PHY–309 L.
Solutions for homework set # 12.
Textbook problem Q.8 at the end of chapter 20:
In Newtonian mechanics, the velocities ~v1 and ~v2 of the same body relative to two different
frames of reference are related to each other as
~v1 = ~v2 + ~u
(1)
where ~u is the velocity of the second frame relative to the first frame. This formula works for
velocities of a boat relative to the water and the dry land, the velocities of a plane relative
to the air and to the ground (cf. the previous problem), the velocities of a spacecraft relative
to the Earth and the Sun, etc., etc.
Eq. (1) for the velocities follows from a simple coordinate transform
~x1 = ~x2 + t~u
(2)
relating position vectors ~x1 and ~x2 of the same body relative to two reference frames. Eq. (2)
— called the Galilean transform — presumes that the time t is the same in both frames of
reference. This assumption — which was though to be obvious before Einstein — is at the
core of Newtonian mechanics.
But the relativity theory has done away with the time universality: Instead, the time
flows at different rates in different frames, and even the chronological order (which event
happens first, which happens second, etc.) may different in different frames. When all
speeds are much smaller than the speed of light, the difference between the times in the
two frames can be neglected, which allows us to simply add velocities according to the
classical formula (1). But when any of the speeds becomes comparable to the light speed
c, the difference between the times t1 and t2 in the two frames becomes important, and the
classical formula (1) does not work any more.
In particular, a key principle of the relativity theory is that the speed of light in the
vacuum is exactly the same in all frames of reference. For example, if you produce a beam
1
of light on Earth, its speed relative to the Earth would be exactly c, and its speed relative
to the Sun would also be exactly c, despite the fact that the Earth moves around the Sun
at speed u = 30 km/s ≈ 10−4 c.
Textbook problem Q.10 at the end of chapter 20:
Before Einstein, physicists believed that the light waves behave similarly to the mechanical
waves such as sound. A sound wave in the air has a fixed speed relative to the air, but its
speed relative to the ground depends on the wind according to the vector formula
plane
plane
~v(ground)
+ ~uwind
= ~v(air)
=⇒
plane
plane
|.
| =
6 |~v(air)
in general, |~v(ground)
(3)
Likewise, it was believed that the light and other electromagnetic waves have a fixed speed c
relative to some mysterious substance called aether that sustained the electric and magnetic
fields. The aether was supposed to permeate the whole space, including the material bodies,
and nobody had a foggiest idea why the aether did not interfere with the mechanical motion
of the material bodies such as people or planets, but that was a separate problem. In any
case, if the Earth moved relative to the aether at some speed −~u, then it should have felt
the ‘aether wind’ at velocity +~u which should be measurable via speed of light relative to the
Earth:
light
light
~v(Earth)
+ ~uaether
= ~v(aether)
light
light
| ≡ c.
| =
6 |~v(aether)
in general, |~v(Earth)
=⇒
(4)
The aether wind ~u itself is a vector difference between the aether’s velocity relative to
the Sun and the Earth orbital velocity relative to the Sun,
~uaether = ~v⊕ (aether) − ~v⊕ (Earth),
(5)
and since the Earth’s velocity changes direction as the Earth goes around the Sun, the aether
wind should vary with the seasons. In particular, for at least half of each year, it should
have speed higher that the Earth’s orbital speed of 30 km/s or 10−4c.
2
Michelson and Morley set up to measure the aether wind with very high precision using
interferometry. Figure 20.9 on textbook page 440 shows the basic idea of their experiment:
A light beam is split in two beams, which bounce of (different) mirrors, come back, and
recombine into a single beam. If the two rays take different times to travel back and forth,
they would be out-of-phase with each other by ∆φ = 2πf × (t1 − t2 ), and this will affect the
way they interfere with each other.
