Solutions Manual Special Relativity a Heuristic Approach

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Solutions Manual

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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

aˆ 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 byp ¯. 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.

β~ fractional velocity of one observer relative to another, β~ = ~v/c 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 eˆx, eˆy, eˆz 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.

γ the Lorentz factor, γ = 1/p1 − β2 = 1/p1 − (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: Z0, 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 electri- cally 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 3 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.

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?

(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.

(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?

(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.

(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

(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.

(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 cβ(cos θ − β) v0 (θ) = tr (β cos θ − 1)2 00 equal to zero to get cos θ = β. The second derivative test shows that vtr(θ) < 0 when cos θ = β. Therefore, vtr is indeed maximum at cos θ = β.

1If you hit one end of a slinky, the other end does not move, at least not immediately. 10 Relativity of Time and Space

(b) sin θ p1 − β2 cβ (vtr)max = cβ = cβ = = cβγ. 1 − β cos θ 1 − β2 p1 − β2 (c) cβγ > c ⇐⇒ β2γ2 > 1 ⇐⇒ β2 > 1 − β2 ⇐⇒ β2 > 1/2.

(d) βγ = 10 ⇐⇒ β2γ2 = 100 ⇐⇒ β2 = 100 − 100β2 ⇐⇒ β2 = 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 2L/c ∆t = ∆t + ∆t = . 1 2 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 L c∆t (1 − β) = L ⇐⇒ ∆t = . 1 1 c(1 − β) On reﬂection, the light and the MM clock move in opposite directions. Therefore, c∆t2 = L − βc∆t2, and L c∆t (1 + β) = L ⇐⇒ ∆t = . 2 2 c(1 + β)

(b) A tick according to O is therefore, L L 2L/c ∆t = ∆t + ∆t = + = . 1 2 c(1 − β) c(1 + β) 1 − β2

But ∆t = γ∆τ = γ(2L0/c). Thus, 2L/c p γ(2L /c) = = γ2(2L/c) ⇐⇒ L = L /γ = L 1 − β2. 0 1 − β2 0 0

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?

Solution: √ 2 (a) On the outbound part,√ the distance for the crew is L0 1 − 0.9 and the time it takes 2 them to get there is L0 1 − 0.9 /(0.9c), where L0 is the distance√ according to Earth 2 observers. Similarly, the time it takes them to come back is L0 1 − 0.95 /(0.95c). So, the entire round trip time is √ √ L 1 − 0.92 L 1 − 0.952 0 + 0 = 4.5 years. 0.9c 0.95c

This gives L0 = 5.54 light years.

(b) √ L 1 − 0.92 ∆τ = 0 = 0.484 years. outbound 0.9c (c) The distance according to Earth is 5.54 light years. So, for outbound trip 5.54 light years ∆t = = 6.15 years. outbound 0.9c Similarly, 5.54 light years ∆t = = 5.83 years. inbound 0.95c Therefore the entire trip takes 6.15 + 0.5 + 5.83 = 12.48 years.

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?

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.

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?

(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.

p 2 (d) Let β2 be the speed of return. Then the distance is 30 1 − β2 light years. And the p 2 time, in terms of speed, is 30 1 − β2 /cβ2. So, we have to solve the equation 30p1 − β2 light years 2 = 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 L L L 0 + 0 = 0 . γ 4γ 4

This gives γ = 5 and √ pγ2 − 1 24 β = = = 0.98, out γ 5 and √ p(4γ)2 − 1 399 β = = = 0.9987, in 4γ 20 (b) The round trip time is 10 years. So, L L L /γ L /(4γ) out + in = 0 + 0 = 10 βout βin βout βin or L L 0 + 0 = 10. 5 × 0.98 20 × 0.9987 This gives L0 = 39.35 light years. 14 Relativity of Time and Space

(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.

0 2.9. A rod of rest length L0 moves with speed v along the positive x -direction of observer 0 O . 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

0 p 2 p 2 0 ∆x = ∆x 1 − (v/c) = L0 1 − (v/c) cos θ0, ∆y = L0 sin θ0

(a) The length in O0 is q p 0 2 0 2 2 2 2 2 2 L = (∆x ) + (∆y ) = L0[1 − (v/c) ] cos θ0 + L0 sin θ0 p 2 2 = L0 1 − β cos θ0, β ≡ v/c.

Note that the answer is consistent with the special cases θ0 = 0 and θ0 = π/2. (b) ∆x0 p1 − β2 cos θ cos θ cos θ = = 0 = 0 p 2 2 p 2 2 L 1 − β cos θ0 γ 1 − β cos θ0 or 0 ∆y sin θ0 tan θ = = = γ tan θ0. 0 p 2 ∆x 1 − (v/c) 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?

(d) How often do you receive the ﬂashes?

Solution: The ﬂasher keeps proper time. 15

(a) 1 s ∆t = √ = 2.294 s. 1 − 0.92 (b) ∆x = 0.9c × 2.294 s = 2.065 light second.

(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 −t/T N(t) = N0e , 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

L0 p L0 0.99 = e−t/T = e−L0/(γcβT ) ⇐⇒ ln(0.99) = − ⇐⇒ γβ = γ2 − 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

β2γ2(β2L2 + b2 + L2/γ2) = β2γ2 b2 + L2 β2 + 1 − β2 = β2γ2(b2 + L2)

This yields (2.8). For (2.9), we have

p L p x = x − L/γ = −β2γL − βγ b2 + L2 − L/γ = −γ β2L + + β b2 + L2 . A c γ2 Equation (2.9) now follows immediately from the deﬁnition of γ in terms of β. Finally,

2 2 2 2 p 2 2 2 2 2 2 2 2 2 2 p 2 2 2 c tA = γ L + β b + L + b = γ L + γ β (b + L ) + 2γ βL b + L + b p = γ2L2 + (γ2 − 1)b2 + γ2β2L2 + 2γ2βL b2 + L2 + b2 p = γ2(L2 + b2) + γ2β2L2 + 2γ2βL b2 + L2 h p i p 2 = γ2 (L2 + b2) + β2L2 + 2βL b2 + L2 = γ2 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 |x0 | (x0 + L/γ)2 + b2 p c = c ⇐⇒ |x0 | = β (x0 + L/γ)2 + b2. v c c c Squaring both sides gives

x0 2 2β2L x0 2 = β2(x0 2 + L2/γ2 + 2x0 L/γ + b2) ⇐⇒ c − x0 − β2(b2 + L2/γ2) = 0. c c c γ2 γ c

Solve this quadratic equation for xc to obtain

0 2 p 4 2 2 2 2 2 2 2 xc = β γL ± β γ L + β γ (b + L /γ ).

0 The negative sign must be chosen because xc < 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 2 2 Solution: Square both sides of |xc| = β (b + L) + xc to get x2 x2 = β2[(b + L)2 + x2] = β2x2 + β2(b + L)2 ⇐⇒ (1 − β2)x2 = c = β2(b + L)2. c c c c γ2

0 Take the square root and chose the negative sign to get the answer. For xc, just change L to −L. 2.15. Derive Equation (2.16).

Solution: For an arbitrary point, you square both sides of |x| = βp(b − z)2 + x2. Then

x2 x2 = β2[(b − z)2 + x2] = β2x2 + β2(b − z)2 ⇐⇒ = β2(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.

0 Solution: A and B have coordinates (xc, 0, −L) and (xc, 0,L), respectively. So, their distances, |ctA| and |ctB| from the camera are given by

2 2 2 2 2 2 0 2 2 c tA = xc + (b + L) , c tB = xc + (b − L) . I’ll ﬁnd the ﬁrst one, leaving the second for you. From (2.14), we have

2 2 2 2 2 2 2 2 2 2 2 c tA = γ β (b + L) + (b + L) = (γ β + 1)(b + L) = γ (b + L) .

Therefore, ctA = −γ(b + L). Similarly, ctB = −γ(b − L). Note that b > L. 2.17. Derive Equation (2.17). 17

Solution: From |x| = βpx2 + y2 + b2, you get

x2 x2 x2 = β2(x2 + y2 + b2) ⇐⇒ = β2(y2 + b2) ⇐⇒ − y2 = 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 c2t2 = x2 + y2 + b2 = = γ2(y2 + b2). P β2 √ p 2 2 2 2 Therefore, ctP = −γ y + b . Thus, ctA = ctB = −γ L + b . 2.19. Derive Equations (2.20) and (2.21).

2 2 2 2 2 2 2 Solution: From xc = β x + a − γ (x − xc) + b , you get

2 2 2 2 2 2 2 2 2 2 xc = β x + a − γ x − γ xc + 2xγ xc + b or 2 2 2 2 2 2 2 2 2 2 2 xc (1 + β γ ) − 2xβ γ xc − β a + b − β γ x = 0 or 2 2 2 2 2 2 2 2 2 2 γ xc − 2xβ γ xc − β a + b − β γ x = 0. The solution is

xβ2γ2 ± px2β4γ4 + β2γ2(a2 + b2 − β2γ2x2) x = c γ2 √ xβ2γ2 ± γβpx2β2γ2 + a2 + b2 − β2γ2x2 xβ2γ2 ± γβ a2 + b2 = = . γ2 γ2

Choosing the negative sign gives (2.20). Then √ β p x + βγ a2 + b2 x − x = x − β2x + a2 + b2 = . c γ γ2

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 x0z0-plane.

(b) What are the slope and the intercept of the line in the x0z0-plane?

0 0 2 x = a0 + a1x + a2z, z = b0 + b1x + b2z

transforms a straight line in the xz-plane into a parabola in the x0z0-plane. Solution: The equation of the line is z = mx + b. (a) Substitute for z in (3.5) to get

0 0 x = a0 + a1x + a2mx + a2b, z = b0 + b1x + b2mx + b2b.

Find x from ﬁrst equation x0 − a − a b x = 0 2 a1 + a2m and substitute it in the second 0 0 x − a0 − a2b z = b0 + b2b + (b1 + b2m) , a1 + a2m or 0 b1 + b2m 0 (b1 + b2m)(a0 + a2b) z = x + b0 + b2b − . a1 + a2m a1 + a2m (b) The slope m0 and the intercept b0 are

0 b1 + b2m 0 (b1 + b2m)(a0 + a2b) m = , b = b0 + b2b − . a1 + a2m a1 + a2m 20 Lorentz Transformation

0 0 2 x = a0 + (a1 + a2m)x + a2b, z = b0 + b1x + b2mx + b2b.

Find x from the ﬁrst and plug it in the second

0 2 0 0 x − a0 − a2b x − a0 − a2b z = b0 + b1 + b2m + b2b. a1 + a2m a1 + a2m

Expanding and collecting terms, you get an expression of the form z0 = 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 e tan α √ = cos α, √ = = sin α. 1 + e2 1 + e2 sec α

3.4. Use Figure 3.5 to prove the coordinate transformation

x0 = a + x cos α − y sin α y0 = 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 = O0A0 = O0A + 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?

Solution: Instead of px2 + y2 = R, we now have p(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 0 0 0 0 0 are two events with coordinates (x1, ct1) and (x2, ct2) in O and (x1, ct1) and (x2, ct2) in O , where the primed coordinates are obtained from the unprimed coordinates by an LT, then

2 0 0 2 0 0 2 2 2 2 c (t2 − t1) − (x2 − x1) = c (t2 − t1) − (x2 − x1) . 21

yʹ y H ʹ P Bʹ x Q R H α B C D b O a x ʹ O ʹ A A ʹ

Figure 3.1: The same point P has diﬀerent pairs of coordinates in diﬀerent coordinate systems. Note that a = O0A and b = O0B.

Solution: To save writing use notations like ∆t21 = t2 − t1, etc. Then

0 0 c∆t21 = γ(c∆t21 + β∆x21), ∆x21 = γ(∆x21 + βc∆t21) and

2 0 2 2 2 2 2 2 c ∆t21 = γ (c ∆t21 + β ∆x21 + 2cβ∆t21∆x21) 0 2 2 2 2 2 2 ∆x21 = γ (∆x21 + β c ∆t21 + 2cβ∆t21∆x21). Subtracting the second from the ﬁrst gives

2 0 2 0 2 2 2 2 2 2 2 2 2 2 c ∆t21 − ∆x21 = γ (c ∆t21 − ∆x21) − γ β (c ∆t21 − ∆x21).

2 2 2 Now invoke the very useful identity γ β = γ − 1. 3.7. Starting with (3.19), show that the origin of O0 has the coordinates

0 0 x0 = γ(−x0 + βct0) 0 0 ct0 = γ(βx0 − ct0) relative to O. Show also that the inverse of Equation (3.19) can be written as

0 0 x = x0 + γ(x − βct ) 0 0 ct = ct0 + γ(βx + ct ).

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 0 = x0 + γ(x0 + βct0) 0 0 = ct0 + γ(βx0 + ct0). Multiply the second equation by β and subtract it from the ﬁrst: x 0 = x0 − βct0 + γ(1 − β2)x ⇐⇒ 0 = x0 − βct0 + 0 . 0 0 0 0 0 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:

0 0 0 0 2 0 0 0 0 x − βct = x0 − βct0 + γ(1 − β )x ⇐⇒ γ(x − βct ) = γ(x0 − βct0) +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 x0 = γ(L + βL) = L, 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 L L p1 − β2L 1 + β 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), (3.1) ∆t0 = 0 immediately gives the length contraction formula:

∆x p ∆x = γ∆x0 ⇐⇒ ∆x0 = = 1 − β2∆x, γ

remembering that ∆x is moving in O0. 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∆t0 c∆τ = γ(c∆t0 − cβ2∆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:

L /γ L /γ E : (0, 0),E : 0, 0 ,E : L , 0 , 1 2 β 3 0 β L L L /γ E : L , 0 , ,E : L , 0 + 0 . 4 0 β 5 0 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 L x = γ 0 − β 0 = −L 2 γβ 0 L L ct = γ 0 − 0 = 0 . 2 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 L x = γ L − β 0 = (γ − 1)L 3 0 γβ 0 L L ct = γ 0 − βL = 0 − γβL . 3 γβ 0 β 0

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

1Assume 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 L x = γ L − β 0 = 0 4 0 β L γL L ct = γ 0 − βL = 0 − γβL = 0 , 4 β 0 β 0 γβ

and ﬁnally the coordinates of E5 in B are L L x = γ L − β 0 + 0 = −L 5 0 β γβ 0 L L L L L L ct = γ 0 + 0 − βL = γ 0 + 0 = 0 + 0 5 β γβ 0 γ2β 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 0 its tail at x = 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

0 0 0 For A (0, 0) (0, t2)(L0, t2)(L0,L0/β)(L0, t5)

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!

(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, |x | γL ∆t = 0 = . cβ cβ

(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) 0 0 x2 = γ(0 + βL/γβ) = L, ct2 = 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 s 1 + β L 1 + β L x = , ct = − 1 − β 2 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 L 1 + β L x = γ − β − = γ(1 + β) = 2 2 2 1 − β 2 s L L L 1 + β 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

L L |x − x2| = γ(1 + β) − γL = γ(1 − β) , 2 2 and light covers it in γ(1 − β)L/2c. (e) The time of arrival is |x − x | L L L t + 2 = −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 − β2) = . 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 x0 = γ 0 + β = L arrive βγ L L ct0 = γ + 0 = . arrive 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 − z0)

0 because b > z and z = z . 3.14. Generalize the discussion of Example 3.4.5 to any clock length L and any speed β by showing that 0 0 c∆t21 = γL(1 + β), and c∆t32 = 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

0 0 0 c∆t21 ≡ c(t2 − t1) = γ(L + βL) = γL(1 + β) and 0 0 0 c∆t32 ≡ c(t3 − t3) = γ[(2L − L) + β(0 − L)] = γL(1 − β). Therefore, 0 0 0 c∆t = c∆t21 + c∆t32 = 2γL, and 2L ∆t0 = γ = 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 x0y0-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?

0 0 (e) Find tA, tB, tA, and tB, 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 of the rod the location of the event is (0, y, 0). So, its distance from the pinhole is py2 + b2, and ct = −py2 + b2. 28 Lorentz Transformation

(b) The spacetime coordinates of the event is (0, −py2 + b2). So, in O0, we have

p p x0 p x0 = γ(0 − β y2 + b2) = −γβ y0 2 + b2 ⇐⇒ = − y0 2 + b2 γβ

because y0 = y.

x0 2 y0 2 − = 1. γ2β2b2 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. √ 2 2 (e) ctA = ctB = − L + b . Therefore,

0 p 2 2 p 2 2 ctA = γ(− L + b + 0) = −γ L + b 0 p 2 2 p 2 2 0 ctB = γ(− L + b + 0) = −γ L + b = ctA

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

0 ctBA = γ[c(tB − tA) − β(xB − xA)] = 5(1 − 4.3 × 0.98) = −16.15 light years.

0 Thus, tBA = −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:

0 ctBA = γ[c(tB − tA) + β(xB − xA)] = 5(1 + 4.3 × 0.98) = 26.2 light years,

0 and tBA = 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:

4.3 light years 4.3 years p 2.5 years = = ⇐⇒ 2.5γβ = 4.3 ⇐⇒ 2.5 γ2 − 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 −t/τ distance of 2911 m, half of them survive. Recall that for any decay process, N(t) = N0e , 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?

(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,

L L 0.5 = exp − 0 ⇐⇒ ln(0.5) = − 0 cβγτ cβγτ or p L0 βγ = γ2 − 1 = − = 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.

(d) The speed of the muons relative to the spaceship is β + β 1.938 β0 = ship = = 0.99969 1 + ββship 1.9386 and 1 γ0 = = 40.2. p1 − β0 2 Therefore, 0 tship = γ 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,

8 −6 c∆tship = γ(c∆t + β∆x) = 3.2(3 × 10 × 9.82 × 10 + 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 − L = βc∆t ⇐⇒ L = − βc∆t γ 0 0 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 L c∆t c∆t0 = γ(c∆t + βL ) = γ c∆t + β − βc∆t = + βL. 0 γ γ It is also instructive to ﬁnd the location of the event in O0, although it is obviously L: L x0 = γ(L + βc∆t) = γ − βc∆t + βc∆t = L. 0 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

0 0 x0 = γ(L0 + 0) = γL0, ct0 = γ(0 + βL0) = γβL0.

0 The coincidence of the right end occurs at ∆t = γ(∆t + βL0). Thus in the time interval 0 0 0 ∆t − t0 the moving ruler covers a distance of L − x0. Thus, 0 0 cβ(∆t − t0) = 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?

0 0 (b) How does t1 compare to t2?

0 0 (d) What is x1 − x2? Is that what you expect?

Solution: Let (0, 0) be the coordinates of E1 in both frames. 0 (a) For both observers x1 = x1 = 0. For Sonya, E2 is on the negative x-axis and has x2 = −L.

0 0 (b) For Sam the two ends coincide simultaneously. Therefore, t1 = t2 = 0. (c) You don’t have to use the length contraction formula! Write the Lorentz transformation for E2:

0 x2 = γ(x2 + βct2) = γ(−L + βct2) 0 ct2 = γ(ct2 + βx2) = γ[ct2 + β(−L)].

0 With t2 = 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 x0 = γ(−L + β2L) = − . 2 γ

0 0 Hence, x1 − x2 = 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

1400 m 2.8 × 108 v = = 2.8 × 108 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

8 −6 c∆t21 = c(t2 − t1) = γ[3 × 10 × 5 × 10 + 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.

0 Solution: Label the events E1 through E6. Let Sonya be observer O and Sam O . 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:

0 0 0 0 0 0 0 0 (x1, 0), (x2, 0), (0, |x1|), (0, |x2|), (x5, ct5), (x6, ct6).

Now write the Lorentz transformation for each event and solve for the unknowns. For E1, we have 0 x1 = γ(−L/2 + βct1), 0 = γ(ct1 − βL/2). 33

0 These two equations give ct1 = βL/2 and x1 = −L/(2γ). For E2, we have

0 x2 = γ(L/2 + βct2), 0 = γ(ct2 + βL/2).

0 0 0 These two equations give ct2 = −βL/2 and x2 = L/(2γ). Note that x2 − x1 = 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

0 x5 = γ 0 + β(1 + β)L/2 = γβ(1 + β)L/2 0 ct5 = γ (1 + β)L/2 + 0 = γ(1 + β)L/2 and

0 x6 = γ 0 + β(1 − β)L/2 = γβ(1 − β)L/2 0 ct6 = 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 0.8 + 0.8 β0 = = 0.9756, 1 + 0.82 and the length contraction formula gives

p 0 2 L = L0 1 − β = 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?

Solution:

(a) Let L0 be the rest length of B. Then q 2 p 2 2 2L0 1 − βA = L0 1 − 0.75 ⇐⇒ 4(1 − βA) = 0.4375 ⇐⇒ βA = 0.9437.

(b) The relative speed of the two spaceships is 0.9437 − 0.75 β = = 0.663. AB 1 − 0.9437 × 0.75 Therefore, p 2 LAB = 2L0 1 − 0.663 = 1.5L0.

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 = ⇐⇒ β2 − 5.2β + 1 = 0, 1 − 0.8β 1 + 0.6β with solutions p √ β = 2.6 ± 2.62 − 1 ⇐⇒ β = 2.6 − 5.76 = 0.2,

because the positive choice gives β > 1.

(b) Her speed relative to Sam is 0.8 − 0.2 β = = 0.714 ss 1 − 0.8 × 0.2 and relative to Pat it is 0.6 + 0.2 β = = 0.714. sp 1 + 0.6 × 0.2

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.

0 0 (b) What is the period according to Sam? Hint: It is not t2 − t1! (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

0 x1 = γ(0 + βct1) = γβct1 0 ct1 = γ(ct1 + 0) = γct1

and

0 x2 = γ(0 + βct2) = γβct2 0 ct1 = γ(ct2 + 0) = γct2.

0 (b) Sam receives the ﬁrst wave front |x1|/c after it was emitted. Thus

rec 0 0 ct1 = ct1 + |x1| = γct1 + γβc|t1| = γct1 − γβct1 = (1 − β)γct1,

because t1 < 0. Similarly

rec 0 0 ct2 = 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 rec ct2 − ct1 = (1 − β)γct2 − (1 − β)γct1 = γ(1 − β)c(t2 − t1) = γ(1 − β)cT.

rec rec 0 Now, t2 − t1 is the period T according to Sam, which can also be written as s 1 − β 1 − β T 0 = T = T. p1 − β2 1 + β

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.

0 (b) Find tref, the time of Eref according to Sonya in terms of t .

0 0 0 (c) Find (xref, tref), the coordinates of Eref according to Sam in terms of t . (d) Show that 1 + β t0 = t0 det 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 = 0 0 γct . Therefore, Eref has coordinates (0, γct ) according to Sonya. (c)

0 2 0 xref = γ(0 + βctref) = γ βct 0 2 0 ctref = γ(ctref + 0) = γ ct

(d) 1 + β 1 + β ct0 = ct0 + |x0 | = γ2ct0 + γ2βct0 = ct0 = ct0 det ref ref 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 = λ. ref 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 1 + β 1 + β 13 9 1300 = 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 0 c∆t to be (L0/γ)/β.

