Question

6) (5 marks) Imagine a stationary rod of length I in reference frame O with ends at XA= l. Show that the length contraction ľ= (v1 – 22 / c2 0 and XB can be derived by calculating the time At an observer in O' measures for the rod to pass her origin and then multiplying by its speed, u.

Answer 1

Sol First we need to find the relation between time in O and o frame. O frame. o- frame to each Let two mitrons parallel other at distanced". B بر سر مر ///. ma u 11///!IL M2 el re d A 7 IR М. m, ast Am unt 2 2 A ray bouncing between The arrangement of minor minior M. & M2 . Distance trand by ray = dtd = 20, in o' frame moving with velocity u with respect to observing frame. st time taken by ray Mitom Now, in AABD, V. Time take to travel, pt = 2d - 0 where ca speed of light. AB²= AD²+BD² [By Pythagoras Theorem]. *= d + (u st) Ja+(1,003 x = IT:

total distance travel by nay in a frame. = 22 Distance 21 d² + ust 2 (ust) We know, Distance - 2 l ďt Speedy time. (Late)* = c xot (cat ( 2 2 'uat =dt 2 2 nat 2 ust 12 =d² 2 2 ((²-1²) At - d² 2 st'? - (2d)? (² u² = (22) c(l U (² st'- (2d)? Nelius 20 C At's ſi ule From ey O. st = 2 eg" 2d с. ot's ot. Vi at's w u? 2

O frame o'frame xo= X 2=0 El BO x=0 he length of rod, ll'=uft ។ Now, Length of rod, l= ust time that actually passed motion seen by o time that t o think passed due to o' frame Dividing © by, - 2 ust Ust' st l From eqh 6, ot's at U² (2 l' at At l ſi-u (² Hence Proned * = 45 u² (2

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