Solutions Manual
For
Special Relativity
A Heuristic Approach
Sadri Hassani
Contents
List of Symbols
1
1 Qualitative Relativity
5
2 Relativity of Time and Space
9
3 Lorentz Transformation
19
4 Spacetime Geometry
49
5 Spacetime Momentum
75
6 Relativity in Four Dimensions
85
7 Relativistic Photography
99
8 Relativistic Interactions
127
9 Interstellar Travel
139
10 A Painless Introduction to Tensors
151
11 Relativistic Electrodynamics
157
12 Early Universe
167
A Maxwell’s Equations
177
B Derivation of 4D Lorentz transformation
181
C Relativistic Photography Formulas
185
List of Symbols, Phrases, and Acronyms
â unit vector in the direction of ~a
anti-particle a particle whose mass and spin are exactly the same as its corresponding
particle, but the sign of all its “charges” are opposite. If a particle is represented
by the letter p, then it is customary to denote its anti-particle by p̄. If a particle is
represented by the letter q − (or q + ), then it is customary to denote its anti-particle
by q + (or q − ).
as arcsecond; an arcsecond is an angle 1/3600 of a degree.
baryon a hadron whose spin is an odd multiple of ~/2. Baryons are composed of three
quarks. Examples of baryons are protons and neutrons.
~ fractional velocity of one observer relative to another, β~ = ~v /c
β
boson a particle whose spin is an integer multiple of ~. All gauge particles are bosons as
are all mesons, as well as the Higgs particle.
causally connected referring to two two events. If an observer or a light signal can be
present at two events, those events are said to be causally connected.
causally disconnected referring to two two events. If an observer or a light signal cannot
be present at two events, those events are said to be causally disconnected.
CBR Cosmic Background Radiation
CM center of mass
CS coordinate system
êx , êy , êz unit vectors along the three Cartesian axes.
EM electromagnetic or electromagnetism, one of the four fundamental forces of nature.
equilibrium temperature temperature of the universe at which matter and radiation
densities are equal
eV electron volt, unit of energy equal to 1.6 × 10−19 J
2
List of Symbols
fermion a particle whose spin is an odd multiple of ~/2. Fermions obey Pauli’s exclusion
principle: no two identical fermions can occupy a single quantum state. Electrons,
protons, and neutrons are fermions, so are all leptons and quarks, as well as all
baryons.
p
p
γ the Lorentz factor, γ = 1/ 1 − β 2 = 1/ 1 − (v/c)2
gauge bosons According to the modern theory of forces, fundamental particles interact
via the exchange of gauge bosons. Excluding gravity, whose microscopic behavior
is not well understood, there are 12 gauge bosons whose exchange explains all the
interactions: Z 0 , W ± and γ (photon) are responsible for electroweak interaction,
while 8 gluons are responsible for strong interaction.
gluons the particles responsible for strong interactions: two or more quarks participate
in strong interaction by exchanging gluons. There are four gluons, which with their
antiparticles comprise the eight gluons whose exchange binds quarks together.
GTR general theory of relativity; the relativistic theory of gravity.
Gyr gigayear, equal to 109 years
hadron a particle capable of participating in strong nuclear interactions. Examples of
hadrons are protons, neutrons and pions. All hadrons are made up of quarks and/or
anti-quarks.
half life the time interval in which one half of the initial decaying particles survive.
LAV Law of Addition of Velocities
lepton a particle that participates only in electromagnetic and weak nuclear interactions,
but not in strong nuclear interactions. Leptons are elementary particles in the sense
that they are not made up of anything more elementary. There are three electrically charged leptons: electron, muon, and tauon. Each charged lepton has its own
neutrino. So, altogether there are six leptons.
LHC Large Hadron Collider
light cone (at an event E) The set of all events that are causally connected to E.
light hour the distance that light travels in one hour, ≈ 1.08 × 1012 m
light minute the distance that light travels in one minute, ≈ 1.8 × 1010 m
light second the distance that light travels in one second, ≈ 3 × 108 m
lightlike referring to two events, when c∆t = ∆x or (∆s)2 = 0.
luminally connected referring to two two events. If a light signal can be present at two
events, those events are said to be luminally connected.
ly light year; one light year is 9.467 × 1015 m.
mean time the time interval in which 1/e of the initial decaying particles survive.
List of Symbols
meson a hadron whose spin is an integer multiple of ~. Mesons are composed of one quark
and one anti-quark. Examples of mesons are pions.
MeV million electron volt, unit of energy equal to 1.6 × 10−13 J
µm micrometer = 10−6 m
Minkowskian distance also called “spacetime distance,”
(∆s)2 = (c∆t)2 − (∆x)2
is an expression involving the coordinates of two events which is independent of the
coordinates used to describe those events.
MM clock sometimes called “light clock” is described on page 13.
Mpc Megaparsec
muon an elementary particle belonging to the group of particles named “leptons,” to which
electron belongs as well. Muon is called a “fat electron” because it behaves very much
like an electron except that it is heavier.
neutrino a neutral lepton with very small mass. Neutrinos participate only in weak nuclear
force. That’s why they are very weakly interacting.
ns nanosecond or 10−9 s
Parsec A distance of about 3.26 light years. One parsec corresponds to the distance at
which the mean radius of the Earth’s orbit subtends an angle of one second of arc.
positron the anti-particle of the electron
quarks elementary particles which make up all hadrons. There are six quarks: up, down,
strange, charm, bottom, top. Quarks participate in all interactions, in particular, the
strong interaction.
RF reference frame
spacelike referring to two events, when c∆t < ∆x or (∆s)2 < 0.
spacetime distance see Minkowskian distance
STR special theory of relativity
tauon an elementary particle belonging to the group of particles named “leptons,” to which
electron belongs as well. It is the heaviest lepton discovered so far.
timelike referring to two events, when c∆t > ∆x or (∆s)2 > 0.
3
CHAPTER
1
Qualitative Relativity
Problems With Solutions
1.1. A rod of length L emits light from all of its points simultaneously (in its rest frame)
when a remote switch is turned on. Its center is on the x-axis and is moving on the axis in
a plane parallel to a very large photographic plate and infinitesimally close to it. When it
reaches the middle of the plate, the switch is turned on.
(a) Compare the length L|| of the image on the photographic plate with L when the rod
is along the x-axis: L|| > L, L|| = L, or L|| < L? Give a reason for your answer.
(b) Compare the length L⊥ of the image on the photographic plate with L when the rod
is perpendicular to the x-axis: L⊥ > L, L⊥ = L, or L⊥ < L? Give a reason for your
answer.
Solution:
(a) Length parallel to the direction of motion shrinks regardless of who sees the events
of light emission simultaneously. See the discussion in Section 1.3 for the reason (as
well as how to capture the length of a moving object).
(b) Length perpendicular to the direction of motion does not change.
Note the importance of the fact that the distance between the rod and the photographic
plate is zero.
1.2. A rod is placed along the x-axis with its center at the origin. A pinhole camera C1 is
located on the z-axis and takes a picture of the stationary rod. Now the rod starts moving
along the x-axis parallel to itself from −∞. Camera C1 is removed and another pinhole
camera C2 replaces it on the z-axis. As soon as the center of the rod reaches the origin
(call it t = 0), C2 takes a picture.
(a) Is the pinhole of C2 collecting the light rays from the two ends of the rod that were
emitted at t = 0?
6
Qualitative Relativity
(b) Is the pinhole collecting the light rays from the two ends of the rod that were emitted
simultaneously, but not at t = 0?
(c) If the answer to (b) is no, which end emitted its light first, the trailing end or the
leading end?
(d) Is it possible for the image of the rod in C2 to be longer than its image in C1 ? Hint:
Consider the location of each end as it emits the light ray captured by C2 .
Solution:
(a) No. It takes time for the light to reach the camera once it leaves its source.
(b) No.
(c) The trailing end is farther away from the camera, so it must emit the light sooner
than the leading end.
(d) The trailing end emits its light, the rod moves a little, then the leading end emits its
light. So, the distance between the source of the light from the trailing end and that
of the leading end is indeed larger than the length of the rod.
Note that the image in camera C2 , which is longer than the image in C1 , has nothing to do
with the actual length of the rod!
1.3. A rod is placed along the y-axis with its center at the origin. A pinhole camera C1 is
located on the z-axis and takes a picture of the stationary rod. Now the rod starts moving
along the x-axis parallel to itself from −∞. Camera C1 is removed and another pinhole
camera C2 replaces it on the z-axis. As soon as the center of the rod reaches the origin
(call it t = 0), C2 takes a picture.
(a) Is the pinhole of C2 collecting the light rays from the two ends of the rod that were
emitted at t = 0?
(b) Is the pinhole collecting the light rays from the two ends of the rod that were emitted
simultaneously, but not at t = 0?
(c) If the answer to (b) is no, which end emitted its light first, the top or the bottom?
(d) Is it possible for the image of the rod in C2 to be longer than its image in C1 ? Hint:
Consider the locations of the ends as they emit their light rays captured by C2 , the
distance between those locations and the pinhole, and the angle they subtend at the
pinhole.
Solution:
(a) No. It takes time for the light to reach the camera once it leaves its source.
(b) Yes. The perpendicular distance does not change. So, the top and bottom of the rod
are equidistant from the pinhole, and to reach it at t = 0, the rays must have been
emitted at the same time in the past.
(c) The answer to (b) is yes!
7
(d) No. Since the locations of the sources of the rays are farther from C2 (they have
negative x-coordinates) than the locations in C1 , they must have a smaller image.
Chapter 7 discusses this in gory mathematical detail!
1.4. A circular ring emits light from all of its points simultaneously (in its rest frame) when
a remote switch is turned on. It is moving in a plane parallel to a photographic plate and
infinitesimally close to it. When it reaches the plate, the switch is turned on. What is the
shape of the image of the photograph? Hint: See Problem 1.1.
Solution: The diameter along the direction of motion shrinks; the diameter perpendicular
to the direction of motion stays the same. So, the shape is an ellipse flattened in the
direction of motion.
1.5. A conveyor belt moving at relativistic speed carries cookie dough. A circular stamp
cuts out cookies as the dough rushes by beneath it. What is the shape of these cookies?
Are they flattened in the direction of the belt, stretched in that direction, or circular?
Solution: The answer is identical to that of the previous problem. So, the cookies are
flattened in the direction of the belt.
1.6. A conveyor belt moving at relativistic speed carries cookie dough. A laser gun one
meter above the belt emits a beam in the shape of the surface of a circular cone that
cuts the dough perpendicularly. Are these cookies flattened in the direction of the belt,
stretched in that direction, or circular? Hint: Concentrate on the two ends of the diameter
of the beam along the dough, and note that their light beams arrive simultaneously at the
stationary bed on which the dough is moving. Now consider how the two events appear
in the RF of the moving dough and what implication it has on the length of the diameter.
The discussion surrounding Figure 1.3 may be helpful.
Solution: The image of the laser beam on the stationary bed is circular and the two
events of the arrival of the beams from the two ends of the horizontal diameter occur
simultaneously. Consider two experiments. In the first experiment, there is no dough and
the laser imprints a circle on the stationary bed. In the second experiment, an observer
riding with the dough records the coincidence of the location of the event in front of her
with the front end of the imprint before the coincidence of the event in the back. She
concludes that for her, the distance between the two events is larger than the two marks
on the stationary bed. So, the image is an ellipse elongated along the direction of the belt.
Thus, the cookies are stretched in the direction of the belt.
1.7. A circular ring is centered at the origin in the xy-plane. A pinhole camera C1 is located
on the z-axis and takes a picture of the stationary ring. Now the ring starts moving along
the x-axis from −∞. Camera C1 is removed and another pinhole camera C2 replaces it on
the z-axis. As soon as the center of the rod reaches the origin (at t = 0), C2 takes a picture.
(a) Is the pinhole of C2 collecting the light rays from the two ends of the horizontal
diameter (along the x-axis) of the ring that were emitted at t = 0? Hint: Look at
Problem 1.2.
(b) Is the pinhole of C2 collecting the light rays from the two ends of the horizontal
diameter of the ring that were emitted simultaneously, but not at t = 0?
8
Qualitative Relativity
(c) If the answer to (b) is no, which end emitted its light first, the trailing end or the
leading end?
(d) Is it possible for the image of the horizontal diameter in C2 to be longer than its
image in C1 ?
(e) Is the image of the vertical diameter (along the y-axis) in C2 equal to, longer than,
or shorter than its image in C1 ? Hint: Look at Problem 1.3.
(f) Can you guess what the shape of the image of the ring is in C2 ?
Solution:
(a) No.
(b) No.
(c) The trailing end.
(d) Yes, it always is.
(e) It is shorter than its image in C1 .
(f) It is an ellipse elongated along the direction of motion.
See Example 2.2.6 for a quantitative analysis of this problem.
CHAPTER
2
Relativity of Time and Space
Problems With Solutions
2.1. Take the most rigid rod you can find, and hit one end of it with the hammer. The
rod as a whole starts to move because it is rigid.1 Actually not! It takes time for the
information that one end of the rod was hit to reach the other end, because of Note 2.1.7.
Now go to the rest frame of the rod which is now moving relative to the hammer. Assume
that the hammer hits the rod in such a way as to cause (the front end of) it to stop. But
the other end knows nothing about the hammer yet. So, it keeps moving! What does this
say about the concept of “rigidity” in relativity?
Solution: Rigidity is not a well defined concept in relativity. The other end of the rod gets
compressed because of its motion.
2.2. In this problem you’ll learn more about superluminal transverse speeds.
(a) Show that the angle that maximizes Equation (2.6) is given by cos θ = β.
(b) Substitute this in (2.6) to obtain (vtr )max = cβγ.
√
(c) Show that (vtr )max is larger than c for any β > 1/ 2.
(d) What speed makes (vtr )max ten times faster than light? What is the angle corresponding to this speed?
Solution:
(a) Set the derivative of vtr
0
vtr
(θ) =
cβ(cos θ − β)
(β cos θ − 1)2
00 (θ) < 0 when
equal to zero to get cos θ = β. The second derivative test shows that vtr
cos θ = β. Therefore, vtr is indeed maximum at cos θ = β.
1
If you hit one end of a slinky, the other end does not move, at least not immediately.
10
Relativity of Time and Space
(b)
(vtr )max
p
sin θ
1 − β2
cβ
= cβγ.
= cβ
= cβ
=p
2
1 − β cos θ
1−β
1 − β2
(c)
cβγ > c ⇐⇒ β 2 γ 2 > 1 ⇐⇒ β 2 > 1 − β 2 ⇐⇒ β 2 > 1/2.
(d)
βγ = 10 ⇐⇒ β 2 γ 2 = 100 ⇐⇒ β 2 = 100 − 100β 2 ⇐⇒ β 2 = 100/101,
or if β = 0.995. The angle corresponding to this β is θ = cos−1 0.995 = 0.0997 or
θ = 5.71◦ .
2.3. Consider an MM clock moving horizontally with speed β relative to observer O. Denote
its length in motion by L and at rest by L0 . Let ∆t1 be the time it takes light to go from
the emitter to the mirror according to O. Let ∆t2 be the time it takes light to go from the
mirror to the emitter according to O.
(a) Show that
c∆t1 = L + βc∆t1 ,
c∆t2 = L − βc∆t2 .
(b) Show that a “tick” according to O is
∆t = ∆t1 + ∆t2 =
2L/c
.
1 − β2
(c) Now use the time dilation formula with ∆τ = 2L0 /c to derive the length contraction
formula.
Solution:
(a) By the time the light that is emitted at the emitter reaches the mirror, the MM clock
has moved. So, the distance that the light covers is L plus the distance that the MM
clock moves in the same time interval. So, c∆t1 = L + βc∆t1 , and
c∆t1 (1 − β) = L ⇐⇒ ∆t1 =
L
.
c(1 − β)
On reflection, the light and the MM clock move in opposite directions. Therefore,
c∆t2 = L − βc∆t2 , and
c∆t2 (1 + β) = L ⇐⇒ ∆t2 =
L
.
c(1 + β)
(b) A tick according to O is therefore,
∆t = ∆t1 + ∆t2 =
L
L
2L/c
+
=
.
c(1 − β) c(1 + β)
1 − β2
But ∆t = γ∆τ = γ(2L0 /c). Thus,
γ(2L0 /c) =
p
2L/c
= γ 2 (2L/c) ⇐⇒ L = L0 /γ = L0 1 − β 2 .
2
1−β
11
2.4. The spaceship Enterprise goes to a planet in a star system far away with a speed of
0.9c, spends 6 months on the planet, and comes back with a speed of 0.95c. The entire trip
takes 5 years for the crew.
(a) How far is the planet according to Earth observers?
(b) How long did it take the crew to get to the planet?
(c) How long did the entire trip take for the Earth observers?
Solution:
√
(a) On the outbound part, √
the distance for the crew is L0 1 − 0.92 and the time it takes
them to get there is L0 1 − 0.92 /(0.9c), where L0 is the distance√
according to Earth
observers. Similarly, the time it takes them to come back is L0 1 − 0.952 /(0.95c).
So, the entire round trip time is
√
√
L0 1 − 0.92 L0 1 − 0.952
+
= 4.5 years.
0.9c
0.95c
This gives L0 = 5.54 light years.
(b)
∆τoutbound
√
L0 1 − 0.92
= 0.484 years.
=
0.9c
(c) The distance according to Earth is 5.54 light years. So, for outbound trip
∆toutbound =
5.54 light years
= 6.15 years.
0.9c
Similarly,
5.54 light years
= 5.83 years.
0.95c
Therefore the entire trip takes 6.15 + 0.5 + 5.83 = 12.48 years.
∆tinbound =
2.5. A rocket ship leaves the Earth at a speed of 0.8c. When a clock on the rocket says 1
hour has elapsed, the rocket ship sends a light signal back to Earth.
(a) According to Earth clocks, when was the signal sent?
(b) According to Earth clocks, how long after the rocket left did the signal arrive back
on Earth?
(c) According to the rocket clock, how long after the rocket left did the signal arrive back
on Earth?
Solution:
(a) The rocket is measuring the proper time of 1 hour. So, for Earth
1 hour
∆t = γ∆τ = √
= 1.67 hours.
1 − 0.82
12
Relativity of Time and Space
(b) The distance of the rocket from Earth when it sends the signal is
1.67 hours × 0.8c = 1.33 light hours.
So, it took 1.33 hours for the signal to arrive at Earth after it was sent. Therefore,
between the rocket leaving and the signal arriving, it took 1.33 + 1.67 = 3 hours
according to Earth. Note that this is proper time for Earth.
(c) The rocket measures coordinate time:
3 hour
∆trocket = √
= 5 hours.
1 − 0.82
It is a good exercise to find the answer to (c) by calculating in the rocket frame. Note that
the Earth moves at 0.8c away from the rocket. So, when the rocket sends the signal, the
Earth is at a distance of 0.8 light hour from rocket. Now the rocket sends a signal that
chases the Earth. When does the light catch up with Earth according to rocket?
2.6. A bicycle wheel of rest radius R is rotating in such a way that the rim has a linear
speed of 0.866c. What is the circumference of the rim? What is the length of the spokes
in motion? But spokes are perpendicular to the direction of motion! Discuss whether in
relativity anything can be considered “incompressible” or “rigid.” See also Problem 2.1.
Solution: For the same reason as Problem 2.1, the spokes are not “rigid.”
2.7. The spaceship Viking goes to a planet in a star system 30 light years away from Earth
with a speed of 0.99c, spends 1 year on the planet, and then returns home. The entire trip
takes 10 years for the crew.
(a) How far is the planet according to crew?
(b) How long does it take the crew to get to the planet?
(c) How long does it take the crew to return to Earth?
(d) What is the speed of the crew on return? Warning! The distance for the crew is not
the same as the distance on their way to the planet.
(e) How far is the Earth from the planet according to crew on their return?
(f) How long did the entire trip take for the Earth observers?
Solution:
√
(a) L = 30 1 − .992 = 4.23 light years.
(b) The distance is 4.23 light years, and they are going at 0.99c, so it takes them 4.23/0.99,
or 4.27 years to get there.
(c) Since the entire trip is 10 years, the return time is 10 − 4.27 − 1 = 4.73 years.
13
p
2
(d) Let β2 be the speed of return.
pThen the distance is 30 1 − β2 light years. And the
2
time, in terms of speed, is 30 1 − β2 /cβ2 . So, we have to solve the equation
p
30 1 − β22 light years
= 4.73 years.
cβ2
The answer comes out to be β2 = 0.988.
√
(e) L = 30 1 − 0.9882 = 4.67 light years.
(f)
30 light years
30 light years
+1+
= 61.67 years.
0.99c
0.988c
2.8. The spaceship Diracus goes to a planet in a star system with a speed whose Lorentz
factor is γ, spends 1 year on the planet, and then returns home with a speed whose Lorentz
factor is 4γ. The captain of the spaceship is 29 years old and has just had a newborn son.
The entire trip takes 11 years for the crew. The odometer of the spaceship shows that the
“milage” for the round trip is a quarter of the Earth-planet distance as measured by Earth
observers.
(a) What is the outbound speed? The inbound speed?
(b) What is the Earth-planet distance according to the Earth observers?
(c) What is the Earth-planet distance according to the crew on their way to the planet?
On their way back?
(d) How long does it take the crew to go to the planet? To return?
(e) Who is older, the son or the father when the ship lands on Earth? By how many
years?
Solution:
(a) Let L0 be the distance according to Earth. Then
L0 L0
L0
+
=
.
γ
4γ
4
This gives γ = 5 and
βout
and
p
√
γ2 − 1
24
=
=
= 0.98,
γ
5
p
√
(4γ)2 − 1
399
βin =
=
= 0.9987,
4γ
20
(b) The round trip time is 10 years. So,
Lout Lin
L0 /γ L0 /(4γ)
+
=
+
= 10
βout
βin
βout
βin
or
L0
L0
+
= 10.
5 × 0.98 20 × 0.9987
This gives L0 = 39.35 light years.
14
Relativity of Time and Space
(c) Lout = L0 /γ = 7.87 light years; Lin = L0 /(4γ) = 1.97 light years.
(d) ∆τout = Lout /cβout = 8.03 years; ∆τin = Lin /cβin = 1.97 years; and these two answers
are consistent with the roundtrip time being 10 years.
(e) ∆tout = L0 /cβout = 40.15 years; ∆tin = L0 /cβin = 39.4. So, the son is 40.15 + 1 +
39.4 = 80.55 years old, while the father is just 29 + 11 = 40 years old.
2.9. A rod of rest length L0 moves with speed v along the positive x0 -direction of observer
O0 . The rod makes an angle θ0 with respect to the x-axis of its rest frame.
(a) Find the length of the rod as measured by O0 .
(b) Find the angle θ the rod makes with the x0 -axis as measured by O0 .
Solution: The projections along the axes in the rest frame are
∆x = L0 cos θ0 ,
∆y = L0 sin θ0 .
In the O0 frame, we have
p
p
∆x0 = ∆x 1 − (v/c)2 = L0 1 − (v/c)2 cos θ0 ,
∆y 0 = L0 sin θ0
(a) The length in O0 is
q
p
(∆x0 )2 + (∆y 0 )2 = L20 [1 − (v/c)2 ] cos2 θ0 + L20 sin2 θ0
p
= L0 1 − β 2 cos2 θ0 , β ≡ v/c.
L=
Note that the answer is consistent with the special cases θ0 = 0 and θ0 = π/2.
(b)
p
1 − β 2 cos θ0
∆x0
cos θ0
cos θ =
=p
= p
2
2
L
1 − β cos θ0
γ 1 − β 2 cos2 θ0
or
tan θ =
∆y 0
sin θ0
=p
= γ tan θ0 .
0
∆x
1 − (v/c)2 cos θ0
2.10. A flasher produces a flash of light every second when at rest. It is moving away from
you at 0.9c.
(a) How frequently does it flash according to you?
(b) By how much does the distance between you and the flasher increase between consecutive flashes?
(c) How long after the emission of a given flash does it reach you?
(d) How often do you receive the flashes?
Solution: The flasher keeps proper time.
15
(a)
∆t = √
1s
= 2.294 s.
1 − 0.92
(b)
∆x = 0.9c × 2.294 s = 2.065 light second.
(c) From its emission, it takes the flash 2.065 seconds to reach you.
(d) The time intervals between flashes is the sum of the time interval between emissions
and the time it takes the flashes to reach you: 2.294 + 2.065 = 4.35 seconds.
Note that this is related to Doppler effect: Think of a flash as a wavefront.
2.11. Charged pions are produced in many collisions in accelerators. They decay in their
rest frame according to
N (t) = N0 e−t/T ,
where T = 2.6×10−8 s is their mean life. A burst of charged pions is produced at the target
of an accelerator and it is observed that only one percent of them decay at a distance of
1 m from the target. What is the Lorentz factor for pions and how fast are they moving?
Solution: Note that t in the decay formula is proper time. So, we have to calculate things
in the rest frame of the pions. The the distance in the rest frame is L0 /γ. Therefore,
t = L0 /(γcβ), and
0.99 = e−t/T = e−L0 /(γcβT ) ⇐⇒ ln(0.99) = −
p
L0
L0
⇐⇒ γβ = γ 2 − 1 = −
.
γcβT
ln(0.99)cT
This gives γ = 12.795 and β = 0.997.
2.12. Derive Equations (2.8), (2.9), and (2.10).
Solution: Since xc < 0, the negative sign in the previous equation must be chosen. The
expression under square root sign can be written as
β 2 γ 2 (β 2 L2 + b2 + L2 /γ 2 ) = β 2 γ 2 b2 + L2 β 2 + 1 − β 2 = β 2 γ 2 (b2 + L2 )
This yields (2.8). For (2.9), we have
2
xA = xc − L/γ = −β γL − βγ
p
b2
+
L2
p
L
2
2
2
− L/γ = −γ β L + 2 + β b + L .
γ
Equation (2.9) now follows immediately from the definition of γ in terms of β. Finally,
2
p
p
c2 t2A = γ 2 L + β b2 + L2 + b2 = γ 2 L2 + γ 2 β 2 (b2 + L2 ) + 2γ 2 βL b2 + L2 + b2
p
= γ 2 L2 + (γ 2 − 1)b2 + γ 2 β 2 L2 + 2γ 2 βL b2 + L2 + b2
p
= γ 2 (L2 + b2 ) + γ 2 β 2 L2 + 2γ 2 βL b2 + L2
h
i
2
p
p
= γ 2 (L2 + b2 ) + β 2 L2 + 2βL b2 + L2 = γ 2 βL + b2 + L2 .
When taking the square root, the negative sign is chosen because the light from A was
emitted in the past.
16
Relativity of Time and Space
2.13. Derive Equations (2.11) and (2.12).
Solution: Just as in the case of A, write
p
p
(x0c + L/γ)2 + b2
|x0c |
=
⇐⇒ |x0c | = β (x0c + L/γ)2 + b2 .
v
c
Squaring both sides gives
x0c2 = β 2 (x0c2 + L2 /γ 2 + 2x0c L/γ + b2 ) ⇐⇒
x0c2 2β 2 L 0
−
xc − β 2 (b2 + L2 /γ 2 ) = 0.
γ2
γ
Solve this quadratic equation for xc to obtain
p
x0c = β 2 γL ± β 4 γ 2 L2 + β 2 γ 2 (b2 + L2 /γ 2 ).
The negative sign must be chosen because x0c < 0. The rest of the solution is identical to
the previous problem. In fact, all the answers are obtained from that problem by changing
the sign of L.
2.14. Derive Equations (2.14) and (2.15).
p
Solution: Square both sides of |xc | = β (b + L)2 + x2c to get
x2c = β 2 [(b + L)2 + x2c ] = β 2 x2c + β 2 (b + L)2 ⇐⇒ (1 − β 2 )x2c =
x2c
= β 2 (b + L)2 .
γ2
Take the square root and chose the negative sign to get the answer. For x0c , just change L
to −L.
2.15. Derive Equation (2.16).
Solution: For an arbitrary point, you square both sides of |x| = β
x2 = β 2 [(b − z)2 + x2 ] = β 2 x2 + β 2 (b − z)2 ⇐⇒
p
(b − z)2 + x2 . Then
x2
= β 2 (b − z)2 .
γ2
Take the square root to get x = ±γβ|b − z|. Choosing the negative sign and noting that
|b − z| = b − z, you obtain the answer.
2.16. Find tA and tB , the times at which A and B emit their light rays when the rod is
oriented along the z-axis as in Example 2.2.4.
Solution: A and B have coordinates (xc , 0, −L) and (x0c , 0, L), respectively. So, their
distances, |ctA | and |ctB | from the camera are given by
c2 t2A = x2c + (b + L)2 ,
c2 t2B = x0c2 + (b − L)2 .
I’ll find the first one, leaving the second for you. From (2.14), we have
c2 t2A = γ 2 β 2 (b + L)2 + (b + L)2 = (γ 2 β 2 + 1)(b + L)2 = γ 2 (b + L)2 .
Therefore, ctA = −γ(b + L). Similarly, ctB = −γ(b − L). Note that b > L.
2.17. Derive Equation (2.17).
17
Solution: From |x| = β
p
x2 + y 2 + b2 , you get
x2 = β 2 (x2 + y 2 + b2 ) ⇐⇒
x2
x2
2 2
2
=
β
(y
+
b
)
⇐⇒
− y 2 = b2 ,
γ2
γ2β2
and the final form follows immediately.
2.18. In Example 2.2.5, find tA and tB , the times at which A and B emit their light rays
when the rod is oriented along the y-axis.
Solution: Let tP be the time that the light from an arbitrary point P of the rod with
coordinates (x, y, 0) was emitted. Then using the results of the example, you have
p
x2
x2 + y 2 + b2 = 2 = γ 2 (y 2 + b2 ).
β
p
√
Therefore, ctP = −γ y 2 + b2 . Thus, ctA = ctB = −γ L2 + b2 .
c2 t2P =
2.19. Derive Equations (2.20) and (2.21).
Solution: From x2c = β 2 x2 + a2 − γ 2 (x − xc )2 + b2 , you get
x2c = β 2 x2 + a2 − γ 2 x2 − γ 2 x2c + 2xγ 2 xc + b2
or
x2c (1 + β 2 γ 2 ) − 2xβ 2 γ 2 xc − β 2 a2 + b2 − β 2 γ 2 x2 = 0
or
γ 2 x2c − 2xβ 2 γ 2 xc − β 2 a2 + b2 − β 2 γ 2 x2 = 0.
The solution is
xβ 2 γ 2 ±
p
x2 β 4 γ 4 + β 2 γ 2 (a2 + b2 − β 2 γ 2 x2 )
γ2
p
√
xβ 2 γ 2 ± γβ x2 β 2 γ 2 + a2 + b2 − β 2 γ 2 x2
xβ 2 γ 2 ± γβ a2 + b2
=
=
.
γ2
γ2
xc =
Choosing the negative sign gives (2.20). Then
√
x + βγ a2 + b2
βp 2
2
a +b =
.
x − xc = x − β x +
γ
γ2
2
Substituting this in the equation of the ellipse (2.18) yields (2.21).
CHAPTER
3
Lorentz Transformation
Problems With Solutions
3.1. A line in the xz-plane has slope m and intercept b.
(a) Show that Equation (3.5) maps this line onto a line in the x0 z 0 -plane.
(b) What are the slope and the intercept of the line in the x0 z 0 -plane?
(c) Show that the transformation
x0 = a0 + a1 x + a2 z,
z 0 = b0 + b1 x2 + b2 z
transforms a straight line in the xz-plane into a parabola in the x0 z 0 -plane.
Solution: The equation of the line is z = mx + b.
(a) Substitute for z in (3.5) to get
x0 = a0 + a1 x + a2 mx + a2 b,
Find x from first equation
x=
z 0 = b0 + b1 x + b2 mx + b2 b.
x0 − a0 − a2 b
a1 + a2 m
and substitute it in the second
z 0 = b0 + b2 b + (b1 + b2 m)
or
z0 =
x0 − a0 − a2 b
,
a1 + a2 m
b1 + b2 m 0
(b1 + b2 m)(a0 + a2 b)
x + b0 + b2 b −
.
a1 + a2 m
a1 + a2 m
(b) The slope m0 and the intercept b0 are
m0 =
b1 + b2 m
,
a1 + a2 m
b0 = b0 + b2 b −
(b1 + b2 m)(a0 + a2 b)
.
a1 + a2 m
20
Lorentz Transformation
(c) Substitute z = mx + b in the equations as before
x0 = a0 + (a1 + a2 m)x + a2 b,
z 0 = b0 + b1 x2 + b2 mx + b2 b.
Find x from the first and plug it in the second
0
x − a0 − a2 b 2
x0 − a0 − a2 b
0
z = b0 + b1
+ b2 m
+ b2 b.
a1 + a2 m
a1 + a2 m
Expanding and collecting terms, you get an expression of the form z 0 = Ax0 2 +Bx0 +C,
which is the equation of a parabola.
3.2. Start with Equation (3.6) and provide all the missing steps that lead to Equation
(3.8).
Solution: Most of the missing steps are actually explained in the textbook.
3.3. Derive Equation (3.10).
Solution: With e = tan α, we get
√
√
1 + e2 = sec α and
1
= cos α,
1 + e2
√
e
tan α
= sin α.
=
sec α
1 + e2
3.4. Use Figure 3.5 to prove the coordinate transformation
x0 = a + x cos α − y sin α
y 0 = b + x sin α + y cos α.
Solution: Refer to Figure 3.1 of the manual. I’ll do the first equation. The second one is
very similar.
x0 = O0 A0 = O0 A + AA0 = a + OD − CD
= a + x cos α − QR = a + x cos α − y sin α.
3.5. In Euclidean space, the locus of points equidistant from the origin of a plane is a circle.
What is the locus of points equidistant (in the spacetime distance sense) from the origin of
a spacetime plane?
p
p
Solution: Instead of x2 + y 2 = R, we now have (ct)2 − x2 = R, or (ct)2 − x2 = R2 ,
which is a hyperbola.
3.6. Verify that LT preserves the spacetime distance (3.13). In other words, if E1 and E2
are two events with coordinates (x1 , ct1 ) and (x2 , ct2 ) in O and (x01 , ct01 ) and (x02 , ct02 ) in O0 ,
where the primed coordinates are obtained from the unprimed coordinates by an LT, then
c2 (t02 − t01 )2 − (x02 − x01 )2 = c2 (t2 − t1 )2 − (x2 − x1 )2 .
21
yʹ
y
Bʹ
P
Hʹ
Q
H
B
Oʹ
α
b
a
O
C
A
Aʹ
x
R
D
xʹ
Figure 3.1: The same point P has different pairs of coordinates in different coordinate systems. Note
that a = O0 A and b = O0 B.
Solution: To save writing use notations like ∆t21 = t2 − t1 , etc. Then
c∆t021 = γ(c∆t21 + β∆x21 ),
∆x021 = γ(∆x21 + βc∆t21 )
and
c2 ∆t0212 = γ 2 (c2 ∆t221 + β 2 ∆x221 + 2cβ∆t21 ∆x21 )
∆x0212 = γ 2 (∆x221 + β 2 c2 ∆t221 + 2cβ∆t21 ∆x21 ).
Subtracting the second from the first gives
c2 ∆t0212 − ∆x0212 = γ 2 (c2 ∆t221 − ∆x221 ) − γ 2 β 2 (c2 ∆t221 − ∆x221 ).
Now invoke the very useful identity γ 2 β 2 = γ 2 − 1.
3.7. Starting with (3.19), show that the origin of O0 has the coordinates
x0 = γ(−x00 + βct00 )
ct0 = γ(βx00 − ct00 )
relative to O. Show also that the inverse of Equation (3.19) can be written as
x = x0 + γ(x0 − βct0 )
ct = ct0 + γ(βx0 + ct0 ).
Solution: Set the left-hand side of (3.19) equal to zero and solve for x and ct, which are
now labeled x0 and ct0 :
0 = x00 + γ(x0 + βct0 )
0 = ct00 + γ(βx0 + ct0 ).
Multiply the second equation by β and subtract it from the first:
0 = x00 − βct00 + γ(1 − β 2 )x0 ⇐⇒ 0 = x00 − βct00 +
x0
.
γ
The first equation now follows trivially. The second equation is obtained similarly.
22
Lorentz Transformation
To obtain the inverse of Equation (3.19), multiply the second equation by β and subtract
it from the first:
x0 − βct0 = x00 − βct00 + γ(1 − β 2 )x ⇐⇒ γ(x0 − βct0 ) = γ(x00 − βct00 ) +x.
{z
}
|
=−x0
This yields the first inverse equation. The second inverse equation is derived similarly. 3.8. A ruler of length L is at rest in O with its left end at the origin. O moves from left
to right with speed β relative to O0 along the length of the ruler. The two origins coincide
at time zero for both, at which time a photon is emitted toward the other end of the ruler.
What are the coordinates in O0 of the event at which the photon reaches the other end?
Solution: The event has coordinates c(L/c), L = (L, L) in O. Therefore, in O0 , it has
coordinates
s
1+β
ct0 = γ(L + βL) =
L
1−β
s
1+β
L,
x0 = γ(L + βL) =
1−β
verifying that light travels with the same speed in both RFs.
There is another way of obtaining the same answer not using Lorentz transformation.
The length of the rod in O0 is L/γ. As the light catches up with the other end of the rod
in time t0 , the rod moves a distance of βct0 . Therefore,
s
p
2L
L
1
−
β
1+β
L
ct0 = + βct0 ⇐⇒ (1 − β)ct0 =
⇐⇒ ct0 =
=
L.
γ
γ
1−β
1−β
The location of the event is where the leading end of the rod is when the light catches up
with it. The leading end was at L/γ at t0 = 0 and it moved a distance of βct0 , so
s
L
1+β
x0 = + βct0 = ct0 =
L.
γ
1−β
I hope you appreciate the power of Lorentz transformation. Once you identify the coordinates of an event, Lorentz transformation takes care of the rest!
3.9. Start with the inverse LT and derive the time dilation (corresponding to ∆x = 0) and
length contraction (corresponding to ∆t0 = 0) formulas.
Solution: With
∆x = γ(∆x0 − βc∆t0 ),
c∆t = γ(c∆t0 − β∆x0 ),
∆t0 = 0 immediately gives the length contraction formula:
∆x = γ∆x0 ⇐⇒ ∆x0 =
remembering that ∆x is moving in O0 .
∆x p
= 1 − β 2 ∆x,
γ
(3.1)
23
For time dilation, we have to set ∆x = 0 and note that then ∆t is the proper time ∆τ .
Then, the first equation in (3.1) gives ∆x0 = βc∆t0 , and the second equation yields
c∆τ = γ(c∆t0 − cβ 2 ∆t0 ) =
c∆t0
⇐⇒ ∆t0 = γ∆τ.
γ
Alternatively, you can set ∆x0 = 0 and note that in that case ∆t0 is the proper time. Then
the second equation in (3.1) yields the answer immediately.
3.10. Two rockets A and B of rest length L0 travel towards each other along their x-axes
with relative speed β.1 According to A, when the tail of B passes the nose of A, a missile
is fired from the tail of A towards B. It will clearly miss due to length contraction of B as
seen by A. But B will see the length of A contracted. So, when the nose of A coincides
with the tail of B, the tail of A is in the middle of B’s rocket and it will hit B. Who is
right? Consider five events: nose of A passes nose of B (call this the origin of spacetime
for both), tail of B passes nose of A, missile is fired, nose of B passes tail of A, and tail of
B passes tail of A. Find the coordinates of these five events in both frames to see who is
right.
Solution: Label the events E1 through E5 . Assume that A has its nose at x = 0 and its
tail at x = L0 , and that B is moving in its positive x direction. Then, considering that the
length of B is L0 /γ for A, these events have the following coordinates according to A:
L0 /γ
L0 /γ
E1 : (0, 0), E2 : 0,
, E3 : L0 ,
,
β
β
L0
L0 L0 /γ
E4 : L0 ,
, , E5 : L0 ,
+
.
β
β
β
A moves in the negative x direction of B, and if the two noses meet at (0, 0), then the tail
of B must be at x = −L0 according to B. Obviously, E1 has the same coordinates in B as
in A. The coordinates of E2 in B are
L0
x2 = γ 0 − β
= −L0
γβ
L0
L0
ct2 = γ
−0 =
.
γβ
β
Of course, we could have guessed this without any equations! But, it is a good idea to use
Lorentz transformation to check our calculations. The coordinates of E3 in B are
L0
= (γ − 1)L0
x3 = γ L0 − β
γβ
L0
L0
ct3 = γ
− βL0 =
− γβL0 .
γβ
β
Thus, t3 < t2 , meaning that the missile is fired before the nose of A passes the tail of B.
The location of the firing of the missile is x3 > 0, which is outside the range of B (remember
1
Assume that their path is slightly shifted in the y-direction so they do not collide.
24
Lorentz Transformation
that the nose of B is at the origin and its tail at −L0 ). So, according to B, the missile
misses the rocket. The coordinates of E4 in B are
L0
x4 = γ L0 − β
=0
β
γL0
L0
L0
− βL0 =
− γβL0 =
,
ct4 = γ
β
β
γβ
and finally the coordinates of E5 in B are
L0
L0
x5 = γ L0 − β
= −L0
+
β
γβ
L0
L0
L0
L0
L0
L0
ct5 = γ
=
+
− βL0 = γ
+
+
2
β
γβ
γ β γβ
γβ
β
Note that the location of E5 is at the tail of either rocket, and the time of occurrence is the
same for both. This makes sense by the symmetry of the problem. Note also that event
E4 for B is the analogue of E2 for A. In both events, the tail of the moving rocket passes
the nose of the stationary one. Since the length of the moving rocket is L0 /γ and its speed
is β, it takes L0 /βγ for the length of the moving rocket to pass the nose of the stationary
one.
Alternative solution: We can avoid using length contraction, as I have done above.
Designate the coordinates of A with primes. Assuming that A has its nose at x0 = 0 and
its tail at x0 = L0 , B must have its nose at x = 0 and its tail at x = −L0 . Note that when
the primed coordinates are on the left-hand side of a Lorentz transformation, β is positive,
and when they are on the right-hand side, β is negative. Now assign coordinates to the five
events in the two RFs, entering as much info as you have for each event. You should be
able to get the following set of coordinates:
E1
E2
E3
E4
E5
For A
(0, 0)
(0, t02 )
(L0 , t02 )
(L0 , L0 /β)
(L0 , t05 )
For B
(0, 0)
(−L0 , L0 /β)
(x3 , t3 )
(0, t4 )
(−L0 , t5 )
I let you do the Lorentz transforming of these fives events and find the coordinates of all
of them in both RFs.
3.11. This problem is the continuation of Example 3.4.4. It is important for you to remember that LT applies only to events. The coincidence of Sonya’s clock with C1 is an event.
But what is the event corresponding to the reading of C2 ? Note 3.4.3 gives the answer:
There is an event E0 at the location of C2 with coordinates (L, 0) in Sam’s RF.
(a) What are the coordinates of E0 in Sonya’s RF? Note that although C2 is at the 12
o’clock position for Sam, it is not for Sonya!
(b) How long does it take her to reach C2 ? Warning: The length Sonya has to travel to
reach C2 is not L/γ, because when Sonya is at C1 , C2 is not at x = L/γ for her!
(c) What time does Sonya’s clock show when she reaches C2 ? This should be L/(cβγ) as
in Example 3.4.4.
25
(d) What are the coordinates (in her RF) of the event at which Sonya reaches C2 ?
(e) Using LT, show that the time coordinate of the same event in Sam’s RF is L/(cβ).
Solution:
(a)
x0 = γ(L − 0) = γL,
ct0 = γ(0 − βL) = −γβL.
(b) The event is at x0 and she is going with speed β toward it. So,
∆t =
|x0 |
γL
=
.
cβ
cβ
(c) The time of Sonya’s arrival at C2 is
ct2 = ct0 + c∆t = −γβL +
γL
γL
L
=
(1 − β 2 ) =
.
β
β
γβ
(d) Sonya is at her origin; and at event E2 , she is also at C2 . So, x2 = 0. Therefore, by
(c), (x2 , ct2 ) = (0, L/γβ).
(e)
x02 = γ(0 + βL/γβ) = L,
ct02 = γ(L/γβ + 0) = L/β.
This agrees with what I found directly in the example.
3.12. To gain more insight into the clock comparisons of Example 3.4.4 and Problem 3.11,
consider the synchronization process of the two clocks C1 and C2 in Sam’s RF as seen by
Sonya. To synchronize the clocks, let two light signals be sent simultaneously to the two
clocks from the midpoint between them. Call this event E.
(a) Show that E has coordinates (L/2, −L/2) in Sam’s RF.
(b) Show that E has coordinates
s
x=
1+βL
,
1−β 2
s
ct = −
1+βL
1−β 2
in Sonya’s RF.
(c) How long does it take this signal to arrive at the (common) origin? What is the time
at which it arrives at the origin? Is that what you expect?
(d) What is the location of C2 relative to Sonya? What is the distance between E and
C2 as measured by Sonya? Show that it takes γ(1 + β)L/2c for the signal to cover
this distance.
(e) From (b) and (d), show that the time of arrival of the signal at C2 is −γβL/c according
to Sonya.
26
Lorentz Transformation
(f) From (d), show that it takes Sonya γL/(cβ) to reach C2 . Therefore, her time of arrival
at C2 is L/(cγβ) after the arrival of the light signal at C2 , as in Example 3.4.4.
(g) Using (f) and LT show that the time shown on C2 is L/(cβ).
Solution:
(a) The event E is equidistant from the clocks; so it has to have x0 = L/2. In order for
the light signals to arrive at the two clocks at t0 = 0, they must have been emitted at
−L/2c.
(b) You have to use the inverse Lorentz transformation with β → −β
s
L
L
1+βL
L
x=γ
= γ(1 + β) =
−β −
2
2
2
1−β 2
s
1+βL
L
L
L
ct = γ − − β
= −γ(1 + β) = −
.
2
2
2
1−β 2
(c) The distance is x; so it takes |x|/c for the light to arrive at the origins. Since it was
emitted at t, it arrives at the origin at (|x|/c) + t = 0, as expected because the origin
of time is the same for Sam and Sonya: if the signal arrives at t0 = 0 according to
Sam, it arrives at t = 0 according to Sonya.
(d) Recall that C2 has coordinates (L, 0) in Sam’s RF. So, its location is x2 = γ(L − 0)
in Sonya’s RF. The distance between E and C2 is
|x − x2 | = γ(1 + β)
L
L
− γL = γ(1 − β) ,
2
2
and light covers it in γ(1 − β)L/2c.
(e) The time of arrival is
t+
L
L
L
|x − x2 |
= −γ(1 + β) + γ(1 − β) = −γβ .
c
2c
2c
c
(f) The arrival of signal at C2 sets that clock. In other words, according to Sonya’s clock,
the time at C2 is −γβL/c when the clock at the origin reads 0. So, when Sonya
arrives at C2 after x2 /cβ = γL/cβ, her clock reads
−
γβL γL
γL
L
+
=
(1 − β 2 ) =
.
c
cβ
cβ
cβγ
Therefore the event of her arrival at C2 has coordinates (0, L/βγ) in her RF.
(g) Lorentz transform the event of Sonya’s arrival to Sam’s RF:
L
0
xarrive = γ 0 + β
=L
βγ
L
L
ct0arrive = γ
+0 = .
βγ
β
The second equation is the time of arrival of Sonya according to Sam, and therefore
what C2 actually reads.
27
3.13. Obtain Equation (2.16) using Lorentz transformation.
Solution: In the rest frame of the rod, where the rod is assumed to lie along the z-axis,
the emission event has coordinates (0, −|b − z|). Lorentz transforming this, you get
x0 = γ(0 − β|b − z|) = −γβ|b − z| = −γβ(b − z 0 )
because b > z and z = z 0 .
3.14. Generalize the discussion of Example 3.4.5 to any clock length L and any speed β
by showing that
c∆t021 = γL(1 + β), and c∆t032 = γL(1 − β).
Solution: In the rest frame of the MM clock, events E1 , E2 , and E3 have coordinates
(0, 0), (L, L), and (0, 2L), respectively. If the clock moves relative to O0 in the positive x
direction with speed β, then
c∆t021 ≡ c(t02 − t01 ) = γ(L + βL) = γL(1 + β)
and
c∆t032 ≡ c(t03 − t03 ) = γ[(2L − L) + β(0 − L)] = γL(1 − β).
Therefore,
c∆t0 = c∆t021 + c∆t032 = 2γL,
and
∆t0 = γ
2L
= γ∆τ.
c
3.15. Reorient the rod of Example 3.4.6 so that now it lies along the y-axis. Let t be the
event at which the light from an arbitrary point of the rod is emitted in such a way that it
reaches P0 of Figure 3.4 at t = 0.
(a) Find ct in terms of y and b.
(b) Find the x0 -coordinate of the event according to O0 , with respect to which the rod is
moving.
(c) Show that the locus of source points whose light rays are collected simultaneously at
the pinhole is a hyperbola in the x0 y 0 -plane as in Example 2.2.5.
(d) Do you expect the rays from the two ends of the rod to have been emitted simultaneously according to O? According to O0 ?
(e) Find tA , tB , t0A , and t0B , the times of occurrence of the emission of the light rays from
the two ends of the rod according to the two observers.
Solution:
(a) In the rest frame
p of the rod the location
p of the event is (0, y, 0). So, its distance from
the pinhole is y 2 + b2 , and ct = − y 2 + b2 .
28
Lorentz Transformation
p
(b) The spacetime coordinates of the event is (0, − y 2 + b2 ). So, in O0 , we have
x0 = γ(0 − β
p
p
p
x0
y 2 + b2 ) = −γβ y 0 2 + b2 ⇐⇒
= − y 0 2 + b2
γβ
because y 0 = y.
(c) Squaring both sides and dividing by b2 yields
x0 2
y0 2
−
= 1.
γ 2 β 2 b2
b2
(d) The two ends of the rod are equidistant from the pinhole in O. So, they must have
been emitted at the same time in O. Since the rod is perpendicular to the direction
of motion, simultaneity is not affected.
√
(e) ctA = ctB = − L2 + b2 . Therefore,
p
p
ct0A = γ(− L2 + b2 + 0) = −γ L2 + b2
p
p
ct0B = γ(− L2 + b2 + 0) = −γ L2 + b2 = ct0A
3.16. The Earth and Alpha Centauri are 4.3 light years apart. Ignore their relative motion.
Events A and B occur at t = 0 on Earth and at 1 year on Alpha Centauri, respectively.
(a) What is the time difference between the events according to an observer moving at
β = 0.98 from Earth to Alpha Centauri?
(b) What is the time difference between the events according to an observer moving at
β = 0.98 from Alpha Centauri to Earth?
(c) What is the speed of a spacecraft that makes the trip from Alpha Centauri to Earth
in 2.5 years according to the spacecraft clocks?
(d) What is the trip time in the Earth RF?
Solution:
(a) Let Earth be the origin O. Then A has coordinates (0, 0) and B has coordinates
(4.3, 1) in units of light years. The observer O0 , is moving in the positive x direction.
Therefore, the Earth is moving in the negative direction, and
ct0BA = γ[c(tB − tA ) − β(xB − xA )] = 5(1 − 4.3 × 0.98) = −16.15 light years.
Thus, t0BA = −16.15 years. The event on Alpha Centauri occurs before that on earth.
This switching of the order of events should not surprise you because the two events
are not causally connected: You can’t send a probe (even a laser beam) that can be
present at the two events, because such a probe must cover a distance of 4.3 light
years in one year.
29
(b) Just change the sign of β:
ct0BA = γ[c(tB − tA ) + β(xB − xA )] = 5(1 + 4.3 × 0.98) = 26.2 light years,
and t0BA = 26.2 years. The event on Alpha Centauri occurs after that on earth.
(c) The distance for the spacecraft is 4.3/γ; so the time is 4.3/(cγβ). Now you can find
the speed:
2.5 years =
p
4.3 years
4.3 light years
=
⇐⇒ 2.5γβ = 4.3 ⇐⇒ 2.5 γ 2 − 1 = 4.3,
cγβ
γβ
or γ = 1.99. Therefore, β = γβ/γ = (4.3/2.5)/1.99 = 0.865.
(d) The trip time is 4.3 light years/(0.865c) = 4.97 years.
3.17. All muons in a group move towards Earth with the same speed. After covering a
distance of 2911 m, half of them survive. Recall that for any decay process, N (t) = N0 e−t/τ ,
where τ is the mean life of the decaying particles as measured in their rest frame, and for
muons τ = 2.2 µs.
(a) What is the speed of the muons?
(b) What is the mean life of the muons in the Earth frame?
(c) According to the muons, how far did they travel?
(d) A spaceship is launched from Earth with a speed of 0.95c. What is the mean life of
the muons in the frame of the spaceship?
Solution:
(a) In the rest frame of the muons, the distance is L0 /γ. So, the time they cover that
distance is L0 /(cβγ). Therefore,
L0
L0
0.5 = exp −
⇐⇒ ln(0.5) = −
cβγτ
cβγτ
or
βγ =
p
γ2 − 1 = −
L0
= 6.36 ⇐⇒ γ = 6.44,
ln(0.5)cτ
and β = βγ/γ = 6.36/6.44 = 0.988.
(b) τ is the proper time. So, in the Earth frame
tEarth = γτ = 6.44 × 2.2 µs = 14.17 µs.
(c) 2911/6.44 = 452 m.
30
Lorentz Transformation
(d) The speed of the muons relative to the spaceship is
β0 =
and
β + βship
1.938
=
= 0.99969
1 + ββship
1.9386
1
γ0 = p
= 40.2.
1 − β0 2
Therefore,
tship = γ 0 τ = 40.2 × 2.2 µs = 88.4 µs.
You can also find the answer to (d) by using Lorentz transformation. In Earth’s RF,
∆x = 2911 m and ∆t = L0 /cβ = 9.82 µs. The spaceship is moving in the negative
x direction according to Earth. Therefore, Earth is moving in the positive direction of
spaceship, and thus,
c∆tship = γ(c∆t + β∆x) = 3.2(3 × 108 × 9.82 × 10−6 + 0.95 × 2911) = 18276.6 m
or ∆tship = 61 µs. This is a fraction of the mean life, because in their rest frame, the muons
cover this distance in 452/(0.988 × 3 × 108 ) = 1.525 µs, which is 0.693 of the mean time.
So, the muon mean life in the spaceship frame is 61/0.693 = 88 µs, as before.
3.18. A ruler moves past another ruler of rest length L and parallel to it with speed β to
the right. Two clocks are attached to the ends of the moving ruler and synchronized in
the RF of that ruler. When the left end of the moving ruler reaches the left end of the
stationary ruler, the left clock reads zero. When the right end of the moving ruler reaches
the right end of the stationary ruler, the right clock reads ∆t.
(a) Find the rest length of the moving ruler in terms of β, L, and ∆t.
(b) Relative to the stationary ruler, how long after the left coincidence does the right
coincidence occur?
Solution: Let (0, 0) be the coincidence of the two left ends in both frames. Let E be the
coincidence of the two right ends. Then in the moving frame, E has coordinates (L0 , c∆t),
with L0 to be determined.
(a) In the moving frame, the stationary ruler has length L/γ. Therefore, the distance
between its right end and the right end of the moving frame is L/γ − L0 , assuming
that ∆t > 0. Therefore,
L
L
− L0 = βc∆t ⇐⇒ L0 = − βc∆t
γ
γ
(b) The moving ruler O is moving in the positive x direction of the stationary ruler O0 .
Thus, Lorentz transforming the coordinates of E, we get
c∆t
L
0
c∆t = γ(c∆t + βL0 ) = γ c∆t + β
− βc∆t
=
+ βL.
γ
γ
It is also instructive to find the location of the event in O0 , although it is obviously
L:
L
0
x = γ(L0 + βc∆t) = γ
− βc∆t + βc∆t = L.
γ
31
We could find (a) by calculating in the stationary ruler’s frame O0 . The presence of a
clocks at an end of the moving ruler constitutes an event. One has coordinates (0, 0) in
both frames, the other E0 , has coordinates (L0 , 0) in the moving frame. Therefore, in the
stationary frame E0 has coordinates
x00 = γ(L0 + 0) = γL0 ,
ct00 = γ(0 + βL0 ) = γβL0 .
The coincidence of the right end occurs at ∆t0 = γ(∆t + βL0 ). Thus in the time interval
∆t0 − t00 the moving ruler covers a distance of L − x00 . Thus,
cβ(∆t0 − t00 ) = L − γL0 ⇐⇒ β(γc∆t + γβL0 − γβL0 ) = L − γL0 ,
or γL0 = L − γβc∆t, which is what we obtained before.
3.19. Sonya (observer O) is holding a long pole of length L in the middle as she runs with
speed β towards a barn. Sam (observer O0 ), standing in the barn, sees that the pole fits
exactly between the entrance and exit doors of the barn. On the other hand, Sonya sees her
pole—which is at rest relative to her—longer than the barn and concludes that there is no
way that the pole can fit in the barn. Who is right? To find out, let E1 be the event when
the front end of the pole exits the barn and E2 the event when the rear end of the pole
enters the barn. Write the general Lorentz transformation for these events, and answer the
following questions:
(a) What are x1 and x2 ?
(b) How does t01 compare to t02 ?
(c) How does t1 compare to t2 ? Conclude that Sam and Sonya are both right!
(d) What is x01 − x02 ? Is that what you expect?
Solution: Let (0, 0) be the coordinates of E1 in both frames.
(a) For both observers x1 = x01 = 0. For Sonya, E2 is on the negative x-axis and has
x2 = −L.
(b) For Sam the two ends coincide simultaneously. Therefore, t01 = t02 = 0.
(c) You don’t have to use the length contraction formula! Write the Lorentz transformation for E2 :
x02 = γ(x2 + βct2 ) = γ(−L + βct2 )
ct02 = γ(ct2 + βx2 ) = γ[ct2 + β(−L)].
With t02 = 0, the second equation gives ct2 = βL. So, E2 occurs after E1 for Sonya,
indicating that the pole is longer than the length of the barn.
(d) Substituting βL for ct2 in the first equation yields
L
x02 = γ(−L + β 2 L) = − .
γ
Hence, x01 − x02 = L/γ, which is the length contraction formula as expected.
32
Lorentz Transformation
3.20. Sam sees a firecracker explode at x = 1.5 km (event E1 ), then after 5 µs, another
firecracker explode at x = 100 m (event E2 ).
(a) What is the velocity relative to Sam of Sonya for whom the events occur at the same
place?
(b) Which event occurs first according to Sonya and what is the time interval between
the explosions for her?
Solution: In order for the two events to occur at the same place for Sonya, she has to be
present at both events.
(a) Therefore, she has to cover the distance of 1400 m in 5 µs. Thus, her speed should be
v=
1400 m
2.8 × 108
8
=
2.8
×
10
m/s
⇐⇒
β
=
= 0.933
5 × 10−6 s
3 × 108
(b) Sonya has to move from the earlier event to the later event. So, she has to move in the
negative x direction of Sam. So, Sam is moving in the positive x direction of Sonya.
In Sam’s RF, the events E1 and E2 have coordinates (1.5 km, 0) and (100 m, 5 µs),
respectively. Obviously, E1 occurs first, and
c∆t21 = c(t2 − t1 ) = γ[3 × 108 × 5 × 10−6 + β(−1400)] m,
or c∆t21 = 539.8 m and ∆t21 = 1.8 µs. Note that Sonya measures proper time. So,
∆t21 is indeed 5 µs/γ.
3.21. Sonya is in the middle of a train car of length L moving relative to Sam with speed
β along Sam’s positive x0 -direction. Firecrackers A in the back and B in the front of the
car explode simultaneously according to Sam at the same time that Sonya passes him.
Let t = t0 = 0 be the time that Sam and Sonya pass each other, and assume that both
observers are at the origins of their coordinate systems. Without using length contraction
or time dilation formulas, find the coordinates of all the following events for both observers:
explosion of A, explosion of B, reception of light from A by Sam, reception of light from B
by Sam, reception of light from A by Sonya, reception of light from B by Sonya.
Solution: Label the events E1 through E6 . Let Sonya be observer O and Sam O0 . Plug in
all the info for each event. So, in Sonya’s RF, the events E1 through E6 have the following
coordinates, respectively:
(−L/2, ct1 ),
(L/2, ct2 ),
(x3 , ct3 ),
(x4 , ct4 ),
(0, ct1 + L/2),
(0, ct2 + L/2),
and in Sam’s RF, they have the following coordinates:
(x01 , 0),
(x02 , 0),
(0, |x01 |),
(0, |x02 |),
(x05 , ct05 ),
(x06 , ct06 ).
Now write the Lorentz transformation for each event and solve for the unknowns. For E1 ,
we have
x01 = γ(−L/2 + βct1 ), 0 = γ(ct1 − βL/2).
33
These two equations give ct1 = βL/2 and x01 = −L/(2γ). For E2 , we have
x02 = γ(L/2 + βct2 ),
0 = γ(ct2 + βL/2).
These two equations give ct2 = −βL/2 and x02 = L/(2γ). Note that x02 − x01 = L/γ,
indicating the contraction of the length of the train car for Sam.
The coordinates of E3 and E4 are the same in Sam’s RF. They are both 0, L/(2γ) .
So, they must also be the same in Sonya’s RF:
x3 = x4 = γ 0 − βL/(2γ) = −βL/2
ct3 = ct4 = γ L/(2γ) − 0 = L/2.
It is important to note that the equality of t3 and t4 does not mean that Sonya receives the
signals simultaneously! It means that Sam receives the two signals simultaneously at L/2c
according to Sonya. Sonya can reason as follows to get this answer: “I know that at my
zero time I passed Sam and at the same time according to him, the two firecrackers, which
I know are at ±L/2 exploded. Therefore, Sam must have gotten the signals, according to
me at L/2c after I passed him.
With ct1 and ct2 determined,
the coordinates of E5 and E6 can
be written in O. E5 has
coordinate 0, (1 + β)L/2 and E6 has coordinate 0, (1 − β)L/2 . Lorentz transforming to
O0 , we get
x05 = γ 0 + β(1 + β)L/2 = γβ(1 + β)L/2
ct05 = γ (1 + β)L/2 + 0 = γ(1 + β)L/2
and
x06 = γ 0 + β(1 − β)L/2 = γβ(1 − β)L/2
ct06 = γ (1 − β)L/2 + 0 = γ(1 − β)L/2.
3.22. Two spaceships of rest length L0 are approaching the Earth from opposite directions
at velocities ±0.8c. How long does one of them appear to the other?
Solution: One spaceship moves relative to the other with speed
β0 =
0.8 + 0.8
= 0.9756,
1 + 0.82
and the length contraction formula gives
p
L = L0 1 − β 0 2 = 0.2195L0 .
3.23. Spaceship A is twice as long as spaceship B when they are at rest. Spaceship B is
moving at three quarters the speed of light relative to Earth. As A overtakes B, an Earth
observer notices that they both have the same length.
(a) How fast is A moving?
34
Lorentz Transformation
(b) What is the length of A relative to B?
(c) What is the length of B relative to A?
Solution:
(a) Let L0 be the rest length of B. Then
q
p
2
2 =L
2
2L0 1 − βA
0 1 − 0.75 ⇐⇒ 4(1 − βA ) = 0.4375 ⇐⇒ βA = 0.9437.
(b) The relative speed of the two spaceships is
βAB =
0.9437 − 0.75
= 0.663.
1 − 0.9437 × 0.75
Therefore,
LAB = 2L0
p
1 − 0.6632 = 1.5L0 .
(c)
LBA = L0
p
1 − 0.6632 = 0.75L0 .
So, A says that the length of B is 37.5% of his rest length and B say that the length of A
is 1.5 times her rest length.
3.24. Sam and Pat move in the same direction at 0.8c and 0.6c relative to Earth, respectively.
(a) How fast should Sonya move relative to Earth so that she observes Sam and Pat
approaching her at the same speed?
(b) What is the speed of Sam (or Pat) relative to Sonya?
Solution:
(a) Sonya must move away from Sam and toward Pat. Let her speed relative to Earth
be β. Then
0.8 − β
0.6 + β
=
⇐⇒ β 2 − 5.2β + 1 = 0,
1 − 0.8β
1 + 0.6β
with solutions
β = 2.6 ±
p
√
2.62 − 1 ⇐⇒ β = 2.6 − 5.76 = 0.2,
because the positive choice gives β > 1.
(b) Her speed relative to Sam is
βss =
0.8 − 0.2
= 0.714
1 − 0.8 × 0.2
βsp =
0.6 + 0.2
= 0.714.
1 + 0.6 × 0.2
and relative to Pat it is
35
3.25. Sonya (observer O) is approaching Sam (observer O0 ) with speed β from the negative
values of his x0 -axis. Sonya has a source of light of wavelength λ sending signals to Sam.
Consider two events: E1 , the source sends a wave crest, and E2 , the source sends the next
wave crest. So, t2 − t1 is the period of the wave according to Sonya.
(a) Write the Lorentz transformation for these two events assuming that Sonya (at her
origin) is holding the source of light.
(b) What is the period according to Sam? Hint: It is not t02 − t01 !
(c) Show that λ0 , the wavelength as measured by Sam is related to λ via the following
formula:
s
1−β
λ0 =
λ.
1+β
Solution: Let (0, 0) be the origin of both coordinate systems. So, the two events have
negative time coordinates. Let E1 and E2 have coordinates (0, ct1 ) and (0, ct2 ), with t1 <
t2 < 0.
(a) According to Sam, E1 and E2 have coordinates
x01 = γ(0 + βct1 ) = γβct1
ct01 = γ(ct1 + 0) = γct1
and
x02 = γ(0 + βct2 ) = γβct2
ct01 = γ(ct2 + 0) = γct2 .
(b) Sam receives the first wave front |x01 |/c after it was emitted. Thus
0
0
ctrec
1 = ct1 + |x1 | = γct1 + γβc|t1 | = γct1 − γβct1 = (1 − β)γct1 ,
because t1 < 0. Similarly
0
0
ctrec
2 = ct2 + |x2 | = γct2 + γβc|t2 | = γct2 − γβct2 = (1 − β)γct2 .
So, the difference in the time of reception of the two wave fronts for Sam is
rec
ctrec
2 − ct1 = (1 − β)γct2 − (1 − β)γct1 = γ(1 − β)c(t2 − t1 ) = γ(1 − β)cT.
rec
0
Now, trec
2 − t1 is the period T according to Sam, which can also be written as
s
1
−
β
1−β
T =
T.
T0 = p
1+β
1 − β2
λ0 = cT 0 and λ = cT should now give the relation between the wavelengths.
36
Lorentz Transformation
3.26. Sonya (observer O) is moving away from Sam (observer O0 ) with speed β along his
positive x0 -axis. At time t0 , Sam sends a light signal to Sonya (event E), which gets reflected
from a reflector at Sonya’s origin (event Eref ), and received by a detector at Sam’s origin
(event Edet ).
(a) Find (x, t), the coordinates of E according to Sonya in terms of t0 .
(b) Find tref , the time of Eref according to Sonya in terms of t0 .
(c) Find (x0ref , t0ref ), the coordinates of Eref according to Sam in terms of t0 .
(d) Show that
t0det =
1+β 0
t
1−β
Solution: Event E has coordinates (0, t0 ) in Sam’s RF, and Sam is moving in the negative
x direction of Sonya.
(a)
x = γ(0 − βct0 ) = −γβct0 ,
t = γ(ct0 − 0) = γct0
(b) The reflection occurs immediately after the signal hits the reflector. Thus, tref = t =
γct0 . Therefore, Eref has coordinates (0, γct0 ) according to Sonya.
(c)
x0ref = γ(0 + βctref ) = γ 2 βct0
ct0ref = γ(ctref + 0) = γ 2 ct0
(d)
ct0det = ct0ref + |x0ref | = γ 2 ct0 + γ 2 βct0 =
1+β 0 1+β 0
ct =
ct
1 − β2
1−β
3.27. Speeder O is in a spacecraft moving away from a policeman (observer O0 ) with speed
β. The policeman sends an EM signal of wavelength λ to O and receives the reflected wave
of wavelength λref .
(a) Use the result of Problem 3.26 for the emission two successive wave crests to show
that
1+β
λ.
λref =
1−β
(b) The speed limit in Metropolis is half the speed of light. Will the speeder get a ticket
if the policeman sends a violet signal of 400 nm and receives an infrared signal of
1.3 µm?
Solution:
(a) The result is immediate.
37
(b) With λ = 400 nm and λref = 1300 nm, we get
1300 =
1+β
1+β
13
9
400 ⇐⇒
=
⇐⇒ β =
> 0.5.
1−β
1−β
4
17
3.28. Sam is on a train of rest length L0 moving at speed β relative to Sonya standing on
the ground. Consider the time interval between the front of the train coinciding with Sonya
and the back of the train coinciding with her.
(a) Find this time interval in Sonya’s frame by calculating in her frame.
(b) Find this time interval in Sonya’s frame by calculating in Sam’s frame.
(c) Find the time interval in Sam’s frame by calculating in Sonya’s frame.
(d) Find the time interval in Sam’s frame by calculating in his frame.
Solution: Let ∆t0 be the time interval according to Sonya and ∆t according to Sam.
(a) The train’s length is L0 /γ according to Sonya and moves at cβ. So, she calculates
c∆t0 to be (L0 /γ)/β.
(b) For Sam the time interval between coincidences is c∆t = L0 /β. He concludes that
since Sonya is measuring proper time, for her the time interval is c∆t0 = (L0 /β)/γ.
(c) Sonya know that she is calculating proper time. So, she concludes that Sam measures
the time interval to be c∆t = γc∆t0 or [(L0 /γ)/β]γ = L0 /β.
(d) Sam sees Sonya move from the front to back with speed cβ. So, he measures the time
to be c∆t = L0 /β.
It is instructive to use Lorentz transformation to find the answers. The two events according
to Sonya are (0, 0) and (0, ∆t0 ) and according to Sam, they are (0, 0) and (−L0 , ∆t). Write
the Lorentz transformation for the second event:
0 = ∆x0 = γ(−L0 + βc∆t),
c∆t0 = γ(c∆t − βL0 ).
The first equation gives c∆t = L0 /β and plugging this in the second equation yields
γL0
L0
L0
0
c∆t = γ
− βL0 =
(1 − β 2 ) =
.
β
β
γβ
This should show you once again the power of Lorentz transformation. Once you specify
your events, you don’t have to second guess. Lorentz transformation will take care of the
rest!
3.29. Sam and Sonya sit in two identical rockets of length L. Sam approaches Sonya from
behind with relative speed β along the length of their rockets. Assume that at time zero
for both, the nose of Sam’s rocket coincides with the tail of Sonya’s. Now consider three
evens occurring in succession: Event A is when the tail of Sam’s rocket coincides with the
tail of Sonya’s; B is when the nose of Sam’s rocket coincides with the nose of Sonya’s; C
is when the tail of Sam’s rocket coincides with the nose of Sonya’s. Using only Lorentz
transformation and inverse Lorentz transformation (no length contraction or time dilation)
find the coordinates of these three events in Sam’s and Sonya’s RFs paying attention to
signs.
38
Lorentz Transformation
Solution: Let Sam be O and Sonya O0 . Since the coincidence of the nose of Sam’s rocket
occurs at the tail of Sonya’s at t = t0 = 0, it is convenient to have Sam’s space origin at
the nose and Sonya’s space origin at the tail. Write all the coordinates in the two frames
inserting all the info given. Then in O, events A, B, and C have coordinates
(−L, ctA ),
(0, ctB ),
(−L, ctC )
and in O0 they have coordinates
(0, ct0A ),
(L, ct0B ),
(L, ct0C ).
Now Lorentz transform the events. For A, we get
0 = x0A = γ(−L + βctA ),
ct0A = γ(ctA − βL).
The first equation gives ctA = L/β, and pugging this in the second equation yields
L
γL
L
ct0A = γ
− βL =
(1 − β 2 ) =
.
β
β
γβ
For B, we get
L = γ(0 + βctB ),
ct0B = γ(ctB − 0).
Finding ctB from first and substituting in the second, we get ctB = L/(γβ) and ct0B = L/β.
Finally, for C, we get
L = γ(−L + βctC ),
ct0C = γ(ctC − βL).
These equations yield
(1 + γ)L
ctC =
,
γβ
ct0C
(1 + γ)L
=γ
− βL .
γβ
The second equation can be simplified:
ct0C =
(1 + γ)L
(1 + γ)L − γβ 2 L
L + γL(1 − β 2 )
(1 + γ)L
− γβL =
=
=
.
β
β
β
γβ
The equality t0C = tC should come as no surprise because of the symmetry of the problem:
They both see nose-tail to tail-nose coincidences during the entire motion (one in reverse
order). You should convince yourself that the results make sense based on length contraction
and time dilation.
3.30. Sam is on a train of rest length L0 moving at speed β relative to Sonya standing on
the ground. He fires a gun from the back of the train to the front. The speed of the bullet
relative to the train is βb .
(a) How much time does the bullet spend in the air before hitting the front of the train
according to Sonya?
(b) What is the distance that the bullet travels according to Sonya?
Solution: I’ll do the problem in two ways. First I use length contraction and the law of
addition of velocities. Then I use Lorentz transformation to show you its power!
39
(a) According to Sonya, the length of the train is L0 /γ and the speed of the bullet is
βb0 = (βb + β)/(1 + βb β). The bullet is catching up with the front of the train which
is moving away from it with speed β, all according to Sonya. So,
L0
L0
+ cβ∆t = cβb0 ∆t ⇐⇒ c∆t =
.
γ
γ(βb0 − β)
But
βb0 − β =
βb + β
βb + β − β − β 2 βb
βb
−β =
= 2
.
1 + βb β
1 + βb β
γ (1 + βb β)
Therefore,
c∆t =
L0 γ(1 + βb β)
L0
=
.
0
γ(βb − β)
βb
(b)
∆x =
cβb0 ∆t
=c
βb + β
1 + βb β
L0 γ(1 + βb β)
βb
=
L0 γ(βb + β)
.
βb
Now let’s do the problem using Lorentz transformation. In Sam’s RF, the event of the bullet
hitting the front of the train has coordinates (L0 , L0 /βb ). Lorentz transforming these, we
get
∆x = γ(L0 + βL0 /βb ), c∆t = γ(L0 /βb + βL0 ).
And that’s it!
3.31. Sam and Sonya are newborn twins. Sam is placed on a rocket that moves at speed
0.99c to a star system 25 light years away. At the moment of his departure a light signal is
sent to the star system and gets reflected.
(a) How old is Sam when he receives the reflected signal? How old is Sonya?
(b) How old is Sonya when she receives the reflected signal? How old is Sam?
(c) How old is Sam when he reaches the star system?
(d) How old is Sonya when Sam reaches the star system?
Hint: See Example 3.4.9.
Solution: The hint makes this just a plug and chug problem! But you should do it anyway
to get a feel for the numbers.
3.32. Multiply both sides of the inequality βb < 1 by the positive quantity 1 − β and show
that β + βb < 1 + ββb . From this conclude that the relativistic law of addition of velocities
never yields a speed larger than the speed of light.
Solution:
βb (1 − β) < 1 − β ⇐⇒ βb − βb β < 1 − β ⇐⇒ βb + β < 1 + βb β.
40
Lorentz Transformation
3.33. A super-ball has the property that when it hits a wall with a given speed relative
to the wall, it bounces back with the same speed in the opposite direction. What do you
measure the speed of the bounced ball to be if you throw it at speed βs towards a wall