(b) For Sam the time interval between coincidences is c∆t = L0/β. He concludes that 0 since Sonya is measuring proper time, for her the time interval is c∆t = (L0/β)/γ. (c) Sonya know that she is calculating proper time. So, she concludes that Sam measures 0 the time interval to be c∆t = γc∆t 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 0 to Sonya are (0, 0) and (0, ∆t ) and according to Sam, they are (0, 0) and (−L0, ∆t). Write the Lorentz transformation for the second event: 0 0 0 = ∆x = γ(−L0 + βc∆t), c∆t = γ(c∆t − βL0).

The ﬁrst equation gives c∆t = L0/β and plugging this in the second equation yields L γL L c∆t0 = γ 0 − βL = 0 (1 − β2) = 0 . β 0 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 0 0 (0, ctA), (L, ctB), (L, ctC ). Now Lorentz transform the events. For A, we get

0 0 0 = xA = γ(−L + βctA), ctA = γ(ctA − βL).

The ﬁrst equation gives ctA = L/β, and pugging this in the second equation yields L γL L ct0 = γ − βL = (1 − β2) = . A β β γβ For B, we get 0 L = γ(0 + βctB), ctB = γ(ctB − 0). 0 Finding ctB from ﬁrst and substituting in the second, we get ctB = L/(γβ) and ctB = L/β. Finally, for C, we get

0 L = γ(−L + βctC ), ctC = γ(ctC − βL). These equations yield

(1 + γ)L (1 + γ)L ct = , ct0 = γ − βL . C γβ C γβ The second equation can be simpliﬁed:

(1 + γ)L (1 + γ)L − γβ2L L + γL(1 − β2) (1 + γ)L ct0 = − γβL = = = . C β β β γβ

0 The equality tC = 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 0 βb = (βb + β)/(1 + βbβ). 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 0 L0 + cβ∆t = cβb∆t ⇐⇒ c∆t = 0 . γ γ(βb − β) But 2 0 βb + β βb + β − β − β βb βb βb − β = − β = = 2 . 1 + βbβ 1 + βbβ γ (1 + βbβ) Therefore, L0 L0γ(1 + βbβ) c∆t = 0 = . γ(βb − β) βb (b) 0 βb + β L0γ(1 + βbβ) L0γ(βb + β) ∆x = cβb∆t = c = . 1 + βbβ βb 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?

(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 − βbβ < 1 − β ⇐⇒ βb + β < 1 + βbβ.

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

βs + βw βsw = . 1 + βsβw 0 Therefore, it bounces back relative to the wall with speed βsw. To find the speed βs of the bounced super-ball relative to you, you have to add the speeds again: 2 0 βsw + βw βs + βw + βw(1 + βsβw) βs + 2βw + βsβw βs = = = 2 . 1 + βswβw 1 + βsβw + (βs + βw)βw 1 + 2βsβw + βw 0 It is interesting to note that βs → 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 β = = = 0.9884 B 1 + 0.8 × 0.9 1.72 0.9884 + 0.95 1.9384 β = = = 0.9997. C 1 + 0.9884 × 0.95 1.939 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: 0.995 + 0.995 β = = 0.999987. ps 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 + βAEβEB 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 √ 2 2βAE 2 1 ± 1 − 0.9 0.9 = 2 ⇐⇒ 0.9βAE − 2βAE + 0.9 = 0 ⇐⇒ βAE = 1 + βAE 0.9 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 β0. Use the relativistic law of addition of velocities (3.26) to obtain γ00 in terms of β, β0, γ, and γ0.

Solution: Substitute β + β0 β00 = 1 + ββ0 in the deﬁnition of γ00:

1 1 + ββ0 γ00 = = r 2 p 0 2 0 2 β+β0 (1 + ββ ) − (β + β ) 1 − 1+ββ0 1 + ββ0 = = γγ0(1 + ββ0). p(1 − β2)(1 − β0 2)

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). 0 (a) Show that O2 moves relative to O with speed β2 given by

+ − 0 P2 − P2 β2 = + − . 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).

0 Use the relativistic law of addition of velocities for β2 and β3 in the form given in (a) 0 to prove that O3 moves relative to O with speed β3 given by

+ − 0 P3 − P3 β3 = + − . P3 + P3

(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) + − = P2 + P2 (1 + β1)(1 + β2) + (1 − β1)(1 − β2) 2(β1 + β2) 0 = = β2. 2(1 + β1β2) 42 Lorentz Transformation

(b)

+ − P2 − P2 0 + − + β3 + − + − 0 β2 + β3 P2 + P2 P2 − P2 + (P2 + P2 )β3 β3 = 0 = + − = + − + − 1 + β2β3 P2 − P2 P2 + P2 + (P2 − P2 )β3 1 + + − β3 P2 + P2 + − + − P2 (1 + β3) − P2 (1 − β3) P3 − P3 = + − = + − P2 (1 + β3) + P2 (1 − β3) P3 + P3

(c)

+ − P3 − P3 0 + − + β4 + − + − 0 β3 + β4 P3 + P3 P3 − P3 + (P3 + P3 )β4 β4 = 0 = + − = + − + − 1 + β3β4 P3 − P3 P3 + P3 + (P3 − P3 )β4 1 + + − β4 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: + − 0 PN−1 − PN−1 βN−1 = + − . PN−1 + PN−1 Now we have to prove that + − 0 PN − PN βN = + − PN + PN is true. To do so, we use the law of addition of velocities:

+ − PN−1 − PN−1 0 + − + βN + − + − 0 βN−1 + βN PN−1 + PN−1 PN−1 − PN−1 + (PN−1 + PN−1)βN βN = 0 = + − = + − + − 1 + βN−1βN PN−1 − PN−1 PN−1 + PN−1 + (PN−1 − PN−1)βN 1 + + − βN PN−1 + PN−1 + − + − PN−1(1 + βN ) − PN−1(1 − βN ) PN − PN = + − = + − PN−1(1 + βN ) + PN−1(1 − βN ) PN + PN

3.39. Recall that hyperbolic functions are deﬁned by

eφ + e−φ eφ − e−φ cosh φ = , sinh φ = . 2 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

1 + β 1/2 cosh φ = γ, sinh φ = βγ, eφ = . 1 − β

Solution:

(a) From e2φ + e−2φ + 2 e2φ + e−2φ − 2 cosh2 φ = , sinh2 φ = 4 2 the ﬁrst equality follows immediately. For the second equality, note that

eφ = cosh φ + sinh φ, e−φ = cosh φ − sinh φ

Therefore,

eφ1+φ2 − e−φ1−φ2 eφ1 eφ2 − e−φ1 e−φ2 sinh(φ + φ ) = = 1 2 2 2 (cosh φ + sinh φ )(cosh φ + sinh φ ) = 1 1 2 2 2 (cosh φ − sinh φ )(cosh φ − sinh φ ) − 1 1 2 2 . 2

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 1 1 − tanh2 φ = ⇐⇒ 1 − β2 = ⇐⇒ cosh φ = γ. cosh2 φ cosh2 φ Then, sinh φ sinh φ β = tanh φ = = ⇐⇒ sinh φ = βγ. cosh φ γ Finally, 1 + β 1 + β 1/2 eφ = cosh φ + sinh φ = γ + βγ = = . p1 − β2 1 − β

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 φ0 + tanh φ tanh φ00 = = 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 → ξ0 = eφξ, so its scale changes by a factor of eφ, 0 −φ −φ and η → η = e η, so its scale changes by a factor of e . 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?

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 0.996 − 0.92 β = = 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.

0 0 0 0 3.44. The acceleration of a “bullet” in O is ab = dvb/dt and in O, moving relative to 0 0 O with constant speed β in the positive direction of the x -axis is ab = dvb/dt. Use (inﬁnitesimal) LT and Equation (3.26) to show that

0 ab ab = 3 3 γ (1 + ββb) Solution: The diﬀerential of Equation (3.26) is

0 (1 + ββb)dβb − (βb + β)βdβb dβb dβb = 2 = 2 2 , (1 + ββb) γ (1 + ββb) and 0 0 cdt = γ(cdt + βdx) = γcdt(1 + ββb) ⇐⇒ dt = γdt(1 + ββb). Therefore, cdβb 0 2 2 0 cdβb γ (1 + ββb) ab ab = 0 = = 3 3 . dt γdt(1 + ββb) γ (1 + ββb) 0 3.45. In Problem 3.44, let βb = 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) cdβ adt = . (1 − β2)3/2 (b) Integrate the equation above and show that if β = 0 at t = 0, then at β = √ . c2 + a2t2 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? (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? 0 Solution: If βb = β, the βb = 0, and a a a0 = b ⇐⇒ cdβ/dt = . b γ3 γ3 (a) Therefore, cdβ adt = γ3cdβ ⇐⇒ adt = . (1 − β2)3/2 (b) Integrate both sides and choose the constant of integration so that β = 0 at t = 0. Then cβ c2β2 at at = ⇐⇒ a2t2 = ⇐⇒ β = √ . p1 − β2 1 − β2 c2 + a2t2 In units of c, the acceleration is 5 m/s2 α ≡ a/c = = 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. 1 − 0.992 0.526 yr−1 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) dx cat catdt cαtdt = √ ⇐⇒ dx = √ = √ . dt c2 + a2t2 c2 + a2t2 1 + α2t2 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 2.94 3.12 − (αt)2 = 1, ⇐⇒ αt = 2.94 ⇐⇒ t = = 5.59 yr. 0.526 yr−1

From (b), we have at αt 2.94 β = √ = = √ = 0.9467. c2 + a2t2 p1 + (αt)2 1 + 2.942

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 2 2 P1P2 = (x2 − x1) + (y2 − y1) , then the axes must be perpendicular.

Solution: Figure 4.1 of the manual shows the two points and their coordinates in a non- perpendicular coordinate system. The law of cosines gives

2 2 2 P1P2 = (x2 − x1) + (y2 − y1) + 2(x2 − x1)(y2 − y1) cos θ. Therefore,

2 2 2 P1P2 = (x2 − x1) + (y2 − y1) ⇐⇒ (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 x 2 θ

P1 y 1 x2 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 x

Figure 4.2: The event E covers a distance of x in time t.

0 4.4. A train of rest length L0 travels at speed β in the positive x -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?

D F

B O ' x'

Figure 4.3: The worldlines of the front and back of the train are the lines O0A and BC, respectively.

Solution: (a) The worldlines of the front and back of the train are, respectively, the lines O0A 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 O0B is that length.

0 0 0 0 (b) By Rule 2 of Note 4.2.6, β = BO /O C or β = (L0/γ)/ct2, where t2 is when the back 0 of the train reaches Sam. Therefore, ct2 = L0/(βγ). (c) By Rule 2 of Note 4.2.6, β = DF/FC. Therefore, FC = DF /β. But DF = O0F 0 0 and, by (b) ct2 = O F + FC. Hence, L O0F O0F L L O0F = ct0 − FC = 0 − ⇐⇒ (1 + β) = 0 ⇐⇒ O0F = 0 , 2 βγ β β βγ (1 + β)γ

0 0 and O F ≡ ct1 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 0 x1 = γ(−L0 + βL0) = −γ(1 − β)L0 γ(1 − β)(1 + β)L L ct0 = γ(L − βL ) = γ(1 − β)L = 0 = 0 , 1 0 0 0 1 + β γ(1 + β) 0 in Sam’s RF, as before. Note that although we didn’t calculate x1 before, the spacetime 0 0 0 diagram clearly shows that x1 = −DF = −O F = −ct1. The second event has coordinates 0 x2 = γ(−L0 + βL0/β) = 0 γ L ct0 = γ(L /β − βL ) = (1 − β2)L = 0 , 2 0 0 β 0 γβ which is identical to what we obtained earlier. 52 Spacetime Geometry

ct ct C D E 1pm C D E 1pm

F ct2 F ct2

ct1 B

A M G ct1 K H

ct0 A G ct0 B x x (a) (b)

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?

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.

(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 0 O 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' E2 ct2 E ct 1 1 x' ' ct1

' ct2

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 E1E2 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 0 0 between those two lines decreases, meaning that the corresponding ct2 − ct1 decreases. As you continue increasing the angle, you reach a line that is parallel to E1E2. For this line, 0 0 ct2 = ct1. It should now be clear that any line that makes an angle with the x-axis larger ◦ 0 than the angle of E1E2 (but less than 45 ) is a good candidate for the x -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?

(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'

O x F1 F2 F3 F4 F5

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.

(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 0 0 connecting E1 and E2. That’s how ct -axis is constructed. The x -axis is then drawn as usual. 56 Spacetime Geometry ct ct'

' ct2 E B 2

' ct1 A E1 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 E1E2 relative to the ct-axis: 300 m − 1500 m v = = −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,

0 0 ct2 − ct1 = AB = γE1E2 = γ(ct2 − ct1),

or t − t p t0 − t0 = 2 1 = 1 − 0.82 × 5 µs = 3 µs. 2 1 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?

'' ct ' ' Lx Lx ct ct '' B x'

A'' A' A

O x '' Lct

' 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:

AA0 AA0 20 β = = = = 0.8 OA OB − AB 45 − 20

(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 00 00 00 00 00 line Lx is parallel to x -axis; if its intersection with ct is a point C , then OC is the 0 00 00 time coordinate of A in Sonya’s RF. Similarly, the line Lct is parallel to ct -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, OC00 > 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 p OA = γOA0 ⇐⇒ OA0 = = 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: 0.8 + 0.8 1 βO0O00 = = 0.9756 ⇐⇒ γO0O00 = √ = 4.56. 1 + 0.8 × 0.8 1 − 0.97562

Since I know OA0, I can invoke Rule 4 of Note 4.2.7:

00 0 OC = γO0O00 OA = 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:

00 00 OD OD 00 βO0O00 = ⇐⇒ 0.9756 = ⇐⇒ OD = 66.67 light minutes, OC00 68.33 light minutes

00 and xA0 = −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 0 t0 +∆t. Sam, in RF O , 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 MF . 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 + HB = γOG + βOD = γ∆x + βγOC = γ∆x + βγ(ct0 + c∆t). |{z} =DC Therefore, AB = γ∆x + βγc∆t = γ(βb + β)c∆t 59

ctʹ ct E F D C x

M E0 G O A H B x ʹ

Figure 4.9: The spacetime diagram for the derivation of the relativistic law of addition of velocities.

Now, let’s calculate MF :

MF = OF − OM = OF − γct0 = OD + DF −γct0 |{z} =GH

= γOC + βOH − γct0 = γ(ct0 + c∆t) + βγ∆x − γct0.

Hence, MF = γ(c∆t + β∆x) = γ(c∆t + βcβb∆t) = γ(1 + ββb)c∆t. Taking the ratio AB/MF 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 ct

' C G x2

E2 E B 2

E1 E1

O x O A x'

' ' ct2 x ' '' H ct ct '' x D x (a) (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. 1 c∆t0 = γ(c∆t − β∆x) = √ (3 − 0.9 × 4) = −1.38 light year. 1 − 0.92

0 Thus, ∆t = −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. 1 c∆t00 = √ (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 E1A E1A β1 = , βps = ⇐⇒ β1OA = βpsAE2, OA AE2

where βps is the speed of the policeman relative to the speeder. Relativistic LAV gives

βpg + βgs βpg − βsg βps = = . 1 + βpgβgs 1 − βpgβsg 61

E1 A

O x

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 remem- bered! 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,

β2 − β1 βps = , 1 − β2β1 and β2 − β1 β2 − β1 β1OA = AE2 ⇐⇒ OA = AE2. 1 − β2β1 β1(1 − β2β1)

We also have cT = OA + AE2. Hence,

2 β2 − β1 β2(1 − β1 ) cT = + 1 AE2 = AE2 β1(1 − β2β1) β1(1 − β2β1) or 2 2 γ1 β1(1 − β2β1)cT γ1 (β2 − β1)cT AE2 = , OA = . β2 β2 The policeman is at a distance

2 γ1 β1(β2 − β1)cT E1A = β1OA = β2 from the speeder when he starts the chase, making his x-coordinate relative to the speeder

2 γ1 β1(β2 − β1)cT x1 = − . β2 The time that the policeman starts chasing the speeder, according to the policeman is OE1. Using Rule 4, we get

0 OA γ1(β2 − β1)cT OA = γ1OE1 ⇐⇒ ct1 = OE1 = = . γ1 β2 62 Spacetime Geometry

Similarly, using γps = γ1γ2(1 − β1β2), we obtain

AE2 β1γ1 AE2 = γpsE1E2 ⇐⇒ E1E2 = = cT. γ1γ2(1 − β1β2) β2γ2 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 = γ1OE1, AE1 = β1OA = β1γ1OE1.

Also by Rule 2,

AE1 β1γ1OE1 β1γ1(1 − β1β2)OE1 AE1 = βpsAE2 ⇐⇒ AE2 = = = , 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

β1γ1(1 − β1β2)OE1 β2OE1 ct = OE2 = OA + AE2 = γ1OE1 + = . β2 − β1 γ1(β2 − β1) Plugging in the numbers, you get 0.995 × 2 minutes ct = = 173.5 minutes 2.294(0.995 − 0.99)

(b) To ﬁnd the time according to the policeman, we need E1E2:

AE2 AE2 β1OE1 AE2 = γpsE1E2 ⇐⇒ E1E2 = = = . γps γ1γ2(1 − β1β2) γ2(β2 − β1) 63

ctʹ ct

A O2

O1 C1 C2 x ʹ

Figure 4.12: Clocks C1 and C2 and the observers O1 and O2.

Thus, 0 β1 ct = OE1 + E1E2 = 1 + 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)

0 0 4.15. Two clocks C1 and C2 separated by a ﬁxed distance L are placed on the x -axis of an observer O0 and synchronized according to O0.1 Reference frame O is moving relative 0 0 to O in the positive direction of x 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?

0 (b) What time does O2 record? Hint: It is the time coordinate in O 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. 0 (a) By Rule 4, C1C2 = γO1O2. Therefore, with x1 = 0, we get x2 = L /γ.

0 0 (b) The time is ct = O2C2 = βL (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

1This is the same problem done in Example 3.4.10. 64 Spacetime Geometry

ctʹ ct

B C Oʹ O x ʹ

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 O0C of Figure 4.13 of the manual. Using Rules 2 and 4, we have

L0 = O0B + BC = γOA + AB = γOA + βO0B s 1 + β = γOA + βγOA = γ(β + 1)L = L. 1 − β

4.17. Sam and Sonya are twins. Sam is put on a spaceship O moving with relative speed 0 0 0 0 β in the positive x direction of Sonya’s reference frame O . At time T0, O sends a light signal toward O. Draw a spacetime diagram showing the relevant axes and the light signal. Use the rules of spacetime geometry.

0 (a) Show that when Sam receives the light signal, the spaceship is at a distance of βT0/(1− β) from Sonya.

p 0 (b) Show that Sam receives the light signal when he is (1 + β)/(1 − β) T0 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 0 0 problem is to ﬁnd OE1 and AE1 in terms of cT0 = O E0. Rule 4 gives

0 0 0 0 0 O A = O E0 + E0A = O E0 + AE1 = O E0 + βO A.

Therefore, cT 0 (1 − β)O0A = O0E ⇐⇒ O0A = 0 . 0 1 − β (a) By Rule 2, βcT 0 OB = AE = βO0A = 0 . 1 1 − β 65

ctʹ ct

A E1

E 0

O Oʹ B xʹ

Figure 4.14: Spacetime diagram for Sonya sending a light signal to her twin brother. Event E0 takes 0 place at T0.

0 (b) By Rule 4, O A = γOE1. Thus,

O0A cT 0 ct = OE = = 0 , 1 1 γ γ(1 − β)

and s βT 0 1 + β t = 0 = T 0 1 γ(1 − β) 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?

ctʹ ct x T

Oʹ 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.

2A 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’ ct

R A B AB x RB At Bx A B A O B x’ A B Bt Ax

(a) (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, O0T 24 β = = = 0.9796. O0O 24.5 It is clear that the speed should be in the negative x0 direction. (b) By Rule 4, O0O = γOT . Therefore,

O0O p OT = = 1 − β2 O0O = 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 Fig- ure 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 BBtRB is an isosceles triangle (show this by drawing a line from Bt parallel to the light world line passing through O). Therefore, BtRB = BBt = OBx, and

ORB = BtRB − OBt = OBx − OBt = γL − βγL = (1 − β)γL.

Similarly, the triangle AAtRA is an isosceles triangle. Therefore, AtRA = AAt = OAx, and

ORA = AtRA + OAt = OAx + OAt = γL + βγL = (1 + β)γL.

Finally, it is clear that ORAB = OA = OB = L. 67

c tʹ ct E2 O ʹ Oʹ x E3 c Δ t31 E1 E2 Δ x x 31 O

E3 E1 (a) (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 0 0 relative to O are just the coordinates of E2 of Figure 4.17 of the manual relative to O . 0 0 0 These are (x0, ct0) ≡ (x2, 0). Now, we calculate x0 and ct0 of Problem 3.7:

0 0 0 0 0 0 x0 = γ(−x0 + βct0) = −γx2, ct0 = γ(βx0 − ct0) = γβx2.

Thus, the inverse Lorentz transformation that includes the coordinates of the origin is

0 0 0 x = − γx2 + γ(x − βct ) 0 0 0 ct =γβx2 + γ(−βx + ct ).

As a check, note that

0 0 0 0 0 0 x2 = −γx2 + γ(x2 − βct2) = 0, ct2 = γβx2 + γ(−βx2 + ct2) = 0.

0 0 0 The coordinates of E1 and E3 in O are (0, −βx2) and (0, ct3), respectively. Therefore, x0 x = −γx0 + γ[0 − β(−βx0 )] = −γx0 (1 − β2) = − 2 1 2 2 2 γ 0 0 ct1 = γβx2 + γ(−0 − βx2) = 0, and

0 0 0 0 x3 = − γx2 + γ(0 − βct3) = −γ(x2 + βct3) 0 0 0 0 ct3 =γβx2 + γ(−0 + ct3) = γ(βx2 + ct3).

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

15.5 E4 E3

10.1 E2

E 1 x O 5 10

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

T1 + T2 βprobe = β + 2 , βγ T2 which is already derived in the text. Manipulate the second term:

2 T1 + T2 T1 1 T1 1 − β T1 1 2 = 2 + 2 = 2 + = 2 + − β. 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?

(d) How old is Sonya when she lands back on Earth? How old is Sam? How old is Pat?

