17 The Twin Paradox
175
t
S
˜
t S
=
c o + v
c o − v
.
(187)
Obviously, it is
c+v
c−v
> 1. Observer B o therefore registers in his reference system
o , t
S > ˜
t S . The hand setting of brother A o ’s clock, the brother who rushed after the
other, goes behind.
Twin A in his reference system
has in the meanwhile also calculated this.
Seen from
, A o enters the super-train at time t
T . Up to this point of time, A o ’s
clock U
A
o moved away with the velocity −v and thus has the hand setting of ˜
t T =
t T = t
T
1 − v 2 /c 2
o when entering the super-train (event T ) due to time dilatation.
Therefore, because of (184), event T in the reference system
takes place at t
T =
x P
v
√
1−v 2 /c 2
o
. Because of the speed at which the train travels u ≈ c o , ˜
t T is also the hand
setting ˜
t S of A o ’s clock U
A
o at the point of arrival. Therefore, as we already know,
˜
t S = t
T
1 −
v 2
c 2
o
=
x P
v
.
Hand setting of the clock U
A
o
at the point of reunion
(185a)
Since departure, brother A o has been moving with the velocity −v and thus finds
himself, with respect to his twin A, during event T at x
T = −v t
T . We thus have the
following coordinates in
for the event T ,
: T
x
T =
−x P
1 − v 2 /c 2
o
, t
T =
x P
v
1 − v 2 /c 2
o
.
(188)
Since brother A o is actually on the super-train, he meets his twin brother A at x
= 0
after a further travelling time of t
d = −x
T /c o =
x P
c o
√
1−v 2 /c 2
o
, so that twin A’s clock
U
A should show the hand setting of t
S = t
T + t
d when they both meet (event S),
thus
t
S =
x P
v
1 − v 2 /c 2
o
+
x P
c o
1 − v 2 /c 2
o
=
x P
1 − v 2 /c 2
o
1
v
+
1
c o
=
x P
v c o
c o + v
1 − v 2 /c 2
o
and once again,
t
S =
x P
v
c o + v
c o − v
.
Hand setting of the clock U
A
at the point of reunion
(186)
Twin A can therefore fully confirm the results made by observer B o . This consequently leads to the unbelievable fact that:
Twin A is older than his arriving twin brother A o .
175
t
S
˜
t S
=
c o + v
c o − v
.
(187)
Obviously, it is
c+v
c−v
> 1. Observer B o therefore registers in his reference system
o , t
S > ˜
t S . The hand setting of brother A o ’s clock, the brother who rushed after the
other, goes behind.
Twin A in his reference system
has in the meanwhile also calculated this.
Seen from
, A o enters the super-train at time t
T . Up to this point of time, A o ’s
clock U
A
o moved away with the velocity −v and thus has the hand setting of ˜
t T =
t T = t
T
1 − v 2 /c 2
o when entering the super-train (event T ) due to time dilatation.
Therefore, because of (184), event T in the reference system
takes place at t
T =
x P
v
√
1−v 2 /c 2
o
. Because of the speed at which the train travels u ≈ c o , ˜
t T is also the hand
setting ˜
t S of A o ’s clock U
A
o at the point of arrival. Therefore, as we already know,
˜
t S = t
T
1 −
v 2
c 2
o
=
x P
v
.
Hand setting of the clock U
A
o
at the point of reunion
(185a)
Since departure, brother A o has been moving with the velocity −v and thus finds
himself, with respect to his twin A, during event T at x
T = −v t
T . We thus have the
following coordinates in
for the event T ,
: T
x
T =
−x P
1 − v 2 /c 2
o
, t
T =
x P
v
1 − v 2 /c 2
o
.
(188)
Since brother A o is actually on the super-train, he meets his twin brother A at x
= 0
after a further travelling time of t
d = −x
T /c o =
x P
c o
√
1−v 2 /c 2
o
, so that twin A’s clock
U
A should show the hand setting of t
S = t
T + t
d when they both meet (event S),
thus
t
S =
x P
v
1 − v 2 /c 2
o
+
x P
c o
1 − v 2 /c 2
o
=
x P
1 − v 2 /c 2
o
1
v
+
1
c o
=
x P
v c o
c o + v
1 − v 2 /c 2
o
and once again,
t
S =
x P
v
c o + v
c o − v
.
Hand setting of the clock U
A
at the point of reunion
(186)
Twin A can therefore fully confirm the results made by observer B o . This consequently leads to the unbelievable fact that:
Twin A is older than his arriving twin brother A o .