which is moving towards you at speed βw ?
Solution: The speed of the super-ball relative to the wall is
βsw =
βs + βw
.
1 + βs βw
Therefore, it bounces back relative to the wall with speed βsw . To find the speed βs0 of the
bounced super-ball relative to you, you have to add the speeds again:
βs0 =
2
βsw + βw
βs + βw + βw (1 + βs βw )
βs + 2βw + βs βw
=
.
=
2
1 + βsw βw
1 + βs βw + (βs + βw )βw
1 + 2βs βw + βw
It is interesting to note that βs0 → 1 when βs → 1, regardless of the speed of the wall! This
is, of course, consistent with the second postulate of relativity.
3.34. In an intergalactic race, team A is moving at speed 0.8c relative to the finish line.
They notice that a faster team B passes them at 0.9c. Team B observes another team C
to pass them at 0.95c. What are the speeds of teams B and C relative to the finish line?
Solution:
0.8 + 0.9
1.7
=
= 0.9884
1 + 0.8 × 0.9
1.72
1.9384
0.9884 + 0.95
=
= 0.9997.
βC =
1 + 0.9884 × 0.95
1.939
βB =
3.35. Sam sees Sonya and Pat flying in opposite directions with a speed β = 0.995. What
is the speed of Pat relative to Sonya?
Solution:
βps =
0.995 + 0.995
= 0.999987.
1 + 0.995 × 0.995
3.36. Two spaceships approach each other with relative speed 0.9c. What are the velocities
of the spaceships relative to Earth assuming that they move with the same speed relative
to Earth?
Solution: Let βAB be the speed of spaceship A relative to spaceship B. Let βAE be the
speed of A relative to Earth and βEB the speed of Earth relative to B. Then the law of
addition of velocities can be written as
βAE + βEB
.
βAB =
1 + βAE βEB
Note the order of subscripts. Assume that A is moving in the positive direction relative to
Earth. Then B is moving in the negative direction. This means that βBE < 0. Therefore,
βEB = −βBE > 0. Furthermore, by assumption, βAE = βEB . So, the above equation
becomes
√
2βAE
1 ± 1 − 0.92
2
0.9 =
⇐⇒ 0.9βAE − 2βAE + 0.9 = 0 ⇐⇒ βAE =
2
0.9
1 + βAE
or βAE = 0.627.
41
3.37. Observer O moves in the positive x0 -direction of observer O0 with speed β. Observer
O0 moves in the positive x00 -direction of observer O00 with speed β 0 . Use the relativistic law
of addition of velocities (3.26) to obtain γ 00 in terms of β, β 0 , γ, and γ 0 .
Solution: Substitute
β 00 =
β + β0
1 + ββ 0
in the definition of γ 00 :
γ 00 = r
1−
=p
1
β+β 0
1+ββ 0
1 + ββ 0
2 = p(1 + ββ 0 )2 − (β + β 0 )2
1 + ββ 0
(1 − β 2 )(1 − β 0 2 )
= γγ 0 (1 + ββ 0 ).
3.38. All motions are in the positive x-direction. Observer O1 moves relative to observer O
with speed β1 . Observer O2 moves relative to O1 with speed β2 . Let P2+ ≡ (1 + β1 )(1 + β2 )
and P2− ≡ (1 − β1 )(1 − β2 ).
(a) Show that O2 moves relative to O with speed β20 given by
β20 =
P2+ − P2−
.
P2+ + P2−
(b) Now introduce another observer O3 moving relative to O2 with speed β3 , and let
P3+ = (1 + β1 )(1 + β2 )(1 + β3 )
P3− = (1 − β1 )(1 − β2 )(1 − β3 ).
Use the relativistic law of addition of velocities for β20 and β3 in the form given in (a)
to prove that O3 moves relative to O with speed β30 given by
β30 =
P3+ − P3−
.
P3+ + P3−
(c) Can you predict what happens if you introduce a fourth observer?
(d) If you are familiar with mathematical induction, prove the formula for N observers.
Solution:
(a)
P2+ − P2−
(1 + β1 )(1 + β2 ) − (1 − β1 )(1 − β2 )
+
− = (1 + β )(1 + β ) + (1 − β )(1 − β )
P2 + P2
1
2
1
2
2(β1 + β2 )
=
= β20 .
2(1 + β1 β2 )
42
Lorentz Transformation
(b)
P2+ − P2−
+ β3
P2+ − P2− + (P2+ + P2− )β3
β20 + β3
P2+ + P2−
0
=
=
β3 =
1 + β20 β3
P + − P2−
P2+ + P2− + (P2+ − P2− )β3
1 + 2+
− β3
P2 + P2
=
P2+ (1 + β3 ) − P2− (1 − β3 )
P3+ − P3−
=
P2+ (1 + β3 ) + P2− (1 − β3 )
P3+ + P3−
(c)
P3+ − P3−
+ β4
P3+ − P3− + (P3+ + P3− )β4
β30 + β4
P3+ + P3−
0
=
=
β4 =
1 + β30 β4
P + − P3−
P3+ + P3− + (P3+ − P3− )β4
β
1 + 3+
4
P3 + P3−
=
P3+ (1 + β4 ) − P3− (1 − β4 )
P4+ − P4−
=
P3+ (1 + β4 ) + P3− (1 − β4 )
P4+ + P4−
(d) In proof by mathematical induction, you show that the equality holds for some particular value, usually for a low value like N = 1 or N = 2. Then assume that it holds
for N − 1 and prove that it holds for N . We have seen that the equation holds for
N = 2. Now assume that the equality holds for N − 1, i.e., assume that the following
is true:
PN+−1 − PN−−1
0
.
βN −1 = +
PN −1 + PN−−1
Now we have to prove that
0
βN
=
PN+ − PN−
PN+ + PN−
is true. To do so, we use the law of addition of velocities:
PN+−1 − PN−−1
0
βN
+ βN
0
PN+−1 + PN−−1
PN+−1 − PN−−1 + (PN+−1 + PN−−1 )βN
βN
−1 + βN
=
=
=
0
1 + βN
PN+−1 − PN−−1
PN+−1 + PN−−1 + (PN+−1 − PN−−1 )βN
−1 βN
1+ +
βN
PN −1 + PN−−1
=
PN+−1 (1 + βN ) − PN−−1 (1 − βN )
PN+−1 (1 + βN ) + PN−−1 (1 − βN )
=
PN+ − PN−
PN+ + PN−
3.39. Recall that hyperbolic functions are defined by
cosh φ =
eφ + e−φ
,
2
sinh φ =
eφ − e−φ
.
2
43
(a) Using these definitions, show the following properties
cosh2 φ − sinh2 φ = 1
sinh(φ1 + φ2 ) = sinh φ1 cosh φ2 + sinh φ2 cosh φ1
cosh(φ1 + φ2 ) = cosh φ1 cosh φ2 + sinh φ1 sinh φ2
tanh φ1 + tanh φ2
tanh(φ1 + φ2 ) =
1 + tanh φ1 tanh φ2
(b) Now define the rapidity φ by tanh φ = β and show that
cosh φ = γ,
sinh φ = βγ,
φ
e =
1+β
1−β
1/2
.
Solution:
(a) From
e2φ + e−2φ + 2
e2 φ + e−2φ − 2
, sinh2 φ =
4
2
the first equality follows immediately. For the second equality, note that
cosh2 φ =
eφ = cosh φ + sinh φ,
e−φ = cosh φ − sinh φ
Therefore,
eφ1 +φ2 − e−φ1 −φ2
eφ1 eφ2 − e−φ1 e−φ2
=
2
2
(cosh φ1 + sinh φ1 )(cosh φ2 + sinh φ2 )
=
2
(cosh φ1 − sinh φ1 )(cosh φ2 − sinh φ2 )
−
.
2
sinh(φ1 + φ2 ) =
The identity now follows immediately. The identity involving cosh(φ1 + φ2 ) can be
derived similarly. For tanh(φ1 + φ2 ), divide both sides of the sinh(φ1 + φ2 ) and
cosh(φ1 + φ2 ) identities to get tanh(φ1 + φ2 ) on the left. Then divide both numerator
and the denominator on the right by cosh φ1 cosh φ2 .
(b) Divide both sides of cosh2 φ − sinh2 φ = 1 by cosh2 φ to get
1 − tanh2 φ =
1
1
⇐⇒ 1 − β 2 =
⇐⇒ cosh φ = γ.
2
cosh φ
cosh2 φ
Then,
β = tanh φ =
sinh φ
sinh φ
=
⇐⇒ sinh φ = βγ.
cosh φ
γ
Finally,
1+β
e = cosh φ + sinh φ = γ + βγ = p
=
1 − β2
φ
1+β
1−β
1/2
.
44
Lorentz Transformation
3.40. Show that in terms of rapidity of Problem 3.39, the LT can be written as
x0 = x cosh φ + ct sinh φ
ct0 = x sinh φ + ct cosh φ.
Solution: With γ = cosh φ and βγ = sinh φ, the result is immediate.
3.41. Redo Problem 3.37 using LTs in terms of rapidities. How is the rapidity φ00 of O00
relative to O related to the rapidity φ0 of O00 relative to O0 and the rapidity φ of O0 relative
to O? The identities in Problem 3.39 may be useful.
Solution: The law of addition of velocities can be written a
tanh φ00 =
tanh φ0 + tanh φ
= tanh(φ0 + φ),
1 + tanh φ0 tanh φ
i.e., rapidities add.
3.42. Derive the following relations among the coordinates of the RFs O and O0 :
x0 + ct0 = eφ (x + ct),
x0 − ct0 = e−φ (x − ct),
where φ is the rapidity as defined in Problem 3.39. Now define two new coordinates (ξ, η)
as
ξ = x + ct, η = x − ct,
and show that they are perpendicular to each other (in the Euclidean sense). Show that
under a LT, ξ and η do not change direction, but their calibration changes. By how much?
Solution: Add the two sides of the Lorentz transformation written in terms of rapidity.
On the left-hand side you get x0 + ct0 and on the right-hand side
x (cosh φ + sinh φ) +ct (sinh φ + cosh φ) = eφ (x + ct).
|
|
{z
}
{z
}
=eφ
=eφ
If you subtract, you get
x0 − ct0 = x(cosh φ − sinh φ) + ct(cosh φ − sinh φ) = e−φ (x − ct).
The coordinate ξ is defined by η = 0 or x = ct, which is the equation of the world line of
light moving in the positive x-direction. The coordinate η is defined by ξ = 0 or x = −ct,
which is the equation of the world line of light moving in the negative x-direction. And
these two world lines make a 90◦ with one another.
Under a Lorentz transformation, ξ → ξ 0 = eφ ξ, so its scale changes by a factor of eφ ,
and η → η 0 = e−φ η, so its scale changes by a factor of e−φ .
3.43. The speed limit in Metropolis is 0.8c. A driver is moving at 0.92c (with respect to
the ground) as he passes a policeman without noticing him. The policeman, accelerating
to 0.996c (also with respect to the ground) instantaneously, catches up with the driver 10
minutes later according to his own clock. The origins of the RFs of both observers are when
and where they pass each other.
(a) What are the coordinates (in the policeman’s RF) of the event at which the policeman
starts chasing the driver?
45
(b) What are the coordinates of that event in the driver’s RF?
(c) How long after the driver passes the policeman does the policeman catch up with him
according to the driver’s clock?
Solution:
(a) The coordinates are (0, 10 light minutes).
(b) In the driver’s RF, with β = −0.92, γ = 2.55,
x = 2.55(0 − 0.92 × 10) = 23.47 light minutes
ct = 2.55(×10 − 0) = 25.5 light minutes.
(c) When the policeman starts chasing the driver, he will be moving in the driver’s
positive direction with speed
β=
0.996 − 0.92
= 0.908.
1 − 0.996 × 0.92
According to the driver, the policeman is 23.47 light minutes away and approaching
him at 0.908c. So, it takes him
23.47 light minutes
= 25.85 minutes
0.908c
to catch up with the driver after he starts chasing him. So, 25.5 + 25.85 = 51.35
minutes after the driver passes the policeman, the policeman catches up with him.
3.44. The acceleration of a “bullet” in O0 is a0b = dvb0 /dt0 and in O, moving relative to
O0 with constant speed β in the positive direction of the x0 -axis is ab = dvb /dt. Use
(infinitesimal) LT and Equation (3.26) to show that
a0b =
ab
γ 3 (1 + ββb )3
Solution: The differential of Equation (3.26) is
dβb0 =
dβb
(1 + ββb )dβb − (βb + β)βdβb
= 2
,
(1 + ββb )2
γ (1 + ββb )2
and
cdt0 = γ(cdt + βdx) = γcdt(1 + ββb ) ⇐⇒ dt0 = γdt(1 + ββb ).
Therefore,
cdβb
0
2
cdβ
ab
γ (1 + ββb )2
b
a0b =
=
= 3
.
0
dt
γdt(1 + ββb )
γ (1 + ββb )3
3.45. In Problem 3.44, let βb0 = β, that is let O be moving with the bullet. Furthermore,
assume that the acceleration ab is constant. Call it a.
46
Lorentz Transformation
(a) Show that (ignoring the prime on t)
adt =
cdβ
.
(1 − β 2 )3/2
(b) Integrate the equation above and show that if β = 0 at t = 0, then
at
.
+ a2 t2
How long do you have to wait for the bullet to reach a speed of 0.99c if the acceleration
is 5 m/s2 ? If you wait long enough, can you accelerate the bullet to a speed larger
than c?
β=√
c2
(c) Ignoring the prime on x as well and noting that cβ = v = dx/dt, find x as a function
of time assuming that x = 0 at t = 0. What kind of a curve do you get on the
xt-plane?
(d) How long does it take the “bullet” to reach Alpha Centauri 4 light years away if the
acceleration of the bullet is 5 m/s2 ? What is its speed when it reaches there?
Solution: If βb0 = β, the βb = 0, and
ab
a
a0b = 3 ⇐⇒ cdβ/dt = 3 .
γ
γ
(a) Therefore,
adt = γ 3 cdβ ⇐⇒ adt =
cdβ
.
(1 − β 2 )3/2
(b) Integrate both sides and choose the constant of integration so that β = 0 at t = 0.
Then
at
cβ
c2 β 2
⇐⇒ β = √
.
at = p
⇐⇒ a2 t2 =
2
2
1−β
c + a2 t2
1 − β2
In units of c, the acceleration is
α ≡ a/c =
5 m/s2
= 1.67 × 10−8 s−1 = 0.526 yr−1 .
3 × 108 m/s
So, using the first equality in the preceding equation, we get
0.99
7.02
0.526 yr−1 = √
= 7.02 ⇐⇒ t =
= 13.35 yr.
2
0.526 yr−1
1 − 0.99
Also it is clear that β → 1 as t → ∞; so you cannot accelerate to the speed of light
in a finite time.
(c) From (b)
cat
catdt
cαtdt
dx
=√
⇐⇒ dx = √
=√
.
2
2
2
2
2
2
dt
c +a t
c +a t
1 + α2 t2
Integration and the condition that x = 0 at t = 0 yield
c p
x = ( 1 + (αt)2 − 1).
α
You can easily show that
αx
2
+ 1 − (αt)2 = 1,
c
which is a hyperbola.
47
(d) Note that
αx
α(4 light years)
=
= 0.526 yr−1 (4 years) = 2.1.
c
c
So, the last equation yields
3.12 − (αt)2 = 1, ⇐⇒ αt = 2.94 ⇐⇒ t =
2.94
= 5.59 yr.
0.526 yr−1
From (b), we have
β=√
αt
2.94
at
=p
=√
= 0.9467.
c2 + a2 t2
1 + 2.942
1 + (αt)2
CHAPTER
4
Spacetime Geometry
Problems With Solutions
4.1. When axes are perpendicular to each other, it does not matter whether you draw lines
that are parallel or perpendicular to the axes to find the coordinates of a point. Why can’t
you draw perpendicular lines when axes are not at right angles to each other? Hint: If a
point lies on an axis, what do you expect its “other” coordinate to be?
Solution: When you drop a perpendicular line from a point of one axis to another axis
it crosses the latter. That cannot happen because points on one axis should have zero
coordinates corresponding to the other axis.
4.2. Consider a non-perpendicular coordinate system with axes x and y. Take any two
points P1 and P2 with coordinates (x1 , y1 ) and (x2 , y2 ), respectively. Show that if
2
P1 P2 = (x2 − x1 )2 + (y2 − y1 )2 ,
then the axes must be perpendicular.
Solution: Figure 4.1 of the manual shows the two points and their coordinates in a nonperpendicular coordinate system. The law of cosines gives
2
P1 P2 = (x2 − x1 )2 + (y2 − y1 )2 + 2(x2 − x1 )(y2 − y1 ) cos θ.
Therefore,
2
P1 P2 = (x2 − x1 )2 + (y2 − y1 )2 ⇐⇒ (x2 − x1 )(y2 − y1 ) cos θ = 0 ⇐⇒ cos θ = 0,
because x2 6= x1 and y2 6= y1 .
4.3. Draw the x and ct axes with acute angles. Draw a wavy line through the origin to
represent the world line of a light signal. Show that if the speed of this light signal is to
be c, then its world line has to make equal angles with both axes. That is, it has to be the
bisector of the angle between the x and ct axes.
50
Spacetime Geometry
y
P2
y
x
θ
2
P1
y
x2
1
θ
x1
Figure 4.1: Two points and their coordinates in a non-perpendicular set of axes.
Solution: I’ll use a dashed line instead of a wavy line. The event E in Figure 4.2 of the
manual, lying on the worldline of a light signal, covers a distance of x in time t. Therefore,
x = ct and the two triangles shown must be isosceles. Therefore all the angles shown must
be equal.
ct
E
ct
x
x
Figure 4.2: The event E covers a distance of x in time t.
4.4. A train of rest length L0 travels at speed β in the positive x0 -direction of Sam. As the
front of the train passes Sam at t0 = 0, a light signal is sent from the front of the train to
the rear.
(a) Draw a spacetime diagram showing the worldlines of the front and rear of the train
and the photon in Sam’s RF.
(b) When does the rear of the train pass Sam?
(c) When does the signal reach the rear of the train according to Sam?
51
ct '
A
C
D
B
F
O'
x'
Figure 4.3: The worldlines of the front and back of the train are the lines O0 A and BC, respectively.
Solution:
(a) The worldlines of the front and back of the train are, respectively, the lines O0 A and
BC of Figure 4.3 of the manual. The train has a length of L0 /γ according to Sam.
So, in the figure, the line segment O0 B is that length.
(b) By Rule 2 of Note 4.2.6, β = BO0 /O0 C or β = (L0 /γ)/ct02 , where t02 is when the back
of the train reaches Sam. Therefore, ct02 = L0 /(βγ).
(c) By Rule 2 of Note 4.2.6, β = DF /F C. Therefore, F C = DF /β. But DF = O0 F
and, by (b) ct02 = O0 F + F C. Hence,
O0 F = ct02 − F C =
O0 F
O0 F
L0
L0
L0
−
⇐⇒
(1 + β) =
⇐⇒ O0 F =
,
βγ
β
β
βγ
(1 + β)γ
and O0 F ≡ ct01 is the time (times c) that the signal reaches the back of the train
according to Sam.
It is instructive to check our answers by using Lorentz transformation. So, place Sonya in
the train and call her observer O. Let the front of the train be x = 0, and assume that
at t = 0 = t0 , the front of the train coincides with Sam. Then according to Sonya, the
light reaches the back of the train at (−L0 , L0 ) and the back of the train reaches Sam at
(−L0 , L0 /β). Therefore, the first event has coordinates
x01 = γ(−L0 + βL0 ) = −γ(1 − β)L0
γ(1 − β)(1 + β)L0
L0
ct01 = γ(L0 − βL0 ) = γ(1 − β)L0 =
=
,
1+β
γ(1 + β)
in Sam’s RF, as before. Note that although we didn’t calculate x01 before, the spacetime
diagram clearly shows that x01 = −DF = −O0 F = −ct01 .
The second event has coordinates
x02 = γ(−L0 + βL0 /β) = 0
γ
L0
,
ct02 = γ(L0 /β − βL0 ) = (1 − β 2 )L0 =
β
γβ
which is identical to what we obtained earlier.
52
Spacetime Geometry
ct
ct
C
D
E
F
ct1
1pm
C
ct2
E
1pm
F
ct2
B
A
M
K
H
ct0
D
ct0
G
A
(a)
x
G
ct1
B
(b)
x
Figure 4.4: The spacetime diagram of Sonya-Sam meeting at 1pm. The worldline of Sonya’s planet is
the ct axis. (a) Sonya’s speed is 0.5c. (b) Sonya’s speed is 0.2c.
4.5. Sonya and Sam live on two different planets one light hour apart. A space station
is located half way between the two planets. The planets and the space station are all
stationary relative to each other. Sonya wants to arrange a meeting with Sam at 1:00
PM on the space station. She decides to send him a light signal so that as soon as he
receives it he starts to move toward the space station to get there at exactly 1:00 PM.
Sonya’s spaceship moves at half the speed of light while Sam’s moves at 0.75c. All times
are according to the common reference frame of the planets and the space station.
(a) At what time should Sonya send the signal?
(b) At what time should she leave her planet?
(c) At what time should Sam leave his planet?
Solution: Figure 4.4(a) of the manual shows the relevant spacetime diagram. Just to let
you know, I drew the lines backward: I started at D and drew Sam and Sonya’s worldlines.
Then from F , I drew a light signal toward the ct axis. I’ll have to do (b) and (c) before I
can do (a)!
(b) By Rule 2 of Note 4.2.6, β = CD/CB, or CB = 0.5 light hour/0.5c = 1 hour. So,
Sonya has to start her trip at 12:00 pm.
(c) Similarly, β = DE/EF gives EF = 0.5 light hour/0.75c or EF = 40 minutes. So,
Sam has to leave at 12:20 pm.
(a) From the equality GF = AG = 1 light hour, we conclude that t0 is one hour before
t2 . So, Sonya has to send the signal at 11:20 am.
4.6. Same as the previous problem except that Sonya’s spaceship moves at 0.2c.
(a) At what time should Sonya leave her planet?
53
(b) At what time should she send the signal?
(c) As she tries to contact Sam, Sonya realizes that she can’t contact him directly. She
decides to send a radio signal to her planet telling them to contact Sam instead. What
is the latest time that Sonya can contact her planet so that Sam receives the message
in time to be able to make it to the meeting?
Solution: Figure 4.4(b) of the manual shows the relevant spacetime diagram.
(a) By Rule 2 of Note 4.2.6, β = CD/CB, or CB = 0.5 light hour/0.2c = 2.5 hours.
So, Sonya has to start her trip at 10:30 am. Similarly, β = DE/EF gives EF =
0.5 light hour/0.75c or EF = 40 minutes. So, Sam has to leave at 12:20 pm.
(b) K is one hour below F (why?). So, K occurs at 11:20 am. Thus BK is 50 minutes.
Again, by rule 2, AM = 0.2AB. Therefore, AK = 0.2AB. Thus,
BK + AK = AB = AK/0.2 = 5AK ⇐⇒ BK = 4AK,
or AK = 12.5 minutes. Therefore, A occurs 62.5 minutes after Sonya’s take-off, or at
11:32.5 am. Since Sonya is measuring proper time, she has to send the signal
62.5 minutes p
= 1 − 0.22 × 62.5 minutes = 61.24 minutes
γ
after her take-off by her own watch.
(c) In order for Sam to receive the signal at F , the planet people have to send it at K.
And in order for them to receive Sonya’s signal at K, she has to send hers at H. With
a reasoning along the same line as in (b), you can show that BK = 6HK, so that
HK = 50/6 = 8.67 minutes, and BH = 5HK = 250/6 = 41.67 minutes. So, Sonya
has to send the stress signal 41.67 minutes after take-off according to her planet’s
time. According to her own time, she has to send the signal
41.67 minutes p
= 1 − 0.22 × 41.67 minutes = 40.82 minutes
γ
after take-off.
4.7. In Example 4.2.4, I showed that when two events are causally disconnected, you can
always find an observer for whom the order of occurrence of the events is switched. I used
algebraic method to prove the statement. Geometry make the argument much easier! Let
O0 have perpendicular axes. Pick two events E1 and E2 that are causally disconnected.
Now choose a set of axes passing through the origin of O0 in such as way that the order of
occurrence of E1 and E2 is switched. Hint: Draw parallel lines from the two events in such
a way that the earlier event cuts ct0 axis at a later time. From this decide what the new
x-axis should be. Make sure that the angle between the x- and x0 -axes is not larger than
45◦ .
Solution: The idea is first to find the x0 -axis. If the two events are causally disconnected,
then the line passing through them must make an angle less that 45◦ with the x-axis. Now
consider candidates for the x0 -axis as lines passing through the origin and making various
54
Spacetime Geometry
ct
ct '
ct2
E1
ct1
E2
x'
ct1'
ct2'
x
Figure 4.5: To find the x0 -axis, draw a line from the origin that makes a larger angle with the x-axis than
the line E1 E2 does.
angles with the x-axis. As you increase the angle from zero (corresponding to the x-axis
itself), and draw two lines from E1 and E2 parallel to the current candidate, the spacing
between those two lines decreases, meaning that the corresponding ct02 − ct01 decreases. As
you continue increasing the angle, you reach a line that is parallel to E1 E2 . For this line,
ct02 = ct01 . It should now be clear that any line that makes an angle with the x-axis larger
than the angle of E1 E2 (but less than 45◦ ) is a good candidate for the x0 -axis. Once the
x0 -axis is determined, the ct0 -axis can be drawn. Figure 4.5 of the manual shows one of the
infinitely many coordinate systems in which the order of occurrence of the two events E1
and E2 has been switched.
4.8. Figure 4.26 shows five firecrackers separated by 4 light seconds from one another in
Sam’s reference frame O. F1 and F3 occur simultaneously 4 seconds into the future; F2 ,
F4 , and F5 occur 16 seconds, 8 seconds, and 20/3 seconds into the future, respectively.
Sonya (reference frame O0 ) moves at 1/3 the speed of light relative to Sam in the positive
x-direction in such a way that at t = t0 = 0 the origins of the two RFs coincide.
(a) With dots, show the events corresponding to the explosion of the firecrackers.
(b) At what times does Sam receive the light signals from the firecrackers?
(c) Draw Sonya’s spacetime axes, x0 and ct0 .
(d) In what order do the firecrackers explode according Sonya?
(e) In what order do the light signals from the firecrackers reach Sonya?
(f) At what times according to Sam do the light signals from the firecrackers reach Sonya?
Solution:
(a) Figure 4.6 of the manual shows the coordinates of the events corresponding to the
explosion of the firecrackers.
(b) Sam receives the light signals from F1 , 8 s into the future, from F2 and F4 , 24 s into
the future, from F3 , 16 s into the future, and from F5 , 26.667 s into the future.
55
ct
ct '
x'
O
F1
F2
F3
F4
F5
x
Figure 4.6: The distance between consecutive firecrackers is 4 light second. The unit on the vertical axis
is also light second. The dashed lines are light worldlines.
(c) Sonya’s spacetime axes, x0 and ct0 , are as shown in the figure.
(d) According to Sonya, F3 and F5 explode simultaneously first, then F1 and F4 also
simultaneously, and finally F2 .
(e) F1 , then F3 , then F2 and F4 simultaneously, and finally, F5 .
(f) F1 , 6 s into the future; F3 , 12 s into the future; F2 and F4 , 18 s into the future; and
finally F5 , 20 s into the future.
4.9. Sonya, who is in reference frame O, sees a flash of red light at x = 1500 m, and after
5 µs, a flash of green light at x = 300 m. Use spacetime diagrams for the problem.
(a) What should Sam’s speed be relative to Sonya so that he sees the two events at the
same point in his RF?
(b) Which event occurs first according to Sam and what is the time interval between the
two flashes?
Solution: Figure 4.7 of the manual shows the details of the problem. In order for Sam’s
RF to be present at both events, his world line (his time axis) must be parallel to the line
connecting E1 and E2 . That’s how ct0 -axis is constructed. The x0 -axis is then drawn as
usual.
56
Spacetime Geometry
ct
ct '
ct2'
B
ct1'
A
E2
E1
x
x'
Figure 4.7: The spacetime diagram for the red and green flashes according to Sonya (observer O) and
Sam (observer O0 ).
(a) Sam’s speed relative to Sonya is the slope of the ct0 -axis relative to the ct-axis, which
is the slope of E1 E2 relative to the ct-axis:
v=
300 m − 1500 m
= −2.4 × 108 m/s ⇐⇒ β = −0.8
5 µs
(b) The red flash occurs first according to Sam. To find the time interval between flashes,
draw lines parallel to the x-axis from E1 and E2 , and let them cut the ct-axis at A
and B. It is then clear that AB = ct2 − ct1 . By Rule 4 of Note 4.2.7,
ct2 − ct1 = AB = γE1 E2 = γ(ct02 − ct01 ),
or
t02 − t01 =
t2 − t1 p
= 1 − 0.82 × 5 µs = 3 µs.
γ
The answer to (b) could also be obtained by noting that Sam measures proper time.
4.10. In a galactic rocket race, Sam and Sonya drive two rockets in opposite directions
towards their respective finish lines as shown in Figure 4.27 (the units are not the same on
the two axes). Sam moves to the right and Sonya to the left. The drivers start their motion
at t = 0 according to all three RFs. When they reach their finish lines—placed at 20 light
minutes on either side of the referee—each driver sends a light signal to the referee relaying
their arrival. The referee receives the signals from Sam and Sonya exactly 45 minutes after
their departure, indicating a tie.
(a) Draw the two events of the arrival of the rockets at the finish lines on the referee’s
RF.
(b) What is the speed of each rocket?
(c) Draw the spacetime axes of both rockets.
57
ct ''
ct
L''x
ct'
L'x
B
A''
A
x'
A'
x
O
L
''
ct
L'ct
x ''
Figure 4.8: The diagram shows all the events. The units are the same for both axes. The prime indicates
Sam and the double-prime indicates Sonya.
(d) Graphically show the coordinates of the two arrival events on the axes of all three
observers.
(e) What is the order of arrivals according to Sam? Does he think he is the winner?
(f) What is the order of arrivals according to Sonya? Does she think she is the winner?
Solution: In Figure 4.8 of the manual the units are the same on the two axes.
(a) The two events labeled A0 and A00 are the events of the arrival of Sam’s and Sonya’s
rockets at the finish lines, respectively.
(b) The speeds are the same. For Sam, it is the slope of ct0 relative to ct:
β=
AA0
AA0
20
=
=
= 0.8
45 − 20
OA
OB − AB
(c) To draw the time axis of each rocket, connect O to its arrival event.
(d) The coordinates of the arrival events are already drawn in O: their x’s are the locations
of their finish lines, and their common ct is OA. To find the coordinates of Sam’s
arrival in Sonya’s RF, draw lines from A0 parallel to Sonya’s axes. In the figure, the
line L00x is parallel to x00 -axis; if its intersection with ct00 is a point C 00 , then OC 00 is the
time coordinate of A0 in Sonya’s RF. Similarly, the line L00ct is parallel to ct00 -axis; if its
intersection with x00 is a point D00 , then OD00 is the x00 coordinate of A0 in Sonya’s RF.
Note that the x00 coordinate is negative, as should be evident. To find the coordinates
of Sonya’s arrival in Sam’s RF, draw lines from A00 parallel to Sam’s axes.
58
Spacetime Geometry
(e) Obviously, OC 00 > OA00 . So, Sam thinks that he reached the finish line earlier than
Sonya, and therefore that he is the winner.
(f) Same as (e) for Sonya.
It is a good idea to put some quantitative flesh on all the qualitative discussion skeleton I
have made so far. First note that OA is the projection on ct-axis of the interval OA0 on
ct0 -axis. Therefore, by Rule 4 of Note 4.2.7,
OA = γOA0 ⇐⇒ OA0 =
OA p
= 1 − 0.82 × 25 light minutes = 15 light minutes.
γ
Next, I want to find the time of Sam’s arrival according to Sonya. For this, I need their
relative speed:
βO0 O00 =
0.8 + 0.8
1
= 0.9756 ⇐⇒ γO0 O00 = √
= 4.56.
1 + 0.8 × 0.8
1 − 0.97562
Since I know OA0 , I can invoke Rule 4 of Note 4.2.7:
OC 00 = γO0 O00 OA0 = 4.56 × 15 light minutes = 68.33 light minutes.
Finally, I want to find the location of A0 in O00 . That’s where Rule 2 of Note 4.2.6 comes
in handy:
βO0 O00 =
OD00
OD00
⇐⇒
0.9756
=
⇐⇒ OD00 = 66.67 light minutes,
68.33 light minutes
OC 00
and x00A0 = −66.67 light minutes.
4.11. Sonya, who is in reference frame O, throws a ball with speed βb at time t0 (event
E0 ) in her positive x-direction, which reaches a point (event E) ∆x from the origin at time
t0 + ∆t. Sam, in RF O0 , with respect to whom Sonya is moving with speed β in the positive
x0 -direction, looks at the same two events. Use a spacetime diagram in which Sam’s axes
are perpendicular to derive the relativistic law of addition of velocities.
Solution: The relevant diagram is depicted in Figure 4.9 of the manual. We need to find
the ratio of AB to M F . I’ll be using Rules 2 and 4 without mentioning them each time.
Let’s start with AB:
AB = OB − OA = OB − βOM = OB − βγct0 .
Now, I calculate OB:
OB = OH + |{z}
HB = γOG + βOD = γ∆x + βγOC = γ∆x + βγ(ct0 + c∆t).
=DC
Therefore,
AB = γ∆x + βγc∆t = γ(βb + β)c∆t
59
ct
ct ʹ
F
D
M
O A
E
C
x
E0
G
H
B
xʹ
Figure 4.9: The spacetime diagram for the derivation of the relativistic law of addition of velocities.
Now, let’s calculate M F :
M F = OF − OM = OF − γct0 = OD + |{z}
DF −γct0
=GH
= γOC + βOH − γct0 = γ(ct0 + c∆t) + βγ∆x − γct0 .
Hence,
M F = γ(c∆t + β∆x) = γ(c∆t + βcβb ∆t) = γ(1 + ββb )c∆t.
Taking the ratio AB/M F yields the relativistic law of addition of velocities.
4.12. Alpha Centauri is about 4 light years away from Earth and does not move significantly
relative to it, so they are both in the same RF. Assume Earth is at the origin of this RF.
Event E1 occurs at t = 0 on Earth. Event E2 occurs on Alpha Centauri 3 years later. Use
spacetime diagrams.
(a) What is the time difference between the two events according to an observer moving
from Earth to Alpha Centauri at 0.9c? Which event occurs first?
(b) What is the time difference between the two events according to an observer moving
from Alpha Centauri to Earth at 0.9c? Which event occurs first?
What happened to the invariance of “earlier” and “later”?
Solution: Figure 4.10 of the manual shows the spacetime diagrams. In Figure 4.10(a) the
Earth RF is drawn with axes perpendicular. The Alpha Centauri worldline is the vertical
line. The RF moving toward Alpha Centauri is primed, and the RF coming toward Earth
is double-primed. To avoid cluttering the figure too much, only the lines constructing
the coordinates of E2 in the primed coordinate system are shown. It is seen that in this
coordinate system, E2 occurs earlier.
(a) In Figure 4.10(b), the Earth and the primed RF are isolated. Furthermore, to ease
calculation, the primed RF is drawn orthogonal. You can derive the coordinates of
E2 in the primed RF as was done in the re-derivation of the Lorentz transformation
in Section 4.4 (see also the solution to the previous problem). What you’ll be doing
is essentially re-deriving the Lorentz transformation. So, I’ll leave the derivation as
an exercise. I’ll just use the Lorentz transformation to calculate the time difference.
60
Spacetime Geometry
ct
ct
x'2
C
ct'
G
E2
B
E1
E1
x
O
x'
ct'
ct'2
E2
A
H
x ''
ct ''
O
(a)
x'
D
x
(b)
Figure 4.10: The Earth RF is unprimed. The RF moving toward Alpha Centauri is primed, and the RF
coming toward Earth is double-primed. (a) All three RFs are shown. (b) Only Earth and the RF moving
toward Alpha Centauri are shown. Earth RF has slanted axes.
The Earth is moving in the negative direction of the primed RF. So, β appears with
a negative sign in the Lorentz transformation.
c∆t0 = γ(c∆t − β∆x) = √
1
(3 − 0.9 × 4) = −1.38 light year.
1 − 0.92
Thus, ∆t0 = −1.38 years. Therefore, E2 occurs before E1 . This is okay, because the
two events are causally disconnected.
(b) The Earth is moving in the positive direction of the double-primed RF. So, β appears
with a positive sign in the Lorentz transformation.
c∆t00 = √
1
(3 + 0.9 × 4) = +15.14 light year,
1 − 0.92
and ∆t00 = 15.14 years.
4.13. Rework Example 3.5.3 using geometric method in which the speeder’s RF has orthogonal axes. Instead of numbers, use β1 for the speed of the speeder and β2 > β1 for the
policeman’s speed, both relative to the ground. Let T be the time according to the speeder
that the policeman catches up with the speeder.
Solution: Figure 4.11 of the manual shows the relevant spacetime diagram. Using Rule 2,
you get
E1 A
E1 A
β1 =
, βps =
⇐⇒ β1 OA = βps AE2 ,
OA
AE2
where βps is the speed of the policeman relative to the speeder. Relativistic LAV gives
βps =
βpg + βgs
βpg − βsg
=
.
1 + βpg βgs
1 − βpg βsg
61
ct
E2
E1
A
x
O
Figure 4.11: Event E1 is when the policeman starts chasing the speeder. Event E2 is when the policeman
catches up with the speeder.
Note the order of subscripts in the first equation. That’s how the LAV should be remembered! Concentrating only on the numerator, it says that the speed of the police relative
to the speeder is equal to the speed of the police relative to the ground plus the speed of
the ground relative to the speeder (divided by the appropriate denominator). Therefore,
βps =
and
β1 OA =
β2 − β1
,
1 − β2 β1
β2 − β1
β2 − β1
AE2 ⇐⇒ OA =
AE2 .
1 − β2 β1
β1 (1 − β2 β1 )
We also have cT = OA + AE2 . Hence,
β2 − β1
β2 (1 − β12 )
+ 1 AE2 =
cT =
AE2
β1 (1 − β2 β1 )
β1 (1 − β2 β1 )
or
AE2 =
γ12 β1 (1 − β2 β1 )cT
,
β2
OA =
γ12 (β2 − β1 )cT
.
β2
The policeman is at a distance
E1 A = β1 OA =
γ12 β1 (β2 − β1 )cT
β2
from the speeder when he starts the chase, making his x-coordinate relative to the speeder
x1 = −
γ12 β1 (β2 − β1 )cT
.
β2
The time that the policeman starts chasing the speeder, according to the policeman is
OE1 . Using Rule 4, we get
OA = γ1 OE1 ⇐⇒ ct01 = OE1 =
OA
γ1 (β2 − β1 )cT
=
.
γ1
β2
62
Spacetime Geometry
Similarly, using γps = γ1 γ2 (1 − β1 β2 ), we obtain
AE2 = γps E1 E2 ⇐⇒ E1 E2 =
AE2
β 1 γ1
cT.
=
γ1 γ2 (1 − β1 β2 )
β 2 γ2
As a check, it’s a good idea to plug in the numbers of Example 3.5.3 and make sure you
get the results of that example.
4.14. This is a variation of Example 3.5.3 for which you are to use spacetime diagrams.
Let the speeder’s RF have orthogonal axes. The speeder passes the policeman with a speed
of 0.99c. Two minutes later (policeman’s time), the policeman starts chasing the speeder
with a speed of 0.995c.
(a) How long does it take the policeman to catch up with the speeder according to the
speeder’s watch?
(b) How long does it take the policeman to catch up with the speeder according to the
policeman’s watch?
To answer these questions you have to find all the following on the speeder’s spacetime
coordinates: the time when the policeman starts the chase, the distance between the two
when the chase starts, the time it takes for the policeman to catch up with the speeder
after the start of the chase.
Solution: We can use Figure 4.11 of the manual for this problem as well. I calculated OE1
in the previous problem.
(a) By Rules 4 and 2,
OA = γ1 OE1 ,
AE1 = β1 OA = β1 γ1 OE1 .
Also by Rule 2,
AE1 = βps AE2 ⇐⇒ AE2 =
AE1
β1 γ1 OE1
β1 γ1 (1 − β1 β2 )OE1
=
=
,
βps
βps
β2 − β1
where βps is the speed of the policeman relative to the driver in the second leg of his
trip, and in the last equality, I used the relativistic LAV. So, the time according to
the speeder is
ct = OE2 = OA + AE2 = γ1 OE1 +
β1 γ1 (1 − β1 β2 )OE1
β2 OE1
=
.
β2 − β1
γ1 (β2 − β1 )
Plugging in the numbers, you get
ct =
0.995 × 2 minutes
= 173.5 minutes
2.294(0.995 − 0.99)
(b) To find the time according to the policeman, we need E1 E2 :
AE2 = γps E1 E2 ⇐⇒ E1 E2 =
AE2
AE2
β1 OE1
=
.
=
γps
γ1 γ2 (1 − β1 β2 )
γ2 (β2 − β1 )
63
ct ʹ
ct
x
A
O1 C1
O2
xʹ
C2
Figure 4.12: Clocks C1 and C2 and the observers O1 and O2 .
Thus,
ct = OE1 + E1 E2 = 1 +
0
β1
OE1 .
γ2 (β2 − β1 )
Plugging in the numbers, you get
0.99
ct0 = 1 +
2 minutes = 41.55 minutes
10.01(0.995 − 0.99)
4.15. Two clocks C1 and C2 separated by a fixed distance L0 are placed on the x0 -axis of
an observer O0 and synchronized according to O0 .1 Reference frame O is moving relative
to O0 in the positive direction of x0 with speed β. Observer O1 is at the origin of O and
observer O2 is placed strategically along the x-axis in such a way that they can read the
two moving clocks at the same time. O1 records the reading as 12:00 (the zero time). Draw
a spacetime diagram with O0 axes perpendicular, and show the world lines of the two clocks
in the O0 reference frame.
(a) What is the location of O2 in O?
(b) What time does O2 record? Hint: It is the time coordinate in O0 of the intersection
of the x-axis with the worldline of C2 .
Solution: The two clocks C1 and C2 and the observers O1 and O2 are shown in Figure 4.12
of the manual.
(a) By Rule 4, C1 C2 = γO1 O2 . Therefore, with x1 = 0, we get x2 = L0 /γ.
(b) The time is ct0 = O2 C2 = βL0 (using Rule 2).
These are the results we obtained in Example 3.4.10.
4.16. Two photons are moving in O at a fixed distance L apart. Show that in O0 with
respect to which O moves with fractional speed β in the negative direction, the two photons
are separated by
s
1+β
L0 =
L.
1−β
1
This is the same problem done in Example 3.4.10.
64
Spacetime Geometry
ct
ct ʹ
Oʹ O
B
C
xʹ
A
x
Figure 4.13: The two photons are a distance of L = OA apart in O.
Solution: The separation of the two photons in O0 is O0 C of Figure 4.13 of the manual.
Using Rules 2 and 4, we have
L0 = O0 B + BC = γOA + AB = γOA + βO0 B
s
1+β
L.
= γOA + βγOA = γ(β + 1)L =
1−β
4.17. Sam and Sonya are twins. Sam is put on a spaceship O moving with relative speed
β in the positive x0 direction of Sonya’s reference frame O0 . At time T00 , O0 sends a light
signal toward O. Draw a spacetime diagram showing the relevant axes and the light signal.
Use the rules of spacetime geometry.
(a) Show that when Sam receives the light signal, the spaceship is at a distance of βT00 /(1−
β) from Sonya.
p
(b) Show that Sam receives the light signal when he is (1 + β)/(1 − β) T00 older than
when he left his sister.
Solution: Figure 4.14 of the manual shows the two RFs of Sonya (O0 ) and Sam (O). The
problem is to find OE1 and AE1 in terms of cT00 = O0 E0 . Rule 4 gives
O0 A = O0 E0 + E0 A = O0 E0 + AE1 = O0 E0 + βO0 A.
Therefore,
(1 − β)O0 A = O0 E0 ⇐⇒ O0 A =
cT00
.
1−β
(a) By Rule 2,
OB = AE1 = βO0 A =
βcT00
.
1−β
65
ct ʹ
A
E1
ct
x
E0
O Oʹ
B
xʹ
Figure 4.14: Spacetime diagram for Sonya sending a light signal to her twin brother. Event E0 takes
place at T00 .
(b) By Rule 4, O0 A = γOE1 . Thus,
ct1 = OE1 =
O0 A
cT00
=
,
γ
γ(1 − β)
and
βT00
t1 =
=
γ(1 − β)
s
1+β 0
T
1−β 0
4.18. Tomorrow is “only one day away;” it is probably not too much to ask the theory of
relativity to help us get there. Utilizing the experience you gained in the case of Bruno’s
death, find observer O who is only 24.5 light hours away.2
(a) What speed should O have? In which direction?
(b) How far away is “tomorrow” taking place from O?
ct ʹ
T
x
Oʹ
ct
O
xʹ
Figure 4.15: Spacetime diagram for “traveling” to the future. T stands for tomorrow. OT must be the
x-axis, because T has to happen NOW for O.
Solution: Figure 4.15 of the manual shows the spacetime diagram for the problem.
2
A light hour is the distance that light travels in one hour. For comparison, Saturn is about 1.25 light
hours away from Sun.
66
Spacetime Geometry
ct’
A
RAB
B
ct
RA
x
At RB
Bx
B
A
A
B
A
Ax
(a)
O
B
x’
Bt
(b)
Figure 4.16: (a) To Sam, the explosion of the two firecrackers occur at the same time that Sonya passes
him by. (b) The spacetime geometry of the events as seen by Sam (O0 ) and Sonya (O).
(a) By Rule 2,
O0 T
24
=
= 0.9796.
0
24.5
OO
It is clear that the speed should be in the negative x0 direction.
β=
(b) By Rule 4, O0 O = γOT . Therefore,
OT =
O0 O p
= 1 − β 2 O0 O = 0.2 × 24.5 light hours = 4.9 light hours.
γ
4.19. Sonya (observer O) moves at 0.99c relative to Sam (observer O0 ) as shown in Figure 4.9. Assume that the length of the train is 50 m. Find the coordinates of all the
points marked as triangle, square, circle, oval, and stars in Figure 4.9(b). Hint: Go through
Example 4.3.1 for warmup.
Solution: I’ve reproduced the figure in Figure 4.16 of the manual, enlarged part (b), and
added some labels in part (b) to facilitate the solution. I’ll just use β for the speed and
2L for the length of the train. From Rule 4, we get OBx = γOB = γL. From Rule 2,
we get OBt = BBx = βOBx = βγL (the time coordinate of B in O is −OBt ). Similarly,
OAx = γL (the space coordinate of A in O is −OAx ) and OAt = AAx = βγL.
The triangle BBt RB is an isosceles triangle (show this by drawing a line from Bt parallel
to the light world line passing through O). Therefore, Bt RB = BBt = OBx , and
ORB = Bt RB − OBt = OBx − OBt = γL − βγL = (1 − β)γL.
Similarly, the triangle AAt RA is an isosceles triangle. Therefore, At RA = AAt = OAx , and
ORA = At RA + OAt = OAx + OAt = γL + βγL = (1 + β)γL.
Finally, it is clear that ORAB = OA = OB = L.
67
c tʹ
ct
E2
O xʹ
Oʹ
c Δ t31
E3
Δ x31
E3
E1
E1
(a)
E2
O
x
(b)
Figure 4.17: (a) The events E1 , E2 , and E3 in the Earth RF. Note that the time of the origin (and thus
the time of all the events on the x0 -axis) is NOW which is the year 2163. (b) The same three events as
seen by the crew of Diracus II.
4.20. Use the result of Problem 3.7 to find the coordinates of E1 , E2 , and E3 of Figure 4.18
in both coordinate systems O and O0 .
Solution: I’ll reproduce the figure for convenience. The coordinates of the origin of O
relative to O0 are just the coordinates of E2 of Figure 4.17 of the manual relative to O0 .
These are (x00 , ct00 ) ≡ (x02 , 0). Now, we calculate x0 and ct0 of Problem 3.7:
x0 = γ(−x00 + βct00 ) = −γx02 ,
ct0 = γ(βx00 − ct00 ) = γβx02 .
Thus, the inverse Lorentz transformation that includes the coordinates of the origin is
x = − γx02 + γ(x0 − βct0 )
ct =γβx02 + γ(−βx0 + ct0 ).
As a check, note that
x2 = −γx02 + γ(x02 − βct02 ) = 0,
ct2 = γβx02 + γ(−βx02 + ct02 ) = 0.
The coordinates of E1 and E3 in O0 are (0, −βx02 ) and (0, ct03 ), respectively. Therefore,
x1 = −γx02 + γ[0 − β(−βx02 )] = −γx02 (1 − β 2 ) = −
x02
γ
ct1 = γβx02 + γ(−0 − βx02 ) = 0,
and
x3 = − γx02 + γ(0 − βct03 ) = −γ(x02 + βct03 )
ct3 =γβx02 + γ(−0 + ct03 ) = γ(βx02 + ct03 ).
You can verify all the results of Section 4.4.1 pertaining to the Kennedy assassination. 4.21. Provide the missing steps leading to the final result of Equation (4.12).
68
Spacetime Geometry
ct
E5
26
20
15.5
E4
E3
10.1
E2
5
E1
O
5
10
x
Figure 4.18: Sonya moves on the heavy black worldline. Sam moves on the grey worldline first and then
joins Sonya to return home. Pat remains on Earth.
Solution: Consider the equation
βprobe = β +
T1 + T2
,
βγ 2 T2
which is already derived in the text. Manipulate the second term:
T1 + T2
T1
1
T1
1 − β2
T1
1
=
+
=
+
=
+ − β.
2
2
2
2
2
βγ T2
βγ T2 βγ
βγ T2
β
βγ T2 β
Substituting this in the previous equation yields the final result.
4.22. Sam, Sonya, and Pat are newly born triplets. Sam and Sonya are put on two different
spaceships that travel to a planet of a star system 10 ly away. Sonya lands on the planet
10.1 years later as seen by observer O, Pat. She waits 4.9 years until Sam, who is traveling
slower, lands on the same planet (see Figure 4.18 of the manual). After six months they
both return home on the same spaceship and land on Earth 26 years after their departure
according to Earth calendar. All times and distances of the figure are given according to
the Earth observers, and all units shown are in light years, and for easier reading most of
the calibration of the ct-axis is made on the worldline parallel to it.
(a) What is the speed of Sonya’s spaceship on her journey to the planet?
(b) What is the speed of Sam’s spaceship on his journey to the planet?
(c) How old is Sonya when she meets Sam? How old is Sam?
(d) How old is Sonya when she lands back on Earth? How old is Sam? How old is Pat?
Solution:
(a) Distance is 10 light years, and time is 10.1 years. Therefore, β1 =
(b) β2 =
10
15
= 0.67.
10
10.1
= 0.99.
69
(c) Find the spacetime distance for each. For Sonya, there are two pieces to calculate:
p
∆s12 = 10.12 − 102 = 1.418 light years ⇐⇒ ∆τ12 = 1.418 years,
and ∆τ23 = 4.9 years. Therefore, she is 6.318 years old when she meets Sam. For
Sam, there is only one piece:
p
∆s13 = 152 − 102 = 11.18 light years ⇐⇒ ∆τ13 = 11.18 years.
Therefore, he is 11.18 years old when he meets Sonya.
(d) We have to add 0.5 + ∆s45 to each, where
p
∆s45 = (26 − 15.5)2 − 102 = 3.2 light years ⇐⇒ ∆τ45 = 3.2 years.
Hence Sonya is 6.318 + 0.5 + 3.2 = 10.02 years old; Sam is 11.18 + 0.5 + 3.2 = 14.88
years old; and Pat is 26 years old.
4.23. Verify that Equation (4.14) is the left half of a hyperbola with center at β0 (κ2 +
√
1)T /2, T /2 , semi-major axis a = β0 κ κ2 + 1 T /2 and semi-minor axis b = κT /2.
Solution: Rewrite (4.14) as
√
cβ0 (κ2 + 1)T
cβ0 κ2 + 1 p 2 2
xs =
−
κ T + (2t − T )2 ,
2
2
or
cβ0 (κ2 + 1)T
cβ0 p 2
=−
(κ + 1)[κ2 T 2 + (2t − T )2 ]
2
2
and note that xs → −∞ as t → ∞, indicating that the graph is in the left half-plane. Now
square both sides and write the result as follows:
2
c2 β02 2
cβ0 (κ2 + 1)T
=
(κ + 1)κ2 T 2 + c2 β02 (κ2 + 1)(t − T /2)2 .
xs −
2
4
xs −
Now transfer the t term to the left and divide both sides by c2 β02 (κ2 + 1)κ2 T 2 /4 to obtain
2
xs − cβ0 (κ2 + 1)T /2
(t − T /2)2
√
−
= 1.
(κT /2)2
(cβ0 κ2 + 1κT /2)2
This is a hyperbola with the said characteristics.
4.24. Show that the absolute value of the speed in (4.15) is always less than c for the
duration of all the three trips.
Solution:
dxs
< c ⇐⇒ β02 (κ2 + 1)(2t − T )2 < κ2 T 2 + (2t − T )2 .
dt
The last inequality is equivalent to
β02 κ2 (2t − T )2 < κ2 T 2 + (1 − β02 )(2t − T )2 .
The second term on the right-hand side is positive, so the minimum of the right hand side
is κ2 T 2 . Since 0 < t < T , the maximum of the left hand side is β02 κ2 T 2 . Since the minimum
of the right hand side is larger than the maximum of the left hand side, the right-hand side
is always larger than the left hand side.
70
Spacetime Geometry
ct
D
C
B
A
O
x
Figure 4.19: Each side of the big triangle is 4 times the corresponding side of each small triangle.
4.25. In Example 4.5.4, assume that Pat makes four identical round trips between t = 0
and t = T with the same speed β0 .
(a) How far does he have to go before turning around in each trip?
(b) How much does he age in each trip?
(c) Compare his total age with his age in the example. Can you give a simple geometric reason for this? Hint: The geometric reason is identical to ordinary Euclidean
geometry.
Solution: Refer to Figure 4.19 of the manual.
(a) Each leg of each trip now lasts T /8. Therefore, the outbound distance travelled is
cβ0 T /8.
p
2
(b) Equation (4.16) of Example 4.5.4 shows that Pat’s age
p is 1 − β0 times the time of
landing. If the time of landing is T /4, then he ages 1 − β02 T /4 during each trip.
p
(c) Hence, during the entire 4 trips, he ages 1 − β02 T , as before. This can be explained
geometrically as shown in Figure 4.19 of the manual. It is obvious that OC = 4OA
and DC = 4AB. Therefore, OD = 4OB.
4.26. In Example 4.5.4, assume that Pat travels with the initial speed of 0.999c to a
destination 40 light years away.
(a) How old is Sonya when Pat returns? How old is Pat?
(b) Sam goes to a destination 30 light years away with the same initial speed as Pat.
What is κ?
(c) How old is Sam when he returns? You’ll have to numerically integrate (4.16) to find
the answer.
71
Solution: It is shown in Example 4.5.4 that xp,max = cβ0 T /2.
(a) Therefore,
0.999cT
80 light years
⇐⇒ T =
= 80.08 years,
2
0.999c
p
and this is Sonya’s age. Example 4.5.4 derived Pat’s age as 1 − β02 T . Thus,
p
TPat = 1 − 0.9992 80.08 years = 3.58 years.
40 light years =
(b) It is shown in Example 4.5.4 that
xs,max =
p
p
cβ0 T 2
κ + 1 − κ κ2 + 1 = xp,max κ2 + 1 − κ κ2 + 1 .
2
Therefore,
p
30 light years = 40 light years κ2 + 1 − κ κ2 + 1
or
p
p
0.75 = κ2 + 1 − κ κ2 + 1 ⇐⇒ κ κ2 + 1 = κ2 + 0.25.
Squaring both sides and simplifying leads to κ2 = 0.125, or κ = 0.3536.
(c) Sam’s age is given by Equation (4.16), or—in years—by
Z 80.08 s
√
0.9992 (1.125)(2t − 80.08)2
TSam = τs (0.999, 0.125) =
dt.
1−
0.125 × 80.082 + (2t − 80.08)2
0
Numerical integration yields TSam = 49.69 years.
4.27. Sonya gets on a spaceship that travels on a parabolic path given by
t2
x(t) = cκ t −
,
T
in Sam’s coordinate system O, where κ is a positive constant. Note that x(0) = x(T ) = 0.
This means that Sonya leaves Sam at t = 0 and meets him again at t = T .
(a) What are Sonya’s velocities relative to Sam at the times of her departure and return?
What restriction does this put on κ?
(b) Show that during the entire Sonya’s trip, her speed is always less than the speed of
light.
(c) When and where does Sonya stop momentarily and come back?
(d) Verify that it take Sonya
T
!
√
sin−1 κ + κ 1 − κ2
2κ
to make her round trip. Show that as κ → 0, this travel time reduces to Sam’s time,
as it should.
72
Spacetime Geometry
(e) Show that no matter how fast Sonya starts to travel, the ratio of her increase in age
to Sam’s can’t be smaller than π/4.
Solution: Sonya’s speed as a function of t is given by
dx
2t
.
= cκ 1 −
dt
T
(a) Her speeds at t = 0 and t = T are
dx
dt
= cκ,
t=0
dx
dt
t=T
2T
= cκ 1 −
T
= −cκ.
Therefore, κ is both the departure and arrival speeds and 0 < κ < 1.
(b)
dx
2t
< c ⇐⇒ κ 1 −
< 1.
dt
T
|1 − 2t/T | is a decreasing function of t for 0 < t < T /2 with a maximum initial value
of 1 and an increasing function of t for T /2 < t < T with a maximum final value of
1. Thus, |1 − 2t/T | < 1 for all t, and therefore, the speed is less than 1 during the
entire trip.
(c) The speed is zero when t = T /2, for which x = cκT /4.
(d)
Z
s = cτ =
T
0
or
Z
τ=
0
T
T
Z
p
2
2
(cdt) − (dx) = c
s
1 − κ2
1−
0
2t
1−
T
2
dt = T
s
dx
cdt
2
dt
!
√
sin−1 κ + κ 1 − κ2
2κ
As κ → 0, each term in the numerator goes to κ, so that the ratio becomes T .
(e) As κ → 1, the first term in the numerator of the right-hand side of the last equation
tends to π/2 and the second term to zero, while the denominator goes to 2. So, the
entire right-hand side goes to (π/4)T .
4.28. Sonya gets on a spaceship that travels on a path given parametrically by
√
√
π
π
x(θ) = cT κ( 2 cos θ − 1), ct(θ) = cT ( 2 sin θ + 1), − ≤ θ ≤
4
4
in Sam’s coordinate system O, where κ is a positive constant less than one. Note that
x(−π/4) = ct(−π/4) = 0 and x(π/4) = 0, ct(π/4) = 2cT . This means that Sonya leaves
Sam at t = 0 and meets him again at t = 2T .
(a) Show that during the entire Sonya’s trip, her speed is always less than the speed of
light.
(b) What are Sonya’s velocities relative to Sam at the times of her departure and return?
73
1.00
0.95
0.90
0.85
0.2
0.4
0.6
0.8
1.0
Figure 4.20: Plot of the ratio of Sam’s round-trip time to Sonya’s.
(c) When and where does Sonya stop momentarily and come back?
(d) Verify that it take Sonya
√
Z
π/4
2T
q
1 − (κ2 + 1) sin2 θ dθ
−π/4
to make her round trip. Show that as κ → 0, this travel time reduces to Sam’s time,
as it should.
(e) Numerically integrate the integral in (d) for various κ and plot the ratio of Sam’s
round-trip time to Sonya’s. What is the limit of this ratio as κ → 1?
Solution:
(a) Take the differential of the two equations and divide dx by cdt:
√
dx
−cT κ 2 sin θdθ
√
β=
=
= −κ tan θ
cdt
cT 2 cos θdθ
and |β| = κ| tan θ|. Now, | tan θ| ≤ 1 for the allowed range of θ, and κ < 1, therefore,
|β| < 1.
(b) Departure corresponds to θ = −π/4. Thus, from (b), β0 = κ. On return, θ = π/4
and the speed is −β0 .
(c) Within the allowed range of θ, the speed is zero at θ = 0, for which
√
x(0) = cT κ( 2 − 1), ct(0) = cT.
74
Spacetime Geometry
(d) Integrating ds = cdτ , we get
Z
π/4
cτ =
Z
p
2
2
(cdt) − (dx) =
−π/4
=
√
π/4
q
√
√
(cT 2 cos θdθ)2 − (−cT κ 2 sin θdθ)2
−π/4
Z
π/4
2cT
Z
p
√
cos2 θ − κ2 sin2 θ dθ = 2cT
−π/4
π/4
q
1 − (κ2 + 1) sin2 θ dθ.
−π/4
As κ → 0, this travel time reduces
√ Z
τ = 2T
π/4
p
√ Z
2
1 − sin θ dθ = 2T
−π/4
π/4
cos θ dθ = 2T.
−π/4
(e) Let R(κ) denote the ratio of Sam’s time to Sonya’s:
R(κ) =
√ Z
2
π/4
q
1 − (κ2 + 1) sin2 θ dθ.
−π/4
Figure 4.20 of the manual shows the plot of this ratio. It is a decreasing function of
κ and its minimum value is R(1) = 0.847213.
CHAPTER
5
Spacetime Momentum
Problems With Solutions
5.1. Show that ubt = γb , and that u2bt − u2bx = 1. Show that for any other observer O0 ,
u0bt2 − u0bx2 = 1 as well.
Solution: First find ubt :
c∆tb
c∆tb
c∆tb
=
=p
c∆τb
∆sb
(c∆tb )2 − (∆xb )2
c∆tb
1
p
=
= γb .
=q
2
c∆tb 1 − (∆xb /c∆tb )
1 − β2
ubt =
b
Now note that
u2bt − u2bx =
(c∆tb )2 (∆xb )2
(c∆tb )2 − (∆xb )2
−
=
= 1.
(∆sb )2
(∆sb )2
(∆sb )2
u0bt2 − u0bx2 =
(c∆t0b )2 (∆x0b )2
(c∆t0b )2 − (∆x0b )2
−
=
= 1,
(∆sb )2
(∆sb )2
(∆sb )2
Furthermore,
because (∆sb )2 = (c∆tb )2 − (∆xb )2 = (c∆t0b )2 − (∆x0b )2 .
5.2. Derive the relativistic law of addition of velocities from the second equation in (5.4).
Solution: The second equation in (5.4) can be written as
γb0 = γ(βγb βb + γb ) = γγb (ββb + 1),
or
(1 − β 2 )(1 − βb2 )
1
(ββb + 1)2
02
=
⇐⇒
1
−
β
=
,
b
(ββb + 1)2
1 − βb0 2
(1 − β 2 )(1 − βb2 )
76
Spacetime Momentum
or
βb0 2 = 1 −
(1 − β 2 )(1 − βb2 )
(ββb + 1)2 − (1 − β 2 )(1 − βb2 )
=
,
(ββb + 1)2
(ββb + 1)2
or
βb0 2 =
2ββb + β 2 + βb2
(β + βb )2
=
.
(ββb + 1)2
(ββb + 1)2
5.3. I defined spacetime velocity by differentiating x and ct with respect to s = cτ . Now
define spacetime acceleration ab by differentiating spacetime velocity with respect to s:
abt =
dubt
,
ds
abx =
dubx
.
ds
(a) Show that
abt ubt − abx ubx = 0.
(5.1)
[Hint: Use Equation (5.2).] You’ll see later that it is possible to define a “dot product”
in two-dimensional relativity by subtracting the product of the two components of the
vectors. More specifically, if A = (Ax , At ) and B = (Bx , Bt ) are two vectors, then
the “dot product” is defined as
A · B = At B t − Ax B x
(5.2)
Equation (5.15) therefore, says that spacetime acceleration is orthogonal to spacetime
velocity.
(b) Verify that spacetime acceleration transforms via LTs. That is, if O measures the
components of the spacetime acceleration of a particle to be (abx , abt ) and O0 measures
them to be (a0bx , a0bt ), then
a0bx = γ(abx + βabt ),
a0bt = γ(βabx + abt ),
where β is the speed of O relative to O0 .
Solution:
(a) Differentiate Equation (5.2) with respect to s and note that the right-hand side vanishes.
(b) Take the differential of Equation (5.4):
du0bx = γ(dubx + βdubt )
du0bt = γ(βdubx + dubt ),
and divide both sides by ds.
5.4. Let y = f (x) describe a curve in the xy-plane on which a particle moves.
(a) Verify that for the velocity vector of the particle to be perpendicular to the position
vector of that particle, f (x) must satisfy the following differential equation
f (x)
df
+ x = 0.
dx
77
(b) Solve this simple equation and show that the solution is
p
f (x) = ± C − x2 ,
where C is the constant of integration. This is a circle of radius
√
C.
(c) Now let x = g(t) describe the world line of a particle in the spacetime plane. Refer to
Equation (5.16) for the definition of the dot product in spacetime plane, and show that
for the spacetime velocity vector of the particle to be perpendicular to the “position”
vector (x, ct) of that particle, g(t) must satisfy the following differential equation
g(t)
dg
− c2 t = 0.
dt
(d) Solve this simple equation and show that the solution is
p
g(t) = ± C + c2 t2 ,
where C is the constant of integration. What kind of a curve is this worldline?
Solution:
(a) The position vector of the particle is hx, f (x)i and its velocity vector is
hẋ, ẏi = hẋ, ẋdf /dxi,
where dot represents differentiation with respect to t. If the position vector is to be
perpendicular to the velocity vector, we must have
hx, f (x)i · hẋ, ẋdf /dxi = 0 ⇐⇒ xẋ + ẋf (x)df /dx = 0
or
f (x)
df
+ x = 0.
dx
(b) Rewrite the equation as f df + xdx = 0, and integrate it to get
1 2
2f
p
+ 21 x2 = constant ⇐⇒ f 2 + x2 = C ⇐⇒ f (x) = ± C − x2 .
(c) The spacetime “position” vector of the particle is hg(t), cti and its spacetime velocity
vector is hdg/ds, cdt/dsi. For the spacetime velocity vector of the particle to be
perpendicular to its spacetime velocity vector, we must have
hg(t), cti · hdg/ds, cdt/dsi = 0 ⇐⇒ g(dg/ds) − c2 t(dt/ds) = 0.
(d) Multiply the previous equation by ds and integrate to get
1 2
2g
p
− 12 c2 t2 = constant ⇐⇒ g 2 − c2 t2 = C ⇐⇒ g(t) = ± C + c2 t2 ,
which is a hyperbola.
5.5. Use Equations (5.7) and (5.8) to derive the relativistic law of addition of velocities.
78
Spacetime Momentum
Solution: Multiply the first equation in (5.8) by c, divide it by the second equation, and
use (5.7):
p0 c
γ(pb c + βEb )
pb c/Eb + β
βb + β
βb0 = b 0 =
=
=
.
Eb
γ(βpb c + Eb )
βpb c/Eb + 1
ββb + 1
5.6. Derive Equation (5.10) directly from the definition of energy and momentum in Equations (5.6) and (5.5). Now use (5.8) to show that Equation (5.10) holds in all RFs.
Solution: From (5.5) and (5.6), we obtain
Eb2 − p2b c2 = (γb mc2 )2 − (mcγb βb )2 c2 = m2 c4 γb2 (1 − βb2 ) = m2 c4 .
From (5.8), we get
Eb0 2 − p0b2 c2 = γ 2 (βpb c + Eb )2 − γ 2 (pb c + βEb )2
= γ 2 (β 2 p2b c2 + Eb2 + 2βpb cEb ) − γ 2 (p2b c2 + β 2 Eb2 + 2βpb cEb )
= γ 2 [Eb2 (1 − β 2 ) − p2b c2 (1 − β 2 )] = Eb2 − p2b c2
5.7. Using the definition of relativistic momentum and the relativistic law of addition of
velocities, show that if relativistic momentum is conserved in all RFs, then relativistic
energy must also be conserved.
Solution: Let two particles of masses m1 and m2 and speeds cβ1 and cβ2 collide to produce
two other particles of masses m3 and m4 and speeds cβ3 and cβ4 . Then, in some RF O,
the conservation of relativistic momentum gives
m1 γ1 cβ1 + m2 γ2 cβ2 = m3 γ3 cβ3 + m4 γ4 cβ4 .
Now let O0 be another RF with respect to which O moves with speed β in the positive x
direction. Let i be one of the numbers 1, 2, 3, or 4. Then, in O0 , the speeds and the Lorentz
factors of the particles are
βi0 =
βi + β
,
1 + βi β
γi0 = γγi (1 + βi β), ⇐⇒ βi0 γi0 = γγi (βi + β),
i = 1, 2, 3, 4.
I have already derived the middle equality, but it is a good exercise for you to derive it
again. In O0 , the conservation of momentum is
m1 γ10 β10 + m2 γ20 β20 = m3 γ30 β30 + m4 γ40 β40 ,
or
m1 γγ1 (β1 + β) + m2 γγ2 (β2 + β) = m3 γγ3 (β3 + β) + m4 γγ4 (β4 + β),
or
m1 γ1 β1 + m1 γ1 β + m2 γ2 β2 + m2 γ2 β = m3 γ3 β3 + m3 γ3 β + m4 γ4 β4 + m4 γ4 β.
Conservation of momentum in O eliminates the sum of the first and third terms on each
side, leaving
m1 γ1 β + m2 γ2 β = m3 γ3 β + m4 γ4 β.
Canceling β and multiplying by c2 yields the conservation of energy.
79
5.8. A particle of mass m moving at speed v collides with another particle of the same
mass at rest. They stick together and move with speed V . What is V in terms of v? What
is the mass of the final combined particle?
Solution: Let M be the mass of the final particle. Then, conservation of momentum and
energy yield the following two equations:
mγv v = M γV V,
mc2 γv + mc2 = M c2 γV .
The second equation gives M γV = m(1 + γv ). Substituting this in the first equation yields
mγv v = m(1 + γv )V ⇐⇒ V =
v
γv v
p
=
1 + γv
1 + 1 − (v/c)2
and
1
γV = s
1−
1 + γv
= p(1 + γ )2 − γ 2 v 2 =
v
v
γv v 2
1 + γv
r
1 + γv
.
2
Now we can calculate M :
M=
p
m(1 + γv )
= m 2(1 + γv ).
γV
Note that when γv → 1, i.e., in the classical limit, M = 2m confirming the conservation of
mass in classical collisions.
5.9. Electrons in projection TV sets are accelerated through a potential difference of 50 kV.
(a) Find the speed of the electrons using the relativistic form of KE (relativistic energy
minus rest energy) assuming that the electrons start from rest.
(b) Find the speed of the electrons using the classical form of KE.
(c) Does the difference in speed have to be taken into account when designing the TV
set?
Solution: By the definition of the electron volt, the potential energy of the electrons is
50 keV.
(a) If electrons start from rest, their KE must equal their potential energy:
me c2 γe − me c2 = 50 keV ⇐⇒ 511(γe − 1) keV = 50 keV ⇐⇒ γe = 1.0978,
because me c2 = 0.511 MeV. This gives βe = 0.413.
(b) From
KEe = 12 me ve2 = 21 me c2 (ve /c)2 = 12 me c2 βe2 ⇐⇒ 50 =
1
2
× 511βe2 ,
we get βe = 0.442.
(c) The difference between classical and relativistic speeds is only 7%, which is within
the tolerance of the operation of a TV set.
80
Spacetime Momentum
5.10. Two identical particles of mass m approaching each other with velocity β collide and
as a result of their collision a particle of mass M and a photon are produced.
(a) Can M remain stationary as in Example 5.3.3?
(b) If the answer to (a) is no, then what are the energies of M and the photon?
(c) What are the momenta of M and the photon?
Solution:
(a) No. The initial momentum is zero. Photon cannot be stationary, so M must have
some momentum to cancel the momentum of the photon.
(b) Conservation of momentum and energy give two equations:
2mc2 γ = E + e,
0 = P + p ⇐⇒ |p| = |P | ⇐⇒ e = |P |c =
p
E 2 − M 2 c4 ,
where upper case letters correspond to M and lower case letters to photon. Substitute
the last equation in the first to get
p
2mc2 γ = E + E 2 − M 2 c4 ⇐⇒ (2mc2 γ − E)2 = E 2 − M 2 c4 ,
which in combination with 2mc2 γ = E + e yields
E = mc2 γ +
M 2 c2
4mγ
and e = mc2 γ −
M 2 c2
.
4mγ
(c) The magnitude of the two momenta are equal:
|p| = e/c = mcγ −
M 2c
= |P |.
4mγ
5.11. Two particles of masses m1 and m2 are moving in opposite directions with the same
relativistic momentum p.
(a) Find their speeds v1 and v2 , in terms of m1 , m2 , and p.
(b) Find their energies E1 and E2 , in terms of m1 , m2 , and p.
(c) The two particles collide and form a particle of mass M at rest. Find the mass of the
final particle.
(d) Verify that the extra mass (times c2 ) is equal to the total initial kinetic energy of the
two particles.
Solution: Equation (5.7) gives speed in terms of momentum and energy.
(a) Therefore,
v1 =
pc2
pc2
pc
=p
=p
,
2
2
2
4
2
E1
p c + m1 c
p + m21 c2
pc
v2 = p
.
2
p + m22 c2
81
(b)
E1 =
q
p2 c2 + m21 c4 ,
E2 =
q
p2 c2 + m22 c4 .
(c) Conservation of energy gives E1 + E2 = M c2 . Therefore,
p
p
p2 + m21 c2 + p2 + m22 c2
E1 + E2
M=
=
c2
c
(d)
(M − m1 − m2 )c2 =
E1 + E2
2
−
m
−
m
1
2 c
c2
= E1 − m1 c2 + E2 − m2 c2 = KE1 + KE2 .
5.12. A particle of mass M at rest decays into two particles of masses m1 and m2 .
(a) What is the sum of the momenta of the two particles?
(b) Find their energies E1 and E2 , in terms of M , m1 , and m2 .
(c) Verify that the common momentum of the particles can be expressed as
p
(M 2 − m21 )2 + (M 2 − m22 )2 − M 4 − 2m21 m22
p=
c.
2M
Note the symmetry of the expression in m1 and m2 .
(d) Show that if the masses are equal, then E1 = E2 and that these are independent of
the common mass of the particles.
(e) Without going through the previous parts, show that if a particle of mass M at rest
decays into two identical particles, the energies of the two particles are equal and this
energy is the same regardless of their mass. In particular, whether M decays into two
massive or massless identical particles, the particles carry the same amount of energy.
What is that energy?
Solution:
(a) Initial momentum is zero. Therefore, the sum of the final momenta is also zero. Let
p denote the magnitude of the common momentum of the two particles.
(b) Conservation of energy is M c2 = E1 + E2 , or
q
q
M c2 = E1 + p2 c2 + m22 c4 = E1 + E12 − m21 c4 + m22 c4
or
(M c2 − E1 )2 = E12 − m21 c4 + m22 c4 ,
yielding
E1 =
M c2 (m21 − m22 )c2
+
.
2
2M
E2 =
M c2 (m22 − m21 )c2
+
.
2
2M
Similarly,
82
Spacetime Momentum
(c) From (b), we have
2 2
p c =
E12
−
m21 c4
M 2 c2 + (m21 − m22 )c2
=
2M
2
− m21 c4 ,
or
M 4 + (m21 − m22 )2 + 2M 2 (m21 − m22 )
p2
=
− m21
c2
4M 2
M 4 + m41 + m42 − 2m21 m22 − 2M 2 (m21 + m22 )
=
4M 2
2
2
2
2
(M − m1 ) + (M − m22 )2 − M 4 − 2m21 m22
=
.
4M 2
(d) If m1 = m2 , then (b) gives
E1 =
M c2
= E2 .
2
(e) Identity of the particles plus the fact that the initial momentum is zero implies that the
particles have the same mass and momentum, therefore the same energy. Conservation
of energy now gives 2E = M c2 , or E = 12 M c2 .
5.13. The neutral pion π 0 has a mass of 2.41 × 10−28 kg. It decays into two photons 98.8%
of the times.
(a) What is the energy of each photon in the rest frame of the pion?
(b) What is the wavelength of each photon? Remember that λ = hc/E where h is the
Planck constant.
(c) Assume that the pion is moving at 0.9c in the lab frame in the same direction as one
of the photons. What are the energies and wavelengths of each photon in the lab?
Solution:
(a) From symmetry (or look at the solution to Problem 5.12), the energies of the two
photons should be equal. Therefore,
e=
mπ0 c2
2.41 × 10−28 × (3 × 108 )2
=
= 1.08 × 10−11 J.
2
2
(b)
6.626 × 10−34 × 3 × 108
hc
=
= 1.84 × 10−14 m.
e
1.08 × 10−11
This is in the short-wave gamma ray region of the EM spectrum.
λ=
(c) Assume that the pion is moving in the positive x direction in the lab. Then (5.8)
gives
s
1+β
e01 = γ(βp1 c + e) = γ(βe + e) =
e = 4.7 × 10−11 J
1−β
s
1−β
e02 = γ(βp2 c + e) = γ(−βe + e) =
e = 2.48 × 10−12 J,
1+β
83
and
6.626 × 10−34 × 3 × 108
= 4.23 × 10−15 m
4.7 × 10−11
6.626 × 10−34 × 3 × 108
λ02 =
= 8.08 × 10−14 m.
2.48 × 10−12
λ01 =
5.14. The negative pion π − has a mass of 2.49 × 10−28 kg. It decays into a muon and
a neutrino 99.988% of the times. Muon has a mass of 1.89 × 10−28 kg and neutrino is
essentially massless.
(a) What are the energies of the muon and the neutrino in the pion’s rest frame?
(b) What is the momentum of the neutrino?
(c) What is the speed of the muon?
(d) Assume that the pion is moving at 0.9c in the lab frame in the same direction as the
neutrino. What are the energies and momenta of the muon and the neutrino in the
lab?
Solution: The formulas are given in the solution of Problem 5.12.
(a) The unit that is more conveniently used for the masses (times c2 ) is MeV. In that
unit mπ− c2 = 139.57 MeV and mµ c2 = 105.66 MeV, and
m2µ c2
mπ− c2
139.57 105.662
+
=
+
= 109.78 MeV
2
2mπ−
2
279.14
m2µ c2
m − c2
139.57 105.662
Eν = π
−
=
−
= 29.79 MeV.
2
2mπ−
2
279.14
Eµ =
(b) Since neutrino is assumed massless, Eν = pν c, and pν = 29.79 MeV/c.
(c) Muon has the same momentum as the neutrino. Thus, by (5.7),
βµ =
pν c
Eν
29.79
=
=
= 0.27.
Eµ
Eµ
109.78
(d) Assume that the pion is moving in the positive x direction in the lab. Then (5.8)
gives
Eµ0 = γ(βpµ c + Eµ ) = 2.29(−0.9 × 29.79 + 109.78) = 190 MeV
Eν0 = γ(βpν c + Eν ) = 2.29(0.9 × 29.79 + 29.79) = 129.6 MeV.
and
p0µ c =
q
p
Eµ0 2 − m2µ c4 = 1902 − 105.662 = 157.9 MeV ⇐⇒ p0µ = 157.9 MeV/c
p0ν c = Eν0 = 129.6 MeV ⇐⇒ p0ν = 129.6 MeV/c.
84
Spacetime Momentum
5.15. This problem continues the discussion of Example 5.3.4. Two 0.5-kg pieces of clay,
each moving at the rate of 2000 m/s toward the other, collide.
(a) What is the total KE distributed among the molecules of the end piece?
(b) Each kilogram of clay has about 4 moles, or about 2.4 × 1024 molecules. What is the
average KE that each molecule receives?
(c) Use KEavg = 23 kB T , where kB = 1.38 × 10−23 J/K is the Boltzmann constant, to find
the temperature of the final piece of clay.
Solution: The problem is entirely classical.
(a)
KEtot = 2KE = mv 2 = 0.5 × 20002 = 2 × 106 J.
(b)
KEavg =
2 × 106
= 8.33 × 10−19 J.
2.4 × 1024
(c)
T =
2KEavg
2 × 8.33 × 10−19
=
= 40258 K.
3kB
3 × 1.38 × 10−23
CHAPTER
6
Relativity in Four Dimensions
Problems With Solutions
6.1. Multiply the matrices in Equation (6.5) to obtain the three equations of (6.6). Solve
these equations to find all matrix elements in terms of a11 .
Solution: Multiply the two matrices on the right:
a11 a21
a11
a12
1 0
.
=
0 −1
a12 a22
−a21 −a22
Continue the multiplication:
a211 − a221
a11 a12 − a21 a22
1 0
.
=
0 −1
a11 a12 − a21 a22
a212 − a222
Now note that two matrices are equal if and only if all their corresponding elements are
equal. This leads to (6.6). The rest is given in the text.
6.2. Show that if β~ is along the x-axis, then (6.14) reduces to (6.13) and (6.16) to (6.12).
Solution: With βx = β and βy = 0 = βz , the 4 × 4 matrix in (6.14) becomes obviously
equal to the matrix in (6.13) except for the element in the second row and second column.
But that element is equal to:
1 + β̂x2 (γ − 1) = 1 + β̂ 2 (γ − 1) = γ
because β̂ 2 = 1.
With β~ = hβ, 0, 0i and ~r = hx, y, zi, we get β~ ·~r = βx. Plugging this in the first equation
of (6.16) gives the first equation in (6.12). Write the second equation of (6.16) as a column
vector:
 0   

