Question

(10 points) Consider a standard Alan and Beth story where Beth's train has proper length of 100 c-ns (light nanoseconds) and moves past Alan at v/c = 3/5. (a) How long does Alan deem the train to be? (b) As they often do, Alan's assistants set off firecrackers simultaneously at either end of the train as it passes by. Use the Lorentz transformation to find the time between the explosions according to Beth. (c) Find the same quantity (the time seperation per Beth) by considering the spacetime interval between the events.

Answer 1

Giveni Lo = 100 cons & Train's proper length i.e, in the Reference frame I of Beth = 100x3x10 m x 10 = 30m = 3c velocity of train with respect to Alan on the ground. @ We know , moving train would appear shorter to Alan. By Length contraction, we have La Lo where r = - - r - 25 : Alan, The length of train acc. to Lubom 4L0 = 80 C-ns 24 m L = 24 m

6 Now, consider firecrackers in the event Alan's events of frame I Train I O (x, t) = (0,0) CL,O) where we make convention (without loss of generality) that assistant's set-off crackers at t=0 de. simultaneously for them.

Let the (0,0) Space - time event Event 1 and that occur at (2,0) event 2, Transformations. Now We have Lorents x = x (x-ut) t=r ( tv) where & x&t xgt are are space-time space-time for for Alan's frame Beth's Shame Event 1 in 0,0) = (xit) x'=r (o-o) =0 ; t'arco-o) = 0. and Event. 2 (Lio) = (xit) as explain in @ x t =r (L-0,0)=rL=Lo; = r Co-LV) (2 02 • For Beth this two events occur at Coo) and CrL , NLV) time Separation - MLY = Ę.24.20 m.my's (m/s)

s Second. = 18 3x108 ot' = 68108 lot = 0.6 nos) © Now we find same but by conservation interval ot' without L-T. of space-time i.e. Dx?_c²at2 = Dne 12 – C² at 2 .. We have according to problem @ Dx=L DX'= Lo st=o ot' =2 . L2-(²,0 = Lo2_c? at 2 i L²= L ² - c² Dt 2 (24)2 = (30)2 - c²ot 2 (² ot2 = 900-576 c²ot 2 = 324 cot = 18 ot'=18 с Second. 107' = 0.6 ms]