Recapitulate. Prof. Shiva Prasad, Department of Physics, IIT Bombay

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2 Recapitulate We discussed two important consequences of Lorentz transformation, Length Contraction and Time Dilation. We gave some examples relating to length contraction. 3

3 Example 1 Measurement of length of a platform by an observer stationary on the platform and another moving in a train with relativistic speed. Assume both frames to be inertial. 4

4 S v S A B 5

5 Define Events E1: The origin of S coinciding with end A of platform. E2: The origin of S coinciding with end B of platform. 6

6 Questions 1. Is the co-ordinate difference equal to length of the platform in any of the two frames S and S? 2. Is the time interval between the two events proper in any frame? 7

7 Discussion 1. The two events occur at two different ends of the platform. 2. However, time for two events is different both in S and S. 3. Hence the co-ordinate difference will be equal to length of the platform only in that frame in which platform is at rest, i.e. S. 8

8 Discussion 1. In S, the two events occur at two different locations. 2. What an observer in S would feel? 3. Time difference is proper in S. 9

9 In S In S, the length is proper and would be given as follows. L x x v t t Time interval is dilated t2 t1 t2 t1 10

10 In S Time interval is proper t2 t1 Difference in x-co-ordinate x2 x1 0 L Length of platform L v t t v

11 L v v t t 2 1 L One thus sees the contracted length in S, as expected. 12

12 Example 2 The incoming primary cosmic rays create -meson in the upper atmosphere. The life time of - mesons at rest is 2 s. If the mean speed of meson is 0.998c, what fraction of mesons created at the height of 20 km reach the sea-level? 13

13 Decay time interval 2x10-6 s is proper in -meson frame. In earth frame it is dilated. c=0.998c gives = Decay time in earth frame =15.82x2x10-6 s = 3.164x10-5 s 14

14 Time to travel to earth in earth frame =20x10 3 /0.998c = 6.68x10-5 s The fraction reaching earth in earth frame can be calculate using N N o e t 15

15 This fraction is Classically it would have been 3.12x What would an observer in meson frame conclude? 16

16 Example 3 Old Problem about light emission from the center of a train compartment. 17

17 c S c v S 18

18 Events E1: Light reaching front wall of the train. E2: Light Reaching the back wall of the train. 19

19 In S frame x x L L ; t 2 2c L L ; t 2 2c No transformation required to write these equations. 20

20 Event 1 in S frame x t 1 L vl L v 1 2 2c 2 c L vl L v 1 2c 2c 2c c 1 2 Inverse Lorentz transformation used. Assume origins were coincident when light was emitted. 21

21 Event 2 in S frame x t L vl L v 1 2 2c 2 c L vl L v 1 2c 2c 2c c 22

22 We see that t2 t1 Hence in S, event 2 occurred before event 1 as was discussed qualitatively earlier. The time difference is given below. t t t Lv c

23 Question Is the co-ordinate difference equal to length of the train compartment in S? 24

24 The difference in co-ordinates of two events. L x x1 x2 2 L 2 25

25 Length in S The x 1 -x 2 is an overestimate of length of the train because t 1 >t 2. Can we get correct length? Yes, if we can find out the distance the train moved in the time difference. 26

26 vδt X 2 X 1 L 27

27 Length in S Lv L x v t L v c 2 L 1 v c 2 2 L 28

28 Summary We discussed some examples of use of Lorentz Transformation. 29