Solution:

10 (a) Distance is 10 light years, and time is 10.1 years. Therefore, β1 = 10.1 = 0.99. 10 (b) β2 = 15 = 0.67. 69

p 2 2 ∆s12 = 10.1 − 10 = 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 2 2 ∆s13 = 15 − 10 = 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 2 2 ∆s45 = (26 − 15.5) − 10 = 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 β (κ2 + √ 0 2 1)T/2,T/2 , semi-major axis a = β0κ κ + 1 T/2 and semi-minor axis b = κT/2. Solution: Rewrite (4.14) as √ 2 2 cβ0(κ + 1)T cβ0 κ + 1p x = − κ2T 2 + (2t − T )2, s 2 2 or 2 cβ0(κ + 1)T cβ0 p x − = − (κ2 + 1)[κ2T 2 + (2t − T )2] s 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: cβ (κ2 + 1)T 2 c2β2 x − 0 = 0 (κ2 + 1)κ2T 2 + c2β2(κ2 + 1)(t − T/2)2. s 2 4 0 2 2 2 2 2 Now transfer the t term to the left and divide both sides by c β0 (κ + 1)κ T /4 to obtain

2 2 2 xs − cβ0(κ + 1)T/2 (t − T/2) √ − = 1. 2 2 2 (cβ0 κ + 1κT/2) (κT/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 2 2 2 2 2 2 < c ⇐⇒ β (κ + 1)(2t − T ) < κ T + (2t − T ) . dt 0 The last inequality is equivalent to 2 2 2 2 2 2 2 β0 κ (2t − T ) < κ T + (1 − β0 )(2t − T ) . The second term on the right-hand side is positive, so the minimum of the right hand side 2 2 2 2 2 is κ T . Since 0 < t < T , the maximum of the left hand side is β0 κ T . 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

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 cβ0T/8.

p 2 (b) Equation (4.16) of Example 4.5.4 shows that Pat’s age is 1 − β0 times the time of p 2 landing. If the time of landing is T/4, then he ages 1 − β0 T/4 during each trip. p 2 (c) Hence, during the entire 4 trips, he ages 1 − β0 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?

Solution: It is shown in Example 4.5.4 that xp,max = cβ0T/2. (a) Therefore, 0.999cT 80 light years 40 light years = ⇐⇒ T = = 80.08 years, 2 0.999c

p 2 and this is Sonya’s age. Example 4.5.4 derived Pat’s age as 1 − β0 T . Thus,

p 2 TPat = 1 − 0.999 80.08 years = 3.58 years.

(b) It is shown in Example 4.5.4 that

cβ T p p x = 0 κ2 + 1 − κ κ2 + 1 = x κ2 + 1 − κ κ2 + 1 . s,max 2 p,max 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 s √ Z 80.08 0.9992(1.125)(2t − 80.08)2 TSam = τs(0.999, 0.125) = 1 − 2 2 dt. 0 0.125 × 80.08 + (2t − 80.08)

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.

(d) Verify that it take Sonya √ ! sin−1 κ + κ 1 − κ2 T 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 dx 2T = cκ, = cκ 1 − = −cκ. dt t=0 dt t=T T 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.

(d) s T T 2 Z p Z dx s = cτ = (cdt)2 − (dx)2 = c 1 − dt 0 0 cdt or s √ ! Z T 2t2 sin−1 κ + κ 1 − κ2 τ = 1 − κ2 1 − dt = T 0 T 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.

(d) Verify that it take Sonya

√ Z π/4 q 2T 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 p Z π/4 q √ √ cτ = (cdt)2 − (dx)2 = (cT 2 cos θdθ)2 − (−cT κ 2 sin θdθ)2 −π/4 −π/4 √ Z π/4 p √ Z π/4 q = 2cT cos2 θ − κ2 sin2 θ dθ = 2cT 1 − (κ2 + 1) sin2 θ dθ. −π/4 −π/4

As κ → 0, this travel time reduces

√ Z π/4 p √ Z π/4 τ = 2T 1 − sin2 θ dθ = 2T cos θ dθ = 2T. −π/4 −π/4

(e) Let R(κ) denote the ratio of Sam’s time to Sonya’s:

√ Z π/4 q R(κ) = 2 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

2 2 0 5.1. Show that ubt = γb, and that ubt − ubx = 1. Show that for any other observer O , 0 2 0 2 ubt − ubx = 1 as well.

Solution: First ﬁnd ubt:

c∆tb c∆tb c∆tb ubt = = = p 2 2 c∆τb ∆sb (c∆tb) − (∆xb) c∆tb 1 = = = γb. p 2 q c∆tb 1 − (∆xb/c∆tb) 2 1 − βb

Now note that

2 2 2 2 2 2 (c∆tb) (∆xb) (c∆tb) − (∆xb) ubt − ubx = 2 − 2 = 2 = 1. (∆sb) (∆sb) (∆sb) Furthermore,

0 2 0 2 0 2 0 2 0 2 0 2 (c∆tb) (∆xb) (c∆tb) − (∆xb) ubt − ubx = 2 − 2 = 2 = 1, (∆sb) (∆sb) (∆sb)

2 2 2 0 2 0 2 because (∆sb) = (c∆tb) − (∆xb) = (c∆tb) − (∆xb) . 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

0 γb = γ(βγbβb + γb) = γγb(ββb + 1), or 2 2 2 1 (ββb + 1) 0 2 (1 − β )(1 − βb ) 0 2 = 2 2 ⇐⇒ 1 − βb = 2 , 1 − βb (1 − β )(1 − βb ) (ββb + 1) 76 Spacetime Momentum

or 2 2 2 2 2 0 2 (1 − β )(1 − βb ) (ββb + 1) − (1 − β )(1 − βb ) βb = 1 − 2 = 2 , (ββb + 1) (ββb + 1) or 2 2 2 0 2 2ββb + β + βb (β + βb) βb = 2 = 2 . (ββb + 1) (ββb + 1) 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: du du a = bt , a = bx . bt ds bx ds (a) Show that abtubt − abxubx = 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 = AtBt − AxBx (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 0 components of the spacetime acceleration of a particle to be (abx, abt) and O measures 0 0 them to be (abx, abt), then 0 0 abx = γ(abx + βabt), abt = 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):

0 dubx = γ(dubx + βdubt) 0 dubt = 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 df f(x) + 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 dg g(t) − c2t = 0. dt

(d) Solve this simple equation and show that the solution is p g(t) = ± C + c2t2,

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

hx,˙ y˙i = hx,˙ xdf/dx˙ i,

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 · hx,˙ xdf/dx˙ i = 0 ⇐⇒ xx˙ +xf ˙ (x)df/dx = 0

or df f(x) + x = 0. dx (b) Rewrite the equation as fdf + xdx = 0, and integrate it to get

1 2 1 2 2 2 p 2 2 f + 2 x = constant ⇐⇒ f + x = C ⇐⇒ f(x) = ± C − x .

(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) − c2t(dt/ds) = 0.

(d) Multiply the previous equation by ds and integrate to get

1 2 1 2 2 2 2 2 p 2 2 2 g − 2 c t = constant ⇐⇒ g − c t = C ⇐⇒ g(t) = ± C + c t , 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): 0 0 pbc γ(pbc + βEb) pbc/Eb + β βb + β βb = 0 = = = . Eb γ(βpbc + Eb) βpbc/Eb + 1 ββb + 1 5.6. Derive Equation (5.10) directly from the deﬁnition of energy and momentum in Equa- tions (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

2 2 2 2 2 2 2 2 4 2 2 2 4 Eb − pb c = (γbmc ) − (mcγbβb) c = m c γb (1 − βb ) = m c . From (5.8), we get

0 2 0 2 2 2 2 2 2 Eb − pb c = γ (βpbc + Eb) − γ (pbc + βEb) 2 2 2 2 2 2 2 2 2 2 = γ (β pb c + Eb + 2βpbcEb) − γ (pb c + β Eb + 2βpbcEb) 2 2 2 2 2 2 2 2 2 = γ [Eb (1 − β ) − pb c (1 − β )] = Eb − pb c

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

m1γ1cβ1 + m2γ2cβ2 = m3γ3cβ3 + m4γ4cβ4. 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

0 βi + β 0 0 0 βi = , γi = γγi(1 + βiβ), ⇐⇒ βiγi = γγi(βi + β), i = 1, 2, 3, 4. 1 + βiβ 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

0 0 0 0 0 0 0 0 m1γ1β1 + m2γ2β2 = m3γ3β3 + m4γ4β4, or m1γγ1(β1 + β) + m2γγ2(β2 + β) = m3γγ3(β3 + β) + m4γγ4(β4 + β), or m1γ1β1 + m1γ1β + m2γ2β2 + m2γ2β = m3γ3β3 + m3γ3β + m4γ4β4 + m4γ4β. Conservation of momentum in O eliminates the sum of the ﬁrst and third terms on each side, leaving m1γ1β + m2γ2β = m3γ3β + m4γ4β. 2 Canceling β and multiplying by c 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:

2 2 2 mγvv = MγV V, mc γv + mc = Mc γV .

The second equation gives MγV = m(1 + γv). Substituting this in the ﬁrst equation yields

γvv v mγvv = m(1 + γv)V ⇐⇒ V = = p 1 + γv 1 + 1 − (v/c)2 and r 1 1 + γv 1 + γv γV = = = . s 2 p 2 2 2 2 γ v (1 + γv) − γv v 1 − v 1 + γv Now we can calculate M:

m(1 + γv) p M = = 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:

2 2 mec γe − mec = 50 keV ⇐⇒ 511(γe − 1) keV = 50 keV ⇐⇒ γe = 1.0978,

2 because mec = 0.511 MeV. This gives βe = 0.413. (b) From

1 2 1 2 2 1 2 2 1 2 KEe = 2 meve = 2 mec (ve/c) = 2 mec βe ⇐⇒ 50 = 2 × 511βe ,

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?

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: p 2mc2γ = E + e, 0 = P + p ⇐⇒ |p| = |P | ⇐⇒ e = |P |c = E2 − M 2c4,

where upper case letters correspond to M and lower case letters to photon. Substitute the last equation in the ﬁrst to get p 2mc2γ = E + E2 − M 2c4 ⇐⇒ (2mc2γ − E)2 = E2 − M 2c4,

which in combination with 2mc2γ = E + e yields

M 2c2 M 2c2 E = mc2γ + and e = mc2γ − . 4mγ 4mγ

M 2c |p| = e/c = mcγ − = |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,

pc2 pc2 pc pc v1 = = = , v2 = . E p 2 2 2 4 p 2 2 2 p 2 2 2 1 p c + m1c p + m1c p + m2c 81

(b) q q 2 2 2 4 2 2 2 4 E1 = p c + m1c ,E2 = p c + m2c .

2 (c) Conservation of energy gives E1 + E2 = Mc . Therefore, E + E pp2 + m2c2 + pp2 + m2c2 M = 1 2 = 1 2 c2 c (d) E + E (M − m − m )c2 = 1 2 − m − m c2 1 2 c2 1 2 2 2 = E1 − m1c + E2 − m2c = 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 − m2)2 + (M 2 − m2)2 − M 4 − 2m2m2 p = 1 2 1 2 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.

2 (b) Conservation of energy is Mc = E1 + E2, or q q 2 2 2 2 4 2 2 4 2 4 Mc = E1 + p c + m2c = E1 + E1 − m1c + m2c or 2 2 2 2 4 2 4 (Mc − E1) = E1 − m1c + m2c , yielding Mc2 (m2 − m2)c2 E = + 1 2 . 1 2 2M Similarly, Mc2 (m2 − m2)c2 E = + 2 1 . 2 2 2M 82 Spacetime Momentum

M 2c2 + (m2 − m2)c2 2 p2c2 = E2 − m2c4 = 1 2 − m2c4, 1 1 2M 1 or p2 M 4 + (m2 − m2)2 + 2M 2(m2 − m2) = 1 2 1 2 − m2 c2 4M 2 1 M 4 + m4 + m4 − 2m2m2 − 2M 2(m2 + m2) = 1 2 1 2 1 2 4M 2 (M 2 − m2)2 + (M 2 − m2)2 − M 4 − 2m2m2 = 1 2 1 2 . 4M 2

(d) If m1 = m2, then (b) gives Mc2 E = = E . 1 2 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 2 1 2 of energy now gives 2E = Mc , or E = 2 Mc . 5.13. The neutral pion π0 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, 2 −28 8 2 m 0 c 2.41 × 10 × (3 × 10 ) e = π = = 1.08 × 10−11 J. 2 2 (b) hc 6.626 × 10−34 × 3 × 108 λ = = = 1.84 × 10−14 m. e 1.08 × 10−11 This is in the short-wave gamma ray region of the EM spectrum. (c) Assume that the pion is moving in the positive x direction in the lab. Then (5.8) gives s 1 + β e0 = γ(βp c + e) = γ(βe + e) = e = 4.7 × 10−11 J 1 1 1 − β s 1 − β e0 = γ(βp c + e) = γ(−βe + e) = e = 2.48 × 10−12 J, 2 2 1 + β 83

and 6.626 × 10−34 × 3 × 108 λ0 = = 4.23 × 10−15 m 1 4.7 × 10−11 6.626 × 10−34 × 3 × 108 λ0 = = 8.08 × 10−14 m. 2 2.48 × 10−12

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?

(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 2 2 unit mπ− c = 139.57 MeV and mµc = 105.66 MeV, and

2 2 2 2 mπ− c mµc 139.57 105.66 Eµ = + = + = 109.78 MeV 2 2mπ− 2 279.14 2 2 2 2 mπ− c mµc 139.57 105.66 Eν = − = − = 29.79 MeV. 2 2mπ− 2 279.14

(b) Since neutrino is assumed massless, Eν = pνc, and pν = 29.79 MeV/c. (c) Muon has the same momentum as the neutrino. Thus, by (5.7),

pνc Eν 29.79 βµ = = = = 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

0 Eµ = γ(βpµc + Eµ) = 2.29(−0.9 × 29.79 + 109.78) = 190 MeV 0 Eν = γ(βpνc + Eν) = 2.29(0.9 × 29.79 + 29.79) = 129.6 MeV. and q 0 0 2 2 4 p 2 2 0 pµc = Eµ − mµc = 190 − 105.66 = 157.9 MeV ⇐⇒ pµ = 157.9 MeV/c 0 0 0 pνc = Eν = 129.6 MeV ⇐⇒ pν = 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?

3 −23 (c) Use KEavg = 2 kBT , where kB = 1.38 × 10 J/K is the Boltzmann constant, to ﬁnd the temperature of the ﬁnal piece of clay.

Solution: The problem is entirely classical.

(a) 2 2 6 KEtot = 2KE = mv = 0.5 × 2000 = 2 × 10 J.

(b) 2 × 106 KE = = 8.33 × 10−19 J. avg 2.4 × 1024 (c) −19 2KEavg 2 × 8.33 × 10 T = = −23 = 40258 K. 3kB 3 × 1.38 × 10

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:

a a a a 1 0 11 21 11 12 = . a12 a22 −a21 −a22 0 −1

Continue the multiplication:

2 2 a11 − a21 a11a12 − a21a22 1 0 2 2 = . a11a12 − a21a22 a12 − a22 0 −1 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:

2 2 1 + βˆx (γ − 1) = 1 + βˆ (γ − 1) = γ because βˆ2 = 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: x0 x γβct 1 y0 = y +  0  + (γ − 1) 0 x, z0 z 0 0 86 Relativity in Four Dimensions

where I used βˆ · ~r = x. Equating the ﬁrst component on either side yields

x0 = x + γβct + (γ − 1)x = γ(βct + x)

0 0 and the equality of the second and third components gives y = y and z = 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 0 0 0 of (6.17) follows immediately. Writing ~r = ~r|| + ~r⊥ as well, the second equation of (6.16) can be expressed as 0 0 ~ ˆ ~ ~r|| + ~r⊥ = ~r|| + ~r⊥ + γβct + (γ − 1) β ~r|| = γ(~r|| + βct) + ~r⊥. | {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 0 ~ cdt = γ(cdt + β · d~rb) 0 ~ ˆˆ d~r b = d~rb + γβcdt + (γ − 1)ββ · d~rb. Divide the second equation by the first to get

d~r + γβcdt~ + (γ − 1)βˆβˆ · d~r β~0 = b b b ~ γ(cdt + β · d~rb) d~r /cdt + γβ~ + (γ − 1)βˆβˆ · (d~r /cdt) = b b ~ γ[1 + β · (d~rb/cdt)] β~ + γβ~ + (γ − 1)βˆβˆ · β~ = b b . ~ ~ γ(1 + β · βb)

0 0 6.5. Observer O moves relative to observer O with velocity β~1. Observer O moves relative 00 00 to observer O with velocity β~2. Therefore, observer O moves relative to observer O with some velocity β~. Using matrices—as given in Equations (6.14) and (6.15)—find the general 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 Λ1Λ2 is sufficient to yield the answer.

0 0 0 Solution: If (ct, ~r) is the 4-vector in O, then Λ1(ct, ~r) is (ct ~r ), the 4-vector in O and 00 00 00 Λ2Λ1(ct, ~r) is (ct , ~r ), the 4-vector in O . Therefore, if Λ1 is parametrized by β~1 and Λ2 is parametrized by β~2, then Λ = Λ2Λ1 is parametrized by β~. This means that         ~ ~ ~ ~e ~ ~e ~e ↔ γ γβe γ2 γ2βe2 γ1 γ1βe1 γ2γ1 + γ2γ1β2β1 γ2γ1β1 + γ2β2 Λ1           =     =   . ~ ↔ ~ ↔ ~ ↔ ~ ↔ ~ ~ ~ ↔ ↔ γβ Λ γ2β2 Λ2 γ1β1 Λ1 γ2γ1β2 + γ1Λ2 β1 γ2γ1β2βe1 + Λ2 Λ1 (6.1) 87

~ ~ ~ ~ Note that β1 and β2 are column vectors and βe1 and βe2 row vectors. So,   β1x ~ ~ ~ ~ βe2β1 = β2x β2y β2z β1y = β2xβ1x + β2yβ1y + β2zβ1z = β2 · β1. β1z Therefore, equating the element in the first row and first column on the left and the right, we get γ = γ1γ2(1 + β~1 · β~2). (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:

     ˆ 2 ˆ ˆ ˆ ˆ    1 0 0 β1x β2x β2xβ2y β2xβ2z β1x ↔ ~ ˆ ˆ ˆ 2 ˆ ˆ Λ2 β1 = 0 1 0 β1y + (γ2 − 1) β2xβ2y β2y β2yβ2z β1y 2 0 0 1 β1z βˆ2xβˆ2z βˆ2yβˆ2z βˆ2z β1z  ˆ ˆ ~    β2xβ2 · β1 βˆ2x ~  ˆ ˆ ~  ~ ˆ ˆ ~ = β1 + (γ2 − 1) β2yβ2 · β1 = β1 + (γ2 − 1) β2y β2 · β1, ˆ βˆ2zβˆ2 · β~1 β2z or ↔ ~ ~ ˆ ˆ ~ Λ2 β1 = β1 + (γ2 − 1)β2β2 · β1. So, the second-row-ﬁrst-column element of the matrix on the right-hand side of (6.1) is

~ ↔ ~ ~ h~ ˆ ˆ ~ i γ2γ1β2 + Λ2 β1 = γ2γ1β2 + γ1 β1 + (γ2 − 1)β2β2 · β1 , and, using (6.2), the corresponding element on the left is

γβ~ = γ1γ2(1 + β~1 · β~2)β.~ Equating the last two equations gives h i γ2γ1β~2 + γ1 β~1 + (γ2 − 1)βˆ2βˆ2 · β~1 β~ = γ1γ2(1 + β~1 · β~2) β~ + γ β~ + (γ − 1)βˆ βˆ · β~ = 1 2 2 2 2 2 1 , 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 s c 2 1 v0 = + v2 1 − . n n2

(b) Show that for v << c, a complete drag—in which the classical LAV holds—yields c nv2 v0 = + . n 2c 88 Relativity in Four Dimensions

(c) Show that relativistically, for low velocity, the drag is not complete but is smaller by v2/(2nc). ~ ~ Solution: In Equation (6.18) of the text, let βb = (c/n)eˆy, and β = βeˆx. (a) Then ~v 0 (1/n)eˆ + γβeˆ 1 = y x = eˆ + βeˆ c γ nγ y x and |~v 0|2 1 1 1 1 = + β2 = (1 − β2) + β2 = + β2 1 − c2 n2γ2 n2 n2 n2 or s c 2 1 |~v 0| = + v2 1 − . n n2

(b) From the classical LAV, we get r c c2 c n2v2 ~v 0 = eˆ + veˆ ⇐⇒ |~v 0|2 = + v2 ⇐⇒ |~v 0| = 1 + . n y x n2 n c2 Approximating the square root by r n2v2 n2v2 1 + ≈ 1 + c2 2c2 because v << c, we get c nv2 |~v 0| = + . n 2c (c) Write the relativistic result in (a) as s c n2v2 1 |~v 0| = 1 + 1 − . n c2 n2

For v << c, this can be approximated as s c n2v2 1 c n2v2 1 |~v 0| = 1 + 1 − ≈ 1 + 1 − . n c2 n2 n 2c2 n2 or c nv2 1 c nv2 v2 |~v 0| ≈ + 1 − = + − . n 2c n2 n 2c 2nc Comparing this with the result in (b), we see that the drag is smaller by v2/(2nc).

6.7. Derive the velocity (6.23) directly from (6.18). ~ ~ Solution: In (6.18), let βb = αeˆy and β = βeˆx. Equation (6.23) follows immediately. 89

6.8. Use (6.25) and the trigonometric identity sin ϕ0 tan( 1 ϕ0) = 2 1 + cos ϕ0 to show that Equation (6.26) holds. Solution: sin ϕ ~ sin ϕ0 γ 1 + β cos ϕ sin ϕ tan( 1 ϕ0) = = = 2 0 1 + cos ϕ β~ + cos ϕ γ(1 + β~ cos ϕ + β~ + cos ϕ) 1 + 1 + β~ cos ϕ q v 1 − β~ 2 u ~ sin ϕ sin ϕ u1 − β 1 = = = t tan( 2 ϕ). γ(1 + β~ )(1 + cos ϕ) 1 + β~ 1 + cos ϕ 1 + 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~ , 0 0 0 0 ~ sin ϕ = 1/γ, and tan ϕ = sin ϕ / cos ϕ = 1/( β γ). 6.10. Use Figure 6.1(d) and the law of sines to obtain c cβ = . cos θ0 sin(θ0 − θ) Show that this can also be written as β + sin θ tan θ0 = 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 = . cos θ0 sin(θ0 − θ) Rewrite the equation above as 1 β = ⇐⇒ sin θ0 cos θ − cos θ0 sin θ = β cos θ0. cos θ0 sin θ0 cos θ − cos θ0 sin θ Divide both sides by cos θ0 and solve for tan θ0 to get β + sin θ tan θ0 = . cos θ 90 Relativity in Four Dimensions

c ʹ c θ ʹ θ

c β

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 θ + tan tan θ0 = tan(θ + ) = , 1 − tan θ tan which, to first order in gives tan θ + tan θ0 ≈ = (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 With γ ≈ 1, the right-hand side of (6.28) becomes tan θ + |β~|/ cos θ. Thus equating the two sides, we get |β~| tan θ + = tan θ + ⇐⇒ = |β~| cos θ, cos2 θ cos θ 0 ~ and θ = θ + = θ + |β| cos θ.