 
x
x
γβct
1
y 0  = y  +  0  + (γ − 1) 0 x,
z0
z
0
0
86
Relativity in Four Dimensions
where I used β̂ · ~r = x. Equating the first component on either side yields
x0 = x + γβct + (γ − 1)x = γ(βct + x)
and the equality of the second and third components gives y 0 = y and z 0 = z.
6.3. Derive Equation (6.17).
Solution: Write ~r = ~r|| + ~r⊥ where ~r|| · β~ = ~r|| β~ and ~r⊥ · β~ = 0. Then the first equation
of (6.17) follows immediately. Writing ~r 0 = ~r||0 + ~r⊥0 as well, the second equation of (6.16)
can be expressed as
~ + ~r⊥ .
~ + (γ − 1) β̂ ~r|| = γ(~r|| + βct)
~r||0 + ~r⊥0 = ~r|| + ~r⊥ + γ βct
| {z }
=~
r||
Equating the parallel and perpendicular components of both sides yield the remaining
equations of (6.17).
6.4. Derive Equation (6.18).
Solution: The differentials of (6.16) for a “ball” are
cdt0 = γ(cdt + β~ · d~rb )
~
d~r 0 = d~rb + γ βcdt
+ (γ − 1)β̂ β̂ · d~rb .
b
Divide the second equation by the first to get
~
d~rb + γ βcdt
+ (γ − 1)β̂ β̂ · d~rb
β~b0 =
γ(cdt + β~ · d~rb )
d~rb /cdt + γ β~ + (γ − 1)β̂ β̂ · (d~rb /cdt)
=
γ[1 + β~ · (d~rb /cdt)]
β~b + γ β~ + (γ − 1)β̂ β̂ · β~b
=
.
γ(1 + β~ · β~b )
6.5. Observer O moves relative to observer O0 with velocity β~1 . Observer O0 moves relative
to observer O00 with velocity β~2 . Therefore, observer O moves relative to observer O00 with
~ Using matrices—as given in Equations (6.14) and (6.15)—find the general
some velocity β.
relativistic law of addition of velocities. Hint: See Example 3.5.1 and note that calculating
the two elements in the first column of the block form of the product Λ1 Λ2 is sufficient to
yield the answer.
Solution: If (ct, ~r) is the 4-vector in O, then Λ1 (ct, ~r) is (ct0~r 0 ), the 4-vector in O0 and
~1 and Λ2 is
Λ2 Λ1 (ct, ~r) is (ct00 , ~r 00 ), the 4-vector in O00 . Therefore, if Λ1 is parametrized by β
~
~
parametrized by β2 , then Λ = Λ2 Λ1 is parametrized by β. This means that