The effect of the aether wind — assuming it exists — on each beam’s travel time depends
on its direction relative to the wind. For example, suppose beam #1 travels parallel to the
aether wind while beam #2 travels ⊥ to it. On its way to the mirror, beam #1 should feel a
‘tailwind’, so its speed relative to the lab should be c + u, so its travel time to the mirror is
L/(c + u) (where L is the distance to the mirror). On its way back, beam #1 should feel a
‘headwind’ reducing its speed (relative to the lab) to c − u and so increasing the travel time
to L/(c − u). Altogether, beam #1 travels back and forth for time
t1 =
L
L
2Lc
+
= 2
.
c+u
c−u
c − u2
(6)
Beam #2 should feel a ‘side wind’, so its heading (relative to the aether) should be slightly
up-wind from its direction to (or from) the mirror relative to the lab. Consequently, in the
√
lab frame, the beam’s velocity component toward the mirror is ± c2 − u2 instead of ±c.
Consequently, the travel time of the second beam should be
t2 = √
L
2L
L
+ √
= √
.
c2 − u 2
c2 − u 2
c2 − u 2
(7)
Note that the times t1 and t2 are different: the first beam takes longer time, which leads to
interference. For a relatively slow wind velocity u ≪ c, the difference is approximately
∆t =
L u 2
×
.
c
c
(8)
In particular, for u similar to the Earth’s orbital velocity,
∆t =
L
× 10−8 ,
c
3
(9)
which leads to the phase difference between the beams
∆φ
L
L
L
= f × ∆t ∼ 10−8 × × f = 10−8 ×
∼
.
8
2π
c
λ
10 λ ∼ 50 m
(10)
Such phase differences can be easily detected by quality interferometers, so Michelson and
Morley were quite confident they would detect the aether wind.
Instead, they got big fat nothing, no matter how hard they have tried. Other physicists
also tried to detect the aether wind, and they all got the same big fat nothing. The aether
wind just wasn’t there!
Several theoretical physicists tried explaining the absence of the aether wind by modifying
the Maxwell equations, but this did not work. Hendrick Lorentz and Joseph Larmor tried
a different approach: keeping the Maxwell equations unchanged and asking what happens
to the atoms and the macroscopic matter if all forces behave like the electric and magnetic
forces on charged particles. They showed that the aether wind slows down all processes and
also shrinks the distances in the direction of the wind, but such changes would not be noticed
by an observer suffering the same time slowdown and distance contraction. Moreover, if the
Michelson–Morley apparatus suffers distance contraction in the direction of the wind, this
would eliminate the time difference between the two beams, and the aether wind would not
be detectable!
At the same time, Albert Einstein — who didn’t even know about the Michelson–Morley
experiment — tried a more radical approach. He postulated that all physics laws — including
the Maxwell equations — must have exactly the same form in all inertial frames of reference,
and then asked how the classical physics needs to be modified to makes this work. His
answer was that the time t is not universal but depends on the reference frame, and from
that followed the rest of the special relativity theory.
Despite a different starting point, Einstein obtained exactly the same formulae as Lorentz
for transforming the space and time coordinates from one reference frame into another; today,
we call them Lorentz transforms. Poincaré and Minkowski showed that Lorenz transforms
are geometric symmetries in a pseudo-Euclidean 4–dimensional geometry combining space
and time, but the mathematics of this geometry is too complicated for this class.
4
Textbook problem SP.2 at the end of chapter 20:
(a)
−1/2
v2
γ = 1, − 2
.
c
(11)
For the pions moving at 90% of light speed, v = 0.9 c,
γ =
1 − 0.92
−1/2
≈ 2.3.
(12)
(b) Pions at rest have half-life time T0 = 17.7 ns (nanoseconds). That is, if you have a bunch
of non-moving pions, then after time T0 , half of them will decay.
For the moving pions, an observer observer that moves with them will see pions at rest
and hence decaying with the same lifetime T0 = 17.7 ns. That is, half of the pions will decay
between t = 0 and t = T0 , then half of the remaining pions decaying will decay between
t = T0 and t = 2T0 , etc., etc. But another observer sitting in the lab on the ground will
see a different life time because his time goes at a different rate than the time of the moving
observer,
tlab = γ × tmoving .
(13)
In particular, half of the pions decay during time tmoving = T0 as measured by the moving
observer, but the observer in the lab will clock this time as
T = γ × T0 = 2.3 × 17.2 ns = 40.6 ns.
