Physics 8.962, Problem Set 1 Solution
1. Let’s call the twin that stays A and the twin on the spaceship B. We work in the (
x, t
)
coordinate system attached to the planet, i.e. A’s coordinate system. A says that the
trip took
Δ
t
≡
T
. Now we want to find the proper time along B’s trajectory: we have
the infinitesimal formula
dτ
2
=
dt
2

dx
2
. Along B’s trajectory
x
B
(
t
) we have
x
B
(
t
) =
(
vt
t <
T
2
vT
2

vt
t >
T
2
(1)
Thus
x
B
(
t
) =
±
v
, depending on the value of
t
; so we find
Δ
τ
B
=
Z
dt
s
1

dx
B
dt
2
=
T
√
1

v
2
(2)
Of course we could have done this immediately using the time dilation formula.
Now, let’s count how many signals
B
receives during the trip. To do this we need the
Doppler shift formula; recall that if a source emits signals at a frequency
f
0
in its rest
frame and an observer is moving
towards
it at velocity
v
, it observes a frequency
f
0
f
0
=
f
0
r
1 +
v
1

v
(3)
It is of course impossible to remember which way the
±
go, but I know that I have it
the right way around because this way it is blueshifted for positive
v
, i.e.
when the
observer is moving
towards
it.
On the way out,
B
is moving away and so sees the signal at a frequency
f
0
=
r
1

v
1 +
v
f
0
(4)
and hence receives
f
0
Δ
τ
B
2
=
T
2
r
1

v
1 +
v
f
0
√
1

v
2
(5)
On the way back, since
B
is moving towards
A
, the doppler shift formula flips and
f
0
Δ
τ
B
2
=
T
2
r
1 +
v
1

v
f
0
√
1

v
2
(6)
Adding those two together, the total number of counts
B
receives is
f
0
T
, which makes
sense; every signal that
A
emitted during his proper time
T
is captured by
B
.
1
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Now, let’s count how many signals
A
receives.
The doppler shifts in this case are
the same. However,
A
sees the receding Doppler shift not only for time
T/
2, but also
for the time it takes for light to reach back from the distant planet,
vT/
2. Thus, from
B
going on the way out,
A
receives
T
2
+
vT
2
r
1

v
1 +
v
f
0
(7)
and on the way back, he receives
T
2

vT
2
r
1 +
v
1

v
f
0
(8)
meaning that in total A receives
f
0
T
√
1

v
2
,
(9)
which again makes sense:
B
was broadcasting at frequency
f
0
for his entire proper time,
which we worked out above was
T
√
1

v
2
. Indeed the number of counts
B
receives
divided by the number of counts
A
received is
γ
, which makes sense as “number of
counts” is a time telling mechanism.
2. Schutz 1.17
For this problem
γ
= 1
/
√
1

.
8
2
= 5
/
3.
(a) The friend on the barn measures the pole to be
l
pole
γ
= (20
m
)(
3
5
) = 12
m
(10)
(b) After the barn door is closed, in the friend’s frame, the front of the pole has 3
m
to travel before hitting the front. Thus the time it takes is
t
=
3
m
.
8
= 3
.
75
m
(11)
The interval between shutting the door and hitting the front is
Δ
s
2
=

Δ
t
2
+ Δ
x
2
=

(3
.
75
m
)
2
+ (15
m
)
2
= 210
.
938
m
2
(12)
which is spacelike.
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 Spring '14
 Physics, Work, General Relativity, Special Relativity, Frame, Lorentz, barn door

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