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ĐẠI HỌC QUỐC GIA HÀ NỘI

KHOA CỎMG NGHỀ
HỆ ĐÀO TẠO CHẤT LƯỢNG CAG
NGÀNH VẬT LÝ KỸ THUẬT

GIÁO lld ữ m VẬT IvÝ I.ƯỢIVG TỂ
Tóm tắt lý thuyết và bài tập


D e d ic a ted to P ro fesso r lUarcus M ainardi.
an extraordinary leacher

PART I TH E S P E C IA L T H EO R Y 0 F R E L A n V I T Y
C H A P TER 1

G A L IL E A N TR A N SP O R M A TIO N S
1.1
1.2
1.3
1.4

1.5

C H A P TER 2

T H E P O S TU LA T ES 0 F E IN S T E IN
2.1
2.2
2.3
2.4

C H A P TER 3

3.4

T he Constancy o f the Speeđ o f Light
The Invariancc o f M axw eU’s Equations
G eneral Considerations in Solvin g Problem s Involving
Lorentz Transformations
Sim ultaneity

RELATIVISnC
4.1

C H A P TER 5

A b solute Space and the Ether
The M ich elson -M orley Experiment
Length and Tim e M easurem ents— A Q uestion o f Principle
T he Postulates o f Einstein

TH E LO R E N T Z C O O R D IN A T E TRA N SFO R M A TIO N S
3.1
3.2
3.3

C H A P TER 4

Events and Coordinates
G alilean Coordinate Transíormations
G alilean V elocity Transformations


G alilean Acceleration Transíormations
Invariance o f an Equation

LEN G TH C O N T R A Q IO N

The D eíìn ition o f Length

R E L A H V I S n C H M E D ILA T IO N
5.1
5.2

Proper Tim e
T im e Dilation

C H A P TER 6

RELATIVISnC

CH A P TER 7

R E L A n V I S n C V E L O C IT Y TR A N SP O R M A n O N S
7.1
7.2
7.3

S P A C E - n M E M EASU R EM EN TS

T he Lorentz V elocity Transformations and the Speed o f Light
G eneral Considerations in S olvin g V elocity Problems
T he R elativistic D oppler E íĩect

10
10
10
10
11

15
15
16

20
20

23
23
23

27

37
37
38
38


CHAPTER 8

M A SS, EN ER G Y , A N D MOMENTUM IN R E L A n V I T Y
8.1
8.2
8.3
8.4
8.5
8.6
8.7

T he N e ed to Redefìne C lassical M om entum
The Variation o f M ass with V elocity
N e w to n ’s Second Law in Rclativity
M ass and Energy Relationship: E = mcP'
M om entum and Energy Relationship
U nits for Energy and M om entum
General Considerations in S olvin g M ass-E n ergy Problem s

45

12,5

45
45
46
46
46
47
47

C H A P TER 13

59

C H A P TER 1 4

12.7

ELEC TR O N S P IN
13.1
13.2
13.3
13.4
13.5

PART I I THE QUANTUM T H EO R Y 0 F ELEC T R O M A G N E T IC R A D IA T IO N

Q uantization o f the Direction o f ửie Orbital Aiigular
M om entum
Explanation o f the Z eem an Eữect

T he S tem -G erlach Experiment
Electron Spin
Spin -O rb it C oupling
Fm e Structure
Total Angular M om entum (The Vector M odel)

120
120

126
126
127
128
128
129

AND M A H ER
PART I V M A N Y-ELEC TR O N ATOMS
CHAPTER 9

ELEC T R O M A G N E T IC R A D IA H O N — PHOTONS
9.1
9.2
9.3
9-4
9.5

T he Theory o f Photons
T he Photoelectric E ffect
The Com pton Effect
Paứ Production and Aiưiihilation
Absorption o f Photons

59
60
61
62
63

14.1
14.2
14.3
14.4

C H A P TER 1 5
CHAPTER 10

M ATTER W AVES
10.1
10.2
10.3
10.4

D e B roglie W aves
Experimental V'erification o f D e B roglìe’s H ypothesis
The Probability Interpretation o f D e B roglie Waves
T he H eisenberg U ncenain ty Principle

80

CHAPTER 11

TH E B O H R ATOM
11.1

The H ydrogen Spectrum

11.2
11.3
11.4
11.5

The Bohr T heory o f the H ydrogcn Atom
E m ission o f Radiation in B o h rs Theory
Energy Level Diagram s
H ydrogenic Atom s

CHAPTER 12 E L E Q R O N O R B IT A L MOTION
12.1
12.2
12.3
12.4
12.5

Orbital Angular M om entum from a C lassical Vievvpoint
C lassical M agnetic D ip ole M om ent
C lassical Energy o f a M agnetic D ip ole M om ent in
an Extem al M agnetic Field
T he Zeem an Experiment
Q uantization o f the M agnitude o f the Orbital Angular
M om entum

C H A P TER 1 6

103
103
104
107
107

117
117
118
119
119

S pectroscopic N otation for E lecừon Coníìgurations in A tom s
T he Periodic Table and an A tom ic Shell M odel
S pectroscopic Notation for A tom ic States
A tom ic E xcited States and L S C oupling
T he A n om alou s Z eem an Efifect

X -R A Y S
16.1
16.2
16.3
16.4
16.5
16.6
16.7

103

Q uantum -M echanical System s with M ore Than One E lecưon
T he Pauli E xclusion Principle
A S in gle Particle in a O ne-D im ensional B ox
M any Particles in a O ne-D im ensional B ox

M A N Y -ELEC TR O N ATOMS AN D TH E P E R IO D IC T A B L E
15.1
15.2
15.3
15.4
15.5

80
81
82
83

PART I I I H Y D R O G EN LIK E ATOMS

T H E P A U L I E X C L U S IO N P R IN C IP L E

X -R ay Apparatus
Production o f Brem sstrahlung
Production o f Characteristic X-R ay Spectra
T he M o se le y Relation
X -R ay A bsorption E đges
A u ger EflFect
X -R ay Pluorescence

135
135
135
135
136

140
140
141
142
142
143

157
157
157
158
160
160
161
161

PART V N U C L E A R P H Y S IC S
CH A P TER 1 7

P R O P E R T IE S 0 F N U C L E I
17.1
17.2
17.3
17.4
17.5
17.6

T h e N u cleon s
N u cleo n Porces
T h e Deuteron
N u clei
T he N u cleu s as a Sphere
N u clear B inding Energy

173
173
174
174
174
175
175


CHAPTER 18

N U C L E A R M O DELS
18.1
18.2

CHAPTER 19

Liquid Drop M odel
Shell M odel

TH E D ECA Y 0 F U N S T A B LE N U C L E I
19.1
19.2
19.3
19.4
19.5

N uclear D ecay
The Slatistical Radioactive D ecay Law
G amma Decav
A lp ha D ecay
Beta D ecay and the Neutrino

181

CH A P TER 2 4

181
182

D IS T R IB U T IO N P U N C n O N S
24.1
24.2
24.3

D iscrete Distribution Punctions
C ontinuous Distribution punctions
Pundamental Distribution Punctìons and Density o f States

N U C L E A R R EA CTIO N S
20.1
20 .2
2 0.3
2 0 .4
2 0.5
20 .6
20 .7

CHAPTER 21

N otaiion
C lassitìcation o f Nuclear R eactions
Laboratory and Center-of-M ass S ystem s
Energetics o f Nuclear R eactions
N u clear Cross Sections
N u clear Pission
N u clear Pusion

P A R T IC L E P H Y S IC S
21.1
2 1 .2
2 1.3
2 1 .4
21.5
2 1 .6
21.7
21.8
21 .9
2 1 .1 0

Particle G enealogy
P an icle Interactions
Conscrvation Laws
Conservation o f Leptons
C onservation o f Baryons
Conservation o f Strangeness
Conservation o f Isotopic Spin and Parity
Short-Lived Particles and the R esonances
The E ightíold Way
Quarks

193
193
194
194
195

C H A P TER 2 5

C L A S S IC A L S T A n S n C S : T H E M A XVVELLBỌLTZM ANN D IS T R IB U n O N

C H A P TER 2 6

207
207
208
209
210
210
211

224
224
224
226
227
227
227
227
228
229
230

CH A P TER 2 7

M O LECU LES
22.1
2 2.2

CHAPTER 23

M olecular Bonding
Excitations o f Diatom ic M olecu les

K IN E T IC TH EO R Y
23.1
23.2

Average Values in a Gas
T he Ideal Gas Law

245
245
246

259
259
259

F erm i-D ừ ac S taóstics
B o se-E in stein Statistics
High-Temperature Lim it
Two U se íu l Integrals
B lackbody Radiation
Free Electron Theory o f M etals
S peciíic H eats o f Crystalline Solids
T he Q uantum -M echanical Ideal Gas
Derivation o f the Q uantum Distribution Punctions

S O L ID S
27.1
27.2

T he Band T heory o f Solids
Superconductivity

287
287
288
288
2S9
289
292
296
301
305

309
309
318

325

A p p e n đ ix
S om e Pundamental Constants in Convenient Units
S om e U seíiil C onversions
M asses o f S om e Particles
M asses o f N euư al A tom s

In d e x

276

QUANTUM S T A H S n C S : P E R M I- D IR A C AND
26.1
26.2
26.3
26.4
26.5
26.6
26.7
26.8
26.9

207

PART V I ATOMIC SYSTEM S
CHAPTER 22

268
269
270

193

B O S E - E IN S T E IN D IS T R IB U n O N S
CHAPTER 20

268

325
325
325
326

333


Galllean
Transformations
I ì
Ể1

E V E ÌST S A N D C O O R D IN A T E S

\Ve b e ^ n by considering the con cept o f a physical cvent. The cvcnt niisiht be the striking o f a tree by a
ligluiiing bó)t or the collision o f tw o particles, and happens at a point in space and at an instant in time. The
particular ẹvent is speciíìed by an obser\’cr by assicn in g to it four coordinales: the three position
coorđinates >r, y , z that measure the distance from the oricin o f a coordinatc sỵstcm \vhere the observer
is located, à id the tim e coordinate í that the obser\'cr rccords wiih his clock.
Consiđer now tu’0 observers, o and 0 \ where o ' travels with a constant velocit\' V \vith respect to o
along th eừ cồm m on .ĩ — x ' axis (Fig. 1-1), Both o b s e n ers are equipped \vlih melcrsticks and clocks so that
they can m ẹ ^ u r e coordinates o f events. Purther, suppose that both observers adjiist their clock s so that
wh en they'*p^s each other at J = y = 0, the clocks rcad' I = i' = 0. Aiiy gÌN cn eveni p will have eight
num bers assợciated with it, the four coordinates
assiuncd by o and tlic four coordinates
ự , / . / , i') aisign ed (to the sam e evem ) by O '.

O

Evenl
(X ./.Í . I)

Fig. 1-1


GALILEAN TRANSPORMATIONS

G a LILEAN

c o o r d in a t e t r

.\n sfo

[CHAP. 1

Ans.'

r .m a t i o n s

The relationship betwecn the measurements (,v.y. /) o f o and the measurements (.r ',y ,
a palicular event is obtaincd by exaniining Pic. 1- 1:
y'

V!

. t ’ = -V -

CHAP. 1]

=

V

r' =

GALILEAN TRANSPORMATIONS

The coordinates assigned to the bird by the man on the station platform are
{x .y ,z , /) = (8 0 0 m ,0 .0 , 2 0 s)

t') o f O '

The passenger measures the distance y to the bird as

z

x' = x - V I = 8 0 0 m - (3 0 m /s)(20s) = 200m
Therefore the birds coordinales as determined by the passenger are

additon, in classical phvsics it is im plicitly assumed that

(,í',y ,z './') = {2 0 0 m. 0, 0,2 0 s)

í' = /
:se fcur equations are called the G alilean coordinate transform atĩons.
1.2.
G ALILEAN V E L O C IT Y T R .\N S F 0 R M A T 1 0 N S
In aidition lo the coorđinates o f an event, the velocity o f a particle is o f interest. Observers o and O'
1 describe the particles velocity by assigning three com ponents to it, with (u ,,
u.) being the velocity
nponents as measured by o. and ịú^, U y, uỊ) being the velocit>' com ponents as meãsured by O '.
The relationship benveen (Uj,
u .) and (u^,
iiỊ) is obtained from the lim e diữerentiation o f the
lilear, coordinate transformations. Thus, from .v' = X — 17,
,

dx

d ^

di

(cix

Reíer to Problem 1.1. Five second s afker making the íìrst coordinate measurement, the man on th
platfom i determines that the bird is 850 m away. From these data find the velocity o f the bir
(assum ed constant) as delcrmined by the man on the platform and by the passenger on the traứi
Ans.