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 d d 1 = 2 . c + v c − v

From this plus d1 + d2 = L ﬁnd d1 and show that 1 x0 = x0 + L(1 + β), R 1 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, d d c − v 1 − β 1 = 2 ⇐⇒ d = d = d , c + v c − v 2 c + v 1 1 + β 1 and d1 + d2 = L yields 1 − β 2d L = d + d = 1 ⇐⇒ d = 1 L(1 + β). 1 1 + β 1 1 + β 1 2 0 0 The ﬁnal answer is obtained once we note that d1 = x1 − xR. 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 = ~α0 · ~α0 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 ~α0 · ~α0 = 1 as well. Solution: (a) Keeping only ﬁrst order terms in Equation (6.41), we get γ ≈ 1 and β~ · ~α ≈ 0, and the equation reduces to ~α0 = ~α + β~, which is the classical LAV.

When all velocities are in the same direction, βˆβˆ · ~α = ~α, and β~ · ~α = β~ ~α . Then Equation (6.41) becomes ~ ~ 0 γ(~α + β) ~α + β ~α = = , γ(1 + β~ ~α ) 1 + β~ ~α which is Equation (3.26) if you ignore the vector signs. (b) The resulting denominator is simply γ2(1+β~ ·~α)2. So, let’s manipulate the numerator Num = [~α + γβ~ + (γ − 1)βˆβˆ · ~α] · [~α + γβ~ + (γ − 1)βˆβˆ · ~α] 2 2 2 2 2 2 = ~α + γ β~ + (γ − 1) (βˆ · ~α) + 2γβ~ · ~α + 2(γ − 1)(βˆ · ~α) + 2γ(γ − 1)β~ · ~α. The sum of the third and ﬁfth terms can be simpliﬁed: sum 3rd+5th = (γ − 1)[(γ − 1)(βˆ · ~α)2 + 2(βˆ · ~α)2] 2 2 2 2 2 2 2 = (γ − 1)(βˆ · ~α) = γ β~ (βˆ · ~α) = γ (β~ · ~α) . So the numerator now becomes 2 2 2 2 2 2 Num = ~α + γ β~ + γ (β~ · ~α) + 2γ β~ · ~α 2 2 2 2 2 = ~α + γ − 1 + γ (β~ · ~α) + 2γ β~ · ~α 2 2 2 = ~α − 1 + γ [1 + (β~ · ~α) + 2β~ · ~α], and dividing it by the denominator leads to 2 ~α − 1 1 |~α0|2 = + 1 = − + 1. 2 ~ 2 2 2 ~ 2 γ (1 + β · ~α) γαγ (1 + β · ~α) 92 Relativity in Four Dimensions

1 1 ~ = ⇐⇒ γα0 = γαγ(1 + β · ~α). 2 2 2 ~ 2 γα0 γαγ (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) = 0 0 ~ ~ (1, 0, 0, 0). Substituting this in Equation (6.45) yields u0 = γ ≡ γβ and ~u = γβ ≡ γββ. 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 E0 = E. 1 + β

Now use E = hc/λ to ﬁnd a formula for the relativistic Doppler shift.

Solution: If ~p is parallel to β~, then ~p · β~ = ~p β~ . Therefore, (6.53) becomes s 0 1 − β E = γ(E − c ~p β~ ) = γ(E − E β~ ) = E, 1 + β

using the notation β ≡ β~ . Writing the last equation in terms of wavelengths, we get s s hc 1 − β hc 1 + β = ⇐⇒ λ0 = λ. λ0 1 + β λ 1 − β

This is the formula for the relativistic Doppler shift. 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:

~p 0 ~p + γβE~ + (γ − 1)βˆβˆ · ~p ~α 0 ≡ = . E0 γ(E + β~ · ~p)

Now divide the numerator and denominator by E:

~α + γβ~ + (γ − 1)βˆβˆ · ~α ~α 0 = . 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 u0a0 − ~u · ~a = 0. Solution: Use dot to denote diﬀerentiation with respect to t: A˙ ≡ dA/dt.

d~u d~u d ~a = = γ = γ (γ ~α) = γ ~αγ˙ + γ ~α˙ . dτ α dt α dt α α α α

3 ˙ Now useγ ˙α = γα~α· ~α obtained in the calculation of a0 just before Equation (6.56) to obtain

h 3 ˙ ˙ i 2 h 2 ˙ ˙ i ~a = γα ~α γα~α · ~α + γα~α = γα γα(~α · ~α)~α + ~α .

From (6.42), we have

h 4 ˙ 4 ˙ 2 ˙ i u0a0 − ~u · ~a = γα(a0 − ~α · ~a) = γα γα~α · ~α − γα(~α · ~α)~α · ~α − γα~α · ~α 3 ˙ 2 2 = γα~α · ~α γα − γα~α · ~α − 1 = 0

2 2 because γα~α · ~α = γα − 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

˙ 3 ˙ ˙ ˙ 2 ˙ 3 ˙ ~α · ~u = γα(~α · ~α)~α · ~α + γα~α · ~α = γα(~α · ~α) γα~α · ~α +γα~α · ~α = γα~α · ~α =u ˙ 0. | {z } 2 =γα−1

6.21. Derive Equation (6.60). 94 Relativity in Four Dimensions

Solution: Using (6.59) and (6.46), we can write

~a 0 ~a + γβa~ + (γ − 1)βˆβˆ · ~a ~u˙ 0 = = 0 . γα0 γγα(1 + β~ · ~α) ˙ Now write a0 = γαu˙ and ~a = γα~u—coming from (6.57)—to rewrite the above as

γ ~u˙ + γ γβ~u˙ + γ (γ − 1)βˆβˆ · ~u˙ ~u˙ 0 = α α 0 α γγα(1 + β~ · ~α) ~u˙ + γβ~α~ · ~u˙ + (γ − 1)βˆβˆ · ~u˙ = , γ(1 + β~ · ~α)

where I used (6.58) foru ˙ 0. 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, 1 − 0.999992 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 E2 = ~p · ~p + m2 with respect to time and use ~p = E~α to obtain (6.65).

Solution: d d dE d~p dE (E2) = (~p · ~p + m2) ⇐⇒ 2E = 2~p · ⇐⇒ = (~p/E) · F.~ dt dt dt dt 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~ + F~ + γβ~ ~α · (F~ + F~ ) + (γ − 1)βˆ βˆ · (F~ + F~ ) ~ 0 ~ 0 || ⊥ || ⊥ || ⊥ F|| + F⊥ = . γ(1 + β~ · ~α)

ˆ ~ ˆ ˆ ~ ~ Now note that β · F⊥ = 0 and β β · F|| = F||. These give us a new equation:

F~ + γβ~ ~α · F~ + γβ~ ~α · F~ + γF~ ~ 0 ~ 0 ⊥ || ⊥ || F|| + F⊥ = γ(1 + β~ · ~α) ~ ~ F~⊥ + γF~ β · ~α + γβ ~α · F~⊥ + γF~ = || || . γ(1 + β~ · ~α)

Notice the change in the second term of the numerator. We can now ﬁnally write this as

~ ~ ~ ~ ~ ~ ~ 0 ~ 0 ~ F⊥ + γβ ~α · F⊥ ~ β ~α · F⊥ F⊥ F|| + F⊥ = 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 E2 = 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 E2 = 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

γ(dp + βdE) dp/dt + βdE/dt F + βαF F 0 = dp0/dt0 = = = = 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 P + β~ · F~ P 0 = , 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 dE0 and dt0, and dividing the first by the second; (b) by noting that P 0 = ~α0 · 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 O moves with velocity β~, we have dE0 = γ(dE + β~ · d~p), dt0 = γ(dt + β~ · d~r). Divide the first by the second: dE0 γ(dE + β~ · d~p) dE/dt + β~ · (d~p/dt) P + β~ · F~ P 0 = = = = . dt0 γ(dt + β~ · d~r) 1 + β~ · (d~r/dt) 1 + β~ · ~α

(b) Use (6.41) and (6.66) in P 0 = ~α0 · F~ 0 to obtain " # " # ~α + γβ~ + (γ − 1)βˆβˆ · ~α F~ + γβ~ ~α · F~ + (γ − 1)βˆ βˆ · F~ P 0 = · . γ(1 + β~ · ~α) γ(1 + β~ · ~α)

The denominator of the right-hand side is simply γ2(1 + β~ · ~α)2. So, let’s calculate the numerator: Num = ~α · F~ + γ~α · β~ ~α · F~ + 2(γ − 1)βˆ · ~αβˆ · F~ + γβ~ · F~ + γ2β~ · β~ ~α · F~ | {z } =γ2−1 + γ(γ − 1)β~ · F~ + γ(γ − 1)~α · β~α~ · F~ + (γ − 1)2βˆ · ~αβˆ · F.~ The sum of the third and last term is 3rd+last = 2(γ − 1)βˆ · ~αβˆ · F~ + γβ~ · F~ + (γ − 1)2βˆ · ~αβˆ · F~ = (γ − 1)(2 + γ − 1)βˆ · ~αβˆ · F~ = (γ2 − 1)βˆ · ~αβˆ · F~ = γ2β2βˆ · ~αβˆ · F~ = γ2β~ · ~αβ~ · F.~ Substituting this in the previous expression and simplifying yields Num = γ2~α · F~ + γ2~α · β~ β~ · F~ + γ2β~ · F~ + γ2~α · β~α~ · F~ = γ2(1 + ~α · β~)(~α · F~ + β~ · F~ ) = γ2(1 + ~α · β~)(P + β~ · F~ ). Dividing this by the denominator gives the result. 97

0 0 (c) By (6.64), f0 = γα0 P . Lorentz transforming this gives 0 ~ ~ γα0 P = γ(f0 + β · f) = γ[γαP + β · (γαF )]

or γγ (P + β · F~ ) P 0 = α . γα0 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

γ2β2 γ2 − 1 (γ − 1)(γ + 1) 1 + = 1 + = 1 + = γ. γ + 1 γ + 1 γ + 1

The ﬁrst line of (6.72) comes from the identity

γ γ(γ + 1) − γ2 = , γ + 1 γ + 1

~ ˆ ~ ~ ~ 2 ˆ ˆ ~ and the second line from β = ββ and therefore, β × (β × F ) = β β × (β × F ).

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 0 1 relative to observer O . Deﬁne the angle θ as sin θ ≡ β. It is argued that—since AbAf is perpendicular to the direction of motion and thus does not change length—it takes light 0 2a/c to travel from Ab to Af according to O . 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 − β2 = 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.

0 0 (c) What is the time diﬀerence tf − tb 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 0 0 hypotenuse of a right triangle whose other two sides are of lengths 2a and β(tf − tb).

1See, for example, Weisskopf, V. F., Phys. Today 13(9), 24–27 (1960). 100 Relativistic Photography

Db y Af Df Ab Ab Af Df Ab Db

Df Af

x θ Df Cb 2a Af Bf Cf Bf C θ Bb f Cf s Bf d s d € (a) (b) (c) (d)

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 0 0 tb = γ(tb − βa), xb = γ(−a + βtb),

and for Ef , we get

0 0 tf = γ(tb + 2a − βa), xf = γ[−a + β(tb + 2a)].

0 0 0 0 0 0 (c) tf − tb = 2aγ and xf − xb = 2aγβ = β(tf − tb). These results do not agree with the argument for rotation!

0 0 (d) The hypotenuse of the right triangle is the path of light. So, it has length tf − tb. Hence,

0 0 2 2 2 0 0 2 0 0 2 2 2 0 0 (tf − tb) = 4a + β (tf − tb) ⇐⇒ (tf − tb) (1 − β ) = 4a ⇐⇒ tf − tb = 2aγ.

This is precisely the kind of reasoning that gave us the time dilation formula (2.3) 0 0 from Figure 2.2. The distance xf − xb is just the speed times this time interval.

0 7.2. Use the space coordinates of (7.7) to calculate |ri −r0| and show that the result agrees 0 with the right-hand side of ti. 0 0 0 0 Solution: With ri − r0 = hxi, yi, zi − bi, we have

0 2 0 2 0 2 0 2 2 2 2 2 |ri − r0| = xi + yi + (zi − b) = γ (xi − β|ri − r0|) + yi + (zi − b) 2 2 2 2 2 2 2 2 = γ xi + γ β |ri − r0| − 2γ βxi|ri − r0| + yi + (zi − b) | {z } 2 2 =|ri−r0| −xi 2 2 2 2 2 2 = (γ − 1)xi + (γ − 1)|ri − r0| − 2γ βxi|ri − r0| + |ri − r0| , 101 where I used the ever useful formula: γ2β2 = γ2 − 1. Now use it backwards in the ﬁrst term and cancel terms to obtain

0 2 2 2 2 2 2 2 |ri − r0| = γ β xi + γ |ri − r0| − 2γ βxi|ri 2 2 = γ (−βxi + |ri − r0|) ,

0 or |ri −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):

2 Den = γ (|r1 − r0| − βx1)(|r2 − r0| − βx2) 2 βx1 βx2 = γ |r1 − r0||r2 − r0| 1 − 1 − . |r1 − r0| |r2 − r0| The numerator can be expressed as

2 Num = γ (x1 − β|r1 − r0|)(x2 − β|r2 − r0|) + y1y2 + (z1 − b)(z2 − b) 2 = γ (x1 − β|r1 − r0|)(x2 − β|r2 − r0|) + |r1 − r0||r2 − r0| cos α12 − x1x2 2 2 2 = (γ − 1)x1x2 + γ β |r1 − r0||r2 − r0| 2 2 − γ x1β|r2 − r0| − γ x2β|r1 − r0| + |r1 − r0||r2 − r0| cos α12.

Invoking the identity γ2β2 = γ2 − 1 in the ﬁrst and the second terms and simplifying, you get

2 2 2 Num = γ β x1x2 + γ |r1 − r0||r2 − r0| − |r1 − r0||r2 − r0| 2 2 − γ x1β|r2 − r0| − γ x2β|r1 − r0| + |r1 − r0||r2 − r0| cos α12 2 = γ (|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 1 − cos α12 cos α12 = 1 − . γ2 1 − βx1 1 − βx2 |r1−r0| |r2−r0|

2 Now use the trigonometric identity 1 − cos θ = 2 sin (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 02 2 q 02 2 y0 xq + b y0 xq + b 0 0 0 b(z0 − b) xqx − b(z − b) = 0 ⇐⇒ x = 0 0 + 0 v xqv xq 102 Relativistic Photography

and substitute both in the ﬁrst equation, you obtain

0 0 0 v 0 0 b 0 0 xq(z0 − b) 0 u = [bx + xq(z0 − b)] = v x + v q 02 2 q 02 2 q 02 2 y0 xq + b y0 xq + b y0 xq + b

 q 02 2  y0 xq + b 0 b 0 b(z0 − b) xq(z0 − b) 0 = q v  0 0 + 0  + q v 02 2 xqv xq 02 2 y0 xq + b y0 xq + b or q 0 2 2 0 2 2 (z0 − b) xq + b 0 0 0 (z0 − b)(xq + b ) 0 xqu = b + v q = b + v , 02 2 y0 y0 xq + b and 0 y0x by v0 = q u0 − 0 . q 0 2 2 q 0 2 2 (z0 − b) xq + b (z0 − b) xq + b

0 Equation (7.14) is obtained by noting that xq = 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

h p 2 2 2i γ x0 − β x0 + y + (z0 − b) u0 = − z0 − b y v0 = − , z0 − b or

γx γβpx2 + y2 + (z − b)2 u0 + 0 = 0 0 z0 − b z0 − b y v0 = − , z0 − b or

2 2 2 2 2 2 0 γx0 γ β [x0 + y + (z0 − b) ] u + = 2 z0 − b (z0 − b) 2 0 2 2 y = v (z0 − b) .

Substituting the second equation in the ﬁrst yields

2 2 2 2 2 0 γx0 2 2 0 2 γ β [x0 + (z0 − b) ] u + = γ β v + 2 . z0 − b (z0 − b)

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 1 γβ γβ cos θ = √ = = = β. 1 + tan2 θ pγ2β2 + 1 pγ2 − 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:

h p 2 2 0 2i γ x0 − β x + y + (y0/v ) 0 0 0 u = 0 y0/v or q 0 0 0 2 2 2 2 y0u = γ x0v − β v (x0 + y0) + y0 or q 0 0 0 2 2 2 2 y0u − γx0v = −γβ v (x0 + y0) + y0. Squaring both sides and simplifying yields (7.23). 7.9. Deﬁne a new pair of photographic plate coordinates by

0 0 0 0 0 0 u = unew cos θ + vnew sin θ, v = −unew sin θ + vnew cos θ with x γy sin θ = 0 , cos θ = 0 . p 2 2 2 p 2 2 2 γ y0 + x0 γ y0 + x0 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 2 0 2 2 2 2 0 2 2 2 2 (x0 + y0)unew − β γ y0vnew − γ β y0 = 0, which is the equation of a hyperbola.

(b) What are the coordinates of the center of this hyperbola in the new coordinate system?

Solution:

(a) Substitute the new pair of photographic plate coordinates in (7.23) and look at each resulting term separately:

2 0 2 2 0 2 2 0 0 1st term = y0(unew cos θ + vnew sin θ + 2unewvnew sin θ cos θ) 2 2 2 2 0 2 2 0 2 2 0 0 2nd term = x0 − γ β y0 (unew sin θ + vnew cos θ − 2unewvnew sin θ cos θ) 0 2 0 0 2 2 0 2 3rd term = −2γx0y0[−unew sin θ cos θ + unewvnew(cos θ − sin θ) + vnew sin θ cos θ]. 104 Relativistic Photography

First consider the new cross term, which can be written as

0 0 2 2 2 2 2 2 2 cross term = 2unewvnew y0 − x0 + γ β y0 sin θ cos θ − γx0y0(cos θ − sin θ) . Let X denote the expression inside the square bracket and use γ2β2 = γ2 − 1 in its ﬁrst term. Then, substituting x γy sin θ = 0 , cos θ = 0 p 2 2 2 p 2 2 2 γ y0 + x0 γ y0 + x0 in that expression yields

γx y γ2y2 − x2 X 2 2 2 0 0 0 0 = γ y0 − x0 2 2 2 − γx0y0 2 2 2 = 0. γ y0 + x0 γ y0 + x0 0 2 The coeﬃcient that multiplies unew is 2 2 2 2 2 2 2 y0 cos θ + x0 − γ β y0 sin θ + 2γx0y0 sin θ cos θ or 2 4 2 2 2 2 γ y0 2 2 2 2 x0 2γ x0y0 2 2 2 + x0 − γ β y0 2 2 2 + 2 2 2 γ y0 + x0 γ y0 + x0 γ y0 + x0 or =1+1/γ2

2 4 4 2 2 2 z }| 2{ 2 4 4 2 2 2 2 2 γ y0 + x0 + γ x0y0 (2 − β ) γ y0 + x0 + γ x0y0 + x0y0 2 2 2 2 2 = 2 2 2 = x0 + y0. γ y0 + x0 γ y0 + x0 0 2 2 2 2 Similarly, you can show that the coeﬃcient of vnew is −β γ y0. (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

2 2 0 2 0 2 (x0 + y0)unew 0 2 unew 0 2 2 2 2 − vnew = 1 ⇐⇒ 2 2 2 2 2 − vnew = 1. γ β y0 γ β y0/(x0 + y0)

p 2 2 This shows that the center is at the origin, the semi-major axis is a = γβy0/ x0 + y0, and the semi-minor axis is b = 1.

p 2 2 0 x0 + y0 0 vnew = ± unew. γβy0 0 Let α be the angle that the line with positive slope makes with the unew axis. Then px2 + y2 1 tan α = 0 0 ⇐⇒ 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 + v0 cos θ = ± tan α(u0 cos θ − v0 sin θ). or cos α(u0 sin θ + v0 cos θ) = ± sin α(u0 cos θ − v0 sin θ). 105

or v0(cos α cos θ ∓ sin α sin θ) = u0(± sin α cos θ − cos α sin θ). or cos(α ± θ)v0 = ∓u0 sin(α ± θ) ⇐⇒ v0 = ∓ 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 = a and their x and y coordinates are ±a. Therefore, their distance from the pinhole is p2a2 + (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

x0 y u0 = − , v0 = − . z − b 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 u0s and v0s is b − a. Similarly, the denominator for all u0s and v0s is b + a for the back corners. To get (7.28) substitute the x0s from (7.26), and to get (7.29), substitute for y. 7.12. Using Equation (7.28), show that u0 > u0 and v0 < v0 for all b and β. Hint: Ab Af Ab Af Show that each of the two terms on the right-hand side of u0 is larger than the corre- Ab sponding term on the right-hand side of u0 . Af Solution: Write u0 and u0 as Ab Af s γa 2a2 u0 = − − γβ + 1 Ab b + a (b + a)2 s γa 2a2 u0 = − − γβ + 1. Af b − a (b − a)2

It is now clear that the magnitude of each term in u0 is smaller than the magnitude of the Ab corresponding term in u0 . Therefore, u0 > u0 . The inequality v0 < v0 is self-evident. Af Ab Af Ab Af 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 = 0 cos ϕ , cos α12 = cos ϕ,e ˆ1 = −βˆ, and β~ · eˆ2 = −β cos ϕ, and (7.5) can be written as 1 − cos ϕ (1 − β)(1 − cos ϕ) cos ϕ0 = 1 − = 1 − γ2(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

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 sin(α0 /2) = , Df Db p 2 2 γ (xc + a) + 2a − β(xc + a) a sin(α0 /2) = , (7.1) Af Ab p 2 2 γ (xc − a) + 2a − β(xc − a)

the angles formed by sides Df Db and Af Ab (and the other three sides of the leading and trailing faces) at the pinhole.

(c) For x < 0 and β > 0, show that α0 > α0 , with the other three angles c Df Db Af Ab satisfying the same kind of inequality. Thus, the trailing face is hidden behind the leading face when the cube is approaching the camera.

(d) For x < 0 and β < 0, show that α0 > α0 if and only if |x |/a > |β|p2γ2 + 1. c Df Db Af Ab c (e) Set β = 0 in (7.1) above to ﬁnd sin(α /2). Then show that, for approach, α0 < Df Db Df 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 C0 image is larger. For the moment assume that and see what condition makes the assumption true. For deﬁniteness, let β > 0 and xc > 0. Then, for α0 > α to hold, you should have Af Df Af Df

p 2 2 p 2 2 γ (xc + a) + 2a − β(xc + a) < (xc + a) + 2a .

Show that this is equivalent to

2 2 [(xc + a) − X−][(xc + a) − X+] > 0,X− < 0 107

2 for certain X+ and X− that you have to ﬁnd. Thus, for the inequality to hold, (xc+a) must be larger than X+. Now show that (x + a)2 c > γ − 1. (7.2) a2 When the cube is suﬃciently far away, this condition holds, but not in general. There- fore, if the cube is close enough to the camera C0 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 sin(α0 /2) Df 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 C0 relative to the image in C.

