PART 6 MODERN PHYSICS
CHAPTER 38
THE THEORY OF RELATIVITY
Section 382:
Matter, Motion, and the Ether
Problem
1.
Consider an airplane flying at 800 km/h airspeed between two points 1800 km apart. What is the roundtrip travel time
for the plane (a) if there is no wind? (b) if there is a wind blowing at 130 km/h perpendicular to a line joining the two
points? (c) if there is a wind blowing at 130 km/h along a line joining the two points? Ignore relativistic
effects. (Why
are you justified in doing so?)
Solution
Since the velocities are small compared to
c
, we can use the nonrelativistic Galilean transformation of velocities in
Equation 310,
u
u
v
=
+
′
, where
u
is the velocity relative to the ground (
S
),
′
u
is that relative to the air (
),
′
S
and
v
is that
of
′
S
relative to
S
(in this case, the wind velocity). We used a notation consistent with that in Equations 3811 and 12. (a) If
v
=
0 (no wind), then
u
u
=
′
(ground speed equals air speed), and the roundtrip travel time is
t
d u
a
=
=
÷
2
2 1800
=
(
)
km
(
/
)
.
800
4 5
km h
h.
=
(b) If
v
is perpendicular to
u
, then
′
=
+
u
u
2
2
2
v
,
or
u
=

=
800
130
789
2
2
km h
km h
/
/
, and the
roundtrip travel time is
t
d u
b
=
=
2
4 56
=
.
h. (c) If
v
is parallel or antiparallel to
u
on alternate legs of the roundtrip, then
u
u
=
′ ±
v
and the travel time is
t
d
u
d
u
c
=
′ +
+
′ 
=
+
+

=
v
v
1800
800
130
1800
800
130
4 62
h
h
h.
.
Note that
t
t
t
a
b
c
<
<
, as mentioned following Equation 382.
Problem
2.
What would be the difference in light travel times on the two legs of the MichelsonMorley experiment if the ether
existed and if Earth moved relative to it at (a) its orbital speed relative to the Sun (Appendix E)? (b)
10
2

c
?
(c) 0 5
.
?
c
(d) 0.99
c
? Assume each light path is exactly 11 m in length, and that the paths are oriented parallel and perpendicular
to the ether wind.
Solution
The difference between Equations 382 and 1 is
∆
t
t
t
cL
c
L
c
L
c
c
c
=

=



=



F
H
G
G
I
K
J
J
⊥

,
2
2
2
1
1
1
1
2
2
2
2
2
2
2
2
v
v
v =
v =
and 2
2 11
3
10
733
8
L c
=
=
=
×
=
(
) (
/
)
.
m
m s
ns for the given interferometer. If
v =
2
2
1
c
¿
, we can expand the denominators
(Appendix A) to obtain
∆
t
L c
c
c
L c
c
¼
(
)[
(
)]
(
)(
).
2
1
1
2
2
2
2
2
2
2
2
2
=
v =
v =
=
v =
+

+
=
(a) For the Earth’s orbital speed
(30 km/s),
v =
2
2
8
10
c
=

, so that
∆
t
=
×
=
×


73 3
10
3 67
10
1
2
8
16
.
.
ns(
)
s (a fraction of the period of visible light). (b) If
v =
2
2
4
12
10
367
10
c
t
=
=
×


,
.
s
∆
(a few thousand periods of visible light). (c) At relativistic speeds, we use the exact
expression for
∆
t
. If
v =
=
v =
=
2
2
1
4
2
2
1
1
2
3
c
c
=
=

=
,
,
g
and
∆
t
=

=
73 3
2
3
131
4
3
.
.
.
ns(
)
ns
=
(d) If
v=
c
=
0 99
.
,
g
=

=
1
1
0 98
7 09
=
.
.
,
and
∆
t
L c
=

=
(
) (
)
.
.
2
1
317
=
g g
m
s
Problem
3.
Figure 3830 shows a plot of James Bradley’s data on the aberration of light from the star
g
Draconis, taken in
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CHAPTER 38
1727–1728. (a) From the data, determine the magnitude of Earth’s orbital velocity. (b) The data very nearly fit a perfect
sine curve. What does this say about the shape of Earth’s orbit?
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 Spring '09
 Miller
 Physics, Theory Of Relativity, Special Relativity, Spacetime

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