Relativistic addition of velocity
Let there are two inertial frames of references S and S’. S is the stationary frame of reference and S’ is the moving frame of reference. At time t=t’=0 that is in the start, they are at the same position that is Observers O and O’ coincides. After that S’ frame starts moving with a uniform velocity v along x axis.
Suppose a particle P is place in frame S’ and it is moving.
The velocity component of particle P from observer O’ in frame S’ will be:
u’x = dx’/dt’ (1a)
u’y = dy’/dt’ (1b)
u’z = dz’/dt’ (1c)
The velocity component of particle P from observer O in frame S will be:
ux = dx/dt (2a)
uy = dy/dt (2b)
uz = dz/dt (2c)
From Lorentz transformation equations:
x’ = (x – vt)/(√1 – v2/c2) (3a)
y’ = y (3b)
z’ = z (3c)
t’ = (t – xv/c2)/(√1 – v2/c2) (3d)
Differentiate equations (3)
dx’ = (dx – vdt)/(√1 – v2/c2) (4a)
dy’ =d y (4b)
dz’ = dz (4c)
dt’ = (dt – dxv/c2)/(√1 – v2/c2) (4d)
Now substitute equations (4a) and (4d) in equation (1a), we get
u’x = (dx – vdt)/ (dt – dxv/c2)
Divide numerator and denominator of R.H.S with dt, we get
u’x = (dx/dt – v)/ (1 – dx/dt(v/c2))
Put equation (2a) in above equation,
u’x = (ux – v)/ (1 – ux(v/c2)) (5)
Similarly by putting equations (4b) and 4d) in equation (1b) and then dividing the numerator and denominator of R.H.S with dt, and then putting equation (2b), we get
u’y = uy(√1 – v2/c2)/ (1 – ux(v/c2)) (6)
Similarly by putting equations (4c) and 4d) in equation (1c) and then dividing the numerator and denominator of R.H.S with dt, and then putting equation (2c), we get
u’z = uz(√1 – v2/c2)/ (1 – ux(v/c2)) (7)
Equations (5, 6 and 7) represent the addition of velocity relations as observed by observers O’ from frame S’.
From the observer O in frame S, the relations (5, 6 and 7) will become:
ux = (u’x + v)/ (1 + u’x(v/c2)) (8)
uy = u’y(√1 – v2/c2)/ (1 + u’x(v/c2)) (9)
uz = u’z(√1 – v2/c2)/ (1 + u’x(v/c2)) (10)
Generally equation (8) is written as
u = (u’ + v)/ (1 + u’(v/c2)) (11)
Special case: If v <<< c, then v/c2 will get neglected and the equation 11 will become
u = u’ + v
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