(i) Verify directly that, when |xc| → ∞, the ratio in (h) reduces to

0 s αD D 1 + β|x |/x f b ≈ c c , 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

Leading face Df :(xc + a, −a, a),Db :(xc + a, a, a),

Cf :(xc + a, −a, −a),Cb :(xc + a, a, −a)

Trailing face Af :(xc − a, −a, a),Ab :(xc − a, a, a),

Bf :(xc − a, −a, −a),Bb :(xc − a, −a, −a)

(b) Substituting from (a) in (7.8) with b = 0, we get

2 xDf xDb + yDf yDb + zDf zDb (xc + a) cos αDf Db = = 2 2 |rDf | |rDb | (xc + a) + 2a and 2 2 2 (xc + a) 2a 2 sin (αDf Db /2) = 1 − cos αDf Db = 1 − 2 2 = 2 2 , (xc + a) + 2a (xc + a) + 2a or a sin(αD D /2) = . f b p 2 2 (xc + a) + 2a Substitute this in (7.10) with b = 0 to obtain √ a 2 2 0 (xc+a) +2a a sin(αD D /2) = = . f b p 2 2 γ 1 − √ β(xc+a) γ (xc + a) + 2a − β(xc + a) 2 2 (xc+a) +2a sin(α0 /2) can be obtained in exactly the same way. Af Ab 108 Relativistic Photography

(c) For xc < 0 and β > 0, (7.1) becomes a sin(α0 /2) = , Df Db p 2 2 γ (|xc| − a) + 2a + β(|xc| − a) a sin(α0 /2) = . Af Ab p 2 2 γ (|xc| + a) + 2a + β(|xc| + a) The denominator of the ﬁrst is smaller than the denominator of the second. Therefore,

sin(α0 /2) > sin(α0 /2) ⇐⇒ α0 > α0 . Df Db Af Ab Df Db Af Ab

(d) For xc < 0 and β < 0, (7.1) becomes a sin(α0 /2) = , Df Db p 2 2 γ (|xc| − a) + 2a − |β|(|xc| − a) a sin(α0 /2) = , Af Ab p 2 2 γ (|xc| + a) + 2a − |β|(|xc| + a) and α0 > α0 if and only if Df Db Af Ab

p 2 2 p 2 2 (|xc| − a) + 2a − |β|(|xc| − a) < (|xc| + a) + 2a − |β|(|xc| + a)

or p 2 2 p 2 2 (|xc| − a) + 2a + 2|β|a < (|xc| + a) + 2a . 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

2 2 2 2 p 2 2 2 2 (|xc| − a) + 2a + 4β a + 4a|β| (|xc| − a) + 2a < (|xc| + a) + 2a .

This can be simpliﬁed to

2 p 2 2 β a + |β| (|xc| − a) + 2a < |xc|

or p 2 2 2 |β| (|xc| − a) + 2a < |xc| − β a.

Both sides are still positive because |xc| > a. (Otherwise the camera will be inside the cube!) So, I square both sides and get

2 2 2 2 2 2 β (|xc| − a) + 2β a < (|xc| − β a) ,

which can be simpliﬁed to

2 2 2 2 2 2 2 2 2 2 2 2 |xc| > γ β a (3 − β ) = γ β a (2 + 1/γ ) = β a (2γ + 1).

(e) With β = 0, we have a sin(αD D /2) = . f b p 2 2 (xc + a) + 2a

On approach, xc < 0 and β > 0, therefore a sin(αD D /2) = . f b p 2 2 (|xc| − a) + 2a 109

and a sin(α0 /2) = . Df Db p 2 2 γ (|xc| − a) + 2a + β(|xc| − a) It is clear that the denominator of sin(α0 /2) is larger than that of sin(α /2). Df Db Df Db Hence, α0 < α . Df Db Df Db (f) Note that both sides of the inequality are positive. Therefore, I can square both sides

2p 2 2 2 2 2 γ (xc + a) + 2a − β(xc + a) < (xc + a) + 2a .

2 2 2 2 2 p 2 2 2 2 γ (xc + a) + 2a + β (xc + a) − 2β(xc + a) (xc + a) + 2a < (xc + a) + 2a .

Using γ2 − 1 = γ2β2, this inequality simpliﬁes to

2 2 2 2 2 2 2 p 2 2 2γ β (xc + a) + 2γ β a < 2γ β(xc + a) (xc + a) + 2a ,

or 2 2 p 2 2 β(xc + a) + βa < (xc + a) (xc + a) + 2a Square both sides to get

2 4 2 4 2 2 2 4 2 2 β (xc + a) + β a + 2a β (xc + a) < (xc + a) + 2a (xc + a)

or 4 2 2 2 2 4 (xc + a) + 2a (xc + a) − β γ a > 0. Find the two roots of the left-hand side by setting it equal to zero and solving the quadratic equation:

2 2 p 4 2 2 4 2 2 (xc + a) = −a ± a + β γ a = −a ± γa .

Then the preceding inequality can be expressed as

2 2 2 2 [(xc + a) + (γ + 1)a ][(xc + a) − (γ − 1)a ] > 0.

The factor on the left is positive, so for the inequality to hold, we must have

2 2 (xc + a) > (γ − 1)a .

(g) If β < 0 and xc < 0, then the starting inequality in (f) becomes

p 2 2 p 2 2 γ (|xc| − a) + 2a − |β|(|xc| − a) < (|xc| − a) + 2a .

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:

2 2 2 2 2 2 (|xc|−a) > (γ −1)(−a) ⇐⇒ (−xc −a) > (γ −1)(−a) ⇐⇒ (xc +a) > (γ −1)a

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 1.0

0 Figure 7.2: The plot of the ratio sin(αDf Db /2)/ sin(αDf Db /2) when xc = 3.

(h) From the solution to part (b), we have

0 p sin(αD D /2) (x + a)2 + 2a2 f b = c . sin(α /2) p 2 2 Df Db γ (xc + a) + 2a − 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 D |x | 1 f b ≈ c = αDf Db γ (|xc| − βxc) γ (1 − βxc/|xc|) s p1 − (βx /|x |)2 1 + βx /|x | = c c = c c . (1 − βxc/|xc|) 1 − βxc/|xc|

2 2 Note that β = (βxc/|xc|) , because xc/|xc| = ±1.

7.15. Derive Equations (7.33) and (7.34).

2 Solution: With r = hx, y, a /bi = hx, y, a sin αi, r0 = h0, 0, bi, and referring to Figure 7.6 of the book, we obtain

2 2 2 2 2 2 2 2 |r0 − r| = x + y + (b − a sin α) = R + b (1 − sin α) = 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

~ ~ ~ 0 β + cos ϕ β − x/(b cos α) β b cos α − x 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 sin ϕ0 = = = . γ 1 + β~ cos ϕ γ 1 − β~ x/(b cos α) γ b cos α − β~ x

7.16. Show that if vˆ of Equation (7.35) is to make an angle ϕ0 with the x-axis, then

2 ηxb cos α 0 ηz = ± tan ϕ . py2 + (b cos2 α)2 Solution: From 2 0 ηxb cos α cos ϕ = vˆ · eˆx = , p 2 2 2 2 2 2 2 ηx(b cos α) + ηz [y + (b cos α) ] we obtain 2 2 2 2 0 ηx(b cos α) cos ϕ = 2 2 2 2 2 2 2 , ηx(b cos α) + ηz [y + (b cos α) ] or 2 2 2 2 0 2 2 2 2 2 0 2 2 2 ηx(b cos α) cos ϕ + ηz [y + (b cos α) ] cos ϕ = ηx(b cos α) or 2 2 2 2 2 0 2 2 2 2 0 ηz [y + (b cos α) ] cos ϕ = ηx(b cos α) sin ϕ or ﬁnally η2(b cos2 α)2 sin2 ϕ0 η2(b cos2 α)2 η2 = x = x tan2 ϕ0. z [y2 + (b cos2 α)2] cos2 ϕ0 y2 + (b cos2 α)2 7.17. With eˆ(φ, β) deﬁned as in Equation (7.40), show that

sin2 α(1 − cos φ) eˆ(φ, β) · eˆ(0, β) = 1 − . γ2(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 eˆ(φ, β) deﬁned as in Equation (7.40), we get * + β~ − sin α cos α eˆ(0, β) = − , 0, − , 1 − β~ sin α γ(1 − β~ sin α) and the dot product becomes ! ! β~ − sin α cos φ β~ − sin α eˆ(φ, β) · eˆ(0, β) = 1 − β~ sin α cos φ 1 − β~ sin α cos2 α + γ2(1 − β~ sin α cos φ)(1 − β~ sin α) 112 Relativistic Photography

γ2( β~ − sin α cos φ)( β~ − sin α) + cos2 α eˆ(φ, β) · eˆ(0, β) = . γ2(1 − β~ sin α cos φ)(1 − β~ sin α) Let’s manipulate the numerator

2 2 2 Num = γ β − γ ( β~ sin α + β~ sin α cos φ) + [(γ2 − 1) sin2 α cos φ + sin2 α cos φ] + cos2 α 2 2 = γ − 1 − γ ( β~ sin α + β~ sin α cos φ) + [γ2β2 sin2 α cos φ + sin2 α cos φ] + cos2 α 2 2 = γ (1 − β~ sin α cos φ)(1 − β~ sin α) − sin α(1 − cos φ).

Dividing this by the denominator, gives the final result. For the purposes of finding the extrema, we just have to differentiate the function 1 − cos φ f(φ) ≡ , 1 − β~ sin α cos φ whose derivative is ~ ~ 0 sin φ(1 − β sin α cos φ) − (1 − cos φ) β sin α sin φ f (φ) = (1 − β~ sin α cos φ)2

(1 − β~ sin α) sin φ = . (1 − β~ sin α cos φ)2

0 So, f (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 eˆ(0, β) = − , 0, − 1 − β~ sin α γ(1 − β~ sin α) * + β~ + sin α cos α eˆ(π, β) = − , 0, − . 1 + β~ sin α γ(1 + β~ sin α) Therefore, 2 cos α D E eˆ + eˆ = − β~ cos α, 0, −1/γ . 0 π 2 1 − β~ sin2 α

Although eâxis is defined as (eˆ0 +eˆπ)/|eˆ0 +eˆπ|, the constant multiplying the angle brackets cancels in the ratio. So, you can define eâxis as the ratio in (7.41). 7.19. Derive Equation (7.42).

Solution: From (7.40) and (7.41), we obtain ! 1 ( β~ − sin α cos φ) β~ cos α cos α eˆ(φ, β) · eˆaxis = q + . 2 2 1 − β~ sin α cos φ γ2(1 − β~ sin α cos φ) 1 − β~ sin α 113

The expression in the big parentheses can be written as

γ2( β~ − sin α cos φ) β~ cos α + cos α , γ2(1 − β~ sin α cos φ) whose numerator is 2 2 2 γ β cos α − γ β~ sin α cos φ cos α + cos α |{z} =γ2−1 or 2 γ cos α(1 − β~ sin α cos φ). Dividing this by the denominator, you get cos α. Therefore, cos α eˆ(φ, β) · eˆaxis = . q 2 1 − β~ sin2 α

7.20. Derive Equation (7.46).

Solution: Use determinant to ﬁnd the cross product.   eˆx eˆy eˆz 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 2 2 2 2 2 2 2 = a sin θ cos ϕeˆx + a sin θ sin ϕeˆy + a (sin θ cos θ cos ϕ + sin θ cos θ sin ϕ)eˆz 2 = a sin θ(sin θ cos ϕeˆx + sin θ sin ϕeˆy + cos θeˆz)

7.21. Using (7.45) and (7.7), derive Equation (7.49).

Solution: With r as given in (7.45), (7.7) yields

0 x = γ a sin θ cos ϕ − β|r − r0| y0 = y = a sin θ sin ϕ z0 = z = a cos θ.

The only thing left to do is to calculate |r − r0|:

2 2 2 2 2 2 2 2 2 2 |r − r0| = x + y + (z − b) = x + y + z +b − 2zb = a + b − 2ab cos θ. | {z } =a2 or p 2 2 |r − r0| = a + b − 2ab cos θ.

7.22. Derive Equations (7.50) and (7.51). 114 Relativistic Photography

Solution: With ∂x0 ∂y0 ∂z0 ∂x0 ∂y0 ∂z0 θ~ 0 ≡ , , , ~ϕ 0 ≡ , , , ∂θ ∂θ ∂θ ∂ϕ ∂ϕ ∂ϕ ~ 0 the only non-obvious component is θ x. So, let’s calculate it: ∂x0 ∂ p = 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:   eˆx eˆy eˆz θ~ 0 × ~ϕ 0 = a2 det γ cos θ cos ϕ − √ βγb sin θ cos θ sin ϕ − sin θ  a2+b2−2ab cos θ  −γ sin θ sin ϕ sin θ cos ϕ 0 2 2 2 2 βγb sin θ cos ϕ = a sin θ cos ϕeˆx + γ sin θ sin ϕeˆy + γ sin θ cos θ − √ eˆz a2 + b2 − 2ab cos θ 2 βγb sin θ cos ϕ = a sin θ sin θ cos ϕeˆx + γ sin θ sin ϕeˆy + γ cos θ − √ eˆz . 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

0 0 0 0 0 0 0 (θ~ × ~ϕ )xx + (θ~ × ~ϕ )yy + (θ~ × ~ϕ )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, y0, z0) of Equation (7.53) satisfy √ p b2 − a2 x02 + y02 + (z0 − b)2 = γ(b − βa cos ϕ). b 115

Solution: a2 |r0 − r |2 ≡ x02 + y02 + (z0 − b)2 = γ2a2 1 − cos2 ϕ + γ2β2(b2 − a2) 0 b2 b2 − a2 a2 (a2 − b2)2 − 2γ2βa cos ϕ + a2 1 − sin2 ϕ + b b2 b2 a2 = (γ2 − 1)a2 1 − cos2 ϕ + (γ2 − 1)(b2 − a2) b2 b2 − a2 a2 (a2 − b2)2 − 2γ2βa cos ϕ + a2 1 − + , b b2 b2 where I wrote sin2 ϕ as 1 − cos2 ϕ and used γ2β2 = γ2 − 1. Let’s continue

b2 − a2 |r0 − r |2 = γ2β2a2 cos2 ϕ + γ2(b2 − a2) − b2 + a2 0 b2 b2 − a2 a2 (a2 − b2)2 − 2γ2βa cos ϕ + a2 1 − + . b b2 b2

The sum of the last two terms is a2 (a2 − b2)2 a2 − b2 a2 1 − + = (a2 − b2 − a2) = b2 − a2, b2 b2 b2 which cancels the last two terms on the ﬁrst line of the previous equation. So, we now have

b2 − a2 b2 − a2 |r0 − r |2 = γ2β2a2 cos2 ϕ + γ2(b2 − a2) − 2γ2βa cos ϕ 0 b2 b b2 − a2 b2 − a2 = γ2 β2a2 cos2 ϕ + b2 − 2βab cos ϕ = γ2 (b − βa cos ϕ)2 . b2 b2

7.25. Show that the line from (0, 0, b) in the direction of eˆaxis of Equation (7.41) cuts the √ x axis at − β~ γ b2 − a2.

Solution: The line in the direction of eˆaxis is teˆaxis, 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 + t0eˆaxis = xeˆx ⇐⇒ 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 x0y0-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 x0 p x0 = γ(x + β a2 + b2), y0 = y ⇐⇒ − β a2 + b2 = x, y0 = y γ or x0 p 2 − β a2 + b2 + y0 2 = a2 γ or √ 2 0 2 2 x − βγ a + b y0 2 + = 1, γ2a2 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

2 2 (θ~ × ~ϕ)xxc = xca sin θ cos ϕ.

Thus, our equation is now

2 2 2 2 0 = a sin θ[a sin θ + cos θ(a cos θ − b)] + xca sin θ cos ϕ = 0, θ 6= 0, π,

or 2 2 2 0 = a sin θ[xc sin θ cos ϕ + a sin θ + a cos 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

2 2 2 2 2 2 2 2 2 2 x + y + (z − b) = xc + a sin θ cos ϕ + 2xca sin θ cos ϕ + a sin θ sin ϕ 2 2 2 2 2 2 + a cos θ + b − 2ab cos θ = xc + a + 2xca sin θ cos ϕ + b − 2ab cos θ 2 2 2 2 2 2 = xc + a + b + 2a (xc sin θ cos ϕ − b cos θ) = xc + b − a . | {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 p 2 2 2i Den = xc xc + a sin θ cos ϕ − β xc + b − a − b(a cos θ − b)

2 p 2 2 2 2 = xc + axc sin θ cos ϕ − βxc xc + b − a − ab cos θ + b 2 2 2 p 2 2 2 = xc + b − a − βxc xc + b − a 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 a cos α sin θ cos α − cos θ sin α + = 0, xc cos ϕ or a cos α sin(θ − α) = − . xc cos ϕ Now define a cos α sin η ≡ xc cos ϕ and show that θ = α − η. Using the trigonometric identity 1 cos α = √ 1 + tan2 α write η in terms of ϕ: a sin η = . p 2 2 2 b + xc cos 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 α a sin θ − cos θ + = 0 cos α xc cos ϕ or a cos α sin θ cos α − cos θ sin α + = 0. xc cos ϕ 118 Relativistic Photography

Therefore, a cos α sin(θ − 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 x cos ϕ cos α = √ = = c 2 p 2 p 2 2 2 1 + tan α 1 + (b/xc cos ϕ) xc cos ϕ + b tan α b/x cos ϕ b sin α = √ = c = , 2 p 2 p 2 2 2 1 + tan α 1 + (b/xc cos ϕ) xc cos ϕ + b sin η has already been calculated. So, I just need cos η: s s q 2 2 2 2 2 2 a b − a + xc cos ϕ cos η = 1 − sin η = 1 − 2 2 2 = 2 2 2 . b + xc cos ϕ b + xc cos 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 + x2 cos2 ϕ = c p 2 2 2 2 2 2 xc cos ϕ + b b + xc cos ϕ ! ! x cos ϕ a − c p 2 2 2 p 2 2 2 xc cos ϕ + b b + xc cos ϕ p 2 2 2 2 b b − a + xc cos ϕ − axc cos ϕ = 2 2 2 . b + xc cos 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

0 b (xc + a sin θ cos ϕ) + xc(a cos θ − b) u = 2 2 2 xc + b − a p 2 2 0 a xc + b sin θ sin ϕ v = 2 2 2 . xc + b − a 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: √ 2 2 2 2 2 b b −a +xc cos ϕ −abxc cos ϕ bxc + a 2 2 2 cos ϕ u = b +xc cos ϕ x2 + b2 − a2 √ c 2 2 2 2 2 xc cos ϕ b −a +xc cos ϕ +abxc a 2 2 2 − bxc b +xc cos ϕ + 2 2 2 , xc + b − a 119 or

2 2 2 2p 2 2 2 2 2 2 bxc(b + xc cos ϕ) + ab b − a + xc cos ϕ cos ϕ − a bxc cos ϕ u = 2 2 2 2 2 2 (xc + b − a )(b + xc cos ϕ) 2 p 2 2 2 2 2 2 2 2 axc cos ϕ b − a + xc cos ϕ + a bxc − bxc(b + xc cos ϕ) + 2 2 2 2 2 2 , (xc + b − a )(b + xc cos ϕ) or 2 2 p 2 2 2 2 2 2 a(b + xc ) cos ϕ b − a + xc cos ϕ + a bxc sin ϕ u = 2 2 2 2 2 2 . (b + xc − a )(b + xc cos 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. The only challenge of the problem is to show that sin φ as defined by (7.62) is indeed p1 − 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 2 2 2 2 2 t = − x + y + (z − b) = − (xc + a sin θ cos ϕ) + a sin θ sin ϕ + (a cos θ − b) q 2 2 2 2 2 2 2 2 2 2 = − xc + a sin θ cos ϕ + 2axc sin θ cos ϕ + a sin θ sin ϕ + a cos θ + b − 2ab cos θ p 2 2 2 = − xc + a + 2axc sin θ cos ϕ + b − 2ab cos θ.

With this t, (7.67) is immediately obtained. 7.34. Derive Equation (7.68). Solution: First find θ~0 and ϕ~0. The first component of θ~0 is ! ∂x0 ax cos θ cos ϕ + ab sin θ = γ a cos θ cos ϕ − β c , p 2 2 2 ∂θ xc + a + b + 2a(xc sin θ cos ϕ − b cos θ) and the first component of ϕ~0 is ! ∂x0 ax sin θ sin ϕ = γ −a sin θ sin ϕ + β c . p 2 2 2 ∂ϕ xc + a + b + 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.   eˆx eˆy eˆz      axc cos θ cos ϕ + ab sin θ  θ~0 × ϕ~0 = det aγ cos θ cos ϕ − γβ a cos θ sin ϕ −a sin θ  |t|      aγ sin θ sin ϕ(βxc/|t| − 1) a sin θ cos ϕ 0

2 2 2 = a sin θ cos ϕ eˆx − a γ sin θ sin ϕ(βxc/|t| − 1)eˆy h γβ + a2γ sin θ cos θ cos2 ϕ − a2x sin θ cos θ cos2 ϕ |t| c 2 2 2 2 i + a b sin θ cos ϕ − (βxc/|t| − 1)a γ sin θ cos θ sin ϕ eˆz.

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

2 2 2 2 2 β (xc + a sin θ cos ϕ) = xc + a + b + 2a(xc sin θ cos ϕ − b cos θ).

(a) Show that this gives a quadratic equation in cos ϕ whose solution is

x /γ2 ± px2/γ2 + β2(a2 + b2 − 2ab cos θ) cos ϕ = c c . β2a 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.

−|x |/γ2 + px2/γ2 + β2[(b − a cos θ)2 + a2 sin2 θ] c c > 1 β2a sin θ is equivalent to the inequality

2 2 2 (|xc| − a sin θ) + γ (b − a cos θ) > 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

x /γ2 − px2/γ2 + β2[(b − a cos θ)2 + a2 sin2 θ] c c < −1. β2a 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 2ax sin θ β2a2 sin2 θ cos2 ϕ − c cos ϕ − (x2/γ2 + a2 + b2 − 2ab cos θ) = 0. γ2 c The quadratic rule gives the solution as

ax sin θ/γ2 ± pa2x2 sin2 θ/γ4 + β2a2 sin2 θ(x2/γ2 + a2 + b2 − 2ab cos θ) cos ϕ = c c c β2a2 sin2 θ x /γ2 ± px2/γ4 + β2(x2/γ2 + a2 + b2 − 2ab cos θ) = c c c . β2a sin θ Now note that x2 β2x2 x2 1 x2 c + c = c + β2 = c . γ4 γ2 γ2 γ2 γ2 This gives x /γ2 ± px2/γ2 + β2(a2 + b2 − 2ab cos θ) cos ϕ = c c , β2a sin θ which is what we are after, but I’ll rewrite it as

x /γ2 ± px2/γ2 + β2[(b − a cos θ)2 + a2 sin2 θ] cos ϕ = c c β2a 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. β2a sin θ β

Similarly, if xc < 0 and I choose the negative sign the magnitude of the numerator is larger than the denominator.

