  t A t B = r A B c + v
where r A B is the length of the moving rod, measured in the stationary system. Now, if besides the rod's length, both time intervals refer to time in the stationary system, so must speed. Given the previous assumption of its constancy, why are terms like c + v and c v in the above equations?
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Answers (1)

Bradley Forbes
Answered 2022-08-13 Author has 12 answers
At time t A , the ray of light has to travel distance r A B to get to the other end. Since the other end is travelling at a velocity v, the ray of light has to travel an additional distance v ( t B t A ) giving a total distance r A B + v ( t B t A ). Since the ray of light travels at velocity c, this distance must also equal c ( t B t A )

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