The coordinates assigned to the bird at the second position by ứie man on the platíorm are
f c . y 2. - 2. ' 2) = (8 5 0 m, 0 ,0 , 25 s)
Hence, the velocity u, o f the bird as measured by the man on the platform is
'

\

Í2 - / |

'

2 5 s-2 0 s

The positive sign ìndicates the bird is Aying in the positive jr-direction. The passeriger finds that at tl
second position the dỉsĩance y*, to the bird ỉs
ogether, the Gaiiìean velocit}- lransfo n n a tio n s are
—V

úy

•rí = ,12 - i-ụ = 850 m - (30m /s)(25s) = 100 m
=

ú. = u.

Thus. (JCÍ.. VỊ.r'^, 4 ) = (lOOm, 0, 0, 25 s), and the velocity u', o f the bird as measured by the passeng
on the train is
4 -^ 1

100i n - 20 0 m

G AL ILE AN A C C E L E R A T IO N T R A iN SPO R M A T IO N S
The acceleration o f a particle is the time derivative o f its vclocity, i.e., a = du /d i , etc. To find the
lilean accelem tion :ransformarions we diíTcrentiate the velocity transformations and use the facts that
= l and L' = constant to obtain

= u, - 1) = lO m /s - 30m /s = -2 0 m /s

a', = a.
JS the measurcd acceleration com poncm s are thc same for all observers moving with uniform rclalive
Dcity.

INVARL-VNCE O F AN E Q U A T IO N

A sam ple o f radioactive material, at rest in the laboratory, ejects two electrons in opposi
direclions, One o f the electrons has a speed o f 0 .6 c and the other has a spced o f 0.7c, as m easun
by a laboratory observer. According to classical velocỉty transfonĩiations, what w ill be the speed (
one electron as measured from the othcr?
Ans.

B y invariance o f an equation it is meant that the equation vvill have the same form vvhen determined by
I observers. In classical theory It is assum ed ihai space and time measurements o f two observers a r é
Ited by the Galilean transformations. Thus, when a particular fom i o f an equation is determined by One
er\'er, the Galilean transformations can be applled to this fonn to determine the form for the other
erver. l f both fom is are thc same, the equation is invariant under thc Galilean transformations. See
blems 1.11 and 1. 12.

S o lv ed P ro b lem s
A passenger in a train moving at 30 m /s passes a man standing on a station platform at í = í" = 0.
Twenty seconds afler the train passes him, the man on the platform determincs that a bird Aying
along the tracks in Ihe same direction as the train is 800 m away. What are Ihe coordinates o f ứie
birđ as determined by Ihe passeimer?

= - 2 0 m /s

so that. as measured by the passenger, the bird is moving in the negative y-direction. Note that th
rcsuil is consistent with that obtained from the Galỉỉean velocity transformation:

Let obsen'er o be at rest with respect to the laboratory and let observer O' be al rest wiửi respect
the panicle moving with specd 0.6c (taken in the positive direetion). Then, from the Galilean veloci
tiansformation,
u', = u , - v = - 0 .7 c - 0.6c = - 1 .3c
This problem demonstrates ihai vclocities greater than Ihe specd of lighl are possible with t
Galilcan transformations, a result that is inconsistent with Special Relali\ ity.

A train m oving with a velocity o f 6 0 m i/h r passes through a railroad station at 12:00, Twcn
seconds later a bolt o f lightning strikes the railroad tracks one m ile from the station in the san
direction that the train is m oving. Find the coordinatcs o f the lightning flash as measured by i
observer at the station and by Ihe engineer o f the train.
Ans.

Both observers measure the limc coordinatc as
/ = /' = (2 0 s)

V3600s^


GALILEAN TRANSPORMATIONS

The-observer at the station measures the spatial coordinate to be .
determined by the engineer of Ihe Irain is

[CHAP. 1

CHAP. 1]

. The spatial coorđinatc as

GALILEAN TRANSPORMATIONS

(a) For the boy in the car the ball ơavels straight up and dovra, so
y =

Vot'

+ ịa í^ = (2 0 f t /s ỵ + ị ( - 3 2

= 20?" - 16f'2

y=V = o

y = ,x - VI = 1 mi - (60 mi/hr)

(Ề) For the stationaiy observer, One obtains (rom the Galilean transformations
1.5.

A hunter on the ground íìres a bullet in the nonheast đirection which sưikes a deer 0.25 miles from
Ihe hunter. The bullet travels with a speed o f 1800m i/hr. At the instam when the bullet is tìred, an
airplane is directiy over the hunter at an allitude h o f one mile and is traveling due east with a
velocity o f 600m i/hr. ViTien the bullet strikes the deer. what are the coordinates as determined by
an observer in the airplane?
Ans.

t = í

1.9.

Using the Galilean transformations,
1800/hr

: I0-' hr

2 = r* = 0

Consider a mass anached to a spring and m ovm g on a horizontal, frictionless suríace. Show, frorr
the classical transíormation laws, that the equations o f motion o f the mass are ứie same ai
detennined by an observer at rest with respect to the suríace and by a second observer moving wiứ
constant velocity along the direction o f the spring.
Ans.

x' = x - v t = (O.:5m ĩ)cos45“ - (600 mi/hr)( 1.39 X 10”* hr) = 0.094 mi
>■' = y = (0.25 mi) sin 45° = 0 .177 mi

The equation o f motion of the mass, as determined by an observer at rest with respect to the surface, i:
F = ma, or

z' = 2 —/ i = 0 — lm i = —Imi

1.6.

y = ý = 2 0 t— l 6 r

I = x ' + p /= 0 + lOOí

- k ịx - X o ) = m ^

u

To determine the equation o f motion as íound by the second observer we use the Galileai
transformations to obtain

An observer, at restwith respect to the ground, observes the following colHsion. A particle o f mass
m, = 3 kg moving with velocity Uị = 4 m / s along the x-axis approaches a second particle o f mass
m, = 1 kg moving with velocity u, = - 3 m /s along the .v-axis. After a head-on collision the
ground observer finds that
has velocity
= 3 m /s along the A:-axis. Find the velocity
o f m,
alter the collision.

d -x
X= /

+ v i'


Xo — Xo + VI

Substitutíng these values in (/) gives

ínitial morĩìenmm = final momentum
m|U, + m,Uj = nt,u’Ị + »12«;
(3 kg)(4 m/s) + (1 kg)(-3 m/s) = (3 kg)«; + (1 kg)(3m /s)

Because (/) and (2) have ứic sarae fomi, the equation o f motíon is invariant under the Galilea
tnuisíormations.

9 kg • m /s = (3 kg)«; + 3 kg • m/s
Solving. u’ị = 2 m/s,
1.10.
1.7.

Show that the electromagnetic wave equation,

A second observcr, O '. who is walking wiih a velocity o f 2 r a /s rclative to thc ground along the ;taxis observcs ihe collision described in Problem 1.6. What are the system momenta before and after
the collision as determined by him?
Ans.

Using ihc Galỉlcan vclocity transformations,
i/, = I/|
= «:
= «;
1/" = uị

- c = 4 m /s - 2 m /s = 2 m/s
- D= - 3 m /s - 2 m/s = - 5 m/s
- c = 2 m /s - 2 m/s = 0

is not invaríant under the Galỉlean tiansfonnations.
Áns.

Tlie equation will be invariaat if it retains the same fonn when expiessed in teiim o f the ftew yariable
y , y , r'. We tìrst find from the Galilean transíormations thai

at “

—t = 3 m /s —2 m /s = 1 m/s

(iniúal momcntum)' =
+ m ,u ; = (3 lcgK 2m i's)-(-{lk gK -5m /s)= Ik g -m /s
(final momentum)' = « 1«;' + m,mV = (3 kg)(0) + (1 kgXl m /s) = 1 kg • m /s

m



ạ y ’^
From the Chain lute uid using the above resutts we have

Thus, as a result of the Galilcan lransformations, o ’ also dctermines that momentum is conserved (but
at a difĩcrent vaíuc from ihal found by O).

1-8.

An open car traveỉing at 100 ft/s has a boy in ìt \vho throws a baịi u p v ^ d with a veỉocity o f 2 0 f t /s
Write thc equation o f motion (giving posiiion as a fiinction o f time) for the ball as w en bỹ (õ) the
boy, (Ế) an observer stationary on the road.

Siitúlariy,
ẾÉ. Ẽ Ề
Ỉb2 * â r '2


GALĩLEAN TR^\NSFORMATIONS

_____,
3f'2
Substituting these exprcssions in the wave equaiion gives
ax'

ẵx'-

[CHAP. 1


; ĩr ậ
ai'2

3í'

3 )'-

í)r^

V

a t 'V

Thercforc the wavc equalion is m l invariant unđer the Galllean tiansformations, for the form of the
equation has changed because of the extra terni on thc left-hand side.
The electromagnetic wave cquation follows from Ma.\weH’s equations o f electromagnetic
thcory. By applying the procedure described here to Maxwell's equations, One finds that Maxwell's
equations also are not invariant under Galilean transformations. Compare with Problem 6,23 which
shows that the electrornagnetic wave cquation is invariant unđer a Lorcntz transformation.

Su p p lem en tary P roblem s
1.11.

A man (O') in ihe back o f a 20-ft Hatcar moving at 30 ft/s records that a Aashbulb is fìrcd in the fiont of the
Aatcar tv,'0 seconds after he has passed a man (O) on the ground. Find the coorđinates of the evenl as
d c te r m in e d b y e a c h o b s e rv c r.

A n s.

ụ . f ) = ( 2 0 ft, 2 s)- ( i í) = ( 8 0 ft 2 s)

1.12.

A boy sees a deer run dlrectly away from him. The deer is running wilh a speed of 20 mi/hr. The boy gives
Chase and nins with a speed of 8 mi/hr. \VhaI is the speed of the decr relativéto the boy?
Ans. 12 mi/hr

1.13.

A boy in a train throws a ball in the fo™-ard direction with a speed of 20mi/hr. If the train is moving with a
speed of 80 mi/hr, whai is the speed o f the ball as mcasuređ by a person on the ground?
Ans. 100 mi/hr

1.14.

A passenger walks backvvarđ along ihe aisle of a train wiih a speed of 2mi/hr as the train moves slong a
^ i g h t track at a constant speed of 60mi/hr with respect to the ground. Wiat is the passengcr’s speed as
measured by an obscrver siandmg on the grounđ?
Arts. 58 nií/hr

1.15.

A conductor standing ọn a railroad platíbmi synchronizes his vvatch with the engineer in the front o f a train
ữaveling at 60 mi/hr. The traín is I/4 milc long. Two minutes after the train leaves the plalfomi a bĩakeman m
the caboose pours a cup o f cofT«. \VhaI arc the coordinates of Ihe brakcman, as dcterminẽd bỹ * e engincér
and by the conductor, w h en the coíTee is poưred?

1.16.

^„ s.

ự . r") = ( - 1 mi, 2 m in ); (x. 0 = (I 2 mĩ. 2 m in)

sitting in a tTdin pours two cups o f coAee One 10 minutes a íte r Ihe oiher. The train is m oving in a

straight line with a velocìty of 2 0 m/s. UTial is the distance separation berween the two pourings as meíured
by a person on the g ro u n d ?

A ns.

1 2.000m

1.17. A one-kilogram ball is constrained to move to ihe north at 3 m/s. It makes a perfectly elastic collision with an
Ịđentical second ball which is al rest, and both balls move on a north-souứi áxis aflcr the coUision. Compute
in the laboratory system. the toul momcntum bcfore and aftcr the collision.
/íns. 3 kg - 111/5
1.Ỉ8. For Problem 1.17, calculatc the total encrgy before and aíter the collision.

