17 The Twin Paradox
177
t
S(u) =
x P
v
u
(u − v)
1 −
v 2
c 2
o
.
(191)
We will now just simply check that, according to (190) and (191) for an arbitrary v
and an arbitrary u, with c o > u > v, brother A o ’s clock U
A
o does in fact go behind
that of twin A’s clock U
A , thus
˜
t s(u)
t
s(u)
< 1 for arbitrary v < u < c o .
The proof lies in the inequality between the geometric and the arithmetic means for
the velocities u and v, i.e.
√
u v <
1
2
(u + v). Let us simplify our conclusion even
more and keep in mind that 0 < u − v,
0 < (u − v)
2
,
2uv < (u
2
+ v
2
) ,
2c
2
o uv < c
2
o (u
2
+ v
2
) ,
−u
2 c
2
o − v
2 c
2
o < −2c
2
o uv ,
c
4
o − u
2 c
2
o − v
2 c
2
o + u
2
v
2
< c
4
o − 2c
2
o uv + u
2
v
2
,
(c
2
o − v
2
)(c
2
o − u
2
) < (c
2
o − uv)
2
,
c 2
o − v 2
c 2
o − u 2 < c
2
− uv ,
2uv
1 −
v 2
c 2
o
1 −
u 2
c 2
o
< 2uv −
2u
2
v
2
c 2
o
,
−2uv <
2u
2
v
2
c 2
o
− 2uv
1 −
v 2
c 2
o
1 −
u 2
c 2
o
,
u
2
− 2uv + v
2
< u
2
−
u
2
v
2
c 2
o
+ v
2
−
u
2
v
2
c 2
o
− 2uv
1 −
v 2
c 2
o
1 −
u 2
c 2
o
,
u − v < u
1 −
v 2
c 2
o
− v
1 −
u 2
c 2
o
,
u − v + v
1 −
u 2
c 2
o
< u
1 −
v 2
c 2
o
,
u − v + v
1 −
u 2
c 2
o
u
1 −
v 2
c 2
o
< 1 ,
and thus, as claimed, the so-called twin inequality, cf. Günther and Müller [],
Chap. 8
˜
t S(u)
t
S(u)
=
1 +
v
u − v
1 −
u
2
c 2
o
u
u − v
1 −
v
2
c 2
o
< 1 .
(192)