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SPACETIME PHYSICS introduction to special relativity Second Edition
Edwin F. Taylor M assachusetts Institute o f Technology
John Archibald Wheeler Princeton U niversity and U niversity o f Texas a t A u stin
W . H. Freeman and Com pany N ew York
Library o f Congress CataloginginPublication Data Taylor, Edwin F. Spacetime physics introduction to special relativity / Edwin F. Taylor, John Archibald Wheeler. — 2nd ed. p. cm. Includes bibliographical references and index. ISBN13: 9780716723271 ISBN10: 0716723271 1. Special relativity (Physics) I. Wheeler, John Archibald, 1911.II. Title. QC173.65T37 1991 530.1T— dc20
92722 CIP
© 1992 by Edwin F. Taylor and John Archibald Wheeler No part o f this book may be reproduced by any mechanical, photographic, or electronic process, or in the form o f a phonographic recording, nor may it be stored in a retrieval system, transmitted, or otherwise copied for public or private use, without written permission from the publisher. Printed in the United States o f America Eleventh printing
Both males andfemales make competent observers. We ordinarily treat the laboratory observer as male and the rocket observer as female. Beyond this, to avoid alternating "his ” and "her” in a single chapter, we use female pronouns fo r an otherwise undesignated observer in oddnumbered chapters and male pronouns in evennumbered chapters.
Epigram, facing page: Einstein remark to his assistant Ernst Straus, quoted in Mainsprings of Scientific Discovery by Gerald Holton in The Nature of Scientific Discovery, Owen Gingerich, Editor (Smithsonian Institution Press, Washington, 1975).
)Nhat I’m really interested in is whether God could have made the world in a different way; that is, whether the necessity of logical simplicity leaves any freedom a t all. — Albert Einstein
Edwin F. Taylor and John Archibald W heeler have w ritten a general relativity sequel to Spacetime Physics, namely: Exploring Black Holes: Introduction to General R elativity Addison Wesley Longman, San Francisco, 2000 ISBN 020138423X
CONTENTS Chapter 1 SPACETIME: OVERVIEW
1
The great u n ity is spacetime; its measure, the spacetime interval, is the same fo r a ll observers. 1.1
Parable of the Surveyors 1
and Intervals Alone! 9 Minute, or Y e a r 11
1 .2 Surveying Spacetim e 5
1 .5 Unity of Spacetim e 15
Acknowledgments 19
1 .3 Events
1 .4 Sam e Unit for Sp ace and Time: M eter, Second, References 18
Introduction to the Exercises 19
Chapter 2 FLOATING FREE
Exercises 20
25
Jum p off the roof: On the w a y down — in free flo a t— we have an (almost!) perfect setting fo r conducting experiments. 2.1
Floating to the Moon 25
2 .3
Local C h aracte r of FreeFloat Frame 30
2 .5 Test Particle 36 2 .7
O b server 39
Frame 41
2 .2 The Inertial (FreeFloat) Frame 26 2 .4 Regions of Spacetim e 34
2 .6 Locating Events With a Latticework of Clocks 37 2 .8 M easuring Particle Speed 4 0
2 .1 0 Summary 43
References 4 4
2 .9 Rocket
Exercises 45
Chapter 3 SAME LAWS FOR ALL
53
W ithout looking out o f the w indow , we cannot tell which freefloat fram e we are in. 3.1
The Principle of Relativity 53
Fram es 56
3 .2 W hat Is NOT the Sam e in Different
3 .3 W hat IS the Sam e in Different Frames 6 0
Simultaneity 62
3 .5 Lorentz Contraction of Length 6 3
3 .4 Relativity of 3 .6 Invariance of
Transverse Dimension 65 3 .7 Invariance of the Interval Proved 6 7 3 .8 Invariance of the Interval for ALL FreeFloat Fram es 71 3 .9 Summary 73 References 76
Acknowledgments 7 7
Exercises 78
Special Topic
LORENTZ TRANSFORMATION
95
Observe an event in the laboratory; predict its space an d time readings in the rocket. L .l Lorentz Transformation; Useful or Not? 95 L .3 First Steps 9 9
L.5 Completing the Derivation 101 Transformation 102 Reference 11 1
L.2 Faster Than Light? 96
L.4 Form of the Lorentz Transformation 100 L .6 Inverse Lorentz
L .7 Addition of Velocities 103
Exercises 1 12
L.8 Summary 111
Chapter 4 TRIP TO CANOPUS
121
T ravel quickly to a d ista n t sta r an d return, to fin d we have traveled into the future. 4.1
Invitation to C anopus 121
4 .3
Faster Than Light? 122
Plan 124 4 .8
4 .2 StrippedDown FreeFloot Frame 121 4 .4 All of Sp ace is Ours! 123
4 .6 Twin Paradox 125
Time Traveler 127
4 .5 Flight
4 .7 Lorentz Contraction 126
4 .9 Relativity of Simultaneity 128
4 .1 0 Experimental Evidence 131
References 134
Exercises 135
Chapter 5 TREKKING THROUGH SPACETIME
137
Move or stan d still; in either case we soar through spacetime. 5.1
Time? No. Spacetim e M ap ? Y es. 137
FreeFloat Frames 139 5 .4
W orldline 143
5 .5 Length Along a Path 147
Along a W orldline 148 Factor 155
5 .2 Sam e Events; Different
5 .3 Invariant H yperbola 143 5 .6 W ristwatch Time
5 .7 Kinked W orldline 152
5 .8 Stretch
5 .9 Touring Spacetim e Without a Reference Frame 160
5 .1 0 Summary 162
References 163
Exercises 163
Chapter 6 REGIONS OF SPACETIME
171
The speed o f light is a m ighty barrier th a t preserves cause an d effect. 6.1
Light Speed : Limit on C ausality 171
Timelike, Spacelike, or Lightlike 172 Spacetim e 177
6 .2 Relation Between Events:
6 .3 Light C on e: Partition in
Exercises 183
Chapter 7 MOMENERGY
189
A second great unity is momentumenergy (momenergy); its measure, mass, is the same fo r a ll observers. 7.1
M om energy: Total C onserved in a Collision 189
Arrow 191 7 .4
7 .2 M om energy
7 .3 M om energy Com ponents and M agnitude 195
Momentum: “ Sp ace Part” of M om energy 199
M om energy 201
7 .5 Energy: “ Time Part" of
7 .6 Conservation of M om energy and its
C onsequences 2 0 7
7 .7 Summary 211
Acknowledgment 213
Exercises 214
C h apters COLLIDE. CREATE. ANNIHILATE.
221
Convert mass to energy and energy to mass. 8.1 The System 221 8 .2 Three M odest Experiments 2 2 2 8 .3 M ass of a System of Particles 22 4 8 .4 Energy Without M ass: Photon 228 8 .5 Photon Used to C rea te M ass 23 3
8 .6 M aterial Particle Used to C re a te M ass 234
8 .7 Converting M ass to U sable Energy: Fission, Fusion, Annihilation 2 3 7 8 .8 Summary 244 References 251
D ia lo g : Use and Abuse of the Concept of M ass 246 Acknowledgments 2 5 2
Vi
Exercises 253
Cliapter 9 GRAVITY: CURVED SPACETIME IN
ACTION
275
G ra vity is not a force reaching across space but a distortion — curvature ! — o f spacetime experienced right where you are. 9.1
G ravity in Brief 275
9 .2 G alileo , Newton, and Einstein 275
Moving O rd ers for M ass 2 7 7 of the Two Travelers 281 9 .7 G ravity W av es 28 8
9 .4 Spacetim e Curvature 2 8 0
9 .3 Local 9 .5 Parable
9 .6 Gravitation a s Curvature of Spacetim e 284 9 .8 Black Hole 2 9 2
9 .9 The Cosm os 2 9 6
References 29 6
ANSWERS TO ODDNUMBERED EXERCISES INDEX
303
v ii
299
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SPAariME: O ur imagination is stretched to the utmost, not, as in fiction, to imagine things which are not really there, hut ju st to comprehend those things which are there. Richard P. Feynman
1.1 PARABLE OF THE SURVEYORS disagree on northward and eastward separations; agree on distance Once upon a time there was a Daytime surveyor who measured off the king’s lands. He took his directions of north and east from a magnetic compass needle. Eastward separations from the center of the town square he measured in meters. The northward direction was sacred. He measured northward separations from the town square in a different unit, in miles. His records were complete and accurate and were often consulted by other Daytimers. A second group, the Nighttimers, used the services of another surveyor. Her north and east directions were based on a different standard of north: the direction of the North Star. She too measured separations eastward from the center of the town square in meters and sacred separations northward in miles. The records of the Nighttime surveyor were complete and accurate. Marked by a steel stake, every corner of a plot appeared in her book, along with its eastward and northward separations from the town square. Daytimers and Nighttimers did not mix but lived mostly in peace with one another. However, the two groups often disputed the location of property boundaries. Why? Because a given corner of the typical plot of land showed up with different numbers in the two record books for its eastward separation from the town center, measured in meters (Figure 11). Northward measurements in miles also did not agree between the two record books. The differences were small, but the most careful surveying did not succeed in eliminating them. No one knew what to do about this single source of friction between Daytimers and Nighttimers. One fall a student of surveying turned up with novel openmindedness. Unlike all previous students at the rival schools, he attended both. At Day School he learned
Daytime surveyor uses magnetic north
Nighttime surveyor uses NorthStar north
CHAPTER 1
SPACETIME: OVERVIEW
magnetic north
NorthStar north
magnetic east NorthStar east
NIGHTTIME: NORTHSTAR NORTH FIGURE 11. The town as plotted by Daytime and Nighttime surveyors. Notice that the line of Daytime magnetic north just grazes the left side of the north gate, while the line of Nighttime NorthStar north just grazes the right side of the same gate. Steel stakes A, B, C, D driven into the ground mark the comers of a disputed plot of land. Ar shown, the eastward separation of stake A from the northsouth line measured by the Daytime surveyor is different from that measured by the Nighttime surveyor.
from one expert his method of recording locations of gates of the town and corners of plots of land based on magnetic north. At Night School he learned the other method, based on NorthStar north. As days and nights passed, the student puzzled more and more in an attempt to find some harmonious relationship between rival ways of recording location. His attention was attracted to a particular plot of land, the subject of dispute between Daytimers and Nighttimers, and to the steel stakes driven into the ground to mark corners of this disputed plot. He carefully compared records of the two surveyors (Figure 11, Table 1 1). Student converts miles to meters
In defiance of tradition, the student took the daring and heretical step of converting northward measurements, previously expressed always in miles, into meters by multi plying with a constant conversion factor k. He found the value of this conversion factor to be ^ = 1609.344 meters/mile. So, for example, a northward separation of 3 miles could be converted to ^ X 3 miles = 1609.344 meters/mile X 3 miles = 4828.032 meters. "At last we are treating both directions the same!” he exclaimed. Next the student compared Daytime and Nighttime measurements by trying various combinations of eastward and northward separation between a given stake and the center of the town square. Somewhere rhe student heard of the Pythagorean Theorem, that the sum of squares of the lengths of two perpendicular legs of a right triangle equals the square of the length of the hypotenuse. Applying this theorem, he discovered that the expression D aytim e / northward \ 2 k X 1 separation I + \ (miles) / _
D aytim e eastward separation (meters)
( 1
1)
PARABLE OF THE SURVEYORS
1.1
 iC^^ ^ B L E 1  1 ^ 
TWO DIFFERENT SETS OF RECORDS; SAME PLOT OF LAND Daytime surveyor’s axes oriented to magnetic north Northward Eastward (miles) (meters) Town square Corner stakes: Stake A Stake B Stake C Stake D
Nighttime surveyor's axes oriented to NorthStar north Eastward Northward (meters) (miles)
4010.1 5010.0 4000.0 5000.0
1.8330 1.8268 1.2117 1.2054
3950.0 4950.0 3960.0 4960.0
1.8827 1,8890 1.2614 1.2676
based on Dayrime measurements of the position of steel stake C had exaaly the same numerical value as the quantity N ig h ttim e
N ig h ttim e 2
/ northward \ k X 1 separation 1 + V (miles) / _
eastward separation (meters)
(
1 2)
computed from the readings of the Nighttime surveyor for stake C (Table 12). He
C j^ B L E
“ INVARIANT DISTANCE” FROM CENTER OF TOWN SQUARE TO STAKE C (Data from Table 11)
Nighttime measurements
Daytime measurements Northward separation 1.2117 miles
Northward separation 1.2614 miles
Multiply by k = 1609.344 meters/mile to convert to meters: 1950.0 meters
Multiply by k = 1609.344 meters/mile to convert to meters: 2030.0 meters
Square the value
3,802,500 (meters)^
Square the value
4,120,900 (meters)^
Eastward separation 3960.0 meters
Eastward separation 4000.0 meters b 16,000,000 (meters)^ = 19,802,500 (meters)^
Square the value and add
+ 15,681,600 (meters)^
Sum of squares
Sum of squares
= 19,802,500 (meters)^
Expressed as a number squared
= (4450 meters)^
Expressed as a number squared
= (4450 meters)^
Square the value and add
This is the square of what measurement?
4450 meters
This is the square of what measurement?
4450 meters
i
i SAME
DISTANCE from center of Town Square
CHAPTER 1
SPACETIME: OVERVIEW
magnetic north
town square
DAYTIME: MAGNETIC NORTH NorthStar north
NIGHHIME; NORTHSTAR NORTH
Discovery: Invariance of distance
FIGURE 12. The distance between stake A and the center of the town square has the same value for Daytime and Nighttime surveyors, even though the northward and eastward sepa rations, respectively, are not the samefor the two surveyors.
cried the same comparison on recorded positions of stakes A, B, and D and found agreement here too. The student’s excitement grew as he checked his scheme of comparison for all stakes at the corners of disputed plots — and found everywhere agreement. Flushed with success, the student methodically converted all northward measure ments to units of meters. Then the student realized that the quantity he had calculated, the numerical value of the above expressions, was not only the same for Daytime and Nighttime measurements. It was also the square of a length: (meters)^. He decided to give this length a name. He called it the d istance from the center of town.
(distance)^ —
northward 2 separarion + (meters)
eastward separarion (meters)
( 1 3)
He said he had discovered the p rin cip le o f invariance o f distance; he reckoned exactly the same value for distance from Daytime measurements as from Nighttime measurements, despite the fact that the two sets of surveyors’ numbers differed significantly (Figure 12). After some initial confusion and resistance, Day timers and Nighttimers welcomed rhe srudent’s new idea. The invariance of distance, along with further results, made it possible to harmonize Daytime and Nighttime surveys, so everyone could agree on the location of each plot of land. In this way the last source of friction between Day timers and Nightrimers was removed.
1.2
SURVEYING SPACETIME
1.2 SURVEYING SPACETIME disagree on separations in space and time; agree on spacetime interval The Parable of the Surveyors illustrates the naive state of physics before the discovery of special relativ ity by Einstein of Bern, Lorentz of Leiden, and Poincare of Paris. Naive in what way? Three central points compare physics at the turn of the twentieth century with surveying before the student arrived to help Daytimers and Nighttimers. First, surveyors in the mythical kingdom measured northward separations in a sacred unit, the mile, different from the unit used in measuring eastward separations. Similarly, people smdying physics measured time in a sacred unit, called the second, different from the unit used to measure space. No one suspected the powerful results of using the same unit for both, or of squaring and combining space and time separations when both were measured in meters. Time in meters is just the time it takes a light flash to go that number of meters. The conversion factor between seconds and meters is the speed of light, c = 299,792,458 meters/second. The velocity of light c (in meters/second) multiplied by time t (in seconds) yields ct (in meters). The speed of light is the only natural constant that has the necessary units to convert a time to a length. Historically the value of the speed of light was regarded as a sacred number. It was not recognized as a mere conversion factor, like the factor of conversion between miles and meters — a factor that arose out of historical accident in human kind’s choice of units for space and time, with no deeper physical significance. Second, in the parable northward readings as recorded by two surveyors did not differ much because the two directions of north were inclined to one another by only the small angle of 1.15 degrees. At first our mythical student thought that small differences between Daytime and Nighttime northward measurements were due to surveying error alone. Analogously, we used to think of the separation in time between two electric sparks as the same, regardless of the motion of the observer. Only with the publication of Einstein’s relativity paper in 1905 did we learn that the separation in time between two sparks really has different values for observers in different states of motion — in different fram es. Think of John standing quietly in the front doorway of his laboratory building. Suddenly a rocket carrying Mary flashes through rhe front door past John, zooms down the middle of the long corridot, and shoots out the back door. An antenna projects from the side of Mary’s rocket. As the rocket passes John, a spark jumps across rhe 1millimeter gap between the antenna and a pen in John’s shirt pocket. The rocket continues down the corridor. A second spark jumps 1 millimeter between the antenna and the fire extinguisher mounted on the wall 2 meters farther down the corridor. Still latet other metal objects nearer the rear receive additional sparks from the passing rocket before it finally exits through the rear door. John and Mary each measure the lapse of time between “pen spark” and “fireextinguisher spark.” They use accurate and fast electronic clocks. John measures this time lapse as 33.6900 thousandmillionths of a second (0.0000000336900 second = 33.6900 X 10“^ second). This equals 33.6900 nanoseconds in the terminology of highspeed electronic circuitry. (One nanosecond = 10~^ second.) Mary measures a slightly different value for the time lapse between the two sparks, 33.0228 nanoseconds. For John the fireextinguisher spark is separated in space by 2.0000 meters from the pen spark. For Mary in the rocket the pen spark and fireextinguisher spark occur at the same place, namely at the end of her antenna. Thus fot her their space separation equals zero. Later, laboratory and rocket observers compare their space and time measurements between the various sparks (Table 13). Space locations and time lapses in both frames are measured from the pen spark.
The second: A sacred unit
Speed of light converts seconds to meters
Time between events: Different for different frames
O n e o bserver uses laboratory frame
Another observer uses rocket frame
6
CHAPTER 1
SPACETIME: OVERVIEW
 C [ ^ B L E l  3 ^ > 
SPACE AND TIME LOCATIONS OF THE SAME SPARKS AS SEEN BY TWO OBSERVERS Distance and time between sparks as measured by observer who is standing in laboratory (John) moving by in rocket (Mary) Distance Time Distance Time (meters) (nanoseconds) (meters) (nanoseconds)
Discovery: Invariance of spacetime interval
Reference spark (pen spark)
Spark A (fireextinguisher spark)
2.0000
33.6900
33.0228
Spark B
3.0000
50.5350
49.5343
Spark C
5.0000
84.2250
82.5572
Spark D
8.0000
134.7600
132.0915
The third point of comparison between the Parable of the Surveyors and the state of physics before special relativity is this: The mythical student’s discovery of the concept of distance is matched by the Einstein  Poincare discovery in 1905 of the in v arian t spacetim e in terv al (formal name L orentz in terv al, but we often say just in te r val), a central theme of this book. Let each time measurement in seconds be converted to meters by multiplying it by the “conversion factor c " the speed of light: c = 299,792,458 meters/second = 2.99792458 X 10* meters/second = 0.299792458 X 10^ meters/second = 0.299792458 meters/nanosecond Then the square of the spacetime interval is calculated from the laboratory observer’s measurements by subtracting the square of the space separation from the square of the time separation. Note the minus sign in equation (14). L aboratory
L aboratory 2
(interval)^ =
/ time \ c X 1 separation 1 — V (seconds) / _
space separation (meters)
( 141
The rocket calculation gives exactly the same value of the interval as the laboratory calculation. R ocket (interval)^ =
/ time \ c X 1 separation 1 V(seconds) / _
R ocket 2
space separation (meters)
(15)
even though the respective space and time separations are not the same. Two observers find different space and time separations, respectively, between pen spark and fireextinguisher spark, but when they calculate the spacetime interval between these sparks their results agree (Table 14). The student surveyor found that invariance of distance was most simply written with both northward and eastward separations expressed in the same unit, the meter. Likewise, invariance of the spacetime interval is most simply written with space and
1.2
SURVEYING SPACETIME
C ^ A B L E
“ INVARIANT SPACETIME INTERVAL” FROM REFERENCE SPARK TO SPARK A (Data from Table 13]
Rocket measurements
Laboratory measurements Time lapse 33.6900 X 10« seconds = 33.6900 nanoseconds
Time lapse 33.0228 X 109 seconds = 33.0228 nanoseconds
Multiply by r = 0.299792458 meters per nanosecond to convert to meters: 10.1000 meters
Multiply by f = 0.299792458 meters per nanosecond to convert to meters: 9.9000 meters
Square the value Spatial separation 2.000 meters
102.010 (meters)^
Square the value
98.010 (meters)^
Spatial separation zero
Square the value and subtract
— 4.000 (meters)^
Square the value and subtraa

Result of subtaction expressed as a number squared
= 98.010 (meters)^
Result of subtaction expressed as a number squared
= 98.010 (meters)^
This is the square of what measurement?
= (9.900 meters)^ 9.900 meters
= (9.900 meters)^
This is the square of what measurement?
9.900 meters
i
i SAME SPACETIME
INTERVAL from the reference event
time separations expressed in the same unit. Time is converted to meters: t (meters) = £• X t (seconds). Then the interval appears in simplified form: 2 time — (interval)^ — separation (meters)
space separation (meters)
(
1 6 )
The in v arian ce o f th e spacetim e in terv al — its independence of the state of motion of the observer — forces us to recognize that time cannot be separated from space. Space and time are part of a single entity, spacetim e. Space has three dimensions: northward, eastward, and upward. Time has one dimension: onward! The interval combines all four dimensions in a single expression. The geometry of spacetime is truly fourdimensional. To recognize the unity of spacetime we follow the procedure that makes a landscape take on depth— we look at it from several angles. That is why we compare space and rime separations between events A and B as recorded by two different observers in relative motion. Why the minus sign in the equation for the interval? Pythagoras tells us to A D D the squares of northward and eastward separations to get the square of the distance. Who tells us to SUBTRACT the square of the space separation between eventsfrom the square of their time separation in order to get the square of the spacetime interval?
Sp a ce and time are part of spacetime
8
CHAPTER 1
SPACETIME: OVERVIEW
Shocked? Then you’re well on the way to understanding the new world of very fast motion! This world goes beyond the threedimensional textbook geometry of Euclid, in which distance is reckoned from a sum of squares. In this book we use another kind of geometry, called Lorentz geometry, more real, more powerful than Euclid for the world of the very fast. In Lorentz geometry the squared space separation is combined with the squared time separation in a new way— by subtraction. The result is the square of a new unity called the spacetime intervalhtvf/ttn events. The numerical value of this interval is invariant, the same for all observers, no matter how fast they are moving past one another. Proof? Every minute of every day an experiment somewhere in the world demonstrates it. In Chapter 3 we derive the invariance of the spacetime interval— with its minus sign— from experiments. They show the finding that no experiment conducted in a closed room will reveal whether that room is “at rest’’ or “in motion” (Einstein’s Principle of Relativity). We won’t wait until then to cash in on the idea of interval. We can begin to enjoy the payoff right now.
SAMPLE
PROBLEM
li;
S P A R K I N G AT A FASTER RATE Another, even faster rocket follows the first, enter ing the ftont door, zipping down the long corridor, and exiting through the back doorway. Each time the rocket clock ticks it emits a spark. As before, the first spark jumps the 1 millimeter from the passing rocket antenna to the pen in the pocket of
John, the laboratory observer. The second flash jumps when the rocket antenna reaches a door knob 4.00000000 meters farther along the hall as measured by the laboratory observer, who records the time between these two sparks as 16.6782048 nanoseconds.
a. W hat is the time between sparks, measured in meters by John, the laboratory
observer? b. W hat is the value of the spacetime interval between the two events, calculated
from John’s laboratory measurements? c. Predict: W hat is the value of the interval calculated from measurements in the
new racket frame? d. W hat is the distance between sparks as measured in this rocket frame? e. W hat is the time (in meters) between sparks as measured in this rocket frame? C om pare w ith the tim e between the sam e sparks as m easured by Jo h n in the
laboratory frame. f. W hat is the speed of this rocket as measured by John in the laboratory?
SOLUTION a. Time in meters equals time in nanoseconds multiplied by the conversion factor,
the speed of light in meters per nanosecond. For John, the laboratory observer, 16.6782048 nanoseconds X 0.299792458 meters/nanosecond == 5.00000000 meters b. The square of the interval between two flashes is reckoned by subtracting the
square of the space separation from the square of the time separation. Using laboratory figures: (interval)^ = (laboratory time)^ — (laboratory distance)^ = (5 meters)^ — (4 meters)^ = 25 (meters)^  16 (meters)^ = 9 (meters)^ ~ (3 meters)^
1.3
EVENTS AND INTERVALS ALONE!
9
Therefore the interval between the two sparks has the value 3 meters (to nine significant figures). c. W e strongly assert in this chapter that the spacetim e in terv al is in v aria n t —
has the same value by whomever calculated. Accordingly, the interval between the two sparks calculated from rocket observations has the same value as the interval (3 meters) calculated from laboratory measurements. d. From the rocket rider’s viewpoint, both sparks jump from the same place, namely the end of her antenna, and so distance between the sparks equals zero for the rocket rider. e. W e know the value of the spacetime interval between two sparks as computed in
the rocket frame (c). And we know that the interval is computed by subtracting the square of the space separation from the square of the time separation in the rocket frame. Finally we know that the space separation in the rocket frame equals zero (d ). Therefore the rocket time lapse between the two sparks equals the interval between them; (interval)^ = (rocket time)^ — (rocket distance)^ (3 meters)^ = (rocket time)^ — (zero)^ from which 3 meters equals the rocket time between sparks. Compare this with 5 meters of lighttravel time between sparks as measured in the laboratory frame. f. Measured in the laboratory frame, the rocket moves 4 meters of distance (state ment of the problem) in 5 merers of lighttravel time (a). Therefore its speed in the laboratory is 4 /5 light speed. Why? Well, light moves 4 meters of distance in 4 meters of time. The rocket takes longer to cover this distance: 5 meters of time. Suppose that instead of 5 meters of time, the rocket had taken 8 meters of time, twice as long as light, to cover rhe 4 meters of disrance. In that case it would be moving at 4 /8 — or half— the speed of light. In the present case the rocket travels the 4 meters of distance in 5 meters of time, so it moves at 4 /5 light speed. Therefore its speed equals (4 /5 ) X 2.99792458 X 10® meters/second 2.3983397 X 10® meters/second
1.3 EVENTS AND INTERVALS ALONE! tools enough to chart matter and motion without any reference frame In surveying, rhe fundamental concept is place. The surveyor drives a steel stake to mark the corner of a plot of land — to mark a place. A second stake marks another corner of the same plot — another place. Every surveyor — no matter what his or her standard of north — can agree on the value of the distance between the two stakes, between the two places. Every stake has its own reality. Likewise the distance between every pair of srakes also has its own teality, which we can experience direcrly by pacing off the straight line from one stake to the other stake. The reading on our pedometer— the distance
Surveying locates a place
lO
Physics locates an event
W ristwatch m easures interval directly
CHAPTER]
SPACETIME: OVERVIEW
between stakes— is independent of all surveyors’ systems, with their arbitrary choice of north. More: Suppose we have a table of distances between every pair of stakes. That is all we need! From this table and the laws of Euclidean geometry, we can constmct the map of every surveyor (see the exercises for this chapter). Distances between stakes: That is all we need to locate every stake, every place on the map. In physics, the fundamental concept is event. The collision between one particle and another is an event, with its own location in spacetime. Another event is the emission of a flash of light from an atom. A third is the impact of the pebble that chips the windshield of a speeding car. A fourth event, likewise fixing in and by itself a location in spacetime, is the strike of a lightning bolt on the rudder of an airplane. An event matks a location in spacetime; it is like a steel stake driven into spacetime. Every laboratory and rocket observer— no matter what his or her relative velocity — can agree on the spacetime interval between any pair of events. Every event has its own reality. Likewise the interval between every pair of events also has its own reality, which we can experience directly. W e carry our wristwatch at constant velocity from one event to the other one. It is not enough just to pass through the two physical locations— we must pass through the actual events', we must be at each event precisely when it occurs. Then the space separation between the two events is zero for us — they both occur at our location. As a result, our wristwatch reads directly the spacetime interval between the pait of events: time (interval)^ — separation (meters) time separation (meters)
‘Do science” with intervals alone
space separation (meters) — [zero]^
time separation (meters)
[wristwatch time}
The time read on a wristwatch carried between two events — the interval between those events — is independent of all laboratory and rocket reference frames. More: To chart all happenings, we need no more than a table of spacetime intervals between every pair of events. That is all we need! From this table and the laws of Lorentz geometry, it turns out, we can construct the space and time locations of events as observed by every laboratory and rocket observet. Intervals between events: That is all we need to specify the location of every event in spacetime. In brief, we can completely describe and locate events entirely without a reference frame. W e can analyze the physical world— we can “do science” — simply by cataloging every event and listing the interval between it and every other event. The unity of spacetime is reflected in the simplicity of entries in our table: intervals only. O f course, if we want to use a reference frame, we can do so. We then list in our table the individual northward, eastward, upward, and time separations between pairs of events. However, these laboratoryframe listings for a given pair of events will be different from the corresponding listings that our rocketframe colleague puts in her table. Nevertheless, we can come to agreement if we use the individual separations to reckon the interval between each pair of events: (interval)^ — (time separation)^ — (space separation)^ That returns us to a universal, frameindependent description of the physical world. When two events both occur at the position of a certain clock, that special clock measures directly the interval between these two events. The interval is called the p r o p e r tim e (or sometimes the local tim e). The special clock that records the proper time directly has the name p r o p e r clo ck for this pair of events. In this book
1.4
SAME UNIT FOR SPACE AND TIME: METER, SECOND, MINUTE, OR YEAR
11
we often call the proper time the w ristw atc h tim e and the proper clock the w ristw atc h to emphasize that the proper clock is carried so that it is “present” at each of the two events as the events occur. In Einstein’s German, the word for proper time is Eigenzeil, or “owntime,” implying “one’s very own time.” The German word provides a more accurate description than the English. In English, the word “proper” has come to mean “ following conventional mles.” Proper time certainly does not do that! Hey! I just thought of something: Suppose two events occur at the same time in myframe but very fa r apart, for example two handclaps, one in New York City and one in San Francisco. Since they are simultaneous in my frame, the time separation between handclaps is zero. But the space separation is not zero— they are separated by the width of a continent. Therefore the square of the interval is a negative number: {interval^ = (time separation^ — (space separation)^ = (zero)^ — (space separation)^ = — (space separation)^ How can the square of the spacetime interval be negative? In most of the situations described in the present chapter, there exists a reference frame in which two events occur at the same place. In these cases time separation predominates in all frames, and the interval squared will always be positive. W e call these intervals tim e lik e in tervals. Euclidean geometry adds squares in reckoning distance. Hence the result of the calculation, distance squared, is always positive, regardless of the relative magni tudes of north and east separations. Lorentz geometry, however, is richer. For your simultaneous handclaps in New York City and San Francisco, space separation between handclaps predominates. In such cases, the interval is called a sp acelik e in te rv al and its form is altered to (interval)^ = (space separation)^ — (time separation)^
[when spacelike]
This way, the squared interval is never negative. The timelike interval is measured directly using a wristwatch carried from one event to the other in a special frame in which they occur at the same place. In contrast, a spacelike interval is measured directly using a rod laid between the events in a special frame in which they occur at the same time. This is the frame you describe in your example. Spacelike interval or timelike interval: In either case rhe interval is invariant— has the same value when reckoned using rocket measurements as when reckoned using laboratory measurements. You may want to skim through Chapter 6 where timelike and spacelike intervals are described more fully.
1.4 SAME UNIT FOR SPACE AND TIME: METER, SECOND, MINUTE, OR YEAR meter for particle accelerators; minute for planets; year for the cosmos The parable of the surveyors cautions us to use the same unit to measure both space and time. So we use meter for both. Time can be measured in meters. Let a flash of light bounce back and forth between parallel mirrors separated by 0.5 meter of
M easure time in meters
12
0.5 meter
FIGURE 13. This twomirror “clock" sends to the eyeflash afterflash, each separatedfrom the next by 1 meter of lighttravel time. A light flash (represented by an asterisk) bounces back and forth between parallel mirrors separated from one another by 0.5 meter of distance. The silver coating of the righthand mirror does not reflect perfectly: It lets 1 percent of the light pass through to the eye each time the light pulse hits it. Hence the eye receives a pulse of light every meter of lighttravel time.
M eter officially defined using light speed
M easure distance in lightyears
CHAPTER 1
SPACETIME: OVERVIEW
distance (Figure 13). Such a device is a “clock” that “ticks” each time the light flash arrives back at a given mirror. Between ticks the light flash has traveled a roundtrip distance of 1 meter. Therefore we call the stretch of time between ticks 1 m e te r o f lig h ttrav el tim e or more simply 1 m e te r o f tim e. One meter of lighttravel time is quite small compared to typical time lapses in our everyday experience. Light travels nearly 300 million meters per second (300,000,000 meters/second = 3 X 10® meters/second, four fifths of the way to Moon in one second). Therefore one second equals 300 million meters of lighrtravel time. So 1 meter of lighttravel time has the small value of one threehundredmillionth of a second. [How come? Because (1) light goes 300 million meters in one second, and (2) one threehundredmillionth of that distance (one meter!) is covered in one threehundredmillionrh of that time.] Nevertheless this unit of time is very useful when dealing with light and with highspeed particles. A proton early in its travel through a particle accelerator may be jogging along at “only” one half the speed of light. Then it travels 0.5 meter of distance in 1 meter of lighttravel time. We, our cars, even our jet planes, creep along at the pace of a snail compared with light. W e call a deed quick when we’ve done it in a second. But a second for light means a distance covered of 300 million meters, seven trips around Earth. As we dance around the room to the fastest music, oh, how slow we look to light! Not zooming. Not dancing. N ot creeping. Oozing! That long slow ooze racks up an enormous number of meters of lighttravel time. That number is so huge that, by the end of one step of our frantic dance, the light that carries the image of the step’s beginning is well on its way to Moon. In 1983 the General Conference on Weights and Measures officially redefined the meter in terms of the speed of light. T h e m e te r is now d efined as th e d istan ce th a t lig h t travels in a v acu u m in th e fractio n 1/299,792,458 o f a second. (For the definition of the second, see Box 32.) Since 1983 the speed of light is, by definition, equal to c = 299,792,458 meters/second. This makes official the central position of the speed of light as a conversion factor between time and space. This official action defines distance (meter) in terms of time (second). Every day we use time to measure distance. “My home is only ten minutes (by car) from work.” “The business district is a fiveminute walk.” Each statement implies a speed — the speed of driving or walking— that converts distance to time. But these speeds can vary— for example, when we get caught in traffic or walk on cmtches. In contrast, the speed of light in a vacuum does not vary. It always has the same value when measured over time and the same value as measured by every observer. We often describe distances to stars and galaxies using a unit of time. These distances we measure in lightyears. One lightyear equals the distance that light travels in one year. Along with the lightyear of space goes the year of time. Here again, space and time are measured in the same units— years. Here again the speed of light is the conversion factor between measures of time and space. From our everyday per spective one lightyear of space is quite large, almost 10,000 million million meters: 1 lightyear = 9,460,000,000,000,000 meters = 0.946 X 10*® meters. Nevertheless it is a convenient unit for measuring distance between stars. For example, the nearest star to our Sun, Proxima Centauri, lies 4.28 lightyears away. Any common unit of space or time may be used as the same unit for both space and time. For example. Table 15 gives us another convenient measure of time, seconds, compared with time in meters. We can also measure space in the same units, lightseconds. Our Sun is 499 lightseconds — or, more simply, 499 seconds — of distance from Earth. Seconds are convenient for describing distances and times among events that span the solar system. Alternatively we could use minutes of time and lightminutes of distance: Our Sun is 8.32 lightminutes from Earth. W e can also use hours of time and lighthours of distance. In all cases, the speed of light is the conversion factor between units of space and time.
1.4
SAME UNIT FOR SPACE AND TIME; METER, SECOND, MINUTE, OR YEAR  d ^ B L E
13

SOME LIGHTTRAVEL TIMES Time in seconds of lighttravel time
Time in meters
Telephone call one way: New York City to San Francisco via surface microwave link
0.0138
4.139,000
Telephone call one way: New York City to San Ftancisco via Earth satellite
0.197
59,000,000
Telephone call one way: New York City to San Francisco bounced off Moon
2.51
752,000,000
Flash of light: Emitted by Sun, received on Earth
499.0
149,600,000.000
Expressing time and space in the same unit m e te r is convenient for describing motion of highspeed particles in the confines of the laboratory. Time and space in the same unit second (or m in u te or h o u r) is convenient for describing relations among events in our solar system. Time and space in the same unit year is convenient for describing relations among stars and among galaxies. In all three arenas spacetime is the stage and special relativity is the spotlight that illuminates the inner workings of Nature. W e are not accustomed to measuring time in meters. So as a reminder to ourselves we add a descriptor: meters of lighttravel time. But the unit of time is still the meter. Similarly, the added words “seconds of distance" and ‘‘lightyears’’ help to remind us that distance is measured in seconds or years, units we usually associate with time. But this unit of distance is really just second or year. The modifying descriptors are for our convenience only. In Nature, space and time form a unity: spacetime! The words sound OK. The mathematics appears straightforward. The Sample Problems seem logical. But the ideas are so strange! Why should I believe them? How can invariance of the interval be proved?
N o wonder these ideas seem strange. Particles zooming by at nearly the speed of light — how far this is from our everyday experience! Even the soaring jet plane crawls along at less than onemillionth light speed. Is it so surprising that the world appears different at speeds a million times faster than those at which we ordinarily move with respect to Earth? The notion of spacetime interval distills a wealth of real experience. W e begin with interval because it endures: It illuminates observations that range from rhe core of a nucleus to the center of a black hole. Understand the spacetime interval and you vault, in a single bound, to the heart of spacetime. Chapter 3 presents a logical proof of the invariance of the interval. Chapter 4 reports a knockdown argument about it. Chapters that follow describe many experiments whose outcomes are rorally incomprehensible unless the interval is invariant. Real verification comes daily and hourly in the ongoing enterprise of experimental physics.
Use convenient units, the sam e for sp ace and time
14
CHAPTER 1
SPACETIME: OVERVIEW
SAMPLE PROBLEM
1 2
P R O T O N , R O C K , AND S T A R S H I P a. A proton moving at 3 /4 light speed (with respect to the laboratory) passes
through two detectors 2 meters apart. Events 1 and 2 are the transits through the two detectors. W hat are the laboratory space and time separations between the two events, in meters? W hat are the space and time separations between the events in the proton frame? b. A speeding rock from space streaks through Earth’s outer atmosphere, creating a
short fiery rrail (Event 1) and continues on its way to crash into Sun (Event 2) 10 minutes later as observed in the Earth frame. Take Sun to be 1.4960 X 10“ meters from Earth. In the Earth frame, what are space and time separations between Event 1 and Event 2 in minutes? W hat are space and time separations between the events in the frame of the rock? c. In the twentythird century a starship leaves Earth (Event 1) and travels at 95
percent light speed, later arriving at Proxima Centauri (Event 2), which lies 4.3 lightyears from Earth. W hat are space and time separations between Event 1 and Event 2 as measured in the Earth frame, in years? W hat are space and time separations between these events in the frame of the starship?