/Ins.

4 5J

1.19. Refcr to Problcm 1.17. Caỉculaic the toial momcntum bcforc and after the collision as measurcd by an
observer moving nonhward at 1.5 m/s.
yífis. 0

CHAP. 1]

GALILEAN TRANSPORMATIONS

Ans.

2.25J

1.20.

For the observer in Problem 1.19, calculale ihe tota! energy before and aỉìer ihe collision,

1.21.

Repeat Problems 1.19 and 1.20 for an observer moving eastuard at 2 m/s.
west; 8.5 J

1.22.

A person is in a boai moving eastwarđ vvith a speed o f 15 f\/s. At the instant ihat ứie boat passes a dock, a
person on the dock throws a rock northward. The rock strikes ihe water 6 s later at a đistance o f 150 ft from thc
đock. Finđ ứie coordinates o f the splash as measured by ihe person in ứie boat.
Ans. (x ,y .0 = (-9 0 ft. I50ft,6s)

1.23.

Consiđer a one-dimensional, elastic collision Ihat takes place along ihe x-axis of o. Show, from ứie classicai
ưansformation equations, ihal kinetic energy will also be conserv'ed as determined by a second observcr, O '
who moves with constant velocity u along the x-axis o f o.

--Í/Ỉ5.

5 kg • m/s 37^ north ol


CHAP. 2]

THE POSTULATES 0 F ErNSTEIN

a well-defined proceđure by which it is measured. lf such a procedure cannot be formuIated, then the
quantity should not be employed in physics.
Einstein could find no way to justify operationally the Galilean transíormation t' = I, i.e., the statement
that two observers can measure the time o f an event to be the same. Consequently, the transformation
t' = I, and with it the rest o f the Galilean transformations, was rẹjected by Einstein.

The Postulates of
Einstein

2.4

T H E PO STULATES O F EIN ST E IN

Einstein’s guiding idea, which he called the Principle o f Reìativịty, was that all nonaccelerating
observers should be treated equally in all respects, even if they are movừig (at constant velocity) relative to
each oứier. This principle can be formalized as follows:
Posruỉate 1:

The Iaws of physics are ứie same (invarianl) for aỉl inerĩial (nonacceleratìng) observers.

Newton's lavvs o f motion are in aecord with the Principle o f Relativity, but M axw ell’s equations
together with the Galilean transformations are in conAict with it. Einstein could see no reason for a basic
difference between dynamical and elecưomagnetic laws. Hence his
Postuỉate 2:

In vacuum the speed of light as measured by all ỉnerlỉal obsen'ers is
c= 1

= 3

X

10* m/s

independent of Ihe motion of the source.
ỈL U T E SPACE A N D THE ETH ER
iquence o f the G alilean velocity tt-ansrormations is that if a certain observer nieasurcs a light
vel with the velocity c = 3 X 10* m /s , Ihen any other observer moving relative to him will
gsame light signal to travel with a velocitv different from c. What determines the particular
ne such that if an o b scn er is at rest rclative to this frame, this privileged observer will
^alue c for the velocity o f light signals?
nstein it was gcnerally believed that this privĩleged observer was the same observer for whom
^ations were valid, MaxweH’s equations describe electromagnetic thcory and predict that
Bc vvavcs will travel with the speed c = 1 / y r ^ = 3 X 10® m /s . The space that was at rest
this privilcged obser\'cr was called “ absolute spacc.” A ny other observer moving wilh
3 absolute space would find the speed o f light to be differenl from c. Sincc light is an
; wave, it was felt by 19th ceniury physicists Ihat a medium must exist through which the
Thus it \vas postulatcd thai the "ether" permeated all o f absolute space.

p lE L S O .N -M O R L E Y E X PE R IM E N T
cxists, then an observer on ihe earth m oving through the eiher should notice an “ether
ptus with the sensitiviiy to measure the earth’s motion through thc hypothesized ether was
ychelson in 1881, and reíìncd by M ichelson and Morley in 1887. The outcome o f the
pat no m otion through the ethcr w as deiecled. S ee Problcms 2.5, 2 .6 and 2.7

Solved Problem s
Suppose that a clock B is located at a distance L from an observer. Describe how this clock can be
synchronized with clock A, which is at the observer’s location.
Ans.

Set the (stopped) clock B to read ÍJ = L/c. At = 0 (as recordeđ by clock A) send a light sìgnal
toward the dislanĩ cỉock B. Start clock B whcn the signal reaches it.

A Aashbulb is located 30 km from an observer. The bulb is fưeđ and theobserver sees the f1ash at
1:00 P.M. What is the actual time that the bulb is fired?
The timc for Ihe iỉght signal to travel 30 km ỉs
..

Ai
c

30 X
3 x lO * m /s

I

X

lO -S

Thcreforc, the Aashbulb was fired 1 X 10“* s bcfore 1:00 P.M.
^ND T IM E M E A S U R E M E N T S — A Q U E S T IO N O F P R IN C IPL E
5ent cotnm on to both the null result o f the M ichelson -M orley experiment and the fact that
Jfons hold only for a pnvileged obsereer is the G alìlean transíomiations. These "obvious”
fw ere rc-exam ined by Einstcin from what might be termed an "opcrational" point o f view.
ne approach that any quaniity relcvant to physical thcories should, at least in principle, have

A rod is moving from left to right. When the left end o f Ihc rod passes a camera, a picture is taken
o f the rod together with a stationary calibrated meterstick. In the developed picturc the left enđ ol
the rod coincides wiưi the zcro mark and the right end coincides \vith the 0,90-m mark on ứiE
mcterstick. If the rod is moving at 0 .8c with respect to the camera, dctermine the acíual length ol
the rod.


THE POSTULATES OF EINSTErN

Ans.

(CHAP. 2

In order th.ll thc light sitỉnaỉ from thé right end of the rod be recorded by the canicra it must have
siancd from the 0.<50-m mark at an earlicr time HÌvcn by

c

0.90 r
= 3
3 X lũ*m /s

X

10“’ s

CHAP. 2]

2.5.

THE POSTLXATES OF EINSTEIN

Pigure 2-3 diagrams a M ichelson-M orley interíerometer oriented with one arm (/^) parallel to the
“ether wind," Show that if the apparatus is rotated through 90“, the number o f íringes, A N , that
move past the telescope crosshairs is, to first order in ( d/ c ) ,

During this time intenal the left cnd of the rod will advance Ihrough a distance As' given by (see
Fig. 2-1)
As' = tAí = (0.8 X 3 X 10*m /s)(3 X 10”’ s) = 0.72 m

C

l ]
(ố) SÌRnal arrivcs ừom rlght end
and is recorded hy op«n caniera
loịỊcther wiih signal from li*n cnd.

(a) Signal starts from rìpht
end; ciimcra shutler closcd.
Fìg. 2-1

Thcrefore. the acmal length of the rod IS I = 0.90 m + 0.72 m = 1.62m. This resull illusirales that
pholographing a moving rod will no! give Ils correct leneth.

I
Fig. 2-3

2.4.

Let tu'0 events occur ai equal distances from an observer. Suppose the obsen'er adopts the
following statement as a deíìnition o f simultaneity o f equidistaiit evems: “The two events are
simullaneoiis i f the light signals em itted from each event reach me at the same time.” Show that
according 10 this dclìnition, if the observ'er detcrmines Ihat two evcnts are simultaneous then
another ob sen xr, moving relative to him, will in general determine that the two events are nol
simultaneous.
Ans.

From Fig. 2-2 ii is secn that if the r»'0 lighl signals reach thc firet observer (O) at the same lime they
will necessarily rcach the second obscrvcr (O') at diíTcrcnl times. Since the two sịgnais slanẽd oul
cquiđistant from 0 \ he will, according lo the above definition, determine thai Ihe Mo events did not
occur simullaneously, bui ihal event fl happcned bcforc cvcnt A.

Ans.

For arm A, ihe lime for light to travel to mirror A is oblained by dividing ừìe path length
by thí
velociry o f light, which from the Galilean velocity transíormaiions is c - t’. On retum the path lengứ
is still /^, but now the velocity is c + t>, so the total tímc for ihc rounđ trip is

c -v^c + v
To travel along the other arm a light ray musl be aimeđ such that its resultanl velocity vecior (velocitỊ
with respect to the ether plus velociry o f Ihe ether with respccl to the intcrierometer) is perpendicula)
to arm A. This gives a speed o f
for both directions along path /j, so the time for the rount
trip ỉs
2lẹ

,
‘ - u(a) SÌ^naLv A and B start.

(b» Slpnol B rcaches 0\

(c) SỉgnaLs.4 and B rcach 0 simulUineously.

2/ , / c

ự l - (v^/v^)

If wc assume v/c ditĩercnce taken:

and tg can be cxpanded to first order ỉn {v/cÝ and thc tinn

(đ) Signal A nrachcs o.
Fig. 2-2

_ 2(1^ - /,)
ì

= Ia - la

21y

ljf_


THE POSTULATES OF EINSTEĨN

Now if the interierometcr is rotated 90°,
time difference. Thus

and /j

[CHAP. 2

: ìnterchanged, and there is also a reversal of the

and the imerfcrence pattem observed would show a ữinge shift of ầ N fringes, where

Here

r

_ ẵ - & ‘ _ c í.ò - ẵ ')
ụ ,+
T
À
Xc^
and X are the period and wavelength o f the light.

The Lorentz
Coordìnate
Transformations

Assume that the earứi’s velocity thiough the ether is the same as its orbital velocity, so that
V = 10“''c. Considcr a M ichelson-M orley experiment where the arms o f the mterferometer are each
lOm long and one arm is in the dừection o f motion o f the earth through the ether, Calculate the
diíĩerence in từne for the two light waves to ữavel along each o f the arms.
Ans.

Refer to Problem 2.5.
2
(3

2.7.

X

10* m/s)c^

(5m ) = 3.33

X

IQ-

The original M ichelson-M orleỵ experiment used an interíerometer with arms o f 11 m and sodium
light o f 5900 A. The experiment would reveal a ÍTÌnge shift o f 0.005 fringes. What upper limit does
a null result place on the speed o f the earth through ửie ether?
Ans.

From Problem 2.5, the number of tringes, AN, seen to pass the telescope crosshairs is
2 o f Section 2.4 requires that the Galilean coordĩnate transfomiations be replaced by the
dinate trans/orm alions. For the two observers o f Fig. 1-1 these are
0.005 = (5900

X

I0-I0m)(3

X

10«m/s)^

Solving, u = 3.47 X 10’ m/s.
The earth's orbital velocity is 3 X 10* m/s, so the interíerometer w: ỉ sensitive enough to dctect
this motion. No fringe shiít was observed.

y' = y

ợ.l)

v/1 i can be inverted to give

Supplem entary P roblem s
■ '" ỉ"
2-8.

Repeat Problcm 2.3 for ứìC case whcre the picture is taken whcn ihe righi cnđ of the rod passcs ưic
caniera.
Ans. 0.18m

2-9.

At thc instant that the midpoini of a moving lĩietersĩick passcs a camcra, the camera shuner opcns and a
picture is lakcn o f ihe meterstick togcthcr with a siationary caìibraicd rule, as in Problcm 2.3. íf its speed
relative to the camcra Is 0.8c, vvhat will be ihc length of ihc moving meterslick as recordcd on the
fi1m?
Ans. 2.778 m

2..10.

Rcfcr to Problem 2.4. If Uic two signals reach O ' simuỉtancousỉy, what is iheir limc sequence as determined
by o?
Ans. A occurs beforc B

2..11.

Assume that thc orbital specd o f the canh, 3 x 1 0 ^ m/s, is cqual to the speed of thc carth through the eứier. If
light takes
seconds to travel ihrough an equaỉ-ann Michclson-Morỉcy apparatus in a direction parallel to
this moiion, calculate how long it U'ill take light to travel pcrpcnđicular to this motion.
Ans. (1 - 0 . 5 X 10-«)/.