−|x |/γ2 + px2/γ2 + β2[(b − a cos θ)2 + a2 sin2 θ] c c > 1 β2a sin θ can be written as q |x | x2/γ2 + β2[(b − a cos θ)2 + a2 sin2 θ] > β2a sin θ + c . c γ2 124 Relativistic Photography

Since both sides are positive, I can square the inequality:

x2 |x |2 c + β2[(b − a cos θ)2 + a2 sin2 θ] > β2a sin θ + c , γ2 γ2 or x2 x2 β2a sin θ|x | c + β2(b − a cos θ)2 + β2a2 sin2 θ > β4a2 sin2 θ + c + 2 c . γ2 γ4 γ2 Multiply both sides by γ4 and collect terms using γ2(1 − β2) = 1 and γ2 − 1 = −β2γ2 to get 2 2 2 4 2 2 2 2 2 2 2 2 γ β xc + γ β (b − a cos θ) + γ β a sin θ > 2γ β a sin θ|xc|, or 2 2 2 2 2 xc + γ (b − a cos θ) + a sin θ − 2a sin θ|xc| > 0, which is the same as

2 2 2 (|xc| − a sin θ) + γ (b − a cos θ) > 0.

(d) For xc > 0 and the negative sign for the square root, multiply both sides of the inequality x /γ2 − px2/γ2 + β2[(b − a cos θ)2 + a2 sin2 θ] c c < −1 β2a sin θ by −1 and change the direction of the inequality

−x /γ2 + px2/γ2 + β2[(b − a cos θ)2 + a2 sin2 θ] c c > 1. β2a 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 2 2 2 p 2 2 2 b b − a + xc − axc xc b − a + xc + ab sin θ0 = 2 2 , cos θ0 = 2 2 b + xc b + xc p 2 2 2 p 2 2 2 b b − a + xc + axc −xc b − a + xc + ab sin θπ = 2 2 , cos θπ = 2 2 , b + xc b + xc and the ﬁrst equation of (7.58) gives

p 2 2 2 bγ xc + a sin θ0 − β xc + b − a + xc(a cos θ0 − b)/γ u0 = 0 2 2 2 p 2 2 2 xc + b − a − βxc xc + b − a p 2 2 2 bγ xc − a sin θπ − β xc + b − a + xc(a cos θπ − b)/γ u0 = . π 2 2 2 p 2 2 2 xc + b − a − βxc xc + b − a 125

0 Adding these two and dividing by 2 gives uc: h i x bγ x + a(sin θ − sin θ )/2 − βpx2 + b2 − a2 + c [a(cos θ + cos θ )/2 − b] c 0 π c γ 0 π u0 = , c 2 2 2 p 2 2 2 xc + b − a − βxc xc + b − a or a2x a2b bγ2 x − c − βpx2 + b2 − a2 + x − b c x2 + b2 c c x2 + b2 u0 = c c c 2 2 2 p 2 2 2 γ xc + b − a − βxc xc + b − a 2 h 3 2 2 2 2 p 2 2 2i 2 2 2 bγ xc + xcb − a xc − β(xc + b ) xc + b − a + xcb a − xc − b = 2 2 2 2 2 p 2 2 2 γ(xc + b ) xc + b − a − βxc xc + b − a or

h 2 2 3 2 2 2 2 2 2 2 2 2 p 2 2 2i b γ β xc + γ β xcb − γ β a xc − γ β(xc + b ) xc + b − a u0 = c 2 2 p 2 2 2p 2 2 2 γ(xc + b ) xc + b − a xc + b − a − βxc h 2 2 2 2 2 2 p 2 2 2i bγ β xc(xc + b − a ) − β(xc + b ) xc + b − a = 2 2 p 2 2 2p 2 2 2 (xc + b ) xc + b − a xc + b − a − βxc h 2 p 2 2 2 2 2 i bγ β xc xc + b − a − β(xc + b ) = . 2 2 p 2 2 2 (xc + b ) xc + b − a − 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 mc2. 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 1 + β2 γ0 = γ γ (1 + β β ) = γ2(1 + β2) = 1 2 1 2 1 − β2 since the two speeds are equal. Now note that the energy is just E0 = γ0mc2. 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

E0 = γ(E + βp) = γ(mγ + βmγβ) = mγ2(1 + β2).

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 q 2 2 2 2 ~ 2 2 M − m1 − m2 2 |P | = E1 − m1 = − m1 2m2 q 1 4 2 2 2 2 2 2 = M + (m1 − m2) − 2M (m1 + m2), 2m2 128 Relativistic Interactions

from (8.9), we get 2 2 2 M − m1 + m2 E = E1 + m2 = 2m2 and from (8.10), we get

~ p 4 2 2 2 2 2 2 |P | M + (m1 − m2) − 2M (m1 + m2) V = = 2 2 2 . E M − m1 + m2 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 2 2 E2cm = E1cm − m1 + m2, which you should derive. Solution: First note that from the ﬁrst equation of (8.13), we have

2 2 2 2 |~p1| E1 − m1 E1m2 + m1 E1 − β~cm · ~p1 = E1 − = E1 − = . 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 2 2 2 2 2 2 E2cm = |~p2| + m2 = |~p1| + m2 = E1cm − m1 + m2.

Now use this and Ecm = E1cm + E2cm in Equation (8.3) to obtain 2 2 2 (E1cm + E2cm) = m1 + m2 + 2m2E1, or 2 2 2 2 2 2E1cm − m1 + m2 + 2E1cmE2cm = m1 + m2 + 2m2E1, or 2 2 E1cm − m1 − m2E1 = −E1cmE2cm, or 2 2 2 2 2 2 2 (E1cm − m1 − m2E1) = E1cm(E1cm − m1 + m2), or 2 2 2 2 2 2 2 (m1 + m2E1) − 2E1cm(m1 + m2E1) = E1cm(−m1 + m2). 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 ~p3 = 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,

~p1 ~p4 ~v1 = = = ~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 = E2 = P 2 + M 2 ⇐⇒ 2PM = 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 M 2 − M 2 (P + M)2 = E2 = P 2 + M 2 ⇐⇒ P = f . f 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

~pγ = −~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 2 2 2 2 2 2 2 2 M = Mf + P = Mf + 4pγ ⇐⇒ Mf = M − 4pγ. 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 2 2 2 2 m4 = m1 + m2 + m3 + 2p1 • p2 − 2p1 • p3 − 2p2 • p3. Assuming that the second particle is at rest, we get

2 2 2 2 m4 = m1 + m2 + m3 + 2m2E1 − 2(E1E3 − ~p1 · ~p3) − 2m2E3, 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 ~p3 and ~p4 making angles θ3 and θ4 with the direction of motion of the incident particle. (a) Use (8.20) to show that

(E1 + m)(E3 − m) cos θ3 = . |~p1||~p3|

(b) Using the square of (a) and a similar result for θ4 show that

2 2m(E1 − E3) tan θ3 = (E1 + m)(E3 − m) and 2 2m(E1 − E4) 2m(E3 − m) tan θ4 = = (E1 + m)(E4 − m) (E1 + m)(E1 − E3) (c) Take the product of the last two results to show that 2m 2 tan θ3 tan θ4 = = E1 + m γ1 + 1

where γ1 is the gamma factor for the incident particle. Solution: With all the masses equal, (8.20) becomes

2 m + mE1 = E3(E1 + m) − ~p1 · ~p3 = 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)

2 2 2 2 2 2 1 − cos θ3 |~p1| |~p3| − (E3 − m) (m + E1) tan θ3 = 2 = 2 2 cos θ3 (E3 − m) (m + E1) 2 2 2 2 2 2 (E1 − m )(E3 − m ) − (E3 − m) (m + E1) = 2 2 . (E3 − m) (m + E1) 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 2 for tan 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

m2E1 ≈ E3E1 − |~p1|| · ~p3| cos θ3 ≈ E3E1(1 − cos θ3) or m2 cos θ3 ≈ 1 − ≈ 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 2 2 pin = 16m − m = 15 m.

Let M be the mass of the ﬁnal particle. Then, conservation of energy and momentum gives √ M 2 = E2 − 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, i = 1, 2 or me(Ee + me) Eγi = , i = 1, 2. Ee + me − |~pe| cos θi So, the two photons have energies,

me(Ee + me) Ee + me + |~pe| Eγ1 = = Ee + me − |~pe| 2 me(Ee + me) Ee + me − |~pe| Eγ2 = = . Ee + me + |~pe| 2

(You should obtain the last equality on each line.) For Ee >> me, as is the case here, we 1 get Eγ1 ≈ Ee and Eγ2 ≈ 2 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.

rE − m |~p| cos θ = e = , 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:

me(E + me) Eγi = , i = 1, 2. E + me − |~pe| cos θi

(b) Clearly, if θ1 = θ2, then Eγ1 = Eγ2.

(E + m )(E − m ) cos θ = e γ e . |~pe|Eγ

Conservation of energy gives E +me = 2Eγ, which also leads to 2(Eγ −me) = E −me. Thus, 2E (E − m ) 2(E − m ) E − m rE − m cos θ = γ γ e = γ e = e = e . |~pe|Eγ |~pe| |~pe| E + me Multiplying the numerator and the denominator of the fraction under the radical sign by E + me yields s 2 2 E − me |~pe| |~pe| cos θ = 2 = = . (E + me) E + me 2Eγ

(d) We can ignore me compared to E and |~pe|. Then |~p | |~p | E cos θ = e ≈ 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 ~p E + me E + me β~cm = , γcm = = . p 2 E + me 2me + 2meE 2me (a) The energy and momentum of the electron in the CM are obtained by Lorentz transforming to the CM using Equation (6.54):

r 2 E + me |~p| Ecm = γcm(E − β~cm · ~p) = E − 2me E + me r rE + m E2 − m2 m (E + m ) = e E − e = e e 2me E + me 2

~pcm = ~p − γcmβ~cmE + (γcm − 1)βˆcmβˆcm · ~p r r ! E + me ~p E + me = ~p − E + − 1 βˆcmβˆcm · ~p 2me E + me 2me | {z } =~p ! rE + m pE~ r m = e − + ~p = e ~p. 2me E + me 2(E + me)

2 2 As a check, you should verify that Ecm − ~pcm · ~pcm = me. Note that since the positron + has no momentum in the lab, its energy and momentum in the CM are Ecm = γcmme + ~ + + and ~pcm = −γcmβcmme. As a further check, show that Ecm = Ecm and ~pcm = −~pcm. 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 · ~pγcm)

~pγ = ~pγcm + γcmβ~cmEγcm + (γcm − 1)βˆcmβˆcm · ~pγcm.

To ﬁnd the energy of each photon in the lab, we need β~cm · ~pγcm. So, take the dot product of the second equation with β~cm:

2 β~cm · ~pγ = β~cm · ~pγcm + γcm|β~cm| Eγcm + (γcm − 1)β~cm · ~pγcm 2 = γcm|β~cm| Eγcm + γcmβ~cm · ~pγcm.

Thus, ~ βcm · ~pγ 2 β~cm · ~pγcm = − |β~cm| Eγcm, γcm and ~ ! βcm · ~pγ 2 Eγcm Eγ = γcm Eγcm + − |β~cm| Eγcm = β~cm · ~pγ + γcm γcm 134 Relativistic Interactions

or Ecm Ecm Eγ = |β~cm||~pγ| cos θ + = |β~cm|Eγ cos θ + γcm γcm or q q me(E+me) 2me E /γ 2 E+me m (E + m ) E = cm cm = = e e . γ |~p| 1 − |β~cm| cos θ 1 − cos θ E + me − |~p| 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 single particle. Show that the mass and the speed of the formed particle are, respectively, p2m(E + m) and p(E − m)/(E + m).

Solution: The 4-momentum conservation gives p1 + p2 = P. Squaring both sides gives 2 2 2 2 2m + 2p1 • p2 = M ⇐⇒ 2m + 2mE = M . Conservation of momentum and energy give p P = p = E2 − m2, E = E + m.

The speed of the ﬁnal particle is therefore √ P E2 − m2 rE − m V = = = . E E + m E + m As a check, note that

P 2 + M 2 = E2 − 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 of the nucleus. Show that the mass and the speed of the excited nucleus are, respectively, pM(2E + M) and E/(E + M).

Solution: The 4-momentum conservation gives p1 + p2 = P. Squaring both sides gives 2 ∗ 2 2 ∗ 2 0 + M + 2p1 • p2 = M ⇐⇒ M + 2EM = M . Conservation of momentum and energy give

P = p = E, E = E + M.

The speed of the ﬁnal particle is therefore P E V = = . E E + m As a check, note that

P 2 + M ∗ 2 = E2 + M 2 + 2ME = (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

2 2 2 M 2 2 M |~p1| = − E1 ⇐⇒ E1 − m1 = − E1 , 2E2 E2 or 4 2 2 2 2 M M M m1E2 −m1 = 2 − E1 ⇐⇒ E1 = + 2 . 4E2 E2 4E2 M 8.17. Show that (8.32) follows from the equation before it.

Solution: From M = Zmp + Nmn − BE, the equation before (8.32) can be written as

(M + BE)2 − m2 − M2 BE2 + 2(BE)M − m2 E = 1 = 1 1 2M 2M BE2 − m2 = BE + 1 . 2M

8.18. Derive Equation (8.34) from (8.5).

Solution: For decays Equation (8.5) becomes E1 + E2 = M and ~p2 = −~p1. So, we have E2 = M − E1 and |~p2| = |~p1|. These two equations lead to

2 2 2 2 2 2 2 2 2 2 E2 = (M − E1) ,E2 − m2 = E1 − m1 ⇐⇒ (M − E1) − m2 = E1 − m1, or 2 2 2 2 2 2 M − 2ME1 − m2 = −m1 ⇐⇒ 2ME1 = M + m1 − m2. 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 2 2 2 M + m1 − 2P • p1 = m2.

The ﬁrst equation of (8.34) follows immediately from P • p1 = ME1. 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. 3 Using the familiar statistical physics formula hKEi = 2 kBT , 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

7 Solution: Use T = 2 × 10 K and mp = 938.272 MeV. Then

3 −23 7 −16 hKEi = 2 (1.38 × 10 )(2 × 10 ) = 4.14 × 10 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 E 938.2746 γ − 1 = − 1 = − 1 = 2.76 × 10−6 ≈ 1 β2. M 938.272 2 This gives a speed of β = 0.00235. 8.21. Derive Equation (8.45). Solution: We want to have the identity

β2E2 + (β~α + ~αβ) · ~pE + (~α · ~p)2 − m2 = E2 − ~p · ~p − m2.

For the equality to hold, the coefficients must match on both sides. E2 has 1 as coefficient 2 on the right. So, β = 1. There is no pxE term on the right. Therefore, its coefficient on the left must be zero. This gives αxβ + βαx = 0; similarly for pyE and pzE 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:

2 (~α · ~p) = (αxpx + αypy + αzpz)(αxpx + αypy + αzpz)

= αxpx(αxpx + αypy + αzpz) + αypy(αxpx + αypy + αzpz)

+ αzpz(αxpx + αypy + αzpz),

2 2 2 (~α · ~p) = αxpx + αxαypxpy + αxαzpxpz 2 2 + αyαxpypx + αypy + αyαzpypz 2 2 + αzαxpzpx + αzαypzpy + αzpz. 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

2 2 2 2 2 2 2 (~α · ~p) = αxpx + αypy + αzpz

+ (αxαy + αyαx)pxpy

+ (αxαz + αzαx)pxpz

+ (αyαz + αzαy)pypz.

2 2 2 2 To get the identity (~α · ~p) = −~p · ~p = −px − py − pz, I must have

2 2 2 αx = αy = αz = −1

αxαy + αyαx = αxαz + αzαx = αyαz + αzαy = 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

4m2 E = 3m + p . min p 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

2 2 (3mp + Nmπ) − m E = 3m + Nm + p . min p π 2M As a check for your answer, make sure you get (a) as a special case.

Solution: Let me denote the numerator of (8.31) by M not to confuse it with the mass of the nucleus. Then

2 2 2 2 2 M = (M + 3mp) − M − mp = 8mp + 6Mmp.

(a) Therefore, M2 4m2 E = = 3m + p min 2M p M (b) If N pions are added to the outcome, then

2 2 2 2 M = (M + 3mp + Nmπ) − M − mp 2 2 = (3mp + Nmπ) + 2M(3mp + Nmπ) − mp,

and 2 2 2 M (3mp + Nmπ) − m E = = 3m + Nm + p min 2M p π 2M (c) (3 × 938.3 + 12 × 140)2 − 938.32 E = 3 × 938.3 + 12 × 140 + , min 2 × 59150 or Emin ≈ 4.7 GeV.

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

0 d(Mγ) + dmgγg = 0 ~ 0 ~ 0 d(Mγβ) + dmgγgβg = 0.

Solution: Write the 4-momentum conservation as Pbef = pgas + Paft or 0 = pgas + dP, and note that

0 0 ~ 0 ~ pgas = (dmgγg, dmgγgβg) and dP = d(Mγ), d(Mγβ) .

9.2. Show that

d(Mγ) = γdM + Mβγ3dβ d(Mγβ) = γβdM + Mγ3dβ.

Solution: Clearly, d(Mγ) = γdM + Mdγ. But

dγ = d(1 − β2)−1/2 = β(1 − β2)−3/2dβ = βγ3dβ

Similarly, d(Mγβ) = γβdM + Md(γβ), with

d(γβ) = γdβ + βdγ = γdβ + β(βγ3dβ) = γ(1 + β2γ2)dβ = γ3dβ.

9.3. Derive Equation (9.3) from the equations preceding it. 140 Interstellar Travel

Solution: Substitute the given expressions in (9.2) to obtain

3 3 (1 ∓ ββg)(γβdM + Mγ dβ) − (β ∓ βg)(γdM + Mβγ dβ) = 0, or 2 3 [(1 ∓ ββg)β − (β ∓ βg)] γdM + [(1 ∓ ββg) − (β ∓ βgβ)] Mγ dβ = 0, | {z } | {z } 2 2 =±βg(1−β ) 1−β or β ± g dM + Mγdβ = 0 ⇐⇒ ±β dM + Mγ2dβ = 0. γ g 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: M 1 p1 − β2 p(1 − β)(1 + β) 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 −2 that (9.7) gives Mf /M0 = [γ0(β0 + 1)] , 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 M 1 1 = . M0 γ0(β0 + 1)

With β(tf ) = 0, the second equation of (9.7) gives M 1 f = . M1 γ0(β0 + 1) Therefore, Mf Mf M1 1 = = 2 2 . M0 M1 M0 γ0 (β0 + 1)

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 2 2 γ0 (β0 + 1) − 1 β0 = 2 2 , γ0 (β0 + 1) + 1 or 2 2 2 2 β0γ0 (β0 + 1) + β0 = γ0 (β0 + 1) − 1, or 2 2 2 1 + β0 = γ0 (β0 + 1) (1 − β0) ⇐⇒ 1 = γ0 (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.

beg Mtot(t) = Mf + Mfuel(t) = Mf + Mfuel − M0t,

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.

beg Mfuel(t) = −M0t/T0 + Mfuel . Then, beg Mfuel(t) = Mf + Mfuel(t) = Mf + Mfuel − M0(t/T0).

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

1 − m2(t) β(t) = . 1 + m2(t)

beg On the other hand, at the beginning of the journey, Mfuel = M0. Thus, (9.10) becomes

M0 m(t) = 1 − m0t, m0 = , t ≤ 1. Mf + M0 142 Interstellar Travel

Now take the derivative of (9.12): dx 2 d −1 2 m0 = −1 − tan (1 − m0t) = −1 + 2 dt m0 dt m0 1 + (1 − m0t) 2 2 1 − (1 − m0t) = −1 + 2 = 2 . 1 + (1 − m0t) 1 + (1 − m0t)

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

2 1 − (1 − m0tacc) 2 β0 = 2 ⇐⇒ (1 − m0tacc) (1 + β0) = 1 − β0, 1 + (1 − m0tacc) or 2 2 1 − β0 1 − β0 (1 − m0tacc) = = 2 , 1 + β0 (1 + β0) or 1 γ0(1 + β0) − 1 1 − m0tacc = ⇐⇒ m0tacc = . 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

dx 2 d h −1 i = 1 + 2 2 tan {γ0(1 + β0) [1 − m0γ0(1 + β0)t]} dt m0γ0 (1 + β0) dt 2 2 2 m0γ0 (1 + β0) = 1 − 2 2 2 2 m0γ0 (1 + β0) 1 + γ0 (1 + β0) [1 − γ0(1 + β0)m0t] 2 = 1 − 2 2 . 1 + γ0 (1 + β0) [1 − γ0(1 + β0)m0t] 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

2 2 2 γ0 (1 + β0) [1 − m0γ0(1 + β0)tdec] − 1 = 0, or 1 1 − m0γ0(1 + β0)tdec = , γ0(1 + β0) or γ0(1 + β0) − 1 m0γ0(1 + β0)tdec = . γ0(1 + β0) This is essentially (9.17). 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

2 2 γ0(1 + β0) − 1 γ0(1 + β0) − 1 γ0 (1 + β0) − 1 1 = + 2 2 = 2 2 , m0γ0(1 + β0) m0γ0 (1 + β0) m0γ0 (1 + β0) or 2 2 1 γ0 (1 + β0) (1 − m0) = 1 ⇐⇒ γ0(1 + β0) = √ , mf because M0 Mf m0 + mf ≡ + = 1. Mf + M0 Mf + M0 Rewrite the last equation as

p 2 s 1 − β 1 − β0 1 0 = = √ . 1 + β0 1 + β0 mf

Square both sides and solve for β to get Equation (9.19). √ 0 Now use γ0(1 + β0) = 1/ mf —obtained above—in (9.13) to get √ √ (1/ mf ) − 1 1 − mf 1 tacc = √ = = √ . (1 − mf )(1/ mf ) 1 − mf 1 + mf √| {z }√ =(1− mf )(1+ mf )

Equation (9.17) now gives √ tacc √ mf tdec = √ = 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): √ π (1/ mf ) − 1 2 −1 √ x(tacc) = − √ − tan mf 2(1 − mf ) (1 − mf )/ mf 1 − mf √ −1 √ π/2 − 1 + mf − 2 tan 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.

