100 Theory of Relativity
the frame 0\, then
A + y\ + *\ = cH\ (2.4.4)By Postulate (ii), and because of relations (2.4.1), (2.4.2) above, equation
(2.4.3) becomes
(axi + (3hf + y\ + z\ = c^2 (7Zi + ah)^2 , or
{a^2 - cV)x^2 + yj + z\ + 2(a0 - cV)zi*i = cV - p^2 /c^2 )t^2.Since the last equality must agree with (2.4.4) for all (xi, j/i, z\, t) values,
it follows that
a^2 - c^272 = 1, (2.4.5)
a/3 - c^2 7<7 = 0, (2.4.6)
a^2 - p^2 /c^2 = 1. (2.4.7)These three equation together with 0/a — v can be solved easily for
a,P, 7,<T. Prom (2.4.6), (5/a — c^2 ^/a = v. So 7 = va/c^2. Substitut-
ing in (2.4.5), we get a = (1 -v^2 /c^2 )~* and so 7 = (v/c^2 )(l — v^2 /c^2 )~2. In
(2.4.7), substitute /? = va and solve for a to get a = a. Therefore, (2.4.1)
and (2.4.2) become y — yi, z = z\ and
v
x =. t =—==£=. (2.4.8)In what follows, we will often use the notations
a=[l--} , 7=^«, (2A9)and write (2.4.8) as
x = ax\ + vati, t = ati+jxi. (2.4.10)Then the inverse transformation becomes
xi — ax - vat, t\ = at - jx. (2.4.11)One can verify this directly or use the physical argument that the inverse
transformation must be obtained from (2.4.10) by switching the roles of
{x,t), {x\,t) and replacing v with —v.