(3-2)

Ị(5.í), V is the velocity o f O' with respect to o along their common axis; V is positive if O'
jitiv e jr-dircction and negative if O ' movcs in the negative í-direction. It has also been
1 origins coincide when the clocks are started, so that f' = f = 0 when y = X = 0. Note
5formations can be obtained fìom the first set o f transformations by interchanging
Ịmed variables and letting t! -> —V. This is to be expected from Postulate 1, since both
Ịipletely equivalent and observer o m oves wiứi velocity - i ! w ith respect to o ' .

ÍS T A N C Y O F T H E S PE E D O F L IG H T
pt at the instant when o and O' pass each other (at í = í" = 0), a light signal is sent from
‘ h in the positive x - x ' direction. If o íìnds that the signaPs spatial and time coordinates




16

THE LORENTZ COORDrNATE TR.\NSFORMATIONS

[CHAP. 3

CHAP. 3]

are relateđ by í = ct, then, according to {3.1), O ' will find ứiat
X

t' =

- 17

ct -

Note tha! if the two events occur at the same spatial location, only o n e clock is needed by each
observer to determine i f the events are simultaneous. On the other hand, if the two evenls are separated
spatially, then each observer needs rn '0 clocks, properly synchronized, to determine vvheứier or not the two
events are simultaneous.

VI

'
V 1-

4 1 -W c )]
Vl -

V[I - (i;/c))[l + (i'/c)]

THE LORENT2 COORDINATE TR.WSF0RMAT10NS

/i - ( u / c ) ^
V 1 + (i>/c)

S olved P roblem s

Thus o ' will find thai y = ư , in agreement with the second postulate o f Einstein. Note also that for this
event on ưie light signal, í' 5^ (, in deíìnite disagreement with the Galilean assumption.
3.1.

Evaluate y i Ans.

3.2

for (o)

V =

ìO ~ -c ; ( b ) V

(1 + x)" = 1 + M +

As discussed in Chapters 1 and 2, MaxweH’s equations by electromagnetic theory are not mvariant
under Galilean transformations. However, as showTi by H. A. Lorentz (before Einstem), they are invariant
under Lorentz fransformations. See Problems 6.21 through 6.23.

(a)

Sening X = - 1 0 ■*and n = ỉ in thc binomial expansion, and, because X is so small, keeping only
the fiisl two terms o f the expansion, we obtain
(1 - lO-")''-

3J

= 0.999Sc.

In ihe following we make use o f the binomial expansion.

TH E INVARIAiNCE O F M Á XW ELL'S EQUATIONS

1 + l ( - 1 0 - ' ) = 1 - 0.00005 = 0,99995

G EN ER AL C O N SID E R A T IO N S IN SOLVING PR O B LE M S INVO LVING L O R E N T Z
T RANSPO RM A TIO NS

ỉữ — lj — ----------- ■ ' ---------- --------

v/rrõ ^ T ?)

y i -(i'2/c^ ) = y/l - (0.9998)- = y/\ - ( 1 -O.OOOIÝ

(6)

When attacking any space-tim e problem, the key concept to keep in múid is that o f “event." Most
problems are concemed with rwo obser%’ers measuring the space and time coordinatcs o f an event (or
events). Thus, each event has eight numbers associated with it: ( x ,y ,z ,t ) , as assigned by o , and
( y , y , / , / ) , as assigned by o The Loreniz coordinate tiansformations express the relationships between
these assignments.
Many times, problems are conceraed with the determination o f the spatial interval and/or the time
interval between two events. In this case a useílil technique is to subtract from each other the appropriate
Lorenti transformations describing each event. For example, suppose observer o ' measures the time and
spatiãl intervals between two events, A and B, and its is đesired to obtain the time interval between these
same two events as measured by o. From {3.2) one obtains upon subtracting
from tg

To evaluale (1 —0.0002) we cmploy the binomial expansion to obtain
(1 - 0.0002)- % 1 - 2(0.0002) = 1 - 0.0004
Using this in the above expression we obiain
v/1 - ( i,“/c2)

3.2.

(3.3)

y i - ( 1 -0 .0 0 0 4 ) = V0.0004 = 0.02

A s measured by o, a Aashbulb goes o f ĩ at X = 1 0 0 km, y = lOkm, r = I km at / = 5 X 10“'' s,
What are the coordinates y , y , z ' , and /' o f this event as đetermined by a second observer, O ',
moving relative to o at - 0 .8c along the com m on x - x ' axis?
Ảns.

From the Lorentz transformations.

Since all the quantities on the right-hand side o f this equation are known, one can determine te - 1^.

IO O k m -(-0 .8
v/1 -

3.4

SIM ULTANEITY
S x lO - ^ s -

Two events are sim uìtaneous to an observer if the observer measures that the two cvents occur at the
same time, With classical physics, when one observer determines that two events are simultaneous then
since r' = í from the Galilean lransformations, every other observer also fìnds that the two events ăre
simultaneous. In relaiivistic physics, on the other hand, two events that are simultaneous to one observer
are, in general, not simultaneous to another observer.
Suppose, for example, that evenls .4 and
are simultaneous as detennined by O ', so that
= 4According to (J.J), observer o measures the time separation o f these same two events as

y /ì

X

3

X

lO^km/sMS

X

10~'s)

y i - (0.8)^
( - 0.8)(100km)
ĩq*km/s

=

12.8 X l O - ^ s

^ 1 - (0.8)=

ý = )■= lOkm

s

3J.

If the two events occur at the same spatial location, so that xỉg =
then the two cvents are also
simultaneous as detcrmined by o. But ifx 'g T Ì x '^ , then o determines ứiat ứie two events are not
simultaneous.

Suppose that a particle moves relative to O ' with a constant velocity o f c /2 in the y y -p la n e such
that its trạjectory makes an angle o f 60“ with the y -a x is . If the velocity o f O ' with respect to o is
0 .6c along the x - x ' axis, find the equations o f motion o f the panicle as determined by o.

1Be 'ẽuuauuiis u f HIUIÌUII« tm eiiii ncd by O' i
Q U Ò C g ia H Ầ N O Ĩ j
(cos 60°)/'
T R Ư N G T Ầ M T H Ô N G TIN ĩ h ỉ í r # N
T

đ ạ i

V

h ọ c

GC

/ 0 2 5 Í 01

y = Uyt' = ^ ( s i n 60 '")r'


THE LORENT2 COORDrNATE TRANSPORMATIONS

[CHAR 3

CHAP. 3]

Substituting from (i.ý) ìn the íìrst expression, we obtain

Ans.

THE LOR£NTZ COORDÍNATE TRANSPORMATIONS

Subtracting Uvo Lorent2 transformations:

■, ^
— = |( c o s 60°)—■

ự i - (t“/c^) 2
v '1 - iv^/c^)
X - (0.6c)í = ^(cos 60°)^/ - —

(I X . 0 - s - 2
______________

X

JT= (0.74C);

ý = >’ = ỉ (sin 60°)
2
V

J

Solving, v/c = - 1 / 2 . Therefore V is in the negative jr-direction.
3.7.

(0-6X0.74í)
= -;(s in 6 0 °)---- = (0.30c)(
2
v> - ( 0-6)^

I

3 x lO S m /s

y i - a -/c ^ )

Substituting in the second expression then gives
I■,

1 0 - s) - í
cV

Refer to Problem 3.6. What is the spatial separation o f the two events as measured by ớ '?
Ans.

Subtracting two Lorentz transformalỉons:
r; - r' =

3.4.

A train i mile long (ạs measured by an obscrver on the ữain) is traveling at a speed o f lOOmi/hr.
Two lightning bolts strike the ends o f ừie ffain simultaneously as determined by an observer 011 the
ground. UTiat is the time separation as measured bv an obsérver on the train?
Ans.

From Problcm 3.6, v /c = —ịo r t; = —1 .5 x 10®m/s.
~

We have

, _ (12 X lO^m- 6 X lO^m)- { - 1 , 5 X 10*m/s)(l X 10-'‘ s - 2 X lO-^s)
------------------------------------- ----- p—
.................................................. = 5.20
- ( - 0 .5 ) '

X

10 m

(100 mi
striking of each lightning
lightni
Let events ^ and B be đetỉncd by the: strikuig
bolt. Wiih o as the ground obser\-er.
we have from (i.J)
(^'b - O +

- <<)

S u p p lem en ta ry P roblem s
3.8.

Obtain 0 .2 ) from (Ì.I).

3.9.

As detennined by O', a lightning bolt strikes at = 60m, >/ = y = 0, ;' = 8 X 10“* s. O' has a velocity of
along the jT-axis of o . What are the space-time coordinates o f the strike as delennined by
ơ?
Ans. ( x . y . z. t) = (93 m. 0. 0, 2 X 1Q-’ s)

3.10.

Observcr O ' has a velocity o f 0.8c relative to o , and clocks are adjusted such ứial / = r" = 0 whenx = y = 0.
If o determines thai a Aashbulb goes off at J = 50m and 1 — 2 X 10“’ s, whal is the time o f this event as
measured by O'?
Ans. 1.11 X IO-’ s

3.11.

Refer to Problcm 3.10. l f a second Aashbulb Aashes at y = lOm and/ ' = 2 X 10"’ s as determineđ by O'
w hat is the tim e in ler\ al b«tw een the two events as m easurcd by ơ ?
Ans. 1.78 X 10“ ’ s

3.12.

Refer to Problem 3.11. Whaĩ is the spatial sepaiation o f the two c\'enls as mcasured by (a) O ' (b) ơ ĩ
Ans. (fl) 6.67 m;(í>) 46.7 m

,+ ^ ị Z L i Ị £ l z : ] i Z i ( o ,5 n ,0
Q ___________ (1.86 X 10’ mi/s)~

Soiving.
3.S.

= -4 .0 2 X 10'’ s. The minus sign denoies that cvent A occurred after event B

Obsei^er 0 notes ihat nvo cvents are scparaled in space and time by 600 m and 8 X IQ-'' s How
fast must an observer O ' be moving relativc to o in order that the évents be simũltaneous to 0 ' ?
Ans.

Subtractỉng two Lorentz transfomiaĩions, we obtain

ÍJ - r; = •

^/|

_

v ị'

600 m

\

Solving, i'/c = 0.4.

3.6.

The space-tịm e coordinatcs o f two events as measured by o are A-, = 6
of o

X

10'* m

^ ° ^
^'2 = ' 2 X 10'* m, >'2 =
= 0 m. Í2 = 1 X 10-* s, W iat must ^
with respect to o if O ' mcasures the tw o evcnts to occur simultaneousiy?

= ’ , = Om
thẽ velocity


CHAP. 4]

RELATIVISTIC LENGTH CONTRACTION

subtraction technique o f Section 3.3; the co ư ect answer will n ol be obtained by multiplying or dividing the
original spatial separation by y i — { v ^ /c ^ ị

Solved P roblem s

Relativistic Length
Contraction

4.1.

How fast does a rocket ship have to go for its lengứi to be contracted to 99% o f its rest length?
Ans.

From the cxprcssion for length cootraction {4.1),
^ = 0.99 = ự \ -(i^ /c ^ )

4.2.

or

u = 0.141c

Calculate the Lorentz contraction o f the earth’s diameter as measured by an observer O ' who is
stationary wiứi respect to the sun.
Ans.

Taking the orbial velocity o f the earth to be 3
expression for the Lorentz contiaction yields
D = D j ì - (v^/c^) = (7.92

X

X

10^ m /s and the diameter o f the earth as 7920 mi, the

10’ m í ) ị i -

« (7.92

X

lo' mi)(l - 0,5

X

IQ-*)

Solving, Do - D = 3.96 X 10"’ mi = 2.51 in. It is seen that relativistic eíTecIs are very small at
speeds that are normally encountered.

ID E P IN IT IO N O F LEN G TH
py is at rest with respect to an observer, its length is determined by measuring the diíĩerence
spatial coordinates o f the endpoints o f the body. Since the body is not moving, ứiese
Jts may be made at any time, and ứie length so deteưnined is called the rest length or proper
Ị body.
ving body, hovvever, the procedure is more complicated, since the spatial coordinates o f ứie
he body m ust be m easured a t the sam e time. The diíTerence berween these coordinates is then
I ứie leiĩgth o f the body.
Inovv a ruler, oriented along ứie x - x ' dircction, that is at rest with respect to observer O'. We
nine how the length measurements o f o and O' are related to each other when O' is moving
bvith a velocity I' in ứie x - x ' direction. Let the ends o f ữie niler be designated by A and B.
|n tz transformation {3.1) we obtain

4.3.