SOLUTION a. The space separation measured in the laboratory equals 2 meters, as given in the
problem. A flash of light would take 2 meters of lighttravel time to travel between the two detectors. Something moving at 1/ 4 light speed would take four times as long: 2 meters/( 1/4 ) = 8 meters of lighttravel time to travel from one detector to the other. The proton, moving at 3 /4 light speed, takes 2 meters/ (3/4) = 8 /3 meters = 2.66667 meters of lighttravel time between events as measured in the laboratory. Event 1 and Event 2 both occur at the position of the proton. Therefore the space separation between the two events equals zero in the proton frame. This means that the spacetime interval — the proper time— equals the time between events in the proton frame. (proton time)^ — (proton distance)^ = (interval)^ = (lab time)^ — (lab distance)^ (proton time)^ — (zero)^ = (2.66667 meters)^ ~ (2 meters)^ = ( 7. 1111— 4) (meters)^ (proton time)^ = 3.1111 (meters)^ So time between events in the proton frame equals the square root of this, or 1.764 meters of time. b. Light travels 60 times as far in one minute as it does in one second. Its speed in
meters per minute is therefore: 2.99792458 X 10® meters/second X 60 seconds/minute = 1.798754748 X 10^® meters/minute So the distance from Earth to Sun is 1.4960 X 10“ meters 1.798754748 X 10^® meters/minute
— 8.3169 lightminutes
1.5
UNITY OF SPACETIME
15
This is the distance between the two events in the Earth frame, measured in lightminutes. The Earthframe time between the two events is 10 minutes, as stated in the problem. In the frame traveling with the rock, the two events occur at the same place; the time between the two events in this frame equals the spacetime interval — the proper time— between these events: (interval)^ = (10 minutes)^ — (8.3169 minutes)^ = (100  69.1708) (minutes)^ = 30.8292 (minutes)^ The time between events in the rest frame of the rock equals the square root of this, or 5.5524 minutes. c. The distance between departure from Earth and arrival at Proxima Centauri is
4.3 lightyears, as given in the problem. The starship moves at 95 percent light speed, or 0.95 lightyears/year. Therefore it takes a time 4.3 lightyears/(0.95 lightyears/year) = 4.53 years to arrive at Proxima Centauri, as measured in the Earth frame. Starship time between departure from Earth and arrival at Proxima Centauri equals the interval: (interval)^ = (4.53 years)^ ~ (4.3 years)^ = (20.52  18.49) (years)2 = 2.03 (years)^ The time between events in the rest frame of the starship equals the square root of this, or 1.42 years. Compare with the value 4.53 years as measured in the Earth frame. This example illustrates the famous idea that astronaut wristwatch time — proper time — between two events is less than the time between these events measured by any other observer in relative motion. Travel to stay young! This result comes simply and naturally from the invariance of the interval.
1.5 UNITY OF SPACETIME time and space: equal footing but distinct nature When time and space are measured in the same unit— whether meter or second or year— the expression for the square of the spacetime interval between two events takes on a particularly simple form: (interval)^ = (time separation)^ — (space separation)^ = ,2 _ „2
[same units for time and space]
This formula shows forth the unity of space and time. Impressed by this unity, Einstein’s teacher Hermann Minkowski (18641909) wrote his famous words, “ Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a union of the two will preserve an independent reality. ’’Today this union of space and time is called spacetime. Spacetime provides the tme theater for
Spacetim e is a unity
16
CHAPTER 1
SPACETIME: OVERVIEW
PAYOFF OF THE PARABLE from distance in space to interval in spacetime DISCUSSION
SURVEYING TOWNSHIP
ANALYZING NATURE
Location marker
Steel stake driven in ground
Collision between two particles Emission of flash from atom Spark jumping from antenna to pen
General name for such a location marker
Point or place
Event
Can its location be staked out for all to see, independent of any scheme of measurement, and independent of all numbers?
Yes
Yes
Simple descriptor of separation between two location markers
Distance
Spacetime interval
Are there ways directly to measure this separation?
Yes
Yes
With enough markers already staked out, how can we tell some one where we want the next one?
Specify points.
Instead of boldly staking out the new marker, or instead of position ing it relative to existing markers, how else can we place the new marker?
By locating point relative to a reference frame
By locating event relative to a ref erence frame
Nature of this reference frame?
Surveyor’s grid yields northward and eastward readings of point (Chapter 1).
Lattice frame of rods and clocks yields space and time readings of event (Chapter 2).
Is such a reference frame unique?
No
No
How do two such reference frames
Tilt of one surveyor’s grid relative to the other
Uniform velocity of one frame rela tive to the other
What are names of two such possi ble reference frames?
Daytime grid: oriented to magnetic north Nighttime grid: oriented to NorthStar north
Laboratory frame Rocket frame
What common unit simplifies analy sis of the results?
The unit meter for both northward and eastward readings
The unit meter for both space and time readings
What is the conversion factor from conventional units to meters?
k=
Converting miles to meters: 1609.344 meters/mile
Converting seconds to meters using the speed of light: c = 299,792,458 meters/second
d iffer from o n e a n o t h e r ?
distances
from
other
Specify spacetime intervals from other events.
1.5
DISCUSSION
UNITY OF SPACETIME
SURVEYING TOWNSHIP
17
ANALYZING NATURE
For convenience, all measurements are referred to what location?
A common origin (center of town)
A common event (reference spark)
How do readings for a single marker differ between two refer ence frames?
Individual northward and eastward readings for one point — for one steel stake — do not have the same values respectively for two survey ors’ grids that are tilted relative to one another.
Individual space and time readings for one event — for one spark — do not have the same values re spectively for two frames that are in motion relative to one another.
When we change from one marker to two, how do we specify the offset between them in referenceframe language?
Subtract: Figure the difference be tween eastward readings of the two points; also the difference in northward readings.
Subtract: Figure the difference be tween space readings of the two events; also the difference in time readings.
How to figure from offset readings a measure of separation that has the same value whatever the choice of reference frame?
Figure the distance between the two points.
Figure the spacetime interval be tween the two events.
Figure how?
(distance)^ =
(interval)^ =
/ difference in V \northw ard readin g s/ _l_ / difference in V \e a stw a rd reading s/
/ difference in V \time read in g s/ _ / difference in \sp a c e readin g s/
Result of this reckoning?
Distance between points as figured from readings using one surveyor’s grid is the same as figured from readings using a second surveyor’s grid tilted with respect to first grid.
Interval between events as figured from readings using one latticework frame is the same as figured from readings using a second frame in steady straightline motion relative to first frame.
Phrase to summarize this identity of separation as figured in two refer ence frames?
Invariance of the distance between points
Invariance of the spacetime inter val between events.
Conclusions from this analysis?
(1) Northward and eastward di mensions are part of a single entity: space.
(1) Space and time dimensions are part of a single entity: spacetime.
(2) Distance is the simple measure of separation between two points, natural because invariant: the same for different surveyor grids.
(2) Spacetime interval is the simple measure of separation between two events, natural because invar iant: the same for different refer ence frames.
18
Difference between time and space
CHAPTER 1
SPACETIME: OVERVIEW
every event in the lives of stars, atoms, and people. Space is different for different observers. Time is different for different observers. Spacetime is the same for everyone. Minkowski’s insight is central to the understanding of the physical world. It focuses attention on those quantities, such as spacetime interval, electrical charge, and particle mass, that are the same for all observers in relative motion. It brings out the merely relative character of quantities such as velocity, momentum, energy, separation in time, and separation in space that depend on relative motion of observers. Today we have learned not to overstate Minkowski’s argument. It is right to say that time and space are inseparable parts of a larger unity. It is wrong to say that time is identical in quality with space. Why is it wrong? Is not time measured in meters, just as space is? In relating the positions of two steel stakes driven into the ground, does not the surveyor measure northward and eastward separations, quantities of identical physical character? By analogy, in locating two events is not the observer measuring quantities of the same nature: space and time separations? How else could it be legitimate to treat these quantities on an equal footing, as in the formula for the interval? Equal footing, yes; same nature, no. There is a minus sign in the formula for the interval squared = (time separation)^ — (space separation)^ that no sleight of hand can ever conjure away. This minus sign distinguishes between space and time. No twisting or turning can ever give the same sign to real space and time separations in the expression for the interval.
The invariarxe of the spacetime interval evidences the unity of space and time while also preserving — in the formula’s minus sign — the distinction between the two. The principles of special relativity are remarkably simple— simpler than the axioms of Euclidean geometry or the principles of operating an automobile. Yet both Euclid and the automobile have been mastered— perhaps with insufficient surprise — by generations of ordinary people. Some of the best minds of the twentieth century stmggled with the concepts of relativity, not because nature is obscure, but because (1) people find it difficult to outgrow established ways of looking at namre, and (2) the world of the very fast described by relativity is so far from common experience that everyday happenings are of limited help in developing an intuition for its descriptions. By now we have won the battle to put relativity in understandable form. The concepts of relativity can now be expressed simply enough to make it easy to think correctly — “to make the bad difficult and the good easy. ’’ This leaves only the second difficulty, that of developing intuition — a practiced way of seeing. W e understand distance intuitively from everyday experience. Box 1.1 applies our intuition for d istance in space to help our intuition for in terv a l in spacetim e. To put so much into so little, to subsume all of Einstein’s teaching on light and motion in the single word spacetime, is to cram a wealth of ideas into a small picnic basket that we shall be unpacking throughout the remainder of this book.
REFERENCES Introductory quote: Richard P. Feynman, The Character of Physical Law (MIT Press, Cambridge, Mass., 1967), page 127. Quote from Minkowski in Section 1.5: H. A. Minkowski, “Space and Time,” in H. A. Lorentz et al., The Principle of Relativity (Dover Publications, New York, 1952), page 75.
ACKNOWLEDGMENTS
19
Quote at end of Section 1.5: “ to make the bad difficult and the good easy, ’’ ‘‘rend le maldifficile et le bien facile.” Einstein, in a similar connection, in a letter to the architect Le Corbusier. Private communication from Le Corbusier. For an appreciation of Albert Einstein, see John Archibald Wheeler, “Albert Einstein,” in The World Treasury of Physics, Astronomy, and Mathematics, Timo thy Ferris, ed. (Little, Brown, New York, 1991), pages 5 6 3 5 7 6 .
ACKNOWLEDGMENTS Many students in many classes have read through sequential versions of this text, shared with us their detailed difficulties, and given us advice. We asked students to write down comments, perplexities, and questions as they read and turn in these reading memos for personal response by the teacher. Italicized objections in the text come, in part, from these commentators. Both we who write and you who read are in their debt. Some readers not in classes have also been immensely helpful; among these we especially acknowledge Steven Bartlett. No one could have read the chapters more meticulously than Eric Sheldon, whose wide knowl edge has enriched and clarified the presentation. William A. Shurcliff has been immensely inventive in devising new ways of viewing the consequences of relativ ity, a few of these are specifically acknowledged in later chapters. Electronicmail courses using this text brought a flood of comments and reading memos from teachers and students around the world. Richard C. Smith originated, organized, and administered these courses, for which we are very grateful. The clarity and simplicity of both the English and the physics were improved by Penny Hull. Some passages in this text, both brief and extended, have been adapted from the book A Journey into Gravity and Spacetime by John Archibald Wheeler (W .H. Freeman, New York, 1990). In turn, certain passages in that book were adapted from earlier drafts of the present text. W e have also used passages, logical arguments, and figures from the book Gravitation by Charles W . Misner, Kip S. Thorne, and John Archibald Wheeler (W. H. Freeman, New York, 1973).
INTRODUCTION TO THE EXERCISES Important areas of current research can be analyzed very simply using the theory of relativity. This analy sis depends heavily on a physical intuition, which develops with experience. Wide experience is not easy to obtain in the laboratory— simple experiments in relativity are difficult and expensive because the speed of light is so great. As alternatives to experiments, the
exercises and problems in this text evoke a wide range of physical consequences of the properties of spacetime. These properties of spacetime recur here over and over again in different contexts: • paradoxes • puzzles
20
EXERCISE 11
COMPARING SPEEDS
• derivations • technical applications • experimental results • estimates • precise calculations • philosophical difficulties The text presents all formal tools necessary to solve these exercises and problems, but intuition — a prac ticed way of seeing — is best developed without hurry. For this reason we suggest continuing to do more and more of these exercises in relativity after you have moved on to material outside this book. The mathematical manipulations in the exercises and problems are very brief: only a few answers take more
than five lines to write down. On the other hand, the exercises require some “ruminarion time.” In some chapters, exercises are divided inro rwo categories. Practice and Problems. The Practice exer cises help you to get used to ideas in the text. The Problems apply these ideas to physical systems, thought experiments, and paradoxes. w heeler ’s first moral principle : Never make a calculation until you know the answer. Make an estimate before every calculation, try a simple physical argument (symmetry! invariance! conservation!) be fore every derivation, guess the answer to every para dox and puzzle. Courage: No one else needs to know what the guess is. Therefore make ir quickly, by instinct. A right guess reinforces this instinct. A wrong guess brings the refreshment of surprise. In eirher case life as a spacetime expert, however long, is more fun!
CHAPTE3R 1 EXERCISES
PRACTICE 11
comparing speeds
Compare the speeds of an automobile, a jet plane, an Earth satellite. Earth in its orbit around Sun, and a pulse of light. Do this by comparing the relative distance each travels in a fixed tim e. Arbitrarily
choose the fixed time to give convenient distances. A car driving at the USA speed limit of 65 miles/hour (105 kilometers/hour) covers 1 meter of distance in about 35 milliseconds = 35 X 10“ ^ second. a How far does a commercial jetliner go in 35 milliseconds? (speed: 650 miles/hour = 1046 kilometers/hour) b How far does an Earth satellite go in 3 5 milli seconds? (speed: 17,000 miles/hour ~ 27,350 kilometers/hour) C How far does Earth travel in its orbit around Sun in 35 milliseconds? (speed: 30 kilometers/second) d How far does a light pulse go in a vacuum in 35 milliseconds? (speed: 3 X 10® meters /second). This distance is roughly how many times the distance from Boston to San Erancisco (5000 kilometers)?
12 images from Neptune At 9:00 P.M. Pacific Daylight Time on August 24, 1989, the planetary probe Voyager 11 passed by the planet Neptune. Images of the planet were coded and rransmitted to Earth by microwave relay. It took 4 hours and 6 minutes for this microwave signal to travel from Neptune to Earth. Microwaves (electromagneric radiation, like light, but of fre quency lower than that of visible light), when propa gating through interplanetary space, move at the ‘‘standard ’’ light speed of one meter of distance in one meter of lighttravel time, or 299,792,458 meters/ second. In the following, neglect any relative motion among Earth, Neptune, and Voyager 11. a Calculate the distance between Earth and Neptune at flyby in units of minutes, seconds, years, meters, and kilometers. b Calculate the time the microwave signal takes to reach Earth. Use the same units as in part a.
13 units of spacetime Light moves at a speed of 3.0 X 10® meters/second. One mile is approximately equal to 1600 meters. One furlong is approximately equal to 200 meters.
EXERCISE 16 a b c d e
How many meters of time in one day? How many seconds of distance in one mile? How many hours of distance in one furlong? How many weeks of distance in one lightyear? How many furlongs of time in one hour?
15 where and when? Two firecrackers explode at the same place in the laboratory and are separated by a time of 3 years as measured on a laboratory clock. a W hat is the spatial distance between these two events in a rocket in which the events are separated in time by 5 years as measured on rocket clocks? b W hat is the relative speed of the rocket and laboratory frames?
A rocket clock emits two flashes of light and the rocket observer records the time lapse (in seconds) between these two flashes. The laboratory observer records the time separation (in seconds) and space separation (in lightseconds) between the same pair of flashes. The results for both laboratory and rocket observers are recorded in the first line of the table. Now a clock in a different rocket, moving at a different speed with respect to the laboratory, emits a different pair of flashes. The set of laboratory and rocket space and time separations are recorded on the
16 mapmaking in space The table shows distances between cities. The units are kilometers. Assume all cities lie on the same flat plane. a Use a ruler and a compass (the kind of compass that makes circles) to construct a map of these cities. Choose a convenient scale, such as one centimeter on the map corresponds to ten kilometers on Earth. D iscussion: How to start? W ith three arbitrary decisions! (1) Choose any city to be at the center of the map. (2) Choose any second city to be “due north” — that is, along any arbitrary direction you select. (3) Even with these choices, there are two places you can locate the third city; choose either of these two places arbitrarily. b If you rotate the completed map in its own plane — for example, turning it while keeping it flat on the table— does the resulting map also satisfy the distance entries above? C Hold up your map between you and a light, with the marks on the side of the paper facing rhe
■
b
20
c
66.8
d
?
e
21
10.72
5.95 99 p
72.9 8.34
6.58 ?
22
21
second line of the table. And so on. Complete the table.
1>4 time stretching and the spacetime interval
Rocket time lapse (seconds)
MAPMAKING IN SPACE
C ^ ^ ^ X E R C IS E

DISTANCES BETWEEN CITIES Distance to city
A
B
C
D
E
F
G
H
20.0
28.3 20.0
28.3 20.0
28.3 44.7
20.0 40.0
28.3 44.7
40.0
40.0
40.0
44.7
56.6
60.0
56.6
44.7
40.0
20.0
40.0
20.0 72.1
20.0
56.6
44.7
from city A B C D E F G H
44.7
22
EXERCISE 17
SPACETIME MAP
light. Does the map you see from the back also satisfy the table entries? D iscussion: In this exercise you use a table con sisting only of distances between pairs of cities to construct a map of these cities from the point of view of a surveyor using a given direction for north. In Exercise 53 you use a table consisting only of spacetime intervals between pairs of events to draw a “spacetime m ap” of these events from the point of view of one freefloat observer. Exercise 17 previews this kind of spacetime map.
17 spacetime map The laboratory space and time measurements of events 1 through 5 are plotted in the figure. Compute the value of the spacetime interval a between event 1 and event 2. b between event 1 and event 3. c between event 1 and event 4. d between event 1 and event 5. e A rocket moves with constant velocity from event 1 to event 2. That is, events 1 and 2 occur at the same place in this rocket frame. W hat time lapse is recotded on the rocket clock between these two events?
event
instruction requires transmission of data from the memory (where data is stored) to the processor (where the computation is carried out) and transmission of the result back to the memory for storage. a W hat is the maximum average distance be tween memory and processor in a “onemegaflop” computer? Is this maximum distance increased or decreased if the signal travels through conductors at one half the speed of light in a vacuum? b Computers are now becoming available that operate at “one gigaflop,” that is, they carry out 10^ sequential instructions per second. W hat is the maxi mum average distance between memory and proces sor in a “onegigaflop” machine? c Estimate the overall maximum size of a “oneteraflop” machine, that is, a computer that can carry out 10*^ sequential instructions per second. d D iscussion question: In contrast with most current personal computers, a “parallel processing” computer contains several or many processors that work together on a computing task. One might think that a machine with 10,000 processors would com plete a given computation task in 1/10,000 the time. However, many computational problems cannot be divided up in this way, and in any case some fraction of the computing capacity must be devoted to coordi nating the team of processors. W hat limits on physi cal size does the speed of light impose on a parallel processing computer?
2 event 4
19 trips to Andromeda by rocket
event '3
t
event
time (meters)
event 1,
1
2
3
4
5
6
space (meters) — ► EXERCISE 17. Spacetime map of some events.
PROBLEMS 18 size off a computer In one second some desktop computers can carry out one million instructions in sequence: One instruction might be, for instance, multiplying two numbers to gether. In technical jargon, such a computer operates at “one megaflop.” Assume that carrying out one
The Andromeda galaxy is approximately two million lightyears distant from Earth as measured in the Earthlinked frame. Is it possible for you to travel from Earth to Andromeda in your lifetime? Sneak up on the answer to this question by considering a series of trips from Earth to Andromeda, each one faster than the one before. For simplicity, assume the EarthAndromeda distance to be exactly two million lightyears in the Earth frame, treat Earth and Andromeda as points, and neglect any relative motion between Earth and Andromeda. a TRIP 1. Your oneway trip takes a time 2 .0 1 X 10^ years (measured in the Earthlinked frame) to cover the distance of 2.00 X 10^ lightyears. How long does the trip last as measured in your rocket frame? b W hat is your rocket speed on Trip 1 as mea sured in the Earthlinked frame? Express this speed as a decimal fraction of the speed of light. Call this fraction, p = where is speed in conven tional units, such as meters/second. D iscussion: If your rocket moves at half the speed of light, it takes
EXERCISE 111
TIME STRETCHING WITH /iMESONS
4 X 10^ years to cover the distance 2 X 10^ lightyears. In this case 2 X 10^ lightyears
1
4 X 10^ years Therefore . . . c TRIP 2. Your oneway EarthAndromeda trip takes 2.001 X 10® years as measured in the Earthlinked frame. How long does the trip last as measured in your rocket frame? W hat is your rocket speed for Trip 2, expressed as a decimal fraction of the speed of light? d TRIP 3. Now set the rocket time for the one way trip to 20 years, which is all the time you want to spend getting to Andromeda. In this case, what is your speed as a decimal fraction of the speed of light? D iscussion: Solutions to many exercises in this text are simplified by using the following approximation, which is the first two terms in the binomial expansion (1 b z)” ~ 1 + nz
l «
1
Here n can be positive or negative, a fraction or an integer; z can be positive or negative, as long as its magnitude is very much smaller than unity. This approximation can be used twice in the solution to part d.
110 trip to Andromeda by Transporter In the Star Trek series a socalled Transporter is used to “beam” people and their equipment from a starship to the surface of nearby planets and back. The Transporter mechanism is not explained, but it ap pears to work only locally. (If it could transport to remote locations, why bother with the starship at all?) Assume that one thousand years from now a Trans porter exists that reduces people and things to data (elementary bits of information) and transmits the data by light or radio signal to remote locations. There a Receiver uses the data to reassemble travelers and their equipment out of local raw materials. One of your descendants, named Samantha, is the first “transporternaut” to be beamed from Earth to the planet Zircon orbiting a star in the Andromeda Nebula, two million lightyears from Earth. Neglect any relative motion between Earth and Zircon, and assume: (1) transmission produces a Samantha iden tical to the original in every respect (except that she is 2 million lightyears from home!), and (2) the time required for disassembling Samantha on Earth and reassembling her on Zircon is negligible as measured
23
in the common rest frame of Transporter and Re ceiver. a How much does Samantha age during her outward trip to Zircon? b Samantha collects samples and makes obser vations of the Zirconian civilization for one Earthyear, then beams back to Earth. How much has Sa mantha aged during her entire trip? C How much older is Earth and its civilization when Samantha returns? d Earth has been taken over by a tyrant, who wishes to invade Zircon. He sends one warrior and has him duplicated into attack battalions at the Receiver end. How long will the Earth tyrant have to wait to discover whether his ambition has been satisfied? e A second transporternaut is beamed to a much more remote galaxy that is moving away from Earth at 87 percent of the speed of light. This time, too, the traveler stays in the remote galaxy for one year as measured by clocks moving with the galaxy before re turning to Earth by Transporter. How much has the transporternaut aged when she arrives back at Earth? (Careful!)
111
time stretching with muons
At heights of 10 to 60 kilometers above Earth, cosmic rays continually strike nuclei of oxygen and nitrogen atoms and produce muons (muons: elementary parti cles of mass equal to 207 electron masses produced in some nuclear reactions). Some of the muons move vertically downward with a speed nearly that of light. Follow one of the muons on its way down. In a given sample of muons, half of them decay to other ele mentary particles in 1.5 microseconds (1.5 X 10~® seconds), measured with respect to a reference frame in which they are at rest. Half of the remainder decay in the next 1.5 microseconds, and so on. Analyze the results of this decay as observed in two different frames. Idealize the rather complicated acmal experi ment to the following roughly equivalent situation: All the muons are produced at the same height (60 kilometers); all have the same speed; all travel straight down; none are lost to collisions with air molecules on the way down. a Approximately how long a time will it take these muons to reach the surface of Earth, as mea sured in the Earth frame? b If the decay time were the same for Earth observers as for an observer traveling with the muons, approximately how many halflives would have passed? Therefore what fraction of those created at a height of 60 kilometers would remain when they
24
EXERCISE 112
TIME STRETCHING WITH 7T+MES0NS
reached sea level on Earth? You may express your answer as a power of the fraction 1/2. c An experiment determines that the fraction 1/ 8 of the muons reaches sea level. Call the rest frame of the muons the rocket frame. In this rocket frame, how many halflives have passed between creation of a given muon and its arrival as a survivor at sea level? d In the rocketframe, what is the space separation between birth of a survivor muon and its arrival at the surface of Earth? (Careful!) e From the rocket space and time separations, find the value of the spacetime interval between the birth event and the arrival event for a single surviving muon.
TIME STRETCHING WITH 7T+MES0NS
Particle
Time for half to decay (measured in rest frame)
muon 1.5 X 10"^ second (207 times electron mass) TT^meson (273 times electron mass)
18 X 10“* second
"Characteristic distance” (speed of light multiplied by foregoing time) 450 meters
5.4 meters
Reference: Nalini Easwar and Douglas A. Macintire, American Jour nal of Physics, Volume 59, pages 5 8 9  5 9 2 (July 1991).
112 time stretching with TT^mesons Laboratory experiments on particle decay are much more conveniently done with TT^mesons (piplus mesons) than with //mesons, as is seen in the table. In a given sample of TT^mesons half will decay to other elementary particles in 18 nanoseconds (18 X 10“^ seconds) measured in a reference frame in which the TT^mesons are at rest. Half of the remainder will decay in the next 18 nanoseconds, and so on. a In a particle accelerator TT^mesons are pro duced when a proton beam strikes an aluminum
target inside the accelerator. Mesons leave this target with nearly the speed of light. If there were no time stretching and if no mesons were removed from the resulting beam by collisions, what would be the greatest distance from the target at which half of the mesons would remain undecayed? b The TT^mesons of interest in a particular ex periment have a speed 0.9978 that of light. By what factor is the predicted distance from the target for halfdecay increased by time dilation over the previous prediction — that is, by what factor does this dilation effect allow one to increase the separation between the detecting equipment and target?
CHAPTER 3 of this uncontracted cube. Interpret this expression for the two limiting cases of cube speed in the laboratory frame: p —* 0 and p —* l . C D iscussion question: Is the word “really” an appropriate word in the following quotations?
EXERCISE 317
CONTRACTION OR ROTATION?
93
Location of cube E
G
(1  v^)'
EXERCISE 3 1 7 . h e ft: Position o f eye o f v is u a l observer w a tc h in g cube pass overhead. R i g h t to p : W h a t the v is u a l observer sees a s she looks u p fro m below. R i g h t b o tto m : H o w the v is u a l observer can interpret the projection o f the second figure.
(1) An observer using the rocket latticework of clocks says, “The stationary cube is really nei ther rotated nor contracted.” (2) Someone riding in the rocket who looks at the stationary cube agrees, “The cube is really nei ther rotated nor contracted.” (3) An observer using the laboratory latticework of clocks says, “The passing cube is really Lorentz contracted but not rotated.” (4) Someone standing in the laboratory frame looking at the passing cube says,' ‘The cube is really rotated but not Lorentz contracted.”
W hat can one rightfully say — in a sentence or two — to make each observer think it reasonable that the other observers should come to different conclu sions? d The analysis of parts b and c assumes that the visual observer looks with one eye and has no depth perception. How will the cube passing overhead be perceived by the viewer with accurate depth percep tion? Reference: For a more complete treatm ent of this topic, see Edwin F. Taylor, Introductory Mechanics (John Wiley and Sons, New York, 1963), pages 3 4 6  3 6 0 .
■'■‘'c'i >;^, ;■■
M0
■>:; +:.‘v^:
'..■V^sv ... 'h'«• ff
•v?
S P E C I A L TOPIC
' i ■'■
•■ ■
LORENTZ TRANSFORMATION
L I LORENTZ TRANSFORMATION: USEFUL OR NOT? related events or lonely events? Events, and the intervals between events, define the layout of the physical world. No latticework of clocks there! Only events and the relation between event and event as expressed in the interval. That’s spacetime physics, lean and spare, as it offers itself to us to meet the needs of industry, science, and understanding. There’s another way to express the same information and use it for the same purposes: Set up a freefloat latticework of recording clocks, or the essential rudiments of such a latticework. The space and time coordinates of that Lorentz frame map each event as a lonesome individual, with no mention of any connection, any spacetime interval, to any other event. This latticebased method for doing spacetime physics has the advantage that it can be mechanized and applied to event after event, wholesale. These regimented space and time coordinates then acquire full usefulness only when we can translate them from the clocklattice frame used by one analyst to the clocklattice frame used by another. This scheme of translation has acquired the name “Lorentz transformation.” Its usefulness depends on the user. Some never need it because they deal always with intervals. Others use it frequently because it regiments records and standardizes analysis. For their needs we insert this Special Topic on the Lorentz transformation. The reader may wish to read it now, or skip it altogether, or defer it until after Chapter 4, 5, or 6. The later the better, in our opinion.
95
Events and intervals only: Spacetim e lean and sp are
O r isolated events described using latticework
Lorentz transformation: Translate event description from lattice to lattice
■^■'jy.^'
w
>
.
5*.
96
SPECIAL TOPIC
LORENTZ TRANSFORMATION
L.2 FASTER THAN LIGHT? a reason to examine the Lorentx transformation No object travels faster than light. So YOU say, but watch ME: I travel in a rocket that you observe to move at 4/5 light speed. Out the front of my rocket I fire a bullet that I observe to fly forward at 4/5 light speed. Then you measure this bullet to streak forward at 4 /5 '\'4 /5 — 8/5 — 1.6 light speed, which is greater than the speed of light. There!
'
No!
Why not? Is it not true that 4/5 + 4/5 — 16?
Velocities do not odd
As a mathematical abstraction: always true. As a description of the world: only sometimes true! Example 1: Add 4 /5 liter of alcohol to 4 /5 liter of water. The result? Less than 8 /5 = 1 .6 liter of liquid! Why? Molecules of water interpenetrate molecules of alcohol to yield a combined volume less that the sum of the separate volumes. Example 2: Add the speed you measure for the bullet (4/5) to the speed I measure for your rocket (4/5). The result? The speed I measure for the bullet is 40/41 = 0.9756. This remains less than the speed of light. Why? And where did you get that number 4 0/4l for the bullet speed you measure? I got the number from the Lorentz transformation, the subject of this Special Topic. The Lorentz transformation embodies a central feature of relativity: Space and time separations typically do not have the same values as observed in different frames. Space an d time separations between w hat?
Between events. W hat events are we talking about here?
Event 1: You fire the bullet out the front of your rocket. Event 2: The bullet strikes a target ahead of you. W hat do these events have to do with speed? We are arguing about speed!
Events define velocities
Let the bullet hit the target four meters in front of you, as measured in your rocket. Then the space separation between event 1 and event 2 is 4 meters. Suppose the time of flight is 5 meters as measured by your clocks, the time separation between the two events. Then your bullet speed measurement is (4 meters of distance)/! 5 meters of time) = 4 /5 , as you said. A nd w hat do YOU measure for the space an d time separations in your laboratory fram e?
For that we need the L orentz co o rd in a te tran sfo rm atio n equations.
L.2
FASTER THAN LIGHT?
9 7
Phooey! I know how to reckon spacetime separations in different frames. We have been doing it for several chapters! From measurements in onefram e we figure the spacetime interval, which has the same value in a ll frames. End of story.
No, not the end of the story, but at least its beginning. True, the invariant interval has the same value as derived from measurements in every frame. That allows you to predict the time between firing and impact as measured by the passenger riding on the bullet — and measured directly by the bullet passenger alone.
Interval: Only a start in reckoning spacetime separations in different frames
Predict how?
You know your space separation x ' = 4 meters (primes for rocket measurements), and your time separation, t' = 'b meters. You know the space separation for the bullet rider, x" = 0 (double primes for bullet measurements), since she is present at both the firing and the impact. From this you can use invariance of rhe interval to determine the wristwatch time between these events for the bullet rider: i t ' y  { x " f = {t'Y  {x'Y or (/")^ — (0)^ = (5 meters)^ — (4 merers)^ — (3 meters)^ so that t” = 3 meters. This is the proper time, agreed on by all observers but measured directly only on the wristwatch of the bullet rider. Fine. C an’t we use the same procedure to determine the space an d time separations between these events in your laboratory frame, an d thus the bullet speed fo r you?
Unfortunately not. We do reckon the same value for the interval. Use unprimed symbols for laboratory measurements. Then f — xd = {?) meters)^. That, however, is not sufficient to determine x or t separately. Therefore we cannot yet find their ratio x /t, which determines the bullet’s speed in our frame.
Need more to compare velocities in different frames
So how can we reckon these x a n d t separations in your laboratory frame, thereby allowing us to predict the bullet speed you measure?
Use rhe Lorentz transformation. This transformation reports that our laboratory space separation between firing and impact is x = 4 0 /3 meters and the time separation is slightly greater: t = 4 1 /3 meters. Then bullet speed in my laboratory frame is predicted to be f = x j t — 40/41 = 0.9756. The results of our analysis in three reference frames are laid out in Table L1. Is the Lorentz transformation generally useful, beyond the specific task o f reckoning speeds as measured in different frames?
Oh yes! Generally, we insert into the Lorentz transformation the coordinates x ', t' of an event determined in the rocket frame. The Lorentz transformation then grinds and whirs, finally spitting out the coordinates x, t of the same event measured in the laboratory frame. Following are the Lorentz transformation equations. Here is the relative velocity between rocket and laboratory frames. For our convenience we lay rhe posirive xaxis along the direction of motion of the rocket as observed in the laboratory frame and choose a common reference event for the zero of time and space for both frames.
Compare velocities using Lorentz transformation
98
SPECIAL TOPIC
LORENTZ TRANSFORMATION C j A B L E L  T ^ 
HOW FAST THE BULLET? Bullet fired (coordinates of this event)
Bullet hits (coordinates of this event)
Speed of bullet (computed from frame coordinates)
Rocket frame (moves at = 4/5 as measured in laboratory)
x' = 0 l' = 0
x' = 4 meters t' = 5 meters
as measured in rocket frame: v' = 4/5 = 0.8
Bullet frame (moves at v' = 4/5 as measured in rocket)
x" = 0 t" = 0
x" = 0 t" = 3 meters (from invariance of the interval)
as measured in bullet frame: r" = 0
Laboratory frame
x= 0 r= 0
k'
X = 40/3 meters t = 41/3 meters (from Lorentz transformation)
as measured in laboratory frame: 1 = 40/41 = 0.9756
rel ‘
(1 x ' + t'
Lorentz transformation preview ed
(1 and Check for yourself that for the impact event of bullet with target (rocket coordi nates: x ' = A meters, /' = 5 meters; rocket speed in laboratory frame: = 4 /5 ) one obtains laboratory coordinates x = 4 0 /3 meters and t = 4 1 /3 meters. Hence v = x f t == 40/41 = 0.9756. You say the Lorentz transformation is general. I f it is so important, then why is this a special topic rather than a regular chapter?
Lorentz transformation: Useful but not fundamental
The Lorentz transformation is powerful; it brings the technical ability to transform coordinates from frame to frame. It helps us predict how to add velocities, as outlined here. It describes the Doppler shift for light (see the exercises for this chapter). On the other hand, the Lorentz transformation is not fundamental; it does not expose deep new features of spacetime. But no matter! Physics has to get on with the world’s work. One uses the method of describing separation best suited to the job at hand. On some occasions the useful fact to give about a luxury yacht is the 50meter distance between bow and stern, a distance independent of the direction in which the yacht is headed. On another occasion it may be much more important to know that the bow is 30 meters east of the stern and 40 meters north of it as observed by its captain, who uses NorthStar north. W hat does the Lorentz transformation rest on? On w hat foundations is it based?
Two foundations of Lorentz transformation
On two foundations: (1) The equations must be linear. That is, space and time coordinates enter the equations to the first power, not squared or cubed. This results from the requirement that you may choose any event as the zero of space and time.
L.3
FIRST STEPS
99
(2) The spacetime interval between two events must have the same value when computed from laboratory coordinate separations as when reckoned from rocket coordinate separations. All right, I 'll reservejudgment on the validity ofwhat you claim, hut show me the derivation itself. Read on!
L.3 FIRST STEPS invariance off the interval gets us started Recall that the coordinates y and z transverse to the direction of relative motion between rocket and laboratory have the same values in both frames (Section 3.6):
yy z = z'
(L1)
where primes denote rocket coordinates. A second step makes use of the difference in observed clock rates when the clock is at rest or in motion (Section 1.3 and Box 33). Think of a sparkplug at rest at the origin of a rocket frame that moves with speed relative to the laboratory. The sparkplug emits a spark at time t' as measured in the rocket frame. The sparkplug is at the rocket origin, so the spark occurs at x ' = 0. Where and when (x and t) does this spark occur in the laboratory? That depends on how fast, v^i, the rocket moves with respect to the laboratory. The spark must occur at the location of the sparkplug, whose position in the laboratory frame is given by
Derive difference in clock rates
X = V^it Now the invariance of the interval gives us a relation between t and t', {t'r  ( x y
=
{ t y  {oy = = f  x ^ = f  { v j f = t w  vij>
from which t' = t { \  ri,)V2 or [when x' = 0]
(L2)
The awkward expression 1/(1 — occurs often in what follows. For simplic ity, this expression is given the symbol Greek lowercase gamma: /. 1
7=
(1 
Because it gives the ratio of observed clock rates, y is sometimes called the tim e stre tc h facto r (Section 5.8). Strictly speaking, we should use the symbol /„i, since the value of y is determined by For simplicity, however, we omit the subscript in the hope that this will cause no confusion. With this substitution, equation (L2) becomes yf
[when x' = 0]
(L3)
Time stretch factor defined
lO O
SPECIAL TOPIC
LORENTZ TRANSFORMATION
Substitute this into the equation x = v ^ ^ t above to find laboratory position in terms of rocket measurements: [when x' = 0]
(L4)
Equations (L1), (L3), and (L4) give the first answer to the question, “If we know the space and time coordinates of an event in one freefloat frame, what are its space and time coordinates in some other overlapping freefloat frame?” These equations are limited, however, since they apply only to a particular situation: one in which both events occur at the same place {x' = 0) in the rocket,
L 4 FORM OF THE LORENTZ TRANSFORMATION any event can be reference event? then transformation is linear W hat general form does the Lorentz transformation have? It has the form that mathematicians call a lin ear tran sfo rm atio n . This means that laboratory coordi nates X and t are related to linear (first) power of rocket coordinates x ' and t ' by equations of the form Lorentz transformation: Linear equations
Arbitrary event as reference event? Then Lorentz transformation must be linear.
/ = fix' f D t' X = Gx' + Ht'
(L5)
where our task is to find expressions for the coefficients B, D, G, and H that do not depend on either the laboratory or the rocket coordinates of a particular event, though they do depend on the relative speed Why must these transfotmations be linear? Because we are free to choose any event as our reference event, the common origin x = y = z = / = 0 in all reference frames. Let our rocket sparkplug emit the flashes at = 1 and 2 and 3 meters. These are equally spaced in rocket time. According to equation (L3) these three events occur at laboratory times t = l y and 2y and 37 meters of time. These are equally spaced in laboratory time. Moving the reference event to the first of these events still leaves them equally spaced in time for both observers: t' = 0 and 1 and 2 meters in the rocket and t = 0 and ly and 2y in the laboratory. In contrast, suppose that equation (L3) were not linear, reading instead t = Kt'^, where K is some constant. Rocket times t' = \ and 2 and 3 meters result in laboratory times t = IK and 4K and 9K meters. These are not equally spaced in time for the laboratory observer. Moving the reference event to the fitst event would result in rocket times t' = 0 and 1 and 2 meters as before, but in this case laboratoty times t = 0 and 1K and 4K metets, with a completely different spacing. But the choice of reference event is arbitrary: Any event is as qualified to be reference event as any other. A clock that runs steadily as observed in one frame must run steadily in the other, independent of the choice of reference event. We conclude that the relation between t and t' must be a linear one. A similar argument requires that events equally separated in space in the rocket must also be equally separated in space as measured in the laboratory. Hence the Lorentz transformation must be linear in both space and time coordinates.