 

 

e~
e
e↔
e
e
e
~
~
~
~
~
~
γ
γ
+
γ
γ
β
β
γ
γ
β
+
γ
β
Λ
γ γβ
γ
γ2 β 2
γ
γ1 β 1
2 1 1
2 2 1
2 1
2 1 2 1

  2
 1
 


=

=
.
↔
↔
↔
e
↔
↔
↔
~
~
~
~
~
~
~
γβ Λ
γ2 β2
Λ2
γ1 β1
Λ1
γ2 γ1 β2 + γ1 Λ2 β1 γ2 γ1 β2 β 1 + Λ2 Λ1
(6.1)
87
e
e
Note that β~1 and β~2 are column vectors and β~ 1 and β~ 2 row vectors. So,
 
β1x
e
β~ 2 β~1 = β2x β2y β2z β1y  = β2x β1x + β2y β1y + β2z β1z = β~2 · β~1 .
β1z
Therefore, equating the element in the first row and first column on the left and the right,
we get
~2 ).
γ = γ1 γ2 (1 + β~1 · β
(6.2)
I’m not going to find all the elements of the product. All I need for the purposes of this
problem is the element in the second row and first column of the product. First I calculate
the matrix product of the second term:
 


 
β̂2x2
β̂2x β̂2y β̂2x β̂2z
β1x
1 0 0
β1x
↔ ~
2


β1y 
Λ2 β1 = 0 1 0 β1y  + (γ2 − 1) β̂2x β̂2y
β̂2y
β̂2y β̂2z
2
β1z
0 0 1
β1z
β̂2x β̂2z β̂2y β̂2z
β̂2z