A meterstick makes an angle o f 30° with respect to the y -a x is o f O '. What must be the value o f V if
the meterstick makes an íUigle o f 45° with respect to the x-axis o f ơ ?
Ans.

We have:
i ; = L' sin ơ” = (1 m) sin 30° = 0.5 m

4 = i ' cos ơ' = (1 m) cos 30° = 0.866 m

Since there will be a length contraction only in the x - x ' dircction,
L, = i y 1 - (i>2/c2) = (0.866m)v/l -

í , = í,; = 0.5 m
Since tan 0 = Ly/L„

0.5 m

tan45° = 1 =

v/1 -

(0 .866m )y i - (1“/
= Lq is the (proper) length o f the ruler as measured by O '. If X g and
are
at the same time, so that Ig — 1^ = 0, then the diíTcrence Xg — Xjf — L will be the length o f
'iured by 0 . Thus we have
L = L j\-ụ ^ lc ^ )
< 1 we have L < io , so Ihat the length o f ữie m oving ruler is measurcd by
esult is called the Lon'nl:~Fir:gerald conlraciion.

4.4.

(4.1)

D=

0.8lóc.

Refer to Problem 4.3. What is the lengUi o f the meterstick as measurcd by ơ?
Àns.

o to be

Ito keep clear the distinclion between the concepts o f “spatial coordinate separation” and
non m istake in solving problems is simply to multiply or divide a given spatial interval by
^ /c ^ ). This approach will work if One is concem ed with íĩnding ửie relations between
he concept o f “ length" is deíìned precisely above. However, if One is interested in the
fbetw een t»'0 evcnts that do not occur simultaneously, Ihen the ansvver is obtaineđ from ứie
20

Solving,

Use the Pythagorean thcorcra or, more simply,
L=

.
sin 45°

sin 45°

= 0.707 m

4.5.

A cube has a (proper) volum e o f lOOOcm’ . Find the volum e as detcrmined by an obscrver O ' who
moves at a velocity o f 0 .8c relative to the cube in a direction parallel to one edge.

'

Ans.

The obsCTver measures an edge o f the cube parallel to the đirection o f moiion to have the contracted
ỉengứi
ỉ. = I.V l

= (lO c m )/! - ( 0.8)= = 6 cm


RELATIVISTIC LENGTH CONTRACTION

[CHAP. 4

The lenglhs o f the other edges are unchanged:
fy = ly = í. = ì. = 10 cm
ThereTore,
= ự x

= (6 a ĩi)(1 0 c m )(1 0 c in ) = 600 cm ’

Relativistic Time
Dìlatìon

S u p p lem en ta ry Problem s
4.6.

An airplane is movmg with respect to the eanh al a speed o f 600 m/s. Its proper length is 50 m. By how much
will il appcar to be shonened to an observer on eanh?
Ans. lO"'” !!! ,

4.7.

Compute the contraction in lengđi o f a ttain ị mile long when it is traveling at
A rs.

lOOmi/hr.

5.58 X 1 0 -'= m i = 3.52 X 1 0 - '“ in,

4.8.

At whal speed must an observer inove past ứie earứi so that ứie eanh appears like an ellipse whose major axis
is sLx tũnes its minor axis?
Ans. 0.985c

4.9.

An obscrver o ' holds a 1.00 m stick atan angle of 30° with respect to thc positive y-axis. o ' is moving in the
positive x - x direction vvith a vclocity 0.8c vvith respect to observer o. What are the length and angle o f the
stick as m easuređ by o ?

A n s.

0.721 m; 43.9°

4.10.

A squarc o f area 100 cm^ is at rest in the reference frame o f o. Observer O ' moves relative to o at 0.8c and
parallcl to One side o f ứie squarc. What does O ' measure for the area?
Ans. 60 cm^

4.11.

For the square of Problem 4.10, finđ the area mcasured by O' if O' is moving at a velocity O.Sc relative to o
and along a diagonal o f the squarc.
Ans. 60 cm^

4.12. Repcat Problem 4.5 if O ' moves with Ihe same spccd parallel to a diagonal o f a face o f the cubc.
A n s.

6 0 0 cm ^

Dserver, say o, determines that two events A and B occur at the same location, the time interval
two events can be determined by o with a single clock. This time interval, t g — A/q, as
o with his single clock, is called the proper tim e interval between the events.

'

sider the same two events A and B as viewed by a second observer, 0 ' , moving with a
Ih respect to o. The second observer will necessarily detemiine that the two events occur at
ũons and will therefore have to use t\vo diíĩerent, properly s^Tichroniĩed clocks to determine
tion t'g — = ầ t ' between A and B. To find the relationship benveen the time separations as
' and o ' we subtract two Lorentz time transformations, obtaining

nines that the two events occur at the same location, Xg A/' = ■ < 1, Aí* > A/ q,

so

= 0. Thus

A/a

that the time intcrval between the two evems as measured by O ' is

bxample ứie single clock was taken to be at rest with respect to 0 . The samc result would
Be clock were takcn to be at rest with respect to O '. Thus, in general, suppose a single
"ough a time interval Alg. l f this clock is moving wilh a velocity V with respect to an
etermine that his two clocks advance through a time intcrval Aí given by


RELATIVISTIC TIME DILATION

[CHAP. 5

CHAP. 5]

SJ.
Clocks I t n d 2
âdvance throuRh ủ í > Afo

Q

CH

í !l

:
Time Dilarion as Vieweđ by Observer

We have
20 m
= 11.1
10* m/sec

distance _
velocity
0.6

5.4.
Fig. 5-1.

Observers o and O' approach each other with a relative velocity o f 0.6c. If o measures the initial
distance to ỡ ' to be 20 m, how much tứne will it take, as determined by o , before the two observers
meet?
Ans.

Single clock A dvuìccs
through ửíQ < A /

RELATIVISTIC TIME DILATION

10-*s

In Problem 5.3, how much time will it take, as determined by O', before the two observers meet?
Ans.

o

X

Time dilation is a very real effect. Suppose in FÌ2. 5-1 cameras are placed at the location o f clock 2 and
at the location o f the single clock, and a picture is taken by each camera when the single clock passes clock
2 . WTièn ứie pictures are developed, each picture will show the same thing— that the single clock has
advanced through AÍQ while clock 2 has advanced Ihrough A t > A/q, with Af and A/q related by the time
dilation expression.

The two events under consideration are: ịA) the position of O' when o makes his initial measurement,
and (B) the coincidence of o and O'. Both o f these events occur at the origin of O'. Thcrefore, the
time lapse measured by O ' is equal to ửie proper time between the two events. From the time dilarion
• expression.
Aío = (A /)Vl -(!.►*/<-) = (11,1

X

10“* ;)v/l - (0.6)- = 8.89 X 10“®s

This problcm can also be solvcd by noting ứiat the inirial distancc
to the đistance measured by o through ứie Lorenlz contraction:

deteriĩimed by O ' is related

L' = L o ự \ - (1-2/c^) = (2 0 m )y i - (0.6)- = 16r
A Warning!
It is importanl to keep clear the distinction between the “time separation" o f two events and the
“proper time interval” berween two e\'ents. l f observers o and O' measure the time separation between two
events ứiat, for both observcrs, occur at different spatial locations, then these time separations are n ol
related by simply multiplying or dividing by

10* m /s
5.S.

Ans.

Ans.

The time measured

Afo = (A r)\/l -

= (6 X lO -S ) ^ ! -(0 .9 5 )2 = I g7 X lQ-‘ s

5.6.

10-* s

We have:
distance = uAí = (0.3

syslem in which the ;i-mesons are at rcsi is the proper lime.

X

Pions have a ha!f-Iife o f 1.8 X 10 *s. A pion beam leaves an accelerator at a speed o f 0.8c.
Classically, what is the expected distance over which h alf the pions shoulđ decay?

Solved P roblem s
The average lifetim e o f /j-m esons with a speed o f 0.9 5 c is mcasured to be 6 X 10“‘ s. Compute the
average lifetime o f /j-m esons in a system in which they are at rest.

= 8.89

X

3

X

10* m/s)( 1.8

10'* s) = 4.32 m

X

Determine the ansvver to Problem 5.5 relativistically.
Am.

The half-life o f 1.8 X 10”* s is determined by an observerat rest with respecl to the pion beam. Froir
the poúit of view o f an observer in the laboratory, ứie half-life has becn increased because o f the timt
dilaiion, and is given by
A;o

^ t = ...= 3 X 1 0 '* s

1.8 x l 0-* s

,

y i - ( 0,8)2
5 .2.

An airplanc is moving vviih respect to the eanh with a speed o f 600 m /s. A s determined by earth
clocks, how long will it take for the airplane’s clock to fall behind by two microseconds?
Ans.

Ạ,

Therefore, the distance travcled is
d = v M = (O.g ; 3 X 10*m /s)(Jx 10-‘ s) = 7.20m

From ửie time dilaiion exprcssion.
_

For an observer at rest wiứi rcspcct to the pion bcam, the distance dp the pions have to ữavcl ìs
shorter than ứie laboratory distance d, by the Lorentz contraction:

^

d, =
(2

X

= 2

X

10-‘ s

= lO^s = ll.ód ays
This result indicatcs the smallncss of relativistic eíTects at ordinary speeds.

= í/,v/l - ( 0.8)2 ^ 0.6í/,

The time elapseđ when this distance is covered is
10-" s = :

0.6d,
: 3 X 10*m/s

Solving. di = 7.20m, which agrecs with the answer determined ftom time dilation.


RELATIVISTIC TIME DILATION

[CHAP. 5

S u p p lem en ta ry Problem s
5.7.

An atom dccays in 2 X 10“^ s. W"hat is ihc decay lime as measured by an observer in a laboratory vvhen the
atom is moving with a speed o f 0.8c?
Ans. 3.33
X 10"^ s

5.8.

How fast woulđ a rocket ship have to go if an obsen-er on the rocket ship aged at half ihe rate of an obsen-er
on the eanh?
Ans. 0.866c

5.9.

A n ia n \v ith 6 0 y ears to live w an ts lo visit a distant galaxy which is 1600Ó0 lieht years away. \Vhat m usl be his
constant sp eed?
A n s. v / c = 1 — (0.7 03 X 10"^)

5.10.

A panicle moving at O.Sc in a laboraiory decays after ưaveling 3 m. How long did it exist as measured bv an
observer in ihe laboratory?
Ans. Ỉ .2 5 x l0 ~ ® s

5.11.

\Miat does an obser\er moving with the particle ofProblem 5.Ỉ0 measure for the time ihe particle lived beforc
Ans. 0.7 5 X 10"® s

đecaying?

Relativistic
Space-Time
Measurements
ous chapters we discussed, more or less separately, relativistic space measurements and
Tie mcasurements. There are, hoNvever, many types o f problems where space and túĩie
; are intemvined and cannot be treated separately.

Solved P roblem s
ick moves with a velocity o f 0.6c relative to you along the direction o f its length. How
tll it take for the meterstick to pass you?
he iengứi o f the meterstick as mcasurcd by you is obiaìncđ from ihe Lorcntz contraciion:
L = io^ /l - (t-2/c2) = (I m)v/l - (0.6)= = 0.8r
: time for the meterstick to pass you is then found from
dìstance = velocity

X

0.8 m = (0.6 X 3

X

timc
10*m/s)

X

Ai

A / = 4 .4 4 X 1 0 " ’ s

ycars for light to reach us from the most distant pans
b. at a constant speed, in 30 years?

or our galaxy.

Coulđ a human

6 dislance traveled by light in 10’ yeare is, according to an obscrver ai resi with respect to the carth,
do = c(A/) = 10’c


RELATIVISTIC SPACE-TIME MEASUREMENTS

[CHAP. 6

CHAP. 6]

»here c is cxpressed in, say, mi/yr. If this obsener now moves with constant speeđ V wiih respect to
the carth. the dístance d that he has to travel is shoncned according to the Lorentz contraction:

Ans.

