class29 - Imagine a measuring rod at rest in the moving...

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example: Time dilation . Consider two observers S and S . Observer S is stationary and observes S is moving away with a constant velocity u . S sends a beam of light towards a mirror L away, and receives the beam back after a time interval 2 Δ t . To S the beam is sent and received at the same place, however for S the beam is sent out at T and received at R , a time interval 2 Δ t later. What is Δ t ? solution: During the time interval 2 Δ t , S moves (as viewed from S ) from T to R through a distance u (2 Δ t ) (13) And the distance of the light path is Δ d = L 2 + ( u Δ t ) 2 (14) Thus, the speed of light in S is c = L Δ t (15) and in S is c = L 2 + ( u Δ t ) 2 Δ t (16) but according to Einstein’s second postulate, these speeds have to be equal and constant: L Δ t = L 2 + ( u Δ t ) 2 Δ t (17) or, solving for the time in S , Δ t = Δ t 1 - ( u/c ) 2 (18) Thus, the observer moving in S measures a longer time interval for the same occurrence. An complementary phenomenon is length contraction
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Unformatted text preview: . Imagine a measuring rod at rest in the moving frame S ± , where its proper length is L . An observer, stationary in S , calculates the length of the rod using the time interval between the passing of the two end of the rod at his origin O in S . From this he measures L = u Δ t (19) in the stationary frame S . The observer in S ± , notices the passing of the two end of the rod at the origin to happen at L ± = L = u Δ t ± (20) 3 Thus L L ± = L L = Δ t Δ t ± (21) But, from our result of time dilation, Δ t ± Δ t = 1 ± 1-( u/c ) 2 (22) therefore L = Δ t Δ t ± L = ± 1-( u/c ) 2 L (23) therefore, the length of the rod is shorter when measured in a frame moving with respect to its stationary frame, with proper length L . 4...
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