1 (a) Show from Equation (9.19) that the maximum attainable speed is 2 m0. 1 (b) From the expansion of (9.21) show that xacc → 8 m0. 1 (c) From (9.22) conclude that xdec → 8 m0. Solution:

(a) With mf = 1 − m0, Equation (9.19) becomes

m0 1 β0 = ≈ 2 m0 for m0 → 0. 2 − m0

2 (b) Since the denominator of (9.21) is m0, we expand the numerator up to m0. First look at each term of the numerator separately:

√ 1 ( 1 − 1) m = (1 − m )1/2 ≈ 1 − 1 m + 2 2 m2 = 1 − 1 m − 1 m2. f 0 2 0 2 0 2 0 8 0 From π x x2 tan−1(1 − x) ≈ − − 4 2 4 and the result above, we get √ π 1 1 tan−1( m ) = tan−1 1 − 1 m − 1 m2 = − ( 1 m + 1 m2) − ( 1 m + 1 m2)2 f 2 0 8 0 4 2 2 0 8 0 4 2 0 8 0 π m m2 m2 π m m2 = − 0 − 0 − 0 = − 0 − 0 , 4 4 16 16 4 4 8 145

2 where I kept terms up to m0 in the ﬁnal expression. Now substitute these in (9.21) and obtain

−1 √ √ −2 tan ( mf ) mf z }| 2{ z 1 }| 1 {2 π m0 m0 1 − m0 − m − + + +π/2 − 1 2 8 0 2 2 4 m0 xacc = = . m0 8

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 = , κ << 1. 1 + κ γf (βf + 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 + κr(t)M 1 + κr(t) m(t) ≡ = las las = . M0 Mlas + κMlas 1 + κ For r(t) = 0.5, the ﬁrst equation in (9.7) gives 1 + 0.5κ 1 = , κ << 1. 1 + κ γf (βf + 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:

2 1 − βf 1 − βf κ κ (1 − 0.5κ) = ⇐⇒ 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

2E0 = 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 0 Thus, 2E = δ(Esail + Psail). 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

nano nano (ti , xi ) = γti(1, −β).

nano (b) How long does it take for this photon to reach the craft? Adding ti to this time, nano, ref find ti . (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 nano nano xi = γ(0 − βti) = −γβti, ti = γ(−0 + ti) = γti. 147

nano (b) The distance from which the photon is emitted is xi (note that it doesn’t matter that the laser doesn’t remain at that position). Therefore, the time at which the photon reaches the craft is

nano, ref nano nano ti = ti + |xi| = γ(1 + β)ti.

nano, ref (c) Since the craft is assumed to be at the origin of its frame, xi = 0. (d) Lorentz transforming the event of the reflection of the ith photon to the laser frame yields t tref = γ tnano, ref + 0 = γ2(1 + β)t = i i i i 1 − β βt xref = γ 0 + βtnano, ref = γ2β(1 + β)t = i , i i i 1 − β with similar equations for i + 1th photon. Equation (9.27) now follows. Note that for ref (9.27), we don’t need xi . 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 δE = sail . 1 + R δPsail q 2 2 (b) Use Esail = Psail + M to show that 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 δE = sail . 1 + R δPsail q 2 2 (b) From Esail = Psail + M , we obtain q 2 2 PsailδPsail Psail δEsail = δ Psail + M = q = δPsail = βδPsail. 2 2 Esail 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 |eˆn · (~r − ~r0)| dΩ = 3 da, |~r − ~r0|

where eˆn 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 ! L2 Ω = 4 tan−1 . p 2 2 2z0 2L + 4z0

(b) Show that if z0 >> L, then L2 Ω ≈ 2 . z0

Solution: In this problem, eˆn = ±eˆz. The ambiguity is due to the fact that a surface has two sides! Let’s take eˆn = −eˆz (away from P0). We also have

~r = hx, y, 0i, ~r0 = h0, 0, z0i ~r − ~r0 = hx, y, −z0i.

Thus, |−eˆ · (~r − ~r )| z dΩ = z 0 dxdy = 0 dxdy, 3 2 2 2 3/2 |~r − ~r0| (x + y + z0) and Z L/2 Z L/2 2 ! dxdy −1 L Ω = z0 = 4 tan . 2 2 2 3/2 p 2 2 −L/2 −L/2 (x + y + z0) 2z0 2L + 4z0

(b) If z0 >> L, then L2 L2 L2 ≈ = , z0 > 0. p 2 2 2z (2|z |) 4z2 2z0 2L + 4z0 0 0 0 Now we know that tan−1(x) ≈ x to ﬁrst order in x. Thus,

L2 L2 Ω ≈ 4 2 = 2 . 4z0 z0

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 ∆p 2∆E 2P F 2P F = = = ⇐⇒ 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

2P 2 × 1011 a = laser = = 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

2P x2 a = laser 0 . Mc x2

(b) Write dv/dt as vdv/dx and integrate the equation in (a) to get 2P x 1 v2 − 1 v2 = laser x 1 − 0 2 2 0 Mc 0 x

assuming that the speed is v0 at x0. (c) Show that the result in (b) can also be written as

r x v = v 2 − 0 , 0 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 2 2 power delivered. In this problem, P = Plaserx0/x . (a) Therefore, the acceleration is 2P x2 a = laser 0 . Mc x2 (b) Write the equation in (a) as

dv dv 2P x2 = v = laser 0 , dt dx Mc x2 or 2x2P dx vdv = 0 laser . Mc x2 150 Interstellar Travel

Integrate this to get Z v 2 Z x 2x0Plaser du udu = 2 , v0 Mc x0 u or 2 1 2 1 2 2x0Plaser 1 1 2x0Plaser x0 2 v − 2 v0 = − = 1 − . Mc x0 x Mc x

(c) The speed v0 is also the speed at the end of the ﬁrst stage. Thus, we can write 2P v2 = 2ax ⇐⇒ 1 v2 = ax = laser x . 0 0 2 0 0 Mc 0 Therefore, the last equation in (b) becomes x 1 v2 − 1 v2 = 1 v2 1 − 0 2 2 0 2 0 x or x r x v2 = v2 2 − 0 ⇐⇒ v = v 2 − 0 . 0 x 0 x

(d) We have r dx x0 dx = v0 2 − ⇐⇒ = v0dt dt x p x0 2 − x or r x dx = v0dt. 2x − x0 Integrating this, we obtain

Z x r u Z t du = v0 du = v0t. x0 2u − x0 0 The integration on the left gives the following complicated and uninteresting ﬁnal result: √ √ √ px(2x − x ) − x 2 x 2 x + 4x − 2x 0 0 0 0 + ln √ √ = v0t. 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 = ∂iijk∂jAk = ijk∂i∂jAk = 0, by Note 10.1.6 and the fact that ∂i∂j = ∂j∂i.

(∇ × ∇f)i = ijk∂j(∂kf) = ijk∂j∂kf = 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:   eˆx eˆy eˆz A × C = det Ax Ay Az = (AyCz − AzCy)eˆx − (AxCz − AzCx)eˆy + (AxCy − AyCx)eˆz. Cx Cy Cz Similarly,

B × D = (ByDz − BzDy)eˆx − (BxDz − BzDx)eˆy + (BxDy − ByDx)eˆz. Therefore,

(A × C)·(B × D) = (AyCz − AzCy)(ByDz − BzDy) + (AxCz − AzCx)(BxDz − BzDx)

+ (AxCy − AyCx)(BxDy − ByDx)

= AyBy(CzDz + CxDx) + AzBz(CyDy + CxDx) + AxBx(CyDy + CzDz)

− AyDy(BzCz + BxCx) − AzDz(ByCy + BxCx) − AxDx(BzCz + ByCy). 152 A Painless Introduction to Tensors

Note that I can write the ﬁrst three terms as

AyByC · D − AyByCyDy + AzBzC · D − AzBzCzDz + AxBxC · D − AxBxCxDx,

or as (A · B)(C · D) − AyByCyDy − AzBzCzDz − AxBxCxDx, (10.1) and the last three terms as

−AyDyB · C + AyDyByCy − AzDzB · C + AzDzBzCz − AxDxB · C + AxDxBxCx

or as −(A · D)(B · C) + AyDyByCy + AzDzBzCz + AxDxBxCx. Combining this with (10.1) above gives the ﬁnal result. To prove Equation (10.8), multiply it on both sides by AjCkBmDn. On the left, you get

ijkimnAjCkBmDn = (ijkAjCk)(imnBmDn) = (A × C)i(B × D)i = (A × C) · (B × D)

and on the right you get

(δjmδkn − δjnδkm)AjCkBmDn = δjmAjBmδknCkDn − δjnAjDnδkmCkBm

= AjBjCkDk − AjDjCkBk = (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 = AiijkBjCk = ijkAiBjCk

= kijCkAiBj = lmnClAmBn = 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:

∇ · (fA) = (∇f) · A + f∇ · A ∇ × (fA) = (∇f) × A + f∇ × A ∇ × (∇ × A) = ∇(∇ · A) − ∇2A

Solution:

∇ · (fA) = ∂i(fA)i = ∂i(fAi) = (∂if)Ai + f(∂iAi) = (∇f) · A + f∇ · A

[∇ × (fA)]i = ijk∂j(fAk) = ijk[(∂jf)Ak + f∂jAk]

= ijk(∂jf)Ak + fijk∂jAk = [(∇f) × A]i + f(∇ × A)i 153

[∇ × (∇ × A)]i = ijk∂j(∇ × A)k = ijk∂j(klm∂lAm) = kijklm∂j∂lAm

= (δilδjm − δimδjl)∂j∂lAm = δilδjm∂j∂lAm − δimδjl∂j∂lAm

= ∂m∂iAm − ∂j∂jAi = ∂i(∂mAm) − (∂j∂j)Ai 2 2 = ∂i(∇ · A) − (∇ )Ai = [∇(∇ · A]i − (∇ A)i.

10.5. Show that the determinant of a 3 × 3 matrix A with elements aij is given by

det A = ijka1ia2ja3k.

Solution: Let   a11 a12 a13 A = a21 a22 a23 . a31 a32 a33 Remember the summation convention:

ijka1ia2ja3k = 1jka11a2ja3k + 2jka12a2ja3k + 3jka13a2ja3k.

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

1jka11a2ja3k = 12ka11a22a3k + 13ka11a23a3k

= 123a11a22a33 + 132a11a23a32

= a11(a22a33 − a23a32); the second term is

2jka12a2ja3k = 21ka12a21a3k + 23ka12a23a3k

= 213a12a21a33 + 231a12a23a31

= −a12(a21a33 − a23a31); and the last term is

3jka13a2ja3k = 31ka13a21a3k + 32ka13a22a3k

= 312a13a21a32 + 321a13a22a31

= a13(a21a32 − a22a31).

When you add all the terms, you get the familiar expansion of the determinant according to its ﬁrst row.

10.6. Let ˆei, i = 1, 2, 3 be the components of a unit vector. Let

aijk = ijmêkêm + imkêjêm + mjkêiêm.

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 ˆei = δi1 and

aijk = ijmδk1δm1 + imkδj1δm1 + mjkδi1δm1

= ij1δk1 + i1kδj1 + 1jkδi1. This implies that a123 = 121δk1 + 11kδ21 + 123δ11 = 123 and

ajik = ji1δk1 + j1kδi1 + 1ikδj1

= −ij1δk1 − 1jkδi1 − i1kδj1 = −aijk, and

aikj = ik1δj1 + i1jδk1 + 1kjδi1

= −i1kδj1 − ij1δk1 − 1jkδi1 = −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)|||]. 0 0 (b) From (6.17) and dt = 0, show that |(d~r 1)||| = |(d~r1)|||/γ, with similar relations for 0 0 |(d~r 2)||| and |(d~r 3)|||. (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.

0 ~ (b) With dt = 0, the ﬁrst equation of (6.17) gives dt = − β ~r|| , and the second equation yields d~r 0 h ~ ~ i ~ 2 || d~r|| = γ d~r|| − β β d~r|| = γ d~r|| − β d~r|| = . 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∂jAk show that ∂jAk − ∂kAj = jkmBm.

Solution: Multiply both sides of Bi = ijk∂jAk by lmi to obtain

lmiBi = lmiijk∂jAk = ilmijk∂jAk = (δljδmk − δlkδmj)∂jAk

= δljδmk∂jAk − δlkδmj∂jAk = ∂lAm − ∂mAl. 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. 1 ˆ ˆ ˆ ˆ W = 2 ijklmn[δkm + (γ − 1)βkβm][δjl + (γ − 1)βjβl]Bn 1 ˆ ˆ 1 ˆ ˆ = 2 ijklmnδkm[δjl + (γ − 1)βjβl]Bn + 2 (γ − 1)ijklmnδjlβkβmBn 1 ˆ ˆ 1 ˆ ˆ = 2 ijklkn[δjl + (γ − 1)βjβl]Bn + 2 (γ − 1)ijkjmnβkβmBn. 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: 1 1 ˆ ˆ W = − 2 (δilδjn − δinδjl)δjlBn − 2 (γ − 1)(δilδjn − δinδjl)βjβlBn 1 ˆ ˆ − 2 (γ − 1)(δimδkn − δinδkm)βkβmBn 1 1 ˆ ˆ ˆ ˆ = − 2 (δijδjn − δinδjj)Bn − 2 (γ − 1)(βjβiBj − βjβjBi) 1 ˆ ˆ ˆ ˆ − 2 (γ − 1)(βkβiBk − βkβkBi). ˆ ˆ ˆ ˆ Now note that δjj = 3 and βkβk = β · β = 1. Then

W = Bi − (γ − 1)(βˆiβˆ · B~ − Bi) = γBi − (γ − 1)βˆi(βˆ · B~ ). 158 Relativistic Electrodynamics

11.3. Show that the sum of the ﬁrst two terms of (11.12) gives γijkβjEk. Solution: This involves simply renaming the dummy indices. I’ll work on the ﬁrst term:

γimkβk∂mA0 = γi♣♠β♠∂♣A0 = −γi♠♣β♠∂♣A0 = −γijkβj∂kA0.

Now I add it to the second term to get

γijkβj(−∂kA0 + ∂0Ak) = γijkβjEk.

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 ~ 0 ~ ~ ~ ~ ~ ˆ ˆ ~ ~ E|| + E⊥ = γ[E|| + E⊥ − β × (B|| + B⊥)] − (γ − 1)β[β · (E|| + E⊥).

Now note that ~ ~ ˆ ~ ˆˆ ~ ~ β × B|| = 0 = β · E⊥, and ββ · E|| = E||. Hence,

~ 0 ~ 0 ~ ~ ~ ~ ~ E|| + E⊥ = γE|| + γE⊥ − γβ × B⊥ − (γ − 1)E|| ~ ~ ~ ~ = E|| + γ(E⊥ − β × B⊥).

The ﬁrst term on the right-hand side is parallel to the direction of motion, so it must be ~ 0 equal to E|| and the second term is perpendicular to the direction of motion, so it must be ~ 0 equal to E⊥ . 11.5. Derive Equation (11.18).

Solution: Dotting both sides of (11.17) by βˆ and rewriting the resulting equation slightly diﬀerently yields

γβ βˆ · ~r = βˆ · ~r 0 − γβ t0 − βˆ · ~r 0 γ + 1 γ2β2 = −γβt0 + 1 + βˆ · ~r = −γβt0 + γβˆ · ~r, γ + 1

2 2 2 because γ β = γ − 1 = (γ − 1)(γ + 1). 11.6. Derive Equation (11.19).

Solution: The equation before (11.19) can be expressed as

γ2β2 ~S = γ~r 0 − γβt~ 0 + γ − γ + 1 βˆβˆ · ~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

γ2β2 ~r = ~r 0 − γβt~ 0 + βˆβˆ · ~r 0 = ~r 0 − γβt~ 0 + (γ − 1)βˆβˆ · ~r 0. γ + 1

Now use (11.19) and substitute for ~r 0:

0 ~ 0 ~ 0 ˆˆ 0 ~ 0 ~r = ~r inst + βt − γβt + (γ − 1)ββ · (~r inst + βt ) 0 ~ 0 ˆˆ 0 ˆˆ ~ 0 = ~r inst − (γ − 1)βt + (γ − 1)ββ · ~r inst + (γ − 1) ββ · β t | {z } =β~ 0 ˆˆ 0 = ~r inst + (γ − 1)ββ · ~r inst.

Now square (dot with itself) both sides

2 0 2 2 ˆ 0 2 ˆ 0 2 r = rinst + (γ − 1) (β · ~r inst) + 2(γ − 1)(β · ~r inst) 0 2 ˆ 0 2 = rinst + (γ − 1)(γ + 1)(β · ~r inst) | {z } =γ2−1=γ2β2 0 2 2 ˆ 0 2 0 2 2 ~ 0 2 = rinst + γ (ββ · ~r inst) = rinst + γ (β · ~r inst) .

11.8. Derive Equation (11.25).

~ ~ 0 Solution: From the equation before (11.25), plus β × ~r = β × ~r inst and (11.22), we get k qγ B~ 0 = e β~ × ~r 0 , h i3/2 inst 0 2 2 ~ 0 2 rinst + γ (β · ~r inst) with

0 2 2 ~ 0 2 0 2 2 2 0 2 2 0 2 2 0 2 2 rinst + γ (β · ~r inst) = rinst + γ β rinst cos θinst = rinst + (γ − 1)rinst cos θinst 0 2 2 2 0 2 2 0 2 2 2 2 = rinst sin θinst + γ rinst cos θinst = rinst(sin θinst + γ cos 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.

(a) Calculate the ﬁelds E~ and B~ in the rest frame of the plate using the elementary 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.

0 (a) From Gauss’s law E~ = 2πkeσβˆ and B~ = 0 in O. Then (11.14) gives E~ = 2πkeσβˆ and B~ 0 = 0 in O0 as well.

(b) I’ll calculate the ﬁeld at the moment that the two origins coincide, so that the plate 0 0 0 0 0 0 0 0 is in the y z -plane of O . Let P0 = (x0, y0, z0) be the ﬁeld point, and Pinst = (0, y , z ) the instantaneous location of the charge element. Then

0 0 0 0 0 0 0 0 0 0 0 ~r inst = hx0, y0, z0i − h0, y , z i = hx0, y0 − y , z0 − z i

and (11.22) becomes

2 0 2 0 0 2 0 0 2 2 2 0 2 r = x0 + (y0 − y ) + (z0 − z ) + γ β x0 2 0 2 0 0 2 0 0 2 = γ x0 + (y0 − y ) + (z0 − z )

using the indispensable identity γ2β2 = γ2 − 1. Plugging this in (11.23) and integrating, we get the electric ﬁeld at P0: Z ∞ Z ∞ hx0 , y0 − y0, z0 − z0i E~ 0 = k σγ 0 0 0 dy0dz0. e 2 0 2 0 0 2 0 0 2 3/2 −∞ −∞ [γ x0 + (y0 − y ) + (z0 − z ) ]

0 0 0 0 Integration over y gives Ey = 0 and integration over z gives Ez = 0. So, the only 0 surviving component is Ex, and that is given by Z ∞ Z ∞ dy0dz0 E0 = k σγx0 = 2πk σ x e 0 2 0 2 0 0 2 0 0 2 3/2 e −∞ −∞ [γ x0 + (y0 − y ) + (z0 − z ) ]

| {z 0 } =2π/(γx0)

0 0 0 0 or E~ = 2πkeσβˆ. Since B~ = β~ × E~ , we get B~ = 0, in agreement with (a).

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 ≡ ∂0Ai − ∂iA0 and conclude that F0i = −Ei. This verifies the first row and the first column. Furthermore, from (11.2) we get

F12 = ∂1A2 − ∂2A1 = B3,F31 = ∂3A1 − ∂1A3 = B2,F23 = ∂2A3 − ∂3A2 = B1.

These verify the rest of the matrix elements. Note how the three indices in Fij = Bk are cyclically permuted. sup µν 11.11. Let F and Fsub denote the matrices of F and Fµν, respectively. Show that the sup matrix form of Equation (11.28) is F = ηFsubη. 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:

µν µα βν µ0 βν µi βν F = η Fαβη = η F0βη + η Fiβη µ0 iν µi βν µ0 iν µi jν = η F0iη + η Fiβη = η F0iη + η Fijη , i < j.

0i 00 ii ij This shows that F = −F0i because η = 1 = −η (no sum over i) and that F = Fij ii jj because η η = 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 only thing you need to know is that Fe = −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 form. Λ is given by (6.14). With obvious notation, we can also deﬁne the block form of F:     γ γβ~e 0 E~e Λ =   , F =   .  ↔    γβ~ Λ −E~ B

Now calculate the product of the three matrices in (11.31). The ﬁrst two give       γ γβ~e 0 E~e −γβ~ · E~ γ(E~e +β~eB)       ΛF =     =   ↔ ↔ ↔ γβ~ Λ −E~ B −Λ E~ γβ~E~e + Λ B and multiplying this by Λ on the right gives     −γβ~ · E~ γ(E~e +β~eB) γ γβ~e 0     F =     . ↔ ↔ ↔ −Λ E~ γβ~E~e + Λ B γβ~ Λ

0 Rather than multiplying out matrices, I’ll look at the elements F(i,j) (with i, j = 1, 2) of 0 the block matrix that I need. Although I don’t need F(1,1), I’ll calculate it to make sure I get zero. 0 2 ~ ~ 2 ~e ~e ~ 2 ~e ~ F(1,1) = −γ β · E + γ (E +βB)β = γ β B β. But, since the matrix B is antisymmetric,

~ ~ ~ ~ ~ βeB β = βi(B β)i = βiBijβj = −βiBjiβj = −β♣B♠♣β♠ = −β♠B♠♣β♣ = −βeB β,

0 0 and F(1,1) = 0 as expected. Next, I’ll calculate F(2,1): 0 ↔ ~ ~ ~e ↔ ~ F(2,1) = −γΛ E + γβE + Λ B γβ. 162 Relativistic Electrodynamics

This is a column vector; so we calculate its ith component:

0 2 ~ ~ F(2,1) = −γΛijEj + γ βiE · β + γΛijBjkβk. i

I’ll evaluate each term separately [see (B.9) for Λij]:

term 1 = −γΛijEj = −γ[δij + (γ − 1)βîβˆj]Ej = −γEi − γ(γ − 1)βîβˆ · E~ 2 2 2 2 term 2 = γ βiE~ · β~ = γ β βîE~ · βˆ = (γ − 1)βîE~ · βˆ ˆ ˆ term 3 = γΛijBjkβk = γ[δij + (γ − 1)βi βj]Bjkβk | {z } =0 ~ = γBikβk = γikmBmβk = γ(β × B~ )i. Adding the three terms, we obtain

0 ˆ ˆ ~ 2 ˆ ~ ˆ ~ ~ F(2,1) = −γEi − γ(γ − 1)βiβ · E + (γ − 1)βiE · β + γ(β × B)i i or 0 ~ ~ ˆ ˆ ~ −Ei = −γ(E − β × B)i + (γ − 1)βiβ · E, which agrees with (11.8). 0 To obtain the transformation of the magnetic ﬁeld, we’ll have to look at F(2,2). 0 ↔ ~ ~e ~ ~e ↔ ↔ F(2,2) = −γΛ Eβ + γβE + Λ B Λ .