L.5
COMPLETING THE DERIVATION
lO l
L5 COMPLETING THE DERIVATION invariance off the interval completes the story Equations (L3) and (L4) provide coefficients D and H called for in equation (L5): t — Bx' + yt' X = Gx' + v^yyt'
(L6)
About the two constants B and G we know nothing, for an elementary reason. All events so far considered occured at point x' = 0 'm the rocket. Therefore the two coefficients B and G could have any finite values whatever without affecting the numerical results of the calculation. To determine B and G we turn our attention from an x' = 0 event to a more general event, one that occurs at a point with arbitrary rocket coordinates x ' and t ' . Then we demand that the spacetime interval have the same numerical value in laboratory and rocket frames for any event whatever;
Demanding invariance of interval . . .
Substitute expressions for t and x from equation (L6): (fix' + y t'Y { G x ' + p^^yt'y = r'2  x'^ On the left side, multiply out the squares. This leads to the rather cumbersome result B2
b 2B yx't' + y V ^  G^x'^  2Gv^{yx't'  vlyyh'^ =
Group together coefficients of x 't' to obtain y \\ 
coefficients of x'^, and coefficients of the crossterm
+ 2y{B  v jG ) x 't'  {G^  B?)x'^ = t'^  x'^
a7)
Now, t' and x ' can each take on any value whatsoever, since they tepresent the coordinates of an arbitrary event. Under these circumstances, it is impossible to satisfy equation (L7) with a single choice of values of B and Gunless they are chosen in a very special way. The quantities B and G must first be such as to make the coefficient of x 't' on the left side of equation (L7) vanish as it does on the tight:
. . . between any pair of events w hatsoever . . .
2y{B  VJG) = 0 But 7 can never equal zero. The value of 7 = 1/(1 ~ ^ ) *^^ equals unity when =0 and is greatet than this fot any othet values of Hence the left side of this equation can be zero only if (B 
=
or
B — v„,G
Second, B and G must be such as to make the coefficient of x right of equation (L7); hence
(L8)
equal on the left and
 B2 = 1
(L9)
Substitute B from equation (L8) into equation (L9): G^  { v ^ fiY = 1
ot
G K \v l^ )= \
. . . lead s to completed form of Lorentz transformation.
102
SPECIAL TOPIC
LORENTZ TRANSFORMATION
Divide through by (1 —
and take the square root of both sides:
1
But the right side is just the definition of the time stretch factor y, so that G=y Substitute this into equation (L8) to find B: B = v ^y
The Lorentz transformation
These results plus equations (L1) and (L6) yield the Lorentz transformation equa tions: t= + yt' x = yx' + v„iyt' (LlOa) y = :/ z^ — z or, substituting for the value of gamma, y = 1/(1 —
x' + Vg,/
yy
and
(L1 Ob)
z —z
In summary, the Lorentz transformation equations rest fundamentally on the re quired linearity of the transformation and on the invariance of the spacetime interval. Invariance of the interval was used twice in the derivation. First, we examined a pair of events both of which occur at the same fixed location in the rocket, so that rocket time between these events— proper time, wristwatch time— equals the spacetime interval between them (Section L.3). Second, we demanded that the interval also be invariant between every possible event and the reference event (the present section).
L.6 INVERSE LORENTZ TRANSFORMATION from laboratory event coordinates, reckon rocket coordinates Equations (L10) provide laboratory coordinates of an event when one knows the rocket coordinates of the same event. But suppose that one already knows the laboratory coordinates of the event and wishes to predict the coordinates of the event measured by the rocket observer. W hat equations should be used for this purpose? An algebraic manipulation of equations (L10) provides the answer. The first two of these equations can be thought of as two equations in the two unknowns x ' and t ' . Solve for these unknowns in terms of the nowknowns x and t. To do this, multiply both sides of the second equation by and subtract corresponding sides of the
L.7
ADDITION OF VELOCITIES
103
resulting second equation from the first. Terms in x cancel to yield
t
y ■ t' vf' = = yt'  vuyt' = nydi K O f == — 3y2 t' = y
Long derivation of inverse Lorentz transformation
2
Here we have useci the definition written
— 1/(1 —
The equation for t' can then be
t' =  ^ r ^ y x + yt A similar procedure leads to the equation for x '. Multiply the first of equations (L10) by i'rel and subtract corresponding sides of the first equation from the second — try it! The y and z components are respectively equal in both frames, as before. Then the inverse L orentz tran sfo rm atio n eq u atio n s become t ' =  v ,,^ y x  V y t /  y x  v^^yt
(L lla )
y =y r_
Z — Z
Inverse Lorentz transformation
Or, substituting again for gamma, y = 1/(1 —
(1 X •
^el t
(Lllb )
and Equations (L11) transform coordinates of an event known in the laboratory frame to coordinates in the rocket frame. A simple but powerful argument from symmetry leads to the same result. The symmetry argument is based on the relative velocity between laboratory and rocket frames. W ith respect to the laboratory, the rocket by convention moves with known speed in the positive xdirection. W ith respect to the rocket, the laboratory moves with the same speed but in the opposite direction, the negative xdirection. This convention about positive and negative directions — not a law of physics! — is the only difference between laboratory and rocket frames that can be observed from either frame. Lorentz transformation equations must reflect this single difference. In consequence, the “inverse” (laboratorytorocket) transformation can be obtained from the “ direct” (rockettolaboratory) transformation by changing the sign of relative velocity, v ^ , in the equations and interchanging laboratory and rocket labels (primed and unprimed coordinates). Carrying out this operation on the Lorentz transformation equations (L10) yields the inverse transformation equations (L11).
L.7 ADDITION OF VELOCITIES add light velocity to light velocity: get light velocity! The Lorentz transformation permits us to answer decisively the apparent contradiction to special relativity outlined in Section L.2, namely the apparent addition of velocities to yield a resultant velocity greater than that of light.
Short derivation of inverse Lorentz transformation
104
SPECIAL TOPIC
Return to velocity addition paradox
LORENTZ TRANSFORMATION
I travel in a rocket that you observe to move at 4(5 light speed. Out the front of my rocket I fire a bullet that I observe to fly forward at 4fb light speed. Then you measure this bullet to streak forward at4l3'k4l5 = 8j5 = 1.6 light speed, which is greater than the speed of light. There!
SAMP L E PRO B L E M L  1
T R A N S F O R M I N G O V E R AND B A C K A rocket moves with speed = 0.866 (so y = 2) along the xdirection in the laboratory. In the rocket frame an event occurs at coordinates x' =
10 meters, y = 1 meters, z = i meters, and t' = 20 meters of lighttravel time with respect to the reference event.
a.
W hat are the coordinates of the event as observed in the laboratory?
b.
Transform the laboratory coordinates back to the rocket frame to verify that the resulting coordinates are those given above.
SOLUTION a.
We already know from Section 3.6 — as well as from the Lorentz transformation, equation (L10) — that coordinates transverse to direction of relative motion are equal in laboratory and in rocket. Therefore we know immediately that y — y' — 1 meters z — z' — i meters The X and t coordinates of the event as observed in the laboratory make use of the first two equations (L10): t
= v^^ifx' + yt' = (0 .866)(2)(10 meters) + (2)(20 meters) = 17.32 + 40 = 57.32 merers
and X — yx' h v^^fyt' = 2(10 meters) + (0.866)(2)(20 meters) = 20 + 34.64 = 54.64 merers So rhe coordinates of the event in the laboratory are t = 57.32 meters, x = 54.64 meters, y = l meters, and z = 3 meters. b.
Use equarion (L11) ro rransform back from laboratory to rocket coordinates. t' = ~ v ^{y x + yt = — (0.866)(2)(54.64 meters) + (2)(57.32 meters) = —94.64 b 114.64 = 20.00 meters and X = y x — v ^ y t = 2(54.64 merers) — (0.866)(2)(57.32 meters) = 109.28  99.28 = 10.00 meters as given in rhe original statement of the problem.
L.7
ADDITION OF VELOCITIES
105
To analyze this experiment, convert statements about the bullet to statements about events, since event coordinates are what the Lorentz transformation transforms. Event 1 is the firing of the gun, event 2 the arrival of the bullet at the target. The Lorentz transformation equations can give locations x,, and X2, ?2 of these events in the laboratory frame from their known locations x \ , t \ and x 2 , t' 2 in the rocket frame. In particular: X2 = yx2 + v^C/t' 2 Xi = y x / + Subtract corresponding sides of these two equations: (X2 — xi) = y (x
2
— x 'l) + v,^{y{t'2 — z'l)
We are inrerested in the differences between the coordinates of the two emissions. Indicate these differences with the Greek uppercase delta. A, for example Ax. Then this xequation and the corresponding /equation become A x = y A x ' + /^ „ iy A / A / = t^reiyAx' + y A /
(L12)
Incremental event separations define velocities
The subscript “tel” distinguishes relative speed between laboratory and rocket frames from other speeds, such as particle speeds in one frame or the other. Bullet speed in any frame is simply space sepatation between two events on its trajectory measured in that frame divided by time between them, observed in the same frame. In the special case chosen, only the xcoordinate needs to be considered, since the bullet moves along the direction of relative motion. Divide the two sides of the first equation (L12) by the corresponding sides of the second equation to obtain labora tory speed:
v^{yiS.t'
Ax
yA x^ +
A/
t „ ,y A x ' + y A /
Then the time stretch factor y cancels from the numerator and denominator on the right. Divide every term in numerator and denominator on the right by A /'. Ax _ At
( A x '/A /) + v ^^ fA x '/A t') + 1
Now, A x '/ A t' is just distance covered per unit time by the patticle as observed in the rocket, its speed — call it v , with a prime. And A x / A t is particle speed in the laboratory — call it simply v. Then (reversing order of terms in the denominatot to give the result its usual form) the equation becomes v' + v„ 1+
V v„
(113)
This is called the Law o f A d d itio n o f V elocities in one dimension. A better name is the Law o f C o m b ination o f V elocities, since velocities do not “add” in the usual sense. Using the Law of Combination of Velocities, we can predia bullet speed in the laboratory. The bullet travels at v' — 4 /5 with respect to the rocket and the rocket moves at v^^ = 4 /5 with tespect to the laboratory. Therefore, speed v of the bullet (continued on page 1 10)
Law of Addition of Velocities
106
SPECIAL TOPIC
LORENTZ TRANSFORMATION
S A M P L E P R O B L E M L2
“ ET TU, S P A C E T I M E I ” Julius Caesar was murdered on March 15 in the year 44 B.c. at the age of 55 approximately 2000 years ago. Is there some way we can use the laws of relativity to save his life? Let Caesar’s death be the reference event, la beled 0: = 0, C ~ 0. Event A is you reading this exercise. In the Earth frame the coordinates of event A are x^ = 0 lightyears, = 2000 years. Simultaneous with event A in your frame, Starship Enterprise cruising the Andromeda galaxy sets off
a firecracker: event B. The Enterprise moves along a straight line in space that connects it with Earth. Andtomeda is 2 million lightyears distant in our frame. Compared with this distance, you can ne glect the orbit of Earth around Sun. Therefore, in our frame, event B has the coordinates Xg = 2 X 10^ lightyears, tg — 2000 years. Take Caesar’s murder to be the reference event for the Enterprise too (x / = 0 , r / = 0).
a.
How fast must the Enterprise be going in the Earth frame in order that Caesar’s murder is happening N O W (that is, = 0) in the Enterprise rest frame? Under these circumstances is the Enterprise moving toward or away from Earth?
b.
If you are acquainted with the spacetime diagram (Chapter 5), draw a spacetime diagram for the Earth frame that displays event 0 (Caesar’s death), event A (you reading this exercise), event B (firectacker exploding in Andromeda), your line of N O W simultaneity, the position of the Enterprise, the worldline of the Enter prise, and the Enterprise N O W line of simultaneity. The spacetime diagtam need not be drawn to scale.
c.
In the Enterprise frame, what are the x and / coordinates of the firecracker explosion?
d.
Can the Enterprise firecracker explosion warn Caesar, thus changing the course of Earth history? Justify your answer.
SOLUTION a.
From the statement of the problem. Xo — x / — 0 L = l/ = 0
=
2000 years
Xg — 2 X 10^ lightyears /g = 2000 years
W e want the speed of the Entetprise such that tg' — 0. The first two Lorentz transformation equations (L10) with tg' = 0 become ^'rel y^B Xg
yX g
W e do not yet know the value of X g '. Solve for by dividing the two sides of the first equation by the respective sides of the second equation. The unknown Xg' drops out (along with y), and we are left with in terms of the known quantities tg and x„: 'B _ 2 X 10^ years —  ^  = 103 = 0.001 X 1 0^ vpar7 transformation of yvelocity A particle moves with uniform speed v'^ = ts.y'/ lS.t' along the;/'axis of the rocket frame. Transform lS.y' and A / to laboratory displacements A x , A y , and A / using the Lorentz transformation equations. Show that the xcomponent and the ycomponent of the velocity of this particle in the laboratory frame are given by the expressions ‘'rel
THE TILTED METER STICK
115
b Show that your answer to Exercise L8 gives the same result when the velocity v ' is given the value unity. c A particle at rest in the rocket ftame emits light uniformly in all directions. Consider the 50 percent of this light that goes into the forward hemisphere in the rocket frame. Show that in the labotatory frame this light is concentrated in a narrow forward cone of halfangle (f)g whose axis lies along the direction of motion of the particle. The halfangle (j)„ is the solu tion to the following equation: cos (/)„ =
< (1
This result is called the h ea d lig h t effect.
L8 transformation of velocity direction A particle moves with velocity v' in the x 'y ' plane of the rocket frame in a direction that makes an angle (f)' with the x'axis. Find the angle (f) that the velocity vector of this particle makes with the xaxis of the laboratory frame. (Hint: Transform space and time displacements rather than velocities.) Why does this angle differ from that found in Exercise L6 on trans formation of angles? Contrast the two results when the relative velocity between the rocket and labora tory frames is very great.
L9 the headlight effect A flash of light is emitted at an angle (f)' with respect to the x'axis of the rocket frame. a Show that the angle (f) the direction of motion of this flash makes with respect to the xaxis of the laboratory frame is given by the equation cos (/) =
cos (/)' + 1+
rel
(f)'
L10 the tilted meter stick N ote: This exercise uses the results of Exercise L7. A meter stick lying parallel to the xaxis moves in the ydirection in the laboratory frame with speed as shown in the figure (left). a
In the rocket fram e the stick is tilted upw ard in
the positive x'direction as shown in the figure (right). Explain why this is, first without using equa tions. b Let the center of the meter stick pass the point X = y = x ' = y ' = 0 at time t = t ' = 0. Calculate the angle ' at which the meter stick is inclined to the x'axis as observed in the rocket frame. D iscussion: Where and when does the right end of the meter stick cross the xaxis as observed in the laboratory frame? Where and when does this event of rightend crossing occur as measured in the rocket frame? W hat is the direction and magnitude of the velocity of the meter stick in the rocket frame (Exercise L7)? Therefore where is the right end of the meter stick at / ' = 0 , when the center is at the origin? Therefore . . .
EXERCISE L10. Left: Meter stick moving transverse to its length as observed in the laboratory frame. Right: Meter stick as observed in rocket frame.
116
EXERCISE L11
THE RISING MANHOLE
L11 the rising manhole
skateboard to fall through the holes in the grid. Y et to
the holes in the grid are much narrower than to the stationary man, and she certainly does not expect her skateboard to fall through them. Which person is correct? The answer hinges on the relativity of rigidity. Idealize the problem as a onemeter rod sliding lengthwise over a flat table. In its path is a hole one meter wide. If the Lorentz contraction factor is ten, then in the table (laboratory) frame the rod is 10 centimeters long and will easily drop into the onemeterwide hole. Assume that in the laboratory frame the meter stick moves fast enough so that it remains essentially horizontal as it descends into the hole (no “ tipping’’ in the laboratory frame). Write an equa tion in the laboratory frame for the motion of the bottom edge of the meter stick assuming that t = / ' = 0 at the instant that the back end of the meter stick leaves the edge of the hole. Eor small vertical velocities the rod will fall with the usual acceleration g. Note that in the laboratory frame we have assumed that every point along the length of the meter stick begins to fall simultaneously. In the meter stick (rocket) frame the rod is one meter long whereas the hole is Lorentzcontracted to a 10centimeter width so that the rod cannot possibly fit into the hole. Moreover, in the rocket frame differ ent parts along the length of the meter stick begin to drop ar different times, due to the relativity of simul taneity. Transform rhe laboratory equations into the rocket frame. Show that the front and back of the rod will begin to descend at different times in this frame. The rod will “droop” over the edge of the hole in the rocket frame — that is, it will not be rigid. Will the rod ultimately descend into the hole in both frames? Is the rod really rigid or nonrigid during the experiment? Is it possible to derive any physical characteristics of the rod (for example its flexibility or compressibility) from the description of its motion provided by rela tivity?
the fast girl her skateboard has its usual length and it is the grid that has the relativistic contraction. To her
Reference: W . Rindler, American Journal o f Physics, Volume 29, page 3 6 5  3 6 6 (1961).
N ote: This exercise uses the results of Exercise L10. A meter stick lies along the xaxis of the laboratory frame and approaches the origin with velocity .A very thin plate parallel to the xz laboratory plane moves upward in the ydirection with speed Vy as shown in the figure. The plate has a circular hole with a diameter of one meter centered on the yaxis. The center of the meter stick arrives at the laboratory origin at the same time in the laboratory frame as the rising plate arrives at the plane y = 0. Since the meter stick is Lorentzcontracted in the laboratory frame it will easily pass through the hole in the rising plate. Therefore there will be no collision between meter stick and plate as each continues its motion. However, someone who objects to this conclusion can make the following argument: “In the rocket frame in which the meter stick is at rest the meter stick is not con tracted, while in this frame the hole in the plate is Lorentzcontracted. Hence the fulllength meter stick cannot possibly pass through the contracted hole in the plate. Therefore there must be a collision between the meter stick and the plate.’’ Resolve this paradox using your answer to Exercise L10. Answer unequiv ocally the question, Will there be a collision between the meter stick and the plate? Reference: R. Shaw, Am erican Jo u rn a l o f Physics, Volume 30, page 72 (1962).
L12 paradox of the skateboard and the grid A girl on a skateboard moves very fast, so fast that the relativistic length contraction makes the skateboard very short. On the sidewalk she has to pass over a grid. A man standing at the grid fully expects the fast short
EXERCISE L 1 1 . W ill the “m eter s tic k ” pass through the “onem eter~diam eter" hole w ith out collision?
EXERCISE L13
PARADOX OF THE IDENTICALLY ACCELERATED TWINS
L13 paradox off the identically accelerated twins N ote: This exercise uses spacetime diagrams, intro duced in Chapter 5. Two fraternal twins, Dick and Jane, own identical spaceships each containing the same amount of fuel. Jane’s ship is initially positioned a distance to the right of Dick’s in the Earth frame. On their twentieth birthday they blast off at the same instant in the Earth frame and undergo identical accelerations to the right as measured by Mom and Dad, who remain at home on Earth. Mom and Dad further observe that the twins run out of fuel at the same time and move thereafter at the same speed v. Mom and Dad also measure the distance between Dick and Jane to be the same at the end of the trip as at the beginning. Dick and Jane compare the ships’ logs of their accelerations and find the entries to be identical. However when both have ceased accelerating, Dick and Jane, in their new rest frame, discover that Jane is older than Dick! How can this be, since they have an identical history of accelerations? a Analyze a simpler trip, in which each spaceship increases speed not continuously but by impulses, as shown in the first spacetime diagram and the event table. How far apart are Dick and Jane at the begin ning of their trip, as observed in the Earth frame? How far apart are they at the end of their accelera tions? W hat is the final speed v (not the average speed) of the two spaceships? How much does each astronaut age along the worldline shown in the dia gram? (The answer is not the Earth time of 12 years.) b The second spacetime diagram shows the two worldlines as recorded in a rocket frame moving with the final velocity of the two astronauts. Copy the figure. On your copy extend the worldlines of Dick and Jane after each has ceased accelerating. Label your figure to show that Jane ceased accelerating before Dick as observed in this frame. Will Dick age the same between events 0 and 3 in this frame as he aged in the Earth frame? Will Jane age the same between events 4 and 7 in this frame as she aged in the Earth frame?
117
(3) W hat is Jane’s age at event 7? (4) W hat is Jane’s age at the same time (in this
frame) as event 3? (5) W hat are the ages of Dick and Jane 20 years after event 3, assuming that neither moves again with respect to this frame? (
6 How far apart in space are Dick and Jane when )
both have ceased accelerating? (7)
Compare this separation with their initial (and final!) separation measured by Mom and Dad in the Earth frame.
d Extend your results to the general case in which Mom and Dad on Earth observe a period of identical continuous accelerations of the two twins. (1) At the two startacceleration events (the two events at which the twins start their rockets), the twins are the same age as observed in the Earth frame. Are rhey the same age at these events as observed in every rocket frame? (2) At the two ceaseacceleration events (the two events at which the rockets run out of fuel), are the twins the same age as observed in the Earth frame? Are they the same age at these events as observed in every rocket frame? (3) The two ceaseacceleration events are simulta
neous in the Earth frame. Are they simulta neous as observed in every rocket frame? (No!) Whose ceaseacceleration event occurs first as observed in the final frame in which both twins come to rest? (Recall the Train Paradox, Sec tion 3.4.) (4)
“ When Dick ceases accelerating, Jane is older than Dick.” Is this statement true according to the astronauts in their final rest frame? Is the statement true according to Mom and Dad in the Earth frame?
(5) Criticize the lack of clarity (swindle?) of the word when in the statement of the problem: ‘‘However when both have ceased accelerat ing, Dick and Jane, in their new rest frame, discover that Jane is older than Dick!”
c Now use the Lorentz transformation to find the space and time coordinates of one or two critical events in this final rest frame of the twins in order to answer the following questions
e Suppose that Dick and J ane both accelerate to the left, so that Dick is in front of Jane, but their history is otherwise the same. Describe the outcome of this trip and compare it with the outcome of the original trip.
(1) How many years earlier than Dick did Jane cease accelerating?
f Suppose that Dick and J ane both accelerate in a direction perpendicular to the direction of their separation. Describe the outcome of this trip and compare it with the outcome of the original trip.
(2) W hat is Dick’s age at event 3? (not the rocket time t ' oi this event!)
118
EXERCISE L13
PARADOX OF THE IDENTICALLY ACCELERATED TWINS
Earth Frame Observations Event number 0
xposition (light years) 0
Time (years) 0 4
1
1
2
3
8
3
6
12
4
12
5
13
4
6
15
8
7
18
12
ROCKET FRAME EXERCISE L13. Top: Worldlines of Dick and Jane as observed in the Earth frame of Mom and Dad. Bottom: Worldlines of Dick and Jane as observed in the “final" rocketframe in which both Dick and Jane come to rest after burnout.
D iscussion: Einstein postulated that physics in a uniform gravitational field is, locally and for small particle speeds, the same as physics in an accelerated frame of reference. In this exercise we have found that two accelerated clocks separated along the direction of acceleration do not remain in synchronism as observed simultaneously in their common frame. Rather, the forward clock reads a later time (“runs faster”) than the rearward clock as so observed. Conclusion from Einstein’s postulate: Two clocks one above the other
in a uniform gravitational field do not remain in synchronism; rather the higher clock reads a later time (“runs faster”) than the lower clock. General relativ ity also predicts this result, and experiment verifies it. (Read about the patrol plane experiment in Section 4.10.) Reference: S. P. Boughn, American Journal o f Physics, Volume 57, pages 7 9 1  7 9 3 (September 1989). Reference to general relativity result: Wolfgang Rindler, Essential Relativity (Springer, New York, 1977), pages 17 and 117.
EXERCISE L14
L14
HOW DO RODS LORENTZCONTRACT?
how do rods Lorenlzcontracl?
N ote: Calculus is used in the solution to this exercise; so is the formula for Lorentz contracrion from Section 5,8. Laboratory observers measure rhe length of a mov ing rod lying along its direction of motion in the laboratory frame. Then the rod speeds up a little. Again laboratory observers measure its length, which they find to be a little shorter than before. They call this shortening of length Lorentz contraction. How did this shortening of length come about.^ As happens so often in relativity, the answer lies in the relativity of simultaneity. First, how much shortening takes place when the rod changes from speed v to speed v + dv} Let be the proper length of the rod when measured at rest. At speed V its laboratorymeasured length L will be shorter than this by the Lorentz contraction factor (Section 5.8): L = (l
dL = 
contraction of a fastmoving rod of proper length L„. More: We want a careful inspecror riding on the fastmoving rod to certify that it has the same proper length L(, as it did when it was at rest in the laboratory frame. To achieve rhis goal, the inspector insists that the pair of accelerating taps be applied to the front and back rod pieces at the same time in the current rest frame of the rod. Otherwise the distance between these pieces would not remain rhe same in the frame of the rod; the rod would change proper length. [Notice that in Exercise L13 the taps occur at the same time in the laboratory (Earth) frame. This leads to results differ ent from those of the present exercise.} b You are the inspector riding along with the front and back pieces of the rod. Consider the two events of tapping the front and back pieces. How far apart A x' are these events along the xaxis in your (rocket) frame? How far apart A /' in time are these events in your frame? Predict how far apart in time Ar these events are as measured in the laboratory frame. Use the Lorentz transformation equation (L10): b it = V yAx'
a Using calculus, show that when the rod speeds up from y to a slightly greater speed v + dv, the change in length dL is given by the expression L^vdv (1  t^2)l/2
The negative sign means that the change is a shorten ing of the rod. We want to explain this change in length. How is the rod to be accelerated from v to v dv^. Fire a rocket attached to the rear of the rod? No, Why not? Because the rocket pushes only against the rear of the rod; this push is transmitted along the rod to the front at the speed of a compression wave — very slow! We want the front and back to change speed “at the same time” (exact meaning of this phrase to be deter mined later). How can this be done? Only by prearrangement! Saw the rod into a thousand equal pieces and tap each piece in the forward direction with a mallet “at exactly 12 noon” as read off a set of synchronized clocks. To simplify things for now, set aside all but the front and back pieces of the rod. Now tap the front and back pieces ‘‘at the same time. ’’ The change in length of the rod dL is then the change in distance between these two pieces as a result of the tapping. So much for how ro accelerare the “rod.” Now the central question: W hat does it mean to tap the front and back pieces of the rod “at the same time”? To answer rhis question, ask another: W hat is our final goal? Answer: To account for the Lorentz
119
+
y l\t'
The relative velocity in equation (L10) is just v, the current speed of the rod. In the laboratory frame is the tap on the rear piece earlier or later than the tap on the front piece? Your answer to part b predicts how much earlier the laboratory observer measures the tap to occur on the back piece than on the front piece of the rod. Let the tap increase the speed of the back end by dv as measured in the laboratory frame. Then during labo ratory time Ar the back end is moving at a speed dv faster than the front end. This relative motion will shorten the distance between the back and front ends. After time interval A t the front end receives the iden tical tap, also speeds up by dv, and once again moves at the same speed as the back end. C Show that the shortening dL predicted by this analysis is dL = ~dvlS.t — —ybsx'vdv = —vjL^dv L jjdv (1  r ; 2)i/2 which is identical to the result of part a, which we wanted to explain. QED. d Now start with the front and back pieces of the rod at rest in the laboratory frame and a distance L^ apart. Tap them repeatedly and identically. As they speed up, be sure these taps take place simultaneously in the rocket frame in which the two ends are currently at rest. (This requires you, the ridealong inspector, to
120
EXERCISE L15
THE PLACE WHERE BOTH AGREE
resynchronize your rodrestframe clocks after each set of frontandback taps.) Make a logically rigorous argument that after many taps, when the rod is mov ing at high speed relative to the laboratory, the length of the rod measured in the laboratory can be reckoned using the first equation given in this exercise. e Now, by stages, put the rod back together. The full thousand pieces of the rod, lined up but not touching, are all tapped identically and at the same time in the current rest frame of the rod. One set of taps increases the rod’s speed from p to p h dp in the laboratory frame. Describe the time sequence of these thousand taps as observed in the laboratory frame. If you have studied Chapter 6 or the equivalent, answer the following questions: W hat kind of interval — timelike, lightlike, or spacelike— separates any pair of the thousand taps in this set? Can this pair of taps be connected by a light flash? by a compression wave moving along the rod when the pieces are glued back together? Regarding the “logic of acceleration,” is there any reason why we should not glue these pieces back together? Done! f During the acceleration process is the reglued rod rigid— unchanging in dimensions — as observed in the rod frame? As observed in the laboratory frame? Is the rigidity property of an object an invariant, the same for all observers in uniform relative motion? Show how an ideal rigid rod could be used to transmit signals instantaneously from one place to another. W hat do you conclude about the idea of a “rigid body” when applied to highspeed phenomena? Reference: Edwin F. Taylor and A. P. French, American Journal of Physics, Volume 51, pages 889893, especially the Appendix
(1983).
L15 the place where both agree At any instant there is just one plane in which both the laboratory and the rocket clocks agree. a By a symmetry argument, show that this plane lies perpendicular to the direction of relative motion. Using the Lorentz transformation equations, show that the velocity of this plane in the laboratory frame is equal to
the binomial expansion (1 + z)” ~ 1 + «z for z «
to show that for low relative velocity, p,=,' p^ / 2 . c W hat isp,=,’ for the extreme relativistic case in which f'rei 1? Show that in this case is com pletely different from ^rel/2. d Suppose we want to go from the laboratory frame to the rocket frame in two equal velocity jumps. Try a first jump to the plane of equal laboratory and rocket times. Now symmetry does work: Viewed from this plane the laboratory and rocket frames move apart with equal and opposite velocities, whose magnitude is given by the equation in part a. A second and equal velocity jump should then carry us to the rocket frame at speed with respect to the laboratory. Verify this directly by using the Law of Addition of Velocities (Section L.7) to show that P ,= ,' + P ,= ,' P r .1 —
L16
—
[1

(1

b Does the expression for p,=,> seem strange? From our everyday experience we might expect that by symmetry the “plane of equal time” would move in the laboratory at half the speed of the rocket. Verify that indeed this is correct for the low relative velocities of our everyday experience. Use the first two terms of
1 + P ,= ,'P ,= ,'
Fizeau experiment
Light moves more slowly through a transparent ma terial medium than through a vacuum. Let t'medium represent the reduced speed of light measured in the frame of the medium. Idealize to a case in which this reduced velocity is independent of the wavelength of the light. Place the medium at rest in a rocket moving at velocity p„^, to the right relative to the laboratory frame, and let light travel through the medium, also to the right. Use the Law of Addition of Velocities (Section L.7) to find an expression for the velocity p of the light in the laboratory frame. Use the first two terms of the binomial expansion (1 f z)” ~ 1 b «z for z «
1
to show that for small relative velocity between the rocket and laboratory frames, the velocity p of the light with respect to the laboratory frame is given approximately by the expression P
=
1
^m edium ^
^ rc lf f
^m edium )
This expression has been tested by Fizeau using water flowing in opposite directions in the two arms of an interferometer similar (but not identical) to the interferometer used later by Michelson and Morley (Exercise 312). Reference: H. Fizeau, Comptes rendus, Volume 33, pages 349355 (1851). A fascinating discussion (in French) of some central themes in relativity theory— delivered more than fifty years before Einstein’s first relativity paper.
‘k/
4.1 INVITATION TO CANOPUS is one lifetime enough? Approximately ninetynine lightyears from Earth lies the star Canopus. The Space Agency asks us to visit it, photograph it, and return home with our records. “But that’s impossible, ”we object. “W e have only a little over forty more years to live. We can spare at most twenty years for the outward trip, and twenty years for the return trip. Even if we could travel at the speed of light, we would need ninetynine years merely to get there.’’ We are greeted with a smile and a cheery, “Think about our request a little longer, won’t you?”
4.2 STRIPPEDDOWN FREEFLOAT FRAME throw away most clocks and rods Troubled thoughts fill us tonight. We dream about invariance of the spacetime interval (Chapter 3). In our dream we find ourselves aboard the rocket used to establish that result (Section 3.7). However, the numbers somehow have changed from meters of distance and meters of lighttravel time to lightyears of distance and years of time. Suddenly we see things in a new perspective. Three revelations crowd in on us.
121
1 22
Retain a single string of Earthlinked clocks
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TRIP TO CANOPUS
The flash of light that got reflected did its work — revelation number one— in establishing the identity of the spacetime interval as measured in either of the two frames. We can remember invariance of the interval and forget about the reflected flash. Eliminating it, we eliminate mirror, photodetector and, most of all, those upwardextended arrays of printout clocks in rocket and laboratory frames whose only purpose was ro track the light flash. The economy goes further. For us aboard the rocket, one reliable calendar clock is enough. As we start our trip from Earth in our dream, that clock by a happy coincidence shows noon on the Fourth of July, 2000 A.D. — and so do clocks at the Space Agency Center on Earth. We celebrate our starr by setting off a firecracker. Later by 6 years — for us — and with a long shipboard program of research and study already completed, our rocket clock — still in our dreams — tells us it is again noon on the Fourth of July and we set off a second firecracker. At that very instant, thanks to the particular speed we had chosen for our rocket relative to Earth, we are passing Lookout Station Number 8. Lonely lighthouse, it has in it little more than a sentry person and a printout clock, one of a series that we have been passing on our trip. They have been stationed out in space, fixed one lightyear apart according to Earth measurements. Each clock is calibrated and synchronized to the reference clock on Earth using a reference flash as described in Section 2.6. The laboratory latticework of Figure 26 has been reduced to a single rightwardstretching string of lookout stations and their clocks. That we can thus simplify our vision of what is going on from three space dimensions to one is our first revelation, vtsr
4.3 FASTER THAN LIGHT? choose your frame, then measure velocity!
Speed ; M easure distance and time in sam e frame
Revelation number two strikes us as— still dreaming — we pass Lookout Station Number 8, 8 lightyears from Earth: W hat speed! W e glance out of our window and see the lookout station clock print out “Fourth of July 2010 A.D.” — 10 years later than the Earth date of our departure. Our rocket clock reads 6 years. We are not shocked by the discrepancy in times for, apart from the change in scale from meters of lighttravel time to years, the numbers are numbers we have seen before. Nor are we astonished at the identity of the spacetime interval as evaluated in the two very different frames. W hat amazes us is our speed. Have we actually covered a distance of 8 lightyears from Earth in a time of 6 years? Can this mean we have traveled faster than light? We have often been told that no one and no object can go faster than light. Yet here we are — in our dream — doing exactly that. Speed, yes, we suddenly say to ourselves, but speed in which frame? Ha! W hat inconsistency! We took the distance covered, 8 lightyears, in the Earthlinked laboratory frame, but the time to cover it, 6 years, in the rocket frame! At this point we recognize that we can talk about our speed in one reference frame or our speed in the other frame, but we get nonsense when we mix together numbers from two distinct reference frames. So we reform. First we pick for reference frame the rocket. But then we get nothing very interesting, because we did not go anywhere with respect to the rocket— we just stayed inside. ' our speed \ relative to I , rocket frame /
distance we cover \
( with respect to rocket/ time we take to cover\ ( it in rocket frame /
(0 lightyears) (6 years)
=
4.4
ALL OF SPACE IS OURS!
123
In contrast, our speed relative to the Earthlinked reference frame, the extended laboratory, equals
' our speed \ relative to 1 , Earth fram e/
/ distance we cover \ \w ith respea to Earth/ / time we take to cover\ \ it in Earth frame /
(8 lightyears) (10 years)
— 0.8 lightspeed
In other words we— and the rocket— travel, relative to Earth, at 80 percent of the maximum possible speed, the speed of light. Revelation number two is our discovery that speed in the abstract makes no sense, that speed has meaning only when referred to a clearly stated frame of reference. Relative to such a frame we can approach arbitrarily close to light speed but never reach it.
4.4 ALL OF SPACE IS OURS! in one lifetime: go anywhere in the cosmos Revelation number three strikes us as — dreaming on — we think more about passing Earthlinked lookout stations. Moving at 80 percent of light speed, we travel 8 lightyears in the Earthlinked frame in 6 years of our rocket time. Continuing at the same rate will get us to Canopus in 74 years of our rocket time. Better than 99 years, but not good enough. Let’s use — in imagination — a faster rocket! We suddenly remember the super rocket discussed in demonstrating the invariance of the spacetime interval (Section 3.8). Converting meters of distance and time to years, we realize that traveling in the superrocket would bring us to Earthlinked Lookout Station Number 20, 20 Earthframe lightyears from Earth, in 6 years of our rocket time. When passing this station, we can see that this station clock reads 20.88 years. Therefore in the Earthlinked frame out superrocket speed amounts to 20/20.88 = 0.958 light speed. Continuing at the same speed would bring us to Canopus in 29.7 years of our rocket time. This is nearly short enough to meet our goal of 20 years. Revelation number three gives us a dizzying new sense of freedom. By going fast enough we can get to Canopus in five minutes of our rocket time if we want! In fact, no matter how far away an object lies, and no matter how short the time allotted to us, nothing in principle stops us from covering the required distance in that time. W e have only to be quite careful in explaining this newfound freedom to our Space Agency friends. Yes, we can go any distance the agency requires, however great, provided they specify the distance in the Earthlinked reference frame. Yes, we can make it in any nonzero time the agency specifies, however short, provided they agree to measure time on the rocket clock we carry along with us. To be sure, the Earthlinked system of lookout stations and printout clocks will record us as traveling at less than the speed of light. Lookouts will ultimately complain to the Space Agency how infernally long we take to make the trip. But when our Space Agency friends quiz the lookouts a bit more, they will have to confess the truth; When they look through our window as we shoot by station after station, they can see that our clock reads much less than theirs, and in terms of our own rocket clock we are meeting the promised time for the trip. Our dream ends with sunlight streaming through the bedroom window. W e lie there savoring the three revelations; economy of description of two events in a reference frame stripped down to one space dimension, speed defined always with respect to a
Five minutes to C a n o p u sor to any star!
124
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TRIP TO CANOPUS
specified reference frame and thus never exceeding light speed, and freedom to go arbittarily far in a lifetime.
