Environmental Engineering Reference
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and we have dropped the subscripts since the motion is all taking place in one
dimension. We want to compute the time elapsed in the Earth frame given that 10
years have elapsed on the spaceship. We can exploit the fact that the astronaut is
instantaneously at rest in S to give
1
d x
d t
2
2
d y
d t
2
d z
d t
1
c 2
d t =
+
+
d t
γ(u) d t ,
i.e. d t
=
(14.8)
where t is the time measured by an observer on the spaceship. Integrating this
equation will deliver the result provided we know how u changes with time, which
we do know from Eq. (14.7), i.e.
γ(u)u
=
gt
(14.9)
which, after squaring both sides and re-arranging, gives that
gt
u
=
1
g 2 t 2 /c 2 .
(14.10)
+
Substituting for γ(u) in Eq. (14.8) gives that
d t
t
1
=
(14.11)
g 2 t 2 /c 2
+
which leads to
sinh 1 gt
c
gt
c
=
,
sinh gt
c
.
gt
c =
i.e.
(14.12)
Putting t =
9 . 81 ms 2
14700 years.
Since the sign of g is unimportant in Eq. (14.11) it follows that the total time
that the spaceship is away from the Earth is 4
10 years and g
=
into this equation gives t
×
14700
59000 years (recall the
astronaut twin has only aged 40 years).
Before leaving the twins paradox we should point out that the fact that the
twin who travels (and hence undergoes an acceleration) always ages more slowly
regardless of the details of their journey. To see this we only need note that the time
elapsed according to the travelling twin (t ) is obtained by integrating Eq. (14.8):
t 2
u(t) 2
c 2
t =
d t
1
<t 2
t 1
(14.13)
t 1
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