This is a 3 × 3 matrix; so let’s ﬁnd it’s ijth element:

0 F(2,2) = −γΛikEkβj + γβiEkΛkj + ΛikBkmΛmj. ij Again I’ll evaluate each term separately: ˆ ˆ ˆ ˆ term 1 = −γΛikEkβj = −γ[δik + (γ − 1)βiβk]Ekβj = −γEiβj − γ(γ − 1)β · E~ βiβj ˆ ˆ ˆ ˆ term 2 = γβiEkΛkj = γβiEk[δkj + (γ − 1)βkβj] = γβiEj + γ(γ − 1)β · Eβ~ iβj ˆ ˆ ˆ ˆ term 3 = ΛikBkmΛmj = [δik + (γ − 1)βiβk]Bkm[δmj + (γ − 1)βmβj] ˆ ˆ ˆ ˆ = [δik + (γ − 1)βiβk][Bkj + (γ − 1)βmβjBkm] ˆ ˆ ˆ ˆ = Bij + (γ − 1)βmβjBim + (γ − 1)βiβkBkj. ˆ ˆ In the third term, the product of the two γ − 1 terms gives zero because βkβmBkm = 0. 0 0 Adding the three terms and noting that F(2,2) = B , we get 0 ˆ ˆ ˆ ˆ Bij = γβiEj − γEiβj + Bij + (γ − 1)βmβjBim + (γ − 1)βiβkBkj. ~ 1 The relation between the matrix B and the magnetic ﬁeld B is Bn = 2 nijBij and Bij = 1 ijnBn. So, multiply both sides of the last equation by 2 nij to obtain 0 1 1 ˆ ˆ ˆ ˆ Bn = γ 2 nij(βiEj − Eiβj) + Bn + (γ − 1) 2 nij(βmβjimkBk + βiβkkjlBl) ~ ~ 1 ˆ ˆ ˆ ˆ = γ(β × E)n + Bn + 2 (γ − 1)(nijimk βmβjBk +nijkjl βiβkBl) | {z } | {z } =δnkδjm−δnmδjk =δnlδik−δnkδil ~ ~ 1 ˆ ˆ ~ ˆ ˆ ~ = γ(β × E)n + Bn + 2 (γ − 1)(Bn − βnβ · B + Bn − βnβ · B) 163 or 0 ~ ~ ~ ˆ ˆ ~ Bn = γ(B + β × E)n − (γ − 1)βn(β · B). This agrees with Equation (11.13). µν 2 2 µν 11.14. Show that F Fµν = 2(|B~ | − |E~ | ). Hint: Prove that F Fµν is the negative of 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.

µν Solution: I let you calculate F Fµν via trace. I’ll calculate it by index manipulation as more exercise in “index gymnastics.”

µν 0ν iν 0i i0 ij 0i 0i i0 i0 ij ij F Fµν = F F0ν + F Fiν = F F0i + F Fi0 + F Fij = −F F − F F + F F 2 = −EiEi − (−Ei)(−Ei) + (ijkBk)(ijmBm) = −2|E~ | + (δjjδjm − δkjδkm)BkBm 2 2 2 2 = −2|E~ | + (3δjm − δjm)BkBm = −2|E~ | + 2BkBk = −2|E~ | + 2|B~ | .

11.15. Show that det F = (E~ · B~ )2, where F is as given by (11.29). 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.

0µ 1ν 2α 3β 0i 1ν 2α 3β det F = µναβF F F F = iναβF F F F 0i 10 2α 3β 0i 1j 2α 3β = i0αβF F F F + ijαβF F F F 0i 10 2j 3β 0i 1j 20 3β 0i 1j 2k 3β = i0jβF F F F + ij0βF F F F + ijkβF F F F 0i 10 2j 3k 0i 1j 20 3k 0i 1j 2k 30 = i0jkF F F F + ij0kF F F F + ijk0F F F F . Here, I used the properties of the 4-dimensional Levi-Civita symbol, which a trivial general- ization of the properties of the 3-dimensional Levi-Civita symbol. Now note that 0ijk = ijk 0i and F = Ei to get

10 2j 3k 1j 20 3k 1j 2k 30 det F = Eiijk(−F F F + F F F − F F F ) 2j 3k 1j 3k 1j 2k = Eiijk(E1F F − E2F F + E3F F ). I’ll evaluate the ﬁrst term in detail, leaving the other two terms—which are obtained in exactly the same way—for you.

2j 3k ijkF F = ijk2jmBm3knBn = jikj2m3knBmBn

= (δi2δkm − δimδk2)3knBmBn = (δi23mn − δim32n)BmBn

= δi2 3mnBmBn −321B1Bi = B1Bi. | {z } =0 Similarly, you should show that

1j 2k 1j 3k ijkF F = B3Bi, ijkF F = −B2Bi. Plugging these results in the expression for the determinant, we get

2 det F = Ei(E1B1Bi + E2B2Bi + E3B3Bi) = (E~ · B~ ) .

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 γ α = γ2 − 1 = 0.0462 ⇐⇒ α = = 0.0461, α0 0 α0 0 1.001 7 and v0 = cα0 = 1.38 × 10 m/s.

q 2 2 −12 (b) From R = p0/qB, and p0 = E − mp = 43.33 MeV/c = 6.9 × 10 J/c, we get

6.93 × 10−12 6.93 × 10−12 R = = = 11.5 cm, cqB (3 × 108)(1.6 × 10−19)1.26 or a little over four inches.

11.17. A K0 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 K0. Solution: First calculate the momentum:

−19 8 −11 p0c = qBRc = (1.6 × 10 )(1.25)(0.55)(3 × 10 ) = 3.3 × 10 J = 206 MeV. √ Then the energy: E = 2062 + 139.572 = 249 MeV. The mass of the K0 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

−19 8 −11 p0c = qBRc = (1.6 × 10 )(0.5)(0.67)(3 × 10 ) = 1.61 × 10 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 2 2 p 2 2 Eπ = 100.5 + 139.57 = 172 MeV,Ep = 100.5 + 938.27 = 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. ∂E ∂ F µi = ∂ F 0i + ∂ F ji = ∂ E + ∂ ( B ) = ∂ E − ∂ B = i − (∇~ × B~ ) . µ 0 j 0 i j jik k 0 i ijk j k ∂t i

σξµi 11.20. Show that ∂σFξµ = 2[∂0Bi + (∇~ × E~ )i].

0ijk Solution: Use F0k = −Ek, Fjk = jkmBm, δkk = 3, and = ijk:

σξµi 0ξµi jξµi 0jµi j0µi jkµi ∂σFξµ = ∂0Fξµ + ∂jFξµ = ∂0Fjµ + ∂jF0µ + ∂jFkµ 0jki j0ki 0ijk = ∂0Fjk + ∂jF0k + jk0i∂jFk0 = (∂0Fjk − ∂jF0k + ∂jFk0)

= ijk(∂0Fjk − 2∂jF0k) = ijk[∂0(jkmBm) + 2∂jEk]

= ∂0(jkijkmBm) + 2ijk∂jEk = ∂0[(δkkδim − δkmδik) Bm] + 2(∇~ × E~ )i | {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 Sµν and Aµν are, respectively symmetric and antisymmetric under the µν interchange of their indices. Show that S 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:

µ (a) ∂µJ = 0.

µν (b) Let SA stand for S Aµν = 0. Then

µν µν ♣♠ ♠♣ SA = S Aµν = −S Aνµ = −S A♠♣ = −S A♠♣ = −SA.

So, SA = 0.

µν ν ∂ν∂µF = µ0∂νJ .

µν Now note that ∂ν∂µ is symmetric and F 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 3 ~ (2πc/λ) 2 uγ = 2 3 (−2πcdλ/λ ) π c ∞ exp (2π~c/λkBT ) − 1 Z ∞ 8πhc 1 = 5 dλ. 0 λ exp (hc/λkBT ) − 1

Equation (12.9) followd from (12.7) in a similar way. 12.2. Consider the reaction γ + γ −→ p +p ¯, which can occur if the initial photons are extremely energetic. Here p is a generic particle andp ¯ its antiparticle.

(a) Use Equation (8.24) to show that

2 mp E1E2 = sin2(θ/2)

where mp is the mass of p (orp ¯), 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 E1E2? 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 E1E2?

(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

2 2 p1 • p2 = 2mp, ⇐⇒ E1E2 − |~p1||~p2| cos θ = 2mp. 168 Early Universe

(a) Since for photons E1 = |~p1| and E2 = |~p2| (speed of light is 1), we get

2 2 2 E1E2(1 − cos θ) = 2mp ⇐⇒ E1E2 sin (θ/2) = mp.

2 (b) E1E2 is minimum when sin (θ/2) = 1 or θ = π/2.

2 (c) E1 + E2 = E1 + mp/E1. Set the derivative of this expression equal to zero and solve for E1: 2 mp 1 − 2 = 0 ⇐⇒ E1 = mp. E1 2 Part (a) now gives mpE2 = mp, 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?

(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 propor- tional 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

4 −8 4 7 2 Iγ = σBT = 5.67 × 10 (5800) = 6.4 × 10 W/m .

(b) N = 6.4 × 107/100 = 640, 000. The Sun is really bright!

2 8 2 7 26 Ptot = 4πR Iγ = 4π(7 × 10 ) (6.4 × 10 ) = 3.94 × 10 W.

(d) Divide (c) by c2: dm 3.94 × 1026 = = 4.4 × 109 kg/s. dt 9 × 1016 169

(e) 9 dm/dt 4.4 × 10 −21 k = = 30 = 2.2 × 10 . M 2 × 10 (f) dm = −km ⇐⇒ m(t) = m e−kt = M e−kt. dt 0 (g) M = M e−kt1/2 ⇐⇒ ln 2 = kt 2 1/2 or ln 2 0.693 t = = = 3.15 × 1020 s, 1/2 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

−33 4 3 −33 4 3 −33 3 ργ = 8.38 × 10 T kg/m = (8.38 × 10 )2.725 kg/m = 4.62 × 10 kg/m , and from (12.27), we obtain

4.62 × 10−33 kg/m3 Ω = = 5.4 × 10−5, γ 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 2 −27 can assume that their energy is just mpc . With mp = 1.67 × 10 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

7 3 3 7 3 3 8 3 nγ = 2 × 10 T photons/m = (2 × 10 )2.725 photons/m = 4 × 10 photons/m

for the present number density of photons. 170 Early Universe

(b) −27 3 −28 3 ρb = Ωbρcrit = 0.0484(8.5 × 10 kg/m ) = 4.1 × 10 kg/m .

(d) Take the ratio of nγ over nb:

8 3 nγ 4 × 10 photons/m 9 = 3 = 1.6 × 10 photons/baryons. 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):

20 2.32 × 10 −11 tα = √ s = 4 × 10 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

−33 4 29 3 ρα = αργ = 32.375(8.38 × 10 T ) = 2.7 × 10 kg/m .

(b) 1/3 M⊕ 3M⊕ −6 ρα = 4 3 ⇐⇒ R⊕ = = 5.3 × 10 m = 5.3 µm. 3 πR⊕ 4πρα 171

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

20 2.32 × 10 −5 tα = √ s = 1.3 × 10 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 π0 (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 π0? 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.

(d) The mean life of π0 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:

9 2 −1 (a) Using Tth = 4.3 × 10 mc MeV K, we have

9 11 Tπ0 = 4.3 × 10 (135) = 5.8 × 10 K, 9 11 Tπ± = 4.3 × 10 (139.57) = 6 × 10 K, 9 12 T3 = 4.3 × 10 (500) = 2.15 × 10 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.

20 2.32 × 10 −5 tα = √ s = 7.9 × 10 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

−6 66 × 10 −2538.5 −1102 r ± = exp − = e = 2.75 × 10 π 2.6 × 10−8 −6 66 × 10 −337216891830 r 0 = exp − = 5.75 × 10 . π 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

20 2.32 × 10 −5 tα = √ s = 8.69 × 10 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)

−33 4 3 ρrel = αργ = 5.375(8.38 × 10 T ) kg/m = 5.375(8.38 × 10−33(1011)4) kg/m3 = 4.5 × 1012 kg/m3.

Using (12.47), the contribution from matter density is

−28 3 3 −28 11 3 3 3 ρm(T ) = 1.32 × 10 T kg/m = 1.32 × 10 (10 ) kg/m = 13200 kg/m ,

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

2 2 2 mn = (mp + ∆m) ≈ mp + 2mp∆m.

Substituting this and letting mp = m, we get

mE1 = m∆m + E3(E1 + m) − E1E3 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 p for E1, 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.

1 (a) Which particles of Table 12.1 are present at this time? 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

1/2 −30 1/2 p4 t p4 10 aα(t) = 4αΩγ = 4(32.375)(0.0000538) 17 tH 4.59 × 10

−26 or aα(t) = 4.26 × 10 .

1Hint: How old is the universe at the beginning of the second epoch? 174 Early Universe

(b) The universe consists of electrons, positrons, neutrinos, and photons. So, α = 5.375, and 1 1/2 a (t) = p4 4(5.375)(0.0000538) = 2.7 × 10−10. α 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. hc (6.626 × 10−34)(3 × 108) λ = = = 3.53 × 10−13 m. E 5.63 × 10−13

(b) Equation (12.16) gives the relation between wavelength and temperature: 5.33 × 10−4 λ T = 5.33 × 10−4 m K ⇐⇒ T = = 1.51 × 109 K. 2 3.53 × 10−13

(d) From the end of the third epoch until tα, the neutrons have 19.68 seconds to decay. Therefore the fraction remaining is

N(t ) α = 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,

−33 9 4 3 ργ = 8.38 × 10 (10 ) = 8380 kg/m and −33 9 4 3 ρν = 0.681(8.38 × 10 )(10 ) = 5707 kg/m .

For ρm we use (12.47):

−28 9 3 3 ρm(T ) = 1.32 × 10 (10 ) = 0.132 kg/m .

This gives a ratio of ρ 8380 γ = = 63485. ρm 0.132 In general, −33 4 ργ(T ) 8.38 × 10 T −5 = −28 3 = 6.35 × 10 T. ρm(T ) 1.32 × 10 T 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 (∇ × B) · da = µ0 x J · da. S S This must be true for all S. Therefore, the integrands must equal: ∇ × B = µ0J. 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) −∇2B = −∇2B, | {z } =0 by (2) and ∂E ∂ ∂ ∂B RHS = µ ∇ × = µ (∇ × E) = µ − . 0 0 ∂t 0 0 ∂t 0 0 ∂t ∂t Now set the two sides equal and obtain

2 2 ∂ B ∇ B = µ00 . ∂t2

√ i(ωt−k·r) i(ωt−k·r) A.3. Consider E = E0e and B = B0e , 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; ω (3) k × E = ωB ; (4) k × B = − E . 0 0 0 c2 0

(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: ∂ ∂ ∂ ∇ · (Cf) = (C f) + (C f) + (C f) ∂x x ∂y y ∂z z ∂f ∂f ∂f = C + C + C = C · (∇f). x ∂x y ∂y z ∂z For the second identity use (A.8) and note that the derivatives operate only on f:     eˆx eˆy eˆz eˆx eˆy eˆz       ∂f ∂f ∂f  ∇ × (Cf) = det  ∂ ∂ ∂  = det   = (∇f) × C.  ∂x ∂y ∂z  ∂x ∂y ∂z          Cxf Cyf Czf Cx Cy Cz (a) Let’s apply the general results obtained above to the speciﬁc function at hand. From k · r = kxx + kyy + kzz, we get ∂ ei(ωt−k·r) = −ik ei(ωt−k·r), ∂x 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

i(ωt−k·r) i(ωt−k·r) 0 = ∇ · E = ∇ · [E0e ] = E0 · [∇e ] i(ωt−k·r) i(ωt−k·r) = E0 · [−ike ] = −iE0 · ke ⇐⇒ 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

i(ωt−k·r) i(ωt−k·r) i(ωt−k·r) i(ωt−k·r) ∇ × [E0e ] = [∇e ] × E0 = [−ike ] × E0 = −iωB0e ,

or k × E0 = ωB0. The fourth equation follows in the same way. 179

A D B C 1

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 2 k × (k × E0) = k (k · E0) −(k · k)E0 = −|k| E0, | {z } =0 and on the right, using (4), you get ω2 ωk × B = − E . 0 c2 0 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 · eˆtBC + E1 · (−eˆt)DA = (E2 · eˆt − E1 · eˆt)BC, C where eˆt 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 · eˆt = E1 · eˆt, 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−1A = 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−1A = AA−1 = 1, we get

A^−1A = AA^−1 = 1˜ ⇐⇒ AeAg−1 = Ag−1Ae = 1.

Therefore, Ag−1 is the inverse of Ae. B.2. Show that AAe = AAe = 1 for a 3×3 matrix implies that each row (or column) of A has 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 AAe is the product of the ith row of Ae and the jth column of A. But the ith row of Ae is the ith column of A. Thus the ijth element of AAe 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 AAe 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 AAe = AAe = 1. −1 B.3. By writing out the 4 × 4 matrices, verify that Λrot as given by Equation (B.4) is the inverse of Λrot as given by Equation (B.3).

Solution: Given that AAe = AAe = 1, this is a simple exercise in matrix multiplication. ˆ ˆ 2 2 ˆ 2 B.4. Using relations such as βxβy + a12a22 + a13a23 = 0 and a12 + a13 = 1 − (βx) , 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   = γ 0 0

 0  ˆ  βx  ˆ Λ12 = γ γβ 0 0   = γββx = γβx a12 a13  0  ˆ  βy  ˆ Λ13 = γ γβ 0 0   = γββy = γβy a22 a23  0  ˆ  βz  ˆ Λ14 = γ γβ 0 0   = γββz = γβz. a32 a33 The second row is only slightly more complicated:

1 ˆ 0 Λ21 = γβx γβx a12 a13   = γβx 0 0

 0  ˆ ˆ  βx  ˆ 2 2 2 ˆ 2 Λ22 = γβx γβx a12 a13   = γβx + a12 + a13 = 1 + βx (γ − 1), a12 | {z } ˆ 2 a13 =1−βx because the ﬁrst row of the submatrix A in (B.3) has unit length.

 0  ˆ ˆ  βy  ˆ ˆ ˆ ˆ Λ23 = γβx γβx a12 a13   = γβxβy + a12a22 + a13a23 = βxβy(γ − 1), a22 | {z } =−βˆ βˆ a23 x y

because the ﬁrst row of the submatrix A in (B.3) is orthogonal to its second row.

 0  ˆ ˆ  βz  ˆ ˆ ˆ ˆ Λ24 = γβx γβx a12 a13   = γβxβz + a12a32 + a13a33 = βxβz(γ − 1), a32 | {z } =−βˆ βˆ a33 x z

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−1Λ = ΛΛ−1 = 1. Solution: First write Λ−1 in block form:   γ −γβ~e Λ−1 =   .  ↔  −γβ~ Λ

Now multiply it on the right by the block form of Λ:      ↔  γ −γβ~e γ γβ~e γ2 − γ2β2 γ2β~e− γβ~e Λ −1       Λ Λ =     =   . ↔ ↔ ↔ ↔↔ −γβ~ Λ γβ~ Λ −γ2β~ + γΛ β~ −γ2β~β~e + Λ Λ

Let’s look at each block element of Λ−1Λ:

−1 2 2 Λ Λ 11 = γ (1 − β ) = 1. ↔ For Λ , I’ll use (B.8). Then

1 0 0 −1 2 ~e ~e↔ 2 Λ Λ 12 = γ β − γβ Λ = γ βx βy βz − γ βx βy βz 0 1 0 0 0 1  2  βˆx βˆxβˆy βˆxβˆz ˆ ˆ ˆ 2 ˆ ˆ − γ(γ − 1) βx βy βz βxβy βy βyβz 2 βˆxβˆz βˆyβˆz βˆz 2 = γ βx βy βz − γ βx βy βz − γ(γ − 1) βx βy βz = 0 0 0 .

Similarly, 0 −1 Λ Λ 21 = 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)2B2, with  2   2   2  βˆx βˆxβˆy βˆxβˆz βˆx βˆxβˆy βˆxβˆz βˆx βˆxβˆy βˆxβˆz 2 ˆ ˆ ˆ 2 ˆ ˆ ˆ ˆ ˆ 2 ˆ ˆ ˆ ˆ ˆ 2 ˆ ˆ B = βxβy βy βyβz βxβy βy βyβz = βxβy βy βyβz = B. 2 2 2 βˆxβˆz βˆyβˆz βˆz βˆxβˆz βˆyβˆz βˆz βˆxβˆz βˆyβˆz βˆz You should go through the matrix multiplication and verify this equation. Now we have ↔↔ Λ Λ = 1 + 2(γ − 1)B + (γ − 1)2B = 1 + (γ2 − 1)B = 1 + γ2β2B. 184 Derivation of 4D Lorentz transformation

−1 This is one of the terms in Λ Λ 22. The other term is   βˆx 2 ~~ 2 2 ˆ ˆ ˆ ˆ 2 2 −γ ββe = −γ β βy βx βy βz = −γ β B. βˆz

Adding this to the previous equation, we get

1 0 0 −1 Λ Λ 22 = 1 = 0 1 0 . 0 0 1

−1 −1 This completes the proof that Λ Λ = 1. You should now show that ΛΛ = 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) = beˆz, the location of the pinhole. C.2. Derive Equation (C.2). Solution: From (C.1), we get

rq0 ≡ r1(tq0 ) = (1 − tq0 )rq + tq0 beˆz ⇐⇒ rq0 − beˆz = rq − beˆz − tq0 rq + tq0 beˆz or rq0 − beˆz = (1 − tq0 )(rq − beˆz).

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

d|rq − beˆz| tp0 = 1 − . (rp − beˆz) · (rq − beˆz) Solution: Rewrite Equation (C.5) as

[(1 − tp0 )rp + tp0 beˆz − (1 − tq0 )rq − tq0 beˆz] · (rq − beˆz) = 0.

Adding and subtracting beˆz, we can re-express this equation as

[(rp − beˆz)(1 − tp0 ) − (rq − beˆz)(1 − tq0 )] · (rq − beˆz) = 0, or 2 (1 − tp0 )(rp − beˆz) · (rq − beˆz) − (1 − tq0 ) |rq − beˆz| = 0, | {z } =d/|rq−beˆz| 186 Relativistic Photography Formulas

or (1 − tp0 )(rp − beˆz) · (rq − beˆz) = d|rq − beˆz|, which yields d|rq − beˆz| 1 − tp0 = . (rp − beˆz) · (rq − beˆz) C.4. Derive Equation (C.7).

Solution: Add and subtract beˆz:

s = r2(tp0 ) − r1(tq0 ) = (1 − tp0 )rp + tp0 beˆz − (1 − tq0 )rq − tq0 beˆz

= (1 − tp0 )rp + tp0 beˆz −beˆz + beˆz −(1 − tq0 )rq − tq0 beˆz | {z } =0

= (1 − tp0 )(rp − beˆz) − (1 − tq0 )(rq − beˆz).

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 − beˆz = hxq, 0, −bi, and with rp = hx, y, zi, we get q 2 2 |rq − beˆz| = xq + b , and (rp − beˆz) · (rq − beˆz) = xxq − b(z − b). Plugging these in (C.2) and (C.6) yields (C.8). Plugging them in (C.7) gives

q 2 2 d xq + b d s = hx, y, z − bi − q hxq, 0, −bi, xxq − b(z − b) 2 2 xq + b

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 2 2 2 t = −|rp − beˆz| = − x + y + (z − b) .

0 C.7. In this problem you’ll ﬁnd xq in a diﬀerent way. 0 (a) If the event of the explosion is at (xq, 0) in O, what is it in O ? (b) What distance does Q travel in O0 before C0 takes the picture of the object?

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

0 0 xqi = γ(xq − 0) = γxq, tq = γ(0 − βxq) = −γβxq.

0 0 0 (b) C takes the picture of the object at t = 0. Therefore, the object travels for tq before 0 2 the picture is taken. The distance is β|tq| = γβ |xq|. (c) So, the ﬁnal location, i.e., the location at which the picture is taken is x x0 ≡ x0 = x0 + β|t0 | = γx + γβ2|x | = γx − γβ2x = q . qf q qi q q q q q γ