4.5 FLIGHT PLAN out and back in 40 years to meet our remote descendents
Round trip: 202 Earth y ears
Round trip: 4 0 astronaut y ears
Wide awake now, we face yesterday’s question: Shall we go to Canopus, 99 lightyears distant, as the Space Agency asks? Yes. And yes, we shall live to retutn and repott. We take paper and pencil and sketch our plan. The numbers have to be different from those we dteamed about. Trial and error gives us the following plan: After a preliminary run to get up to speed, we will zoom past Earth at 99/101 = 0.9802 light speed. We will continue at that speed all the 99 lightyears to Canopus. We will make a loop around it and record in those few minutes, by highspeed cameta, the features of that strange star. We will then retutn at unalteted speed, flashing by our finish line without any letup, and as we do so, we will toss out our bundle of records to colleagues on Earth. Then we will slow down, turn, and descend quietly to Earth, our mission completed. The fitst long run takes 101 Earth years. W e have already decided to travel at a speed of 9 9/101, or 99 lightyears of distance in 101 years of time. Going at that speed for 101 Earth years, we will just cover the 99 lightyears to Canopus. The return trip will likewise take 101 Earth years. Thus we will deliver our records to Earth 202 Earthclock years after the start of our trip. Even briefer will be the account of our trip as it will be perceived in the freefloat rocket frame. Relative to the ship we will not go anywhere, either on the outbound or on the return trip. But time will go on ticking away on our shipboard clock. Moreover our biological clock, by which we age, and all other good clocks carried along will tick away in concord with it. How much time will that rocket clock rack up on the outbound trip? Twenty years. How do we know? We reach this answer in three steps. First, we already know from records in the Earthlinked laboratory frame that the spacetime interval — the proper time— between departure from Earth and artival at Canopus will equal 20 years: L aboratory (interval)^ = = = =
L aboratory
(time separation)^ — (space separation)^ (101 years)^ — (99 years)^ 10,201 years^ — 9801 years^ 400 years^ = (20 years)^
Second, as the saying goes, “interval is interval is interval” : The spacetime interval is invariant between frames. The interval as registered in the rocket frame must therefore also have this 20year value. Third, in the rocket frame, separation between the two events (depatture ftom Earth and atrival at Canopus) lies all in the time dimension, zero in the space dimension, since we do not leave the rocket. Therefore separation in rocket time itself between these two events is the proper time and must likewise be 20 years: R ocket (interval)^ = = = =
R ocket
(time separation)^ — (space separation)^ (time separation)^ — (zero)^ (rocket time)^ = (proper time)^ (20 years)^
4.6
TWIN PARADOX
12 5
We boil down our flight plan to bare bones and take it to the Space Agency for approval: Speed 99/101 = 0.9802 light speed; distance 99 lightyears out, 99 lightyears back; time of return to Earth 202 years after start; astronaut’s aging during trip, 40 years. The responsible people greet the plan with enthusiasm. They thank us for volunteering for a mission so unprecedented. They ask us to take our proposal before rhe Board of Direcrors for final approval. We agree, not realizing what a hornets’ nest we are walking into. The Board of Direcrors consists of people from various walks of life, set up by Congress to assure that major projects have support of the public at large. The media have reported widely on our proposal in the weeks before we meet with the board, and many people with strong objecrions to relativity have written to voice their opinions. A few have met with board members and talked to them at length. We are unaware of this as we enter rhe paneled board room. At the request of the chairman we summarize our plan. The majority appear to welcome it. Several of their colleagues, however, object.
4.6 TWIN PARADOX a kink in the path explains the difference “Your whole plan depends on relativity,’’ stresses James Fastlane, “but telativity is a swindle. You can see for yourself that it is selfcontradictory. It says that the laws of physics are identical in all freefloat frames. Very well, here’s your rocket frame and here’s Earth frame. You tell me that identical clocks, started near Earth at identical times, each in one of these freefloat frames, will read very different time lapses. You go away and return only 40 years older, while we and our descendants age 202 years. But if there’s any justice, if relativity makes any sense at all, it should be equally possible to regard you as the stayathome. Relative to you, we speed away in the opposite direction and return. Hence we should be younger than you when we meet again. In contrast, you say you will be younger than we are. This is a flat contradiction. Nothing could show more conclusively that neither result can be right. Aging is aging. It is impossible to live long enough to cover a distance of 99 lightyears twice— going and coming. Forget the whole idea.” “Jim ,” we reply, “your description is the basis for the famous Twin Paradox, in which one twin stays on Earth while the other takes the kind of round rrip we have been describing. Which twin is older when they come together again? I would like to leave this question for a minure and consider a similar trip across the United States. “We all know, Jim, that every July you drive straight north on Interstate Highway 35 from Laredo, Texas, on the Mexican border, to Duluth, Minnesota, near the Canadian border. Your tires roll along a length of roadway equal to 2000 kilometers and the odometer on your car shows it. ‘'I too drive from Laredo to Duluth, but last year I had to make a stop in Cincinnati, Ohio, on the way. I drove northeast as straight as I could from Laredo to Cincinnati, 1400 kilometers, and northwest as straight as I could from Cincinnati to Duluth, another 1400 kilometers. Altogether, my tires rolled out 2800 kilometers. When we left Laredo you could have said that my route was deviating from yours, and I could have said with equal justice that yours was deviating from mine. The great difference between our travels is this, that my course has a sharp turn in it. That’s why my kilometerage is greater than yours in the ratio of 2800 to 2000.” Fastlane interrupts: “Are you telling me that the turn in the rocket trajectory ar Canopus explains the smaller aging of the rocket traveler? The turn in your trip to Duluth made your travel distance longer, not shorter.”
Which twin travels?
Curved path in space is a longer path
126
Astronaut who turns around a g e s less . . .
CHAPTER 4
TRIP TO CANOPUS
“That is the difference between path length in Euclidean space geometry and wrisrwatch time in Lorentz spacetime geometry,” we reply. “In Euclidean geometry the shortest path length between two points is achieved by the traveler who does not change direction. All indirect paths are longer than this minimum. In spacetime the greatest aging between two events is experienced by the traveler who does not change direction. For all travelers who change direction, the total proper time, the total wristwatch time, the total aging is less than this maximum. “The distinction between distance in Euclidean geometry and aging in spacetime comes ditectly from the contrast between plus sign in the expression for distance between two locations and minus sign in the expression for interval between two events. In going to Duluth by way of Cincinnati I use the plus sign:
(
distance: V Laredo to 1 = Cincinnati /
/ northward V / eastward separation: 1 __ I separation: 1 Laredo to I » 1 Laredo to I \ Cincinnati / T \ Cincinnati /
“Contrast this with motion in spacetime. In analyzing my trip to Canopus, I use the minus sign:
. becau se of a minus sign!
(
proper time:' Earth to Canopus ,
(
rocket time:' Earth to Canopus ,
(
Earth time:' Earth to Canopus ,
t
(
Earth distance:' Earth to Canopus ,
“The contrast between a plus sign and a minus sign: This is the distinction between distance covered during travel in space and time elapsed — aging— during travel in spacetime."
4.7 LORENTZ CONTRACTION go a shorter distance in a shorter time
Canopus much closer for astronaut
As James Fastlane ponders this response. Dr. Joanne Short breaks in. “The Twin Paradox is not the only one you have to explain in order to convince us of the correctness of your analysis. Look at the outward trip as observed by you yourself, the rocket traveler. You reach Canopus after just 20 years of your time. Yet we know that Canopus lies 99 lightyears distant. How can you possibly cover 99 lightyears in 20 years?” “That is exactly what I dreamed about, Joanne!” we reply. “ First of all, it is confusing to combine distances measured in one reference frame with time measured in another reference frame. The 99lightyear distance to Canopus is measured with respect to the Earthlinked frame, while the 20 years recorded on the outward traveler’s clock refers to the rocket frame. No wonder the result appears to imply a rate of travel faster than light. Why not take what I paid for fuel for my car last week and divide it by the number of gallons you bought today foryo«rcar, to figure the cost of a gallon of fuel? A crazy, mixedup, wrong way to work out cost— but no crazier than that way to figure speed! “ But your question about time brings up a similar question about distance: distance between Earth and Canopus measured in the frame in which they are at rest does not agree with the distance between them measured from a rocket that moves along the line connecting them.
4.8
TIME TRAVELER
“Any freefloat frame is as good as any other for analyzing motion — that is the Principle of Relativity! So think of the entire outward trip in terms of rocket measure ments. At the starting gun (or firecracker) Earth is rushing past the rocket at speed 9 9 /101. Twenty years later Canopus arrives at the rocket, Canopus also traveling at that speed, 99/101 in that rocket frame. This means that for the rocket traveler the EatthCanopus distance is only about 20 lightyears. In fact it is just the fraction (99/101) of 20 lightyears, so that at speed 99/101 this distance is covered in exactly 20 years.’’ “O f course. We are dealing with L orentz c o n tra c tio n ,” huffs Professor Bright, who thinks any objection to relativity is a waste of time. He has no head for politics, so does not appreciate how important it is for the public to accept the expenditures proposed for this project. He continues, ‘'Think of a very long stick lying with one end at Eatth, the othet end at Canopus. Each observer, with the help of colleagues, measures the position of the two ends of this stick at the same time in his or her frame. By this means the outward rocket traveler measures a shorter length of the stick — a smaller Earth  Canopus distance — than does an observer in the Earthlinked frame in which the stick lies at test. “The factor by which the stick appears contracted in the rocket frame is just the same as the ratio of rocket time to Earth time for the outwatd trip. This ratio is (20 years)/(101 years). Hence the rocket observer measures the Earth  Canopus distance to be (99 lightyears)(20/101) = 19.6 lightyears — just a bit less than 20 lightyears, as you said. “Everybody has a satisfactory picture: The astronaut can get to Canopus in 20 years of rocket time because the astronaut’s measurements show Canopus to be slightly less than 20 lightyears distant. We on Earth agree that the time lapse on the rocket clock is 20 years, but our ‘explanation’ rests on the invariance of the interval between the events of departure from Earth and arrival at Canopus.” Professor Bright pounds the table: “Why are you giving this poor astronaut such a hard time, when relativity is so utterly simple?” He is surprised by the outburst of laughter from other board members and the audience in the room.
12 7
Lorentz contraction
4.8 TIME TRAVELER visit the future, don't come back. Laura Long has been thoughtfully following the argument. She comments, “You know, we have been discussing you as a space travelet. But you are a time traveler as well. Do you realize that by traveling to Canopus and back at 99/101 of light speed, you journey six generations forward in time: 202 years at 33 years per generation? So you will be able to visit your greatgreatgreatgreatgreatgrandchildren at a cost of only 40 years of your life.” “Yes, I did think of that,” we reply. “Time and space are not so different in this respect. Just as we can travel to as great an Earthlinked distance as we want in as short a rocket time as we want, so we can also ttavel as fat forward into Earth’s future as we wish. “While I was trying various numbers in making up the proposed plan, I realized that if we traveled not at 99/101 light speed but at 9999/10,001 light speed, then a round trip would take not 40 rocket years but only 3.96 rocket years and 198 Earth years. Ten such round trips will age us 39.6 years and bring us back finally at an Earth time about two thousand years in the future, or some year in the fortieth century. That
Travel to Earth's future
128
Time travel is one w ay
CHAPTER 4
TRIP TO CANOPUS
is not six generations ahead, but sixty generations, an additional time equal to one third of recorded history on Earth.” “Why stop there?” pursues Laura Long excitedly. “Why not go even faster, make more round trips, and learn the ultimate fate of Earth and its solar system — or even the still more remote future of the Universe as a whole? Then you could report back to us whether the Universe expands forever or ends in a crunch.” “Sorry, but no report back to our century is possible,” smiles Professor Bright. “There are differences between travel in time and travel in space. To begin with, we can stand still on Earth if we choose and go nowhere in space with respect to that frame. Concerning travel through time, however, we have no such choice! Even when we stand stock still on Earth, we nevertheless travel gently but inevitably forward in time. Time proceeds inexorably! “Second, time travel is one way. You may be able to buy a roundtrip ticket to Canopus, but you can get only a oneway ticket to the fortieth century. You can’t go backward in time. Time won’t reverse.” Turning to us he adds, “As for the fate of the solar system and the end of the Universe, our descendants may meet you there as fellow observers, but we ourselves will have to bid you a fitm and final ‘goodbye’ as you leave us on any of the trips we have been discussing. The French au revoir— until we meet again — will not do.”
4.9 RELATIVITY OF SIMULTANEITY we turn around; our changing colleagues say Earth's clock flies forward
Rocket observer: Fewer Earthclock ticks on outward trip . . .
also few er Earthclock ticks on return trip
By this time James Fastlane has gotten his second wind. “I am still stuck in this Twin Paradox thing. The time for the outward trip is less as measured in the rocket frame than as measured in the Earth frame. But if relativity is correct, every freefloat frame is equivalent. As you sit on the rocket, you feel yourself to be at rest, stationary, motionless; you measure our Earth watchstation clocks to be zipping by you at high velocity. Who cares about labels? For you these Earth clocks are in motion! Therefore the time for the ourwatd trip should be less as measured on the (‘moving’) Earth clock than as measured on your (‘stationary’) rocket clock.” We nod assent and he continues. “ Nothing prevents us from supposing the existence of a series of rocket lookout stations moving along in step with your rocket and strung out at separations of one lightyear as measured in your rocket frame, all with clocks synchronized in your rocket frame and running at the same rate as your rocket clock. Now, as Earth passes each of these rocket lookout stations in turn, won’t those stations read and record the times on the passing Earth clock to be less than their own times? Otherwise how can relativity be correct?” “Yes, your prediction is reasonable,” we reply. “And on the return trip will not the same be true: Returningrocket lookout stations will measure and record time lapses on the passing Earth clock to be less than on their own clocks?” “That conclusion is inevitable if relativity is consistent.” “ Aha!” exclaims Mr. Fastlane, “ Now I’ve got you! If Earth clock is measured by rocket lookout stations to show smaller time lapses during the outward trip — and also during the return trip — then obviously total Earth time must be less than rocket roundtrip time. But you claim just the opposite: that total rocket time is less than Earth time. This is a fundamental contradiction. Your relativity is wrong!” Folding his arms he glowers at us.
4.9
RELATIVITY OF SIMULTANEITY
There is a long silence. Everyone looks at us except Professor Bright, who has his head down. It is hard to think with all this attention. Yet our mind runs over the trip again. Going out . . . coming back . . . turning around . . . that’s it! “ All of us have been thinking the wrong way!’’ we exclaim. “We have been talking as if there is only a single rocket frame. True, the same vehicle, with its traveler, goes out and returns. True, a single clock makes the round trip with the traveler. But this vehicle turns around— reverses its direction of travel — and that changes everything. “Maybe it’s simpler to think of two rockets, each moving without change of velocity. We ride on the first rocket going out and on the second rocket coming back. Each of these two is really a rocket frame: each has its own long train of lookout stations with recording clocks synchronized to its reference clock (Figure 41). The traveler can be thought of as 'jumping trains’ at Canopus — from outwardbound rocket frame to inwardbound rocket frame — carrying the calendar clock. “Now follow Mr. Fastlane’s prescription to analyze the trip in the rocket frame, but with this change: make this analysis using two rocket frames — one outward bound, the other inward bound. “It is 20 yeats by outwardrocket time when the traveler arrives at Canopus. That is the reading on all lookour station clocks in that outwardrocket frame. One of rhose lookout stations is passing Earth when this rocket time arrives. Its clock, synchronized to the clock of the outward traveler at Canopus, also reads 20 years. W hat time does that rocket lookoutstation guard read on the passing Earth clock? For the rocket observer Earth clock reads less time by the same factor that rocket clocks read less time (20 years at arrival at Canopus) for Earth observers (who read 101 years on their own clocks). This factor is 2 0/101. Elence for the outwardrocket observer the Earth clock must read 20/101 times 20 years, or 3.96 years.’’ “W hat!” explodes Fastlane. “According to your plan, the turnaround at Canopus occurs at 101 years of Earth time. Now you say this time equals less than 4 years on Earth clock.” “No sir, I do not say that,” we reply, feeling confident at last. “ I did say that at the same time as the outgoing rocket arrives at Canopus, Earth clock reads 3.96 years as measured in that outgoing rocket frame. An equally true statement is that at the same time as the outgoing rocket arrives at Canopus, Earth clock reads 101 years as measured in the Earthbound frame. Apparently observers in different reference frames in relative motion do not agree on what events occur at the same time when these events occur far apart along the line of relative motion.” Once again Professor Bright supplies the label. “Yes, that is called relativity o f sim u ltan eity . Events that occur at the same time— simultaneously— judged from
returnrocket lookout stations
V ....V V .... V..... V V .... V  V....V < ^ V ..... A....A A A > outgoingrocket lookout stations
O
Earth
.........................Q
 O
.......................... 
Earth lookout stations
o
Canopus
FIGURE 41. Schematic plot in the E arthlinked fra m e showing the outgoing rocket a n d the return rocket used in the round trip between E arth a n d Canopus. The two rockets meet at Canopus, where the traveler jumps from outgoing rocket to return rocket. Each reference frame has its own string of lookout stations, at rest and synchronized in that frame, shown by small squares, triangles, and inverted triangles. In this figure the outgoing and return rocket lines of motion are displaced vertically for purposes of analysis; tn reality, all motion lies along the single line between Earth and Canopus. The figure is not to scale!
129
Astronaut jumps from outgoing frame to returning frame
Outgoing rocket: As it arrives at Canopus, Earth clock read s 3 .9 6 y ea rs
130
Returning rocket: As it leaves Canopus, Earth clock reads 198.04 years
Forward “jump” in Earth clock results from frame change
CHAPTER 4
TRIP TO CANOPUS
one freefloat frame but far apart along the line of relative motion do not occur simultaneously as judged from another freefloat frame. “As an example of relativity of simultaneity, consider either chain of lookout stations strung along the line of relative motion. If all clocks in the lookout stations of one frame strike exactly at noon in that frame, these strikes are not simultaneous as measured in another frame in relative motion with respect to the first. This is called relative synchronization o f clocks. “Incidentally, most of the socalled ‘paradoxes’ of relativity, one of which we are considering now, turn on misconceptions about relativity of simultaneity.” Dr. Short breaks in. “W hat about the returning rocket? W hat time on the Earth clock will the returning rocket lookout station measure as the traveler starts back?” “That shouldn’t be too difficult to figure out,” we reply. “W e know that the clock on the returning rocket reads 40 years when we arrive home on Earth. And the Earth clock reads 202 years on that return. Both of these readings occur ar the same place (Earth), so we do not need to worry about relativity of simultaneity of that reading. And during the return trip Earth clock records less elapsed time than rocket clocks’ 20 years by the same factor, 20/101, or a total elapsed time of 20 X 20/101 = 3.96 years according ro return rocket observations. Therefore at the earlier turnaround, return rocket observers will see Earth clock reading 202 — 3.96 = 198.04 years.” “Wait a minute!” bellows Eastlane. “First you say that the rocket observer sees the Earth clock reading 3.96 years at turnaround in the outwardbound frame. Now you say that the rocket observer sees the Earth clock read 198.04 years at turnaround in the inwardbound frame. Which one is right?” “ Both are right,” we reply. “The rwo observations are made from two different frames. Each of these frames has a duly synchronized system of lookoutstation clocks, as does the Earthlinked frame (Figure 41). The socalled Twin Paradox is resolved by noticing that between the Earthclock reading of 3.96 years, taken from the outward rocket lookout station at turnaround and the Earthclock reading of 198.04 years, taken by the returningrocket lookout station at turnaround, there is a difference of 194.08 years. “This ‘jump’ appears on no single clock but is the result of the traveler changing frames at Canopus. Yet this jump, or difference, resolves the paradox: For rhe traveler, the Earth clock reads small time lapses on the outward leg — and also small time lapses on the return leg — but it jumps way ahead at turnaround. This jump accounts for the large value of Earthaging during the trip: 202 years. In conrrast rhe traveler ages only 40 years during the trip (Table 41). “And notice that the traveler is unique in the experience of changing frames; only the traveler suffers the terrible jolt of reversing direction of motion. In contrast, the
OBSERVATIONS OF EVENTS ON CANOPUS TRIP Earthclock reading observed by
Event Depart Earth
Time measured in Earthlinked frame
Time measured by traveler
outgoingrocket lookout stations passing Earth
0 years
0 years
Arrive Canopus
101 years
20 years
20 years X 20/101 = 3.96 years
Depart Canopus
101 years
20 years
3.96 years
Arrive Earth
202 years
40 years
returnrocket lookout stations passing Earth
0 years
202  3.96 = 198.04 years 202 years
4.10
EXPERIMENTAL EVIDENCE
Earth observer stays relaxed and comfortable in the same frame during the astronaut’s entire trip. Therefore there is no symmetry between rocket traveler and Earth dweller, so no genuine contradiction in their differing time lapses, and the story of the twins is not a paradox. “ In fact, the observer in each of the three frames — Earthlinked, outwardrocket, and inwardrocket— has a perfectly consistent and nonparadoxical interpretation of the sequence of events. However, in accounting for disagreements between his or her readings and those of observers in other freefloat frames, each observer infers some misbehavior of measuring devices in these other frames. Each observes less elapsed time on clocks in the other frame than on his or her own clocks (time stretching or time dilation). Each thinks that an object lying along the line of relative motion and at rest in another frame is contracted (Lorentz contraction). Each thinks that lookoutstation clocks in other frames are not synchronized with one another (relative synchronization of clocks). As a result, each cannot agree with other observers as to which events far apart along the line of relative motion occur at the same time (relativity of simultane ity).’’ “ Boy,” growls Fastlane, “all these different reference frames sure do complicate the story!’’ “ Exactly!” we exclaim. “These complications arise because observations from any one frame are limited and parochial. All disagreements can be bypassed by talking only in the invariant language of spacetime interval, proper time, wristwatch time. The proper time from takeoff from Earth to arrival at Canopus equals 20 years, period. The proper time from turnaround at Canopus to rearrival at Earth equals 20 years, period. The sum equals 40 years as experienced by the astronaut, period. On the Earth clock, the proper time between departure and return is 202 years, period. End of story. Observers in all freefloat frames reckon proper times— spacetime intervals between these events — using their differing space and time measurements. However, once the data are translated into the common language of proper time, every observer agrees. Proper times provide a universal language independent of reference frame.”
131
All observers agree on result, disagree on reason
Spacetime interval is universal language
4.10 EXPERIMENTAL EVIDENCE objects large and small, slow and fast: many witnesses for the Canopus trip Alfred Missouri has remained silent up to this point. Now he declares, “All this theory is too much for me. I won’t believe a word you say unless you can show me an experimental demonstration.” We reply, “Atomic clocks have been placed on commercial airliners and carried around Earth, some in an eastward direction, others in a westward direction. In each case the airliner clocks were compared with reference clocks at the U.S. Naval Observatory before and after their trips. These clocks disagreed. Results were consist ent with the velocityrelated predictions of special relativity. “This verification of special relativity has two minor difficulties and a major one. Minor difficulties: (1) Each leg of a commercial airliner’s trip may be at a different speed, not always accurately known and for which the timestretching effect must be separately calculated. Also, temperature and pressure effects on airborne clocks are hard to control in a commercial airliner. (2) More fundamentally. Earth rotates, cartying the reference Naval Observatory clocks eastward around the center of Earth. Earth center can be regarded as the inertial point in freefloat around Sun. With
“Airliner” test of twin effect
1 32
CHAPTER 4
TRIP TO CANOPUS
DO WE NEED GENERAL RELATIVITY? NO! The group takes a break and mills around the conference room, chatting and eating refreshments. Joanne Short approaches us juggling coffee, a donut, and her notes. ‘‘I didn’t want to embarrass you in public,” she says, “ but isn’t your plan faulty because of the turnaround? You can’t be serious about leaping from one highspeed rocket to another rocket going in the opposite direction. That means certain death! Be realistic: You and your rocket will have to slow down over some time period, come to rest at Canopus, then speed up again, this time headed back toward Earth. During this change of velocity you will be thrown against the front of the rocket ship, as I’m thrown when I slam on my car brakes. Release a test particle from rest and it will hurtle forward! Surely you are not in an inertial (freefloat) frame. Therefore you cannot use special relativity in your analysis of this time period. What does that do to your description of the ‘jump ahead’ of Earth clocks as you slow down and speed up again? Don’t you need general relativity to analyze events in accelerated reference frames?” “ Oh yes, general relativity can describe events in the accelerated frame,” we reply, “ but so can special relativity if we take it in easy steps! I like to think of a freight yard with trains moving at different speeds along parallel tracks. Each train has its own string of recording clocks along its length, each string synchronized in that particular train frame. Each adjacent train is moving at a slightly different speed from the one next to it. Now we can change frames by walking a c r o s s the trains, stepping from the top of one freight car to the top of the freight car rolling next to it at a slightly different speed. “ Let these trains become rocket trains in space. Each train then has an observer passing Earth as we step on that train. Each observer, by prearran gement, reads the Earth clock a t th e s a m e tim e that we step onto his train (‘at the same time’ as recorded in that frame). When you assemble all these data later on, you find that the set of observers on the sequence of trains see the Earth clock jumping forward in time much faster than would be expected. The net result is similar to the single horrible jerk as you jump from the outgoing rocket to the incoming rocket. “ Notice that it takes a whole set of clocks in different frames, all reading the single Earth clock, to establish this result. So there is never any contradiction between a single clock in one frame and a single clock in any other frame. In this case special relativity can do the job just fine.” The directors reassemble and Joanne Short, smiling, takes her place with them.
respect to this center, one airborne clock moves even faster eastward than Earth’s surface, while the other one— heading west with respect to the surface— with respect to Earth’s center also moves eastward, but more slowly. Taking account of these various relative velocities adds further complication to analysis of results. “We overcome these two minor difficulties by having an airplane fly round and round in circles in the vicinity of a single groundbased reference atomic clock.
4.10
EXPERIMENTAL EVIDENCE
Then — to a high accuraq'— only relative motion of these two clocks enters into the specialrelativity analysis. On N ovember 22, 197 5 ,aU .S .N avy P3 C antisubmarine patrol plane flew back and forth for 15 hours at an altitude of 25,000 to 35,000 feet (7600 to 10,700 meters) over Chesapeake Bay in an experiment arranged by Carroll Alley and collabo rators. The plane carried atomic clocks that were compared by laser pulse with identical clocks on the ground. Traveling at an average speed of 270 knots (140 meters per second), the airborne clocks lost an average of 5.6 nanoseconds = 5.6 X 10“^ seconds due to velocityrelated effects in the 15hour flight. The expected specialrelativity difference in clock readings for this relative speed is 5.7 nanoseconds. This result is remarkably accurate, considering the low relative velocity of the two clocks: 4.7 X 10“ ^ light speed. “The major difficulty with all of these experiments is this: A highflying airplane is significantly farther from Earth’s center than is the groundbased clock. Think of an observer in a helicopter reading the clocks of passing airplanes and signaling these readings for comparison to a groundbased clock directly below. These two clocks — the helicopter clock and the Earthbound clock— are at rest with respect to one another. Are they in the same inertial (freefloat) frame? The answer is No. “We know that a single inertial reference frame near Earth cannot extend far in a vertical direction (Section 2.3). Even if the two clocks — helicopter and Earthbound — were dropped in free fall, they could not both be in the same inertial frame. Released from rest 30,000 feet one above the other, they would increase this relative distance by 1 millimeter in only 0.3 second of free fall — too rapid a change to be
133
“Circling airplane” test of twin effect
Trouble: Large frame is not inertial
ignored. B ut the experim ent w ent on not for 0.3 second b u t for 15 hours!
“Since the helicopter clock and Earthbound clock are not in the same inertial frame, their behavior cannot be analyzed by special relativity. Instead we must use general relativity — the theory of gravitation. General relativity predicts that during the 15hour flight the higheraltitude clock in the Chesapeake Bay experiment will record greater elapsed time by 52.8 nanoseconds due to the slightly reduced gravitational field at altitudes at which the plane flew. From this must be subtracted the 5.7 nanoseconds by which the airborne clock is predicted to record less elapsed time due to effects of relative velocity. These velocity effects are predicted by both special relativity and general relativity and were the only results quoted above. The overall predicted result equals 52.8 — 5.7 = 47.1 nanoseconds net gain by the highaltitude clock compared with the clock on the ground. Contrast this with the measured value of 47.2 nanoseconds. “ Hence for airplanes flying at conventional speeds and conventional altitudes, tidalgravitational effects on clocks can be greater than velocitydependent effects to which special relativity is limited. In fact, the Chesapeake Bay experiment was conducted to verify the results of general relativity: The airplane pilot was instructed to fly as slowly as possible to reduce velocity effects! The P3C patrol plane is likely to stall below 200 knots, so a speed of 270 knots was chosen. “In all these experiments the timestretching effect is small because the speed of an airplane is small compared to the speed of light, but atomic clocks are now so accurate that these speed effects are routinely taken into account when such clocks are brought together for direct comparison.” Professor Bright chimes in. “W hat the astronaut says is correct: We do not have large clocks moving fast on Earth. On the other hand, we have a great many small clocks moving very fast indeed. When particles collide in highspeed accelerators, radioactive fragments emerge that decay into other particles after an average lifetime that is well known when measured in the rest frame of the particle. When the radioactive particle moves at high speed in the laboratory, its average lifetime is significantly longer as measured on laboratory clocks than when the patticle is at rest. The amount of lengthening of this lifetime is easily calculated from the particle speed in the same way the astronaut calculates time stretching on the way to and from
Solution: Use general relativity
“Highspeed radioactive particle' test of twin effect
134
Earth frame: Freefloat for particle experiments
"Oscillating iron nucleus” test of twin effect
Twin effect verified!
CHAPTER 4
TRIP TO CANOPUS
Canopus. The timestretch factor can be as great as 10 for some of these particles: the fastmoving particles are measured to live 10 times longer, on average, than their measured lifetime when at rest! The experimental results agree with these calculations in all cases we have tried. Such time stretching is part of the everyday experience of highenergy particle physicists. “And for these increasedlifetime experiments there is no problem of principle in making observations in an inertial, freefloat frame. While rhey are decaying, particles cover at most a few tens of meters of space. Think of the flight of each particle as a separate experiment. An individual experiment lasts as long as it takes one highspeed particle to move through the apparatus— a few tens of meters of lighttravel time. Ten meters of lighttravel time equals about 33 nanoseconds, or 33 X 10“ ^ seconds. “Can we construct an inertial frame for such happenings? Two ball bearings released from rest say 20 meters apart do not move together very far in 33 nanosec onds! Therefore these increasedlifetime experiments could be done, in principle, in freefloat frames. It follows that special relativity suffices to describe the behavior of the ‘radioactivedecay clocks’ employed in these experiments. W e do not need the theory of gravitation provided by general relativity. “O f course, in none of these highspeed particle experiments do particles move back and forth the way our astronaut friend proposes to do between Earth and Canopus. Even that backandforth result has been verified for certain radioactive iron nuclei vibrating with thermal agitation in a solid sample of iron. Atoms in a hotter sample vibrate back and forth faster, on average, and thus stay younger, on average, than atoms in a cooler sample. In this case the ‘tick of the clock’ carried by an iron atom is the period of electromagnetic radiation (‘gamma ray’) given off when its nucleus makes the transition from a radioactive state to one that is not radioaaive. For detailed reasons that we need not go into here, this particular ‘clock’ can be read with very high accuracy. Beyond all such details, the experimental outcome is simply stated: Clocks that take one or many round trips at higher speed record a smaller elapsed time than clocks that take one or many round trips at lower speed. “These various results— plus many others we have not described— combine to give overwhelming experimental support for the predictions of the astronaut concern ing the proposed trip to Canopus.’’ Dr. Bright sits back in his chair with a smile, obviously believing that he has disposed of all objections singlehandedly. “Yes,” we conclude, “about the reality of the effect there is no question. Therefore if you all approve, and the Space Agency provides that new and very fast rocket, we can be on our way.” The meeting votes approval and our little story ends.
REFERENCES The “airliner check” of time stretching (Section 4.10) is reported in J. C. Hafele and Richard E. Keating, Science, Volume 177, pages 1 6 6 1 6 7 and 1 6 8 1 7 0 (14 July 1972). The “patrol plane” check of general relativity (Section 4.10) is reported by Carroll O. Alley in Quantum Optics, Experimental Gravity, and Measurement Theory, edited by Pierre Meystre and Marian O. Scully (Plenum, New York, 1983). See also 1976 physics Ph.D. theses by Robert A. Reisse and Ralph E. Williams, University of Maryland. The “radioactive nuclei” check of time stretching (Section 4.10) is reported in R. V. Pound and G. A. Rebka, Jr., Physical Review Letters, Volume 4, pages llA  T V b (1960).
EXERCISE 43
A RELATIVISTIC OSCILLATOR
135
CHAPTER 4 EXERCISES N ote: The following exercises are related to the story line of this chapter. Additional exercises may be se lected from Chapter 3 or the Special Topic on the Lorentz Transformation following Chapter 3.
41
practical space travel
In 2200 A.D. the fastest available interstellar rocket moves ^x: V = 0.75 of the speed of light. James Ab bott is sent in this rocket at full speed to Sirius, the Dog Star (the brightest star in the heavens as seen from Earth), a distance D = 8.7 lightyears as mea sured in the Earth frame. James stays there for a time T = 7 years as recorded on his clock and then returns to Earth with the same speed p — 0. 75. Assume Sirius is at rest relative to Earth. Let the departure from Earth be the reference event (the zero of time and space for all observers). According to Earthlinked observers: a At what time does the rocket arrive at Sirius? b At what time does the rocket leave Sirius? c At what time does the rocket arrive back at Earth? According to James’s observations: d At what time does he arrive at Sirius? e At what time does he leave Sirius? f At what time does he arrive back at Earth? g As he moves toward Sirius, James is accompa nied by a string of outgoing lookout stations along his direction of motion, each one with a clock synchro nized to his own. W hat is the spatial distance between Earth and Sirius, according to observations made with this outgoing string of lookout stations? h One of James’s outgoing lookout stations, call it Q, passes Earth at the same time (in James’s outgo ing frame) that James reaches Sirius. W hat time does Q's clock read at this event of passing? W hat time does the clock on Earth read at this same event? i As he moves back toward Earth, James is ac companied by a string of incoming lookout stations along his direction of motion, each one with a clock synchronized to his own. One of these incoming look out stations, call it Z, passes Earth at the same time (in James’s incoming frame) that James leaves Sirius to return home. W hat time does Z ’s clock read at this event of passing? W hat time does the clock on Earth read at this same event? To rea/ly understand the contents of Chapter 4, repeat this exercise many times with new values of p, D, and T that you choose yourself.
42 oneway twin paradox? A worried student writes, “I still cannot believe your solution to the Twin Paradox. During the outward trip to Canopus, each twin can regard the other as moving away from him; so how can we say which twin is younger? The answer is that the twin in the rocket makes a turn, and in Lorentz spacetime geom etry, the greatest aging is experienced by the person who does not turn. This argument is extremely unsat isfying. It forces me to ask: W hat if the rocket breaks down when I get to Canopus, so that I stop there but cannot turn around? Does this mean that it is no longer possible to say that I have aged less than my Earthbound twin? But if not, then I would never have gotten to Canopus alive.” Write a halfpage response to this student, answering the questions politely and decisively.
43 a relativistic oscillator In order to test the laws of relativity, an engineer decides to construct an oscillator with a very light oscillating bob that can move back and forth very fast. The lightest bob known with a mass greater than zero is the electron. The engineer uses a cubical metal box, whose edge measures one meter, that is warmed slightly so that a few electrons “boil off” from its surfaces (see the figure). A vacuum pump removes air from the box so that electrons may move freely inside without colliding with air molecules. Across the mid dle of the box — and electrically insulated from it— is a metal screen charged to a high positive voltage by a power supply. A voltagecontrol knob on the power supply can be turned to change the DC voltage V„ between box and screen. Let an electron boiled off from the inner wall of the box have very small velocity initially (assume that the initial velocity is zero). The electron is attracted to the positive screen, increases speed toward the screen, passes through a hole in the screen, slows down as it moves away from the attract ing screen, stops just short of the opposite wall of the box, is pulled back toward the screen; and in this way oscillates back and forth between the walls of the box. a In how short a time T can the electron be made to oscillate back and forth on one round trip between the walls? The engineer who designed the equipment claims that by turning the voltage control knob high enough he can obtain as high a frequency of oscilla tion / = 1 /T as desired. Is he right? b For sufficiently low voltages the electron will be nonrelativistic— and one can use Newtonian me
136
EXERCISE 43
A RELATIVISTIC OSCILLATOR
chanics to analyze its motion. For this case the fre quency of oscillation of the electron is increased by what factor when the voltage on the screen is doubled? D iscussion: At corresponding points of the elec tron’s parh before and afrer voltage doubling, how does the Newtonian kinetic energy of rhe electron compare in the two cases? How does its velocity com pare in rhe two cases? c W hat is a definite formula for frequency/as a funcrion of volrage in the nonrelativistic case? W ait as late as possible to substitute numbers for mass of elecrron, charge of electron, and so forrh. d W hat is the frequency in the extreme relativis tic case in which over most of its course rhe elecrron is moving . . . (rest of sentence suppressed!) . . . ? Call this frequency . e On rhe same graph, plot two curves of the dimensionless quantity f / f ^ as functions of the di mensionless quantity qW^/{2mc'^), where q is the charge on the electron and m is its mass. First curve: the nonrelativistic curve from parr c to be drawn
heavily in the region where it is reliable and indicated by dashes elsewhere. Second curve: the extreme relarivisric value from part d, also with dashed lines where not reliable. From the resulting graph estimate quantitatively the voltage of transition from the nonrelativistic to the relativistic region. If possible give a simple argument explaining why your resulr does or does nor make sense as regards order of mag nitude (that is, overlooking factors of 2, 7T, etc.). f Now think of the roundtrip “proper period’’ of oscillation T experienced by the electron and logged by its recording wristwatch as it moves back and forth across rhe box. Ar low electron speeds how does rhis proper period compare wirh rhe laboratory period recorded by the engineer? W hat happens at higher electron speeds? At extreme relativistic speeds? How is this reflected in the “proper frequency’’ of oscillarion Tproper experienced by rhe elecrron? On rhe graph of parr e draw a rough curve in a different color or shading showing qualitatively the dimensionless quantity/p„p,^//„„ as a function of qV„/{2mc^).
TREKKING THROUGH SPACETIME
5.1 TIME? NO. SPACETIME MAP? YES. no such thing as the unique time off an event! Events are the sparkling grains of history. They define spacetime. Spacetime, yes. Time, no. ‘‘Time, no”? How come? Time here in Tokyo, at this enthronement of the successor of the Emperor Hirohito? Where is any meter to be seen that shows any such quality of location as time? Meter to measure the temperature here and now? Yes, this thermom eter. Meter to measure atmospheric pressure here and now? Yes, this barometer. But look as we will, nowhere can we see any meter that we can poke into the space hereabouts to measure its “time.” The time of an event? Impossible! No such thing. Time is not “meterable.” Anything with which to compare time? Yes. Odometer reading, whether miles or kilometers, on the dashboard of our car. There’s no such thing as the odometer reading of Tokyo. Try every gadget one can, thrust it out into this Tokyo air, not one will register anything with the slightest claim to be called the odometer reading of these hereabouts. W hat about looking at the dashboards of the cars in this neighborhood? N ot all of them; that would be nonsense. Only the cars that were new, with odometer reading zero, at the time of Hirohito’s own enthronement. Now at last we are getting into a line of questioning that shows some prospect of clearing up what we mean by “time.” W e ask our companion, “W hat do all those dayandyearcounting wristwatches now read that were set to zero at the time of that earlier ceremony?” “Sixtytwo years, two days,” is her first reply. But then we ask, “W hat about that team that zoomed out to the nearest eyecatching star. Alpha Centauri, and back with almost the speed of light? Didn’t they get back ten years younger than we stayathomes?”