RELATrVISTlC SPACE-TIME MEASUREMENTS

From ứie time dilation expression,
30 min

d = ư„v/l

= (10’c ) / r ^ õ ? 7 ? )

y i - (0.6)^

The time inienal available to travel this distance is 50 years, so thai
Therefore, the time at the space station is 12;37.5 P.M,
lO'cv/1
50

6.6.

Solving,
- = V l - 2.5 X 10'- a 0.999999 S75

In Problem 6.5, what is the distance from
pilot? (Ế) by an observer on the earth?
Ans.

6 .3.

A ;i-m eson with an average lifetjm e o f 2 X 10"* s is created in the upper acmosphere at an elevation
o f 6000 m. \Vhen it is created it has a velocity o f 0.998c ứi a direction toward the earth. What is the
average distance ửiat it will travel before decaying, as determined by an observer on the earth?
{Classically, this distance is

6.7.

Ảns.

distance = velocity X tim e = (0.6 X 3 X 10* m/s)(30min X óOs/min)
dìstance = velocity X dme = (0.6 X 3 X 10*m/s)(37.5min X 60s/min) = 4.05

= 3.24 X10"
X lo" m

(a) According to an earth observer,
distance 4.05 X 10" m
1 min
time = — , . = ——
^------- X
= 22,5 min
velocity
3 X 10*m/s
60 s
Thus the signal arrives, according to an observer on the earth, at

so ihat //-m esons would not, on thc average. reach the earth.)

12:37.5 P.M. + 22.5 min = 1:00 P.M.

As detennined by an observer on the eanh, the !ifetime is increased because o f time dilalion:


detennined (a) by

Refer to Problems 6,5 and 6 .6 . When the rocket ship passes ứie space station, the pilot reports to
earth by radio. When does the earth receive the signal, (a) by eartìi tứne? (b) by rocket túĩie?

= 1' A f = (0 .9 98 X 3 X 10*m /s)(2 X 10“S ) = 599 m

.•1/15.

(a)
(i)

Therefore a human traveling at ihis speed will finđ ửiat when he completes ửie ưip he has aged 50
years.

the earth to the space station as

(6) Accordỉng to the pilot,

2 x l0 -‘ s
: = 3 1 .6 x 10-‘ s
, / r r -; (0.998)-'

distance 3.24 X 10" m
1 min
.
time = - ■, . = —-----—^— — X , " = 18 min
vclocity

The avcrage distance travelcd, as determineđ by an earth observ'er, is

3 X 1 0 * m /s

60 s

Thus, according to the pilot, the signal anives at ứie earth at

= (0.998 X 3 X 10* m/s)(31.6 X 10-‘ s) = 9470 m

12:30P.M. + 18min = 12:48P.M.

Thus, anobsener on thc carth delermines that, on the average, a ;j-meson will rcach the earth.

6 .4 .

Consider an ob sen.er at resi with rcspect to the íi-m eson o f Problem 6.3. How far will he measure
the earth 10 approach him before ữie /j-meson disintegrates? Compare ứiis distance with the
diitance he measures from the poinl o f creaiion o f the p-m eson to the earth.
Am.

As deiennined by an obscr\’er at rest with respect to the /j-meson, the distance travcled by the earth is
d = v à t„ = (0.998 X 3 X 10*m/s)(2 X lQ-‘ s) = 599m
The iniiialdisiance,

6.8.

Suppose an observer o detennines that two events are separated by 3.6 X 10* m and occur 2 s apart
What is the proper time interval between the occurrence o f thesc two events?
Ans.

Thcre exists a sccond observer, ơ \ moving relative to ứie fifst observcr who will ổctcrmỉne that thí
two events occur at the Sãiĩie spatỉal locatỉon. The propcr time intcrval betwccn the two cvcnls ỉs thí
tiine interval measured by this observer. Denoting the two events by A and B, wc obtain upoi
subtracting two Lorentz transformations

i , to the eanh, however, is shortencđ because o f the Lorcntz contraciion:
^

i = io > /l -{i>-’ /í^) = (6 x 10’ m )^ ! -(0.998)^ = 379m

3 .6 x 1 0 « m -i;(2 s)
v/1 - (é ^ ịé )

Thus, an obscr\'er on thc íi-meson dctermines that, on ứie average, it will reach the earth, in agreement
with the result o f Problcm 6.3.

u = 1.8

X

10* m /s = 0.6c

Agaìn subtracting two Lorentt ứ^sformations, wc obtain ihc proper timc inicrval as
<6.5.

A pilot in a rocket ship traveling wiứi a velocity o f 0 .6 c passes the earth and adjusts his clock so ứiat
it coincides wiih 12:00 P.M. on earth. At 12;30P.M., as determined by ứie pilot, ứie rocket ship
passes a space station that is stationary wiih respect to the earth. What time is it at the station when
the rocket passcs?

,

\

= -------- = 4

^ )

\

^

0.6 X 3.6 X 10* m/s

= = ------ = ------------3 x lO » m /s
7 i - ( 0.6)'


30

RELATIVISTIC SPACE-TIME MEASUREMENTS

[CHAP. 6

Another way to soh e ứiis problem is 10 use u and the time dilation expression;
Aio = (Aí)

CHAP. 6]

6.U ,

= (2 s)^ /r^ "{ã67 = 1.6 s

Refer to Problem 6 .Í0 . Suppose that every year, as determined by ơ , o sends a iight signal to ỡ '
How many signals are received by O ' on each leg o f his joum ey? (In other words, what would t\vừ
O ' actually see if he looked at his sister o through a telescope?)
Ans.

6.9.

RELATIVISTIC SPACE-TIME MEASUREMENTS

For obser\'er o , rvvo events are simultaneous and occur 600 km apart. What is the time difFerence
ber\veen ihese tsvo events as detennined by O ', who measures ứìeir spatial separation to be

As determined by o , brother O ' reaches a-Centauri at f = 5 yr. In order for a light signal to rcach
a-Centauri simultaneously with ỡ ', it must have been sent by o at an earlier time, determừied by
distance _ 4 yr X (distance traveled by lighưỵr)
= 4yr
velocity
disiance traveled by lighưyr

1200km ?

Let A and B designate the tv.'0 events. Subưacting tv.'0 Lorentz transíormations, we obiain
^3 "

Thereíore, a signal sent by ơ at í = I yr reaches a-Centauri simultaneously wiứi O'. Since o sends í
toial o f 10 signals, the remaining 9 signals all reach O' on ihe retum joumey.

V T m PT ?)

12 X ỈO^m =

6.12.

6 X 10^ m - 1'(0)

v 'T ^ F 7 ? )

Ans.

Again subtraciine rwo Loreníz transformations:
t



As determineđ by o , brother O ' reaches a-Centauri at / = 5yr. A lightsignal sent by O'
a-Centauri will reach o in a time interva! (as determineđ by 0 ) of

0.866(6 X 10’ r

^

3 x lO » m /s
i : ầ i £ ^
y i1 - (0.866)^

~

v'l -

Refer to Problems 6 .Ỉ 0 and 6.11. Suppose that every year, as determined by O ', O ' senđs a ỉighi
signal to o . Consider the signal sent by O ' just as he reaches CỂ-Centauri. What is the time, aí
đetermmed by o , when this signal is received. (That is, whaí \vouId twin o see if she looked at hei
brother O ' through a telescope?)

^ __ distance
velocity

= - 3 .4 6 x l 0 - > s

ữorr

_ 4 yr X (distance traveleđ by light/>T)
distance traveled by light/yr^

Therefore, ứiis signal reaches o at / = 5 yr + 4yr = 9 yr. Hence, of the six signals sent by o three ol
ứiem arereceived by o during the first nine years (one cvery threc years) and the remaining three ar<
receiveđ by o during the lasl year.

The minus sicn denoies that evenl A occurred aíter event B as determined by O'.
Problems 6 .] 0 -6 . ỉ 2 illustrate the íamous “nv’in eữ ect” in special Reìativity.
6.10.

A man in the back o f a rocket shoots a high-speeđ buỉlet toward a target in the front o f the rocket,
The rocket is 60 m iong and the bullet’s speed is 0.8c, both as measured by the man. Find the time
that ửie bullet is in Aight as measured by the man.

Observer o \ moving with a speeđ o f 0.8c relative to a space platíorm, travels to a-Centauri,which,
at a distance o f 4 light years, is the nearest star to the piatform. When he reaches the star he
immcđiatcly tums around and retums to the platform at the same speed. W hen O ' reaches the space
p!aĩform, compare his age with ửiat o f his twin sister ơ , who has stayed on the platform.
Ans.

Ans.

We have

According to o ứie tìme elapsed during ứìe trip from the space pIatform to a-Cenlauri is
^

_ disiance

velocity

velocity

4 y r X {distance traveled by lighưyr)

0.8

X

(distance traveled by lighưyr)

0.8 X 3 X 10* m/s

= 2.50 X 10"’ s

^

Sincc thc rctum trip lakcs placc wiih ứìc same speed, ứie total time eiapseđ, as measured by ứìe
plaiỉoưn observcr o , is

6.14,

Refer to Problem 6.13. I f the rocket moves with a speed o f 0.6c relaiíve to the earth, fìnd the time
that the bullet is in Aight as measured by an observer on the earth.
Subtracting tw'0 invcrse Lorenlz transformations:

'^'rounđ trip = 'Oyr
ƯB

0 ' measures the propcr time inter\'al betwecn thc departurc from Ihe plaiíorm and ứic anrival ai
thc siar. Hence, from the timc dilation cxpression.

~ Q

~

^a)

' 2.5 X 10“ '^ s +

(0.6X60 m)
3 X 10*m/s
.
,„ _ 7
, ---- — = 4.63 X 10 s

1 - ( 0 .6 )'

Aío = (Aí)v/1 -

= (5 yr)^! - (0.8)= = 3 yr
6.15.

and the toial timc clapsed, as rricasuređ by 0 ', is
= 6 yr
Thcrcfore, O' is 4 ycarĩ younger ứian o when ứiey meet. Tliis resuit illustrates the íamous ‘*twin
effect” in Spccial Relativity. Note thai the motion o f the twins is dcíìnitcly noi symmetrical. In ordcr
to get back home ihc ưavcling twin must tum around. This tuming arounđ is real (O' cxpcricnccs
mcasurable accelerations), in contras! lo thc apparcnt tuming around that O ' obscrves o f o (who
expencnces no acceleration during her entirc history). Thus the motion o f O' is equivalcnt to that of
two diíTerent inertial obscrvers, One moving wiih V = +0.8c and ứie other moving wilh V = —O.Sc.
Twin o , on the oứier hanii. is cquivalenỉ lo One singlc incrtial obscrver.

The rest lengths o f spaceships A ánd B arc 90 m and 200 m, rcspectively. A s Ihey travel in opposite
directions a pilot in spaceship A determines that the nose o f spaceship B requires 5 X 10"’ s to
ttaverse the lengủi o f A . What is ứie relative vclocity o f the two spaceships?
Ans.

As determined by pilot A,
A/

6.16.

5 x l O 1-7
- ’ «s

1.8 x 10'm /s
' = 0 .6c

In Problem 6,15, what is the time interval, as determined by a pilot in the nose o f B, bctween
passing the íiront and rear ends o f A?


[CHAP. 6

RELATIVISTIC SPACE-TIME MEASUREMENTS

ÀỈĨS.

CHAP. 6}

The relaiive specds are Ihe same as đetermined by each observer. Pilot B measurcs ứie lengih of
spaceship .4 to be contracted according 10

RELATIMSTIC SPACE-TIME MEASUREMENTS

The light signal travels in the negative direclion at speed c , so thai
= -c{ỈB - U)

L = L o ự l - (r^/c^) = ( 9 0 m ) y i^ ^ .6 ) * = 72m

Subsiituting from above,

The time intcrval as measured by B is Ihen
V

6.17.

0.6

X

0 - (4.0 X 10* m /s)í; = { - 3 x 1 0*m/s) ^30 s - - Ì j

72 m
3 X 10*m/s '

A rocket ship 90 m long travels at a constant velocity o f 0 .8 c relative to the ground. A s ứie nose o f
ứie rocket ship passes a ground observer, ửie pilot ừi the nose o f the ship shines a Aashlight toward
ửie tail o f the ship, What tứne does ứie signal reach ửie tail o f the ship as recordeđ by (a) the pilot,
(b) the ground observer?
Ẩns.