(14)
Half of the remaining pions will decay during the second 17.7 ns by the moving observer’s
clock, but by the lab’s clock this will take another 40.6 ns. Likewise, one half of the pions
that still remain will decay during the third 17.7 ns period by the moving clock, i.e. during
the third 40.6 period by the lab clock, etc., etc. In other words, in the lab frame, the pions
decay with half-life time
T = γ × T0 = 40.6 ns.
5
(15)
(c) In the lab frame, the pions move at speed v = 0.9 c = 2.7 · 108 m/s. In time T = 40.6 ns
by the lab clock, half of the pions will decay. During this time, the remaining pions will
move through distance
L = v × T = (2.7 · 108 m/s) × (40.6 · 10−9 s) ≈ 11 m.
(16)
(d) Strictly speaking, in the frame that moves with the pions, the pions do not move at
all. However, in the pion frame, the ground and the lab move towards the pions at speed
v = 0.9 c = 2.7 · 108 m/s, so the distance between the lab and the pions decreases with time
as
x = x0 − v × t.
(17)
By the pion’s frame clock, half of the pions decay in time T0 = 17.7 ns, and during this time,
the distance between the pions and the lab shrinks by
L′ = x0 − x = v × T0 = (2.7 · 108 m/s) × (17.7 · 10−9 s) ≈ 4.78 m.
(18)
Note that this distance is measured in the frame of the moving pions, so it is shorter than
the distance (16) measured in the lab frame because of Lorentz contraction. Indeed, the
ration between the two distances is precisely the γ factor (12),
L
11.0 m
=
= 2.3 = γ.
L′
4.78 m
(19)
Textbook problem E.9 at the end of chapter 20:
The length of a fast-moving object — like the spaceship in question — depends on the
observer who makes the measurement. More accurately, the length depends on the relative
velocity v between the object and the observer. Specifically,
r
v 2
1
≡ L0 × 1 −
L = L0 ×
γ
c
(20)
where L0 is the object’s length measured in its own frame of reference, i.e. by the observer
who moves together with the object, so their relative velocity is zero. Note: it’s the relative
velocity which matters here, not the absolute velocities of the object or the observer.
6
Now let’s apply eq. (20) to the spaceship in question. The astronauts move together with
the ship, so regardless of its huge velocity relative to the Earth, relative to the astronauts
the ship is at rest. Consequently, the 50 meter length they measure is the ship’s length in
its own frame of reference, L0 = 50 m. On the other hand, the observers on Houston see
the ship moving at a very high speed v = 0.8 c, so to them the ship appears shorter by the
Lorentz factor
1
≡
γ
r
1−
v 2
c
=
p
1 − 0.82 = 0.6.
(21)
Thus, the observers at Houston measure the spaceship’s length to be
L = L0 ×
1
= (50 m) × 0.6 = 30 m.
γ
(22)
Textbook problem E.10 at the end of chapter 20:
Similar to the previous problem E.9, in this problem we must compare the same length —
the distance between two cities — in two different frame of reference, the frame of the Earth
and the frame of a spaceship flying by our planet. Again, the key formula for this problem
is
1
≡ L0 ×
L = L0 ×
γ
r
1−
v 2
c
(20)
where L0 is the length (or distance) measured by the observer at rest relative to the object in
question while L is the length measure by the observer which moves at high speed v relative
to the object.
Note: it does not matter who moves, the object or the observer, the only thing that
matters is the relative velocity between them. This an observer on Earth will see the spaceship shorter than the same ship as seen by the astronauts on board. At the same time, the
astronauts looking back at Earth will see its squashed in the direction of the relative motion,
and to them the distances on Earth would appear shorter than the same distances as seen
by the Earth-side observers.
The specific formula relating the distances between the same two cities as measured by
different observer is eq. (20), but this time it’s the L0 which is measure by the ground-side
7
observers while L is measured by the astronauts on board the spaceship. Thus, the 600 km
distance (370 miles) we are given is L while the L0 follows by solving eq. (20):
1
L0 = L × γ = L × p
1 − (v/c)2
1
= (600 km) × √
= 750 km.
1 − 0.62
(23)
In other words, while the astronauts measure the distance between two cities as 600 km, the
ground-side observers see the two cities being 750 km apart.
