Question

. A car begins to drive i the positive direction at0 s. The speed of the car is 0.8 c, and the initial spatial position of the vehicle s 1.0,0.0, 2.0) where spatia coordinates are given i meters. The car drives for 10 nanoseconds (1 nanosecond 10 s), and is halted by a sudden loss of tire pressure. Note that the car starting to move and the tire blow-out are both considered space-time events within Special Relativity. (a) Write down four-vectors for the event in which the car starts, and the event in which there is a tire failure (b) What is the space time inteva between the second and the frst event? Note that the interval will be a four-vector determined by calculating the difference of two four-vectors culate the relativistic invariant associated with the space time interval.

Answer 1

At t = 0 (x, y, z) & (x^1, y^1, z^1) are same (a) x = x^1 = 1.0 y = y^1 = 0.0 z = z^1 = 2.0 as car start moving prime coordinate (x^1, y^1, z^1) start moving along car then x = r(x^1 + vt) y = y^1 z = z^1 t = r(t + vx^1/cv) r = root1/1 - v/c^2 bv = 0.8c r = 1/root1 - (0.8)^2 = 5/3 event fire failure, t = 10 ns = 10^-8 sec x = 5/3 (1 + 0.8 times 3 times 10^8 times 10^-8) = 5 times 3.4/3 = 17/3 y = 0.0 z = 2. 0 t = 5/3 (10^-8 + 0.8 times 3 times 10^8 times 1/(3 times 10^5)^2) = 19/9 times 106-8 sec \r\n b) space time interval = roote^2(delta t)^2 - (alpha^2 + delaty^2 + delatz^2) delta t = time interval b/w two event = 19/9 times 10^-8 sec