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 2Δ 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 2Δ t ± later. What is Δ t ± ? solution: During the time interval 2Δ t ± , S moves (as viewed from S ± ) from T to R through a distance u (2Δ 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 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 t (20) 3 Thus L L = L L = t t (21) But, from our result of time dilation, t t = 1 1-( u/c ) 2 (22) therefore L = t 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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This note was uploaded on 09/20/2010 for the course AMATH 261 taught by Professor Rogermelko during the Spring '10 term at Waterloo.

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

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