Page 306 -

′
u ( cos( )θ′ + v)
utcos( )θ = t′ (9.32)
1 − β 2
utsin( )θ = ′′ (9.33)
u t sin( )θ′

′
t = [1 + ( uv c/ 2 )cos(θ ′)] t′ (9.34)
1 − β 2

Dividing Eqs. (9.32) and (9.33) by Eq. (9.34), we obtain:

θ′
′
u ( cos( ) + v)
ucos( ) θ = (9.35)
[ +1 ( ′ / 2 θ′
uv c )cos( )]
′
u sin( )θ′ 1 − β 2
usin( )θ = (9.36)
uv c/
[ +1 ( ′ 2 )cos( )]θ′
From this we can deduce the magnitude and direction of the velocity of the
particle, as measured in the unprimed system:

′
u′ + v + 2 u v ′ cos(θ ′ −) ( u v / c )sin (θ ′)
2
2
2
2
2
2
u = (9.37)
2
2
[ 1+ ( uv c )cos(θ ′)] 2
′ /
′ u sin( )θ′ 1 − β 2
tan( )θ = (9.38)
′ u cos( )θ′ + v
Preparatory Exercises
Pb. 9.22 Find the velocity of a photon (the quantum of light) in the
unprimed system if its velocity in the primed system is u′ = c.
(Note the constancy of the velocity of light, if measured from either the
primed or the unprimed system. As previously mentioned, this constituted
one of only two postulates in Einstein’s formulation of the theory of special
relativity, which determined uniquely the form of the dynamical boost trans-
formation.)
Pb. 9.23 Show that if u′ is parallel to the x′-axis, then the velocity addition
formula takes the following simple form:

u′ + v
u = uv ′
1 + 2
c

© 2001 by CRC Press LLC