Question

A cosmic ray proton streaks through the lab with velocity 0.82c at an angle of 52° with the +x direction (in the xy plane of the lab). Compute the magnitude and direction of the proton's velocity when viewed from frame S' moving with β = 0.74.

Answer 1

x-component of the speed:

    u_x = (0.82c) cos52⁰   ......... (1)

y-component of the speed:

    u_y = (0.82c) sin52⁰ ............ (2)

from given problem, we have

β = v/c = 0.74

   v = 0.74 c

factor γ = 1/√[1-(v²/c²)] = 1.4867

from Lorentz transformation,(relativistic velocity transformation),

              u_x' = (u_x-v)/[1-(vu_x/c²)] .......... (3)

              u_y' = (u_y)/(γ)[1-(vu_x/c²)]   .......... (4)

substitute the eq (1), in eq (3), we get

   u_x' = (((0.82c) cos52⁰)-(0.74c))/[1-((0.74c)((0.82c) cos52⁰ )/c²)]

        = -0.3754 c

substitute the eq (2), in eq (4), we get

   u_y' = (((0.82c) sin52⁰)/(1.4867)[1-((0.74c)((0.82c) sin52⁰ )/c²)]

        = 0.8329 c

.................................................

therefore velocity u = √( u_x')² + (u_y')²

                              = 0.9136 c

............................................................

angle θ = tan^-1( u_y'/u_x')

            = -65.7^o

            = 180-65.7

            = 114.3^o (along +ve x-direction)