Solving, (, = lO.Os.
This result, and that o f Problem 6.20, point out the distinction betvveen seeing an event and
measuring the coordinates o f the same event

6.20.

(a) Let events A and 3 be deíĩned by the emission o f the Hght signal and the light signal’s striking the
tail of ứie rocket, respcciively. Since the signal travels al speed c in ửie negative đừection,
-3

“ C

X

Refer to Problem 6.19. If O ' looks at 0 's clock through a telescope, what time does his o\vn clock
read when he sees 0 's clock reading 30 s?
Ans.

Let events A and B be delĩned by the emission o f the light signal from o and reception o f the same
signal by O', respectively. Our problem is to find (g. Applying the Lorentz transfonriations to event A
gives

10“ m/s

í'. = •
^v'' - (‘^/^) ~

(t) Subtraction of tv.'0 inverse LorenLz ưansformations gives

0 - ( (3
3

X

10*m/sK30s)
10'm/sK

■ = - 1 5 0 X 10*m

yi-(0.8)^

y i -(0.8)^

- ( 0 .8 )=

As measured by O', the light signal travels in the positive direction with speed c, so that
= c(l's - i'^)

x'b -

6.18.

Refer to Problem 6.17. VVhen does the tail o f the rocket pass the ground observer, (a) according to
ứie ground observer? (è ) according to ứie pilot?
Ans.

Substituling from above,
0 - ( - I 5 0 x 10*m) = (3

(a) As determined by the ground observer, the lengứi, L, of the rocket is

X

10* m/s)(íá - 50s)

Solving, ÍJ = lOOs.
L = L j \ - (t-7c^) = (90m )v/l -(0 .8 )^ = - 5 4 r
6.21.
■0.8
ih)

” 0.8

X

X

3

X

10*m/s

The equation for a spherical pulse o f light starting from the origin at / = í' = 0 is

= 2.25 X 10-’ s

90 m
= 3.75
3 X 10*m/s

+
X

10-" s

Ans.
6.19.

The speed o f a
the rocket and
í = / = 0 when
does he see on
Ans.

-cV = 0

Show from the Lorentz transformations that o ' w ill also measure this same pulse to be spherical, in
accord with Einstein's second postulate stating that the velocity o f light is ứie same for all observers.
From the inverse Lorentz MnsfonnatÌĐns

rockct vviứi respect to a space station ịs 2,4 X 10* m /s , and observers O ' and 0 in
the space station, respectively, synchronize ứieừ clocks in the usual íashion (i.e.,
JT = y = 0). Suppose that o looks at 0"% clock ửirough a telescope. What time
ỡ ' ’s clock when his ow n clock reads 30 s?

1

y 1-

3- 3 - (x'^ + I^r- + 2vx'r')

Let evenls A and
B bedetined, respectively, by the emission o f the light signal from O ' and the
reception of ứie
samesignal by o, Our problem is to find 1^. Applying the inverse Lortntz
transfomiations 10 cvcnt A, we obtaỉn
Substituting, one íìnds that
,

<,

+ Wc^K.

<,+Wc^)(0)

<,

+>^ +z= Therefore sincc

x '1 + vl’.

- (‘ /<^)

0 + (0.8 X 3 X 10* m /s)(^
ự l - (O .sỹ

+ /

= 0, we also have

x'^ + / “ + 1'^
so that the pulse as dctcrmined by o ' is also spherical.

=0


34

6.22.

RELATIVISTIC SPACE-TIME MEASUREMENTS

[CHAP. 6

CHAP. 6)

Show ứiat the diíĩerential expression

RELATIVĨSTIC SPACE-TIME MEASUREMENTS

Subsiituting these in ứìe wave equation, we obtain
dx^ + d \^ +

- c 'd r

Bx^

ạr

dr^

^

dr'-

is invariant under a Lorentz transformation.
so thai the equation is invariant under Lorentz transfomiations. Recall that the wave equation is not
invariant under Galilean transíormations (Problem 1.10).

If Ihe cxpression is ỉnvanant, H will retain the same form when expressed in terms of the primed
coordinates. From the invose Lorentz traJisformation One fìnds
(d x ' +

d p + 2ti dx' dí’)

yi -

S u p p lem en tary Problem s

dí' + (u/c^Kv'

1

d)T = dv'^

dx^ +

6.24. An unstable particle wiứi a mean liĩetìme o f 4 ịis is fonned by a high-energy acceierator and projected through
a laboratory Wiửì a speed o f 0.6c. (a) What is the mean lifetime of the particle as determined by an observer in
the laboratory? (ố) What is the average distance that the particle travels in the laboratory before đismtegrating?
(c) How far does an observer al rest with respect to the panicle determine that he ữ^vels before the particle
đisintegrates?
Ans. {a)5ịiS-, (6)900m; (c)720m

dz^ = dz'^

Subsiituting these expressions one íinds
í i r + d ) p + í/r^ - C"

6 .23.

= d x '^ + d y ^ + i i : " — (T d t’"

6.25.

A /í-meson with a lifetime o f 8 X 10“^ s is formed lOOOOm high in the upper atmosphere and is movừi^
directly toNvard the eanh. íf the /í-meson decays just as it reaches the earth’s surTace, what is its speed relative
to the earth?
Ans. 0.972c

6.26.

A meterstick moves along the x-axis with a velocĩty o f 0.6c. The miđpoint o f the meterslick passes ơ at / = 0.
As đetennined by 0 , wherc are the enđs of the meterstick at / = 0?
Ans. 40 cm and - 4 0 cm

6.27.

Obser\'er o measures the area o f a circle at rest in his j;>'-plane to be 12 cm". An observer o ' moving relative
to o at 0.8c also obser\'es the figure.
area does 0 measure?
Ans. 7.2 cnr

6.28.

As đetennined by observ'er o a ređ light Aashes, and, 10“®s later, a bluc ỉight Aashes 600 m farther out on the
x-axis. What are ửie magnitude and direction o f the velocity o f a second observer, 0 \ if he measures the red
and blue Aashes to occur simultaneously?
Ans. +0.5c

6.29.

Reícr to Problem 6.28. What is the spatial separation o f the ređ and blue Aashes as đeiermined by
ơ'?
Àns. 520 m

6.30.

A rocket ship ! 50 m long travels at a specd o f 0.6f. As the tail of the rocket passes by a man on a staíionary
space platform, he shines a Aashlight in the đirection o f ứie nosc. (a) How far ÍTom ihe platform is ửie nose
when thc light reaches it? (6) As measured by Ihe observer on the space phiíform, how much time elapses
betwcen the emission and arrival o f tíie lighi signal? (c) What is the tìme inlcrval bctwcen emission and
recepiion of thc sígnal as detemũned by an observer in Ihc nose o f the rocket?
Ans. (a)300m: ( è )!0 “^s; (c)0.5 X 10" S

6.31.

Two cvcnts occur at the same place and are separatcd by a 4 s time intcrval as deiemiìned by onc obscrver. ỉf a
second observer measures the time separation between thcse two events to be 5 s, what is his đetermination of
their spaiia! separation?
Ans. 9 X 10*m

6.32.

An observcr sets off two Aashbulbs that arc on his x-axis. Hc rccords ihat the íìrst bulb is sct off at his origin at
1 o*ciock and the second bulb is set off 20 s laicr at X = 9 X 10* m. A sccond observcr is moving atong ihe
common X - / axis with a speed o f —o.óc with respect to the first observcr. What are thc úme and spatial
separations betwccn Ihe lwo fiashes as measured by ihc sccond observcr?

Show that ihe elecưom agnetic wave equation,

d^ậ l ctộ
Or-

_

c- dí^

is invariant Iinder a Loreniz tnmsformation.
Ans.

The equation will be invariant i f it reiains the same form when expressed in terms of ứie new variables
To express the wave equation in terms of the primcđ variables we íìrst find from the
Lorentz transformations that
ò/

\

dx'

V

^ ~ y i - ự^/c^)
3í'
dx

v/1

v /ir
v'!

^ = — =1
ạv &

By

à:

Bx

From the chain rule, and using the above resulls, vve havc
Bộ

Bộít^
dộBt'
' ãac'
? a tt Ó Ịạ>'
> ''S
a r '' ' ' ĩar"
t ' aã i ĩ ã 3/'
? Far

-v /( T
^ '1 -

(i~/c^) aí'

3ự>

at-

DiiTerenliating again with respect to X. we havc
ĨT(I> _
Bx'
i

1
3

íd^tịi
f

v^ír<ị>
^

2vă^<ị>\
c2at'3r'j

Similarly we havc
Sộ
-n
3ự>
1
Sệ
8' “ y i - ( 1-/C = )a x ''^ y i _(p 2 /c:)3 F

ỷậ _

1

/ , á^ệ

Br- ~ i - (i'V<^) v ’ ac^
d}'^

Srệ
dz~

ỷệ
dz^

ỷệ
3/'^

“ax' di')

Ans.

27.3s; 56.3 X IO*m


RELATIVISTIC SPACE-TIME MEASUREMENTS

[CHAP, 6

33

The relalivc spced o f o and O' is 0.8c. At í' = 2 X 10"’ s, a super bullet is fired from = 100 m. Traveling
in the negative y-direciion with a constant specd, it strikcs a target at ihe origin of O' at r" = 6 X lO '’ s. As
determincd bv o , what is the speed of the bullet and how far did it travel?
Ans. 3 X 10’ m /s ; 6 .6 7 m

34.

A ground obsener determincs that it takes 5 X 10"'’ s, for a rocket 10 travel betvveen two markers in
ground that are 90 m apan. \Vhat is the speed of the rocket as detennineđ by the ground obser%'er?

35.

Refer to Problem 6,34. As determined by an observer in the rocket, what is ửie distance between the two
markers and ửie time mterval between passing the two markers?
Ans. 72m; 4 X 10“'' s

,36.

A laser beam is rotated ai 150 rev/min and ứirow5 a beam on a screen 50000 miles away. What is the svveep
speed of ứie bearaacross the screen?
Ảns. 7.85 X 10^ m i/s (note; since c = 1.86 X 10’ mi/s, the sweep
speed is larger than c.)

.37.

Shovv that ửie expressions r
transformations.

th e

Relativistic Velodty
Transformations

+ T - írr^ m d dx^ + d}~ + dz^ - c ĩd r arc not invariant under Galilean

velocity ữansfonnations, \ve consider an arrangement identical lo that for the Lorentz
■ansformations (Fig, 1-1). One obser\’er, O ', moves along ihe com mon x -x ' axis at a constant
[iửi respect to a second observer, 0 . Each observer measures the velocity o f a súigle particle,
ding (u ,, u,„ Uj) and Q ' recording (i4, u'ỵ, uỊ) for the components o f the particle’s velocity.
entz coordinate transformations One finds the following L orenl: velocity ư ansform ations (see

1 - (v!ỷ)u .

u j \

=

-

1 - (v/c~)u.

(7.1)

Ị velocity V is positive if O ' moves in the positive x-direction and negative if O' moves in the
Ection. When these equations are inverted, one obtains

» ĩV l - (v^/c^)

u'j, + V

' 1 + (v/c^X

(7.2)

l+ ( r / c ^ K

i with the Lorentz coordinate transforTnatíons, the inverse velocity ừansformations can be
: velocity transfomiations (7.1) by interchanging primeđ and unprimed variables and
This is to be expected from symmetry, since from Postulate 1 o f Section 2.4 both
Bmpletely equivalent, and observer o m oves with a velocity o f - V with respecl to O'.

