17 The Twin Paradox
173
as soon as possible and can bring everything concerning the farm to a conclusion’.
(Here of course we assume that c T = c o ).
Brother A o takes the super-train, and his clock U
A
o reads ˜
t P = t P = x P /v. We
will call this event T with the coordinates in o according to
o : T
x T = 0, t T =
x P
v
.
(184)
In Fig. 17.3, we have illustrated events T and P together in order to distinguish them
from the already defined events P and R in Fig. 17.2.
Twin A is located at x P (compare to the upper plot of Fig. 17.2) at time x P /v in the
reference system o . The observer B o in o observes the following: Twin A rushes
away with the velocity v, brother A o rushes after him with the velocity c T . Thus,
A o approaches to his twin brother with the velocity (c T − v) and the time t c that B o
in o measures for the super-train from its starting point up to the meeting point is
t c =
x P
c T −v
. The settings of the personal clocks of both twins result in the following
calculation:
The hand setting of brother A o ’s clock U
A
o was at ˜
t P = x P /v when he stepped into
the train. He then moves with the velocity u ≈ c T , which results in the Lorentz factor
1 − u 2 /c 2
o ≈ 0 for the pace of his clock, because we assumed that u ≈ c T = c o .
The hand of brother A o ’s clock therefore does not move (or not enough to count) due
to this highest possible velocity u of the super-train. We define the happy meeting
reunion of the brothers after the journey in the super-train as event S. The hand setting
of brother A o ’s clock U
A
o therefore still has the hand setting of ˜
t S = ˜
t P = x P /v,
˜
t S =
x P
v
.
Hand setting of the clock U
A
o
at the point of reunion
(185)
Observer B o also discovers that: Whilst brother A o stepped on to the train, twin
A was at the location x P in o , and that his clock U
A had the hand setting of
t
P =
x P
v
1 − v 2 /c 2
o according to Eq. (179). Twin A moved at the velocity v during
A o ’s time of journey. The hand setting of U
A changed, during the above measured
journey time of brother A o in the super-train by B o in the reference system o , from
t c =
x P
c T −v
by the amount t
c =
x P
c T −v
1 − v 2 /c 2
o due to time dilatation (121). Thus,
the hand setting of A’s clock U
A is at t
S = t
P + t
c when the twins meet. If we insert
c T = c o , we find
t
S =
x P
v
1 −
v 2
c 2
o
+
x P
c o − v
1 −
v 2
c 2
o
=
1 −
v 2
c 2
o
x P
v
+
x P
c o − v
,
t
S = x P
1
v
+
1
c o − v
1 −
v 2
c 2
o
=
x P
v
c o
c o − v
1 −
v 2
c 2
o
=
x P
v
c 2
o − v 2
(c o − v) 2 ,
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