β̂2x β̂2 · β~1
β̂2x

 ~
~
~

= β1 + (γ2 − 1) β̂2y β̂2 · β1  = β1 + (γ2 − 1) β̂2y  β̂2 · β~1 ,
β̂2z
β̂2z β̂2 · β~1
or
↔ ~
Λ2 β1 = β~1 + (γ2 − 1)β̂2 β̂2 · β~1 .
So, the second-row-first-column element of the matrix on the right-hand side of (6.1) is
h
i
↔
γ2 γ1 β~2 + Λ2 β~1 = γ2 γ1 β~2 + γ1 β~1 + (γ2 − 1)β̂2 β̂2 · β~1 ,
and, using (6.2), the corresponding element on the left is
~
γ β~ = γ1 γ2 (1 + β~1 · β~2 )β.
Equating the last two equations gives
β~ =
=
h
i
γ2 γ1 β~2 + γ1 β~1 + (γ2 − 1)β̂2 β̂2 · β~1
γ1 γ2 (1 + β~1 · β~2 )
~1
β~1 + γ2 β~2 + (γ2 − 1)β̂2 β̂2 · β
,
γ2 (1 + β~1 · β~2 )
which is the relativistic LAV (6.18).
6.6. In Fizeau’s experiment (see Example 3.5.2), light moves perpendicular to the direction
of motion of the transparent medium.
(a) Show that
0
v =
s
c 2
n
+
v2
1
1− 2 .
n
(b) Show that for v << c, a complete drag—in which the classical LAV holds—yields
v0 =
c
nv 2
+
.
n
2c
88
Relativity in Four Dimensions
(c) Show that relativistically, for low velocity, the drag is not complete but is smaller by
v 2 /(2nc).
Solution: In Equation (6.18) of the text, let β~b = (c/n)êy , and β~ = βêx .
(a) Then
(1/n)êy + γβêx
~v 0
1
êy + βêx
=
=
c
γ
nγ
and
1
|~v 0 |2
1
1
1
2
2
2
2
= 2 2 + β = 2 (1 − β ) + β = 2 + β 1 − 2
c2
n γ
n
n
n
or
0
|~v | =
s
c 2
n
+
v2
1
1− 2 .
n
(b) From the classical LAV, we get
c
c2
c
~v = êy + vêx ⇐⇒ |~v 0 |2 = 2 + v 2 ⇐⇒ |~v 0 | =
n
n
n
r
0
Approximating the square root by
r
1+
1+
n2 v 2
.
c2
n2 v 2
n2 v 2
≈
1
+
c2
2c2
because v << c, we get
|~v 0 | =
c
nv 2
+
.
n
2c
(c) Write the relativistic result in (a) as
s
c
n2 v 2
1
0
|~v | =
1+ 2
1− 2 .
n
c
n
For v << c, this can be approximated as
s
c
n2 v 2
1
c
n2 v 2
1
0
1+ 2
|~v | =
1− 2 ≈
1+
1− 2
.
n
c
n
n
2c2
n
or
|~v 0 | ≈
c
nv 2
+
n
2c
1−
1
n2
=
c
nv 2
v2
+
−
.
n
2c
2nc
Comparing this with the result in (b), we see that the drag is smaller by v 2 /(2nc).
6.7. Derive the velocity (6.23) directly from (6.18).
~b = αêy and β~ = βêx . Equation (6.23) follows immediately.
Solution: In (6.18), let β
89
6.8. Use (6.25) and the trigonometric identity
tan( 12 ϕ0 ) =
sin ϕ0
1 + cos ϕ0
to show that Equation (6.26) holds.
Solution:
sin ϕ0
=
1 + cos ϕ0
sin ϕ
γ 1 + β~ cos ϕ
sin ϕ
=
~
~
γ(1 + β cos ϕ + β~
β + cos ϕ
1+
1 + β~ cos ϕ
q
v
u
2
~
u1 −
1
−
β
sin ϕ
sin ϕ
=
=
=t
γ(1 + β~ )(1 + cos ϕ)
1 + β~ 1 + cos ϕ
1+
tan( 12 ϕ0 ) =
+ cos ϕ)
β~
tan( 12 ϕ).
~
β
6.9. The headlights of a car send light only in the forward hemisphere as seen in the
rest frame of the car. Show that the maximum angle as seen by a ground observer is
tan−1 [1/(γβ)], or cos−1 β.
Solution: The maximum angle in the rest frame is π/2. Therefore, by (6.25), cos ϕ0 = β~ ,
sin ϕ0 = 1/γ, and tan ϕ0 = sin ϕ0 / cos ϕ0 = 1/( β~ γ).
6.10. Use Figure 6.1(d) and the law of sines to obtain
c
cβ
=
.
0
cos θ
sin(θ0 − θ)
Show that this can also be written as
tan θ0 =
β + sin θ
cos θ
which is the non-relativistic version of (6.28).
Solution: Figure 6.1 of the manual labels the appropriate angles. With those labels, the
law of sines immediately leads to
c
cβ
=
.
0
cos θ
sin(θ0 − θ)
Rewrite the equation above as
1
β
=
⇐⇒ sin θ0 cos θ − cos θ0 sin θ = β cos θ0 .
0
0
cos θ
sin θ cos θ − cos θ0 sin θ
Divide both sides by cos θ0 and solve for tan θ0 to get
tan θ0 =
β + sin θ
.
cos θ
90
Relativity in Four Dimensions
θʹ
cʹ
c
θ
cβ
Figure 6.1: The non-relativistic LAV for aberration.
6.11. Derive Equation (6.29). Hint: Substitute θ0 = θ + on the left-hand side of (6.28),
expand both sides keeping only the first powers of and β; then equate the small terms on
both sides.
Solution: The left-hand side is
tan θ0 = tan(θ + ) =
tan θ + tan ,
1 − tan θ tan which, to first order in gives
tan θ0 ≈
tan θ + = (tan θ + )(1 − tan θ)−1 .
1 − tan θ
Use binomial expansion to first order in to get
tan θ0 ≈ (tan θ + )(1 + tan θ) ≈ tan θ + + tan2 θ. = tan θ +
.
cos2 θ
~ cos θ. Thus equating the two
With γ ≈ 1, the right-hand side of (6.28) becomes tan θ + |β|/
sides, we get
~
|β|
~ cos θ,
tan θ +
=
tan
θ
+
⇐⇒ = |β|
cos2 θ
cos θ
~ cos θ.
and θ0 = θ + = θ + |β|
6.12. From a classical perspective, since satellite S2 of Figure 6.3 is moving away from R,
the light it emits moves slower than c while the light from S1 moves faster. To simplify the
classical derivation of (6.39) and (6.40), let y = 0. Let d1 and d2 be the distances of S1 and
S2 from R, respectively. Since time is universal classically, the emission of signals occur at
the same time according to O0 as well as O. Use this to show that
d1
d2
=
.
c+v
c−v
From this plus d1 + d2 = L find d1 and show that
1
x0R = x01 + L(1 + β),
2
which is (6.40) for γ ≈ 1 and y = 0. From this you can also find (6.39).
91
Solution: The light from S1 moves at c + v and covers the distance of d1 in the same time
interval that the light from S2 moves at c − v and covers the distance of d2 . Therefore,
d2
c−v
1−β
d1
=
⇐⇒ d2 =
d1 =
d1 ,
c+v
c−v
c+v
1+β
and d1 + d2 = L yields
L = d1 +
1−β
2d1
d1 =
⇐⇒ d1 = 12 L(1 + β).
1+β
1+β
The final answer is obtained once we note that d1 = x01 − x0R .
6.13. Show that
(a) Equation (6.41) reduces to the non-relativistic law when all velocities are small and
to (3.26) when all velocities are in the same direction.
(b) Take the dot product of both sides of Equation (6.41) to find |~
α0 |2 = α
~0 · α
~ 0 in terms
of unprimed quantities.
(c) From (b) calculate γα0 and verify Equation (6.46).
(d) Show also that if α
~ is the velocity of light, so that α
~ ·α
~ = 1, then α
~0 · α
~ 0 = 1 as well.
Solution:
(a) Keeping only first order terms in Equation (6.41), we get γ ≈ 1 and β~ · α
~ ≈ 0, and
0
~
the equation reduces to α
~ =α
~ + β, which is the classical LAV.
When all velocities are in the same direction, β̂ β̂ · α
~ =α
~ , and β~ · α
~ = β~ α
~ . Then
Equation (6.41) becomes
α
~0 =
~
α
~ + β~
γ(~
α + β)
=
,
~ )
~
γ(1 + β~ α
1 + β~ α
which is Equation (3.26) if you ignore the vector signs.
(b) The resulting denominator is simply γ 2 (1+ β~ · α
~ )2 . So, let’s manipulate the numerator
N um = [~
α + γ β~ + (γ − 1)β̂ β̂ · α
~ ] · [~
α + γ β~ + (γ − 1)β̂ β̂ · α
~]
2
~ + γ 2 β~
= α
2
+ (γ − 1)2 (β̂ · α
~ )2 + 2γ β~ · α
~ + 2(γ − 1)(β̂ · α
~ )2 + 2γ(γ − 1)β~ · α
~.
The sum of the third and fifth terms can be simplified:
sum 3rd+5th = (γ − 1)[(γ − 1)(β̂ · α
~ )2 + 2(β̂ · α
~ )2 ]
2
= (γ 2 − 1)(β̂ · α
~ )2 = γ 2 β~ (β̂ · α
~ )2 = γ 2 (β~ · α
~ )2 .
So the numerator now becomes
2
N um = α
~ + γ 2 β~
2
+ γ 2 (β~ · α
~ )2 + 2γ 2 β~ · α
~
2
= α
~ + γ 2 − 1 + γ 2 (β~ · α
~ )2 + 2γ 2 β~ · α
~
2
= α
~ − 1 + γ 2 [1 + (β~ · α
~ )2 + 2β~ · α
~ ],
and dividing it by the denominator leads to
2
|~
α0 |2 =
α
~ −1
1
+1=−
+ 1.
2
2
2
2
~
γ (1 + β · α
~)
γ γ (1 + β~ · α
~ )2
α
92
Relativity in Four Dimensions
(c) The last equation in (b) leads to
1
1
⇐⇒ γα0 = γα γ(1 + β~ · α
~ ).
=
2
2
2
2
~
γ α0
γα γ (1 + β · α
~)
(d) This trivially follows from the first equality of the last equation in (b).
6.14. What are the coordinates of a particle’s 4-velocity in its own rest frame? Use this
and Equation (6.45) to obtain (6.42), the particle’s 4-velocity in an arbitrary frame.
Solution: The coordinates of a particle’s 4-velocity in its own rest frame is (u0 , ~u) =
~
(1, 0, 0, 0). Substituting this in Equation (6.45) yields u00 = γ ≡ γβ and ~u 0 = γ β~ ≡ γβ β.
These are the components of (6.42), except for the fact that now β~ is the velocity of the
particle.
6.15. Provide the details of the proof of the statement: a particle is massless if and only
if it moves at light speed.
Solution: From Equation (6.52), we get
~v = c ⇐⇒ p~ /E = 1/c ⇐⇒ E = p~ c ⇐⇒ m = 0.
These are all if-and-only-if implications. In the last implication, I used (6.50).
~
6.16. Apply (6.53) to a photon moving in the β-direction
and use |~
p| = E/c to show that
s
1−β
0
E =
E.
1+β
Now use E = hc/λ to find a formula for the relativistic Doppler shift.
~ then p~ · β~ = p~ β~ . Therefore, (6.53) becomes
Solution: If p~ is parallel to β,
s
0
E = γ(E − c p~ β~ ) = γ(E − E β~ ) =
1−β
E,
1+β
using the notation β ≡ β~ . Writing the last equation in terms of wavelengths, we get
hc
=
λ0
s
1 − β hc
⇐⇒ λ0 =
1+β λ
This is the formula for the relativistic Doppler shift.
s
1+β
λ.
1−β
6.17. Use (6.52)—which holds for any particle in any frame—and (6.54) to find the general
relativistic law of addition of velocities (6.41).
93
Solution: Write (6.52) as c~
p/E = α
~ . Then divide the left-hand side of the second equation
in (6.54) by the first one, noting that c = 1:
α
~0≡
~ + (γ − 1)β̂ β̂ · p~
p~ 0
p~ + γ βE
=
.
E0
γ(E + β~ · p~)
Now divide the numerator and denominator by E:
α
~0=
α
~ + γ β~ + (γ − 1)β̂ β̂ · α
~
.
~
γ(1 + β · α
~)
This is precisely (6.41).
6.18. Show that the 4-acceleration is orthogonal to the 4-velocity.
Solution: From Equation (6.43) we have u • u = 1. Now differentiate this with respect to
proper time and note that the right-hand side gives zero:
a • u + u • a = 2u • a = 0.
6.19. Differentiate the space component of the 4-velocity with respect to τ to get the second
equation in (6.56). Now show directly that u0 a0 − ~u · ~a = 0.
Solution: Use dot to denote differentiation with respect to t: Ȧ ≡ dA/dt.
~a =
d~u
d~u
d
= γα
= γα (γα α
~ ) = γα α
~ γ˙α + γα α
~˙ .
dτ
dt
dt
~ ·α
~˙ obtained in the calculation of a0 just before Equation (6.56) to obtain
Now use γ̇α = γα3 α
h i
h
i
~a = γα α
~ γα3 α
~ ·α
~˙ + γα α
~˙ = γα2 γα2 (~
α·α
~˙ )~
α+α
~˙ .
From (6.42), we have
h
i
u0 a0 − ~u · ~a = γα (a0 − α
~ · ~a) = γα γα4 α
~ ·α
~˙ − γα4 (~
α·α
~˙ )~
α·α
~ − γα2 α
~ ·α
~˙
= γα3 α
~ ·α
~˙ γα2 − γα2 α
~ ·α
~ −1 =0
because γα2 α
~ ·α
~ = γα2 − 1.
6.20. Derive Equation (6.58).
Solution: If you dot the first equation on the second line of (6.57) with α
~ you get
α
~ · ~u˙ = γα3 (~
α·α
~˙ )~
α·α
~ + γα α
~ ·α
~˙ = γα (~
α·α
~˙ ) γα2 α
~ ·α
~ +γ α
~ ·α
~˙ = γα3 α
~ ·α
~˙ = u̇0 .
| {z } α
2 −1
=γα
6.21. Derive Equation (6.60).
94
Relativity in Four Dimensions
Solution: Using (6.59) and (6.46), we can write
~ 0 + (γ − 1)β̂ β̂ · ~a
~a + γ βa
~a 0
=
~u˙ 0 =
.
γ α0
γγα (1 + β~ · α
~)
Now write a0 = γα u̇ and ~a = γα ~u˙ —coming from (6.57)—to rewrite the above as
γα ~u˙ + γα γ β~ u̇0 + γα (γ − 1)β̂ β̂ · ~u˙
~u˙ 0 =
~·α
γγα (1 + β
~)
~ α · ~u˙ + (γ − 1)β̂ β̂ · ~u˙
~u˙ + γ β~
=
,
γ(1 + β~ · α
~)
where I used (6.58) for u̇0 .
6.22. How long does it take a particle to attain a speed of 0.999c, if its acceleration is 10
m/s2 ? What is the answer based on Newtonian mechanics? How do the answers change if
the ultimate speed of the particle is 0.99999c?
Solution: The appropriate equation is that before (6.68), with F/m = 10m/s2 . For α =
0.999, we get
0.999 × 3 × 108
√
= 10t ⇐⇒ t = 6.7 × 108 s = 21.28 years.
1 − 0.9992
Note that I had to restore the factor of c to make units right on both sides. If you use
Newtonian mechanics, you get
cα = F t/m ⇐⇒ 0.999 × 3 × 108 = 10t ⇐⇒ t = 2.997 × 107 s = 0.95143 year.
For 0.99999c, relativity gives
0.99999 × 3 × 108
√
= 10t ⇐⇒ t = 6.7 × 1010 s = 2130 years,
2
1 − 0.99999
while classical mechanics gives
0.99999 × 3 × 108 = 10t ⇐⇒ t = 2.99997 × 107 s = 0.95237 year.
The difference between this and the previous classical time is only 8.25 hours!
6.23. Differentiate both sides of E 2 = p~ · p~ + m2 with respect to time and use p~ = E~
α to
obtain (6.65).
Solution:
d 2
d
dE
(E ) = (~
p · p~ + m2 ) ⇐⇒ 2E
= 2~
p·
dt
dt
dt
d~
p
dt
⇐⇒
dE
= (~
p/E) · F~ .
dt
This is the result we are after, because p~/E = α
~.
6.24. Write F~ as F~|| + F~⊥ on both sides of (6.66) and derive (6.67).
95
Solution: Write (6.66) as
F~||0 + F~⊥0 =
F~|| + F~⊥ + γ β~ α
~ · (F~|| + F~⊥ ) + (γ − 1)β̂ β̂ · (F~|| + F~⊥ )
.
γ(1 + β~ · α
~)
Now note that β̂ · F~⊥ = 0 and β̂ β̂ · F~|| = F~|| . These give us a new equation:
F~||0 + F~⊥0 =
F~⊥ + γ β~ α
~ · F~|| + γ β~ α
~ · F~⊥ + γ F~||
γ(1 + β~ · α
~)
=
F~⊥ + γ F~|| β~ · α
~ + γ β~ α
~ · F~⊥ + γ F~||
.
γ(1 + β~ · α
~)
Notice the change in the second term of the numerator. We can now finally write this as
F~⊥ + γ β~ α
~ · F~⊥
β~ α
~ · F~⊥
F~⊥
F~||0 + F~⊥0 = F~|| +
= F~|| +
+
.
γ(1 + β~ · α
~)
1 + β~ · α
~
γ(1 + β~ · α
~)
The sum of the first two terms on the right-hand side is the parallel component, and the
third term is the perpendicular component of the right-hand side. Setting the corresponding
components equal on both sides yields (6.67).
6.25. Confine yourself to one dimension.
(a) Use E 2 = p2 + m2 for a moving object of mass m to show that dE/dt = F α, where
F = dp/dt is the force and α is the speed of the object.
(b) Lorentz transform E, p, t, and x to another reference frame and use the result obtained
in (a) to show that in one dimension force is invariant, i.e., it does not change when
you go from one RF to another.
Solution: Remember that α = p/E.
(a) With that in mind, differentiating E 2 = p2 + m2 gives
2EdE/dt = 2pdp/dt ⇐⇒ EdE/dt = pF ⇐⇒ dE/dt = (p/E)F = αF.
(b) Let F act on an object for dt to displace it by dx and change it momentum by dp and
its energy by dE, all in reference frame O. In another RF O0 with respect to which
O moves in positive x direction, we have
dp0 = γ(dp + βdE),
dt0 = γ(dt + βdx).
Now divide the first by the second to get
F 0 = dp0 /dt0 =
γ(dp + βdE)
dp/dt + βdE/dt
F + βαF
=
=
= F,
γ(dt + βdx)
1 + βdx/dt
1 + βα
where I used the result in (a).
96
Relativity in Four Dimensions
6.26. Define the power P consumed by an object as P ≡ dE/dt = α
~ · F~ , where E is the
relativistic energy of the object and F~ and α
~ are, respectively, the force acting on the object
(the source of the power) and the velocity of the object. If P and P 0 are powers in the
reference frames O and O0 , respectively, then
P0 =
P + β~ · F~
,
1 + β~ · α
~
where β~ is the velocity of O relative to O0 . Prove this in three different ways:
(a) by writing the Lorentz transformation of dE 0 and dt0 , and dividing the first by the
second;
(b) by noting that P 0 = α
~ 0 · F~ 0 and using the relativistic law of addition of velocities and
the transformation rule for the three-force F~ ;
(c) by Lorentz transforming the zeroth component of the 4-force.
Solution:
(a) In a reference frame O, let F~ act on an object for dt to displace it by d~r and change
its momentum by d~
p and its energy by dE. In another RF O0 with respect to which
~ we have
O moves with velocity β,
dE 0 = γ(dE + β~ · d~
p),
dt0 = γ(dt + β~ · d~r).
Divide the first by the second:
P0 =
dE 0
γ(dE + β~ · d~
p)
dE/dt + β~ · (d~
p/dt)
P + β~ · F~
=
=
=
.
~ · (d~r/dt)
dt0
γ(dt + β~ · d~r)
1+β
1 + β~ · α
~
(b) Use (6.41) and (6.66) in P 0 = α
~ 0 · F~ 0 to obtain
# "
#
"
α
~ + γ β~ + (γ − 1)β̂ β̂ · α
~
F~ + γ β~ α
~ · F~ + (γ − 1)β̂ β̂ · F~
0
P =
·
.
γ(1 + β~ · α
~)
γ(1 + β~ · α
~)
~·α
The denominator of the right-hand side is simply γ 2 (1 + β
~ )2 . So, let’s calculate
the numerator:
N um = α
~ · F~ + γ~
α · β~ α
~ · F~ + 2(γ − 1)β̂ · α
~ β̂ · F~ + γ β~ · F~ + γ 2 β~ · β~ α
~ · F~
| {z }
=γ 2 −1
~ α · F~ + (γ − 1) β̂ · α
+ γ(γ − 1)β~ · F~ + γ(γ − 1)~
α · β~
~ β̂ · F~ .
2
The sum of the third and last term is
3rd+last = 2(γ − 1)β̂ · α
~ β̂ · F~ + γ β~ · F~ + (γ − 1)2 β̂ · α
~ β̂ · F~
= (γ − 1)(2 + γ − 1)β̂ · α
~ β̂ · F~ = (γ 2 − 1)β̂ · α
~ β̂ · F~
~ · F~ .
= γ 2 β 2 β̂ · α
~ β̂ · F~ = γ 2 β~ · α
~β
Substituting this in the previous expression and simplifying yields
~ α · F~
N um = γ 2 α
~ · F~ + γ 2 α
~ · β~ β~ · F~ + γ 2 β~ · F~ + γ 2 α
~ · β~
~ α · F~ + β~ · F~ ) = γ 2 (1 + α
~
= γ 2 (1 + α
~ · β)(~
~ · β)(P
+ β~ · F~ ).
Dividing this by the denominator gives the result.
97
(c) By (6.64), f00 = γα0 P 0 . Lorentz transforming this gives
γα0 P 0 = γ(f0 + β · f~) = γ[γα P + β · (γα F~ )]
or
P0 =
γγα (P + β · F~ )
.
γ α0
Using Equation (6.46) gives the final answer.
6.27. Provide the missing steps of (6.71) and (6.72) to arrive at (6.73).
Solution: To go from the second line of (6.71) to the last line, just note that
1+
γ2β2
γ2 − 1
(γ − 1)(γ + 1)
=1+
=1+
= γ.
γ+1
γ+1
γ+1
The first line of (6.72) comes from the identity
γ
γ(γ + 1) − γ 2
=
,
γ+1
γ+1
~ = β β̂ and therefore, β~ × (β~ × F~ ) = β 2 β̂ × (β̂ × F~ ).
and the second line from β
CHAPTER
7
Relativistic Photography
Problems With Solutions
7.1. Consider the cube of Figure 7.10(a) which moves with speed β along the positive x-axis
relative to observer O0 . Define the angle θ as sin θ ≡ β. It is argued1 that—since Ab Af is
perpendicular to the direction of motion and thus does not change length—it takes light
2a/c to travel from Ab to Af according to O0 . During this time, the cube moves a distance
of
d = (βc)(2a/c) = 2aβ ≡ 2a sin θ,
exposing the trailing face of the cube to the camera. Furthermore, the top and bottom
sides, Af Df and Bf Cf , shrink from 2a to
p
p
s = 2a 1 − β 2 = 2a 1 − sin2 θ = 2a cos θ.
As Figures 7.10(c) and 7.10(d) indicate, the combination of these two effects manifests itself
as the appearance of a rotation on the photographic plate. To see what is wrong with this
argument, do the following.
(a) Consider two events: emission of light from Ab and its arrival at Af . Write the
spacetime coordinates of these two events in O where the cube is at rest, assuming
that the emission event takes place at time tb .
(b) Lorentz transform these two events to O0 .
(c) What is the time difference t0f − t0b between the emission at Ab and arrival at Af
according to O0 ? How far does the cube move during this time according to O0 ? Do
these results agree with the argument for rotation?
(d) You can derive the result of (c) without using Lorentz transformation. Simply note
that according to O0 , the emission and arrival events occur at the two ends of the
hypotenuse of a right triangle whose other two sides are of lengths 2a and β(t0f − t0b ).
1
See, for example, Weisskopf, V. F., Phys. Today 13(9), 24–27 (1960).
100
Relativistic Photography
y
Ab
Af
Db
Df
Db
Ab
Df
Af
Ab
Df
Af
Cb
Bf
x
Cf
Cf
Bf
€
(a)
Df
θ
2a
d
Af
s
θ
Bb
d
Bf
Cf
(d)
(c)
(b)
s
Figure 7.1: (a) The distant cube of side length 2a is centered at the origin of O, its rest frame. (b) The
cube as it appears on the photographic plate of camera C in O. (c) The orientation of the top face of the
cube when the cube is rotated counterclockwise by an angle θ = sin−1 β about the y-axis. (d) The image
of the rotated cube on the photographic plate of camera C in O.
Solution: The figure has been reproduced in Figure 7.1 of the manual.
(a) Let Eb denote the emission of light from Ab and Ef its arrival at Af . Then, with
c = 1, Eb has coordinates (tb , −a, a, −a) and Ef has coordinates (tb + 2a, −a, a, a).
(b) For Eb , we get
t0b = γ(tb − βa),
x0b = γ(−a + βtb ),
and for Ef , we get
t0f = γ(tb + 2a − βa),
x0f = γ[−a + β(tb + 2a)].
(c) t0f − t0b = 2aγ and x0f − x0b = 2aγβ = β(t0f − t0b ). These results do not agree with the
argument for rotation!
(d) The hypotenuse of the right triangle is the path of light. So, it has length t0f − t0b .
Hence,
(t0f − t0b )2 = 4a2 + β 2 (t0f − t0b )2 ⇐⇒ (t0f − t0b )2 (1 − β 2 ) = 4a2 ⇐⇒ t0f − t0b = 2aγ.
This is precisely the kind of reasoning that gave us the time dilation formula (2.3)
from Figure 2.2. The distance x0f − x0b is just the speed times this time interval.
7.2. Use the space coordinates of (7.7) to calculate |r0i − r0 | and show that the result agrees
with the right-hand side of t0i .
Solution: With r0i − r0 = hx0i , yi0 , zi0 − bi, we have
|r0i − r0 |2 = x0i 2 + yi0 2 + (zi0 − b)2 = γ 2 (xi − β|ri − r0 |)2 + yi2 + (zi − b)2
= γ 2 x2i + γ 2 β 2 |ri − r0 |2 − 2γ 2 βxi |ri − r0 | + yi2 + (zi − b)2
|
{z
}
=|ri −r0 |2 −x2i
= (γ 2 − 1)x2i + (γ 2 − 1)|ri − r0 |2 − 2γ 2 βxi |ri − r0 | + |ri − r0 |2 ,
101
where I used the ever useful formula: γ 2 β 2 = γ 2 − 1. Now use it backwards in the first
term and cancel terms to obtain
|r0i − r0 |2 = γ 2 β 2 x2i + γ 2 |ri − r0 |2 − 2γ 2 βxi |ri
= γ 2 (−βxi + |ri − r0 |)2 ,
or |r0i −r0 | = γ(|ri −r0 |−βxi ), where the sign is chosen to make the right-hand side positive.
7.3. Substitute Equations (7.7) and (7.8) in (7.9) and follow the steps similar to the ones
that led to (7.5) and (7.6) to obtain (7.10).
Solution: The first equation in (7.7) provides the denominator of (7.9):
Den = γ 2 (|r1 − r0 | − βx1 )(|r2 − r0 | − βx2 )
βx2
βx1
2
1−
.
= γ |r1 − r0 ||r2 − r0 | 1 −
|r1 − r0 |
|r2 − r0 |
The numerator can be expressed as
N um = γ 2 (x1 − β|r1 − r0 |)(x2 − β|r2 − r0 |) + y1 y2 + (z1 − b)(z2 − b)
= γ 2 (x1 − β|r1 − r0 |)(x2 − β|r2 − r0 |) + |r1 − r0 ||r2 − r0 | cos α12 − x1 x2
= (γ 2 − 1)x1 x2 + γ 2 β 2 |r1 − r0 ||r2 − r0 |
− γ 2 x1 β|r2 − r0 | − γ 2 x2 β|r1 − r0 | + |r1 − r0 ||r2 − r0 | cos α12 .
Invoking the identity γ 2 β 2 = γ 2 − 1 in the first and the second terms and simplifying, you
get
N um = γ 2 β 2 x1 x2 + γ 2 |r1 − r0 ||r2 − r0 | − |r1 − r0 ||r2 − r0 |
− γ 2 x1 β|r2 − r0 | − γ 2 x2 β|r1 − r0 | + |r1 − r0 ||r2 − r0 | cos α12
= γ 2 (|r1 − r0 | − βx1 )(|r2 − r0 | − βx2 ) − |r1 − r0 ||r2 − r0 |(1 − cos α12 ).
Dividing the first term by the first form of the denominator and the second term by the
second form of the denominator gives
0
cos α12
=1−
γ2
1 − cos α12
1
1−
1 − |r1βx
−r0 |
βx2
|r2 −r0 |
.
Now use the trigonometric identity 1 − cos θ = 2 sin2 (θ/2) to obtain the final result.
7.4. Derive Equations (7.14) and (7.15).
Solution: It is actually more convenient to use the first equality in (7.12). If you write the
second equation of (7.12) as
q
q
2
2
y0 x02
y0 x02
q +b
q +b
b(z0 − b)
0
x0q x0 − b(z − b) =
⇐⇒
x
=
+
v0
x0q v 0
x0q
102
Relativistic Photography
and substitute both in the first equation, you obtain
x0q (z0 − b) 0
b
v0
0
0
0 0
q
q
[bx + xq (z0 − b)] =
vx + q
v
u =
2
02 + b2
2
y0 x02
+
b
y
x
y0 x02
0
q
q
q +b
 q

02 + b2
y
x
0
q
x0q (z0 − b) 0
b
b(z0 − b) 
q
= q
v0 
+
+
v
x0q v 0
x0q
y x02 + b2
y x02 + b2
0
0
or
0
q
q
q
0 2 + b2 )
(z
−
b)
x0q2 + b2
0
(z
−
b)(x
0
q
q
x0q u0 = b + v 0
= b + v0
,
y0
2
y0 x02
+
b
q
and
y0 x0q
by0
q
q
u0 −
.
(z0 − b) x0q2 + b2
(z0 − b) x0q2 + b2
v0 =
Equation (7.14) is obtained by noting that x0q = xq /γ.
Derivation of (7.15) is very straightforward, and I’ll leave it for you to do.
7.5. Let xq = 0 and derive (7.17) from (7.16).
Solution: With xq = 0, (7.16) becomes
i
h
p
γ x0 − β x20 + y 2 + (z0 − b)2
u0 = −
z0 − b
y
0
v =−
,
z0 − b
or
p
x20 + y 2 + (z0 − b)2
z0 − b
y
0
v =−
,
z0 − b
γβ
γx0
u +
=
z0 − b
0
or
γx0
u +
z0 − b
0
2
=
γ 2 β 2 [x20 + y 2 + (z0 − b)2 ]
(z0 − b)2
y 2 = v 0 2 (z0 − b)2 .
Substituting the second equation in the first yields
γx0 2
γ 2 β 2 [x20 + (z0 − b)2 ]
0
= γ 2β 2v0 2 +
.
u +
z0 − b
(z0 − b)2
Dividing both sides by the last term gives (7.17).
7.6. Derive Equation (7.19) from Equation (7.18).
Solution: Transfer γx0 to the left of Equation (7.18) and square both sides.
103
7.7. Show that the line of Equation (7.20) makes an angle θ with the x0 -axis given by
cos θ = β.
Solution: The slope is 1/(γβ). Therefore, tan θ = 1/(γβ), and
cos θ = √
1
1+
tan2 θ
=p
γβ
γβ
=p
= β.
γ2β2 + 1
γ2 − 1 + 1
7.8. Derive Equation (7.23) from Equation (7.22).
Solution: Substitute for b − z from the second equation of (7.22) in the first:
i
h
p
γ x0 − β x20 + y02 + (y0 /v 0 )2
u0 =
y0 /v 0
or
q
2
2
2
0
2
y0 u = γ x0 v − β v (x0 + y0 ) + y0
0
0
or
y0 u0 − γx0 v 0 = −γβ
q
v 0 2 (x20 + y02 ) + y02 .
Squaring both sides and simplifying yields (7.23).
7.9. Define a new pair of photographic plate coordinates by
0
u0 = u0new cos θ + vnew
sin θ,
with
x0
sin θ = p
,
2
γ y02 + x20
0
v 0 = −u0new sin θ + vnew
cos θ
γy0
cos θ = p
.
2
γ y02 + x20
Note that the new axes are obtained from the old by a clockwise rotation of angle θ.
(a) Substitute these values in Equation (7.23) and simplify to show that that equation
reduces to
2
02
(x20 + y02 )u0new
− β 2 γ 2 y02 vnew
− γ 2 β 2 y02 = 0,
which is the equation of a hyperbola.
(b) What are the coordinates of the center of this hyperbola in the new coordinate system?
(c) What are the equations of the axes of the hyperbola in the new coordinate system?
In the old coordinate system?
Solution:
(a) Substitute the new pair of photographic plate coordinates in (7.23) and look at each
resulting term separately:
2
02
0
1st term = y02 (u0new
cos2 θ + vnew
sin2 θ + 2u0new vnew
sin θ cos θ)
02
2
2 2 2
2
02
2
0
2nd term = x0 − γ β y0 (unew sin θ + vnew cos θ − 2u0new vnew
sin θ cos θ)
2
0
02
3rd term = −2γx0 y0 [−u0new
sin θ cos θ + u0new vnew
(cos2 θ − sin2 θ) + vnew
sin θ cos θ].
104
Relativistic Photography
First consider the new cross term, which can be written as
2
0
cross term = 2u0new vnew
y0 − x20 + γ 2 β 2 y02 sin θ cos θ − γx0 y0 (cos2 θ − sin2 θ) .
Let X denote the expression inside the square bracket and use γ 2 β 2 = γ 2 − 1 in its
first term. Then, substituting
x0
,
sin θ = p
γ 2 y02 + x20
γy0
cos θ = p
γ 2 y02 + x20
in that expression yields
X=
γ 2 y02
−
x20
γx0 y0
− γx0 y0
2
γ y02 + x20
γ 2 y02 − x20
γ 2 y02 + x20
= 0.
2 is
The coefficient that multiplies u0new
y02 cos2 θ + x20 − γ 2 β 2 y02 sin2 θ + 2γx0 y0 sin θ cos θ
or
or
γ 2 y04
x20
2γ 2 x20 y02
2
2 2 2
+
x
−
γ
β
y
+
0
0
γ 2 y02 + x20
γ 2 y02 + x20 γ 2 y02 + x20
=1+1/γ 2
z }| {
γ 2 y04 + x40 + γ 2 x20 y02 (2 − β 2 )
γ 2 y04 + x40 + γ 2 x20 y02 + x20 y02
=
= x20 + y02 .
2
2
2
2
2
2
γ y0 + x0
γ y0 + x 0
0 2 is −β 2 γ 2 y 2 .
Similarly, you can show that the coefficient of vnew
0
(b) Transfer the constant term of the equation in part (a) of the problem to the right-hand
side and divide both sides by that term to obtain
02
2
(x20 + y02 )unew
u0new
02
02
− vnew
= 1.
−
v
=
1
⇐⇒
new
2
2
2
2
2
2
2
2
γ β y0
γ β y0 /(x0 + y0 )
p
This shows that the center is at the origin, the semi-major axis is a = γβy0 / x20 + y02 ,
and the semi-minor axis is b = 1.
(c) The equations of the axes of the hyperbola in the new coordinate system are
p
x20 + y02 0
0
unew .
vnew = ±
γβy0
Let α be the angle that the line with positive slope makes with the u0new axis. Then
p
x20 + y02
1
tan α =
⇐⇒ cos α = √
= β cos θ
γβy0
1 + tan2 α
To obtain the equations in the old coordinate system, substitute for the new in terms
of the old:
u0 sin θ + v 0 cos θ = ± tan α(u0 cos θ − v 0 sin θ).
or
cos α(u0 sin θ + v 0 cos θ) = ± sin α(u0 cos θ − v 0 sin θ).
105
or
v 0 (cos α cos θ ∓ sin α sin θ) = u0 (± sin α cos θ − cos α sin θ).
or
cos(α ± θ)v 0 = ∓u0 sin(α ± θ) ⇐⇒ v 0 = ∓ tan(α ± θ)u0 .
Thus, the axes of the old hyperbola are rotated by θ compared to the new axes.
This is what we should expect, because the new photographic plate coordinates are
obtained from the old by the same rotation.
7.10. Derive Equations (7.25) and (7.26).
Solution: All the front-face corners have z =
p a and their x and y coordinates are ±a.
Therefore, their distance from the pinhole is 2a2 + (b − a)2 . With c = 1, and the fact
that t is negative, we get the first equation in (7.25). The second equation follows the same
way. The only difference is that now z = −a.
To get the equations in (7.26), note that the only coordinates that matter are x and t.
Therefore, all the front corners have the same t-term in the Lorentz transformation. So, the
left front corners have the same x of −a, and their Lorentz transformations are identical.
Similarly for the rest of the Lorentz transformations.
7.11. Derive Equations (7.28) and (7.29).
Solution: With xq = 0, (7.12) becomes
u0 = −
x0
,
z−b
v0 = −
y
.
z−b
So, to get (7.28) and (7.29) just substitute for x0 , y, and z. Since all the front corners have
z = a, the denominator for all u0 s and v 0 s is b − a. Similarly, the denominator for all u0 s
and v 0 s is b + a for the back corners. To get (7.28) substitute the x0 s from (7.26), and to
get (7.29), substitute for y.
0 < v0
7.12. Using Equation (7.28), show that u0Ab > u0Af and vA
Af for all b and β. Hint:
b
0
Show that each of the two terms on the right-hand side of uAb is larger than the corresponding term on the right-hand side of u0Af .
Solution: Write u0Ab and u0Af as
s
u0Ab
γa
=−
− γβ
b+a
2a2
+1
(b + a)2
s
u0Af
γa
=−
− γβ
b−a
2a2
+ 1.
(b − a)2
It is now clear that the magnitude of each term in u0Ab is smaller than the magnitude of the
0 < v 0 is self-evident.
corresponding term in u0Af . Therefore, u0Ab > u0Af . The inequality vA
Af
b
106
Relativistic Photography
7.13. Recall that the aberration formula involves the angle with the direction of motion
of a ray of light from a single point source. Therefore, if the aberration formula is to be
interpreted as the angle between two light rays originating from an object (as is the case
when image formation of the object is being considered), then one of those rays ought to
be along the direction of motion. Show that when that is the case, then (7.5) reduces to
the aberration formula (6.25).
0 =
Solution: Assume that the first photon is in the direction of motion. Then, cos α12
cos ϕ0 , cos α12 = cos ϕ, ê1 = −β̂, and β~ · ê2 = −β cos ϕ, and (7.5) can be written as
1 − cos ϕ
(1 − β)(1 − cos ϕ)
=1−
γ 2 (1 + β)(1 + β cos ϕ)
1 + β cos ϕ
1 + β cos ϕ − (1 − β)(1 − cos ϕ)
β + cos ϕ
=
=
.
1 + β cos ϕ
1 + β cos ϕ
cos ϕ0 = 1 −
7.14. Let’s see what a cube looks like when it moves toward or away from the camera. Place
the camera at the origin (i.e., let b = 0) and point it to the right face of the approaching
cube as shown in Figure 7.3(b). Let the center of the cube have coordinates (xc , 0, 0) in O.
(a) Write down the coordinates of all the eight corners of the cube in O.
(b) From (a), (7.8), and (7.10) with b = 0, obtain
a
0
sin(αD
/2) = p
,
f Db
2
γ (xc + a) + 2a2 − β(xc + a)
a
0
sin(αA
/2) = p
,
f Ab
γ (xc − a)2 + 2a2 − β(xc − a)
(7.1)
the angles formed by sides Df Db and Af Ab (and the other three sides of the leading
and trailing faces) at the pinhole.
0
0
(c) For xc < 0 and β > 0, show that αD
> αA
, with the other three angles
f Db
f Ab
satisfying the same kind of inequality. Thus, the trailing face is hidden behind the
leading face when the cube is approaching the camera.
p
0
0
(d) For xc < 0 and β < 0, show that αD
> αA
if and only if |xc |/a > |β| 2γ 2 + 1.
f Db
f Ab
0
(e) Set β = 0 in (7.1) above to find sin(αDf Db /2). Then show that, for approach, αD
<
f Db
0
αDf Db , indicating that the picture in C is smaller than in C.
(f) The case of recession is more complicated. The limiting case of Equation (7.11) may
lead you to believe that the C 0 image is larger. For the moment assume that and see
what condition makes the assumption true. For definiteness, let β > 0 and xc > 0.
0
Then, for αA
> αAf Df to hold, you should have
f Df
γ
p
p
(xc + a)2 + 2a2 − β(xc + a) < (xc + a)2 + 2a2 .
Show that this is equivalent to
[(xc + a)2 − X− ][(xc + a)2 − X+ ] > 0,
X− < 0
107
for certain X+ and X− that you have to find. Thus, for the inequality to hold, (xc +a)2
must be larger than X+ . Now show that
(xc + a)2
> γ − 1.
a2
(7.2)
When the cube is sufficiently far away, this condition holds, but not in general. Therefore, if the cube is close enough to the camera C 0 when the pictures are taken, the
image in C can be larger.
(g) Show that (7.2) above holds also when β < 0 and xc < 0.
(h) Plot the ratio
0
sin(αD
/2)
f Db
sin(αDf Db /2)
for xc = +3a as a function of β for −1 < β < 1 to get a feel for the magnification of
the image in C 0 relative to the image in C.
(i) Verify directly that, when |xc | → ∞, the ratio in (h) reduces to
s
0
αD
1 + β|xc |/xc
f Db
≈
,
αDf Db
1 − β|xc |/xc
as in Equation (7.11).
Solution:
(a) Referring to Figure 7.3, with the center of the cube at (xc , 0, 0), we get
Df : (xc + a, −a, a),
Leading face
Cf : (xc + a, −a, −a),
Af : (xc − a, −a, a),
Trailing face
Bf : (xc − a, −a, −a),
Db : (xc + a, a, a),
Cb : (xc + a, a, −a)
Ab : (xc − a, a, a),
Bb : (xc − a, −a, −a)
(b) Substituting from (a) in (7.8) with b = 0, we get
cos αDf Db =
xDf xDb + yDf yDb + zDf zDb
(xc + a)2
=
|rDf | |rDb |
(xc + a)2 + 2a2
and
2 sin2 (αDf Db /2) = 1 − cos αDf Db = 1 −
or
(xc + a)2
2a2
=
,
(xc + a)2 + 2a2
(xc + a)2 + 2a2
a
.
sin(αDf Db /2) = p
(xc + a)2 + 2a2
Substitute this in (7.10) with b = 0 to obtain
√
0
sin(αD
/2) =
f Db
a
(xc +a)2 +2a2
γ 1− √
β(xc +a)
(xc +a)2 +2a2
=
a
p
.
γ
(xc + a)2 + 2a2 − β(xc + a)
0
sin(αA
/2) can be obtained in exactly the same way.
f Ab
108
Relativistic Photography
(c) For xc < 0 and β > 0, (7.1) becomes
a
0
sin(αD
/2) = p
,
f Db
γ (|xc | − a)2 + 2a2 + β(|xc | − a)
a
0
sin(αA
/2) = p
.
f Ab
2
γ (|xc | + a) + 2a2 + β(|xc | + a)
The denominator of the first is smaller than the denominator of the second. Therefore,
0
0
0
0
sin(αD
/2) > sin(αA
/2) ⇐⇒ αD
> αA
.
f Db
f Ab
f Db
f Ab
(d) For xc < 0 and β < 0, (7.1) becomes
a
0
sin(αD
/2) = p
,
f Db
2
γ (|xc | − a) + 2a2 − |β|(|xc | − a)
a
0
sin(αA
/2) = p
,
f Ab
γ (|xc | + a)2 + 2a2 − |β|(|xc | + a)
0
0
and αD
> αA
if and only if
f Db
f Ab
p
p
(|xc | − a)2 + 2a2 − |β|(|xc | − a) < (|xc | + a)2 + 2a2 − |β|(|xc | + a)
or
p
p
(|xc | − a)2 + 2a2 + 2|β|a < (|xc | + a)2 + 2a2 .
Note how I arranged terms on either side so that both sides are always positive. So,
I can square both sides and keep the direction of the inequality. Doing so gives
p
(|xc | − a)2 + 2a2 + 4β 2 a2 + 4a|β| (|xc | − a)2 + 2a2 < (|xc | + a)2 + 2a2 .
This can be simplified to
p
β 2 a + |β| (|xc | − a)2 + 2a2 < |xc |
or
p
|β| (|xc | − a)2 + 2a2 < |xc | − β 2 a.
Both sides are still positive because |xc | > a. (Otherwise the camera will be inside
the cube!) So, I square both sides and get
β 2 (|xc | − a)2 + 2β 2 a2 < (|xc | − β 2 a)2 ,
which can be simplified to
|xc |2 > γ 2 β 2 a2 (3 − β 2 ) = γ 2 β 2 a2 (2 + 1/γ 2 ) = β 2 a2 (2γ 2 + 1).
(e) With β = 0, we have
a
sin(αDf Db /2) = p
.
(xc + a)2 + 2a2
On approach, xc < 0 and β > 0, therefore
a
sin(αDf Db /2) = p
.
(|xc | − a)2 + 2a2
109
and
a
0
sin(αD
/2) = p
.
f Db
γ (|xc | − a)2 + 2a2 + β(|xc | − a)
0
It is clear that the denominator of sin(αD
/2) is larger than that of sin(αDf Db /2).
f Db
0
Hence, αDf Db < αDf Db .
(f) Note that both sides of the inequality are positive. Therefore, I can square both sides
p
2
γ 2 (xc + a)2 + 2a2 − β(xc + a) < (xc + a)2 + 2a2 .
or
p
γ 2 (xc + a)2 + 2a2 + β 2 (xc + a)2 − 2β(xc + a) (xc + a)2 + 2a2 < (xc + a)2 + 2a2 .
Using γ 2 − 1 = γ 2 β 2 , this inequality simplifies to
2γ 2 β 2 (xc + a)2 + 2γ 2 β 2 a2 < 2γ 2 β(xc + a)
p
(xc + a)2 + 2a2 ,
or
p
β(xc + a)2 + βa2 < (xc + a) (xc + a)2 + 2a2
Square both sides to get
β 2 (xc + a)4 + β 2 a4 + 2a2 β 2 (xc + a)2 < (xc + a)4 + 2a2 (xc + a)2
or
(xc + a)4 + 2a2 (xc + a)2 − β 2 γ 2 a4 > 0.
Find the two roots of the left-hand side by setting it equal to zero and solving the
quadratic equation:
p
(xc + a)2 = −a2 ± a4 + β 2 γ 2 a4 = −a2 ± γa2 .
Then the preceding inequality can be expressed as
[(xc + a)2 + (γ + 1)a2 ][(xc + a)2 − (γ − 1)a2 ] > 0.
The factor on the left is positive, so for the inequality to hold, we must have
(xc + a)2 > (γ − 1)a2 .
(g) If β < 0 and xc < 0, then the starting inequality in (f) becomes
p
p
γ (|xc | − a)2 + 2a2 − |β|(|xc | − a) < (|xc | − a)2 + 2a2 .
This differs from the previous case only by the change in the sign of a. So, without
going through any calculation, we can write the final inequality by changing a to −a:
(|xc | − a)2 > (γ − 1)(−a)2 ⇐⇒ (−xc − a)2 > (γ − 1)(−a)2 ⇐⇒ (xc + a)2 > (γ − 1)a2
which is the inequality in (f).
110
Relativistic Photography
3.0
2.5
2.0
1.5
1.0
0.5
-1.0
0.5
-0.5
Figure 7.2: The plot of the ratio
0
sin(αD
/2)/ sin(αDf Db /2)
f Db
1.0
when xc = 3.
(h) From the solution to part (b), we have
0
sin(αD
/2)
f Db
sin(αDf Db /2)
p
(xc + a)2 + 2a2
=
γ
p
.
(xc + a)2 + 2a2 − β(xc + a)
The plot of this ratio is shown in Figure 7.2 of the manual.
(i) When |xc | → ∞, we can ignore a in the equation above. Therefore, since angles get
small, the ratio of sines becomes the ratio of the angles:
0
αD
f Db
αDf Db
|xc |
1
=
γ (|xc | − βxc )
γ (1 − βxc /|xc |)
s
p
1 − (βxc /|xc |)2
1 + βxc /|xc |
=
.
=
(1 − βxc /|xc |)
1 − βxc /|xc |
≈
Note that β 2 = (βxc /|xc |)2 , because xc /|xc | = ±1.
7.15. Derive Equations (7.33) and (7.34).
Solution: With r = hx, y, a2 /bi = hx, y, a sin αi, r0 = h0, 0, bi, and referring to Figure 7.6
of the book, we obtain
|r0 − r|2 = x2 + y 2 + (b − a sin α)2 = R2 + b2 (1 − sin2 α)2
= a2 cos2 α + b2 cos4 α = b2 sin2 α cos2 α + b2 cos4 α = b2 cos2 α.
The rest of (7.33) follows easily. For (7.34), use (6.25) and (7.33):
111
β~ − x/(b cos α)
β~ b cos α − x
β~ + cos ϕ
=
=
1 + β~ cos ϕ
1 − β~ x/(b cos α)
b cos α − β~ x
√
√
sin ϕ
b2 cos2 α − x2 /(b cos α)
b2 cos2 α − x2
0
=
= .
sin ϕ = γ 1 + β~ cos ϕ
γ 1 − β~ x/(b cos α)
γ b cos α − β~ x
cos ϕ0 =
7.16. Show that if v̂ of Equation (7.35) is to make an angle ϕ0 with the x-axis, then
ηx b cos2 α
ηz = ± p
tan ϕ0 .
y 2 + (b cos2 α)2
Solution: From
ηx b cos2 α
cos ϕ0 = v̂ · êx = p
,
ηx2 (b cos2 α)2 + ηz2 [y 2 + (b cos2 α)2 ]
we obtain
cos2 ϕ0 =
ηx2 (b cos2 α)2
,
2
2
ηx (b cos α)2 + ηz2 [y 2 + (b cos2 α)2 ]
or
ηx2 (b cos2 α)2 cos2 ϕ0 + ηz2 [y 2 + (b cos2 α)2 ] cos2 ϕ0 = ηx2 (b cos2 α)2
or
ηz2 [y 2 + (b cos2 α)2 ] cos2 ϕ0 = ηx2 (b cos2 α)2 sin2 ϕ0
or finally
ηz2 =
ηx2 (b cos2 α)2
ηx2 (b cos2 α)2 sin2 ϕ0
=
tan2 ϕ0 .
[y 2 + (b cos2 α)2 ] cos2 ϕ0
y 2 + (b cos2 α)2
7.17. With ê(φ, β) defined as in Equation (7.40), show that
ê(φ, β) · ê(0, β) = 1 −
sin2 α(1 − cos φ)
γ 2 (1 − β~ sin α cos φ)(1 − β~ sin α)
.
Now prove that the dot product has a maximum at φ = 0 and a minimum at φ = π.
Solution: With ê(φ, β) defined as in Equation (7.40), we get
+
*
β~ − sin α
cos α
, 0, −
,
ê(0, β) = −
1 − β~ sin α
γ(1 − β~ sin α)
and the dot product becomes
β~ − sin α cos φ
1 − β~ sin α cos φ
ê(φ, β) · ê(0, β) =
+
!
β~ − sin α
1 − β~ sin α
!
cos2 α
γ 2 (1 − β~ sin α cos φ)(1 − β~ sin α)
112
Relativistic Photography
or
ê(φ, β) · ê(0, β) =
γ 2 ( β~ − sin α cos φ)( β~ − sin α) + cos2 α
.
γ 2 (1 − β~ sin α cos φ)(1 − β~ sin α)
Let’s manipulate the numerator
N um = γ 2 β 2 − γ 2 ( β~ sin α + β~ sin α cos φ)
+ [(γ 2 − 1) sin2 α cos φ + sin2 α cos φ] + cos2 α
~ sin α cos φ)
= γ 2 − 1 − γ 2 ( β~ sin α + β
+ [γ 2 β 2 sin2 α cos φ + sin2 α cos φ] + cos2 α
~ sin α cos φ)(1 − β~ sin α) − sin2 α(1 − cos φ).
= γ 2 (1 − β
Dividing this by the denominator, gives the final result.
For the purposes of finding the extrema, we just have to differentiate the function
f (φ) ≡
1 − cos φ
,
1 − β~ sin α cos φ
whose derivative is
f 0 (φ) =
=
sin φ(1 − β~ sin α cos φ) − (1 − cos φ) β~ sin α sin φ
(1 − β~ sin α cos φ)2
(1 − β~ sin α) sin φ
.
(1 − β~ sin α cos φ)2
So, f 0 (φ) = 0 if sin φ = 0 or if φ = 0, π.
7.18. Provide all the missing steps leading to Equation (7.41).
Solution: Equation (7.40) gives
*
+
β~ − sin α
cos α
ê(0, β) = −
, 0, −
1 − β~ sin α
γ(1 − β~ sin α)
*
+
~ + sin α
β
cos α
ê(π, β) = −
, 0, −
.
1 + β~ sin α
γ(1 + β~ sin α)
Therefore,
ê0 + êπ =
D
E
2 cos α
~ cos α, 0, −1/γ .
−
β
2
1 − β~ sin2 α
Although êaxis is defined as (ê0 + êπ )/|ê0 + êπ |, the constant multiplying the angle brackets
cancels in the ratio. So, you can define êaxis as the ratio in (7.41).
7.19. Derive Equation (7.42).
Solution: From (7.40) and (7.41), we obtain
1
ê(φ, β) · êaxis = q
2
1 − β~ sin2 α
( β~ − sin α cos φ) β~ cos α
cos α
+
1 − β~ sin α cos φ
γ 2 (1 − β~ sin α cos φ)
!
.
113
The expression in the big parentheses can be written as
γ 2 ( β~ − sin α cos φ) β~ cos α + cos α
,
γ 2 (1 − β~ sin α cos φ)
whose numerator is
γ 2 β 2 cos α − γ 2 β~ sin α cos φ cos α + cos α
| {z }
=γ 2 −1
or
γ 2 cos α(1 − β~ sin α cos φ).
Dividing this by the denominator, you get cos α. Therefore,
cos α
ê(φ, β) · êaxis = q
.
2
1 − β~ sin2 α
7.20. Derive Equation (7.46).
Solution: Use determinant to find the cross product.