Non-textbook problem #I:
In vector notations, the velocity ~vg of a plane relative to the ground and the velocity ~va of
the same plane relative to the moving air are related as
~vg = ~va + ~u
(24)
where ~u is the wind’s velocity (i.e., the velocity of the air relative to the ground). Note the
vector nature of this formula: The relation between plane’s groundspeed ~vg and airspeed ~va
is more complicated and depends on the directions of the plane and of the wind.
(a) The first plane flies in the same direction — due East — as the wind, so the wind’s speed
adds to its airspeed:
~vg
~va
~u
for ~va ↑↑ ~u,
|~va + ~u| = |~va | + |~u|
thus for a plane in a tailwind, groundspeed = airspeed + wind-speed,
|~vg | = |~va | + |~u| = 570 MPH + 100 MPH = 670 MPH.
8
(25)
(b) The second plane flies in the opposite direction from the wind, so the wind speed subtracts
from its airspeed:
~vg
~u
for ~va ↑↓ ~u,
~va
|~va + ~u| = |~va | − |~u|
thus for a plane in a headwind, groundspeed = airspeed − wind-speed,
|~vg | = |~va | − |~u| = 570 MPH − 100 MPH = 470 MPH.
(26)
(c) The third plane flies at an angle to the wind, so we need a vector diagram to understand
its motion:
~u
90◦
~va
~vg
θ
~vg = ~va + ~u.
(27)
Note that the vectors ~va and ~u are connected head-to tail, while the ~vg = ~va + ~u vector
connect the tail of ~va to the head of ~u: that’s how we add up vectors graphically.
The direction of ~va — the heading of the plane — is due North, while the direction of
the wind velocity ~u is due East. These two vectors are ⊥ to each other, so the triangle made
from the three vectors on the diagram (27) is a right triangle. This fact gives us a simple
formula for the angle θ between the ~va and ~vg — i.e., between the heading of the plane and
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the direction of its flight relative to the ground:
θ = arctan
|~u|
100 MPH
= arctan
≈ 10◦ .
|~va |
570 MPH
(28)
Thus, the direction of the third plane’s flight relative to the ground is 10◦ east from due
north.
(d) On the diagram (27), the groundspeed ~vg of the third plane is the length of the long side
of the right triangle. By the Pythagoras theorem,
q
q
2
2
|~vg | =
|~va | + |~u| =
(570 MPH)2 + (100 MPH)2 ≈ 580 MPH.
(29)
Non-textbook problem #II:
The rest mass of a body is its mass measured by an observer moving at the same velocity as
the body, so in his frame of reference the body is at rest. Unlike the relativistic mass defined
by Einstein, the rest mass of a body does not depend on its velocity. Thus all electrons in
Nature have the same rest mass me = 9.11 · 10−31 kg regardless of how fast they are moving,
all protons have the same rest mass mp = 1.67 · 10−27 kg regardless of velocity, etc., etc.. On
the other hand, in nuclear or sub-nuclear reactions the net rest mass is not conserved. For
example, when an electron and a positron annihilate each other and turn into photons, the
net rest mass drops from 2me before the reaction to zero after the reaction.
In terms of the rest mass of a body, its mechanical momentum is not p~ = m~v but rather
p~ = γm × ~v
where γ = p
1
1 − (v/c)2
.
(30)
In particular, for the space probe in question
1
1
= 1.67,
= √
γ = p
2
1 − 0.802
1 − (v/c)
(31)
hence the momentum is
p = γmv = 1.67 × (1200 kg) × (0.80 × 3.00 · 108 m/s) = 4.8 · 1011 kg m/s.
(32)
Also, the net energy of a relativistic body — including the kinetic energy and the rest
10
energy — is related to the rest mass as
Enet = Ekin + Erest = γ × mc2
(33)
where γ is precisely as in eq. (30). Subtracting the rest energy
Erest = mc2
(34)
from the net energy (33), we obtain the kinetic energy
Ekin = γ × mc2 − mc2 = (γ − 1) × mc2 .
(35)
For the space probe in question
Ekin = (1.67 − 1) × (1200 kg) × (3.00 · 108 m/s)2 = 7.2 · 1019 J.
11
(36)