137
“ Time” of an event has no unique meaning
C a r mileage d epends on car's path between places
138
W ristwatch reading d epends on its history of travel between events
G eo g rap h ic mop assigns kilometer coordinates to places
Spacetim e map assigns sp ace and time coordinates to events
Limit attention to one sp a ce dimension plus one time dimension
CHAPTER 5
TREKKING THROUGH SPACETIME
“Yes,” she agrees, “surely their wristwatches now read fiftytwo years, not sixtytwo. So let me draw the lesson. There is no such thing as time. There is only totalized interval of time, time as that interval is racked up between the enthronement of Hirohito and the enthronement of the new Emperor Akihito, between event A and event B, on a wristwatch that has undergone its own individual history of travel from A to B.” “I agree. The concept of time does not apply to location in spacetime. It applies to individual history of travel through spacetime.” “How apt the comparison with odometer reading. Each dashboard shows, not the kilometerage of Akihito, but the kilometers traveled by that particular car between the one imperial ceremony and the other.” Yes, it is nonsense to attribute a kilometer reading to Tokyo. However, it is not at all nonsense to make a map showing where Tokyo lies relative to all the towns roundabout, a map in which kilometers do appear, kilometers north and south, kilometers easr and west. Likewise the term “the time” of an event is totally without meaning. However, that event — and every event near it — lends itself to display on a spacetime diagram (Figure 51), with distance (the locator of latticework clock) running in one direction, and in another direction time (the reading printed out by that clock on the occasion of that event). Time as employed in this sense acquires meaning only because it serves as a measure on a latticeworkdefined map. A different latticework? A different set of clocks, different readings on those clocks, a different map — but same events, same spacetime, same tools to measure the historydependent interval between event and event. Only on such a spacetime plot does one see at a glance the layout of all nearby events, and how one history of travel from event A to event B differs from another. One problem in making our map: Spacerime has four dimensions— three space dimensions plus time. W e picture our event points most readily when they occupy a twodimensional domain and let themselves be dotted in on a twodimensional page. Therefore for the present we limit attention to time and one space dimension; to events, whatever their timing, that occur on one line in space. All events that do not occur on this line we ignore for now. The space location of each event on this line we plot along a ho rizo n tal axis on the page. The latticeclock time at which an event occurs we plot along a vertical axis, from bottom to top of the page. Space and time we measure in the same unit, for example meters of distance and meters of time — or lightyears of distance and years of time. We call the result a spacetim e m ap or a spacetim e diagram . Each spacetime map represents data from a particular reference frame, for example “the laboratory frame.” Figure 51 shows such a spacetime map. Five sample event points appear on the laboratory spacetime map of Figure 51, events labeled 0, A, B, C, and D. • E vent 0 is the referen ce event, the firing of the starting gun, which we take to locate zero position in space and the zero of time. For our own convenience, we place point 0 at the origin of the spacetime map and measure space and time locations of all other events with respect to it.
same place
space ■
FIGURE 51. Laboratory spacetim e map, showing the reference event O, other events A, B, C, a n d D, a horizontal dashed line of si m ultaneity in time, a n d a verti c al dashed line o f equal position in space.
5.2
SAME EVENTS; DIFFERENT FREEFLOAT FRAMES
139
• E vent B stands on the vertical time axis, directly above reference event 0. Therefore event B occurs at a later time than event 0. Event B lies neither to the right of the reference event nor to the left; its horizontal (space) location is zero. Therefore it occurs at the same place as the reference event 0 in the laboratory but later in time. • E vent A lies on the horizontal space axis, directly to the right of reference event 0. Therefore event A occurs at a different space location than event 0. It is neither above nor below event 0; its vertical (time) location is zero. Therefore it occurs at the same time as reference event 0 as observed in the laboratory. • E vent C rests above and to the right of the reference event. Standing higher than the reference event on the map, event C occurs later in time than 0 in this frame. Since it lies to the right, event C occurs at a positive space location with respect to event 0 in this frame. • E vent D reposes above and to the left of the reference event. It also occurs later in time than reference event 0 but at a negative space location with respect to event 0 as observed in the laboratory. Scatter other event points on the spacetime map. Each event point can represent an important happening. Then a single glance at the spacetime map gives us, in principle, a global picture of all significant events that have occurred along one line in space and as far back in time as we wish to look. The spacetime map puts all this history at our fingertips! In exploring history, w e may want to know which events occurred at the same time as others in the laboratory freefloat frame. Two events that occur at the same time have the same vertical (time) location on the spacetime map. A horizontal line drawn through one event point passes through all events simultaneous with that event in the given frame. In Figure 5 1, the dashed horizontal line shows that events B and D are simultaneous as observed in the laboratory frame, although they occur at different locations in space. Similarly, events 0 and A are simultaneous as observed in this frame. When we wish to “retell history,” we draw a sequence of horizontal lines above one another on the spacetime map. We mimic the advance of time by stepping in imagination from one horizontal line to the next horizontal line above it, noting which events occur at each time. Vertical lines on the spacetime map indicate which events occur at the same place along the single line in space. Events A and C in Figure 51 occur at the same space location as measured in the laboratory, but at different times as measured in this frame. Similarly, events 0 and B occur at the same place as one another in the laboratory.
5.2 SAME EVENTS; DIFFERENT FREEFLOAT FRAMES different frames: different points for an event on their spacetime maps, but same spacetime interval between two events Figure 51 demonstrates two great payoffs of the spacetime map: (1) It places space and time on an equal footing, thus recognizing a basic symmetry of nature. (2) It allows us to review at a single glance the whole history of events and motions that have occurred along the given line in space.
Horizontal line on spacetime diagram picks out events that a re simultaneous in this frame
140
Sam e events, difFerent frames: Different spacetim e maps
CHAPTERS
TREKKING THROUGH SPACETIME
We want to take advantage of a third payoff of the spacetime map; Plot the same events on two, three, or more spacetime maps based on two, three, or more different freefloat frames in uniform relative motion. Compare. In this way analyze the various space and time relations among these events as measured in difFerent frames. Why do this? In order to find out what is difFerent in the difFerent frames and what remains the same. Figure 53 shows three spacetime maps — for laboratory, rocket, and superrocket freefloat frames. The superrocket moves faster than the rocket with respect to the laboratory (but not faster than light!). On each of the three spacetime maps we plot the same two events: the events of emission E and reception f? of a light flash. These are the two events analyzed in Chapter 3 to derive the expression for the spacetime interval. As a reminder of the physical phenomena behind events E and R, refer to Figure 52. The light flash is emitted (event E) from a sparkplug attached to the reference clock of the first rocket. Take event E as the reference event, called event 0 in Figure 51. By prearrangement the sparkplug fires at the instant when both the rocket reference clock and the superrocket reference clock pass the laboratory reference clock. All three
LABORATORY PLOT
A 0
0 ®
000®
0 0 0 0 ®
0 ®
0 ®
© ® ® ® ® E R
ROCKET PLOT
A
AAAAAAAAA AAAAA AA
AAAAAAAAA AAAAAA AA
\1 SUPERROCKET PLOT FIGURE 52 (Figure 35 repeated). The flash p a th as recorded in three differentfram es, showing event E, emission of the flash, a n d event R, its reception a fte r reflection. Squares, circles, and triangles represent the latticework of recording clocks in laboratory, rocket, and superrocket frames, respec tively. The superrocket frame moves to the right with respect to the rocket, so that the event of reception, R, occurs to the left of the event of emission, E, as measured in the superrocket frame. The reflecting mirror is fixed in the rocket, hence appears to move from left to right in the laboratory and from right to left in the superrocket.
5.2 laboratory time
SAME EVENTS; DIFFERENT FREEFLOAT FRAMES FIGURE 53. Spacetime maps fo r three fram es, showing emission o f the reference flash a n d its reception a fte r reflection. The hyperbola drawn in each map satisfies the equa tion for the invariant interval (or proper time), which has the same value in all three frames: (interval)^ = (time)^ — (space)^.
LABORATORY SPACETIME MAP rocket time
SPACETIME MAP superrocket time
SUPERROCKET SPACETIME MAP
reference clocks are set to read zero at this reference event, whose event point is placed at the origin of all three spacetime maps. Now use the latticework of meter sticks and clocks in each freefloat frame (clocks pictured in Figure 52) to measure the position and time of every other event with respect to the reference event. In particular, record the position and time of the reception (event R) of the flash in each of the three frames. The reception of the light ray (event R) occurs at different locations and at different times as measured in the three frames. In the rocket the reception of the reflected flash occurs back at the reference clock (the zero of position) and 6 meters of time later, as
141
142
CHAPTERS
TREKKING THROUGH SPACETIME
seen in Figure 52 and more directly in Figure 53 (center): Same events, different frames: Different space and time coordinates
R ocket: (position of reception, event R) — 0 R ocket: (time of reception, event R) = 6 meters Emission and reception occur at the same place in the rocket frame. Therefore the rocket time, 6 meters, is just equal to the interval, or proper time, between these two events: R ocket (proper time)'
_ / time of y \ reception / R o ck et
R ocket position o f y
( reception / R ocket
_ / time of y _ (zero)^ — (6 meters)^ \ reception/ In the laboratory the reception event R occurs at a time greater than 6 meters, as can be seen from the expression for interval: L aboratory / time of y \ reception/
L aboratory / p o s .r io n o f y _ \ reception /
,
In this equation the square of 6 meters results from subtracting a positive quantity from the square of the laboratory time of reception. Therefore the laboratory time of reception itself must be greater than 6 meters: L aboratory: (position of reception, event R) = 8 meters L aboratory: (time of reception, event /?) = 10 meters
Same events, different frames: Sam e spacetim e interval
In the laboratory frame, reception appears to the right of the emission, as seen in Figure 52. Hence it is plotted to the right of the origin in the laboratory map (Figure 53, top). In the superrocket frame, moving faster than the rocket with respect to the laboratory, the event of reception appears to the left of the emission (Figure 52). Therefore the space separation is called negative and plotted to the left of the origin in the superrocket map (Figure 53, bottom). The time separation in the superrocket is greater than 6 meters, by the same argument used for the time of reception in the laboratory frame: S u p erro ck et S uperrocket / time of y _ / position o f y . ■(6 meters)^ \reception/ \ reception / In this equation, the space separation is a negative quantity. Nevertheless its square is a positive quantity. So the equation says that the square of 6 meters results from subtracting a positive quantity from the square of the superrocket time of reception. Therefore the superrocket time separation must also be greater than 6 meters: S uperrocket: (position of reception, event R) — — 20 meters S uperrocket: (time of reception, event R) = 20.88 meters
5.4
WORLDLINE
143
5.3 INVARIANT HYPERBOLA all observers agree: ''event point lies somewhere on this hyperbola" Different reception points marked R in different spacetime maps all refer to the same event. W hat do these different sepatations of the same event from the reference event have in common? They all satisfy invariance of the interval, reflected in the equation (time separation)^ — (space sepatation)^ = (interval)^
constant
Constant? Constant w ith respect to w hat?
— W ith respect to freefloat frame. Record different space and time measurements in different frames, but figure out from rhem always the same interval.
Curves drawn on the three maps conform to this equation. This kind of curve, in which the difference of two squares equals a constant, is called a hyperbola. Somewhere on this hyperbola is recorded the time and position of one and the same reception event as measured in every possible rocket and supettocket frame. Same reception event, different frames, all summarized in one hyperbola, the in v arian t hyperbola. Spacetime arrows in all three maps connect the same pair of events. They imply the identical invariant interval. They embody the same spacetime reality. In a deep sense these thtee attows on the page tepresent the same artow in spacetime. Spacetime maps of different observets show different projections — different petspectives — of the same atrow in spacetime. The same arrow? The same magnitude fo r the spacetime arrow pictured in a ll three maps of Figure 5  3 ? Then why do the three arrows have obviously different lengths in the three maps?
Because the paper picture of spacetime is a lie! The length of an arrow on a piece of paper is Euclidean, related to the sum of squares of the space separations of the endpoints in two perpendicular directions. Euclidean geometry works fine if what is being represented is flat space, for example the map of a township. But Euclidean geometry is the wrong geometry and betrays us when we try to lay out time along one direction on the page. Instead we need to use Lorentz geometry of spacetime. In Lorentz geometry, time must be combined with space through a difference of squares to find the correct magnitude of the resulting spacetime vector— the interval. That is why the arrows in the different spacetime maps of Figure 53 seem to be of different lengths. The reality that these lengths represent, however— rhe value of the interval between two events— is the same in all three spacetime maps.
5.4 WORLDLINE the moving particle traces out a line — its woHdline— on the spacetime diagram We describe the world by listing events and showing how they relate to one another. Until now we have focused on pairs of events and spacetime intervals between them. Now we turn to a whole chain of events, events that track the passage of a particle
Invariant hyperbola; Locus of sam e event in all rocket frames
144
String of event pearls: Worldline!
Worldline versus line on spacetime map
Exam ples of worldlines
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TREKKING THROUGH SPACETIME
through spacetime. Think of a speeding sparkplug that emits a spark every meter of time read on its own wrisrwatch. Each spark is an event; the collection of spark events forms a chain that threads through spacetime, like pearls. String the pearls together. The thread connecting the pearl events, tracing out the path of a particle through spacetime, has a wonderfully evocative name: w o rld lin e. The sparkplug travels through spacetime trailing its worldline behind it. The speeding sparkplug is only an example. Every particle has a worldline that connects events along its spacetime path, events such as collisions or nearcollisions (close calls) with other particles. Events — pearls in spacerime — exist independent of any reference frame we may choose to describe them. A worldline strings these event pearls together. The world line, too, exists independent of any reference frame. A particle traverses spacetime— follows a worldline — totally oblivious to our poor efforts to describe its motion using one or another freefloat frame. Yet we are accustomed to using a freefloat frame and its associated latticework of rods and clocks. One clock after another records its encounter with the particle. The worldline of the particle connects this chain of encounter events. We can draw this worldline of a particle on the spacetime map for this reference frame. Such worldlines are shown in Figure 55 and in later figures of this chapter. Strictly speaking, the line drawn on the spacetime map is not the worldline itself. It is an image of the worldline— a strand of ink printed on a piece of paper. When we use a highway map, we often refer to a line drawn on the paper as “the highway.” Yet is not the highway itself, but an image. Ordinarily this causes no confusion; no one tries to drive a car across a highway map! Similarly, we loosely refer to the line drawn on the spacetime map as the worldline, even though the worldline in spacetime stands above and beyond all our images of it. The worldline is seen in no way more clearly than through example. Particle 1 starts at the laboratory reference clock at zero time and moves to the right with constant speed (Figure 54). As particle 1 zooms along a line of laboratory latticework of clocks, each clock it encounters records the time at which the particle passes. Each clock record shows where the clock is located and the time at which particle 1 coincides with the clock. “Where and when” determines an event, the event o f coincidence of particle and recording clock. Afterwards the chief observer travels throughout the lattice of clocks, collecting the records of these coincidence events. She plots these events as points on her spacetime map. She then draws a line through event points in sequence — the worldline of particle 1 (Figure 55). Particle 1 moves w ith constant speed along a single direction in space. T he distance it covers is equal for each tick o f the laboratory clocks. T he w orldline o f particle 1 shows equal changes in space during equal lapses o f tim e by being straight on the spacetim e m ap.
Particle 2 moves to the right faster than particle 1 and so covers a greater distance in the same time lapse (Figure 54). Lattice clocks record their events of coincidence with particle 2, and the observer collecrs rhese records and plots the worldline of particle 2 on the same spacetime map (worldline shown in Figure 55). And so it goes: Particle 3 is a light flash and moves to the right in space (Figure 54) with maximum speed: one meter of distance per meter of time. W ith horizontal and vertical axes calibrated in meters, the lightflash worldline rises at an angle of 45 degrees (Figure 55). Particle 4 does not move at all in laboratory space; it rests quietly next to the laboratory reference clock. Like you sitting in your chair, it moves only along the time dimension; in the laboratory spacetime map its worldline is vertical (Figure 55). Particle 5 moves not to the right but to the left in space according to the laboratory observer (Figure 54), so its worldline angles up and leftward in the laboratory spacetime map (Figure 55). Each of these particles moves with constant speed, so each traces out a straight worldline. After 3 meters of time as measured in the laboratory frame, different
5.4
^ 1
5 < P 
WORLDLINE
*►2 
►3
■
4
145 FIGURE 54. Trajectories in space {not in spacetime!) o f p articles 1 through 3 durin g 3 meters o f time. Each particle starts at the reference clock (the square) at zero of time and moves with a constant velocity.
FIGURE 55. W orldlines in space time o f the particles shown in Fig ure 34, plotted fo r the laboratory fram e. Only the worldline for particle 1 includes a sample set of event points that are connected to make up the worldline.
particles have moved different distances from the starting point (Figure 54). In the laboratory spacetime map their space positions after 3 meters of time lie along the upper horizontal line o f sim ultaneity, shown dashed in Figure 55. Particle 4 is not the only object stationary in space. Every laboratory clock lies at rest in the laboratory frame; it moves neither right nor left as time passes. Nevertheless each laboratory clock moves forward in time, tracing out its own vertical worldline in the laboratory spacetime map. The background vertical lines in Figure 55 are worldlines of rhe row of laboratory clocks. W hat is the difference between a “p a th in space" a n d a “worldline in spacetime”?
The transcontinental airplane leaves a jet trail in still air. That trail is the plane’s path in space. Take a picture of that trail and you have a space map of the motion. From that space map alone you cannot tell how fast the jet is moving at this or that different point on its path. The space map is an incomplete record of the motion. The plane moves not only in space but also in time. Its beacon flashes. Plot those emissions as events on a spacetime map. This spacetime map has not only a horizontal space axis but also a vertical time axis. Now connect those event points with a worldline. The worldline gives a complete description of the motion of the jet as recorded in that frame. For example, from the worldline we can reckon the speed of the plane at every event along its path. Worldline gives spacetime map of the journey of the jet. Likewise a worldline drawn on a spacetime map images the journey of any particle through spacetime. A worldline is not a physical path, not a trajectory, not a line in space. An object at rest in your frame has, for you, no path at all through space; it stays always at one space point. Yet this stationary particle traces out a "vertical” worldline in your spacetime map (such as line 4 in Figure 55). A particle always has a worldline in spacetime. As you sit quietly in your chair reading this book, you glide through spacetime on your own unique worldline. Every stationary object lying near you also traces out a worldline, parallel to your own on your spacetime map.
Not all particles move with constant speed. When a patticle changes speed with respect to a freefloat frame, we know why: A force acts on it. Think of a train moving
Path in space versus worldline in spacetime
146
Changing speed means curving worldline
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on a straight sttetch of track. A force applied by the locomotive speeds up all the cars. Small speed: small distance covered in a given time lapse; worldline inclined slightly to the vertical in the spacetime map. Great speed: great distance covered in the same stretch of time; wotldline inclined at a gteater angle to the vertical in the spacetime map. Changing speed: changing distances covered in equal time periods; worldline that changes inclination as it ascends on the spacetime map— a curved worldline! Wait a minute! The train moves along a straight track. Yet you say its worldline is curved. Straight or curved? Make up your mind!
Straight in space does not necessarily mean straight in spacetime. Place your finger on the straight edge of a table near you. Now move your finger rapidly back and forth along this edge. Clearly this motion lies along a straight line. As your fingertip changes speed and direction, however, it travels different spans of distance in equal time periods. During a spell in which it is at rest on the table edge, your fingertip traces out a vertical portion of its worldline on the spacetime map. When it moves slowly to the right on the table, it traces out a worldline inclined slightly to the right of vertical on the map. When it moves rapidly to the left, your fingertip leaves a spacetime trail inclined significantly to the left on the map. Changing inclination of the worldline from point to point results in a curved worldline. Your finger moves straight in space but follows a curved worldline in spacetime!
Limit on worldline slope: speed of light
Figure 56 shows a curved worldline, not for a locomotive, but for a particle constrained to travel down the straight track of a linear accelerator. The particle starts at the reference clock at the time of the reference event (0 on the map). Initially the particle moves slowly to the right along the track. As time passes — advancing upward on the spacetime map — the particle speed increases to a large fraction of the speed of light. Then the particle slows down again, comes to rest at event Z, with a vertical tangent to its worldline at that event. Thereafter the particle accelerates to the left in space until it arrives at event P. W hat possible worldlines are available to the particle that has arrived at event P? A material particle must move at less than the speed of light. In other words, it travels less than one meter of distance in one meter of time. Its future worldline makes an “angle with the vertical’’ somewhere between plus 45 degrees and minus 45 degrees when space and time are measured in the same units and plotted to the same scale along horizontal and vettical axes on the graph. These limits of slope— which apply to every point on a particle worldline— are shown as dashed lines emerging from event P in Figure 56 (and also from event 0). limits on worldline slope
FIGURE 56. C urved laboratory w orldline o f a p article th a t changes speed as it moves back a n d fo rth along a straight line in space. Some possible worldlines avail able to the particle after event P.
5.5
LENGTH ALONG A PATH
The worldline gives a complete description of particle motion in spacetime. As drawn in the spacetime map for any frame, the worldline tells position and velocity of the particle at every event along its trail. In contrast, the trajectory or orbit or path shape of a particle in space does not give a complete description of the motion. To complete the description we need to know when the particle occupies each location on that trajectory. A worldline in a spacetime map automatically displays all of this information. The spacetime map provides a tool for retrospective study of events that have already taken place and have been reported to the freefloat observer who plots them. Once she plots these event points, this analyst can trace already plotted worldlines backward in time. She can examine at a single glance event points that may have occurred lightyears apart in space. These features of the spacetime map do not violate our experience that time moves only forward or that nothing moves faster than light. Everything plotted on a spacetime map is history; it can be scanned rapidly back and forth in the space dimension or the time dimension or both. The spacetime map supplies a comprehensive tool for recognizing patterns of events and teasing out laws of nature, but it is useless for influencing the events it represents.
147
Spacetime map displays only already detected events
5.5 LENGTH ALONG A PATH straight line has shortest length between two given points in spare Distance is a central idea in all applications of Euclidean geometry. For instance, using a flexible tape measure it is easy to quantify the total distance along a winding path that starts at one point (point 0 in Figure 57) and ends at another point (point B). Another way to measure distance along the curved path is to lay a series of short straight sticks end to end along the path. Provided the straight sticks are short enough to conform to the gently curving path, total distance along the path equals the sum of lengths of the sticks. The length of a short stick laid between any two nearby points on the path — for instance, points 3 and 4 in Figure 57 — can also be calculared using the northward separation and the eastward separation between the two ends of the stick as measured by a surveyor.
Measure length of curved path with tape measure . . .
. . . or with short straight sticks laid end to end along path
(length)^ = (northward separation)^ b (eastward separation)^ Distance is invariant for surveyors. Therefore the length of this stick is the same when calculated by any surveyor, even though the northward and eastward separations between two ends of the stick have different values, respectively, for different survey ors. The length of another stick laid elsewhere along the path is also agreed on by all surveyors despite their use of different northward direcrions. Therefore the sum of the lengths of all short sticks laid along the path has the same value for all surveyors. This sum equals the value of the total length of the path, on which all surveyors agree. And this total length is just the length measured using the flexible tape. It is possible to proceed from 0 to B along quite another path — for example along straight line OB in Figure 57. The length of this alternative path is evidently different from that of the original curved path. This feature of Euclidean geometry is so well known as to occasion hardly any comment and certainly no surprise: In Euclidean geometry a curved path between two specified points is longer than a straight path between them. The existence of this difference of length between two paths violates no law. No one would claim that a tape measure fails to perform properly when laid along a curved path.
All surveyors agree on length of path
148
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1211/2
FIGURE 57. Length along a w inding p a th startin g a t the town square. Notice that the total length along the winding path from point O to point B is greater than the length along the straight northward axis from O /o B.
Straight path in space has shortest length
Among all possible paths between two points in space, the straightline path is unique. All surveyors agree that this path has the shortest length. When we speak of “the distance between two points,” we ordinarily mean the length of this straight path.
5.6 WRISTWATCH TIME ALONG A WORLDLINE straight worldline has longest proper time between two given events in spaeetime
Measure proper time along curved worldline with wristwatch . . .
A curved path in Euclidean space is determined by laying down a flexible tape measure and recording distance along the path’s length. A curved worldline in Lorentz spacetime is measured by carrying a wristwatch along the worldline and recording what it shows for the elapsed time. The summed spacetime interval — the proper time read directly on the wristwatch — measures the worldline in Lorentz geometry in the same way that distance measures path length in Euclidean geometry. A particle moves along the worldline in Figure 58. This particle carries a wrist watch and a sparkplug; the sparkplug fires every meter of time (1, 2 , 3 , 4 , . . . ) as read off the particle’s wristwatch. The laboratory observer notes which of his clocks the
5.6
WRISTWATCH TIME ALONG A WORLDLINE
14 9
WORLDLINE IN SPACETIME FIGURE 58. Proper time along a curved worldline. Notice that the total proper time along the curved worldline from event O to event B is smaller than the proper time along the straight line from O to B.
traveling particle is near every time the sparkplug fires. He plots that location and that lattice clock time on his spacetime map, tracing out the worldline of the particle. He numbers spark points sequentially on the resulting worldline, as shown in Figure 58, knowing that these numbers register meters of time recorded on the moving wristwatch. Consider the spacetime interval between two sequential numbered flashes of the sparkplug, for instance those marked 3 and 4 in the figure. In the laboratory frame these two sparks are separated by a difference in position and also by a difference in time (the time between them). The squared interval— the proper time squared — between the sparks is given by the familiar spacetime relation: (proper time)^ = (difference in time)^ — (difference in position)^ W hat about the proper time between sparks 3 and 4 calculated from measurements made in the sparkplug frame? In this frame, both sparks occur at the same place, namely at the position of the sparkplug. The difference in position between the sparks equals zero in this frame. As a result, the time difference in the sparkplug frame — the “wristwatch time” — is equal to the proper time between these two events: (proper time)^ “ (1 meter)^
(zero)^ ~ (1 meter)^
[recorded on traveling wristwatch]
This analysis assumes that sparks are close together in both space and time. For sparks close enough together, the velocity of the emitting particle does not change much from one spark to the next; the particle velocity is effectively constant between sparks; the piece of curved wbrldline can be replaced with a short straight segment. Along this straight segment the particle acts like a freefloat rocket. The proper time is
. . . or as sum of intervals between adjacent events
150
All observers a g ree on proper time along worldline
Straight worldline has longest proper time
Principle of M aximal Aging predicts motion of free particle
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invariant in freefloat rocket and freefloat laboratory frames. Thus the laboratory observer can compute the value of the proper time between events 3 and 4 and predict the time lapse — one meter— on the traveling wristwatch, which measures the proper time directly. Elsewhere along the worldline the particle moves with a different speed. Neverthe less the proper time between each consecutive pair of sparks must also be independent of the freefloat frame in which that interval is reckoned. For sparks close enough rogether, this proper time equals the time read directly on the wristwatch. All observers agree on the proper time between every sequential pair of sparks emitted by the sparkplug. Therefore the sum of of all individual proper times has the same value for all observers. This sum equals the value of the total proper time, on which all freefloat observers agree. And this total proper time is just the wristwatch time measured by the traveling sparkplug. In brief, proper time is the time registered in a rocket by its own clock, or by a person through her own wristwatch or her own aging. Like aging, proper time is cumulative. To obtain total proper time racked up along a worldline between some marked starting event and a designated final event, we first divide up the worldline into segments so short that each is essentially “straight” or “freefloat.” For each segment we determine the interval, that is, the lapse of proper time, the measurement of aging experienced on that segment. Then we add up the aging, the proper time for each segment, to get total aging, total wristwatch time, total lapse of proper time. An automobile may travel the most complicated route over an entire continent, but the odometer adds it all up and gives a wellunderstood number. The traveler through the greater world of spacetime, no matter how many changes of speed or direction she undergoes, has the equivalent of the odometer with her on her journey. It is her wristwatch and her body— her aging. Your own wristwatch and your biological clock automatically add up the bits of proper time traced out on all successive segments of your worldline. It is possible to proceed from event 0 to event B along quite another worldline— for example, along the straight worldline OB in Figures 58 and 59 (bottom). The proper time from Ot oB along this new worldline can be measured directly by a flashing clock that follows this new worldline. It can also be calculated from records of flashes emitted by the clock as recorded in any laboratory or rocket frame. Total proper time along this alternative worldline has a different value than total proper time along the original worldline. In Lorentz geometry a curved worldline berween two specified events is shorter than the direct worldline between them — shorter in terms of total proper time, total wristwatch time, total aging. T otal proper tim e, the aging along any given worldline, straight or curved, is an invariant: it has the same value as reckoned by observers in all overlapping freefloat frames. This value correctly predicts elapsed time recorded directly on the wristwatch of the particle that travels this worldline. It correctly predicts the aging of a person or a mouse that travels this worldline. A different worldline between the same two events typically leads to a different value of aging — a new value also agreed on by all freefloat observers: Aging is maximal along the straight worldline between two events. This uniqueness of the straight worldline is also a matter of complete agree ment among all freefloat observers. All agree also on this: The straight worldline is the one actually followed by a free particle. Conclusion: Between two fixed events, a free particle follows the worldline of maximal aging. This more general prediction of the worldline of a free particle is called the P rin cip le o f M axim al Aging. It is true not only for “straight” particle worldlines in the limited regions of spacetime described by special relativiry but also, with minor modification, for the motion of free particles in wider spacetime regions in the vicinity of gravitating mass. The Principle of Maximal Aging provides one bridge between special relativity and general relativity. The stark contrast between Euclidean geometry and Lorentz geometry is shown in Figure 59. In Euclidean geometry distance between nearby points along a curved
5.6
WRISTWATCH TIME ALONG A WORLDLINE
151
FIGURE 59. P ath in space: In Euclidean geometry the curved path has greater length. W orldline in spacetime: In Lorentz geometry the curved worldline is traversed in shorter proper time.
path is always equal to or greater than the northward separation between those two points. In contrast, proper time between nearby events along a curved worldline is always equal to or less than the corresponding time along the direct worldline as measured in that frame.
Stark contrast between Euclidean and Lorentz geom etries
152
Proper times com pare worldlines
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The difference of proper time between two alternative worldlines in spacetime violates no law, just as the difference of length between two alternative paths in space violates no law. There is nothing wrong with a wristwatch that reads different proper times when carried along different worldlines between events 0 and B in spacetime, just as there is nothing wrong with a tape measure that records different lengths for different paths between points 0 and B in space. In both cases the measuring device is simply giving evidence of the appropriate geometry: Euclidean geometry for space, Lorentz geometry for spacetime. In brief, the determination of cumulative interval, proper time, wristwatch time, aging along a worldline between two events is a fundamental method of comparing different worldlines that connect the same two events. Among all possible worldlines between two events, the straight worldline is unique. All observers agree that this worldline is straight and has the longest proper time — greatest aging — of any possible worldline connecting these events.
5.7 KINKED WORLDLINE kink in the worldline decreases aging along that worldline
Accelerationproof clocks
Simplify: W orldlines with straight segments
The change in slope of the worldline from event to event in Figures 58 and 59 (bottom) means that the clock being carried along this worldline changes velocity: It accelerates. Different clocks behave differently when accelerated. Typically a clock can withstand a great acceleration only when it is small and compact. A pendulum clock is not an accurate timepiece when carried by car through stopandgo traffic; a wrist watch is fine. A wristwatch is destroyed by being slammed against a wall; a radioactive nucleus is fine. Typically, the smaller the clock, the more acceleration it can withstand and still register properly, and the sharper can be the curves and kinks on its worldline. In all figures like Figures 58 and 59 (bottom), we assume the ideal limit of small (accelerationprooO clocks. W e are now free to analyze a motion in which particle and clock are subject to a great acceleration. In particular, consider the simple special case of the worldline of Figure 58. That worldline gradually changes slope as the particle speeds up and slows down. Now make the period of speeding up shorter and shorter (great driving force!); also make the period of slowing down shorter and shorter. In this way come eventually to the limiting case in which episodes of acceletation and deceleration — curved portions of the worldline — are too short even to show up on the scale of the spacetime map (worldline OQB in Figure 510). In this simple limiting case the whole history of motion is specified by (1) initial event 0, (2) final event B, and (3) turnaround event Q, halfway in time between 0 and B. In this case it is particularly easy to see how the lapse of proper time between 0 and B depends on the location of the halfway event— and thus to compare three worldlines, OPB, OQB, and ORB. Path OPB is the worldline of a particle that does not move in space; it stays next to the referenceframe clock. Proper time from 0 to B by way of P is evidently equal to time as measured in the freefloat frame of this reference clock: (total proper time along OPB) = 1 0 meters of time In contrast, on the way from 0 to B via R, for each segment the space separation equals the time separation, so the proper time has the value zero:
5.7
KINKED WORLDLINE
153
FIGURE 510. Three alternative worldlines connecting events O a n d B. The sharp changes of velocity at events Q and R have been drawn for the ideal limit of small clocks that tol erate great acceleration. The boldface number j is the proper time along the segment O Q , reckonedfrom the differ ence between the squared time separa tion and the squared space separation:
= 5"  4^.
(proper time along leg ORY — = = (total proper time along ORB) = =
(time)^ — (space)^ (5 meters)^ “ (5 meters)^ 0 2 X (proper time along OR) 0
Z ero p ro p er time for light
As far as we know, only three things can travel 5 meters of distance in 5 meters of time; light (photons), neutrinos, and gravitons (see Box 81). No material clock can travel at light speed. Therefore the worldline ORB is not actually attainable by a material particle. However, it can be approached arbitrarily closely. One can find a speed sufficiently close to light speed — and yet less than light speed — so that a trip with this speed first one way then the other will bring an ideal clock back to the reference clock with a lapse of proper time that is as short as one pleases. In the same way we can, in principle, go to the star Canopus and back in as short a roundtrip rocket time as we choose (Section 4.8). As distinguished from the limiting case ORB, worldline OQB demands an amount of proper time that is greater than zero but still less than the 10 meters of proper time along the direct worldline OPB: (proper time along leg OQ)' = = = =
(5 meters)^ — (4 meters)^ 25 (meters)^ — 16 (meters)^ 9 (meters)^ (3 meters)^
(proper time along leg OQ) = 3 meters and (total proper time along both legs OQB) — 2 X (proper time along OQ) = 6 meters This is less proper time than (proper time along OPB) = 1 0 meters thar characterized the “direct” worldline OPB. Our trip to Canopus and back described in Chapter 4 follows a worldline similar to OQB.
Reduced proper time along kinked worldline
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SAMPLE PROBLEM
51
MORE IS LESS In the spacetime map shown, time and space are measured in years. A table shows space and time locations of numbered events in this frame.
SPACE AND TIME LOCATIONS OF EVENTS Space (years) Event Event Event Event
1 2 3 4
Time (years)
1  0 .5
2
 space — ►
T wo alternative worldlines between events I and 4
One traveler moves along the solid straight worldline segments from event 1 to events 2 ,3 , and 4. Calculate the time increase on her clock between event 1 and event 2; between event 2 and event 3; between event 3 and event 4. Calculate total proper time — her aging — along worldline 1, 2, 3, 4. b.
Another traveler, her twin brother, moves along the straight dotted worldline from event 1 directly to event 4. Calculate the time increase on his clock along the direct worldline 1, 4.
c.
Which twin (solidline traveler or dottedline traveler) is younger when they rejoin at event 4?
SOLUTION From the table next to the map, space separation between events 1 and 2 equals 0. Time separation equals 1 year. Therefore the interval is reckoned from (interval)^ = P — 0 ^= 1 ^. Thus the proper time lapse on a clock carried between events 1 and 2 equals 1 year. Space separation between event 2 and event 3 equals 1 — (—0.5) = 1.5 lightyears. Time separation equals 2 years. Therefore the square of the interval is 2^ — (1.5)^ = 4 — 2.25 = 1.75 (years)^ and the advance of proper time equals the square root of this, ot 1.32 years. Between event 3 and event 4 space separation equals 2.5 lightyears and time separation 3 years. The square of the interval has the value 3^ — (2.5)^ = 9 — 6.25 = 2.75 (years)^ and proper time between these two events equals the square root of this, or 1.66 years.
5.8
STRETCH FACTOR
1 55
Total proper time— aging — along worldline 1, 2, 3, 4 equals the sum of proper times along individual segments: 1 + 1.32 + 1.66 = 3.98 years. b.
Space separation between events 1 and 4 equals 1 lightyear. Time separation is 6 years. The squared interval between them equals 6^ — 1^ = 36 — 1 = 35 (years)^. A traveler who moves along the direct worldline from event 1 to event 4 records a span of proper time equal to the square root of this value, or 5.92 years.
c.
The brother who moves along straight worldline 1, 4 ages 5.92 years during the trip. The sister who moves along segmented worldline 1, 2, 3, 4 ages less: 3.98 years. As always in Lorentz geometry, the ditect worldline (shown dotted) is longer— that is, it has more elapsed proper time, greater aging — than the indirect worldline (shown solid).
5.8 STRETCH FACTOR ratio off frameclock time to wristwatch time A speeding beacon emits two flashes, F and S, in quick succession. These two flashes, as recorded in the rocket that carties the beacon, occur with a 6meter separation in time but a zero separation in space. Zero space separation? Then 6 metets is the value of the interval, the proper time, the wristwatch time between F and S. As registered in the laboratory, in contrast, the second flash S occurs 10 meters of time later than the first flash F. The ratio between this frame time, 10 meters, and the proper time, 6 meters, between the two events we call the time stretch factor, or simply stretch factor. Some authors use the lowercase Greek letter gamma, y, for the stretch factor, as we do occasionally. We will also use the Greek letter tau, T , for proper time. The same two events register in the superrocket frame that overtakes and passes the beacon — register with a separation in time of 20.88 meters. In this frame, the time stretch factor between the two events is (20.88)/6 = 3.48. In the beacon frame the stretch factor is unity: 6 /6 = 1. Why? Because in this beacon frame flashes F and S occur at the same place, so beaconframe clocks record the proper time directly. This proper time is less than the time between the two flashes as measuted in either laboratory or superrocket frame. The larger value of time observed in laboratory and superrocket frames shows up in Figure 511 (center and right). Among all conceiv able frames, the separation in time between the two flashes evidently takes on its minimum value in the beacon frame itself, the value of the proper time T. Hold it! In Sections 3.6 an d 5 ■ 7 you insisted th at the time along a straight worldline is a MAXIM UM . Now you show us a straight worldline along which the time is — you say — a M INIM UM . Maximum or minimum? Please make up your mind!
— The worldline taken by the beacon wristwatch from F to S is straight. It is straight ^ whether mapped in the beacon frame itself or in the rocket or superrocket frame. The beacon racks up 6 meters of proper time regardless of the frame in which we reckon this time. When we turn from this wristwatch time to what different
Different reference frames: different times between two events
Time lapse minimum for frame in which events occur at sam e place
156
CHAPTER 5
TREKKING THROUGH SPACETIME rocket time
laboratory time
LABORATORY
ROCKET
SUPERROCKET
f, = 10 meters
f', = 6 meters
t", = 20.88 meters
FIGURE 511. Spacetime maps o f Figure 53, modified to show the worldline o f the speeding beacon (heavy dashed line) and the segment o f this line between emission F o f thefirst flash and the secondflash S (solid section o f worldline). Emission F is taken as the zero ofspace and time. Time tj of the second emission S is different as recorded in differentframes. The shortest time is recorded in thatframe in which the two events occur at the same place— in this case the rocket frame.
freefloat frames show for the separation in map time (latticework time, frame time) between the two flashes, however, the record displays a minimal value for that separation in time only in the beacon frame itself In contrast. Figure 512 (Figure 510 in simplified form) shows two different worldlines that join events 0 and B mapped in the same reference frame. In this case we compare two different proper times: a proper rime of 10 meters racked up by a wristwatch carried along the direct course from 0 to B, and a proper time of 6 meters recorded by the wristwatch carried along on the kinked worldline OQB. In every such comparison made in the context of flat spacetime, the direct worldline displays maximum proper time. Caution: Conditions can be different in curved spacetime (Chapter 9). In summary, two points come to the fore in these comparisons of the time between two events. (1) Are we comparing map time (frame time, latticework time) between those two events, pure and simple, free of any talk about any wotldline that might connect those events? Then separation in time between those events is least as mapped in the freefloat frame that shows them happening at the same place. (2) Or are we directing our arrention to a worldline that connects the two events? More specifically, to the time racked up by a wrisrwarch toted along that worldline? Then we have to ask, is that worldline straight? Then it registers maximal passage of proper time. Or does it have a kink? Then the proper time racked up is not maximal.
When we find ourselves in a freefloat frame and see a beacon zooming past in a straight line with speed v, how much is the factor by which our frameclock time is stretched relative to the beacon wristwatch time? Answer: The stretch factor is (stretch factor) = y ■
FIGURE 512. Figure 510 stripped down
to emphasize total proper time (wristwatch time), pninted boldface along two different worldlines between the same two events O and B in a given referenceframe. Among all possible worldlines connecting events O and B, the straight worldline registers maxi mal lapse of proper time.