êx
êy
êz
θ~ × ϕ
~ = det  a cos θ cos ϕ a cos θ sin ϕ −a sin θ
−a sin θ sin ϕ a sin θ cos ϕ
0
= a2 sin2 θ cos ϕêx + a2 sin2 θ sin ϕêy + a2 (sin θ cos θ cos2 ϕ + sin θ cos θ sin2 ϕ)êz
= a2 sin θ(sin θ cos ϕêx + sin θ sin ϕêy + cos θêz )
7.21. Using (7.45) and (7.7), derive Equation (7.49).
Solution: With r as given in (7.45), (7.7) yields
x0 = γ a sin θ cos ϕ − β|r − r0 |
y 0 = y = a sin θ sin ϕ
z 0 = z = a cos θ.
The only thing left to do is to calculate |r − r0 |:
|r − r0 |2 = x2 + y 2 + (z − b)2 = x2 + y 2 + z 2 +b2 − 2zb = a2 + b2 − 2ab cos θ.
{z
}
|
=a2
or
|r − r0 | =
p
a2 + b2 − 2ab cos θ.
7.22. Derive Equations (7.50) and (7.51).
114
Relativistic Photography
Solution: With
~0
θ ≡
∂x0 ∂y 0 ∂z 0
,
,
∂θ ∂θ ∂θ
,
0
ϕ
~ ≡
∂x0 ∂y 0 ∂z 0
,
,
∂ϕ ∂ϕ ∂ϕ
,
the only non-obvious component is θ~ 0x . So, let’s calculate it:
p
∂x0
∂ =γ
a sin θ cos ϕ − β a2 + b2 − 2ab cos θ
∂θ
∂θ
ab sin θ
= γ a cos θ cos ϕ − β √
a2 + b2 − 2ab cos θ
For the cross product, use determinant:


êx
êy
êz


βγb sin θ
θ~ 0 × ϕ
~ 0 = a2 det γ cos θ cos ϕ − √a2 +b2 −2ab cos θ cos θ sin ϕ − sin θ
−γ sin θ sin ϕ
sin θ cos ϕ
0
βγb sin2 θ cos ϕ
2
2
2
êz
= a sin θ cos ϕêx + γ sin θ sin ϕêy + γ sin θ cos θ − √
a2 + b2 − 2ab cos θ
βγb sin θ cos ϕ
= a2 sin θ sin θ cos ϕêx + γ sin θ sin ϕêy + γ cos θ − √
êz .
a2 + b2 − 2ab cos θ
7.23. Obtain Equation (7.52) by substituting (7.49) and (7.51) in (C.17).
Solution: In the present context, (C.17) can be written as
(θ~ 0 × ϕ
~ 0 )x x0 + (θ~ 0 × ϕ
~ 0 )y y + (θ~ 0 × ϕ
~ 0 )z (z − b) = 0.
Ignoring the factor a2 sin θ in the expression for θ~ 0 × ϕ
~ 0 , we have
p
0 = γ sin θ cos ϕ a sin θ cos ϕ − β a2 + b2 − 2ab cos θ + aγ sin2 θ sin2 ϕ
βγb sin θ cos ϕ
+ γ cos θ − √
(a cos θ − b)
a2 + b2 − 2ab cos θ
p
= a sin2 θ − β sin θ cos ϕ a2 + b2 − 2ab cos θ
βb sin θ cos ϕ(a cos θ − b)
+ a cos2 θ − b cos θ − √
a2 + b2 − 2ab cos θ
or
β sin θ cos ϕ[(a cos θ − b)b + (a2 + b2 − 2ab cos θ)]
√
0 = a − b cos θ −
a2 + b2 − 2ab cos θ
βa sin θ cos ϕ
= (a − b cos θ) 1 − √
.
a2 + b2 − 2ab cos θ
7.24. Show that (x0 , y 0 , z 0 ) of Equation (7.53) satisfy
√
p
b2 − a2
x02 + y 02 + (z 0 − b)2 =
γ(b − βa cos ϕ).
b
115
Solution:
a2
|r − r0 | ≡ x + y + (z − b) = γ a 1 − 2 cos2 ϕ + γ 2 β 2 (b2 − a2 )
b
2
2
b −a
a2
(a2 − b2 )2
2
2
− 2γ βa
cos ϕ + a 1 − 2 sin2 ϕ +
b
b
b2
a2
= (γ 2 − 1)a2 1 − 2 cos2 ϕ + (γ 2 − 1)(b2 − a2 )
b
2
2
b −a
a2
(a2 − b2 )2
2
2
− 2γ βa
cos ϕ + a 1 − 2 +
,
b
b
b2
0
2
02
02
0
2
2 2
where I wrote sin2 ϕ as 1 − cos2 ϕ and used γ 2 β 2 = γ 2 − 1. Let’s continue
2
2
0
2
2 2 2 b −a
cos2 ϕ + γ 2 (b2 − a2 ) − b2 + a2
|r − r0 | = γ β a
b2
b2 − a2
a2
(a2 − b2 )2
− 2γ 2 βa
.
cos ϕ + a2 1 − 2 +
b
b
b2
The sum of the last two terms is
a2
(a2 − b2 )2
a2 − b2 2
2
a 1− 2 +
=
(a − b2 − a2 ) = b2 − a2 ,
b
b2
b2
which cancels the last two terms on the first line of the previous equation. So, we now have
2
2
b2 − a2
0
2
2 2 2 b −a
2
2 2
2
2
|r − r0 | = γ β a
cos
ϕ
+
γ
(b
−
a
)
−
2γ
βa
cos ϕ
b2
b
b2 − a2 2
b2 − a2 2 2 2
2
2
=
γ
β
a
cos
ϕ
+
b
−
2βab
cos
ϕ
=
γ (b − βa cos ϕ)2 .
b2
b2
7.25. Show that
√the line from (0, 0, b) in the direction of êaxis of Equation (7.41) cuts the
~
x axis at − β γ b2 − a2 .
Solution: The line in the direction of êaxis is têaxis , where −∞ < t < ∞. A vector parallel
to this line with its tail at r0 = h0, 0, bi cuts the x-axis for some t = t0 . So, we have to solve
the equation
D
E
r0 + t0 êaxis = xêx ⇐⇒ h0, 0, bi + −t0 β~ cos α, 0, −t0 /γ = hx, 0, 0i
or
D
E
−t0 β~ cos α, 0, b − t0 /γ = hx, 0, 0i.
The equality of the last component gives t0 = γb. Plugging this in the equality of the first
component yields
p
p
x = −bγ β~ cos α = −bγ β~ 1 − a2 /b2 = −γ β~ b2 − a2 .
116
Relativistic Photography
7.26. A point source produces a circular cone of light. It is placed at (0, 0, b) with its axis
along the z-axis in reference frame O, moving with speed β in the positive direction of
the x0 -axis of O0 . At t = t0 = 0, when the two origins coincide, the source emits a pulse,
producing a circular image of radius a in the xy-plane of O. What is the shape of the image
in the x0 y 0 -plane of O0 ? Hint: Look at the events of the arrival of light beams to the planes.
Solution: √
In the reference frame O, the event of a beam arriving at (x, y) has time coordinate t = a2 + b2 , because the beam was emitted at t = 0. Lorentz transform to the RF
O0 :
p
p
x0
x0 = γ(x + β a2 + b2 ), y 0 = y ⇐⇒
− β a2 + b2 = x, y 0 = y
γ
or
0
2
p
x
2
2
−β a +b
+ y 0 2 = a2
γ
or
2
√
x0 − βγ a2 + b2
y0 2
+
= 1,
γ 2 a2
a2
√
which is an ellipse centered at (βγ a2 + b2 , 0) with semi-major axis γa and semi-minor
axis a.
7.27. Derive Equation (7.57).
Solution: The cross product is identical to that given in the solution of Problem 7.20. Let
rc ≡ hxc , 0, 0i, r0 ≡ h0, 0, bi, and r ≡ hx, y, zi. With these notations, we can use the result
of Problem 7.20 and find what we are looking for with minimal effort. We want to find the
solution to
0 = (θ~ × ϕ
~ ) · (r + rc − r0 ) = (θ~ × ϕ
~ ) · (r − r0 ) + (θ~ × ϕ
~ ) · rc .
The first term on the right has been calculated and is given as the left-hand side of the
equation after Equation (7.46). The second term is
(θ~ × ϕ
~ )x xc = xc a2 sin2 θ cos ϕ.
Thus, our equation is now
0 = a2 sin θ[a sin2 θ + cos θ(a cos θ − b)] + xc a2 sin2 θ cos ϕ = 0,
θ 6= 0, π,
or
0 = a2 sin θ[xc sin θ cos ϕ + a sin2 θ + a cos2 θ − b cos θ],
θ 6= 0, π,
or
0 = xc sin θ cos ϕ + a − b cos θ = 0,
θ 6= 0, π.
7.28. Starting with Equation (7.57), provide all the missing steps leading to (7.58).
117
Solution: First note that
x2 + y 2 + (z − b)2 = x2c + a2 sin2 θ cos2 ϕ + 2xc a sin θ cos ϕ + a2 sin2 θ sin2 ϕ
+ a2 cos2 θ + b2 − 2ab cos θ = x2c + a2 + 2xc a sin θ cos ϕ + b2 − 2ab cos θ
= x2c + a2 + b2 + 2a (xc sin θ cos ϕ − b cos θ) = x2c + b2 − a2 .
{z
}
|
=−a by previous problem
The denominators of (7.12) are all the same. With xq = xc and using the result obtained
above, they can be written as
h
i
p
Den = xc xc + a sin θ cos ϕ − β x2c + b2 − a2 − b(a cos θ − b)
p
= x2c + axc sin θ cos ϕ − βxc x2c + b2 − a2 − ab cos θ + b2
p
= x2c + b2 − a2 − βxc x2c + b2 − a2
by the result obtained in the previous problem.
7.29. Define tan α ≡ b/(xc cos ϕ) and show that (7.57) can be written as
sin θ cos α − cos θ sin α +
or
sin(θ − α) = −
Now define
sin η ≡
a cos α
= 0,
xc cos ϕ
a cos α
.
xc cos ϕ
a cos α
xc cos ϕ
and show that θ = α − η. Using the trigonometric identity
cos α = √
write η in terms of ϕ:
1
1 + tan2 α
a
sin η = p
.
2
b + x2c cos2 ϕ
Since b > a, this shows that our definition of sin η is consistent with the fact that | sin η| < 1.
With both α and η given in terms of ϕ, derive (7.59).
Solution: Divide (7.57) by xc cos ϕ to write it as
b
a
sin θ −
cos θ +
=0
xc cos ϕ
xc cos ϕ
|
{z
}
=tan α
or
sin θ −
sin α
a
cos θ +
=0
cos α
xc cos ϕ
or
sin θ cos α − cos θ sin α +
a cos α
= 0.
xc cos ϕ
118
Relativistic Photography
Therefore,
sin(θ − α) = −
a cos α
= sin(−η),
xc cos ϕ
and θ − α = −η, or θ = α − η.
Now, I want to express everything in terms of the original parameters. For this, I need
the sines and cosines of the newly defined angles.
1
1
xc cos ϕ
=p
=p
2
2
1 + (b/xc cos ϕ)
xc cos2 ϕ + b2
1+
tan α
b/xc cos ϕ
b
sin α = √
,
=p
=p
2
2
2
1 + (b/xc cos ϕ)
xc cos2 ϕ + b2
1 + tan α
cos α = √
tan2 α
sin η has already been calculated. So, I just need cos η:
s
s
q
2
a
b2 − a2 + x2c cos2 ϕ
cos η = 1 − sin2 η = 1 − 2
=
.
2
2
b + xc cos ϕ
b2 + x2c cos2 ϕ
Now we are ready to calculate sin θ and cos θ:
sin θ = sin(α − η) = sin α cos η − cos α sin η
! s
!
b
b2 − a2 + x2c cos2 ϕ
= p
b2 + x2c cos2 ϕ
x2c cos2 ϕ + b2
!
!
a
xc cos ϕ
p
− p
x2c cos2 ϕ + b2
b2 + x2c cos2 ϕ
p
b b2 − a2 + x2c cos2 ϕ − axc cos ϕ
.
=
b2 + x2c cos2 ϕ
This is the first equation of (7.59). The second equation can be derived similarly.
7.30. Show that with β = 0, Equations (7.58) and (7.59) give (7.60).
Solution: With β = 0, (7.58) becomes
b (xc + a sin θ cos ϕ) + xc (a cos θ − b)
x2c + b2 − a2
p
a x2c + b2 sin θ sin ϕ
v0 =
.
x2c + b2 − a2
u0 =
I calculate the first equation, leaving the easier second equation for you. Substitute for sin θ
and cos θ from (7.59) in the first equation:
√
b2 b2 −a2 +x2c cos2 ϕ −abxc cos ϕ
bxc + a
cos ϕ
b2 +x2c cos2 ϕ
u=
x2c + b2 − a2
√
x2 cos ϕ b2 −a2 +x2c cos2 ϕ +abxc
a c
− bxc
b2 +x2c cos2 ϕ
+
,
x2c + b2 − a2
119
or
p
bxc (b2 + x2c cos2 ϕ) + ab2 b2 − a2 + x2c cos2 ϕ cos ϕ − a2 bxc cos2 ϕ
u=
(x2c + b2 − a2 )(b2 + x2c cos2 ϕ)
p
ax2 cos ϕ b2 − a2 + x2c cos2 ϕ + a2 bxc − bxc (b2 + x2c cos2 ϕ)
+ c
,
(x2c + b2 − a2 )(b2 + x2c cos2 ϕ)
or
p
a(b2 + x2c ) cos ϕ b2 − a2 + x2c cos2 ϕ + a2 bxc sin2 ϕ
u=
.
(b2 + x2c − a2 )(b2 + x2c cos2 ϕ)
7.31. Show that with cos φ defined as (7.61), Equation (7.62) follows and u and v of (7.60)
could be expressed as in (7.63).
Solution: Verifying (7.63) once (7.61) and (7.62) are given is trivial. Verifying the definition of cos φ by (7.61) is also very straightforward.
p The only challenge of the problem is to
show that sin φ as defined by (7.62) is indeed 1 − cos2 φ. Instead of deriving (7.62) it is
easier to verify it: square the numerator of (7.62) and add it to the square of the numerator
of (7.61) and show that the sum is the square of the common denominator. I’ll leave that
for you.
7.32. Derive Equation (7.66) from Equations (7.59), (7.64), and (7.65).
Solution: Rather than go through the algebra, which happens to be very messy, I show you
how to obtain the results using Mathematica. Figure 7.3 of the manual shows the outline
of the procedure. Note how I first found the semi-major axis and center of the ellipse, and
then used the first equation in (7.65) to find cos ϑ.
7.33. Substitute (7.56) in (7.7) and simplify to obtain (7.67).
Solution: The time of emission for the event at (x, y, z) is the negative of the distance
from that event to the pinhole. Thus
q
p
2
2
2
t = − x + y + (z − b) = − (xc + a sin θ cos ϕ)2 + a2 sin2 θ sin2 ϕ + (a cos θ − b)2
q
= − x2c + a2 sin2 θ cos2 ϕ + 2axc sin θ cos ϕ + a2 sin2 θ sin2 ϕ + a2 cos2 θ + b2 − 2ab cos θ
p
= − x2c + a2 + 2axc sin θ cos ϕ + b2 − 2ab cos θ.
With this t, (7.67) is immediately obtained.
7.34. Derive Equation (7.68).
~0 . The first component of θ~0 is
Solution: First find θ~0 and ϕ
∂x0
=γ
∂θ
axc cos θ cos ϕ + ab sin θ
!
axc sin θ sin ϕ
!
a cos θ cos ϕ − β p
x2c + a2 + b2 + 2a(xc sin θ cos ϕ − b cos θ)
,
~0 is
and the first component of ϕ
∂x0
=γ
∂ϕ
−a sin θ sin ϕ + β p
x2c + a2 + b2 + 2a(xc sin θ cos ϕ − b cos θ)
.
120
Relativistic Photography
Figure 7.3: Outline of solution using Mathematica.
121
The other components are also trivially calculate. To save space I’ll use |t| for the square
root in the calculation of the cross product using determinant.


êx
êy
êz






ax
cos
θ
cos
ϕ
+
ab
sin
θ
c
~0 = det aγ cos θ cos ϕ − γβ
a cos θ sin ϕ −a sin θ
θ~0 × ϕ


|t|




aγ sin θ sin ϕ(βxc /|t| − 1)
a sin θ cos ϕ
0
= a sin2 θ cos ϕ êx − a2 γ sin2 θ sin ϕ(βxc /|t| − 1)êy
h
γβ 2
a xc sin θ cos θ cos2 ϕ
+ a2 γ sin θ cos θ cos2 ϕ −
|t|
i
+ a2 b sin2 θ cos ϕ − (βxc /|t| − 1)a2 γ sin θ cos θ sin2 ϕ êz .
A little algebra in the z-component yields the desired result.
7.35. Substitute Equations (7.67) and (7.68) in Equation (C.17) to get one expression.
Multiply out the parentheses in Equation (7.69) to get another expression. Now show that
the two expressions are equal.
Solution: This is the messiest problem in the book if you were to do it by hand! It begs
for a solution using computer algebra. So, I have shown my calculation in Mathematica in
Figure 7.4 of the manual.
7.36. For the second parentheses of (7.69) to be zero, you should have
β 2 (xc + a sin θ cos ϕ)2 = x2c + a2 + b2 + 2a(xc sin θ cos ϕ − b cos θ).
(a) Show that this gives a quadratic equation in cos ϕ whose solution is
p
xc /γ 2 ± x2c /γ 2 + β 2 (a2 + b2 − 2ab cos θ)
cos ϕ =
.
β 2 a sin θ
(b) Verify that when xc is positive (negative) and you choose the positive (negative) sign
for the square root, | cos ϕ| > 1 is trivially satisfied.
(c) For xc < 0 and the positive sign for the square root, show that the inequality
p
−|xc |/γ 2 + x2c /γ 2 + β 2 [(b − a cos θ)2 + a2 sin2 θ]
>1
β 2 a sin θ
is equivalent to the inequality
(|xc | − a sin θ)2 + γ 2 (b − a cos θ)2 > 0,
which is obviously true. Therefore, cos ϕ > 1.
122
Relativistic Photography
Figure 7.4: Outline of solution of Problem 7.35 using Mathematica.
123
(d) For xc > 0 and the negative sign for the square root cos ϕ is negative. Therefore, you
have to prove the inequality
p
xc /γ 2 − x2c /γ 2 + β 2 [(b − a cos θ)2 + a2 sin2 θ]
< −1.
β 2 a sin θ
Show that this leads to the same inequality as in (c), proving that cos ϕ < −1.
Solution:
(a) Squaring the left-hand side and collecting terms, you get
β 2 a2 sin2 θ cos2 ϕ −
2axc sin θ
cos ϕ − (x2c /γ 2 + a2 + b2 − 2ab cos θ) = 0.
γ2
The quadratic rule gives the solution as
p
axc sin θ/γ 2 ± a2 x2c sin2 θ/γ 4 + β 2 a2 sin2 θ(x2c /γ 2 + a2 + b2 − 2ab cos θ)
cos ϕ =
β 2 a2 sin2 θ
p
xc /γ 2 ± x2c /γ 4 + β 2 (x2c /γ 2 + a2 + b2 − 2ab cos θ)
=
.
β 2 a sin θ
Now note that
x2c
β 2 x2c
x2c
+
=
γ4
γ2
γ2
1
+ β2
γ2
=
x2c
.
γ2
This gives
cos ϕ =
xc /γ 2 ±
p
x2c /γ 2 + β 2 (a2 + b2 − 2ab cos θ)
,
β 2 a sin θ
which is what we are after, but I’ll rewrite it as
p
xc /γ 2 ± x2c /γ 2 + β 2 [(b − a cos θ)2 + a2 sin2 θ]
cos ϕ =
β 2 a sin θ
and note that every term under the radical sign is positive.
(b) It is now clear that if xc > 0 and I choose the positive sign the numerator is larger
than the denominator. In fact, if you keep just the last term under the radical sign
the ratio of the numerator to the denominator will be
βa sin θ
1
= > 1.
2
β a sin θ
β
Similarly, if xc < 0 and I choose the negative sign the magnitude of the numerator is
larger than the denominator.
(c) For xc < 0 and the positive sign for the square root, the inequality
p
−|xc |/γ 2 + x2c /γ 2 + β 2 [(b − a cos θ)2 + a2 sin2 θ]
>1
β 2 a sin θ
can be written as
q
|xc |
x2c /γ 2 + β 2 [(b − a cos θ)2 + a2 sin2 θ] > β 2 a sin θ + 2 .
γ
124
Relativistic Photography
Since both sides are positive, I can square the inequality:
x2c
+ β 2 [(b − a cos θ)2 + a2 sin2 θ] >
γ2
or
|xc | 2
2
β a sin θ + 2
,
γ
x2c
x2c
β 2 a sin θ|xc |
2
2
2 2
2
4 2
2
+
β
(b
−
a
cos
θ)
+
β
a
sin
θ
>
β
a
sin
θ
+
+
2
.
γ2
γ4
γ2
Multiply both sides by γ 4 and collect terms using γ 2 (1 − β 2 ) = 1 and γ 2 − 1 = −β 2 γ 2
to get
γ 2 β 2 x2c + γ 4 β 2 (b − a cos θ)2 + γ 2 β 2 a2 sin2 θ > 2γ 2 β 2 a sin θ|xc |,
or
x2c + γ 2 (b − a cos θ)2 + a2 sin2 θ − 2a sin θ|xc | > 0,
which is the same as
(|xc | − a sin θ)2 + γ 2 (b − a cos θ)2 > 0.
(d) For xc > 0 and the negative sign for the square root, multiply both sides of the
inequality
p
xc /γ 2 − x2c /γ 2 + β 2 [(b − a cos θ)2 + a2 sin2 θ]
< −1
β 2 a sin θ
by −1 and change the direction of the inequality
p
−xc /γ 2 + x2c /γ 2 + β 2 [(b − a cos θ)2 + a2 sin2 θ]
> 1.
β 2 a sin θ
Noting that xc = |xc | (because xc is positive), this is identical to the inequality in (c).
7.37. Derive Equation (7.70) from Equation (7.58).
Solution: For ϕ = 0 and ϕ = π, with obvious notation, (7.59) gives
p
p
b b2 − a2 + x2c − axc
xc b2 − a2 + x2c + ab
sin θ0 =
,
cos θ0 =
b2 + x2c
b2 + x2c
p
p
−xc b2 − a2 + x2c + ab
b b2 − a2 + x2c + axc
,
cos
θ
=
,
sin θπ =
π
b2 + x2c
b2 + x2c
and the first equation of (7.58) gives
p
bγ xc + a sin θ0 − β x2c + b2 − a2 + xc (a cos θ0 − b)/γ
p
u00 =
x2c + b2 − a2 − βxc x2c + b2 − a2
p
bγ xc − a sin θπ − β x2c + b2 − a2 + xc (a cos θπ − b)/γ
p
u0π =
.
x2c + b2 − a2 − βxc x2c + b2 − a2
125
Adding these two and dividing by 2 gives u0c :
h
i x
p
c
bγ xc + a(sin θ0 − sin θπ )/2 − β x2c + b2 − a2 + [a(cos θ0 + cos θπ )/2 − b]
γ
p
,
u0c =
x2c + b2 − a2 − βxc x2c + b2 − a2
or
2
p
a b
a2 xc
2
2
2
xc − 2
− β xc + b − a + xc 2
−b
xc + b2
xc + b2
p
u0c =
γ x2c + b2 − a2 − βxc x2c + b2 − a2
h
i
p
bγ 2 x3c + xc b2 − a2 xc − β(x2c + b2 ) x2c + b2 − a2 + xc b a2 − x2c − b2
p
=
γ(x2c + b2 ) x2c + b2 − a2 − βxc x2c + b2 − a2
bγ 2
or
h
i
p
b γ 2 β 2 x3c + γ 2 β 2 xc b2 − γ 2 β 2 a2 xc − γ 2 β(x2c + b2 ) x2c + b2 − a2
p
p
u0c =
γ(x2c + b2 ) x2c + b2 − a2
x2c + b2 − a2 − βxc
h
i
p
bγ β 2 xc (x2c + b2 − a2 ) − β(x2c + b2 ) x2c + b2 − a2
p
p
=
(x2c + b2 ) x2c + b2 − a2
x2c + b2 − a2 − βxc
h
i
p
bγ β 2 xc x2c + b2 − a2 − β(x2c + b2 )
p
=
.
(x2c + b2 ) x2c + b2 − a2 − βxc
I leave the calculation of the semi-major axis A for you.
CHAPTER
8
Relativistic Interactions
Problems With Solutions
8.1. Two identical particles of mass m approach each other along a straight line with speed
v = βc as measured in the lab frame. Show that the energy of one particle as measured in
the rest frame of the other is
1 + β2 2
mc .
1 − β2
Solution: There are two ways to do the problem. In the first, I use the relativistic LAV.
Recall that the Lorentz factor associated with the combined velocity is given by
γ 0 = γ1 γ2 (1 + β1 β2 ) = γ 2 (1 + β 2 ) =
1 + β2
1 − β2
since the two speeds are equal. Now note that the energy is just E 0 = γ 0 mc2 .
In the second way, I use Lorentz transformation of energy. In the lab, with c = 1, the
energy and momentum of one of the particles are E = mγ and p = mγβ. In the rest frame
of one of the particles with respect to which the lab moves with speed β, the energy is
E 0 = γ(E + βp) = γ(mγ + βmγβ) = mγ 2 (1 + β 2 ).
8.2. Find an expression for the magnitude of the momentum (8.8), energy (8.9), and velocity
(8.10) of the produced particle in the lab frame all in terms of only masses m1 , m2 , and M .
Solution: Since everything is given in terms of E1 and E1 is already given in terms of m1 ,
m2 , and M in (8.7), the problem reduces to substituting (8.7) in the relevant equations.
From (8.8), we get
s
2
2
q
M − m21 − m22
2
2
~
|P | = E1 − m1 =
− m21
2m2
q
1
=
M 4 + (m21 − m22 )2 − 2M 2 (m21 + m22 ),
2m2
128
Relativistic Interactions
from (8.9), we get
E = E1 + m2 =
M 2 − m21 + m22
2m2
and from (8.10), we get
|P~ |
V =
=
E
p
M 4 + (m21 − m22 )2 − 2M 2 (m21 + m22 )
.
M 2 − m21 + m22
8.3. Insert (8.13) in (8.14) to derive (8.15). Another way of obtaining this result is to use
Equation (8.3) with Ecm = E1cm + E2cm , and the fact that
2
2
E2cm
= E1cm
− m21 + m22 ,
which you should derive.
Solution: First note that from the first equation of (8.13), we have
E1 − β~cm · p~1 = E1 −
|~
p1 |2
E2 − m21
E1 m2 + m21
= E1 − 1
=
.
E1 + m2
E1 + m2
E1 + m2
This, in combination with the second equation of (8.13), yields the final result.
In the CM, both initial particles have the same momentum. Thus
2
2
E2cm
= |~
p2 |2 + m22 = |~
p1 |2 + m22 = E1cm
− m21 + m22 .
Now use this and Ecm = E1cm + E2cm in Equation (8.3) to obtain
(E1cm + E2cm )2 = m21 + m22 + 2m2 E1 ,
or
2
2E1cm
− m21 + m22 + 2E1cm E2cm = m21 + m22 + 2m2 E1 ,
or
2
E1cm
− m21 − m2 E1 = −E1cm E2cm ,
or
2
2
2
(E1cm
− m21 − m2 E1 )2 = E1cm
(E1cm
− m21 + m22 ),
or
2
2
(m21 + m2 E1 )2 − 2E1cm
(m21 + m2 E1 ) = E1cm
(−m21 + m22 ).
A tiny amount of further algebra gives the desired result.
8.4. Suppose that in the elastic scattering of two identical relativistic particles in the lab
frame, one of the final particles is produced at rest. Use (8.22) and the conservation of
4-momentum to show that the other final particle moves with the same velocity as the
initial incident particle.
Solution: Assume that the third particle is produced at rest. Then p~3 = 0 and the first
equation in (8.22) gives mE1 = mE4 or E1 = E4 . Since the total initial momentum is that
of the first particle, the fourth particle carries that momentum. Therefore,
~v1 =
p~1
p~4
=
= ~v4 .
E1
E4
129
8.5. A photon of momentum pγ hits a macroscopic object of mass M initially at rest, gets
absorbed, and sets the object in motion. All motions are in one dimension. Write (8.18)
for this process assuming that the mass of the macroscopic object does not change. What
is the final momentum of the macroscopic object? What is wrong? How can you resolve
the issue?
Solution: For this problem, with Eγ , E, and P denoting the initial energy of the photon
and the final energy and momentum of the macroscopic object, Equation (8.18) becomes
Eγ + M = E,
pγ = P ⇐⇒ Eγ = P ⇐⇒ P + M = E.
or
(P + M )2 = E 2 = P 2 + M 2 ⇐⇒ 2P M = 0 ⇐⇒ P = 0,
which violates momentum conservation. This means that the mass of the macroscopic
object must change. If Mf is its final mass, then the last equation yields
(P + M )2 = E 2 = P 2 + Mf2 ⇐⇒ P =
Mf2 − M 2
2M
.
8.6. A photon of momentum pγ hits a perfectly reflecting (i.e., the photon does not lose
any of its initial energy upon reflection) macroscopic object of mass M initially at rest. All
motions are in one dimension. Write (8.18) for this process. Can you assume that the mass
of the macroscopic object does not change? If not, write an expression connecting the final
mass with M and pγ .
Solution: With obvious notation, Equation (8.18) becomes
Eγ + M = Eγ + E
p~γ = −~
pγ + P~ .
or, ignoring the vector signs,
M = E,
2pγ = P.
If you assume that the mass of the macroscopic object does not change, then the first
equation implies that P = 0, which contradicts the second equation. So assume that the
final mass is Mf . Then
M 2 = Mf2 + P 2 = Mf2 + 4p2γ ⇐⇒ Mf2 = M 2 − 4p2γ .
So, the object must lose some of its mass. What this really means is that the final object
is a different object than the initial one.
8.7. Derive (8.20) by transferring p3 to the left-hand side of (8.16) and squaring both sides.
Solution: Squaring both sides of p1 + p2 − p3 = p4 , yields
m24 = m21 + m22 + m23 + 2p1 • p2 − 2p1 • p3 − 2p2 • p3 .
Assuming that the second particle is at rest, we get
m24 = m21 + m22 + m23 + 2m2 E1 − 2(E1 E3 − p~1 · p~3 ) − 2m2 E3 ,
which is a rearranged version of (8.20).
130
Relativistic Interactions
8.8. A particle of mass m and energy E1 collides with an identical stationary particle. The
two particles scatter with momenta p~3 and p~4 making angles θ3 and θ4 with the direction
of motion of the incident particle.
(a) Use (8.20) to show that
cos θ3 =
(E1 + m)(E3 − m)
.
|~
p1 ||~
p3 |
(b) Using the square of (a) and a similar result for θ4 show that
tan2 θ3 =
and
tan2 θ4 =
2m(E1 − E3 )
(E1 + m)(E3 − m)
2m(E1 − E4 )
2m(E3 − m)
=
(E1 + m)(E4 − m)
(E1 + m)(E1 − E3 )
(c) Take the product of the last two results to show that
tan θ3 tan θ4 =
2m
2
=
E1 + m
γ1 + 1
where γ1 is the gamma factor for the incident particle.
Solution: With all the masses equal, (8.20) becomes
m2 + mE1 = E3 (E1 + m) − p~1 · p~3 = E3 (E1 + m) − |~
p1 ||~
p3 | cos θ3
or
(a)
|~
p1 ||~
p3 | cos θ3 = E3 (E1 + m) − m(m + E1 ) = (E3 − m)(m + E1 ).
Similarly,
|~
p1 ||~
p4 | cos θ4 = (E4 − m)(m + E1 ).
(b)
1 − cos2 θ3
|~
p1 |2 |~
p3 |2 − (E3 − m)2 (m + E1 )2
=
cos2 θ3
(E3 − m)2 (m + E1 )2
(E12 − m2 )(E32 − m2 ) − (E3 − m)2 (m + E1 )2
.
=
(E3 − m)2 (m + E1 )2
tan2 θ3 =
Write the first term of the numerator as
(E1 − m)(E1 + m)(E3 − m)(E3 + m),
factor out the common terms and simplify to get the final result. The final expression
for tan2 θ4 comes from the conservation of energy: E1 + m = E3 + E4 .
(c) Multiplying the two tangent terms in (b) immediately gives you the first expression
on the right-hand side. The second expression follows from E1 = mγ1 .
131
8.9. Show that in the ultra-relativistic case, Equation (8.20) leads to the angle between
the final two particles being zero.
Solution: Actually, Equation (8.20) gives more than that. It shows that both particles
move in the original direction of the projectile. Ultra-relativistic means that we can ignore
masses compared with energies. Therefore, all momentum magnitudes are equal to the
corresponding energies. With these in mind, Equation (8.20) becomes
m2 E1 ≈ E3 E1 − |~
p1 || · p~3 | cos θ3 ≈ E3 E1 (1 − cos θ3 )
or
cos θ3 ≈ 1 −
m2
≈ 1,
E3
with a similar result for θ4 .
8.10. A particle of mass m and relativistic energy 4mc2 collides with another stationary
particle of mass 2m and sticks to it. What is the mass of the resulting composite particle.
Solution: Set c = 1. Initially, the energy is 6m, and the momentum is
p
√
pin = 16m2 − m2 = 15 m.
Let M be the mass of the final particle. Then, conservation of energy and momentum gives
√
M 2 = E 2 − P 2 = 36m2 − 15m2 = 21m2 ⇐⇒ M = 21 m
8.11. An electron of kinetic energy 1 GeV strikes a positron (antielectron) at rest and the
two particles annihilate each other and produce two photons, one moving in the forward
direction (the direction that electron had before collision) and the other in the backward
direction. What are the energies of the two photons. The mass (times c2 ) of electron and
positron are the same and equal to 0.511 MeV.
Solution: Using Equation (8.20) with obvious notation, you can write
m(Ee + me ) = Eγi (Ee + me ) − |~
pe ||~
pγi | cos θi
= Eγi (Ee + me ) − |~
pe |Eγi cos θi ,
or
Eγi =
me (Ee + me )
,
Ee + me − |~
pe | cos θi
i = 1, 2
i = 1, 2.
So, the two photons have energies,
me (Ee + me )
Ee + me + |~
pe |
=
Ee + me − |~
pe |
2
me (Ee + me )
Ee + me − |~
pe |
=
=
.
Ee + me + |~
pe |
2
Eγ1 =
Eγ2
(You should obtain the last equality on each line.) For Ee >> me , as is the case here, we
get Eγ1 ≈ Ee and Eγ2 ≈ 12 me . I let you plug in the numbers.
132
Relativistic Interactions
8.12. An electron of energy E and momentum p~ strikes a positron (antielectron) at rest.
The two particles annihilate each other and produce two photons of energies Eγ1 and Eγ2 .
(a) Use (8.20) to express the energy of each photon in terms of E, the mass of the electron
(or positron) me , and the photon’s scattering angle.
(b) Show that the photons have the same scattering angle if and only if they have the
same energy.
(c) Prove that when the photons have the same scattering angles θ, then
r
|~
p|
E − me
cos θ =
=
,
E + me
2Eγ
where Eγ is the energy of either photon.
(d) Show that for an ultra-relativistic incident electron, both photons move in the forward
direction.
Solution:
(a) This is done in the previous problem. I reproduce the result here:
Eγi =
me (E + me )
,
E + me − |~
pe | cos θi
i = 1, 2.
(b) Clearly, if θ1 = θ2 , then Eγ1 = Eγ2 .
(c) Solve (b) for the common cos θ in terms of the common Eγ :
cos θ =
(E + me )(Eγ − me )
.
|~
pe |Eγ
Conservation of energy gives E +me = 2Eγ , which also leads to 2(Eγ −me ) = E −me .
Thus,
r
2Eγ (Eγ − me )
2(Eγ − me )
E − me
E − me
cos θ =
=
=
=
.
|~
pe |Eγ
|~
pe |
|~
pe |
E + me
Multiplying the numerator and the denominator of the fraction under the radical sign
by E + me yields
s
E 2 − m2e
|~
pe |
|~
pe |
cos θ =
=
=
.
(E + me )2
E + me
2Eγ
(d) We can ignore me compared to E and |~
pe |. Then
cos θ =
|~
pe |
|~
pe |
E
≈
≈
= 1.
E + me
E
E
8.13. An electron of energy E, momentum p~, and mass me strikes a positron (antielectron)
at rest. The two particles annihilate each other and produce two photons. Transfer the
collision to the center of mass and do the following:
133
(a) Find the energy and momentum of each photon in the CM in terms of E, p~, and me .
(b) Transfer back to the rest frame of the positron and show the results in Problem 8.12.
Solution: The velocity and γ of the center of mass are given in (8.13):
r
E + me
E + me
p~
~
=
βcm =
, γcm = p
.
2
E + me
2me
2me + 2me E
(a) The energy and momentum of the electron in the CM are obtained by Lorentz transforming to the CM using Equation (6.54):
r
E + me
|~
p|2
~
E−
Ecm = γcm (E − βcm · p~) =
2me
E + me
r
r
2
2
E − me
E + me
me (E + me )
E−
=
=
2me
E + me
2
~
p~cm = p~ − γcm βcm E + (γcm − 1)β̂cm β̂cm · p~
!
r
r
E + me
p~
E + me
E+
− 1 β̂cm β̂cm · p~
= p~ −
| {z }
2me E + me
2me
=~
p
r
=
E + me
2me
!
−
~
pE
+ p~
E + me
r
=
me
p~.
2(E + me )
2 −p
As a check, you should verify that Ecm
~cm · p~cm = m2e . Note that since the positron
+ =γ m
has no momentum in the lab, its energy and momentum in the CM are Ecm
cm e
+ = −γ β
+
+ = −~
~
and p~cm
~cm
pcm .
cm cm me . As a further check, show that Ecm = Ecm and p
By energy conservation, the sum of the energies of the two photons should be 2Ecm .
Since photons have equal and opposite momenta, their energies should be equal. So,
Eγcm = Ecm , and since they are massless, |~
pγcm | = Eγcm . The direction of the photon
momenta cannot be determined.
(b) Now transform the energy and momentum of the photons back to the lab frame:
Eγ = γcm (Eγcm + β~cm · p~γcm )
p~γ = p~γcm + γcm β~cm Eγcm + (γcm − 1)β̂cm β̂cm · p~γcm .
To find the energy of each photon in the lab, we need β~cm · p~γcm . So, take the dot
product of the second equation with β~cm :
β~cm · p~γ = β~cm · p~γcm + γcm |β~cm |2 Eγcm + (γcm − 1)β~cm · p~γcm
= γcm |β~cm |2 Eγcm + γcm β~cm · p~γcm .
Thus,
β~cm · p~γ
β~cm · p~γcm =
− |β~cm |2 Eγcm ,
γcm
and
Eγ = γcm
β~cm · p~γ
Eγcm +
− |β~cm |2 Eγcm
γcm
!
= β~cm · p~γ +
Eγcm
γcm
134
Relativistic Interactions
or
or
Ecm
Ecm
= |β~cm |Eγ cos θ +
Eγ = |β~cm ||~
pγ | cos θ +
γcm
γcm
q
q
me (E+me )
2me
2
E+me
Ecm /γcm
me (E + me )
Eγ =
=
=
.
|~
p|
~
E + me − |~
p| cos θ
1 − |βcm | cos θ
1−
cos θ
E+me
This is exactly what we had in Problem 8.12.
8.14. A particle of mass m and energy E collides with an identical particle at rest. The
collision results in the formation of a p
single particle. Show
p that the mass and the speed of
the formed particle are, respectively, 2m(E + m) and (E − m)/(E + m).
Solution: The 4-momentum conservation gives p1 + p2 = P. Squaring both sides gives
2m2 + 2p1 • p2 = M 2 ⇐⇒ 2m2 + 2mE = M 2 .
Conservation of momentum and energy give
p
P = p = E 2 − m2 ,
E = E + m.
The speed of the final particle is therefore
r
√
P
E 2 − m2
E−m
V =
=
=
.
E
E+m
E+m
As a check, note that
P 2 + M 2 = E 2 − m2 + 2m2 + 2mE = (E + m)2 ,
which is the square of the energy of the final particle.
8.15. A photon of energy E is absorbed by a stationary nucleus of mass M . The collision
results in an excitation ofpthe nucleus. Show that the mass and the speed of the excited
nucleus are, respectively, M (2E + M ) and E/(E + M ).
Solution: The 4-momentum conservation gives p1 + p2 = P. Squaring both sides gives
0 + M 2 + 2p1 • p2 = M ∗ 2 ⇐⇒ M 2 + 2EM = M ∗ 2 .
Conservation of momentum and energy give
P = p = E,
E = E + M.
The speed of the final particle is therefore
V =
P
E
=
.
E
E+m
As a check, note that
P 2 + M ∗ 2 = E 2 + M 2 + 2M E = (E + M )2 ,
which is the square of the energy of the excited nucleus.
135
8.16. Derive Equation (8.26).
Solution: From (8.25), we have
M2
− E1 ⇐⇒ E12 − m21 =
|~
p1 | =
2E2
or
−m21 =
M2
− E1
E2
2
,
M4
M2
m21 E2
M2
E
⇐⇒
E
=
+
.
−
1
1
E2
4E2
M2
4E22
8.17. Show that (8.32) follows from the equation before it.
Solution: From M = Zmp + N mn − BE, the equation before (8.32) can be written as
(M + BE)2 − m21 − M2
BE 2 + 2(BE)M − m21
=
2M
2M
2
2
BE − m1
= BE +
.
2M
E1 =
8.18. Derive Equation (8.34) from (8.5).
Solution: For decays Equation (8.5) becomes E1 + E2 = M and p~2 = −~
p1 . So, we have
E2 = M − E1 and |~
p2 | = |~
p1 |. These two equations lead to
E22 = (M − E1 )2 ,
E22 − m22 = E12 − m21 ⇐⇒ (M − E1 )2 − m22 = E12 − m21 ,
or
M 2 − 2M E1 − m22 = −m21 ⇐⇒ 2M E1 = M 2 + m21 − m22 .
This is the first equation of (8.34). The second equation follows similarly.
8.19. Obtain Equation (8.34) by transferring p1 or p2 to the left-hand side of P = p1 + p2
and squaring both sides.
Solution: This problem demonstrates the advantage of using 4-vectors. Square both sides
of P − p1 = p2 to get
M 2 + m21 − 2P • p1 = m22 .
The first equation of (8.34) follows immediately from P • p1 = M E1 . To get the second
equation of (8.34), write the 4-momentum conservation as P − p2 = p1 .
8.20. The interior of a star like the Sun has a temperature of about 15-20 million Kelvin.
Using the familiar statistical physics formula hKEi = 23 kB T , with kB the Boltzmann constant, estimate the kinetic energy of a proton in the interior of the Sun. What is the speed
β of this proton?
136
Relativistic Interactions
Solution: Use T = 2 × 107 K and mp = 938.272 MeV. Then
hKEi = 32 (1.38 × 10−23 )(2 × 107 ) = 4.14 × 10−16 J,
which is about 2.6 keV or 0.0026 MeV. This is the KE of the proton. The total energy of
the proton is thus 938.2746 MeV. The Lorentz factor for the proton is given by
γ−1=
E
938.2746
−1=
− 1 = 2.76 × 10−6 ≈ 12 β 2 .
M
938.272
This gives a speed of β = 0.00235.
8.21. Derive Equation (8.45).
Solution: We want to have the identity
β 2 E 2 + (β~
α+α
~ β) · p~E + (~
α · p~)2 − m2 = E 2 − p~ · p~ − m2 .
For the equality to hold, the coefficients must match on both sides. E 2 has 1 as coefficient
on the right. So, β 2 = 1. There is no px E term on the right. Therefore, its coefficient on
the left must be zero. This gives αx β + βαx = 0; similarly for py E and pz E terms. The
terms that are of second order in momentum are (~
α · p~)2 on the left and −~
p · p~ on the right.
So, they too must equal.
8.22. Derive Equations (8.46) and (8.47).
Solution: Let’s evaluate (~
α · p~)2 , being careful not to commute the components of α
~:
(~
α · p~)2 = (αx px + αy py + αz pz )(αx px + αy py + αz pz )
= αx px (αx px + αy py + αz pz ) + αy py (αx px + αy py + αz pz )
+ αz pz (αx px + αy py + αz pz ),
or
(~
α · p~)2 = αx2 p2x + αx αy px py + αx αz px pz
+ αy αx py px + αy2 p2y + αy αz py pz
+ αz αx pz px + αz αy pz py + αz2 p2z .
In the final product on the right-hand side I haven’t commuted the p terms. I can! Because
they are ordinary functions. Commuting the ps and collecting terms, I get
(~
α · p~)2 = αx2 p2x + αy2 p2y + αz2 p2z
+ (αx αy + αy αx )px py
+ (αx αz + αz αx )px pz
+ (αy αz + αz αy )py pz .
To get the identity (~
α · p~)2 = −~
p · p~ = −p2x − p2y − p2z , I must have
αx2 = αy2 = αz2 = −1
αx αy + αy αx = αx αz + αz αx = αy αz + αz αy = 0.
137
8.23. A proton of mass mp = 938.3 MeV collides with a stationary nucleus of mass M
to produce antiprotons. The minimum number of particles at the end is two protons, one
antiproton, and the nucleus.
(a) Show that the minimum amount of energy for this production is
Emin = 3mp +
4m2p
.
M
(b) If pions of mass mπ ≈ 140 MeV are also produced, this minimum energy increases.
Show that if N pions are added to the outcome, then
Emin = 3mp + N mπ +
(3mp + N mπ )2 − m2p
.
2M
As a check for your answer, make sure you get (a) as a special case.
(c) Find Emin for N = 12 when the initial proton impinges on a copper nucleus of mass
M = 59150 MeV.
Solution: Let me denote the numerator of (8.31) by M not to confuse it with the mass of
the nucleus. Then
M2 = (M + 3mp )2 − M 2 − m2p = 8m2p + 6M mp .
(a) Therefore,
Emin =
4m2p
M2
= 3mp +
2M
M
(b) If N pions are added to the outcome, then
M2 = (M + 3mp + N mπ )2 − M 2 − m2p
= (3mp + N mπ )2 + 2M (3mp + N mπ ) − m2p ,
and
Emin
(3mp + N mπ )2 − m2p
M2
=
= 3mp + N mπ +
2M
2M
(c)
Emin = 3 × 938.3 + 12 × 140 +
or Emin ≈ 4.7 GeV.
(3 × 938.3 + 12 × 140)2 − 938.32
,
2 × 59150
CHAPTER
9
Interstellar Travel
Problems With Solutions
9.1. Assume that M >> dmg and consider the ejection of dmg from the rocket as a
decay. Then use the energy and 3-momentum components of the 4-momentum conservation
P = p1 + p2 to show that
d(M γ) + dmg γg0 = 0
~ + dmg γ 0 β~ 0 = 0.
d(M γ β)
g g
Solution: Write the 4-momentum conservation as Pbef = pgas + Paft or 0 = pgas + dP,
and note that
~ .
pgas = (dmg γg0 , dmg γg0 β~g0 ) and dP = d(M γ), d(M γ β)
9.2. Show that
d(M γ) = γdM + M βγ 3 dβ
d(M γβ) = γβdM + M γ 3 dβ.
Solution: Clearly, d(M γ) = γdM + M dγ. But
dγ = d(1 − β 2 )−1/2 = β(1 − β 2 )−3/2 dβ = βγ 3 dβ
Similarly, d(M γβ) = γβdM + M d(γβ), with
d(γβ) = γdβ + βdγ = γdβ + β(βγ 3 dβ) = γ(1 + β 2 γ 2 )dβ = γ 3 dβ.
9.3. Derive Equation (9.3) from the equations preceding it.
140
Interstellar Travel
Solution: Substitute the given expressions in (9.2) to obtain
(1 ∓ ββg )(γβdM + M γ 3 dβ) − (β ∓ βg )(γdM + M βγ 3 dβ) = 0,
or
[(1 ∓ ββg )β − (β ∓ βg )] γdM + [(1 ∓ ββg ) − (β 2 ∓ βg β)] M γ 3 dβ = 0,
|
{z
}
|
{z
}
=±βg (1−β 2 )
or
±
1−β 2
βg
dM + M γdβ = 0 ⇐⇒ ±βg dM + M γ 2 dβ = 0.
γ
9.4. Consider the acceleration part of the photon propulsion (9.6) with initial speed zero.
Show that you get Equation (9.4) with βg = 1.
Solution: The acceleration part of the photon propulsion (9.6) with initial speed zero
reads:
p
p
(1 − β)(1 + β)
M
1 − β2
1
1 − β 1/2
=
=
=
=
.
M0
γ(1 + β)
1+β
1+β
1+β
This is the accelerating version of Equation (9.4) with βg = 1.
9.5. Suppose that a rocket accelerates from rest to β0 upon launch and decelerates from
β0 to zero upon landing. If M0 is the mass at launch and Mf is the mass at landing, show
that (9.7) gives Mf /M0 = [γ0 (β0 + 1)]−2 , in agreement with (the first half of) Equation
(9.5).
Solution: With β0 = 0 and β(t1 ) = β0 , the first equation of (9.7) gives
1
M1
.
=
M0
γ0 (β0 + 1)
With β(tf ) = 0, the second equation of (9.7) gives
Mf
1
.
=
M1
γ0 (β0 + 1)
Therefore,
Mf
=
M0
Mf
M1
M1
M0
=
1
.
+ 1)2
γ02 (β0
9.6. Show that both equations in (9.8) are consistent with m(0) = 1 for β(0) = β0 .
Solution: In (9.8), replace β(t) with β0 on the left and set m(t) = 1 on the right and show
that the equality holds. For the first equation, we get
β0 =
γ02 (β0 + 1)2 − 1
,
γ02 (β0 + 1)2 + 1
or
β0 γ02 (β0 + 1)2 + β0 = γ02 (β0 + 1)2 − 1,
or
1 + β0 = γ02 (β0 + 1)2 (1 − β0 ) ⇐⇒ 1 = γ02 (β0 + 1)(1 − β0 ),
which holds identically. The second equation is very similar.
141
9.7. The first equation in (9.8) yields β = 1 when m(t) is completely depleted. Is that a
violation of relativity or a confirmation of it? Think about it for a while before you look
up the answer in Note 5.2.3.
Solution: It is a confirmation of the theory. When m(t) = 0, the rocket is massless, so by
Note 5.2.3, it has to move at the speed of light.
9.8. Let −k be the rate of depletion of the fuel of a rocket.
(a) Solve dMfuel /dt = −k, when k is a positive constant subject to the condition that
beg
beg
Mfuel (0) = Mfuel
. Here Mfuel
is the amount of fuel at the beginning (which is not
necessarily the take-off) of the part of motion under consideration.
(b) Assuming that the fuel mass at take-off is M0 , determine k from the condition that
Mfuel (T0 ) = 0, where T0 is the time it takes for the fuel to be completely depleted.
(c) Show that if time is measured in units of T0 , then
beg
Mtot (t) = Mf + Mfuel (t) = Mf + Mfuel
− M0 t,
where Mf is the mass left over at the end of the trip.
Solution:
(a) The solution is Mfuel (t) = −kt + C, where C is the constant of integration. With
beg
beg
Mfuel (0) = Mfuel
, we get C = Mfuel
. Therefore, the complete solution can be written
as
beg
.
Mfuel (t) = −kt + Mfuel
(b) We have 0 = −kT0 + M0 . Thus, k = M0 /T0 .
(c) With k given in (b), write Mfuel (t) of (a) in terms of T0 :
beg
.
Mfuel (t) = −M0 t/T0 + Mfuel
Then,
beg
− M0 (t/T0 ).
Mfuel (t) = Mf + Mfuel (t) = Mf + Mfuel
9.9. Derive Equation (9.11) and by taking the time derivative of (9.12) show that the latter
is the integral of (9.11).
Solution: With β0 = 0, the first equation of (9.8) becomes
β(t) =
1 − m2 (t)
.
1 + m2 (t)
beg
On the other hand, at the beginning of the journey, Mfuel
= M0 . Thus, (9.10) becomes
m(t) = 1 − m0 t,
m0 =
M0
,
Mf + M 0
t ≤ 1.
142
Interstellar Travel
Now take the derivative of (9.12):
m0
dx
2 d −1
2
= −1 −
tan (1 − m0 t) = −1 +
dt
m0 dt
m0 1 + (1 − m0 t)2
2
1 − (1 − m0 t)2
= −1 +
=
.
1 + (1 − m0 t)2
1 + (1 − m0 t)2
9.10. Show that the time required to accelerate the rocket from rest to β0 is given by (9.13).
Solution: By (9.11), we have to solve the equation
β0 =
1 − (1 − m0 tacc )2
⇐⇒ (1 − m0 tacc )2 (1 + β0 ) = 1 − β0 ,
1 + (1 − m0 tacc )2
or
(1 − m0 tacc )2 =
or
1 − m0 tacc =
1 − β0
1 − β02
=
,
1 + β0
(1 + β0 )2
γ0 (1 + β0 ) − 1
1
⇐⇒ m0 tacc =
.
γ0 (1 + β0 )
γ0 (1 + β0 )
9.11. Show that x(tacc ) is given by Equation (9.14).
Solution: Just substitute the result of the previous problem in (9.12). No algebra required!
9.12. Differentiate Equation (9.16) to get (9.15).
Solution: The derivative of (9.16) is
i
dx
2
d h −1
=1+
tan
{γ
(1
+
β
)
[1
−
m
γ
(1
+
β
)t]}
0
0
0
0
0
dt
m0 γ02 (1 + β0 )2 dt
2
m0 γ02 (1 + β0 )2
=1−
m0 γ02 (1 + β0 )2 1 + γ02 (1 + β0 )2 [1 − γ0 (1 + β0 )m0 t]
2
.
=1−
2
2
1 + γ0 (1 + β0 ) [1 − γ0 (1 + β0 )m0 t]
This can be trivially shown to equal to (9.15).
9.13. Show that the time required to decelerate the rocket from β0 to rest is given by (9.17)
and the distance covered by (9.18).
Solution: Setting (9.15) equal to zero yields
γ02 (1 + β0 )2 [1 − m0 γ0 (1 + β0 )tdec ]2 − 1 = 0,
or
1 − m0 γ0 (1 + β0 )tdec =
or
m0 γ0 (1 + β0 )tdec =
This is essentially (9.17).
1
,
γ0 (1 + β0 )
γ0 (1 + β0 ) − 1
.
γ0 (1 + β0 )
143
9.14. The maximum possible speed attainable is when no fuel is left upon landing. Show
that this speed is given by Equation (9.19), and for such a speed, tacc and tdec are given by
Equation (9.20).
Solution: Setting the sum of (9.13) and (9.17) equal to one yields
1=
γ02 (1 + β0 )2 − 1
γ0 (1 + β0 ) − 1 γ0 (1 + β0 ) − 1
=
,
+
m0 γ0 (1 + β0 )
m0 γ02 (1 + β0 )2
m0 γ02 (1 + β0 )2
or
γ02 (1 + β0 )2 (1 − m0 ) = 1 ⇐⇒ γ0 (1 + β0 ) = √
because
m0 + mf ≡
1
,
mf
Mf
M0
+
= 1.
M f + M0 Mf + M 0
Rewrite the last equation as
s
p
2
1 − β0
1 − β0
1
=
=√ .
1 + β0
1 + β0
mf
Square both sides and solve for β0 to get Equation (9.19).
√
Now use γ0 (1 + β0 ) = 1/ mf —obtained above—in (9.13) to get
tacc
√
(1/ mf ) − 1
=
=
√
(1 − mf )(1/ mf )
√
1 − mf
1 − mf
| {z }
=
1
√ .
1 + mf
√
√
=(1− mf )(1+ mf )
Equation (9.17) now gives
tdec
√
mf
tacc
√
= √
= mf tacc =
√ .
1/ mf
1 + mf
9.15. Derive Equations (9.21) and (9.22). Plot each separately as well as the sum xacc +xdec
and verify that the sum is a decreasing function with a maximum of 0.57 light second
occurring when mf = 0.
√
Solution: Substitute γ0 (1 + β0 ) = 1/ mf and m0 + mf = 1 in (9.14):
√
(1/ mf ) − 1
π
2
√ x(tacc ) =
−
−
tan−1
mf
√
2(1 − mf ) (1 − mf )/ mf
1 − mf
√
√ π/2 − 1 + mf − 2 tan−1
mf
=
.
1 − mf
This is (9.21). I let you derive (9.22). Figure 9.1 of the manual shows xacc , xdec , and
xacc + xdec as functions of mf .
9.16. I discussed the case of maximum attainable speed when mf is small. What about
the case when mf ≈ 1? Write mf = 1 − m0 and assume that m0 is very small.
144
Interstellar Travel
0.6
0.5
0.4
0.3
0.2
0.1
0.2
0.4
0.6
0.8
1.0
Figure 9.1: The lowest plot is xdec (mf ), the middle plot is xacc (mf ), and the top plot is the sum of the
two.
(a) Show from Equation (9.19) that the maximum attainable speed is 21 m0 .
(b) From the expansion of (9.21) show that xacc → 81 m0 .
(c) From (9.22) conclude that xdec → 81 m0 .
Solution:
(a) With mf = 1 − m0 , Equation (9.19) becomes
β0 =
m0
≈ 21 m0 for m0 → 0.
2 − m0
(b) Since the denominator of (9.21) is m0 , we expand the numerator up to m20 . First look
at each term of the numerator separately:
√
mf = (1 − m0 )1/2 ≈ 1 − 12 m0 +
From
tan−1 (1 − x) ≈
1 1
2(2
− 1) 2
m0 = 1 − 12 m0 − 18 m20 .
2
π x x2
− −
4
2
4
and the result above, we get
π 1
1
√
tan−1 ( mf ) = tan−1 1 − 12 m0 − 18 m20 = − ( 12 m0 + 18 m20 ) − ( 12 m0 + 18 m20 )2
4 2
4
π m0 m20 m20
π m0 m20
= −
−
−
= −
−
,
4
4
16
16
4
4
8
145
where I kept terms up to m20 in the final expression. Now substitute these in (9.21)
and obtain
√
xacc
mf
√
−2 tan−1 ( mf )
z
}|
{
z
}|
{ π m
2
m
0
1 − 12 m0 − 81 m20 − +
+ 0 +π/2 − 1
m0
2
2
4
=
=
.
m0
8
(c) I leave this part for you.
9.17. Problem 9.16 is based on the assumption of uniform depletion of the fuel. The result,
however, turns out to be general, at least for the case of a laser-propelled rocket as discussed
on page 206.
(a) Suppose you use half the fuel (saving the other half for landing), i.e., the lasing
photons, on a massless rocket (Mf = 0) to accelerate the rocket from rest to a speed
βf . Using the first equation in (9.7), show that βf is given by
1 + 0.5κ
1
=
,
1+κ
γf (βf + 1)
κ << 1.
(b) Solve this equation for βf in terms of κ to show that βf ≈ 0.5κ.
Solution:
(a) The fuel mass at time t is κMlas (t), where Mlas (t) is the mass of the lasing material
at t. Define r(t) ≡ [Mlas (t)/Mlas ] as the ratio of the fuel mass at t to the initial fuel
mass. Then the fuel mass at time t is κr(t)Mlas , and with Mf = 0, the ratio of the
mass at t to the initial mass is
m(t) ≡
M (t)
Mlas + κr(t)Mlas
1 + κr(t)
=
=
.
M0
Mlas + κMlas
1+κ
For r(t) = 0.5, the first equation in (9.7) gives
1 + 0.5κ
1
=
,
1+κ
γf (βf + 1)
κ << 1.
(b) Approximate the left-hand side to first order in κ:
1 + 0.5κ
= (1 + 0.5κ)(1 + κ)−1 ≈ (1 + 0.5κ)(1 − κ) ≈ 1 − 0.5κ;
1+κ
p
rewrite the right-hand side as (1 − βf )/(1 + βf ) and square both sides:
(1 − 0.5κ)2 =
1 − βf
1 − βf
κ
κ
⇐⇒ 1 − κ =
⇐⇒ βf =
≈ .
1 + βf
1 + βf
2−κ
2
146
Interstellar Travel
9.18. Lorentz transform both the energy E of a photon and the sum Esail + Psail of the
energy and momentum of the light sail to another RF and show that you get the same
equality (9.25) in the new RF.
Solution: Write Equation (9.24) as
0
E − Eref = Esail
− Esail = δEsail
0
p + pref = Psail
− Psail = δPsail ,
which is more general than (9.24) because I am not assuming that M is constant. Add
these two equations to obtain
2E = δ(Esail + Psail ).
Let O be the old RF and assume it moves in the positive x direction of the new frame O0 .
Then the left-hand side of the equation above transforms to
2E 0 = 2γ(E + βp) = 2γ(E + βE) = γ(1 + β)(2E).
The right-hand side transforms to
0
0
δ(Esail
+ Psail
) = δ[γ(Esail + βPsail ) + γ(Psail + β(Esail )]
= δ[γ(1 + β)(Esail + Psail )] = γ(1 + β)δ(Esail + Psail ).
0 ).
0
+ Psail
Thus, 2E 0 = δ(Esail
9.19. In this problem you are going to verify Equation (9.27).
(a) Lorentz transform the event of the emission of the ith photon to the instantaneous
reference frame of the spacecraft and show that
(tnano
, xnano
) = γti (1, −β).
i
i
(b) How long does it take for this photon to reach the craft? Adding tnano
to this time,
i
ref
find tnano,
.
i
(c) From (a) and (b) and the fact that reflections occur at the spacecraft, conclude that
the events of the reflection of the ith and i + 1th photons in the spacecraft’s RF are
γ(1 + β)ti , 0 , and γ(1 + β)ti+1 , 0 .
(d) Now Lorentz transform this back to the laser RF and derive (9.27).
Solution:
(a) The spacecraft is moving in the positive direction of laser’s frame. So, the laser is
moving in the negative direction of the spacecraft frame. Assuming that the laser is
at the origin, and the origin of time is the same for both RFs, the coordinates of the
ith photon in the laser frame is (ti , 0). Lorentz transforming to the nanocraft’s RF,
we get
xnano
= γ(0 − βti ) = −γβti , tnano
= γ(−0 + ti ) = γti .
i
i
147
(b) The distance from which the photon is emitted is xnano
(note that it doesn’t matter
i
that the laser doesn’t remain at that position). Therefore, the time at which the
photon reaches the craft is
ref
tnano,
= tnano
+ |xi |nano = γ(1 + β)ti .
i
i
ref
(c) Since the craft is assumed to be at the origin of its frame, xnano,
= 0.
i
(d) Lorentz transforming the event of the reflection of the ith photon to the laser frame
yields
ti
nano, ref
tref
=
γ
t
+
0
= γ 2 (1 + β)ti =
i
i
1−β
βti
nano, ref
xref
= γ 2 β(1 + β)ti =
,
i = γ 0 + βti
1−β
with similar equations for i + 1th photon. Equation (9.27) now follows. Note that for
(9.27), we don’t need xref
i .
9.20. Why reflection coefficient R makes no sense in relativity! Write Equation (8.18) as
0
E − Eref = Esail
− Esail = δEsail
0
p + pref = Psail
− Psail = δPsail ,
(9.1)
where no approximation is made regarding the mass of the spacecraft.
(a) Using R, show that these equations yield
1−R
δEsail
=
.
1+R
δPsail
(b) Use Esail =
q
2 + M 2 to show that
Psail
1−R
= β,
1+R
where β is the instantaneous speed of the craft.
Solution:
(a) With Eref = pref = Rp = RE and taking the ratio of the first to the second equation
above, we get
1−R
δEsail
=
.
1+R
δPsail
q
2 + M 2 , we obtain
(b) From Esail = Psail
δEsail = δ
q
Psail
2 + M 2 = qPsail δPsail
Psail
=
δPsail = βδPsail .
E
2
2
sail
Psail + M
This gives the result we are after.
148
Interstellar Travel
9.21. Recall that the solid angle subtended at point P0 by an infinitesimal area da located
at point P is given by
|ên · (~r − ~r0 )|
da,
dΩ =
|~r − ~r0 |3
where ên is the unit vector normal to da and ~r and ~r0 are the position vectors of P and
P0 , respectively. Now center a square of side L in the xy-plane and let P0 have coordinates
(0, 0, z0 ).
(a) Show that the total solid angle Ω subtended at P0 by the square is
!
2
L
p
Ω = 4 tan−1
.
2z0 2L2 + 4z02
(b) Show that if z0 >> L, then
Ω≈
L2
.
z02
Solution: In this problem, ên = ±êz . The ambiguity is due to the fact that a surface has
two sides! Let’s take ên = −êz (away from P0 ). We also have
~r = hx, y, 0i,
~r0 = h0, 0, z0 i ~r − ~r0 = hx, y, −z0 i.
Thus,
dΩ =
|−êz · (~r − ~r0 )|
z0
dxdy,
dxdy = 2
3
2
(x + y + z02 )3/2
|~r − ~r0 |
and
Z
L/2
Z
L/2
Ω = z0
−L/2
−L/2
dxdy
= 4 tan−1
(x2 + y 2 + z02 )3/2
L2
2z0
p
2L2 + 4z02
!
.
(b) If z0 >> L, then
L2
L2
L2
p
=
≈
,
2z0 (2|z0 |)
4z02
2z0 2L2 + 4z02
z0 > 0.
Now we know that tan−1 (x) ≈ x to first order in x. Thus,
2
L
L2
=
.
Ω≈4
4z02
z02
9.22. This problem treats the first stage of the journey of a spacecraft nonrelativistically.
Assume perfect reflection of a beam of photons from the light sail.
(a) What is the momentum transfer to the light sail by a beam of photons carrying energy
∆E? Show that the force exerted on it is 2P/c, where P is the fraction of laser power
hitting the light sail. What is the acceleration if the light sail has a mass M ?
(b) For the first leg of the trip, show that the acceleration is uniform. Find the value of
this acceleration for a nanocraft with a mass of 10 grams driven by a 100 GW ground
laser.
149
Solution:
(a) The momentum carried by the beam of photons is ∆E/c. The momentum transfer is
∆p = 2∆E/c, assuming perfect reflection. The force is
F =
∆p
2∆E
2P
F
2P
=
=
⇐⇒ a =
=
.
∆t
c∆t
c
M
Mc
(b) For the first stage, P = Plaser . Therefore the acceleration is constant. For a nanocraft
with a mass of 10 grams driven by a 100 GW ground laser, the acceleration is
a=
2Plaser
2 × 1011
= 6.7 × 104 m/s2 .
=
Mc
0.01 × 3 × 108
9.23. This problem treats the second stage of the journey of a spacecraft nonrelativistically.
Assume perfect reflection of a beam of photons from the light sail.
(a) For the second leg of the trip, show that the acceleration is
a=
2Plaser x20
.
M c x2
(b) Write dv/dt as vdv/dx and integrate the equation in (a) to get
1 2
2v
− 12 v02 =
x0 2Plaser x0 1 −
Mc
x
assuming that the speed is v0 at x0 .
(c) Show that the result in (b) can also be written as
r
x0
v = v0 2 − ,
x
√
giving v∞ = 2 v0 . Hint: Find x0 in terms of v0 from Problem 9.22.
(d) Integrate (c) to find t as a function of x with the constant of integration determined
by the condition that x = x0 at t = 0.
Solution: The difference between this and the previous problem is the fraction of laser
power delivered. In this problem, P = Plaser x20 /x2 .
(a) Therefore, the acceleration is
a=
2Plaser x20
.
M c x2
(b) Write the equation in (a) as
dv
dv
2Plaser x20
=v
=
,
dt
dx
M c x2
or
vdv =
2x20 Plaser dx
.
M c x2
150
Interstellar Travel
Integrate this to get
v
Z
v0
or
1 2
2v
−
1 2
2 v0
2x2 Plaser
udu = 0
Mc
2x2 Plaser
= 0
Mc
1
1
−
x0 x
Z
x
x0
=
du
,
u2
2x0 Plaser x0 1−
.
Mc
x
(c) The speed v0 is also the speed at the end of the first stage. Thus, we can write
v02 = 2ax0 ⇐⇒
1 2
2 v0
= ax0 =
2Plaser
x0 .
Mc
Therefore, the last equation in (b) becomes
1 2
2v
x0 − 12 v02 = 12 v02 1 −
x
or
2
v =
v02
x0 2−
⇐⇒ v = v0
x
r
2−
x0
.
x
(d) We have
dx
= v0
dt
r
x0
dx
2−
⇐⇒ p
x
2−
or
r
Integrating this, we obtain
Z xr
x0
x0
x
= v0 dt
x
dx = v0 dt.
2x − x0
u
du = v0
2u − x0
Z
t
du = v0 t.
0
The integration on the left gives the following complicated and uninteresting final
result:
p
√
√
√
x(2x − x0 ) − x0
2 x0
2 x + 4x − 2x0
√ √
+
ln
= v0 t.
2
4
(2 + 2) x0
Note that if x = x0 , the left-hand side gives zero, as it should.
CHAPTER
10
A Painless Introduction to Tensors
Problems With Solutions
10.1. Using indices, show that the divergence of the curl of any 3-vector is zero. Similarly,
show that the curl of the gradient of any function is zero.
Solution:
∇ · (∇ × A) = ∂i (∇ × A)i = ∂i ijk ∂j Ak = ijk ∂i ∂j Ak = 0,
by Note 10.1.6 and the fact that ∂i ∂j = ∂j ∂i .
(∇ × ∇f )i = ijk ∂j (∂k f ) = ijk ∂j ∂k f = 0,
again by Note 10.1.6 and the fact that ∂j ∂k = ∂k ∂j .
10.2. Using the elementary determinant way of calculating cross products, show that
(A × C) · (B × D) = (A · B)(C · D) − (A · D)(B · C).
Now use this vector identity and Note 10.1.10 to prove Equation (10.8).
Solution:


