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Unformatted text preview: PHYS 230 Fall 2007 Assignment 6 1. French 3.6 The space and time coordinates of two events as measured in a frame S are as follows: Event 1: x 1 = x , t 1 = x /c , ( y 1 = z 1 = 0). Event 2: x 2 = 2 x , t 2 = x / 2 c , ( y 2 = z 2 = 0). (a) There exists a frame in which both events occur at the same time. Find the velocity of this frame with respect to S . Using the (inverse) Lorentz transformation t = γ ( t Ux/c 2 ) where U is unknown: t 1 = γ ( x /c Ux /c 2 ) = t 2 = γ ( x / 2 c 2 Ux /c 2 ) Carrying out the algebra, we obtain U = c/ 2 . (b) What is the value of t at which both events occur in the new frame? From part (a): t 1 = t 2 = x /c (1 + 1 / 2) / (1 1 / 4) 1 / 2 so that the time is √ 3 x /c . 2. McCall 5.6 Playing with time dilation. A clock at rest in frame S is observed by an observer at rest in frame S to run slow. Similarly a clock at rest in S is observed to run slow by an observer at rest in S . But, the (“proper times” of the) two clocks are identical. How can this be? Follow these steps: (a) Choose the origin of x, x and of t, t to be where and when the two clocks are at the same place. This is event 1, with coordinates (0,0) and (0,0). (b) Event 2 is the time at which the clock in S completes one period T . What are the coordinates of this event in S ? In S ? This event is at x = 0 , t = T , so, using the Lorentz transformation: x = γ (0 βcT ) = γβcT ct = γ ( cT 0) = γcT 1 (c) Event 3 is the time at which the clock in S completes one period T . What are the coordinates of this event in S ? In S ? This event is defined in S . It has x = 0 , t = T . Thus: x = γ (0 + βcT ) = + γβcT ct = γ ( cT 0) = γcT (d) Event 4 is an event simultaneous, in frame S , with event 3, but happening at the origin of coordinates. What are the coordinates of this event in S ? In S ? This event has x = 0 , t = γT . Therefore x = γ (0 βγcT ) = γ 2 βcT ct = γ ( γcT 0) = γ 2 cT (e) The observer in frame S sends a light beam signal towards the origin as soon as his clock completes one period (event 3). When does this light beam reach the observer in S , who is at the origin of his coordinate system? The light beam starts in S at the coordinates of event 3, namely x = γβcT and t = γT . It travels the distance to the origin in time γβT , and therefore arrives there at time γT (1 + β ) = T q (1 + β ) / (1 β ) ....
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This note was uploaded on 04/29/2008 for the course PHYS 230 taught by Professor Harris during the Fall '07 term at McGill.
 Fall '07
 Harris

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