1
(1 r.2)i/2
(51)
How can we derive this famous formula? If you do not cover up the following lines and derive this answer on your own, here is the reasoning: Start with measurements in the laboratory frame. We know that for this rocket (advance in proper time)^ = (advance in lab time)^ ~ (lab distance covered)^ However, we want to compare lapses in laboratory time and proper time; laboratory distance covered is not of interest. For the laboratory observer the proper clock moving
5.8
STRETCH FACTOR
157
along a straight worldline covers the distance between the two events in the time between the events. Therefore this distance and time are related by particle speed: Stretch factor = frame time/proper time
(lab distance covered) = (speed) X (advance in lab time) Substitute this expression into the equation for proper time: (proper time)^ = (lab time)^ ~ (speed)^ X (lab time)^ = (lab time)^ [1 ~ (speed)^} This leads to an expression for the square of the stretch factor: (lab time)^ (proper time)^
. . _ (stretch factor)^
1 1 — (speed)^
1 1
where we use the symbol v — v ^ ^ /c for speed. The equation for the stretch factor becomes (stretch factor) — y —
1 (1  «^2) 1/2
Stretch factor derived (5  1 )
The stretch factor has the value unity when p = 0. For all other values of v the stretch factor is greater than unity. For very high relative speeds, speeds close to th a t o f light
(v —* 1), the value of the stretch factor increases without limit. The value of the stretch factor does not depend on the direction of motion of the rocket that moves from first event to second event: The speed is squared in equation (51), so any negative sign is lost. The stretch factor is the ratio of frame time to proper time between events, where speed ( = p) is the steady speed necessary for the proper clock to pass along a straight worldline from one event to the other in that frame. The stretch factor also describes the Lorentz contraction, the measured shortening of a moving object along its direction of motion when the observer determines the distance between the two ends a t the same time. For example, suppose you travel at speed v between Earth and a star that lies distance L away as measured in the Earth frame. Your trip takes time t = Ljv in the Earthlinked frame. Proper time T— your wristwatch tim e— is smaller than this by the stretch factor: T = L /\v X (stretch faaor)} = (L/v) (1 — iqow think of a very long rod that reaches from Earth to star and is at rest in the Earth frame. How long is that rod in your rocket frame? In your frame the rod is moving at speed V. One end of the rod, at the position of Earth, passes at speed v. A time T later in your frame the other end of the rod arrives — along with the star — also moving at speed V according to your rocket measurements. From these data you calculate that the length of the rod in your rocket frame— call it L' — is equal to L' = pT = viLjp) (1 — r^)’''^ = L (1 — r>^)Y2 'This is a valid measure of length. By this method the rod is measured to be shorter.
Finally, the stretch factor is often used as an alternative measure of particle speed: A particle moves with a speed such that the stretch factor is 10. This statement assumes that the particle is moving with constant speed, so that the separation between any pair of events on the particle worldline has the same stretch factor as the separation between any other pair. This way of describing particle speed can be both convenient and powerful. We will see (Chapter 7) that the total energy of a particle is proportional to the stretch factor.
Lorentzcontraction by sam e “ stretch" factor
Stretch factor as a measure of speed
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TREKKING THROUGH SPACETIME
SAMPLE PROBLEM
52
R O U N D TRIP O B S E R V E D IN DI F F E RE NT FRAME Return to the alternative worldlines between events 0 and B, shown in Figure 510 and the spacetime maps in this sample problem. Measure these worldlines from a rocket frame that moves outward with the particle from 0 to Q and keeps on going forever at the same constant velocity. Show that an observer in this outwardrocket frame predicts the same proper time— wristwatch
time — for worldline OQB as that predicted in the laboratory frame. Similarly show that this outwardrocketframe observer predicts the same proper time along the direct worldline OPB as does the laboratory observer. Finally, show that both observers predict the elapsed wristwatch time along OQB to be less than along OPB.
SOLUTION Here are laboratory and rocket spacetime maps for these round trips, simplified and drawn to reduced scale.
t time 1 V
10
p
\
q
A c5 space
space ■
— >■
lABORATORY SPACETIME MAP
OUTWARDROCKET SPACETIME MAP
Laboratory a n d outwardrocket spacetime maps, each showing altern ativ e worldlines (direct
OPB a n d indirect OQB) between events O a n d B. Laboratory spacetime map: Figure 3W , redrawn to a different scale. Proper times are shown on the laboratory spacetime map. Outwardrocket spacetime map: The rocket in which the outgoing particle is at rest. Portions of two invariant hyperbolas show how events Q and B transform. The direct worldline OPB has longer total proper time— greater aging— as computed using measurements from either frame.
F ind x ' q a n d t ' q. First compute space and time locations of events Q and B in the outgoing rocket frame — righthand map. (Event 0 is the reference event, x = 0 and t = 0 in all frames by convention.) We choose the rocket frame so that the worldline segment OQ lies vertical and the outbound rocket does not move in this frame. As a result, event Q occurs at rocket space origin: x ' q — 0. (Primes refer to measurements in the outwardrocket frame.) The rocket time Cq for this event is just the wristwatch time between 0 and Q, because the wristwatch is at rest in this frame: Cq = 3 meters. In summary, using a prime for rocket measurements: X q= 0 3 meters
5.8
STRETCH FACTOR
159
F in d x ' b a n d In the laboratory frame, the particle moves to the right from event 0 to event Q, covering 4 meters of distance in 5 meters of time. Therefore its speed is the fraction t' = 4 /5 = 0.8 of light speed. As measured in the rocket frame, the laboratory frame moves to the left with speed = 0.8 , by symmetry. Use equation (51) with v = 0.8 to compute the value of the stretch factor;
1
1
1
1
10
[1  (0.8)2}*/2
[1  0 .6 4 ] V 2
[0.36} i/2
0.6
T
1
This equals the ratio of rocket time period t'g to proper time along the direct path OPB. Hence elapsed rocket time t'g = (5 /3 ) X 10 meters = 50/3 meters of time. In this time, the laboratory moves to the left in the rocket frame by rhe distance x'g = — vt' g = ■“ (4 /5 ) (5 0 /3 ) = ~ 200/15 = ~ 4 0 / 3 meters. In summary for outgoing rocket: 40 —— meters ■
13~ meters 3
50 Imeters — 16— meters of time 3 3 Events Q and B ate plotted on the rocket spacetime map. C o m pare W ristw atch Tim es: Now compute the total proper time — wristwatch time, aging — along alternative wotldlines OPB and OQB using rocket measurements. Direct wotldline OB has proper time Tgg given by the regular expression for interval:
2500
1600
900 =  = 1 0 0 (meters)^
whence Toe = 1 0 meters computed from rocket measutements. This is the same value as computed in the laboratory frame (in which proper time equals laboratory time, since laboratory separation in space is zero). Worldline OQB has two segments. On the first segment, OQ, proper time lapse is just equal to the rocket time span, 3 meters, since the space separation equals zero in the rocket ftame. For the second segment of this wotldline, QB, we need to compute elapsed time in this ftame: ^ QB
50 50 9 41 , = 3    — meters 3 3 3 3
^
40 —— meters Therefore, I^qb)^
qb)^
1681
{x 1600
81 ■— = 9 (meters)^ 9
whence Tqb = 3 meters. So the total increase in proper time— the total aging— along worldline OQB sums to 3 + 3 = 6 meters as reckoned from outwardrocket measure ments. This is the same as figured from laboratory measurements.
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TREKKING THROUGH SPACETIME
How can these weird results be true? In our everyday lives why don’t we have to take account of clocks that record different elapsed times between events, and rods that we measure to be contracted as they speed by us?
In answer, consider two events that occur at the same place in our frame. The proper clock moving in spacetime between these two events has speed zero for us. In this case the stretch factor has the value unity; the frame clock is the proper clock. The same is approximately true for events that are much closer together in space (mea sured in meters) than the time between them (also measured in meters). In these cases the proper clock moving between them has speed v — measured in meters/ meter— that is very much less than unity. That is, the proper clock moves very much slower than the speed of light. For such slow speeds, the stretch factor has a value that approaches unity; the proper clock records very nearly the same time lapse between two events as frame clocks. This is the situation for all motions on earth that we can follow by eye. For all such “ordinaryspeed” motions, moving clocks and stationary clocks record essentially the same time lapses. This is the assumption of Newtonian mechanics: “Absolute, true, and mathematical time, of itself, and from its own nature, flows equably without relation to anything external . . .” A similar argument leads to the conclusion that Lorentz contraction is negligible for objects moving at everyday speeds. Newton’s mechanics— with its unique measured time between events and its unique measured length for an object whether or not it moves— gives correct results for objects moving at everyday speeds. In contrast, for particle speeds approaching light speed (approaching one meter of distance traveled per meter of elapsed time in the laboratory frame), the denomina tor on the right of equation (51) approaches zero and the stretch factor increases without limit. Increased without limit, also, is the laboratory time between ticks of the zooming particle’s wristwatch. This is the case for highspeed particles in accelerators and for cosmic rays, very highenergy particles (mostly protons) that continually pour into our atmosphere from space. Newton’s mechanics gives results wildly in error when applied to these particles and theit interactions; the laws of relativistic mechanics must be used. More than one cosmic ray has been detected (indirectly by the resulting shower of particles in the atmosphere) moving so fast that it could cross our galaxy in 30 seconds as recorded on its own wristwatch. During this trip a thousand centuries pass as recorded by clocks on Earth! (See Exercise 77.)
5.9 TOURING SPACETIME WITHOUT A REFERENCE FRAME all you need is worldlines and events
Events and worldlines exist independent of any reference frame
An explosion is an explosion. Your birth was your birth. An event is an event. Every event has a concreteness, an existence, a reality independent of any reference frame. So, too, does a worldline that connects the trail of event points left by a highspeed sparkplug that flashes as it streaks along. Events mark worldlines, independent of any reference frame. Worldlines also locate events. The intersection of two worldlines locates an event as clearly and sharply as the intersection of two straws specifies the place of a dust speck in agreat barn full of hay (Eigure 513). To say that an event marks a collision between two particles is identification enough. The worldlines of those two particles are rooted in the past and stretch out into the future. They have a rich texture of connections with nearby worldlines. The nearby worldlines in turn are linked in a hundred ways with
5.9
TOURING SPACETIME WITHOUT A REFERENCE FRAME
161
FIGURE 513. The crossing of straws in a bam full of bay is a symbol for the worldlines that fill up spacetime. By their crossings and jogs, these worldlines mark events with a uniqueness beyond all need of reference frames. Straight worldlines track particles with mass; wiggly worldlines trace photons. Typical events symbolized in the map (black dots) from left to right: absorption of a photon; reemission of a photon; collision between a particle and a particle; collision between a photon and another particle; another collision between a photon and a particle; explosion of a firecracker; collision ofa particle from outside with one of the fragments of that firecracker.
worldlines more remote. How then does one tell the location of an event? Tell first what worldlines thread the event. Next follow each of these worldlines. Name additional events that they encounter. These events pick out further worldlines. Eventually the whole barn of hay is cataloged. Each event is named. One can find one’s way as surely to a given intersection as the London dweller can pick her path to the meeting of St. James’s Street and Piccadilly. No numbers giving space and time location of an event in a given reference frame. No reference frame at all! Most streets in Japan have no names and most houses no numbers. Yet mail is delivered just the same. Each house is named after its senior occupant, and everyone knows how the streets interconnect these named houses. Now print the map of Japanese streets on a rubber sheet and stretch the sheet this way and that. The postal carrier is not fooled. Each house has its unique name and the same interconnections with neighbor houses as on the unstretched map. So dispense with all maps! Replace them with a catalog or directory that lists each house by name, notes streets passing the house, and tabulates the distance to each neighboring house along the streets. Similarly, the visual pattern of event dots on a spacetime map (spacetime diagram) and the apparent lengths of worldlines that connect them depend on the reference frame from which they are observed (for example, compare alternative spacetime maps of the same worldline shown in the figure in Sample Problem 52). However, each named event is the same for every observer; the event of your birth is unique to you and to everyone connected with you. Moreover, the segment of a worldline that
Locate house at intersection of streets
Locate event at intersection of worldlines
162
Events and worldlines alone can describe Nature
CHAPTERS
TREKKING THROUGH SPACETIME
connects one event with the next has a unique magnitude— the interval or proper time — also the same for every observer. Therefore dispense with reference frames altogether! Replace them with a catalog or directory that lists each event by name, notes each worldline that threads the event, and tabulates the interval that connects the event with the next event along each worldline. With this directory in hand we can say precisely how all events are interconnected with each other and which events caused which other evenrs. That is the essence of science; in principle we need no reference frames. But reference frames are convenient. We are accustomed to them. Most of us prefer to live on named streets with numbered houses. Similarly, most of us speak easily of space separations between events and time separations between the same events as if space and time separations were unconnected. In this way we enjoy the concreteness of using our latticework of rods and clocks while suffering the provinciality of a single reference frame. So be it! Nevertheless, with worldlines Nature gives us power to relate events — to do science— without reference frames at all.
5.10 SUMMARY straighter worldline? greater aging! Events? Yes. Each event endowed with its own location in that great fabric we call spacetime? Yes. But time? No point in all that fabric displays any trace of anything we can identify with any such thing as the “time” of that event. Label that event with a “time” anyway? Sure. No one can stop us. Moreover, such labeling often proves quite useful. But it is our labeling! A different reference frame, a different wrisrwatch brought to that event along a different worldline yields a different time label for that event. For our own convenience, then, we plot events on a spacetim e m ap (spacetim e diagram ) for a particular freefloat frame and its latticework of rods and clocks. This map can be printed on the page of a book if events are limited to one line in space. Distance along this line is plotted horizontally on the spacetime map, with time of the event plotted vertically (Section 5.1). The time and space values of an event are measured with respect to a common referen ce event, plotted at the origin of the spacetime map. The invariance of the interval: (interval)^ = (time)^ — (distance)^ between an event and the reference event corresponds to the equation of a hy p erbola, the same hyperbola as plotted on the spacetime map of every overlapping freefloat frame. The event point lies somewhere on the same in v arian t h y p erb o la as plotted on every one of these spacetime maps (Sections 5.2 and 5.3). Billions of events sparkle like sand grains scattered over the spacetime map. A given event is unconnected to most other events on the map. Here we pay attention to particular strings of events that are connected. The w o rld lin e of a particle connects in sequence events that occur at the particle (Section 5.4). The “ length” of a worldline between an initial and a final event is the elapsed time measured on a clock carried along the worldline between the two events (Section 5.6). This is called the proper time, wrisrwatch time, or aging along this worldline. The lapse of proper time is given the symbol T, in contrast to the symbol t for the frame time read on the latticework clocks in a given freefloat frame. Carry a wrisrwatch (or grow old!) along a worldline: This is one way to measure the total proper time along it from some initial event (such as the birth of a person or a particle) to some final event (such as death of a person or annihilation of a particle). This method is direct, experimental, simple. A second method? Calculate the interval between each pair of adjacent events that make up the worldline, and then add up all
EXERCISE 51
MORE IS LESS
163
these intervals, assuming that each tiny segment is short enough to be considered straight. This method seems more bothersome and detailed, but it can be carried out by the observer in any freefloat frame. All such observers will agree wirh one another— and with the clockcarrier — on the value of the total proper time from the initial event to the final event on the worldline (Section 5.6). Among all possible worldlines between two given events, the straight line is the worldline of m axim al aging. This is the acmal worldline followed by a free particle that travels from one of these two events to the other (Section 5.6). As measured in a given freefloat ftame, the stre tc h fa cto r = 1 / ( 1 — equals the ratio of elapsed frame rime t to elapsed proper time T along a segment of worldline in which the particle moves with speed v in that frame. The stretch factor is also the Lorentz contraction factor (Section 5.8): Locate, at the same time, the front and back ends of an object moving in a given freefloat frame. These end locations will be (1 — t/2)V2 as far apart in that frame as they are in a frame in which the object is at rest. Worldlines connect events. Like events, they exist independent of any reference frame. In principle, worldlines allow us to relate events to one another — to do science— without using reference frames at all (Section 5.9).
REFERENCES Newton quotation toward the end of Section 5.8: Sir Isaac Newton, Mathemati cal Principles of Natural Philosophy and His System of the World (Philosophiae Naturalis Principia Mathematica), Joseph Streater, London, July 5, 1686; translated from Latin — the scholarly language of Newton’s time— by Andrew Motte in 1729, revised and edited by Florian Cajori and published in two paperback volumes (University of California Press, Berkeley, 1962). Section 5.9 uses slightly modified passages from Charles W. Misner, Kip S. Thorne, and John Archibald Wheeler, Gravitation (W .H. Freeman, New York, 1973), pages 5  8 . Figure 513 is taken directly from this reference, its caption slightly altered from the original.
CHAPTER 5 EXERCISES
PRACTICE 51 more is less The spacerime diagram shows two alternative world lines from event A to event D. The table shows co ordinates of numbered events in this frame. Time and space are measured in years.
a One traveler moves along the solid segmented worldline from evenr A to events B, C, and D. Calcu late the time inctease on his wristwatch (proper clock) between event A and event B. (2) between event B and event C (3) between event C and event D. (4) Also calculate the total proper time along worldline A, B, C, D. (1)
164
EXERCISE 52
TRANSFORMING WORLDLINES
worldline from event A directly to event D. Calculate the time increase on her wristwatch between events A and D. c Which twin (solidline or dottedline traveler) is younger when they rejoin at event D?
Event 0
x =0
t= 0
Event 1
x = 3 .0 0 0
t= 4 .0 0 0
Event 2
x = 1 .7 5 0
f= 7 .0 0 0
Event 3
X = 5 .0 0 0
t= 1 1 .0 0 0
EXERCISE 5 2 . Two worldlines as recorded in the laboratory frame. Numbers on the segmented worldline are proper times along each straight segment.
52 transforming worldlines The laboratory spacetime diagram in the figure shows two worldlines. One, the vertical line labeled B, is the worldline of an object that is at rest in this frame. The other, the segmented line that connects events 0, 1,2, and 3, is the worldline of an object that moves at different speeds at different times in this frame. The proper time is written on each segment and invariant hyperbolas are drawn through events 1,2, and 3. The event table shows the space and time locations in this frame of the four events 0, 1, 2, and 3. a Trace the axes and hyperbolas onto a blank piece of paper. Sketch a qualitatively correct spacetime diagram for the same pair of worldlines observed in a frame in which the particle on the segmented worldline has zero velocity between event 1 and event
2.
b W hat is the velocity, in this new frame, of the particle moving along worldline B? C On each straight portion of the segmented worldline for this new frame write the numerical value of the interval between the two connected events.
53
mapmaking in spacetime
N ote: Recall Exercise 16, the corresponding map making exercise in Chapter 1. Here is a table of timelike intervals between events, in meters. The events occur in the time sequence ABCD in all frames and along a single line in space in all frames. (They do not occur along a single line on the spacetime map.) INTERVAL to event
A
B
c
D
1.0 0
3.161 2.0 0
5.196 4.0 2.0 0
from event A B C D
a Use a ruler and the hyperbola graph to con struct a spacetime map of these events. Draw this map
EXERCISE 53
MAPMAKING IN SPACETIME
165
 space —■>EXERCISE 53. Template of hyperbolas for converting intervals into a spacetime map.
on thin paper so you can lay it over the hyperbola graph and see the hyperbolas. D iscussion: How to start? W ith three arbitrary decisions! (1) Choose event A to be at the origin of the spacetime map. (2) Choose event B to occur at the same place as event A. That is, event point B is located on the positive time axis with respect to event point A. After plotting B, use your ruler to draw this straight time axis through event points A and B. Keep this line parallel to the vertical lines on the hyperbola graph in all later constructions. (3) Even with these choices, there are two spacetime locations (x, t) at which you can locate the event point C; choose either of these two
spacetime locations arbitrarily. Then go on to plot event D. Analogy to surveying: In surveying (using Eu clidean geometry) you locate all points a given dis tance from some stake by using that stake as origin and drawing a circle of radius equal to the desired distance. In a spacetime map (using Lorentz geome try) you locate all event points a given interval from some event by using that event point as origin and drawing a hyperbola with nearest point equal to the desired interval. b Now take a new piece of paper and draw a spacetime map for another reference frame. Choose
166
EXERCISE 54
THE POLE AND BARN PARADOX
event D to be at the origin of the spacetime map. This means that all other events occur before D. Hence turn the hyperbola plot upside down, so that the hyperbolas open downward. Choose event B to occur at the same place as D. Now find the locations of A and C using the same strategy as in part a. e Find an approximate value for the relative speed of the two frames for which you have made spacetime plots. d Hold one of your spacetime maps up to the light with the marks on the side of the paper facing the light. Does the map you see from the back also satisfy the table entries?
PROBLEMS 54 the pole and barn paradox A worried student writes, “Relativity must be wrong. Consider a 20meter pole carried so fast in the direc tion of its length that it appears to be only 10 meters long in the laboratory frame of reference. Let the runner who carries the pole enter a barn 10 meters long, as shown in the figure. At some instant the farmer can close the front door and the pole will be entirely enclosed in the barn. However, look at the same situation from the frame of reference of the runner. To him the barn appears to be contraaed to half its length. How can a 20meter pole possibly fit into a 5meter barn? Does not this unbelievable con clusion prove that relativity contains somewhere a fundamental logical inconsistency?”
Write a reply to the worried student explaining clearly and carefully how the pole and barn are treated by relativity without internal contradiaion. Use the following outline or some other method. a Make two carefully labeled spacetime dia grams, one an xt diagram for the barn rest frame, the other an x 't ' diagram for the runner rest frame. Re ferring to the figure, take the event “Q coincides with A” to be at the origin of both diagrams. In both plot the worldlines of A, B, P, and Q. Pay attention to the scale of both diagrams. Label both diagtams with the time (in meters) of the event “ Q coincides with B” (derived from Lorentz transformation equations or otherwise). Do the same for the times of events “ P coincides with A” and “ P coincides with B.” b D iscussion question: Suppose the barn has no back door but rather a back wall of steelreinforced concrete. W hat happens after the farmet closes the front door on the pole? c Replace the pole with a line of ten tennis balls the same length as the pole and moving together with the same velocity as the pole. The farmer’s ten chil dren line up inside the barn, and each catches and stops one tennis ball at the same time as the farmer closes the front door of the barn. Describe the stop ping events as recorded by the observer riding on the last tennis ball. Plot them on your two diagrams.
55
radar speed trap
A highway patrolman aims a stationary radar trans mitter backward along the highway toward oncom ing traffic. A detector mounted next to the transmitter analyzes the radar wave reflected from an approach ing car. An internal computer uses the shift in fre quency of the reflected wave to reckon and display the car’s speed. Analyze this shift in frequency as in parrs a  e or with some other method. Treat the car as a simple mirror and assume that the radar signals move back and forth along one line on the highway. Radar is an electromagnetic wave that moves with the speed of light. The figure shows the worldline of the car, world lines of two adjacent maxima of the radar wave, and the wavelength A of incident and reflected waves. a From the 4 5degree right triangle ABC, show that A t = tsAt +
EXERCISE 54. Fast runner with “20meter" pole enclosed in a “10meter" ham. In the next instant he will hurst through the hack door, which is made of paper.
From the 4 5degree right triangle DBF, show that
A/ = 4dd«,e 
fA i
EXERCISE 55
RADAR SPEED TRAP
167
second
EXERCISE 55. Worldlines of approaching car and two radar wave maxima that reflect from the car. The speed of the car is greatly exaggerated.
Eliminate At from these two equations to find an expression for in terms of A;„adent and the automobile speed v. b The frequenq? / of radar (in cycles/second) is related to its wavelength A in a vacuum by the for mula / = c/A, where c is the speed of light (~ the speed of radar waves in air). Derive an expression or frequency yrefleaed the reflected radar signal in terms of f r e q u e n c y o f the incident wave and the speed V of the oncoming automobile. Show that the result is
fte& ecad
f\nad
C For an automobile moving at a speed v = t'conv/r that is a small fraction of the speed of light, assume that the fractional change in frequency of
reflected radar is small. Under this assumption, use the first two terms of the binomial expansion (1 — z)” ^ 1 — «z for lz «
1
to show that the fractional change of frequency is given by the approximate expression — ~ 2v f Substitute the speed of a car moving at 100 kilometers/hour (= 2 7 .8 meters/second = 60 miles/hour) and show that your assumption about the small frac tional change is justified. d One radar gun used by the Massachusetts Highway Patrol operates at a frequency of 10.525 X 10^ cycles/second. By how many cycles/
168
EXERCISE 56
A SUMMER EVENING'S FANTASY
second is the reflected beam shifted in frequency when reflected from a car approaching at 100 kilometers/ hour? e W hat discrimination between different fre quency shifts must the unit have if it can distinguish the speed of a car moving at 100 kilometers/hour from the speed of one moving at 101 kilometers/ hour?
a
Plot EVENTS labeled with the following NUMBERS. 0. your location when the aliens land (at the origin) 1. Sun explodes 2. light from Sun explosion reaches you 3. Venus’s atmosphere blown away 4. light from event 3 reaches you 5. you and aliens depart Earth (you hope!) 6. Earth atmosphere blown away
b
Plot WORLDLINES labeled with the follow ing CAPITAL LETTERS. A. your worldline B. worldline of Earth C. aliens’ worldline D. worldline of Sun E. worldline of Venus E. worldline of light from Sun’s explosion G. worldline of the “speedonehalf” pulse of particles from Sun’s explosion H. worldline of light emitted when Venus loses atmosphere J. terminal part of the worldline of the laser cannon pulse fired at Sun by the aliens
Reference: T. M. Kalotas and A. R. Lee, AmericanJournal ofPhysics, Volume 58, pages 187188 (February 1990).
56 a summer evening's fantasy You are standing alone outdoors at dusk on the first day of summer. You see Sun setting due west and the planet Venus in the same direction. On the opposite horizon the full Moon is rising due east. An alien ship approaches from the east and lands beside you. The occupants inform you that they are from Proxima Centauri, which lies due east beyond the rising Moon. They say they have been traveling straight to Earth and that their reduced approach speed within the solar system was such that the time stretch factor gamma during the approach was 5/3. At the same instant that the aliens land, you see Sun explode. The aliens admit to you that earlier, on their way to Earth, they shot a laser light pulse at Sun, which caused this explosion. They warn that Sun’s explosion emitted an immense pulse of particles moving at half the speed of light that will blow away Earth’s atmosphere. In confirmation, shortly after the aliens land you notice that the planet Venus, lying in the direction of Sun, suddenly changes color. You grab a passing human of the opposite sex and plead with the aliens to take you both away from Earth in order to establish the human gene pool else where. They agree and set the dials to flee in an easterly direction away from Sun at top speed, with time stretch factor gamma of 25/7. The takeoff is to be 7 minutes after the alien landing on Earth. Do you make it? Draw a detailed Earth spacetime diagram showing the events and worldlines of this story. Use the fol lowing information. • Sun is 8 lightminutes from Earth. • Venus is 2 lightminutes from Earth. • Assume that Sun, Venus, Earth, and Moon all lie along a single direction in space and are rela tively at rest during this short story. The incom ing and outgoing paths of the alien ship lie along this same line in space. • All takeoffs and landings involve instantaneous changes from initial to final speed. • 5"  3 ' = 42 and (25)2  {1? = (24)2
c Write numerical values for the speed v = i'conv/'^ on every segment of all worldlines.
57 the runner on the train paradox A letter sent to the Massachusetts Institute of Tech nology by HsienYen Tsao of Los Angeles poses the following paradox, which he asserts disproves the theory of relativity. The Chairman of the Physics Department sends the inquiry along to you, asking you to respond to Mr. Tsao. You determine to make the answer clear, concise, decisive, and polite — a personal test of your diplomacy and grasp of relativ ity. T h e setting: A train travels at high speed. A runner on the train sprints toward the back of the train with the same speed (with respect to the train) as the train moves forward (with respect to Earth). There fore the runner is not moving with respect to Earth. T h e paradox: We know that, crudely speaking, clocks on the train run “slow” compared to the Earth clock. We also know that the runner’s clock runs “slow” compared to the train clocks. Therefore the runner’s clock should run “doubly slow” with respect to the Earth clock. But the runner is not moving with respect to Earth! Therefore the runner’s clock must run at the same rate as the Earth clock. How can it possibly be that the runner’s clock runs "doubly slow” with respect to the Earth clock and also runs at the same rate as the Earth clock?
EXERCISE 58
58
THE TWIN PARADOX PUT TO REST—A WORKED EXAMPLE
the twin paradox put to rest— a worked example
Motto: The swinging line of simultaneity tells alH Combine the Lorentz transformation with the spacetime diagram to clear up — once and for all! — the solution to the Twin Paradox. An astronaut travels from Earth to Canopus (Chapter 4) at speed = 99 /1 0 1 , arriving at Canopus t ' = 20 years later ac cording to her rocket clock, t = 101 years later ac cording to Earthlinked clocks — which means that the stretch factor y has the value 101/20. The key idea is “lines of simultaneity” (boxed labels in the figure). A line of simultaneity connects events that occur “at the same time.” But events simultaneous in the Earth (“laboratory”) frame are typically not simultaneous in the rocket frame (Sec tion 3.4). Horizontal is the line of simultaneity on the Earth (“laboratory”) spacetime map that connects
169
events occurring at the same time in the Earth frame. Totally different— not a horizontal line! — is a line of simultaneity on the Earth spacetime map that con nects events simultaneous in the outgoing astronaut frame. To draw this line of outgoingastronaut simul taneity, start with the inverse Lorentz transformation equation for time:
t' =
+ yt
For the outgoing astronaut, = 99/101 and y = 101/20. W e want the line of simultaneity that passes through turnaround event T. So let t ' = 20 years. Then: 20 =  (99/101)(101/20)x b (101/20) t Multiply through by 20/101: 400/101 =  ( 9 9 /1 0 1 ) x + r
time
EXERCISE 58. Earth spacetime map of the trip to Canopus and hack. A t the astronaut arrives at Canopus, her colleagues in her outgoing reference frame record along line AT events simultaneous with this arrival, including Earthclock reading of 396 years at A. At Canopus the astronaut changesframes, thus changing the line
of simultaneity, which swings to BT. Ar she leaves Canopus, her new colleagues take an Earthclock reading of 1 9 8 .0 4 years at B. At turnaround, the ticks on the Earth clock along worldline segment AB gofrom the outwardmoving astronaut's future to the incoming astronaut’s past.
170
EXERCISE 58
THE TWIN PARADOX PUT TO REST  A WORKED EXAMPLE
which yields / = 0.980 x  f 3.96 This is the equation for a straight line passing through event points A and T in the spacetime dia gram. It is the line of simultaneity for the outgoing astronaut, connecting all events simultaneous with the arrival of the rocket at Canopus (simultaneous in that frame). Among these events is event A, the Earth clock reading of 3.96 years, which occurs ar Earth position x = 0. In brief, at the moment the rocket arrives at Canopus, the Earth clock reads 3.96 years as observed in the outgoing rocket frame. Now the astronaut jumps to the incoming rocket frame. This reverses the velocity of the astronaut with respect to the Earthlinked frame— and so reverses the slope of the line of astronaut simultaneity. This new line of astronaut simultaneity passes through event points B and T in the figure. Event B is the Earth clock reading of 202 — 3.96 = 198.04 years. To go back over the astronaut trip while looking at the spacetime map is (finally!) to solve the Twin Paradox. As the astronaut travels outward toward Canopus, many colleagues follow her at the same speed, with clocks synchronized in her frame. As they whiz past Earth, each records the reading on the Earth clock. Later analysis leads them to agree that the time between ticks of Earth’s clock is longer than the time between ticks of their own outwardmoving clocks. (They say, “The Earth clock mns slow.’’) At any event point on her outward worldline, the astronaut’s line of simultaneity slopes upward to the right in the Earth spacetime diagram, as shown in the figure. Simultaneous with astronaut arrival at Canopus (event T, when all outwardmoving clocks read 20 years), one of her colleagues reads a time 3.96 years on the Eanh clock (event A). Now the astronaut jumps from the outwardmov ing rocket to a returning rocket. She inherits a com pletely new set of colleagues, with a new set of synchro nized clocks. The astronaut’s new line of simultaneity slopes upward to the left in the Earth spacetime dia gram. Simultaneous with her deparmre from Cano pus (event T, when all inwatdmoving clocks read 20 years), one of her new colleagues reads a time 202 — 3.96 = 198.04 years on the passing Earth clock (event B). Thereafter new colleague after new col league streaks past Earth, recording the fact that Earth clock ticks are farther apart in time than the ticks on their own clocks. (They say, “The Earth clock runs slow.’’). The analysis so far accounts for the short time segments OA and BC recorded by the Earth clock on its vertical wotldline AC W hat about the omitted
time lapse AS? This is recorded, sure enough, by the Earth clock plowing forward along worldline OC in its comfortable single freefloat frame. However, the story of time AB is quite different for the turnaround astronaut. Before she reaches turnaround at T, events on line AB are in herfuture. All those Earth clock ticks are yet to be recorded by her outgoing colleagues. These events lie above her line of simultaneity A T a.s she arrives at Canopus at T. However, as she turns around, her line of simultaneity also slews forward, swinging from line A T to line BT. Suddenly the events on line AB— all those intermediate ticks of the Earth clock— are in the astronaut’s past. These events lie below the line of simultaneity B T as she starts back at T. Her outwardmoving colleague reads 3.96 years on the Earth clock as she teaches Canopus; an instant later on her clock, her new inwardmoving colleague reads 198.04 on the Earth clock. Shall we say that the Earth clock “jumps ahead’’ as the astronaut turns around? No! Utterly ridiculous! Eor what single observer does it jump ahead? Not for the Earth observer. Not for the outgoing set of clockreaders. Not for the returning set of clock readers. For whom then? Nobody! A t the same time as she reaches Canopus— old meaning of simultaneous! — the as tronaut’s outgoing colleague records 3.96 years for the Earth clock. A t the same time as she leaves Canopus— new meaning of simultaneous! — her new ingoing colleague records 198.04 years on the Earth clock. The astronaut has nobody but herself to blame for her misperception of a “jump” in the Earth clock reading. The “lost Earth time” AB in the figure makes consistent the story each observer tells about the clocks. Simple is the story told by the Earth observer: “ My clock ticked along steadily at the ‘proper’ rate from astronaut departure to astronaut return. In con trast, ticks on the astronaut clock were far apart in time on both the outgoing and incoming legs of her trip. W e agree that her total ticks are less than my total ticks: she is younger than I when we meet again.” More complicated is the astronaut account of clock behavior: “Ticks on the Earth clock were far apart in time as I traveled to Canopus; ticks on the Earth clock were also far apart as I traveled home again. But as I turned around, a whole bunch of Earth clock ticks went from my future to my past. This accounts for the larger number of total ticks on the Earth clock than on my clock during the trip. We agree that I am younger when we meet again.” So saying, the astronaut renounces her profession and becomes a standup comedian. Reference: E. Lowry, American Journal o f Physics, Volume 31, page
59 (1963).
6.1 LIGHT SPEED: LIMIT ON CAUSAUTY no signal reaches us faster than light Nineyearold Meredith waves her roy magician’s wand and shouts, "Sun is exploding right now!” Is she right? We have no way on Earth of knowing— at least not for a while. Sun lies 150,000 million meters from Earth. Therefore it will take 150,000 million meters of lightrravel time for the first light flash from the explosion to reach us. This equals 500 seconds— 8 minutes and 20 seconds. W e will just have to wait and see if Meredith is correct . . . When 8 minutes and 20 seconds pass, we have evidence that Meredith was mistaken; Looking through our special dark glasses, we see no exploding Sun. But Meredith’s wand has started us thinking. W hat in the laws of nature prohibirs rhe wave of her wand from being the signal fot Sun to explode at that same instant? Or — more reasonably, given the awesome event— what prevents Meredith from having instanraneous warning, so that she raises her wand simultaneously with Sun’s explosion in order to give us (in light of later developments) a false impression of her power? Both questions have the same answer; “The speed of light.” Whatever her powers, Meredith cannot affect Sun in less than 500 seconds; neither can a warning signal reach us from Sun in less time than that. All during that intervening 500 seconds we would see the accustomed round shape of Sun, apparently healthy as ever. More generally, one event cannot cause another when their sparial separation is greater rhan the distance light can travel in the time between these events. Light speed sets a limit on causality. No known physical process can overcome this limit; not gravity, not some other field, not a zooming particle of any kind. "Spacetime interval” quantifies this limit on causality. Interval between faraway events — unlike distance between faraway points — can be zero. In this and other ways rhe spacerime geometry of the real world differs fundamentally from the space geometry of Euclid’s 2300yearold rextbook.
171
Signal Sun with super speed?
No, just speed of light
172
CHAPTER 6
REGIONS OF SPACETIME
6.2 RELATION BETWEEN EVENTS: TIMELIKE, SPACELIKE, OR LIGHTLIKE minus sign yields three possible relations between pairs of events Using Euclidean geometry, a surveyor reckons the distance between two steel stakes from the sum of the squates of the northward and eastward separations of these stakes: uared distance: Positive or zero
(distance)^ = (northwatd separation)^ + (eastward separation)^ In consequence, in Euclidean geometry a distance— or its square — always has a positive value or zero. In contrast, the spacetime interval between events in Lorentz geometry arises from the difference of squares of time and space separations:
Squared interval: Positive, zero, or negative
Timelike interval: Time part dominates
(interval)^ = (separation in time)^ — (separation in space)^ In consequence of the minus sign, this equation yields a number that may be positive, negative, or zero, depending on whether the time or the space separation predomi nates. Moreover, whichever of these three descriptions characterizes the interval in one freefloat frame also characterizes the interval in any other freefloat frame. Why? Because the spacetime interval between two events has the same value in all overlap ping freefloat frames. In the threefold possibilities for an interval, nature reveals the causal relation between events. An interval between two events earns the name tim e lik e or spacelike or lig h t like depending on whether the time part predominates, the space part predominates, or the time and space parts are equal, respectively, as shown in Table 61. Eor convenience, the minus sign is placed so that the resulting squared interval is greater than or equal to zero. T im elik e Interval: W e picture the sequence of sparks emitted by a moving sparkplug. Points representing these sparks on the spacetime map trace out the worldline of the particle (Chapter 5). No material particle has ever been measured to travel faster than light. Every material particle always travels less than one meter of distance in one meter of lighttravel time. The sparks emitted by the particle have a greater time separation than their separation in space. In other words, the worldline of a particle consists of events that have a timelike relation with one another and with the initial event. We say that a material particle follows a tim e lik e w o rld lin e. The interval T between two timelike events reveals itself to the observer in any freefloat frame: (timelike interval)^ =
= (time separation)^ — (space separation)^
(61)
6 T ^
CLASSIFICATION OF THE RELATION BETWEEN TWO EVENTS Description
Squared interval is named and reckoned
Time part of interval dominates space part Space part of interval dominates time part Time part of interval equals space part
(timelike interval)^ = = (time)^ — (distance)^ (spacelike interval)^ = = (distance)^ — (time)^ (lightlike interval)^ = 0 = (time)^ — (distance)^
6.2
RELATION BETWEEN EVENTS; TIMELIKE, SPACELIKE, OR LIGHTLIKE
leftrocket time
173
laboratory time
rightrocket time
LABORATORY FRAME
RIGHTMOVING ROCKET FRAME
A
leftrocket space
LEFTMOVING ROCKET FRAME
FIGURE 61. Events A a n d Bform a timelike p a ir (w ith event A a rb itra rily chosen as reference event), here recorded in the spacetime maps o f three freefloat fram es, Point B lies on a hyperbola opening along the time axis in each frame. The shortest time between events A and B is recorded in the laboratory frame, the frame in which the two events occur at the same place.