êx êy êz
A × C = det Ax Ay Az  = (Ay Cz − Az Cy )êx − (Ax Cz − Az Cx )êy + (Ax Cy − Ay Cx )êz .
Cx Cy Cz
Similarly,
B × D = (By Dz − Bz Dy )êx − (Bx Dz − Bz Dx )êy + (Bx Dy − By Dx )êz .
Therefore,
(A × C)·(B × D) = (Ay Cz − Az Cy )(By Dz − Bz Dy ) + (Ax Cz − Az Cx )(Bx Dz − Bz Dx )
+ (Ax Cy − Ay Cx )(Bx Dy − By Dx )
= Ay By (Cz Dz + Cx Dx ) + Az Bz (Cy Dy + Cx Dx ) + Ax Bx (Cy Dy + Cz Dz )
− Ay Dy (Bz Cz + Bx Cx ) − Az Dz (By Cy + Bx Cx ) − Ax Dx (Bz Cz + By Cy ).
152
A Painless Introduction to Tensors
Note that I can write the first three terms as
Ay By C · D − Ay By Cy Dy + Az Bz C · D − Az Bz Cz Dz + Ax Bx C · D − Ax Bx Cx Dx ,
or as
(A · B)(C · D) − Ay By Cy Dy − Az Bz Cz Dz − Ax Bx Cx Dx ,
(10.1)
and the last three terms as
−Ay Dy B · C + Ay Dy By Cy − Az Dz B · C + Az Dz Bz Cz − Ax Dx B · C + Ax Dx Bx Cx
or as
−(A · D)(B · C) + Ay Dy By Cy + Az Dz Bz Cz + Ax Dx Bx Cx .
Combining this with (10.1) above gives the final result.
To prove Equation (10.8), multiply it on both sides by Aj Ck Bm Dn . On the left, you
get
ijk imn Aj Ck Bm Dn = (ijk Aj Ck )(imn Bm Dn ) = (A × C)i (B × D)i = (A × C) · (B × D)
and on the right you get
(δjm δkn − δjn δkm )Aj Ck Bm Dn = δjm Aj Bm δkn Ck Dn − δjn Aj Dn δkm Ck Bm
= Aj Bj Ck Dk − Aj Dj Ck Bk = (A · B)(C · D) − (A · D)(B · C).
Since Equation (10.8) leads to a true vector identity, it holds.
10.3. Express A · (B × C) in index form. Then using it show the cyclic property of this
triple product:
A · (B × C) = C · (A × B) = B · (C × A).
Solution: Just use the properties of the Levi-Civita symbol, for example, ijk = kij .
A · (B × C) = Ai (B × C)i = Ai ijk Bj Ck = ijk Ai Bj Ck
= kij Ck Ai Bj = lmn Cl Am Bn = Cl (A × B)l = C · (A × B).
In the last line I changed the dummy indices from ijk to lmn. I leave the remaining cyclic
property for you.
10.4. Using indices, prove the following vector identities:
∇ · (f A) = (∇f ) · A + f ∇ · A
∇ × (f A) = (∇f ) × A + f ∇ × A
∇ × (∇ × A) = ∇(∇ · A) − ∇2 A
Solution:
∇ · (f A) = ∂i (f A)i = ∂i (f Ai ) = (∂i f )Ai + f (∂i Ai ) = (∇f ) · A + f ∇ · A
[∇ × (f A)]i = ijk ∂j (f Ak ) = ijk [(∂j f )Ak + f ∂j Ak ]
= ijk (∂j f )Ak + f ijk ∂j Ak = [(∇f ) × A]i + f (∇ × A)i
153
[∇ × (∇ × A)]i = ijk ∂j (∇ × A)k = ijk ∂j (klm ∂l Am ) = kij klm ∂j ∂l Am
= (δil δjm − δim δjl )∂j ∂l Am = δil δjm ∂j ∂l Am − δim δjl ∂j ∂l Am
= ∂m ∂i Am − ∂j ∂j Ai = ∂i (∂m Am ) − (∂j ∂j )Ai
= ∂i (∇ · A) − (∇2 )Ai = [∇(∇ · A]i − (∇2 A)i .
10.5. Show that the determinant of a 3 × 3 matrix A with elements aij is given by
det A = ijk a1i a2j a3k .
Solution: Let


a11 a12 a13
A = a21 a22 a23  .
a31 a32 a33
Remember the summation convention:
ijk a1i a2j a3k = 1jk a11 a2j a3k + 2jk a12 a2j a3k + 3jk a13 a2j a3k .
Now I’ll calculate each term separately by summing over the remaining indices, using the
properties of the Levi-Civita symbol. The first term is
1jk a11 a2j a3k = 12k a11 a22 a3k + 13k a11 a23 a3k
= 123 a11 a22 a33 + 132 a11 a23 a32
= a11 (a22 a33 − a23 a32 );
the second term is
2jk a12 a2j a3k = 21k a12 a21 a3k + 23k a12 a23 a3k
= 213 a12 a21 a33 + 231 a12 a23 a31
= −a12 (a21 a33 − a23 a31 );
and the last term is
3jk a13 a2j a3k = 31k a13 a21 a3k + 32k a13 a22 a3k
= 312 a13 a21 a32 + 321 a13 a22 a31
= a13 (a21 a32 − a22 a31 ).
When you add all the terms, you get the familiar expansion of the determinant according
to its first row.
10.6. Let êi , i = 1, 2, 3 be the components of a unit vector. Let
aijk = ijm êk êm + imk êj êm + mjk êi êm .
Show that a123 = 1, ajik = −aijk , and aikj = −aijk . Therefore, aijk = ijk .
154
A Painless Introduction to Tensors
Solution: Without loss of generality, choose your axes so that the unit vector lies along
the x-axis. Then êi = δi1 and
aijk = ijm δk1 δm1 + imk δj1 δm1 + mjk δi1 δm1
= ij1 δk1 + i1k δj1 + 1jk δi1 .
This implies that
a123 = 121 δk1 + 11k δ21 + 123 δ11 = 123
and
ajik = ji1 δk1 + j1k δi1 + 1ik δj1
= −ij1 δk1 − 1jk δi1 − i1k δj1 = −aijk ,
and
aikj = ik1 δj1 + i1j δk1 + 1kj δi1
= −i1k δj1 − ij1 δk1 − 1jk δi1 = −aijk .
10.7. By substituting the components of d~r1 , d~r2 , and d~r3 parallel and perpendicular to β~
in dV = d~r1 · (d~r2 × d~r3 ),
(a) show that
dV = |(d~r1 )|| ||(d~r2 )⊥ × (d~r3 )⊥ | + |(d~r2 )|| ||(d~r3 )⊥ × (d~r1 )⊥ |
+ |(d~r3 )|| ||(d~r1 )⊥ × (d~r2 )⊥ |.
(10.2)
This shows that the volume element is the product of an area perpendicular to the
direction of motions [terms like |(d~r2 )⊥ × (d~r3 )⊥ |] times a height in the direction of
motion [terms like |(d~r1 )|| |].
(b) From (6.17) and dt0 = 0, show that |(d~r 01 )|| | = |(d~r1 )|| |/γ, with similar relations for
|(d~r 02 )|| | and |(d~r 03 )|| |.
(c) Insert these results in the expression for dV 0 as given by (10.2) and derive the formula
dV 0 = dV /γ.
Solution:
(a) First calculate the cross product
d~r2 × d~r3 = [(d~r2 )|| + (d~r2 )⊥ ] × [(d~r3 )|| + (d~r3 )⊥ ]
= (d~r2 )|| × (d~r3 )⊥ + (d~r2 )⊥ × (d~r3 )|| + (d~r2 )⊥ × (d~r3 )⊥ .
As you dot d~r1 = (d~r1 )|| +(d~r1 )⊥ with the expression above, note that the dot product
of (d~r1 )|| with any cross product containing parallel components gives zero. Also note
that (d~r2 )⊥ ×(d~r3 )⊥ is parallel to the direction of motion, so it gives zero when dotted
with (d~r1 )⊥ . With these remarks in mind, we get
dV = (d~r1 )|| · [(d~r2 )⊥ × (d~r3 )⊥ ] + (d~r1 )⊥ · [(d~r2 )|| × (d~r3 )⊥ + (d~r2 )⊥ × (d~r3 )|| ]
= (d~r1 )|| · [(d~r2 )⊥ × (d~r3 )⊥ ] + (d~r2 )|| · [(d~r3 )⊥ × (d~r1 )⊥ ]
+ (d~r3 )|| · [(d~r1 )⊥ × (d~r2 )⊥ ],
155
where in the last step, I used the cyclic property of the mixed dot and cross product.
Finally note that the participants in the dot product are in the same direction, so you
can replace the dot product with the product of absolute values.
(b) With dt0 = 0, the first equation of (6.17) gives dt = − β~ ~r|| , and the second equation
yields
h
i
d~r
~ 2 d~r|| = || .
d~r||0 = γ d~r|| − β~ β~ d~r|| = γ d~r|| − β
γ
(c) When you prime (10.2), the cross products don’t change [see the third equation of
(6.17)], and by (b), the parallel components get divided by γ. Therefore, the entire
expression is divided by γ.
CHAPTER
11
Relativistic Electrodynamics
Problems With Solutions
11.1. From Bi = ijk ∂j Ak show that ∂j Ak − ∂k Aj = jkm Bm .
Solution: Multiply both sides of Bi = ijk ∂j Ak by lmi to obtain
lmi Bi = lmi ijk ∂j Ak = ilm ijk ∂j Ak = (δlj δmk − δlk δmj )∂j Ak
= δlj δmk ∂j Ak − δlk δmj ∂j Ak = ∂l Am − ∂m Al .
11.2. Go through the details of the manipulation of the last line of (11.10) and derive
(11.11).
Solution: Multiply the two square brackets and note that the product of the two β terms
will contain symmetric indices which give zero when multiplied by the Levi-Civita symbol.
W = 12 ijk lmn [δkm + (γ − 1)β̂k β̂m ][δjl + (γ − 1)β̂j β̂l ]Bn
= 12 ijk lmn δkm [δjl + (γ − 1)β̂j β̂l ]Bn + 12 (γ − 1)ijk lmn δjl β̂k β̂m Bn
= 12 ijk lkn [δjl + (γ − 1)β̂j β̂l ]Bn + 12 (γ − 1)ijk jmn β̂k β̂m Bn .
Now use Note 10.1.7 to rewrite the products of the Levi-Civita symbols. You have to
rearrange indices to bring them to the same order as they appear in the Note. Once you
do that, you’ll end up with the following expression:
W = − 12 (δil δjn − δin δjl )δjl Bn − 12 (γ − 1)(δil δjn − δin δjl )β̂j β̂l Bn
− 12 (γ − 1)(δim δkn − δin δkm )β̂k β̂m Bn
= − 12 (δij δjn − δin δjj )Bn − 12 (γ − 1)(β̂j β̂i Bj − β̂j β̂j Bi )
− 12 (γ − 1)(β̂k β̂i Bk − β̂k β̂k Bi ).
Now note that δjj = 3 and β̂k β̂k = β̂ · β̂ = 1. Then
~ − Bi ) = γBi − (γ − 1)β̂i (β̂ · B).
~
W = Bi − (γ − 1)(β̂i β̂ · B
158
Relativistic Electrodynamics
11.3. Show that the sum of the first two terms of (11.12) gives γijk βj Ek .
Solution: This involves simply renaming the dummy indices. I’ll work on the first term:
γimk βk ∂m A0 = γi♣♠ β♠ ∂♣ A0 = −γi♠♣ β♠ ∂♣ A0 = −γijk βj ∂k A0 .
Now I add it to the second term to get
γijk βj (−∂k A0 + ∂0 Ak ) = γijk βj Ek .
11.4. Derive Equation (11.14) from (11.8) and (11.13).
Solution: Since (11.8) and (11.13) are identical (except for the change in sign in the first
term), I’ll do (11.8). Inserting the parallel and perpendicular components on both sides of
the equation, I get
~ ||0 + E
~ ⊥0 = γ[E
~ || + E
~ ⊥ − β~ × (B
~ || + B
~ ⊥ )] − (γ − 1)β̂[β̂ · (E
~ || + E
~ ⊥ ).
E
Now note that
~ || = 0 = β̂ · E
~ ⊥,
β~ × B
~ || = E
~ || .
and β̂ β̂ · E
Hence,
~ ||0 + E
~ ⊥0 = γ E
~ || + γ E
~ ⊥ − γ β~ × B
~ ⊥ − (γ − 1)E
~ ||
E
~ || + γ(E
~ ⊥ − β~ × B
~ ⊥ ).
=E
The first term on the right-hand side is parallel to the direction of motion, so it must be
~ ||0 and the second term is perpendicular to the direction of motion, so it must be
equal to E
~ ⊥0 .
equal to E
11.5. Derive Equation (11.18).
Solution: Dotting both sides of (11.17) by β̂ and rewriting the resulting equation slightly
differently yields
γβ
0
0
0
β̂ · ~r
β̂ · ~r = β̂ · ~r − γβ t −
γ+1
γ2β2
= −γβt0 + 1 +
β̂ · ~r = −γβt0 + γ β̂ · ~r,
γ+1
because γ 2 β 2 = γ 2 − 1 = (γ − 1)(γ + 1).
11.6. Derive Equation (11.19).
Solution: The equation before (11.19) can be expressed as
2 2
~S = γ~r 0 − γ βt
~ 0 + γ γ β − γ + 1 β̂ β̂ · ~r 0 .
γ+1
|
{z
}
=0
The fact that the expression in parentheses is zero can be readily shown.
159
11.7. Derive Equation (11.21). Now take the dot product of the equation with itself to
derive (11.22).
Solution: Rewrite (11.17) as
2 2
~ 0 + γ β β̂ β̂ · ~r 0 = ~r 0 − γ βt
~ 0 + (γ − 1)β̂ β̂ · ~r 0 .
~r = ~r 0 − γ βt
γ+1
Now use (11.19) and substitute for ~r 0 :
~ 0 − γ βt
~ 0 + (γ − 1)β̂ β̂ · (~r 0 + βt
~ 0)
~r = ~r 0inst + βt
inst
~ 0 + (γ − 1)β̂ β̂ · ~r 0 + (γ − 1) β̂ β̂ · β~ t0
= ~r 0inst − (γ − 1)βt
inst
| {z }
~
=β
= ~r 0inst + (γ − 1)β̂ β̂ · ~r 0inst .
Now square (dot with itself) both sides
02
r2 = rinst
+ (γ − 1)2 (β̂ · ~r 0inst )2 + 2(γ − 1)(β̂ · ~r 0inst )2
02
= rinst
+ (γ − 1)(γ + 1)(β̂ · ~r 0inst )2
|
{z
}
=γ 2 −1=γ 2 β 2
02
02
= rinst
+ γ 2 (β β̂ · ~r 0inst )2 = rinst
+ γ 2 (β~ · ~r 0inst )2 .
11.8. Derive Equation (11.25).
Solution: From the equation before (11.25), plus β~ × ~r = β~ × ~r 0inst and (11.22), we get
~0 = h
B
ke qγ
02
rinst
+
γ 2 (β~
· ~r 0inst )2
0
i3/2 β~ × ~r inst ,
with
02
02
02
02
02
rinst
+ γ 2 (β~ · ~r 0inst )2 = rinst
+ γ 2 β 2 rinst
cos2 θinst = rinst
+ (γ 2 − 1)rinst
cos2 θinst
02
02
02
= rinst
sin2 θinst + γ 2 rinst
cos2 θinst = rinst
(sin2 θinst + γ 2 cos2 θinst ).
11.9. Consider an infinite plate charged uniformly with surface density σ. The plate is
moving uniformly with speed β perpendicular to its surface. There are two ways to calculate
the electric and magnetic fields of this charge distribution.
~ and B
~ in the rest frame of the plate using the elementary
(a) Calculate the fields E
method of Gauss’s law. Now transform those fields to the frame in which the plate is
moving.
(b) Use (11.24) and (11.25) to write the contribution from each element of charge on the
surface and integrate over the plate to get the fields.
160
Relativistic Electrodynamics
Solution: I’ll assume that the plate is infinitely thin. This precludes a conductor because
a conductor is really two infinite sheets with the same surface charge density (that’s how
the field inside becomes zero). Let O be the rest frame of the plate and place it in the
yz-plane of O. Consider the side facing the positive x direction.
~ = 2πke σ β̂ and B
~ = 0 in O. Then (11.14) gives E
~ 0 = 2πke σ β̂ and
(a) From Gauss’s law E
0
0
~
B = 0 in O as well.
(b) I’ll calculate the field at the moment that the two origins coincide, so that the plate
is in the y 0 z 0 -plane of O0 . Let P0 = (x00 , y00 , z00 ) be the field point, and Pinst = (0, y 0 , z 0 )
the instantaneous location of the charge element. Then
~r 0inst = hx00 , y00 , z00 i − h0, y 0 , z 0 i = hx00 , y00 − y 0 , z00 − z 0 i
and (11.22) becomes
r2 = x002 + (y00 − y 0 )2 + (z00 − z 0 )2 + γ 2 β 2 x002
= γ 2 x002 + (y00 − y 0 )2 + (z00 − z 0 )2
using the indispensable identity γ 2 β 2 = γ 2 − 1. Plugging this in (11.23) and integrating, we get the electric field at P0 :
Z ∞Z ∞
hx00 , y00 − y 0 , z00 − z 0 i
~ 0 = ke σγ
E
dy 0 dz 0 .
2 x0 2 + (y 0 − y 0 )2 + (z 0 − z 0 )2 ]3/2
[γ
−∞ −∞
0
0
0
Integration over y 0 gives Ey0 = 0 and integration over z 0 gives Ez0 = 0. So, the only
surviving component is Ex0 , and that is given by
Z ∞Z ∞
dy 0 dz 0
0
0
Ex = ke σγx0
= 2πke σ
0
0
2 02
0 2
0 2 3/2
−∞ −∞ [γ x0 + (y0 − y ) + (z0 − z ) ]
{z
}
|
=2π/(γx00 )
~ 0 = 2πke σ β̂. Since B
~ 0 = β~ × E
~ 0 , we get B
~ 0 = 0, in agreement with (a).
or E
11.10. Verify that the elements of the matrix (11.27) are indeed the electric and magnetic
fields as given there.
Solution: Compare (11.5) with F0i ≡ ∂0 Ai − ∂i A0 and conclude that F0i = −Ei . This
verifies the first row and the first column. Furthermore, from (11.2) we get
F12 = ∂1 A2 − ∂2 A1 = B3 ,
F31 = ∂3 A1 − ∂1 A3 = B2 ,
F23 = ∂2 A3 − ∂3 A2 = B1 .
These verify the rest of the matrix elements. Note how the three indices in Fij = Bk are
cyclically permuted.
11.11. Let Fsup and Fsub denote the matrices of F µν and Fµν , respectively. Show that the
matrix form of Equation (11.28) is Fsup = ηFsub η. Now carry out the multiplication of the
three matrices to come up with Equation (11.29).
161
Solution: Look at a general element of Fsup , and since the matrix is antisymmetric, assume
that µ < ν:
F µν = η µα Fαβ η βν = η µ0 F0β η βν + η µi Fiβ η βν
= η µ0 F0i η iν + η µi Fiβ η βν = η µ0 F0i η iν + η µi Fij η jν ,
i < j.
This shows that F 0i = −F0i because η 00 = 1 = −η ii (no sum over i) and that F ij = Fij
because η ii η jj = 1 (no sum over i or j).
11.12. Verify Equation (11.30) by going through each step of its derivation.
Solution: There’s really nothing to verify because all the steps are in the equation. The
e = −F.
only thing you need to know is that F
11.13. Noting that F0 is given by the same matrix as (11.29) except that its elements carry
a prime, use Equation (11.31) to derive the transformation rules of (11.8) and (11.13) for
electric and magnetic fields.
Solution: It is convenient to write the matrices in block
obvious notation, we can also define the block form of F:



e
γ γ β~
0



Λ=
, F = 
↔
~
−E
γ β~ Λ
form. Λ is given by (6.14). With
~e
E


.
B
Now calculate the product of the three matrices in (11.31). The first two give

