Question

In an inertial frame two events occur simultaneously at a distance of 3 m apart. In a frame moving with respect to the laboratory frame, one event occurs later than the other by 10s. By what spatial distance are the two events separated in the moving frame? Solve this problem in two ways: first by finding the Lorentz boost that connects the two frames and second by making use of the invariance of the spacetime distance between the two events.

Answer 1

(i) Using Lorentz Transformation since x prime = y (x middot vt) y = -1/squareroot 1 - v^2/c^2 and t prime = y (t - xv/c^2) then t prime_A = y (t_A - x_A V/C^2) and t prime_B = y (t_B - x_B v/C^2) t prime_B - t prime_A = y [(t_B - t_A) + (x_A - x_B V/C^2)] Given: t_B - t_A = 0 {since events occur simultaneously in S} x_A - x_B = 3m t prime_B - t prime_A = 10^-8 sec. So, 10^-8 = y (3v/C^2) = (1 - v^2/C^2)^-1/2 (-3V/9 times 10^8) doublerightarrow (1 - V^2/C^2)^1/2 = v/3 times 10^8 doublerightarrow 1 - V^2/9 times 10^16 = V^2/9 times 10^6 doublerightarrow V = 3/squareroot 2 times 10^8 m/s then, y = (1 - v^2/C^2)^-1/2 = (1 - 1/2)^-1/2 = squareroot 2 Now, x prime_A = y (x_A - V t_A) = squareroot 2 (x_A - C/squareroot 2 t_A) and x prime_B = y (x_B - V t_A) = squareroot 2 (x_B - C/squareroot 2 t_B) So, x prime_B - x prime_A = squareroot 2 (x_B - x_A) + 0 = squareroot 2 (x_B - x_A) x prime_B - x prime_A = 3 squareroot 2 m