Same two sparks registered in different frames? Different records for the separation in time between those sparks. Different records for the separation in space. Same figure for the timelike interval between them! Nobody can keep us from tracing out on one and the same diagram (Figure 61) the very different records for the separation AB that observers get in different freefloat frames. One frame? One point on the diagram. Another frame? Another point on the diagram. And so on. These many records for the same pair of events AB trace out a hyperbola. This hyperbola opens out in the time direction. The two sparks, A and B — definite locations though they occupy in spacetime— nevertheless register in different frames of reference as having different separations in referenceframe time. Among the many conceivable frames, which one records this separation in time as smallest? Answer: The frame in which spark B occurs at the same place as spark A. In other words, the frame that happens to move along in sync with the sparkplug, even if only briefly. In that frame the clock records a separation in time between A and B identical with the timelike interval AB. As seen in the leftmoving rocket frame in Figure 61, spark B lies to the right of spark A. In contrast, spark B occurs to the left of spark A in the rightmoving rocket. The position of B relative to A depends on the reference frame from which it is measured. For a pair of events separated by a timelike interval, labels “right” and “left” have no invariant meaning: they are framedependent. S pacelike Interval: The interval between two events A and D is spacelike when the space part predominates over the time part. Such was the case for a possible explosion of Sun (event A) and Meredith’s wand waving (event D), simultaneous with A as recorded in the Earth frame (Section 6.1). Events A and D, if they occurred, would be separated in the EarthSun frame by a distance of 150,000 million meters and separated by a time of zero meters. Clearly the space part predominates over the time part! Whenever the space part predominates, we call the relation between the two events spacelike. The interval s (sometimes called by the Greek letter sigma, (T) between two spacelike events reveals itself to the observer in any freefloat frame: (sp a celik e interval)^ =
= (sp ace separation)^ — (tim e separation)^
(6  2 )
Timelike interval: Invariant hyperbola opens along time axis
Spacelike interval: Space part dominates
174
Spacelike interval: Invariant hyperbola opens along space axis
CHAPTER 6
REGIONS OF SPACETIME
Events A and D registered in different frames? Then different records for the separa tion in time between those events. Also different records for the separation in space. Same numerical value for rhe spacelike interval between rhem! We plot on another spacetime diagram (Figure 62) all of the very differenr records for the separation AD that observers get in different freefloat frames. One frame? One point on the diagram. Another frame? Another poinr on the diagram. And so on. These many records for the same pair of evenrs AD trace out a hyperbola. This hyperbola opens out in the space direction. The two events, A and D — definite locations though they occupy in spacetime — nevertheless register in different frames of reference as having different separations in referenceframe space. Among the many conceivable frames, which one records this separation in space as smallest? Answer: The frame in which spark D occurs at the same time as spark A. In that frame a long srick records a separation in space between A and D identical with the spacelike interval, AD. This is called the p ro p e r d istan ce between the two spacelike events. In the Earth  laboratory frame in Figure 62, Meredith waves her wand (event D) at the same time as Sun explodes (event A). In the rightmoving rocket frame Sun explodes after Meredirh waves her wand. In the leftmoving rocket frame Sun explodes before the wand wave. For a pair of events separated by a spacelike interval, labels “before” and “after” have no invariant meaning: they are framedependent. To allow the wand to control Sun would be to scramble cause and effect! No particle — not even a flash of light — can move between two events connected by a spacelike interval. To do so would require it to cover a distance greater than the time available to cover rhis disrance (space separation greater than time separation). In brief, it would have to travel faster than light. This is alternative evidence that two events separated by a spacelike interval cannot be causally connected: one of them cannot “get at” the other one by any possible signal.
LEFTMOVING ROCKET FRAME
LABORATORY FRAME
RIGHTMOVING ROCKET FRAME
FIGURE 62. The spacelike p a ir o f events A a n d D {with event A a rb itra rily chosen as reference event) as recorded in the spacetime maps o f three freefloat fram es. Point D lies on a hyperbola opening along the space axis in every rocket and laboratory frame. The shortest distance between these events is recorded in the laboratory frame, the frame in which the two events occur at the same time. A heavy line represents the spacetime separation AD. No particle can travel along this line; the speed would be greater than light speed— and would be infinitely great as measured in the laboratory frame, since the particle would have to cover the distance from A /o D in zero time!
6.2
RELATION BETWEEN EVENTS: TIMELIKE, SPACELIKE, OR LIGHTLIKE
175
E PR
RELATIONS BETWEEN EVENTS Events 1,2, and 3 all have laboratory locations y — z — 0. Their x and t measurements are plotted on the laboratory spacetime map. a.
Classify the interval between events 1 and 2; timelike, spacelike, or lightlike.
b.
Classify the interval between events 1 and 3.
c.
Classify the interval between events 2 and 3.
event
2 7
6
t time (meters)
1
event 3,
' 3
2
event 1,
1 1
2
3
4
5
 space (meters) —
SOLUTION a.
For event 1, / = 2 meters and x — I meter. For event 2, t = 7 meters and x = 4 meters. The squared interval between them: (interval)^ — { 7 ~ 2Y — (4 — 1)^ = 5^“ 3^ = 2 5 “ 9== 16 (meters)^. The time part is greater than the space part, so the interval between these two events is timelike: T = 4 meters.
b.
For event 1, t = 2 meters and x = I meter. For event 5, t = 5 meters and x = 6 meters. The squared interval between them: (interval)^ = (5 “ 2)^ — (1 — 6)^ = 3^ — 5^ = 9 ~ 2 5 = “ 16 (meters)^. The space part is greater than the time part, so the interval is spacelike: s = 4 meters. (For spacelike intervals, we subtract the squared time part from the squared space parr before taking the square root.)
c.
For event 2, / = 7 meters and x = 4 meters. For event 3 ,^ = 5 meters and x = 6 meters. The squared interval between them: (interval)^ ~ { 7 ~ ~ 5)^ — (4 — 6)^ = 2^ — 2^ = 4 — 4 = 0 (meters)^. The time part equals the space part, so the interval is lightlike, it is a null interval.
L ightlike In terv al (N ull Interval): Two events stand in a lightlike relation when the interval between them is zero: Lightlike interval; Time separation equa sp a ce separation
(time separation)^ — (space separation)^ — 0 or magnitude of (separation in time) — (distance in space)
[for lightlike interval]
(63)
176 Lightlike interval: Plotted along ±45 degree lines
CHAPTER 6
REGIONS OF SPACETIME
An interval that is lightlike? A separation in time between two events, A and G, identical to the distance in space between them? W hat does this condition mean? This: A pulse of light can fly directly from event A and arrive with perfect timing at event G. How come? Distance in meters between the two locations measures the meters of time required for light to fly from one place to the other. Separation in time between the two events represents the time available for the trip. Time available equals time needed? Guarantee that the pulse from A arrives in coincidence with event G\ More generally, whenever the influence of one event, spreading out at the speed of light, can directly affect a second event, then the interval between those two events rates as lightlike, zero, null. Only light (“photons”), neutrinos, and gravitons can move directly between two events connected by a lightlike interval. Only by means of one of these lightspeed particles can the one event in a lightlike pair cause the other. The spherical outgoing pulse of light from an event. A, may trigger two widely separated events, E and G(Figure 63). Does this common genesis imply that E and G occur at the same time? Yes and no! Yes, there’s always a freefloat reference frame in which the two daughter events appear as simultaneous. That frame — for no good reason — we call the laboratory frame in Figure 63. In other frames of reference — for example, the leftmoving rocket frame in Figure 63 — the clocks show that E occurs before G. There are still other frames — the rightmoving rocket frame is one — in which the clocks register E and G in the opposite order of time. But no frame shows either £ or G in the past of A. Hold it! Aren't spacelike separations impossible? I understand timelike a n d lightlike separations between two events, because a p article— or a t least a light fla sh — can travel between them. Not even a light flash, however, can travel from one event to a second event separated from the first by an interval th a t is spacelike. The first event cannot possibly cause the second event in the spacelike case. Therefore a spacelike interval cannot arise in nature. So why talk about it?
leftrocket time
laboratory time
rightrocket time
A
A
G .'
laboratory space
LEFTMOVING ROCKET FRAME
LABORATORY FRAME
rightrocket space
RIGHTMOVING ROCKET FRAME
FIGURE 63. Two lightlike p a irs of events AE a n d AG {with event A a rb itra rily chosen as reference event) as recorded in spacetime maps o f three freefloat fram es. A flash originates at A and spreads outwardfrom the center ofa rod at rest in the laboratoryframe. Events E and G are receptions of this flash at the two ends of the rod as recorded by different observers. In the laboratoryframe, reception events E and G occur at the same time. In the rightmoving rocketframe, the rod moves to the left, so event G occurs sooner than event E. In the leftmoving rocket frame, the rod moves to the right, so event E occurs sooner than event G.
6.3
LIGHT CONE: PARTITION IN SPACETIME
177
Oops! A spacelike interval between two events certainly can and does arise in nature. Signals from the supernova labeled 1987A reported that event to us in 1987, which was 150,000 years after the explosion occurred. Yet occur it did! N o astron omer of Babylonian, Egyptian, or Greek days reported it, nor could they even know of it. Yet it had already happened for them. That event separated itself from each of them by a spacelike interval. Only the advance of time to the year 1987 brought down the interval between that explosion and Earthbound observers from spacelike to lightlike. In that year a light pulse carried the earliest possible report of that explosion to our eyes. And look today? See no explosion at that location in the sky. The light from it has passed us by. Our present relation to that event? Timelike!
6.3 LIGHT CONE: PARTITION IN SPACETIME invariance of the interval preserves cause and effect Thus far in dealing with the interval between two events, A and B, we have considered primarily the situation in which these events lie along a single direction in space — on the reference line where the laboratory and rocket reference clocks are located. In contrast, the surveyors in our imaginary kingdom made use of two space dimensions — northward and eastward. We know, however, that Euclidean space is truly threedimensional. A surveyor measuring hilly terrain soon appreciates the need for a third dimension: the direction vertically upward! The measure of distance in three dimen sions requires a simple extension of the expression for distance in two dimensions: The square of the distance becomes the sum of the squares of three mutually perpendicular separations: (distance)^ = (north separation)^ + (east separation)^ + (up separation)^ Euclidean space requires three dimensions. In contrast, spacetime, which includes the time dimension, demands four. The expression for the square of a timelike interval now has four terms: a positive term (the square of the time separation) and three negative terms (the squares of the separations in three space dimensions). (interval)^ = (time separation)^ — (north separation)^ — (east separation)^ — (up separation)^ The three space terms can be represented by the single distance term in the equation above, yielding (timelike interval)^ = (time separation)^ — (distance)^ (spacelike interval)^ = (distance)^ ^ (time separation)^ (lightlike interval)^ = 0 = (time separation)^ — (distance)^ or, for the lightlike interval, magnitude of (separation in time) = (distance in space)
tiightlike interval]
(63)
For pairs of events with lightlike separation, the interval equals zero. The zero interval is a unique feature of Lorentz geometry, new and quite different from
Interval generalized to three space dimensions
178
CHAPTER 6
REGIONS OF SPACETIME
p r o
b l e m
62
E X P L E T I V E DELETED At 12:00 noon Greenwich Mean Time (GMT) an astronaut on Moon drops a wrench on his toe and shouts “Damn!” into his helmet microphone (event A), carried by a radio signal toward Earth. At one second after 12:00 noon GMT a short
circuit (event D) temporarily disables the receiving amplifier at Mission Control on Earth. Take Earth and Moon to be 3.84 X 10® meters apart in the Earth frame and assume zero relative motion.
a.
Does Mission Control on Earrh hear the astronaut’s expletive?
b.
Could the astronaut’s strong language have caused the short circuit on Earth?
c.
Classify the spacetime separation between events A and D: timelike, spacelike, or lightlike.
d.
Find the proper distance or proper time between events A and D.
e.
For all possible rocket frames passing between Earth and Moon, find the shortest possible distance between events A and D. In the rocket frame for which this distance is shortest, determine the time between the two events.
SOLUTION a.
In one second, electromagnetic radiation (light and radio waves) travels 3.0 X 10® meters in a vacuum. Therefore the radio signal does not have time to travel the 3.84 X 10® meters between Moon and Earth in the one second available between the events A and D as measured in the Earth frame. So Mission Control does nor hear the exclamation.
b.
No signal travels faster than light. So the astronaut’s strong language cannot have caused the short circuit.
c.
The space part of the separation between events (3.84 X 10® meters) dominates the time part (one second = 3.0 X 10® meters). Therefore the separation is spacelike.
d.
The square of the proper distance s comes from the expression f2 = (space separation)^ — (time separation)^ — (3.84 X 10® meters)^ — (3.00 X 10* meters)^ = (14.75  9.00) X 10*6 (meters)^ = 5.75 X 10*6 (meters)^ The proper distance equals the square root of this value: r = 2.40 X 10® meters
e.
The proper distance equals the shortest distance between two spacelike events as measured in any rocket frame moving between them (Figure 62, laboratory map). Hence 2.40 X 10® meters equals the shortest possible distance between events A and D. In the particular rocket frame for which the distance is shortest, the time between the two events has the value zero — events A and D are simultaneous in this frame.
6.3
LIGHT CONE: PARTITION IN SPACETIME
179
SAMPLE PROBLEM
SUNSPOT Bradley grabs his sister’s wand and waves it, shouting “Sunspot!” At that very instant his fa ther, Lloyd, who is operating a home solar obser vatory, sees a spot appear on the face of Sun. Let event E be Bradley waving the wand and event A
be eruption of the sunspot at the surface of Sun itself The EatthS un distance equals approxi mately 1.5 X 10“ meters. Neglect relative motion between Earth and Sun.
a.
Is it possible that Bradley’s wand waving caused the sunspot to erupt on Sun?
b.
Is it possible that the sunspot erupting on Sun caused Bradley to wave his wand?
c.
Classify the spacetime separation between events A and E: timelike, spacelike, or lightlike.
d.
Find the value of ptoper distance ot propet time between events A and E.
e.
For all possible rocket frames passing between Earth and Sun, find the shortest possible distance or the shortest possible time between events A and E.
SOLUTION a.
Light travels 1 meter ot distance in 1 meter of time — or 1.5 X 10“ meters of distance in 1.5 X 10^* meters of time. Hence in the EarthSun frame, eruption of the sunspot (event A) occurted 1.5 X 10“ meters of time before Bradley waved the wand (event E). So Bradley’s wand waving could not have caused the eruption on Sun.
b.
On the other hand, it is possible that eruption of the sunspot caused Bradley to wave his wand: He raises the wand in the air, looks over his father’s shoulder, and waves the wand as the spot appears on the projection screen. (We neglect his reaction time.)
c.
Events A and E are connected by one light pulse; their space and time separations both have the value 1.5 X 10“ meters in the Eatth frame. Therefore the spacetime separation between them is lightlike.
d.
Space and time separations between events A and E are equal. Therefore the interval between them has value zero. Hence proper time between them — equal to ptoper distance between them — also has value zero.
e.
The interval is invariant. Thetefore all possible fteefloat rocket frames passing between Earth and Sun reckon zero interval between events A and E. This means each of them measures space separation between events A and E equal to the time separation between these events. The common value of the space and time separations are not the same for all rocket frames, but they are equal to one another in every individual rocket frame. We are asked to find the shottest possible value for this time. Think of a tocket just passing Sun as the sunspot erupts, the rocket headed towatd Earth at nearly light speed with respect to Earth. Rocket lattice clocks record the light flash from the sunspot moving away from the rocket at standard speed unity. However, these clocks recotd that Earth lies very close to Sun (Lorentz contraction of distance) and that Earth rushes toward the rocket at nearly light speed. Therefore light does not travel far to get to Earth in this rocket frame; neither does it take much time. For a rocket moving arbitrarily close to light speed, this distance between A and E approaches zero, and so does the time
180
CHAPTER 6
REGIONS OF SPACETIME
SA M P L E P R O B L E M 6 3 between A and E. Hence the shortest possible distance between A and E— equal to the shortest possible time between A and E — has the value zero. But this constitutes a limiting case, since rocket speed may approach but cannot equal the speed of light in any freefloat frame.
Light flash traces out light cone in spacetime diagram
anything in Euclidean geometry. In Euclidean geometry it is never possible for distance A G between two points to be zero unless all three of the separations (northward, eastward, and upward) equal zero. In contrast, interval A G between two events can vanish even when separation in space and separation in time are individually quite large. Equation (63) describes the separation between lightlike events, but now separation in space may show up in two or three space dimensions as well as one time dimension. The distance in space is always positive. It is interesting to plot on an appropriate map locations of all events, G, Gj, G^, Gj, , that can be connected with one given event A by a single spreading pulse of light. Every such future event has a distance in space from A identical to its delay in time after A. Only so can it satisfy the requirement (63) for a null interval. For it: (future time with respect to A) = d (distance in space from A)
[lightlike interval]
(64)
It is equally interesting to display — and on the same diagram — all the events H, Hj, H 2 , Hj, . . . that can send a light pulse to A. Every such event fulfills the condition (past time relative to A) = — (distance in space from A)
(for lightlike interval]
(65)
Both of these equations satisfy the magnitude equation (63). In Figure 64 we suppress display of a third space dimension in the interest of simplicity. We limit attention to future events G, G„ G2 , . . . and past events H, Hj, H 2 , . . . that lie on a northsouth/eastw est plane in space. A flash emitted from event A expands as a circle on this space plane. As it spreads out from event A, this circle of light traces out a cone opening upward in the spacetime map of Figure 64. This is called the fu tu re lig h t cone of event A. The cone opening downward traces the history of an incoming circular pulse of radiation so perfectly focused that it converges toward event A, collapsing exactly at event A at time zero. This downward opening cone has the name past lig h t cone of event A. All the events G, G^, G2 , . . . lie on the future light cone of event A, all events H, H„ H 2 , . . . on its past light cone. Numerous as the events may be that lie on the light cone, typically there are many more that don’t! Look, for example, at all the events that occur 7 meters of time later than the zero time of event A. On the spacetime map, these events define a plane 7 meters above the r = 0 plane in which event A lies, and parallel to that plane. The light cone intersects this plane in a circle (circle in the present map; a sphere in a full spacetime map with three space dimensions). An event on the plane falls into one or another of three categories, relative to event A, according as it lies inside the circle (as does B in Figure 64), on it (as does G), or outside it (as does D). The light cone is unique to Lorentz geometry. It gives nature a structure beyond any power of Euclidean geometry. The light cone does more than divide events on a single plane into categories. It classifies every event, everywhere in spacetime, into one or another of five distinct categories according to the causal relation that event bears to the chosen event. A:
6.3
LIGHT CONE: PARTITION IN SPACETIME
181
FIGURE 64. Light cone as p artitio n in spacetime; perspective threedim ensional spacetime map showing eastw ard, northw ard, a n d time locations of events occurring on a f la t p lan e in space. Events G, Gj, Gj, and Gj are on the future light cone of event A; events H, H,, H 2 , and Hj are on its past light cone. See also Figure 65.
1. Can a material p artic le emitted at A affect what is going to happen at El If so, B lies inside the fu tu r e light cone of A and forms a timelike pair with event A. 2. Can a lig h t ray emitted at A affect— with no time to spare— what is going to happen at G? If so, d i e s on the fu tu r e light cone of A and forms a lightlike pair with event A. 3. Can no effect w h atev er produced at A affect what happens at D? If so, D lies outside the future and past light cones of A and forms a spacelike pair with event A. It lies in the absolute elsewhere of A. 4. Can a material p artic le emitted at J affect what is h a p p e n in g at A? If so, J lies inside the p a st light cone of A and forms a timelike pair with event A. 5. Can a lig h t ray emitted at H affect — with no time to spare — what is h ap p e n in g at A? If so, H lies on the p a st light cone of A and forms a lightlike pair with event A. Nature reveals a causeandeffect structure beyond the vision of Euclidean geome try. The causal relation between an event B and another event A falls into one or the
C au se and effect preserved by light cone
182
CHAPTER 6
REGIONS OF SPACETIME
E
FIGURE 65. Exploded view o f the regions in classified w ith respect to a selected event A.
EXERCISE 61
RELATIONS BETWEEN EVENTS
183
other of five categories picked out by the light cone of A. That light cone and those categories have an existence in spacetime quite apart from any space and time measurements that may be used to describe them. Zero interval between events in one freefloat frame means zero interval between the same events in every overlapping freefloat frame. The light cone is the light cone is the light cone! Event A appears at the origin of every spacetime map in this chapter. What’s so special about event A?
1L Nothing whatever is special about event A! On the contrary, we have not captured the full story of the causal structure of spacetime until for every event A {Aj, A 2 , Aj, . . . ) we have classified every event B (Bj, B2, Bj, . . . ) into the appropri ate category— timelike! lightlike! spacelike!— with respect to that event. Figure 65 summarizes the relations between a selected event A and all other events of spacetime.
CHAPTER 6 EXERCISES
PRACTICE 61
Is it possible to find a rocket frame in which the temporal order of the two events is reversed? That is, is it possible to find a rocket frame in which the event that occurs before the other event in the laboratory frame occurs after the other event in the rocket frame?
(4)
relations between events
This is a continuation of Sample Problem 61. Events 1, 2, and 3 all have the laboratory coordinates y = z = 0. Their x and /coordinates are plotted on the laboratory spacetime diagram. a Answer the following questions three times: once for the timelike pair of events 1 and 2, once for the spacelike pair of events 1 and 3, and once for the lightlike pair of events 2 and 3. (
1) W hat is the proper time (or proper distance)
between the two events? (2) Is it possible that one of the events caused the other event? (3) Is it possible to find a rocket frame in which the spatial order of the two events is teversed? That is, is it possible to find a rocket frame in which the event that occurs to the right of the other event in the laboratory frame will occur to the left of the other event in the rocket frame?
event
2
7 6 ■'t 4' time (meters) ^
I
2
event
event 1,
1 0
1 2 3 4 5 6  space (meters) —►
EXERCISE 61. Laboratory spacetime map.
184
EXERCISE 62
TIMELIKE, LIGHTLIKE, OR SPACELIKE?
b For the timelike pair of events, find the speed and direction of a rocket frame with respect to which the two events occurred at the same place. For the spacelike pair of events, find the speed and direction of a rocket frame with respect to which the two events occurred at the same time.
62 timelike, lightlike, or spacelike?
63
The first table lists the space and time coordinates of three events plus the reference event (event 0) as observed in rhe laboratory frame.
LABORATORY COORDINATES OF THREE EVENTS t
X
y
(years)
(years)
(years)
Event 1
3
4
Event 2
6
5
Event 3
8
8
3
Event 0
c Find the speed (with respect to the laboratory frame) of a rocket frame in which evenr 1 and event 2 in the first table occur at the same place. d Find the speed (with respect to the laboratory frame) of a rocket frame moving along the xaxis in which event 2 and event 3 in the first table occur at the same time.
a Copy the second table. In the top half of each box in the second table, write the nature of the interval — timelike, lightlike, or spacelike — between the two corresponding events. b In the bottom half of each box in the second table, write ‘‘yes” if it is possible that one of the events caused the other and “no” if it is not possible.
proper time and proper distance
N ote: This exercise uses the Lorentz transformation equations. a Two events P and Q have a spacelike separa tion. Show in general that a rocket frame can be found in which the two events occur at the same time. Also show that in this rocket frame the distance between the two events is equal to the proper distance between them. (One method: assume that such a rocket frame exists and then use the Lotentz transformation equa tions to show that the relative velocity of this rocket frame is less than the speed of light, thus justifying the assumption made.) b Two events P and R have a timelike separa tion. Show in general that a rocket frame can be found in which the two events occur at the same place. Also show that in this rocket frame the time between the two events is equal to the proper time between them.
PROBLEMS 64 autobiography of a photon A photon emitted by a star on one side of our galaxy is absorbed near a star on the other side of our galaxy.
■/2
(73) (716)
E
(712, 20)
f^Newton
^
pNewton
XnV
Px Newton ~ PxNewton
[1  (v_A)^]‘/2
(72, 5) (72, 5)
/
x
(77, 9)
IT
^ c o n v Newcon
r c o n v Newton
__ _
2
,2 ^ ^ ^ conw
" ‘^^conv
Pxconv Newton Pyconv Newton
fflV y conv
conv Newton
^ ^ 2 conv
VtV^
2 . Total xmomentum of the system is the same before and after the interaction. 3. Total ymomentum of the system is the same before and after the interaction. 4. Total zmomentum of the system is the same before and after the interaction.
In this chapter we have developed expressions that relate energy, momentum, mass, and velocity. Which of these expressions is useful depends upon circumstances and the system we are trying to analyze. Figure 77 summarizes these equations and circum stances under which they may be useful. Table 71 compares energy and momentum in units of mass and in conventional units,
ACKNOWLEDGMENT The authors are grateful to William A. Shurcliff for the idea of a “handle” that displays on a Euclidean page the invariant magnitude of a particle’s momenergy 4vector — a magnitude equal to the mass of the particle.
conv
214
EXERCISE7I
MOMENERGV 4VECTOR
CHAPTE3R 7 EXERCISES
PRACTICE 71
8 there is a lot more discussion about the mass of a system of particles.)
momenergy 4veclor
73 much ado about little
For each of the following cases, write down the four components of the momentumenergy (momen ergy) 4vector in the given frame in the form {E,p^,py, p ^. Assume that each particle has mass m .Y o n may use square roots in your answer. a A particle moves in the positive xdirection in the laboratory with total energy equal to five times its rest energy. b Same particle as observed in a frame in which it is at rest. c Another particle moves in the zdirection with momentum equal to three times its mass. d Yet another particle moves in the negative ydirection with kinetic energy equal to four times its mass. e Still another particle moves with total energy equal to ten times its mass and x, y, and zcomponents of momentum in the ratio 1 to 2 to 3.
72
system mass
Determine the mass of the system of particles shown in Figure 76. Is this system mass equal to the sum of the masses of the individual particles in the system? Does the mass of this system change as a result of the interaction? Does the momenergy 4vector of the sys tem change as a result of the interaction? (In Chapter
Two freight trains, each of mass 5 X 10® kilograms (5000 metric tons) travel in opposite directions on the same track with equal speeds of 42 meters/second (about 100 miles/hour). They collide head on and come to rest. a Calculate in milligrams the kinetic energy for each train (1 /2)m v^ before the collision. (Newtonian expression OK for 100 mph!) (1 milligram = 10~^ gram = 10~® kilogram) b After the collision, the mass of the trains plus the mass of the track plus the mass of the roadbed has increased by what number of milligrams? Neglect energy lost in the forms of sound and light.
74 fast protons Each of the protons described in the table emits a flash of light every meter of its own (propet) time dT. Between successive flash emissions, each proton trav els a distance given in the left column. Complete the table. Take the rest energy of the proton to be equal to 1 GeV = 10’ eV and express momentum in the same units. Hints: Avoid calculating or using the speed V in relativistic particle problems; it is too close to unity to distinguish between protons of radically different energies. An accuracy of two significant fig
< C l^ j^ X E R C IS E 7 4 ^ ^
FAST PROTONS Lab distance Ax traveled between flashes (meters) 0
0.1
10 103
10^
Momentum mdx/dx (GeV)
Energy (GeV)
Time stretch factor y
Lab time between flashes (meters)
EXERCISE 7 7 ures is fine; don’t give more. Recall: and E = mdt/d% = my [note tau!}.
= nE
PROBLEMS 75
Lorentx transformation for momenergy components
The rocket observer measures energy and momentum components of a particle to have the values E' and p j , py , and p^'. W hat are the corresponding values of energy and momentum measured by the laboratory observer? The answer comes from the Lorentz trans formation, equation (L10) in the Special Topic fol lowing Chapter 3. The moving particle emits a pair of sparks closely spaced in time as measured on its wristwatch. The rocket latticework of clocks records these emission events; so does the laboratory latticework of clocks. The rocket observer constmcts components of particle momentum and energy, equation (72), from knowl edge of particle mass m, the spacetime displacements d t ', d x ',d y ', and d z' derived from the event record ings, and the proper time dx computed from these spacetime components. Laboratory momenergy com ponents come from transforming the spacetime dis placements. The Lorentz ttansformation, equation (L10), for incremental displacements gives dt dx dy dz
= = = =
vydx' + ydt' ydx' b vydt' dy' dz'
a Multiply both sides of each equation by the invariant mass m and divide through by the invariant proper time dx. Recognizing the components of the momenergy 4vector in equation (72), show that the transformation equations for momenergy are E = vyp'^ b yE' p . = yp'. + Py = P'y A = /. b Repeat the process for particle displacements dt, dx, dy, and dz recorded in the laboratory frame to derive the inverse transformations from laboratory to rocket. E' = — vyp„ + yE p \ = yp.  vyE P' y=Py
76
SUPER COSMIC RAYS
2 15
fast electrons
The TwoMile Stanford Linear Accelerator accelerates electrons to a final kinetic energy of 47 GeV (47 X 10^ electronvolts; one electronvolt = 1.6 X 10“ ^^ joule). The resulting highenergy electrons are used for experiments with elementary particles. Electro magnetic waves produced in large vacuum tubes (“klystron tubes’’) accelerate the electrons along a straight pipelike structure 10,000 feet long (approxi mately 3000 meters long). Take the rest energy of an electron to be w ^ 0.5 MeV = 0.5 X 10^ electronvolts. a Electrons increase their kinetic energy by ap proximately equal amounts for every merer traveled along the accelerator pipe as observed in the labora tory frame. W hat is this energy gain in MeV/meter? Suppose the Newtonian expression for kinetic energy were correct. In this case how far would the electron travel along the accelerator before its speed were equal to the speed of light? b In reality, of course, even the 47GeV elec trons that emerge from the end of the accelerator have a speed v that is less than the speed of light. W hat is the value of the difference {I ~ p) between the speed of light and the speed of these electrons as measured in the laboratory frame? [Hint: For v very near the value unity, 1 (1 + v){\  v ) ^ 2(1  v).\ Let a 4 7GeV electron from this accelerator race a flash of light along an evacuated tube straight through Earth from one side to the other (Earth diameter 12,740 kilometers). How far ahead of the electron is the light flash at the end of this race? Express your answer in millimeters. C How long is the “ 3000meter’’ accelerator tube as recorded on the latticework of rocket clocks moving along with a 47GeV electron emerging from the acceletatot?
77 super cosmic rays The Haverah Park extensive air shower array near Leeds, England, detects the energy of individual cos mic ray particles indirectly by the resulting shower of particles this cosmic ray creates in the atmosphere. Between 1968 and 1987 the Haverah Park array detected more than 25,000 cosmic rays with enetgies greater than 4 X 10^^ electronvolts, including 5 with an energy of approximately 10^° electronvolts, (rest energy of the proton = 1 0 ^ electronvolts = 1.6 X 10~^“ joule) a Suppose a cosmic ray is a proton of energy 10^° electronvolts. How long would it take this proton to cross our galaxy as measured on the proton’s wristwatch? The diameter of our galaxy is approximately
216
EXERCISE 78
ROCKET NUCLEUS
10’ lightyears. How many centuries would this trip take as observed in our Earthlinked frame? b The research workers at Haverah Park find no evidence of an upper limit to cosmic ray energies. A proton must have an energy of how many times its rest energy for the diameter of our galaxy to appear to it Lorentzcontracted to the diameter of the proton (about 1 femtometer, which is equal to 10“ ^’ meters)? How many metric tons of mass would have to be converted to energy with 100percent efficiency in order to give a proton this energy? One metric ton equals 1000 kilograms. Reference: M. A. Lawrence. R. J. O, Reid, and A. A. Watson, Journal of Physics G: Nuclear and Particle Physics, Volume 17, pages 733757 (1991).
78
rocket nucleus
A radioactive decay or “inverse collision” is observed in the laboratory frame, as shown in the figure. Suppose that = 20 units, m Q = 2 units, and £ (  = 5 units. a W hat is the total energy £^ of particle A? b From the conservation of energy, find the total energy (rest plus kinetic) of particle D. c Using the expression = ni^ find the momentum pc of particle C. d From the conservation of momentum, find the momentum p^, of particle D. e W hat is the mass of particle D? f Does me + m^, after the collision equal before the collision? Explain your answer. g Draw three momenergy diagrams for this re action similar to those of Figure 76: BEFORE, SYS TEM, and AFTER. Plot positive and negative mo mentum along the positive and negative horizontal direction, respectively, and energy along the vertical direction. On the AFTER diagram draw the momen ergy vectors for particles C and D head to tail so that they add up to the momenergy vector for the system. Place labeled mass handles on the arrows in all three diagrams, including the arrow for the system.
Q particle A (at rest)
BEFORE EXERCISE 7 8 .
particle C
particle D
AFTER Radioactive decay of a particle.
particle A
particle C (at rest)
particle 6
AFTER
BEFORE EXERCISE 7 9 .
Two particles collide to form a third at rest in the
laboratory frame.
79 sticky collision An inelastic collision is observed in the laboratory frame, as shown in the figure. Suppose that = 2 units, £^ = 6 units, ~ 15 units. a From the conservation of energy, what is the energy Eg of particle B? b W hat is the momentum p^ of particle A? Therefore what is the momentum pg of particle B? c From m^ = E^ — p^ find the mass mg of par ticle B. d Quick guess: Is the mass of particle C after the collision less than or greater than the sum of the masses of particles A and B before the collision? Vali date your guess from the answer to part c.
710
colliding putty balls
A ball of putty of mass m and kinetic energy K streaks across the frozen ice of a pond and hits a second identical ball of putty initially at rest on the ice. The two stick together and skitter onward as one unit. Referring to the figure, find the mass of the combined particle using parts a  e or some other method. a W hat is the total energy of the system before the collision? Keep the kinetic energy K explicitly, and don’t forget the rest energies of both particles A and B . Therefore what is the total energy E c of particle C after the collision? b Using the equation m^ — E^ — p^ = {m  \K) 2 — p 2 momentum p^^ of particle A before the collision. W hat is the total momentum of the system before the collision? Therefore what is the momentum pc of particle C after the collision?
o—
particle
A
particle B (at rest)
particle C
BEFORE EXERCISE 7 1 0 .
Two putty balls stick together.
AFTER
EXERCISE 712
DERIVATION OF RELATIVISTIC EXPRESSION FOR MOMENTUM
c Again use the equation to find the mass me o f particle C Show that the result satisfies the equation
te — {2my + 2mK = {2mY
(9
d Examine the result of part c in two limiting cases. (1) The value of mc 'tn the Newtonian lowve locity limit in which kinetic energy is very much less than mass: K /m « 1. Is this what one expects from everyday living? (2) W hat is the value of in the highly relativistic limit in which K /m » 1? W hat is the upper limit on the value of m ^ D iscussion: Submicroscopic particles moving at extreme relativis tic speeds rarely stick together when they collide. Rather, their collision often leads to creation of addi tional particles. See Chapter 8 for examples. e D iscussion q uestion: Are the results of part c changed if the resulting blob of putty rotates, whir ling like a dumbbell about its center as it skitters along?
711
limits off Newtonian mechanics
a One electronvolt (eV) is equal to the increase of kinetic energy that a singly charged particle experi ences when accelerated through a potential difference of one volt. One electronvolt is equal to 1.60 X 10“ *^ joules. Verify the rest energies of the electron and the proton (masses listed inside the back cover) in units of million electronvolts (MeV). b The kinetic energy of a particle of a given velocity v is not correctly given by the expression 1/2 mv^. The error relativistic expressions
expressions ( for kinetic energy / (Newtonian for kinetic energy / Newtonian expressions ( for kinetic energy I is one percent when the Newtonian kinetic energy has risen to a certain fraction of the rest energy. W hat fraction? Hint: Apply the first three terms of the binomial expansion (1 + Z ) ’’ 
1 + «z H— «(«
2
1)
+ . . .
to the relativistic expression for kinetic energy, an accurate enough approximation if z « 1. Let this point— where the error is one percent— be arbitrar ily called the “limit of Newtonian mechanics.” W hat
217
is the speed of the particle at this limit? At what kinetic energy does a proton reach this limit (energy in MeV)? An electron? c An electron in a modern color television tube is accelerated through a voltage as great as 25,000 volts and then directed by a magnetic field to a particular pixel of luminescent material on the inner face of the tube. Must the designer of color television tubes use special relativity in predicting the trajectories of these electrons?
712
derivation off the reiativistic expression ffor momentum— a worked example
A very fast particle interacts with a very slow particle. If the collision is a glancing one, the slow particle may move as slowly after the collision as before. Reckon the momentum of the slowmoving particle using the Newtonian expression. Now demand that momen tum be conserved in the collision. From this derive the relativistic expression for momentum of the fastmoving particle. The top figure shows such a glancing collision. After the collision each particle has the same speed as before the collision, but each particle has changed its direction of motion. Behind this figure is a story. Ten million years ago, and in another galaxy nearly ten million lightyears distant, a supernova explosion launched a proton toward Earth. The energy of this proton far exceeded anything we can give to protons in our earthbound particle accelerators. Indeed, the speed of the proton so nearly approached that of light that the proton’s wristwatch read a time lapse of only one second be tween launch and arrival at Earth. W e on Earth pay no attention to the proton’s wristwatch. For our latticework of Earthlinked ob servers, ages have passed since the proton was launched. Today our remote outposts warn us that the streaking proton approaches Earth. Exactly one second on our clocks before the proton is due to arrive, we launch our own proton at the slow speed one meter/second almost perpendicular to the direction of the incoming proton (BEFORE part of the top figure). Our proton saunters the one meter to the impact point. The two protons meet. So perfect is our aim and timing that after the encounter our proton simply reverses direction and returns with the same speed we gave it originally (AFTER part of the top figure). The incoming proton also does not change speed, but it is deflected upward at the same angle at which it was originally slanting downward.
218
EXERCISE 712
DERIVATION OF RELATIVISTIC EXPRESSION FOR MOMENTUM
i Earth Frame: BEFORE
I Earth Frame: AFTER
EXERCISE 712. Top; A symmetric elastic collision between a fast proton and a slow proton in which each proton changes direction hut not speed as a result of the encounter. Center: Events and separations as observed in Earth frame before the collision. Here x = 10 million lightyears and y = I meter, so these figures are not to scale! Bottom; Events and separations as observed in the rocket frame before the collision.