γ

ΛF = 
γ β~
e
γ β~
↔
Λ



0
~
−E
~e
E


~
−γ β~ · E
 
=
B
↔~
−Λ E
~e
~e +βB)
γ(E



~e + ↔
ΛB
γ β~ E
and multiplying this by Λ on the right gives



e
~e
~e +βB)
~ γ(E
−γ β~ · E
γ γ β~



F0 = 

.
↔
↔~
↔
e
~
~ +ΛB
γβ Λ
−Λ E γ β~ E
Rather than multiplying out matrices, I’ll look at the elements F0(i,j) (with i, j = 1, 2) of
the block matrix that I need. Although I don’t need F0(1,1) , I’ll calculate it to make sure I
get zero.
~e β~ = γ 2 β~e B β.
~
~ + γ 2 (E
~e +βB)
F0
= −γ 2 β~ · E
(1,1)
But, since the matrix B is antisymmetric,
e
~ i = βi Bij βj = −βi Bji βj = −β♣ B♠♣ β♠ = −β♠ B♠♣ β♣ = −β~e B β,
~
β~ B β~ = βi (B β)
and F0(1,1) = 0 as expected. Next, I’ll calculate F0(2,1) :
F0(2,1)
↔~
↔
e
~
~ + Λ B γ β.
= −γΛ E + γ β~ E
162
Relativistic Electrodynamics
This is a column vector; so we calculate its ith component:
~ · β~ + γΛij Bjk βk .
F0(2,1) = −γΛij Ej + γ 2 βi E
i
I’ll evaluate each term separately [see (B.9) for Λij ]:
~
term 1 = −γΛij Ej = −γ[δij + (γ − 1)β̂i β̂j ]Ej = −γEi − γ(γ − 1)β̂i β̂ · E
~ · β~ = γ 2 β 2 β̂i E
~ · β̂ = (γ 2 − 1)β̂i E
~ · β̂
term 2 = γ 2 βi E
term 3 = γΛij Bjk βk = γ[δij + (γ − 1)β̂i β̂j ]Bjk βk
| {z }
=0
~ i.
= γBik βk = γikm Bm βk = γ(β~ × B)
Adding the three terms, we obtain
~ + (γ 2 − 1)β̂i E
~ · β̂ + γ(β~ × B)
~ i
F0(2,1) = −γEi − γ(γ − 1)β̂i β̂ · E
i
or
~ i + (γ − 1)β̂i β̂ · E,
~
−Ei0 = −γ(E − β~ × B)
which agrees with (11.8).
To obtain the transformation of the magnetic field, we’ll have to look at F0(2,2) .
↔ ~ ~e
e ↔ ↔
0
~
~
F(2,2) = −γΛ E β + γ β E + Λ B Λ .
This is a 3 × 3 matrix; so let’s find it’s ijth element:
F0(2,2)
= −γΛik Ek βj + γβi Ek Λkj + Λik Bkm Λmj .
ij
Again I’ll evaluate each term separately:
~ β̂i βj
term 1 = −γΛik Ek βj = −γ[δik + (γ − 1)β̂i β̂k ]Ek βj = −γEi βj − γ(γ − 1)β̂ · E
~ i β̂j
term 2 = γβi Ek Λkj = γβi Ek [δkj + (γ − 1)β̂k β̂j ] = γβi Ej + γ(γ − 1)β̂ · Eβ
term 3 = Λik Bkm Λmj = [δik + (γ − 1)β̂i β̂k ]Bkm [δmj + (γ − 1)β̂m β̂j ]
= [δik + (γ − 1)β̂i β̂k ][Bkj + (γ − 1)β̂m β̂j Bkm ]
= Bij + (γ − 1)β̂m β̂j Bim + (γ − 1)β̂i β̂k Bkj .
In the third term, the product of the two γ − 1 terms gives zero because β̂k β̂m Bkm = 0.
Adding the three terms and noting that F0(2,2) = B0 , we get
B0ij = γβi Ej − γEi βj + Bij + (γ − 1)β̂m β̂j Bim + (γ − 1)β̂i β̂k Bkj .
~ is Bn = 1 nij Bij and Bij =
The relation between the matrix B and the magnetic field B
2
ijn Bn . So, multiply both sides of the last equation by 21 nij to obtain
Bn0 = γ 12 nij (βi Ej − Ei βj ) + Bn + (γ − 1) 12 nij (β̂m β̂j imk Bk + β̂i β̂k kjl Bl )
~ n + Bn + 1 (γ − 1)(nij imk β̂m β̂j Bk +nij kjl β̂i β̂k Bl )
= γ(β~ × E)
2
| {z }
| {z }
=δnk δjm −δnm δjk
~ n + Bn +
= γ(β~ × E)
1
2 (γ
=δnl δik −δnk δil
~ + Bn − β̂n β̂ · B)
~
− 1)(Bn − β̂n β̂ · B
163
or
~ + β~ × E)
~ n − (γ − 1)β̂n (β̂ · B).
~
Bn0 = γ(B
This agrees with Equation (11.13).
~ 2 − |E|
~ 2 ). Hint: Prove that F µν Fµν is the negative of
11.14. Show that F µν Fµν = 2(|B|
the trace (the sum of the diagonal elements) of the product of the two matrices in (11.29)
and (11.27). Then find that trace.
Solution: I let you calculate F µν Fµν via trace. I’ll calculate it by index manipulation as
more exercise in “index gymnastics.”
F µν Fµν = F 0ν F0ν + F iν Fiν = F 0i F0i + F i0 Fi0 + F ij Fij = −F 0i F 0i − F i0 F i0 + F ij F ij
~ 2 + (δjj δjm − δkj δkm )Bk Bm
= −Ei Ei − (−Ei )(−Ei ) + (ijk Bk )(ijm Bm ) = −2|E|
~ 2 + (3δjm − δjm )Bk Bm = −2|E|
~ 2 + 2Bk Bk = −2|E|
~ 2 + 2|B|
~ 2.
= −2|E|
~ · B)
~ 2 , where F is as given by (11.29).
11.15. Show that det F = (E
Solution: You can use the evaluation of the determinant of the matrix according to one
of its rows or columns, or use indices. I’ll use the latter.
det F = µναβ F 0µ F 1ν F 2α F 3β = iναβ F 0i F 1ν F 2α F 3β
= i0αβ F 0i F 10 F 2α F 3β + ijαβ F 0i F 1j F 2α F 3β
= i0jβ F 0i F 10 F 2j F 3β + ij0β F 0i F 1j F 20 F 3β + ijkβ F 0i F 1j F 2k F 3β
= i0jk F 0i F 10 F 2j F 3k + ij0k F 0i F 1j F 20 F 3k + ijk0 F 0i F 1j F 2k F 30 .
Here, I used the properties of the 4-dimensional Levi-Civita symbol, which a trivial generalization of the properties of the 3-dimensional Levi-Civita symbol. Now note that 0ijk = ijk
and F 0i = Ei to get
det F = Ei ijk (−F 10 F 2j F 3k + F 1j F 20 F 3k − F 1j F 2k F 30 )
= Ei ijk (E1 F 2j F 3k − E2 F 1j F 3k + E3 F 1j F 2k ).
I’ll evaluate the first term in detail, leaving the other two terms—which are obtained in
exactly the same way—for you.
ijk F 2j F 3k = ijk 2jm Bm 3kn Bn = jik j2m 3kn Bm Bn
= (δi2 δkm − δim δk2 )3kn Bm Bn = (δi2 3mn − δim 32n )Bm Bn
= δi2 3mn Bm Bn −321 B1 Bi = B1 Bi .
| {z }
=0
Similarly, you should show that
ijk F 1j F 2k = B3 Bi ,
ijk F 1j F 3k = −B2 Bi .
Plugging these results in the expression for the determinant, we get
~ · B)
~ 2.
det F = Ei (E1 B1 Bi + E2 B2 Bi + E3 B3 Bi ) = (E
164
Relativistic Electrodynamics
11.16. The second cyclotron that Lawrence’s group built could accelerate a proton to a
kinetic energy of 1 MeV in a magnetic field of 1.26 Tesla.
(a) What is γα0 and v0 for such a proton?
(b) What was the diameter of the cyclotron?
Solution: The energy of the proton is
E = KE + mp = 1 + 938.272 = 939.272 MeV.
(a) γα0 = E/mp = 1.001, and
q
0.0462
γα0 α0 = γα2 0 − 1 = 0.0462 ⇐⇒ α0 =
= 0.0461,
1.001
and v0 = cα0 = 1.38 × 107 m/s.
q
(b) From R = p0 /qB, and p0 = E 2 − m2p = 43.33 MeV/c = 6.9 × 10−12 J/c, we get
R=
6.93 × 10−12
6.93 × 10−12
=
= 11.5 cm,
cqB
(3 × 108 )(1.6 × 10−19 )1.26
or a little over four inches.
11.17. A K 0 meson at rest decays into two charged pions π + and π − in a bubble chamber in
which a magnetic field B = 1.25 T is present. The pions have equal mass mπ = 139.57 MeV.
If the radius of curvature of the pions is 55 cm, determine the momenta and speeds of the
pions and the mass of the K 0 .
Solution: First calculate the momentum:
p0 c = qBRc = (1.6 × 10−19 )(1.25)(0.55)(3 × 108 ) = 3.3 × 10−11 J = 206 MeV.
√
Then the energy: E = 2062 + 139.572 = 249 MeV. The mass of the K 0 meson is twice
this or 498 MeV, and the speed of each pion is α0 = p0 /E = 0.827.
11.18. A Λ particle at rest decays into a proton and a π − in a bubble chamber in which a
magnetic field B = 0.5 T is present. The mass of the pion is mπ = 139.57 MeV and that of
the proton is mp = 938.27 MeV. If the radius of curvature of the pion is 67 cm, determine
its speed, the radius of curvature of the proton, the speed of the proton, and the mass of
Λ.
Solution: The momentum of the pion is
p0 c = qBRc = (1.6 × 10−19 )(0.5)(0.67)(3 × 108 ) = 1.61 × 10−11 J = 100.5 MeV.
Since the Λ particle is at rest, this is also the momentum of the proton. Therefore, the
radius of curvature of the proton is also 67 cm. The energies of the pion and proton are
p
p
Eπ = 100.52 + 139.572 = 172 MeV, Ep = 100.52 + 938.272 = 943.6 MeV,
and the mass of the Λ particle is the sum of these two energies: mΛ = 943.6 + 172 =
1115.6 MeV. The speed of the pion is απ = 100.5/172 = 0.54, and that of the proton is
αp = 100.5/943.6 = 0.107.
165
11.19. Verify Equation (11.40) for i = 2 and i = 3.
Solution: I’ll verify it for general i as more practice in index manipulation.
∂µ F µi = ∂0 F 0i + ∂j F ji = ∂0 Ei + ∂j (jik Bk ) = ∂0 Ei − ijk ∂j Bk =
∂Ei
~ × B)
~ i.
− (∇
∂t
~ × E)
~ i ].
11.20. Show that σξµi ∂σ Fξµ = 2[∂0 Bi + (∇
Solution: Use F0k = −Ek , Fjk = jkm Bm , δkk = 3, and 0ijk = ijk :
σξµi ∂σ Fξµ = 0ξµi ∂0 Fξµ + jξµi ∂j Fξµ = 0jµi ∂0 Fjµ + j0µi ∂j F0µ + jkµi ∂j Fkµ
= 0jki ∂0 Fjk + j0ki ∂j F0k + jk0i ∂j Fk0 = 0ijk (∂0 Fjk − ∂j F0k + ∂j Fk0 )
= ijk (∂0 Fjk − 2∂j F0k ) = ijk [∂0 (jkm Bm ) + 2∂j Ek ]
~ × E)
~ i
= ∂0 (jki jkm Bm ) + 2ijk ∂j Ek = ∂0 [(δkk δim − δkm δik ) Bm ] + 2(∇
|
{z
}
=2δim
11.21. In this problem, you’ll show that Maxwell’s equations imply the conservation of the
electric charge.
(a) Write the continuity equation (see Note A.1.2) in terms of indices in four-dimensional
spacetime.
(b) Suppose that S µν and Aµν are, respectively symmetric and antisymmetric under the
interchange of their indices. Show that S µν Aµν = 0.
(c) Differentiate the first equation in Note 11.3.2 and use (b) to prove that Maxwell’s
equations imply charge conservation.
Solution:
(a) ∂µ J µ = 0.
(b) Let SA stand for S µν Aµν = 0. Then
SA = S µν Aµν = −S µν Aνµ = −S ♣♠ A♠♣ = −S ♠♣ A♠♣ = −SA.
So, SA = 0.
(c) Differentiate the first equation in Note 11.3.2 with respect to xν to get
∂ν ∂µ F µν = µ0 ∂ν J ν .
Now note that ∂ν ∂µ is symmetric and F µν antisymmetric. So, by (b) the left-hand
side is zero.
CHAPTER
12
Early Universe
Problems With Solutions
12.1. Derive (12.6) from (12.4) and (12.9) from (12.7).
Solution: Substitute ω = 2πc/λ and dω = −2πcdλ/λ2 in (12.4) to obtain
Z 0
~
(2πc/λ)3
uγ = 2 3
(−2πcdλ/λ2 )
π c ∞ exp (2π~c/λkB T ) − 1
Z ∞
8πhc
1
=
dλ.
5
λ exp (hc/λkB T ) − 1
0
Equation (12.9) followd from (12.7) in a similar way.
12.2. Consider the reaction γ + γ −→ p + p̄, which can occur if the initial photons are
extremely energetic. Here p is a generic particle and p̄ its antiparticle.
(a) Use Equation (8.24) to show that
E1 E2 =
m2p
sin2 (θ/2)
where mp is the mass of p (or p̄), E1 and E2 are the energies of the photons, and θ is
the angle between the direction of motion of the initial photons.
(b) What value of θ minimizes E1 E2 ? If E1 > E2 , what is the total momentum of the
pair on the right-hand side (in terms of E1 and mp ) for the θ that minimizes E1 E2 ?
(c) Using the θ in (b), show that the value of E1 that minimizes the total energy E1 + E2
is E1 = mp . What is the total momentum of the pair on the right-hand side now?
Solution: With m1 = m2 = 0 and m3 = m4 = mp , Equation (8.24) yields
p1 • p2 = 2m2p , ⇐⇒ E1 E2 − |~
p1 ||~
p2 | cos θ = 2m2p .
168
Early Universe
(a) Since for photons E1 = |~
p1 | and E2 = |~
p2 | (speed of light is 1), we get
E1 E2 (1 − cos θ) = 2m2p ⇐⇒ E1 E2 sin2 (θ/2) = m2p .
(b) E1 E2 is minimum when sin2 (θ/2) = 1 or θ = π/2.
(c) E1 + E2 = E1 + m2p /E1 . Set the derivative of this expression equal to zero and solve
for E1 :
m2p
1 − 2 = 0 ⇐⇒ E1 = mp .
E1
Part (a) now gives mp E2 = m2p , or E2 = mp . So both particles are produced at rest.
Hence, the total momentum is zero.
12.3. In Example 12.1.3, I calculated the surface temperature of the Sun assuming that it
was (approximately) a black body radiator. Using the result of that example and Equation
(12.17),
(a) find the brightness of the Sun.
(b) How many 100-Watt light bulbs do you have to put on a square meter to give this
brightness?
(c) The radius of the Sun is 700,000 km. What is the power output of the Sun?
(d) If the mass of the Sun is the source of this energy, how many kilograms of its mass
does the Sun lose every second?
(e) The mass of the Sun is 2 × 1030 kg. Assuming that its mass depletion rate is proportional to its mass, find the present proportionality constant.
(f) Assuming that the constant in (e) does not change with time, find m(t), the Sun’s
mass as a function of time.
(g) How many years do you have to wait before the Sun loses half of its mass (and is no
longer a shining star)? This is a huge overestimate! The Sun dies far sooner than this
due to nuclear processes that give off colossal explosive (and implosive) energies.
Solution: The temperature was found to be 5800 K.
(a) The brightness is given by (12.17), which, in conjunction with (12.18), gives
Iγ = σB T 4 = 5.67 × 10−8 (5800)4 = 6.4 × 107 W/m2 .
(b) N = 6.4 × 107 /100 = 640, 000. The Sun is really bright!
(c) Brightness is power per square meter. So, the total power output of the Sun is
Ptot = 4πR2 Iγ = 4π(7 × 108 )2 (6.4 × 107 ) = 3.94 × 1026 W.
(d) Divide (c) by c2 :
dm
3.94 × 1026
=
= 4.4 × 109 kg/s.
dt
9 × 1016
169
(e)
k=
4.4 × 109
dm/dt
=
= 2.2 × 10−21 .
M
2 × 1030
(f)
dm
= −km ⇐⇒ m(t) = m0 e−kt = M e−kt .
dt
(g)
M
= M e−kt1/2 ⇐⇒ ln 2 = kt1/2
2
or
t1/2 =
0.693
ln 2
=
= 3.15 × 1020 s,
k
2.2 × 10−21
or about 1013 years.
12.4. Using the present temperature of the universe and Equations (12.27) and (12.36),
calculate Ωγ and compare it with the value given in (12.29).
Solution: From (12.36), we get
ργ = 8.38 × 10−33 T 4 kg/m3 = (8.38 × 10−33 )2.7254 kg/m3 = 4.62 × 10−33 kg/m3 ,
and from (12.27), we obtain
Ωγ =
4.62 × 10−33 kg/m3
= 5.4 × 10−5 ,
8.5 × 10−27 kg/m3
which compares very well with (12.29).
12.5. In this problem you are asked to compare the present number of photons and baryons
in the universe.
(a) Using the present temperature of the universe, calculate nγ , the present photon number density.
(b) From (12.27) and the value of Ωb in (12.29), calculate the present ρb .
(c) Ignore the negligible mass difference between a proton and a neutron and assume
that only protons contribute to ρb . Moreover, since protons are not moving fast, you
can assume that their energy is just mp c2 . With mp = 1.67 × 10−27 kg, find nb , the
present baryon number density.
(d) How many photons are there in the universe for each baryon?
Solution:
(a) Equation (12.11) yields
nγ = 2 × 107 T 3 photons/m3 = (2 × 107 )2.7253 photons/m3 = 4 × 108 photons/m3
for the present number density of photons.
170
Early Universe
(b)
ρb = Ωb ρcrit = 0.0484(8.5 × 10−27 kg/m3 ) = 4.1 × 10−28 kg/m3 .
(c)
nb =
ρb
4.1 × 10−28 kg/m3
=
= 0.25 baryons/m3 .
mp
1.67 × 10−27 kg
(d) Take the ratio of nγ over nb :
nγ
4 × 108 photons/m3
9
photons/baryons.
=
3 = 1.6 × 10
nb
0.25 baryons/m
12.6. Decide which particles of Table 12.1 were present at the beginning of the second
epoch.
(a) Calculate α assuming that whatever particle you have is relativistic.
(b) How old was the universe at the beginning of the second epoch?
Solution: All the particles were present at the beginning of the second epoch, because all
the particles in Table 12.1 have threshold temperatures below 1015 K.
(a) So, α is the sum of all the α’s in Table 12.1: α = 32.375.
(b) Now that we have α, we can calculate the age of the universe using (12.49):
2.32 × 1020
tα = √
s = 4 × 10−11 s = 40 ps.
32.375 1030
12.7. In this problem, you’ll calculate the density of the universe at the beginning of the
second epoch.
(a) Find α and from that, the density of the universe.
(b) The Earth has a mass of 5.97 × 1024 kg. What would the radius of the Earth be if it
were composed of such a dense material?
(c) At the confinement temperature, the density of the universe dropped to 6.5×1017 kg/m3 .
What would the radius of the Earth be if it were composed of this material?
Solution: All the particles were present at the beginning of the second epoch, because all
the particles in Table 12.1 have threshold temperatures below 1015 K.
(a) So, α is the sum of all the α’s in Table 12.1: α = 32.375. The density is
ρα = αργ = 32.375(8.38 × 10−33 T 4 ) = 2.7 × 1029 kg/m3 .
(b)
ρα =
M⊕
⇐⇒ R⊕ =
4
3
3 πR⊕
3M⊕
4πρα
1/3
= 5.3 × 10−6 m = 5.3 µm.
171
(c)
R⊕ =
3M⊕
4πρconf
1/3
=
3(5.97 × 1024 )
4π(6.5 × 1017 )
1/3
= 0.04 m = 4 cm.
12.8. In this problem you’ll calculate the age of the universe when quarks confined at a
temperature of 2 × 1012 K.
(a) Look at Table 12.1 and decide which particles were present just before that temperature was reached. Calculate α.
(b) Now use (12.49) to calculate the age of the universe.
Solution:
(a) All particles with a threshold temperature less that 2 × 1012 K were present in the
universe. This gives α = 20.275.
(b) Now from (12.49), we get
tα = √
2.32 × 1020
s = 1.3 × 10−5 s = 13 µs.
20.275 (2 × 1012 )2
12.9. Assume that at a temperature of 2 × 1012 K, quarks and gluons are bagged into
hadrons, the lightest of which are pions with spin zero (therefore α = 0.5, just like the
Higgs boson). There are three kinds of pion: neutral pions π 0 (with no antiparticle) with a
mass of 135 MeV and two charged pions (one being the antiparticle of the other) π ± with
a mass of 139.57 MeV. The next lightest hadron has a mass of approximately 500 MeV.
(a) What is the threshold temperature for π 0 ? For π ± ? For the next lightest hadron?
(b) Determine the particle content of the universe at the beginning of the third epoch,
and from that, α.
(c) Now use (12.49) to calculate the age of the universe at the beginning of the third
epoch.
(d) The mean life of π 0 is 8.5 × 10−17 s, and that of π ± is 2.6 × 10−8 s. What fraction of
charged pions decay between their production at confinement time and the beginning
of the third epoch? What fraction of neutral pions?
(e) In light of (d), do you have to reconsider (b) and (c)? If so, recalculate them and find
the new value for the age of the universe at the beginning of the third epoch.
Solution:
(a) Using Tth = 4.3 × 109 mc2 MeV−1 K, we have
Tπ0 = 4.3 × 109 (135) = 5.8 × 1011 K,
Tπ± = 4.3 × 109 (139.57) = 6 × 1011 K,
T3 = 4.3 × 109 (500) = 2.15 × 1012 K.
172
Early Universe
(b) The third epoch begins with a temperature below quark confinement. So, no free
quarks or gluons are there, and (a) shows that out of all hadrons, only pions will
be present. From Table 12.1, only photons, neutrinos, electrons and muons have
temperatures below 1012 K. The α for each pion is 0.5. Thus the three of them
contribute 1.5 to the total α; and the contribution of Table 12.1 is 7.125. Hence,
α = 8.625.
(c) From (12.49), we get
2.32 × 1020
tα = √
s = 7.9 × 10−5 s = 79 µs.
8.625 (1012 )2
(d) From Problem 12.8, quarks confine 13 µs after the big bang. From (c), the third
epoch occurs 79 µs after the big bang. So, pions have 66 µs to decay. The fraction
r of any decaying particle remaining at time t is given by r = e−t/τ , where τ is the
mean life of the particle. Thus, the fraction of pions remaining is
66 × 10−6
= e−2538.5 = 2.75 × 10−1102
rπ± = exp −
2.6 × 10−8
66 × 10−6
rπ0 = exp −
= 5.75 × 10−337216891830 .
8.5 × 10−17
In other words, zero pion will see the light of the third epoch!
(e) We have to subtract the contribution of pions to α. So, α = 7.125 and
2.32 × 1020
tα = √
s = 8.69 × 10−5 s = 86.9 µs.
7.125 (1012 )2
12.10. Right after µ+ -µ− annihilation, the temperature is 1011 K.
(a) Which particles are present then? Ignore the baryon “contamination.” What is the
value of α?
(b) How old is the universe?
(c) What is the density of the universe assuming that it is all ρrel ? Do you have to worry
about the contribution from matter density? Use (12.47) to find out!
(d) Assume that a grain of sand is a cube of side one millimeter. What would the mass
of this grain be if it had a density you found in (c)?
Solution:
(a) The particles present are photons, neutrinos, and electrons. Therefore, From Table
12.1, α = 5.375.
(b) From (12.49),
2.32 × 1020
tα = √
s = 0.01 s.
5.375 (1011 )2
173
(c)
ρrel = αργ = 5.375(8.38 × 10−33 T 4 ) kg/m3
= 5.375(8.38 × 10−33 (1011 )4 ) kg/m3 = 4.5 × 1012 kg/m3 .
Using (12.47), the contribution from matter density is
ρm (T ) = 1.32 × 10−28 T 3 kg/m3 = 1.32 × 10−28 (1011 )3 kg/m3 = 13200 kg/m3 ,
much smaller than ρrel .
12.11. Derive Equation (12.51) from the equations preceding it.
Solution: On the right-hand side of the equation for neutron production we have
m2n = (mp + ∆m)2 ≈ m2p + 2mp ∆m.
Substituting this and letting mp = m, we get
mE1 = m∆m + E3 (E1 + m) − E1 E3 cos θ,
or
E1 (m − E3 + E3 cos θ) = m∆m + mE3 ,
which is the first equation in (12.51). The only difference between the two equations is that
for Ep1 , the roles of mn and mp are switched, introducing a minus sign for ∆m.
12.12. Things happen very quickly when the universe is very young. To see how quickly,
suppose two points of the universe are one millimeter apart 10−30 second after the big bang.
(a) Which particles of Table 12.1 are present at this time?1 Use (12.48) to find aα (t).
(b) One second after the big bang, the fourth epoch starts. What is α now? What is
aα (t)?
(c) How far apart are the two points now? Recall that the ratio of distances is the same
as the ratio of the scale factors.
(d) Compare the distance in (c) with the Earth-Sun distance. From 1 mm to this distance
in just one second! Now you can appreciate why we call it a “bang,” a “BIG” bang!
Solution:
(a) All particles in Table 12.1 are present. Thus, α = 32.375, and (12.48) gives
p
aα (t) = 4 4αΩγ
t
tH
1/2
=
p
4
4(32.375)(0.0000538)
or aα (t) = 4.26 × 10−26 .
1
Hint: How old is the universe at the beginning of the second epoch?
10−30
4.59 × 1017
1/2
174
Early Universe
(b) The universe consists of electrons, positrons, neutrinos, and photons. So, α = 5.375,
and
1/2
p
1
4
aα (t) = 4(5.375)(0.0000538)
= 2.7 × 10−10 .
4.59 × 1017
(c)
R
2.7 × 10−10
=
⇐⇒ R = 6.4 × 1012 m.
0.001
4.26 × 10−26
(d) The Earth-Sun distance is 1.5 × 1011 m, so this distance is more than 42 times the
Earth-Sun distance!
12.13. In this problem, you are asked to estimate the helium-proton abundance if e− +D →
2n + ν were responsible for deuteron dissociation. I have already calculated the minimum
energy for the electron. It is 3.517 MeV.
(a) What is the wavelength associated with this energy?
(b) Assuming that electrons (and positrons) outnumber protons and neutrons 1.6 billion
to 1, use Example 12.1.2 to show that the threshold temperature for deuteron formation is 1.5 × 109 K. However, there is a problem here. This temperature is less
than e+ -e− annihilation temperature. At 1.5 × 109 K, the electrons do not outnumber
baryons 1.6 billion to 1. This means that 3.517 MeV is not sufficient for deuteron
production. Let’s assume that the minimum energy corresponds to the e+ -e− annihilation temperature of 2.2 × 109 K.
(c) If the content of the universe is electron, positron, photons, and neutrinos, what is
α? Use (12.49) to estimate the age of the universe when deuterons start to form at
2.2 × 109 K.
(d) Show that from the end of the third epoch, about 2.2% of the neutrons decay into
protons. Even if the other contributions to neutron depletion remain at 28% (the
actual number is smaller than this because 28% corresponds to the entire fifth epoch),
the total depletion rate is 30.2%. Now show that the helium-proton abundance is
34.9% and 65.1%, respectively.
Solution:
(a) The energy is 3.517(1.6 × 10−13 ) = 5.63 × 10−13 J.
λ=
hc
(6.626 × 10−34 )(3 × 108 )
=
= 3.53 × 10−13 m.
E
5.63 × 10−13
(b) Equation (12.16) gives the relation between wavelength and temperature:
λ2 T = 5.33 × 10−4 m K ⇐⇒ T =
5.33 × 10−4
= 1.51 × 109 K.
3.53 × 10−13
(c) Table 12.1 gives α = 5.375, and (12.49) yields
tα = √
2.32 × 1020
= 20.68 s.
5.375 (2.2 × 109 )2
175
(d) From the end of the third epoch until tα , the neutrons have 19.68 seconds to decay.
Therefore the fraction remaining is
N (tα )
= e−tα /τ = e−19.68/880.3 = 0.978.
N0
Therefore, about 2.2% of the neutrons decay into protons. Add this to the other decay
contribution of 28% to get 30.2%. So, the fraction of neutrons reduces from 25% to
25 × 0.698 = 17.45%. These neutrons combine with an equal number of protons to
form helium. So, 34.9% of the baryons are in the form of helium nuclei and 65.1% in
the form of protons.
12.14. Calculate the photon and neutrino mass densities at the beginning of the fifth epoch
and compare them with matter density. What was the ratio ργ /ρm then? Do the same for
the beginning of the last epoch.
Solution: Table 12.1 gives αν = 0.681 after electron-positron annihilation. Therefore,
ργ = 8.38 × 10−33 (109 )4 = 8380 kg/m3
and
ρν = 0.681(8.38 × 10−33 )(109 )4 = 5707 kg/m3 .
For ρm we use (12.47):
ρm (T ) = 1.32 × 10−28 (109 )3 = 0.132 kg/m3 .
This gives a ratio of
ργ
8380
=
= 63485.
ρm
0.132
In general,
ργ (T )
8.38 × 10−33 T 4
=
= 6.35 × 10−5 T.
ρm (T )
1.32 × 10−28 T 3
For the last epoch, T = 108 K; thus, the ratio is 6350, which is a tenth of the ratio at the
beginning of the fifth epoch.
APPENDIX
A
Maxwell’s Equations
Problems With Solutions
A.1. Derive the differential form of the fourth equation in (A.1).
Solution: Use Stokes’ Theorem on the left and the definition of current in terms of current
density on the right:
x
x
(∇ × B) · da = µ0
J · da.
S
S
This must be true for all S. Therefore, the integrands must equal: ∇ × B = µ0 J.
A.2. Starting with Maxwell’s equations, show that the magnetic field satisfies the same
wave equation as the electric field. In particular, that it, too, propagates with the same
speed.
Solution: Take the curl of the 4th equation in (A.12) and use the last equation in (A.9).
You’ll get
LHS of (4) = ∇ × (∇ × B) = ∇ (∇ · B) −∇2 B = −∇2 B,
| {z }
=0 by (2)
and
RHS = µ0 0 ∇ ×
∂E
∂t
∂
∂
= µ0 0 (∇ × E) = µ0 0
∂t
∂t
∂B
−
.
∂t
Now set the two sides equal and obtain
∇2 B = µ0 0
∂2B
.
∂t2
√
A.3. Consider E = E0 ei(ωt−k·r) and B = B0 ei(ωt−k·r) , where i = −1, E0 , B0 , k, and ω
are constants. The E and the B so defined represent plane waves moving in the direction
of the vector k.
178
Maxwell’s Equations
(a) Show that they satisfy Maxwell’s equations in free space if:
(1) k · E0 = 0;
(2) k · B0 = 0;
(3) k × E0 = ωB0 ;
(4) k × B0 = −
ω
E0 .
c2
(b) In particular, show that k, the propagation direction, and E and B form a mutually
perpendicular set of vectors.
(c) By taking the cross product of k with an appropriate equation, show that |k| = ω/c.
Solution: The problem becomes a lot easier if we first derive two more general identities.
If C is a constant vector and f is a function, then
∇ · (Cf ) = C · (∇f )
and ∇ × (Cf ) = (∇f ) × C.
The first identity is derived as follows:
∂
∂
∂
(Cx f ) +
(Cy f ) +
(Cz f )
∂x
∂y
∂z
∂f
∂f
∂f
= Cx
+ Cy
+ Cz
= C · (∇f ).
∂x
∂y
∂z
∇ · (Cf ) =
For the second identity use (A.8) and

êx
êy


 ∂
∂
∇ × (Cf ) = det  ∂x
∂y


Cx f Cy f
note that the derivatives


êx êy
êz




 ∂f ∂f
∂ 
=
det

 ∂x ∂y
∂z 



Cz f
Cx Cy
operate only on f :

êz


∂f 
= (∇f ) × C.
∂z 


Cz
(a) Let’s apply the general results obtained above to the specific function at hand. From
k · r = kx x + ky y + kz z, we get
∂ i(ωt−k·r)
e
= −ikx ei(ωt−k·r) ,
∂x
with similar results for y and z. Therefore,
∇ei(ωt−k·r) = −ikei(ωt−k·r) .
Now let’s apply what we have obtained so far to Maxwell’s equation in free space.
The first equation yields
0 = ∇ · E = ∇ · [E0 ei(ωt−k·r) ] = E0 · [∇ei(ωt−k·r) ]
= E0 · [−ikei(ωt−k·r) ] = −iE0 · kei(ωt−k·r) ⇐⇒ k · E0 = 0,
with a similar result for the magnetic field. It is obvious that time differentiation
introduces a multiplicative iω. So, the third Maxwell equation becomes
∇ × [E0 ei(ωt−k·r) ] = [∇ei(ωt−k·r) ] × E0 = [−ikei(ωt−k·r) ] × E0 = −iωB0 ei(ωt−k·r) ,
or k × E0 = ωB0 . The fourth equation follows in the same way.
179
E
A
D
B
C
1
2
Figure A.1: The long sides of the rectangle ABCD are infinitesimals. The short sides are infinitesimals
compared to the long sides.
(b) (3) shows that B0 is perpendicular to both E0 and k. (4) shows that E0 is perpendicular to both B0 and k. So, all three vector are mutually perpendicular to each
other.
(c) Take the cross product of (3) with k. Then, using the “bac cab rule,” on the left you
get
k × (k × E0 ) = k (k · E0 ) −(k · k)E0 = −|k|2 E0 ,
| {z }
=0
and on the right, using (4), you get
ωk × B0 = −
ω2
E0 .
c2
The last two equations yield |k| = ω/c.
A.4. Take a rectangular loop with two long sides on either side of a boundary surface. All
sides are infinitesimal, but the short side is infinitesimal even compared to the long side.
Apply the third Maxwell equation in integral form (A.1) to this loop. Since everything is
infinitesimal, you can forget about the integrals. The magnetic flux is zero because the area
is triply infinitesimal. The contribution from the two short sides to the left-hand side is
zero. Now complete the proof of the equality of the tangential components of the electric
field on the two sides of the boundary.
Solution: Figure A.1 of the manual shows a rectangle with the long sides in two different
media. Applying the third Maxwell equation in (A.1) to this loop, for the right-hand side we
get zero because the area is infinitesimal to the third power, and therefore, the flux, which
is the product of B and this area, can be neglected. The left-hand side can be evaluated as
follows:
I
E · dr ≈ E2 · êt BC + E1 · (−êt )DA = (E2 · êt − E1 · êt )BC,
C
where êt is a unit vector tangent to the boundary, which I have taken to be pointing from
right to left in the figure. I have neglected the contributions from the smaller sides. Since
BC 6= 0, we have to assume that E2 · êt = E1 · êt , i.e., that the tangential components of
the electric field are equal on both sides.
APPENDIX
B
Derivation of 4D Lorentz transformation
Problems With Solutions
B.1. Take the transpose of A−1 A = AA−1 = 1 to show that the inverse of the transpose is
the transpose of the inverse. You need both matrix products because an inverse must be a
left inverse as well as a right inverse.
Solution: Taking the transpose of A−1 A = AA−1 = 1, we get
g
g
^
^
−1 A = AA
−1 = 1̃ ⇐⇒ A
−1 = A
−1 A
eA
e = 1.
A
g
−1 is the inverse of A.
e
Therefore, A
e = AA
e = 1 for a 3 × 3 matrix implies that each row (or column) of A has
B.2. Show that AA
a unit length and the dot product of any two different rows (or different columns) vanishes.
Solution: Consider the rows and columns of A as vectors. Let ~ai denote the ith row and
~bi the ith columns of A. The ijth element of AA
e is the product of the ith row of A
e and the
e
e
jth column of A. But the ith row of A is the ith column of A. Thus the ijth element of AA
~
~
is bi · bj . This is equal to the ijth element of the unit matrix. Therefore, if i = j, we get
~bi · ~bi = 1 and if i 6= j, we get ~bi · ~bj = 0. Evaluating the ijth element of AA
e implies that if
i = j, we get ~ai · ~ai = 1 and if i 6= j, we get ~ai · ~aj = 0. Note that the assumption that A is
3 × 3 never entered the derivation. So, the conclusion holds for any n × n matrix satisfying
e = AA
e = 1.
AA
B.3. By writing out the 4 × 4 matrices, verify that Λ−1
rot as given by Equation (B.4) is the
inverse of Λrot as given by Equation (B.3).
e = AA
e = 1, this is a simple exercise in matrix multiplication. Solution: Given that AA
B.4. Using relations such as β̂x β̂y + a12 a22 + a13 a23 = 0 and a212 + a213 = 1 − (β̂x )2 , obtained
from Problem B.2, derive Equation (B.5).
182
Derivation of 4D Lorentz transformation
Solution: I’ll calculate the first two rows of Equation (B.5) from the equation preceding
it. You provide the calculation for the third and fourth row. From the equation preceding
(B.5), for the first row, we have
 
1
0

Λ11 = γ γβ 0 0 
0 = γ
0

Λ12

0
 β̂x 

= γ γβ 0 0 
a12  = γβ β̂x = γβx
a13
 
0
 β̂y 

Λ13 = γ γβ 0 0 
a22  = γβ β̂y = γβy
a23
 
0
 β̂z 

Λ14 = γ γβ 0 0 
a32  = γβ β̂z = γβz .
a33
The second row is only slightly more complicated:
Λ21 = γβx γ β̂x a12
 
1
0

a13 
0 = γβx
0

Λ22 = γβx γ β̂x a12

0
 β̂x 
2
2

+ a2 = 1 + β̂x2 (γ − 1),
a13 
a12  = γ β̂x + a
| 12 {z 13}
=1−β̂x2
a13
because the first row of the submatrix A in (B.3) has unit length.

0
 β̂y 

a + a a = β̂x β̂y (γ − 1),
a13 
a22  = γ β̂x β̂y + a
| 12 22 {z 13 23}
=−β̂x β̂y
a23

Λ23 = γβx γ β̂x a12
because the first row of the submatrix A in (B.3) is orthogonal to its second row.

Λ24 = γβx γ β̂x a12

0
 β̂z 

a + a a = β̂x β̂z (γ − 1),
a13 
a32  = γ β̂x β̂z + a
| 12 32 {z 13 33}
=−β̂x β̂z
a33
because the first row of the submatrix A in (B.3) is orthogonal to its third row.
183
B.5. Multiply the matrix of Equation (B.5) by the column 4-vector of r to obtain Equation
(B.6).
Solution: This is a trivial exercise in matrix multiplication.
B.6. Show directly that the matrices of (B.5) and (B.11) satisfy Λ−1 Λ = ΛΛ−1 = 1.
Solution: First write Λ−1 in block form:

γ

Λ−1 = 
−γ β~
e
−γ β~
↔
Λ


.
Now multiply it on the right by the block form of Λ:


 

e
e
e
~e ↔
γ2 − γ2β2
γ 2 β~ − γ β
Λ
γ
−γ β~
γ γ β~


 

Λ−1 Λ = 

=
.
↔
↔
e
↔
↔
↔
−γ β~
Λ
γ β~ Λ
−γ 2 β~ + γΛ β~ −γ 2 β~ β~ + Λ Λ
Let’s look at each block element of Λ−1 Λ:
Λ−1 Λ 11 = γ 2 (1 − β 2 ) = 1.
↔
For Λ , I’ll use (B.8). Then

1
e
e↔
Λ−1 Λ 12 = γ 2 β~ − γ β~ Λ = γ 2 βx βy βz − γ βx βy βz 0
0
 2

β̂x β̂y β̂x β̂z
β̂x

− γ(γ − 1) βx βy βz
β̂x β̂y β̂y2 β̂y β̂z 
β̂x β̂z β̂y β̂z β̂z2
= γ 2 βx βy βz − γ βx βy βz − γ(γ − 1) βx βy

0 0
1 0
0 1
βz = 0 0 0 .
Similarly,
Λ−1 Λ
21
 
0
= 0 .
0
For the last element, first write (B.8) as Λ = 1 + (γ − 1)B, with obvious definition for B.
↔↔
Now, let’s evaluate Λ Λ first:
↔↔
Λ Λ = [1 + (γ − 1)B] [1 + (γ − 1)B] = 1 + 2(γ − 1)B + (γ − 1)2 B2 ,
with
 2
  2

β̂x
β̂x β̂y β̂x β̂z
β̂x
β̂x β̂y β̂x β̂z
β̂x2 β̂x β̂y β̂x β̂z
B2 = β̂x β̂y β̂y2 β̂y β̂z  β̂x β̂y β̂y2 β̂y β̂z  = β̂x β̂y β̂y2 β̂y β̂z  = B.
β̂x β̂z β̂y β̂z β̂z2
β̂x β̂z β̂y β̂z β̂z2
β̂x β̂z β̂y β̂z β̂z2

You should go through the matrix multiplication and verify this equation. Now we have
↔↔
Λ Λ = 1 + 2(γ − 1)B + (γ − 1)2 B = 1 + (γ 2 − 1)B = 1 + γ 2 β 2 B.
184
Derivation of 4D Lorentz transformation
This is one of the terms in Λ−1 Λ
22
. The other term is
 
β̂x
e
−γ 2 β~ β~ = −γ 2 β 2 β̂y  β̂x β̂y β̂z = −γ 2 β 2 B.
β̂z
Adding this to the previous equation, we get


1 0 0
Λ−1 Λ 22 = 1 = 0 1 0 .
0 0 1
This completes the proof that Λ−1 Λ = 1. You should now show that ΛΛ−1 = 1 as well. APPENDIX
C
Relativistic Photography Formulas
Problems With Solutions
C.1. Verify that Equation (C.1) is a line passing through Q and the pinhole of the camera.
Solution: The equation gives r1 (0) = rq , the location of Q and r1 (1) = bêz , the location
of the pinhole.
C.2. Derive Equation (C.2).
Solution: From (C.1), we get
rq0 ≡ r1 (tq0 ) = (1 − tq0 )rq + tq0 bêz ⇐⇒ rq0 − bêz = rq − bêz − tq0 rq + tq0 bêz
or
rq0 − bêz = (1 − tq0 )(rq − bêz ).
I can now take the absolute value of both sides and note that |1 − tq0 | = 1 − tq0 because
1 − tq0 > 0. This leads to Equation (C.2).
C.3. Substitute (C.1), (C.2), and (C.4) in (C.5) to show that
tp0 = 1 −
d|rq − bêz |
.
(rp − bêz ) · (rq − bêz )
Solution: Rewrite Equation (C.5) as
[(1 − tp0 )rp + tp0 bêz − (1 − tq0 )rq − tq0 bêz ] · (rq − bêz ) = 0.
Adding and subtracting bêz , we can re-express this equation as
[(rp − bêz )(1 − tp0 ) − (rq − bêz )(1 − tq0 )] · (rq − bêz ) = 0,
or
(1 − tp0 )(rp − bêz ) · (rq − bêz ) − (1 − tq0 ) |rq − bêz |2 = 0,
| {z }
=d/|rq −bêz |
186
Relativistic Photography Formulas
or
(1 − tp0 )(rp − bêz ) · (rq − bêz ) = d|rq − bêz |,
which yields
1 − tp0 =
d|rq − bêz |
.
(rp − bêz ) · (rq − bêz )
C.4. Derive Equation (C.7).
Solution: Add and subtract bêz :
s = r2 (tp0 ) − r1 (tq0 ) = (1 − tp0 )rp + tp0 bêz − (1 − tq0 )rq − tq0 bêz
= (1 − tp0 )rp + tp0 bêz −bêz + bêz −(1 − tq0 )rq − tq0 bêz
| {z }
=0
= (1 − t )(rp − bêz ) − (1 − tq0 )(rq − bêz ).
p0
Now use (C.2) and (C.6) to get the final answer.
C.5. Derive Equations (C.8) and (C.9).
Solution: When Q is on the x-axis, then rq − bêz = hxq , 0, −bi, and with rp = hx, y, zi, we
get
q
|rq − bêz | = x2q + b2 , and (rp − bêz ) · (rq − bêz ) = xxq − b(z − b).
Plugging these in (C.2) and (C.6) yields (C.8). Plugging them in (C.7) gives
q
d x2q + b2
d
s=
hx, y, z − bi − q
hxq , 0, −bi,
xxq − b(z − b)
x2 + b2
q
whose components are expressed in (C.9).
C.6. Derive Equation (C.12) from Equation (C.11).
Solution: The only thing you need to remember is that x0 is the Lorentz transform of the
x coordinate of the event whose time t is the negative of the distance between the location
of the event and the pinhole. Thus,
p
t = −|rp − bêz | = − x2 + y 2 + (z − b)2 .
C.7. In this problem you’ll find x0q in a different way.
(a) If the event of the explosion is at (xq , 0) in O, what is it in O0 ?
(b) What distance does Q travel in O0 before C 0 takes the picture of the object?
(c) From (a) and (b) determine where Q is according to O0 when C 0 takes a picture and
show that x0q = xq /γ.
187
Solution: (a) Assume that the object is approaching the origin from negative values of
x. Thus, xq < 0, and the initial location of the object when the explosion occurs and
its time of occurrence are
x0qi = γ(xq − 0) = γxq ,
t0q = γ(0 − βxq ) = −γβxq .
(b) C 0 takes the picture of the object at t0 = 0. Therefore, the object travels for t0q before
the picture is taken. The distance is β|t0q | = γβ 2 |xq |.
(c) So, the final location, i.e., the location at which the picture is taken is
x0qf ≡ x0q = x0qi + β|t0q | = γxq + γβ 2 |xq | = γxq − γβ 2 xq =
xq
.
γ