ENTZ V E L O C IT Y T R A N S P O R M A T IO N S A N D T H E S PEE D 0 F L IG H T
the experiment discussed in Section 3.1 where a light signal is sent in the x - x !
j com mon origin when o and o ' pass each other at f = /' = 0 . If o measures the signars
|n ts to be u , = c, Uy =
= 0. then, by (7 ./) , O' w ill measurc
,

u, ~ V

c — V

,

,

^

Ịíteimincs that the light signal travels vvith speed c, in accord with the second postulate o f


RELATĨVISTIC VELOCITY TR.\NSFORMATIONS

7.2

[CHAP. 7

G E N E R A L C O N S ID E R .\T IO N S IN SOLVĨNG V E L O C IT Y PRO B LE M S

In velocity problems there are three objects involved: two observers, o and O ', and a particle, p. The
particle p has two velocities (and, hence, six numbers) associated with it: its velocity vvith respect to o ,
(u ,,
u,).
its velocity with respect to 0 ‘, (u^,
uỊ). The quamity V appearing in ứie velocity
transformations is the veIocit>' o f O' wilh respect to o.
\M ien anacking a velocity problem, one should íìrst determine which objects in the problem are to be
identiíied with o. O ', and p. Sometimes this identification is dicated; other times the identiíìcation can be
made arbiưarily (see, for example, Problem 7.3). Once the identiíìcation has been made, one ữien uses ửie
appropriate Lorentz velocity transfomiations to achieve the answer.
In dealing vvith velocity problems, the best way to avoid mistakes is not to forget the phrase “with
respect to.” The phase “ velocity o f an object" is meaningless (both classically and relativistically) because
a velocity is always measured with respect to something.

CHAP. 7]

RELATIVISTIC VELOCITY TRANSFORMATIONS

Because all observers measure the speed o f light as c, the above equations also aIlow ứie change in
wavelength to be obtained via À = c/v .

Solved P roblem s
7.1.

Derive the Lorentz velocity ơansfonnation for the x-direction.
À ns.

Taking the differentials o f the Lorentz coorđinate transformations

one fìnđs

.MỄh

dx — vdĩ

Dividing dx! by dt' gives
7.3

T H E R E L A T IV IST IC D O PPL E R EFFE C T

dx

Consider a source that emits electromaanetic radiation with a Irequency V(| as measured by an observer
who is at rest with respect to the source. Suppose this same source is in motion wiih respect to anoứier
obser\er, who measures the ữequency V o f the radiation received from the source. \Vith the angie 0 and
velocit>' V o f the source as detìned in Fig. 7-1, the ửequency V as measured by observer o is given by the
D oppler equation-.

dyf

,

7.2.

dx — V dí

At what speeds w ill the Galilean and Lorentz expressions for
Ans.

Let ứie Galilean transformation be

điffer by 2%?

i! and the Lorenư transfoưnation be
1 - (u/c2)u.

1-

Rearranging,

Thus, if the product vUg cxceeds 0.02c^, ửíe error in using ưie Galilean tnmsformation instead o f ứie
Lorentz ữansíormation will excceđ 2%.

Fig. 7-1

^

y/1 - (V-Jc-)

7.3.

“ 1 - (p /c )c o s O

Rocket A ữavels to the right and rocket B travels to the left, with velocities 0 ,8 c and 0.6c,
respectively, relative to the earth. What is the velocity o f rocket A measured from rocket B I
Ans.

If thc source and obscrser arc moving tow ard each other, 0 = 0 and we have

Lct observere ỡ, O ' and ứie panicle bc associated with the earth, rocket B, and rocket A, respective!y.
Then
,

u ^-v

0.8c - ( - 0 .6C)
.6c)

“ 1 - (i>/c=K " J _ ( - 0 M (0-8c)
In ứiis case, V > Vg.
ỉ f Ihe source and observer are mo\'ing

ữw 'í7v fro m

The problem can also be solved wiứi other associations. For example, let obscrvm o , o ' and the
paiticle be associated with rocket A, rocket B, and tìie eaith, respectively. Then

each other, ớ = 180° and we have

u, - lí
“í =

0.6c =

-0 ,8 c - V
1 - (i./c=)(-0.8c)

Solving, V = —0.94ÓC, which agrees with the abovc answer. (The minus sign ap[>ears because V Ì5 the
velocitỹ o f O' with respect to o , which, with the prcsent association, is the velocity o f rocket B with
respect to rocket A.)

In this case, V < vq .
ỉ f the radiation is obscn'cd íransverse to the đirection o f m oiion, 0 — 90® and we have
V = V o ^ l - (1-2/c 2 )

7.4.
Thus, V < v„.

Repeat Problem 7.3 if rocket yi travels with a velocity o f 0.8c in the +.v-direction relative to the
earth. (Rockct B stills travels in the —A:-direction.)


[CHAP. 7

RELATIVISTIC VELOCITY TR.ANSPORMATIONS

CHAP. 7)

Lei observers o , O ' and the panicle be associaieđ with the earth, rocket B, and rocket A, respeciively.
Then
-

t'

“ 1 ~{v/(r)u^ ~
,

u ,yi -

j4ns.

RELATIVISTIC VELOCrTY TR.-KNSHORMATIONS

Wiưi ửie same assocỉation as in Probỉem 7.6,

u - í,-u ,0

1-0

O ne

has

< + !)

0 + 0.5C

“ 1 + (v /c ^ X ~

1+0

( 0 .8 c ) y i -(0.6)2
^ ------Ĩ^T5------=

"

(0.9c)ự l - (0 .5 f


= ------Ĩ T Õ ------=

which give ihe magniĩude and direction of ứie desired velocity as
ú = ^ u ĩ + u ỹ = v/(0.6c)2 + (0.64c)2 = 0.8

ui
7.5.

u=

0.5ŨC

tan

A panicle moves with a speed o f O.Str at an angle o f 30° to the JT-axis, as determúied by o. What is
the velocity o f the particle as determined by a sccond observer, 0 \ moving with a speed o f - 0 ,6c
along the common x - x ' axis?
/Ins.

7.8.

For observer o we have
u , = (O.Sc) cos 30“ = 0.693C

= (0.8c) sin 30° = 0.400c

+ uỊ = ự ( 0 .5 c f + (Q.779cf = 0.92ỖC

u

0 779í?

u,

0.5c

= 1.56

or

ậ = 5 1 .y

At í = 0 observer o emits a photon ưaveling at speeđ c in a dừection o f 60° with ứie x-axis. A
second observer, o ữ a v e l s wiứi a speed o f 0.6c along the com mon x - x ' axis. What angle does the
photon make with the y -a x is o f 0 "ì
Ans.

We have

Using ứie Lorenc velocity ữansformations, we have for observer o '
u ,-v

_

0.693C - ( - 0 .6 c )

u^ = c cos 60° = 0.50ŨC
= 0.913c

u ,-v


(0 .4 c )v /l-(0 .6 )^

u^ = c sin 60° = 0.86ÓC

0 .5 c - 0 .6 c

1 - l v / c ‘ )u ^ -

^

l-(f/c 2 K



lo .6 c )( o ìc ) -

,,,,

........... =

(0 .866c ) y i - (0 .6)'

'( 0 .6 9 3 c )

The speed measured by observer O ' is
u' = J u Ị + uỹ = V(0.913c)ỉ + (0.226c)2 =: 0.941c

,

(0.6c)(0.5c)

Thus

and the anglc ệ ' thc velocity makes with the y-axis is

ian

u;

7.6.

0,226c
0 .9 13c

ui

= 0.248

ò' = 13.9“

Consider a radioactive nucleus ihat moves vvith a constant speed o f O.Sc relative to the laboratory.
The nucleus decays and emits an electton with a speed o f 0 .9 c relative to ửie nucleus along the
direction o f motion. Find the velocity o f ứie electron in the laboratory frame.
Àns.

u' = ự u ỉ + uỹ = y(-0.143c)^ + (0.990c)^ = c
as Ì5 necessary.

Lct the laboraiory observer, Ihe mdioactỉve nuclcus and ư>e electron be respectively associaled wiih o ,
O ' and the panicle. Then
_

< + f

_

0.9c + 0.5c

_ ^

(0 .5c)(0.9c) ■

7.7.

- 0 .1 4 3 C

and ựi' = 81.8° above the negative y-M is. The magnimde o f ihe velocity of the photon as measured
by Ó' is

Refer to Problem 7.6. Suppose that the nucleus decays by emitting an electron with a speed o f 0.9c
in a direction perpendicular to the direction o f (the laboratory’s) motion as determined by an
observer at rest with respect to the nucleus. Find the velocity o f the electron as measured by an
observer in the laboratory ữanie.

7.9.

The speed o f light in still water is c / n , where the index o f retiaction for water is approxúnately
n - 4 /3 . Fizeau, in 1851, found that the speed (relative to the laboratory) o f light in water moving
with a speed y (relative to the laboratory) could be expressed as

u = - + if'
n
where the “dragging coeíỉìcient” was measured by him to be * s: 0.44, Determine the value o f k
predicted by the Lorentz velocity transformations.


RELATIVISTIC VELOCITY TR..\NSF0RMAT10NS

••Íní-

[CHAP, 7

CHAP. 7]

An observer ai rest relative to the water will measure the speed of light to be u’^ = c/n. Treating the
light as a particle, ihe laboratory observer vviil find its speed 10 be

Ans.

RELATIVISTIC VELOCITY TRANSPORMATIONS

The Doppler equation gives
TX = /q
• JY ------£T+ y

or

-l = •*!!

Hence, AX = 5920 Ả — 5890 Ằ=: 30 Ằ. The shift is to a greater wave]engứi {red shifỉ).
For small values of y ứie approximation
7.13.

Suppose that the Doppler shift Ln the sodium D 2 ĩine (5890 Ả) is lOOẢ when the light is observed
from a distaní slar. Determine ứie star’s velocity o f recession.

yields
[\ + (v/c)
i ĩ ^

vvhere lerms of order y~ /c have been neglected. Thus

Solving,

7.14.

(4/3)which agrces wiih Fizeau’s experimental result.

7.10.

.

í

5 9 9 0 Ằ = (5 8 9 0 Ả ự .

= 0 .0 17c.

A man in a rocket ship m oving wiứi a speed o f 0.6c away from a space platform shines a light o f
wavelength 5000 A tovvard the platform. What is the ữequency o f the light as seen by an observer
on the platform?

Esaluate the Doppler equation to íìrst order in v / c when the source and obser\’er are receding from
each other.

_

7.15.

.

)

/ l —(tĩ/c)
1 + (t!/c) 5

X

3 X 10^ m /s /1 —0.6
10-’ m V 1 + 0.6 “ ■

Refer to Problem 7.14. What is the frequency o f the light as seen by a passenger in a second rocket
ship that moves m ứie opposite direction with a speed o f 0 .8c relative to ứie space platform?
Ans.

The velocity o f the íìnt rocket relative to the second is foimd from the Lorentz vclocity transfonnation:

which is Ihe cỉassỉcal expression for the Doppler efFect vvhen the receíver is stationary with respect to
the medium.

u ,- v

_

0.6í - ( - 0.8c)
(-0.8cK 0.6c) ■

1-

7.11.

A car is approaching a radar speed trap at 80m i/hr. If the radar set works at a írequency o f
20 X 10’ Hz, what frcqucncy shift is observeđ by the palrolman at the rađar set?
Ans.

The trcquency observcd by ứie second rocket is then given by
/l - (ư jc )
3 X 10*m/s /1 - 0.946
‘'“v ì + w / c ) ~ 5 x 10- ^ m V Ì + 0'946
í

To first order in v/c, the frequency received by the car is



„14 , ,

S u p p lem en ta ry P roblem s

c
The car ihen acts as a moving source with this frequency. The frequency received back al the rađar set

7.16.

A rocket moves with a velocity o f c/3 with respect lo a man Holding a lantem. The pilot o f the rocket
measures the speed o f light reaching him firom the lantem. Determine this speed fix)m the Lorentz velocity
ưansformations.
Ans. c

from which (80m i/hr=:35m /s)

7.17.

The pilot o f a rocket m oving a ta velocity o f0 .8 c r e la tiv e to the earth observes a second rocket approaching in

the opposite directíon at a velocity o f 0.7c. What does an observer on earth measure for the sêrônd rockẻtls
vclocity?
Ans. 0.227c

X ‘0’
7.18.
7.1 2 .

A star is receding from the earth at a speed o f 5 X 1 0 -’ c. What is the vvavelength shift for the
sodium Dj line (5890 Ả)?

c

An observer in rocket A finds that rockets and B are moving away from him in opposite đirections at spccds
of 0.6c and 0.8í, respectively. What is the speed o f as measured by
Ans.

0.946c (classicallỵ, 1,4c)

c


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