How much does ^/momentum of our slowmov ing proton change during this encounter? Newton can tell us. At a particle speed of one meter/second, his expression for momentum, mv, is accurate. Our pro ton simply reverses its direction. Therefore the change in its momentum is just 2 mv, twice its original mo mentum in the ydirection. W hat is the change in the ymomenmm of the incoming proton, moving at extreme relativistic speed? W e demand that the change in jmomentum of the fast proton be equal in magnitude and opposite in direction to the change in ymomentum of our slow proton. In brief, ymomentum is conserved. This de mand, plus a symmetry argument, leads to the rela tivistic expression for momentum. Key events in our story are numbered in the center figure. Event 1 is the launching of the proton from the supernova ten million years (in our frame) before the impact. Event 2 is the quiet launch of our local proton one second (in our frame) before the im paa. Event 0 is the impact itself. The xdirection is chosen so that ydisplacements of both protons have equal magni tude between launch and impact, namely one meter. Now view the same events from a rocket moving along the xaxis at such a speed that events 1 and 0 are
vertically above one another (botrom figure). For the rocket observer the transverse yseparations are the same as for the Earth observer (Section 3.6), soy = 1 meter in both frames. The order of events 1 and 2, however, is exactly reversed in time: For the racket observer, we released our proton at high speed ten million years before impact and she releases hers one second before the collision. Otherwise the diagrams are symmetrical: To make the bottom figure look like the center one, exchange event numbers 1 and 2, then stand on your head! Rocket observer and Earth observer do not agree on the time between events 1 and 0, but they agree on the proper time Tm between them, namely one sec ond. They also agree on the propet time T20 between events 2 and 0. Moreover, because of the symmetry between the center and bottom figures, these two proper times have the same value: For the case we have chosen, the wristwatch (proper) time for each proton is one second between launch and impact. no
> •2 0
W e can use these quantities to construct expres sions for the ymomenta of the two protons. Both are
EXERCISE 712
DERIVATION OF RELATIVISTIC EXPRESSION FOR MOMENTUM
protons, so their masses m are the same and have the same invariant value for both observers. Because of the equality in magnitude of theydisplacements and the equality of Tjo and T,o, we can write
m
y _
y
 M 
"fio
[both frames]
'^20
The final key idea in the derivation of the relativis tic expression for momentum is that the slowmoving proton travels between events 2 and 0 in an Earthmeasured time that is very close in value to the proper time between these events. The vertical separation y between events 2 and 0 is quite small; one meter. In the same units, the time between them has a large value in the Earth frame: one second, or 300 million meters of lighttravel time. Therefore, for such a slowmoving proton, the proper time X20 between events 2 and 0 is very close to the Earth time (2 0 between these events; T 20
^20
^
[Earth frame only]
Hence rewrite the bothframes equation for the Earth frame: y M
_

^10
_ tn
y

[Earth frame only]
^20
The right side of this equation gives the ymomenturn of the slow proton before the collision, correctly cal culated using the Newtonian formula. The change in momentum of the slow proton during the collision is twice this magnimde. Now look at the left side. We claim that the expression on the left side is the ymomentum of the very fast proton. Theymomentum of the fast proton also reverses in the collision, so the change is just twice the value of the left side. In brief, this equation embodies the conservation of the ycomponent of total momentum in the collision. Con
219
clusion: The left side of this equation yields the rela tivistic expression for ymomentum: mass times displacement divided by proper time for this displace ment. W hat would be wrong with using the Newtonian expression for momentum on the left side as well as on the right? That would mean using earth time instead of proper time T,o in the denominator of the left side. But is the time it took the fast proton to reach Earth from the distant galaxy as recorded in the Earth frame — ten million years or 320 million mil lion seconds! W ith this substitution, the equation would no longer be an equality; the left side would be 320 million million times smaller in value than the right side (smaller because r,o would appear in the denominator). Nothing shows more dramatically than this the radical difference between Newtonian and relativistic expressions for momentum — and the correctness of the relativistic expression that has proper time in the denominator. This derivation of the relativistic expression for momentum deals only with its ycomponent. But the choice of ydirection is arbitrary. We could have in terchanged y and x axes. Also the expression has been derived for particles moving with constant velocity before and after the collision. When velocity varies with time, the momentum is better expressed in terms of incremental changes in space and time. For a parti cle displacement dr between two events a proper time dx apart, the expression for the magnitude of the momentum is dr p = m—
^
dx
Onesentence summary; In order to preserve con servation of momentum for relativistic collisions, simply replace Newton’s “universal time” t in the expression for momentum with Einstein’s invariant proper time T .
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8.1 THE SYSTEM an isolated island of violence Particle physics is one of the gteat adventures of our time. No one can venture into the heart of it without momenergy as guide and lamp. Particles clash, yes. But however cataclysmic the encounter, it always displays one great simplicity. It takes place on a local stage, an island of violence, apart from all happenings in the outside world. In other isolated arenas of action football players form a team, actors a troupe, soldiers a platoon; but in a battle of matter and energy, the participants receive the name system . W hat the action starts with, what particles there are, what speeds they have, what directions they take: that’s the story of the system at the start of the action. W e may or may not pursue in all detail every stage of every encounter, as we view the scenes of a play or watch the episodes of a game. However, nothing that claims to be an account of the clash, brief though it may be, is worthy of the name unless it reports every participant that leaves the scene with its speed and its direaion. Departing, they still belong to the system. Moreover, at every step of the way from entry to departure we continue to use for the collection of participants the name system. The child keeps count of who wins and who loses in the shootout before he or she learns to ask questions of right and wrong, of why and wherefore. W e likewise keep tabs on what goes into an encounter and what comes out only to the extent of broadcasting the participants’ momenergies before and after the act of violence. We do not open up in this book the more complex story of the forces, old and new, that govern the chances for this, that, and the other outcome of a given encounter. W e limit outselves to the ground mles of momenergy conservation in an isolated system,
221
Keep score of momenergy for the system
222
CHAPTER 8
COLLIDE. CREATE. ANNIHILATE.
8.2 THREE MODEST EXPERIMENTS elastic glass balls; inelastic wads off gum; weighing heat
Elastic collision: Momenergy automatically conserved
A collision does not have to be violent to qualify for attention nor be exotic to make momenergy scorekeeping interesting. It is fun to begin with momenergy scorekeeping for three encounters of everyday kinds before strolling out onto the laboratory floor of highenergy particle physics. First E xperim ent: Elastic Collision. Suspend two identical glass marbles from the ceiling by two threads of the same length so that the marbles hang, at rest, just barely touching. Draw one back with the finger and release it (Figure 81). The released marble gathers speed. The speed peaks just as the first marble collides with the second. The collision is elastic: Total kinetic energy before the collision equals total kinetic energy after the collision. The elastic collision brings the first marble to a complete stop. The impact imparts to the second all the momentum the first one had. Conservation of momentum could not be clearer: /
total momentum \ to the right just b efo re the collision, all of it resident on ' the first marble '
Inelastic collision: Momenergy also conserved
/ total momentum \ to the right just a fte r the collision, all of it resident on 'th e second marble/
And energy? In the collision the two particles exchange roles. The first patticle comes to a halt. The second particle moves exactly as the first one did before the collision. Hence energy too is clearly conserved. Just before the collision and just after: How do conditions compare? Same total momentum. Same total energy. Therefore same total momenergy. Second E xperim ent: Inelastic Collision. Replace the two glass marbles by two identical balls of putty, wax, or chewing gum (Figure 82). Pull them aside by equal amounts and release. Both released balls of chewing gum gather speed, moving toward one anorher. The equal and opposite velocities peak just before they collide with each other. By symmetry, the momentum of the rightmoving particle has the same magnitude as the momentum of the leftmoving particle. However, these momenta point in opposite directions. Regarded as vectors, they sum to zero. The momentum of the system therefore equals zero just before the collision. Just after rhe collision? The two balls have stuck together. They are both at rest; each has zero momentum. Their combined momentum is also zero. In other words.
LAB RECORDS, INC. i
FIGURE 81. One marble collides elastically with another.
8.2
THREE MODEST EXPERIMENTS
223
I I I I I I I I I I I I I I I I I I I I I I I I I I lab r e c o r d s ,
FIGURE 82. One ball of chewing gum locks onto the other.
the momentum of the system is zero after the collision. Zero it was also before the collision. Thus the momentum of the system is conserved. For system energy the outcome is more perplexing. Just before the collision, each ball has an energy consisting of its mass m and its kinetic energy K. These energies add to make the total energy of the system: = 2m + 2K. After the collision? Both balls of chewing gum are at rest, stuck together as a single blob, which now constitutes the entire system. The energy of that stationary blob must be its rest energy, equal to the mass of the system: Fsystem“ ^rest ~ ■^system W hat is the value of that system energy? It must be the same as the energy of the system before the collision, equal to 2m + 2K, where m is the mass of each ball before the collision. Hence, if energy is conserved, = 2w + 2K. This is greater than the sum of masses of the incoming particles. Where does this extra mass come from? The energy of relative motion of the incoming particles gets converted, during the collision, into energy of plastic deforma tion and heat. Each of these forms of lockedin energy yields an increment of mass. In consequence the mass of the pair of balls, stuck together as one, exceeds the sum of masses of the two balls before impact. T h ird E xperim ent: W eighing H eat. If warmed and distorted balls of gum have more mass than cool and undistorted balls, then maybe we can measure directly the increased mass simply by heating an object and weighing it. In this case the system consists of a single large object, such as a tub of water, stationary and therefore with zero total momentum. System energy consists of the summed individual masses of all water molecules plus the summed kinetic energies of their random motions. This summed kinetic energy increases as we add heat to the water; hence its mass should increase. Can we detect the corresponding increase in weight as we heat the water in the tub? Alas, never yet has anyone succeeded in weighing heat. In 1787 Benjamin Thomp son, Count Rumford (1 7 5 3  1814), tried to detect an increase in weight of barrels of water, mercury, and alcohol as their temperature rose from 29° E to 6 l ° E (in which range ice melts). He found no effect. He concluded “that ALL ATTEMPTS TO DISCOVER ANY EFFECTS OF HEAT UPON THE APPARENT WEIGHTS OF BODIES WILL BE ERUITLESS” (capital letters his). Professor Vladimir Bra ginsky of the University of Moscow once described to us a new idea for weighing heat. Let a tiny quartz pellet hang on the end of a long thin nearhorizontal quartz fiber, like a reeledin fish at the end of a long supple fishing rod. A fly that settles on the fish increases its weight; the fishing rod bends a little more. Likewise heat added to the pellet will increase its mass and will bend the quartzfiber “fishing rod” a little more. That is the idea. The sensitivity required to detect a bending so slight unfortunately surpasses the present limit of technology. Braginsky himself already has invented, published, and made available to workers all over the world a now widely applied scheme to measure very small effects. There is real hope that he— or someone else — will weigh heat and confirm what we already confidently expect.
Kinetic energy converted to moss
Con we weigh heat? Not yet!
8.3 MASS OF A SYSTEM OF PARTICLES energies add. momenta add. masses do not add.
8
8
mass
a
Tiass v= 3/5
3/5
20 mass
O
20 energy
0 momentum
CONSIDERED AS A SINGLE SYSTEM
20 mass
CO
■o,)
FIGURE 83. Two noninteracting particles, each of mass 8, are in relative motion. Taken together, they constitute a system of mass 20. Where does the mass 2 0 reside? In the system!
No one with any detective instincts will rest content with the vague thought that heat has mass. Where within our stucktogether wads of chewing gum or Rumford’s barrel of water or Braginsky’s quartz pellet is that mass located? In random morions of the atoms? Nonsense. Each atom has mass, yes. But does an atom acquire additional mass by virtue of any motion? Does motion have mass? No. Absolutely not. Then where, and in what form, does the extra mass reside? Answer: Not in any part, but in the system. Heat resides not in the particles individually but in the system of particles. Heat arises not from motion of one particle but from relative motions of two or more particles. Heat is a system pro p erty . The mass of a system is greatet when system parts move telative to each other. O f this central point, no simplet example offers itself than a system composed of a single pait of masses. Our example? Two identical objects (Figure 83). Each has mass 8. Relative to the laboratory frame of reference each object has momentum 6, but the two momenta are opposite in direction. The energy of each object is E = (jrP f = (8" + 6^)V2 = 10. The total momentum of the twoobject system is / ’system— 6 — 6 = 0. The energy of the system is = 10 + 10 = 20. Therefore the mass of the system is Msystem “ (Esystem^ ~ / ’system^) ~ [(20)^ — 0^]*/^ = 20. Thus the mass of the system exceeds the sum of the masses of the two parts of the system. The mass of the system does not agree with the sum of the masses of its parts. Energy is additive. Momentum is additive. But mass is ttol additive. Ask where the extra 20 — 16 = 4 units of mass are located? Silly question, any answer to which is also silly! Ask where the 20 units of mass are located? Good question, with a good answer. The 20 units of mass belong to the system as a whole, not to any part individually. Where is the life of a puppy located? Good question, with a good answet. Life is a property of the system of atoms we call a puppy, not a property of any part of the puppy. Where is the extra ingredient added to atoms to yield a live puppy? Unacceptable question, any answer to which is also unacceptable. Life is not a property of any of the individual atoms of which the puppy is constituted. Nor is it a property of the space between the atoms. Nor is it an ingredient that has to be added to atoms. Life is a ptoperty of the puppy system. Life is remarkable, but in one respect the twoobject system that we are talking about is even more remarkable. Life requires organization, but the twoobject system of Figure 83 lacks organization. Neither mass interacts with the other. Yet the total energy of the twoobject system, and its total momentum, regarded from first one frame of reference, then another, then another, take on values identical in every respect to the values they would have were we dealing throughout with a single object of mass 20 units. Totally unlinked, the two objects, viewed as a system, possess the dynamic attributes — energy, momentum, and mass — of a single object. This wider idea of mass — the mass of an isolated system composed of disconnected objects: what right have we to give it the name “mass”? Nature, for whatever reason, demands conservation of total momenergy in every collision. Each collision, no matter how much it changes the momenergy of each participant, leaves unchanged the sum of their momenergies, regarded as a directed arrow in spacetime— a 4vector. Encounter or no encounter, and however complex any encounter, system momenergy does not alter. Neither in spacetime direction nor in magnitude does it ever change. But the magnitude — the length of the arrow of total momenergy, figured as we figure any spacetime interval — is system mass. Whether the system consists of a single object or
8.3
MASS OF A SYSTEM OF PARTICLES
of many objects, and whether these objects do or do not collide or otherwise interact with each other, this system mass never changes. That’s why the concept of system mass makes sense! An example? Again, two objects of mass 8, again each moving toward a point midway between them z.tv = (momentum)/(energy) = ( p = 6 ) /( E = 10) = 3 /5 the speed of light. Now, however, we analyze the two motions in a frame moving with the righthand object (Figure 84). In this new frame the righthand object is at rest: mass, m = 8; momentum, p = 0\ energy, E = [m^ b p^V^^ = 8. The lefthand object is approaching with a speed (addition of velocities: Section L.7 of the Special Topic following Chapter 3; also Exercise 311) 3/5 + 3/5
6/5
1 + (3 /5 )(3 /5 )
34/25
17
It has energy E = m /[l — — 8 / [ l — (15/17)^}*/^ — 17 and momentum p — vE = 15. So much for the parts of the system! Now for the system itself. For the system the energy is E^^^^ — 8 + 1 7 — 25 and the momentum is p^ 0 + 15 = 15. Now for the test! Does the concept of system mass make sense? In other words, does system mass turn out to have the same value in the new frame as in the original frame? It does: ^ s y s te m =
^2)1/2 = (25)2  (i5)2]i/2 = i;625  225]!/^
(^ s,
= I400y/^  20
8
mass
8 energy
25 energy
BEFORE
SYSTEM
AFTER
(before one/after!)
8
8
mass
O 
20 mass
15/17
'= 3 /5
FIGURE 84. System of Figure 83 observedfrom a fra m e moving w ith the righthand object. The righthand object is therefore initially at rest. Before: Arrows of momenergyfor two objects before collision. Each object has a mass of eight units (shaded handles). The upper, vertical, arrow belongs to the particle originally at rest, the slanted arrow to the incoming particle. System: Addition of the two momenta (one of them zero!) gives the total momentum before collision. Similarly, addition of the two energies gives the total energy. Mass of the system — even before the two particles interact!— comes from the expression for the "hypotenuse” of a spacetimelike triangle. Result: 2 0 units of mass (shaded handle on center 4vector): (mass)^ = (energy)^ — (momentum)^ = (25F ~ (D F ~ 625 — 22 5 = 4 0 0 = (20)^ After: The two particles now collide and amalgamate to form one particle. Arrow of total momenergy after the amalgamation is identical to arrow of total momenergy before the collision. Mass of this twoobject system exceeds the mass of one object plus the mass of the other, not only after the collision but also before. Mass is not an additive quantity.
225
Different freefloat frames. Same system mass.
226
CHAPTER 8
COLLIDE. CREATE. ANNIHILATE.
S A M P L E P R O B L E M 81
MA S S OF A S Y S T E M OF MATERIAL PARTICLES Compute Msystemfor each of the following systems. The particles that make up these systems do not interact with one another. Express the system mass
in terms of the unit mass m, do not use momenta or velocities in your answers. [Note: In the following diagrams, arrows represent (3vector) momenta.}
System a (kinetic energy
= K = 3m]
 ►
m
(at rest)
System b (kinetic energy
= K = 5m)
' O  ^
(kinetic energy = K = 5m)
'O  ^
System c 3m
o
(energy = E = 7m) m(3
rest)
System d
o (E = 6m)
'O
(E = 6m)
SOLUTION System a: System energy equals the rest energy of the two particles (the sum of their masses) plus the kinetic energy of the moving particle: = (»z h w) f 3»z = 5m. Squared momentum of the system equals that of the moving particle: ~ P ^~ E? — m^ = {AmY — mA= \ 5m^. Mass of the system is reckoned from the difference between the squares of energy and momentum: ^system
•
~
= [lOY/^m = 3.162 m
Moreover, if the two objects collide and amalgamate, the system energy remains at the value 25, the system momentum remains at the value 15, and the system mass remains 20, as illustrated in Figure 84. In summary, the mass of an isolated system has a value independent of the choice of frame of reference in which it is figured. System mass remains unchanged by en counters between the constituents of the system. And why? Because the system mass is the length (in the sense of spacetime interval) of the arrow of total momentumenergy. This momenergy total is unaffected by collisions among the parts or by any transfor mations, decays, or annihilations they may undergo. System mass does make sense! System! System! You keep talking about “system,” even when the particles do not interact, as in the system o f Figure 83 It seems to me th a t you are totally arbitrary in the way you define a system. Who chooses which particles are in the system?
8.3
MASS OF A SYSTEM OF PARTICLES
System b: System energy equals rest energy of the two particles plus kinetic energy of the two particles; = 2w + lOm = \2m. Squared momentum of each particle is ^ = E^ — np ' ( v= 1
FIGURE 811 . Momenergy conservation in the twophoton electronpmsitron annihilation process. Before'. Before annihilation each oppositely charged particle has rest energy and no momentum. After: The two particles have annihilated, creating two highenergy photons (gamma rays). The two photons fly apart in opposite directions; total momentum remains zero.
CONVERTING MASS TO USABLE ENERGY: FISSION, FUSION, ANNIHILATION
ANALYZING A PARTICLE ENCOUNTER Conservation of total momenergy! In any given freefloat frame that means conservation of total energy and conservation of each of three components of total momentum. In no way does the power and scope of this principle make itself felt more memorably than the analysis of simple encounters of this, that, and the other kind in an isolated system of particles. “Analyzing an encounter” means using conservation laws and other relations to find un known masses, energies, and momenta of particles in terms of known quanti ties. Sometimes a complete analysis is not possible; the information provided may be insufficient. Here are suggested steps in analyzing an encounter. Sample Problems 83 and 84 illustrate these methods. 1. Draw a diagram of particles before and particles a ft e r the interaction. Label particles entering with numbers or letters and particles leaving with different numbers or letters (even if they are the same particles). Use arrows to show particle directions of motion and label with symbols their masses, energies, and momenta, whether initially known or unknown. 2. Write down algebraically the conservation of total energy. Do not forget to include the rest energy — the mass m — of any particle not moving in the chosen freefioat frame. 3. Write down algebraically the conservation of total momentum. Do not forget that momentum is a vector. In general this means demanding con servation of each of three components of total momentum. 4. Try to solve for unknowns in terms of knowns, still using symbols. a. Make liberal use of the relation m^ = — p^, where p^ = p„^ + p^^ + p^^. For a photon or neutrino, mass equals zero and £ = p (in magni tude: Pay attention to the direction of the momentum vector p — or its sign if motion is in one space dimension). b. Do NOT use speed v of a particle unless forced to by requirements of the problem. Relativistic particles typically move with speeds very close to light speed, so speed proves to be a poor measure of signifi cance. Increase by one percent the speed of a particle moving at v = 0.99 and you increase its energy by a factor of almost 10. c. Substitute numerical values into resulting equations as late as possible. Before substituting numerical values, check that all values are ex pressed in concordant units. 5. Check your result. Check units of the solution. Is the order of magnitude of numerical results reasonable? Substitute limiting values, for example let ting energy of an incoming particle become very large (and very small). Is the limitingcase result reasonable? Is there any general conclusion you can draw from your specific solution? Does this exercise illustrate a deep principle or lead to an even more inter esting application of conservation laws?
239
240
CHAPTER 8
COLLIDE. CREATE. ANNIHILATE.
S A M P L E P R O B L E M 83
SYMMETRIC ELASTIC C O L L IS IO N A proton of mass m and kinetic energy K in the laboratory frame strikes a proton initially at rest in that frame. The two protons undergo a symmetric elastic collision; the outgoing protons move in di rections that make equal and opposite angles 0 /2 with the line of motion of the original incoming particle. Find energy and momentum of each out going particle and angle 0 between their outgoing directions of motion for rhis symmetric case.
SOLUTION/ 1.
H istorical note; When impact speed is small compared to the speed of light, this separation of directions, 0, is 90 degrees, according to Newto nian mechanics. Early cloudchamber tracks sometimes showed symmetric collisions with angles of separation substantially less than 90 de grees, thereby giving evidence for relativistic me chanics and providing the first reliable measure ments of impact energy.
following steps in Box 82
D raw a diag ram and label all four particles with letters: 'P c
Symmetry of this diagram implies that the two outgoing particles have equal energy and equal magnitude of momentum; that is, — Ej and (in magnitude) Pc =
Pd
2. C onservation o f energy: Energy of each particle equals mass plus kinetic energy. And the masses don’t change in this reaction. Therefore rotal kinetic energy after the encounter (divided equally between the two particles) equals the (known) total kinetic energy before the encounter, all localized on one particle. In brief; K ^= K j — K J 2 = K /2. Simple answer to one of the three questions we were asked! 3. C onservation o f m o m en tu m : By symmetry, rhe vertical components of momenta of the outgoing particles cancel. Horizontal components add, leading to the relation A ot
Pa ^
Pc
cos(0/2)
b
p j cos(0/2) = 2pj cos(0/2)
or, in brief. p ^
—
2 p j
C O s (0 / 2 )
[conservation of momentum]
4. Solve fo r th e u n k n o w n angle 0: Along the way find the other requested
quantity, the magnitude p c ~ pd of the momenta after the collision. To that end, first find the momentum p^ before the collision, using the general formula for the momentum of an individual particle:
8.7
CONVERTING MASS TO USABLE ENERGY: FISSION, FUSION, ANNIHILATION
P = { E ^  w2]V2 = = (K^ + 2m K y/^
+ m y  »?2}1/2 = (^^2 + 2mK +
241
 m^y/^
Therefore = (K^ + 2/wK)*/2 From conservation of energy, K^ = K j = K /2. Therefore p , = [{K /2 y + 2m {K /2)y/^ Substitute these expressions for p^ and p j into the equation for conservation of momentum: (fC2 + 2« fO '/2 = 2[(fC/2)2 + 2m (K /2)y/^cos(9/2) Square both sides and solve for cos^(0/2) to obtain „
K + 2m
Now apply to this result the trigonometric identity , _ (cos 0 + 1 ) cos2(0/2) = ^ After some manipulation, obtain the desired result: cos e =
(K/m) (K/m) + 4
Here K is the kinetic energy of the incoming particle, m the mass of either particle, and 6 the angle between outgoing particles. This result assumes (1) an elastic collision (kinetic energy conserved), (2) one particle initially at rest, (3) equal masses of the two particles, and (4) the symmetry of outgoing paths shown in the diagram. 5a. L im iting case; Low energy. In the case of low energy (Newtonian limit), the incoming particle has a kinetic energy K very much less than its rest energy m, so the ratio K/m approaches zero. In the limit, cos 6 becomes zero and 9 = 90 degrees. This is the accepted Newtonian result for low velocities (except for an exactly headon collision, in which case the incoming particle stops dead and the struck particle moves forward with the same speed and direction as the original incoming particle). 5b. L im itin g case: H igh energy. For extremely highenergy elastic collisions, the incident particle has a kinetic energy very much greater than its rest energy, so the ratio K/m increases without limit. In this case the quantity 4 in the denominator becomes negligible compared with K/m, so numerator and denominaror both approach the value K/m, with the result cos 0 ^ 1 and 0  ^ 0 . This means that in the special symmetric case discussed here both resulting particles go forward in the same direction as the incoming particle, sharing equally the kinetic energy of the incoming particle. For an incoming particle of very high energy, the elastic collision described here is only one of several possible outcomes. Alternative processes include creation of new particles.
S A M P L E P R O B L E M 84
ANNIHILATION A positron of mass m and kinetic energy equal to its mass strikes an electron at rest. They annihilate, creating two highenergy photons. One photon enters a detector placed at an angle of 90 degrees
SOLUTION/ 1.
with respect to the direction of the incident posi tron. W hat are the energies of both photons (in units of mass of the electron) and direction of motion of the second photon?
following steps in Box 82
D raw a diagram and label the particles with letters. Pb = 0 Eb = m m
Ill
Ec, Pc
^a/ pa
b Ed, Pd
BEFORE 2.
AFTER
C onservation o f energy expressed ii^ the symbols of the diagram, and includ ing the rest energy of the initial stationary particle: m '■ E .+ E,
3.
C onservation o f each co m p o n en t o f to tal m o m en tu m : P x tot
Pa
[horizontal momentum]
P c COS 0
Pywc = 0 = p , s i n 6  Pj 4.
[vertical momentum]
Solve; First of all, the problem states that the kinetic energy K of the incoming positron equals its rest energy m. Therefore its total energy E^ = m\~ K = m\ m = 2m. Second, the outgoing particles are photons, for w h i c h a n d pj = Ej in magnitude, respectively. W ith these substitutions, the three conservation equations become E^
m — 2m m Pa = E, cos 6 Ej = sin 6
^ tn
~
E^
[conservation of energy]
E j
[conservation of horizontol momentum] [conservation of vertical momentum]
S A M P L E P R O B L E M 8 5
C O N V E R S I O N OF M A S S TO E N E R G Y IN SUN Luminous energy from Sun pours down on the outer atmosphere of Earth at a rate of 1372 watts per square meter of area that lies perpendicular to the direction of this radiation. The figure 1372 watts per square meter has the name so lar co n
stant. The radius of Earth equals approximately 6.4 X 10^ meters and the EarthSun distance equals 1.5 X 10“ meters. The mass of Sun is approximately 2.0 X 10^° kilograms.
243 These are three equations in three unknowns and £^and 6. Square both sides of the second and third equations, add them, and use a trigonometric identity to get rid of the angle 9: p j + E / = £/(cos^ 6 + sin^ 9) = E / Substitute p / = E / — nE on the left side of this equation and again use E^ = 2m to obtain a first expression for £ /: E^ = E ^ — rrp
E / = ArtE — rrp
E / = ^np + £ /
Now solve the equation of conservation of enetgy for E^ and square it to obtain a second expression for £ /; £ / = {5m  EjY = 9m^  6mEj + £ / Equate these two expressions for E / and subtract E / from borh sides to obtain 5rrP = 9m^ — 6mEj Solve for unknown Ej. 9nP — 5nP 6m^ Ej =  m Gm Gm This yields our first unknown. Use this result and conservation of energy to find an expression for E/. E^ = 5m — Ej = 5m — m = 2m Finally, angle 9 comes from conservation of vertical momentum. For a photon p = £, so sin 9 '
£^
2m
2
from which 9 — 50 degrees. W e have now solved for all unknowns: £„ — 2m, Ej = m, and 9 = 50 degrees. 5.
L im iting cases: There is no limiting case here, since the energy of the incoming positron is specified fully in terms of the mass m. common to electron and positron.
a.
How much mass is converted to energy every second in Sun to supply the luminous energy that falls on Earth?
b.
W hat total mass is convened to energy every second in Sun to supply luminous energy?
c.
Most of Sun’s energy comes from burning hydrogen nuclei (mostly protons) into helium nuclei (mostly a twoproton  twoneutron combination). Mass of the proton equals 1.67262 X 10“ ^^ kilogram, while the mass of a helium nucleus of this kind equals 6.64648 X 10“ ^^ kilogram. How many metric tons of hydrogen
S A M P L E P R O B L E M 8 5 must Sun convert to helium every second to supply its luminous output? (One metric ton is equal to 1000 kilograms, or 2200 pounds.) d.
Estimate how long Sun will continue to warm Earth, neglecting all other processes in Sun and emissions from Sun.
SOLUTION a.
One watt equals one joule per second = one kilogram meter^/second^. We want to measure energy in units of mass— in kilograms. Do this by dividing the number of joules by the square of the speed of light (Section 7.5 and Table 71): 1372 joules _ 1.372 X 10^ kilogram metersVsecond^
9.00 X 10^®meters^/second^ = 1.524 X 10“ ''' kilograms Thus every second 1.524 X 10“ *'' kilogram of luminous energy falls on each square meter perpendicular to Sun’s rays. The following calculations are based on a simplified model of Sun (see last paragraph of this solution). Therefore we use the approximate value 1.5 X 10“ *'' kilogram per second and twodigit accuracy. W hat total luminous energy falls on Earth per second? It equals the solar constant (in kilograms per square meter per second) times some area (in square meters). But what area? Think of a huge movie screen lying behind Earth and perpendicular to Sun’s rays (see the figure). The shadow of Earth on this screen forms a circle of radius equal to the radius of Earth. This shadow represents the zone of radiation removed from that flowing outward from Sun. Call the area of this circle the crosssectional area A of Earth. Earth’s radius r = 6.4 X 10^ meters, so the crosssectional area A seen by incoming Sunlight equals A = = 1.3 X 10*'' meters^. Hence a total luminous energy equal to (1.5 X 10“ *'' kilograms/ meter^) X (1.3 X 10*'' meters^) = 2.0 kilograms fall on Earth every second. This equals the mass converted every second in Sun to supply the light incident on Earth.
light from Sun
8.8 SUMMARY mass: the magnitude off the 4vector called momenergy Mass can be converted into energy and energy can be converted into mass” — this is a loose and sometimes misleading way to summarize some consequences of the two
245 b.
Assume that Sun delivers sunlight at the same ‘‘solarconstant rate” to every part of a sphere surrounding Sun of radius equal to the Earth  Sun distance. The area of this large sphere has the value 471/?^ where i? = 1.5 X 10“ meters, the average distance of Earth from Sun. This area equals 2.8 X 10^^ meters^. Therefore Sun converts a total of 2.8X 10^^ meters^ X 1.5 X 10“ kilograms/meter^ (from a) = 4.2 X 10^ kilograms of mass into luminous energy every second, or about 4 million metric tons per second.
c.
Through a series of nuclear processes not described here, four protons transform into a helium nucleus consisting of two protons and two neutrons. The four original protons have a mass 4 X 1.67262 X 10“ ^^ = 6.69048 X 10“ ^^ kilogram. The helium nucleus has a mass 6.64648 X 10~^^ kilogram. The difference, 0.04400 X 10"^^ kilogram, comes out mostly as light. (We cannot use twodigit accuracy here, because the important result is a difference between nearly equal numbers.) The ratio of hydrogen burned to mass converted equals 6.69048/0.04400 = 150 (back to twodigit accuracy!). So for each kilogram of mass converted to electromagnetic radiation, 150 kilograms of hydrogen burn to helium. In other words, about 0.7 percent of the rest energy (mass) of the original hydrogen is converted into radiation. Hence in order to convert 4.2 X 10^ kilograms per second into radiation. Sun burns 1 5 0 X 4 .2 x 1 0 ^ kilograms per second = 6.3 X 10“ kilograms of hydrogen into helium per second — about 630 million metric tons each second.
d. W e can reckon Sun’s mass by figuring how much Sun gravity it takes to guide our planet around in an orbit of 8 lightminute radius and one year time of circuit. Result: about 2.0 X 10^° kilograms. If Sun were all hydrogen, then the process of burning to helium at the present rate of 6.3 X 10“ kilograms every second would take (2.0 X 10^® kilograms)/(6.3 X 10“ kilograms/second) = 3.2 X 10'® seconds. At 32 million seconds per year, this would last about 10“ years, or 100 billion years. O f course the evolution of a star is more complicated than the simple conver sion of hydrogen into heliumplusradiation. Other nuclear reactions fuse helium into more massive nuclei on the way to the most stable nucleus, iron56 (Section 8.7). These other reactions occur at higher temperatures and typically proceed at faster rates than the hydrogentohelium process. Sun emits a flood of neutrinos (invisible; detected with elaborate apparatus; amount presently uncertain by a factor of 2, carry away less than 1 percent of Sun’s output). Sun also loses mass as particles blown away from the surface, called the so lar w ind. And stars do not convert all their hydrogen to helium and other nuclei — or live for 100 billion years. According to current theory, the lifetime of a star like Sun equals approxi mately 10 billion years ( lO'" years). We believe Sun to be 4 to 5 billion years old. The remaining 6 billion years (6 X 10^ years) or so should be sufficient time for our descendants to place themselves in the warmth of nearby stars.
principles that ate basic and really accurate: (1) The total momenergy of an isolated system of particles remains unchanged in a reaction; (2) The invariant magnitude of the momenergy of any given particle equals the mass of that particle. How much sound infotmation about physics can be extracted from these basic principles? W hat troubles sometimes atise from accepting a too loose formulation of the ‘‘principle of equivalence of mass and energy”? Some answers to these questions appeat in the dialog that follows, which serves also as a summary of this chaptet.
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CHAPTER 8
COLLIDE. CREATE. ANNIHILATE.
DIALOG: USE AND ABUSE OF THE CONCEPT OF MASS Does an isolated system have the same mass as observed in every inertial (freefloat) reference frame?
Yes. Given in terms of energy E and momentum p by ■— in one frame, by rrp = {E'Y (p 'Y in another frame. Mass of an isolated system is thus an invariant.
Does its energy have the same value in every inertial frame?
No. Enetgy is given by E = {nY +
or
E = m /{\ — or E — (mass) + (kinetic energy) = m\ K Value depends on the frame of reference from which the patticle (or isolated system of particles) is ob served. Value is lowest in the frame of reference in which the particle (ot system) has zero momenmm (zero total momentum in the case of an isolated system of particles). In that frame, and in that frame only, energy equals mass. Does energy equal zero for an object of zeto mass, such as a photon or neutrino or graviton?
No. Energy has value £ = (0^ + pYY^^ = p (or in conventional units = cp^oaY Alternatively one can say— formally— that the entire energy resides in the fotm of kinetic energy {K = p in this special case of zero mass), none at all in the form of rest energy. Thus, £ = (mass) + (kinetic energy) = ^
K= K= p
(case of zero mass only!). Can a photon — that has no mass— give mass to an absorber?
Yes. Light with energy £ transfers mass m = Eiy= Econv/^) to a heavy absorber (Exercise 8.5).
Invariance of mass; Is that feature of nature the same as the principle that all electrons in the universe have the same mass?
No. It is tme that all elementary particles of the same kind have the same mass. However, that is a fact totally distinct from the principle that the mass of an isolated system has identical value in whatever freefloat frame it is figured (invariance of system mass).
Invariance of mass: Is that the same idea as the conservation of the momenergy of an isolated sys tem?
No. Conservation of momenergy— the principle valid for an isolated system — says that the momenergy 4vector figured before the constiments of a system have interacted is identical to the momen ergy 4vector figured after the constituents have intetacted. In contrast, invariance of mass — the magnitude of the momenergy 4vector— says that that mass is the same in whatever freefloat frame it is figured.
8.8
SUMMARY: USE AND ABUSE OF THE CONCEPT OF MASS
247
Momenergy: Is that a richer concept than mass?
Yes. Momenergy 4vector reveals mass and more: the motion of object or system with the mass
Conservation of the momenergy of an isolated sys tem: Does this imply that collisions and interactions within an isolated system cannot change the sys tem’s mass?
Yes. Mass of an isolated system, being the magni tude of its momenergy 4vector, can never change (as long as the system remains isolated).
Conservation of the momenergy of an isolated sys tem: Does this say that the constituents that enter a collision are necessarily the same in individual mass and in number as the constituents that leave that collision?
No! The constituents often change in a highspeed encounter. Exam ple 1: Collision of two balls of putty that stick together— after collision hotter and therefore very slightly more massive than before. E xam ple 2: Collision of two electrons (e~) with sufficient violence to create additional mass, a pair consisting of one ordinary electron and one positive electron (positron: e'^): e (fast) + e (at rest)
+ 3e.
Exam ple 3: Collision that radiates one or more photons: e” (fast) + e~ (at rest) ~
(
electrons of \ intermediate speed /
1
^ electromagnetic \ energy (photons)  b emitted in the I \ collision process /
In all three examples the system momenergy and system mass are each the same before as after. Can I figure the mass of an isolated system composed of a number, n, of freelymoving objects by simply adding the masses of the individual objects? Exam ple: Collection of fastmoving molecules.
Ordinarily N O , but yes in one very special case: Two noninteracting objects move freely and in step, side by side. Then the mass of the system does equal the sum of the two individual masses. In the general case, where the system parts are moving relative to each other, the relation between system mass and mass of parts is not additive. The length, in the sense of interval, of the 4vector of total momenergy is not equal to the sum of the lengths of the individual momenergy 4vectors, and for a simple reason: In the general case those vectors do not point in the same spacetime direction. Energy however, does add and momentum does add: ^system
and
P x, system
i= 1
P x ,i i= 1
From these sums the mass of the system can be evaluated: M 2 ^ system
Can we simplify this expression for the mass of an isolated system composed of freely moving objects when we observe it from a freefloat frame so chosen as to make the total momentum be zero?
=
P
2 system
_
*2
r x, system
—
/)2
r y, system
—
*2
r z, system
Yes. In this case the mass of the system has a value given by the sum of energies of individual particles: M system — '^system F
[in zerototal momentum frame]
248
CHAPTER 8
COLLIDE. CREATE. ANNIHILATE. Moreover, rhe energy of each particle can always be expressed as sum of rest energy m plus kinetic energy K: Ei = nii\ K:
a = 1 ,2 ,3 ,
, «)
So the mass of the system exceeds the sum of the masses of its individual particles by an amount equal to the total kinetic energy of all particles (but only as observed in the frame in which total momentum equals zero); =
[in zerototal momentum fromel
+ 1= 1
For slow particles (Newtonian lowvelocity limit) the kinetic energy term is negligible compared to the mass term. So it is natural that for years many thought that the mass of a system is the sum of the masses of its parts. However, such a belief leads to incorrect results at high velocities and is wrong as a matter of principle at all velocities. W hat’s the meaning of mass for a system in which the particles interact as well as move?
The energies of interaction have to be taken into account. They therefore contribute to the total en ergy, fisystem , that gives the mass M ^ ^system = (E^ '•''system
■r
)l/2
system/
How do we find out the mass of a system of particles (Table 81) that are held — or stick — together?
Weigh it! Weigh it by conventional means if we are here on Earth and the system is small enough, other wise by determining its gravitational pull on a satel lite in freefloat orbit about it.
Does mass measure “amount of matter’’?
Nature does not offer us any such concept as “amount of mattet. ’’ History has struck down every proposal to define such a term. Even if we could count number of atoms or by any other counting method try to evaluate amount of matter, that number would not equal mass. First, mass of the specimen changes with its temperature. Second, atoms tightly bonded in a solid weigh less — are less massive — than the same atoms free. Third, many of nature’s atoms undergo radioactive decay, with still greater changes of mass. Moreover, around us occasionally, and continually in stars, the number of atoms and number of particles themselves undergo change. How then speak honestly? Mass, yes; “amount of matter,’’ no.
Does the explosion in space of a 20megaton hydro gen bomb convert 0.93 kilogram of mass into en ergy (fusion, Section 8.7)? conv/^^ “ (20 X 10^ tons TNT) X (10^ grams/ton) X (10^ calories/gram of “TN T equivalent’’) X (4.18 joules/calorie)/c^ = (8.36 X 10*^ joules)/(9 X 10^® meters^/second^) = 0.93 kilogram}
Yes and no! The question needs to be stated more carefully. Mass of the system of expanding gases, fragments, and tadiation has the same value imme diately after explosion as before; mass M of the system has not changed. However, hydrogen has been transmuted to helium and other nuclear trans formations have taken place. In consequence the makeup of mass of the system
8.8
SUMMARY: USE AND ABUSE OF THE CONCEPT OF MASS
=
+
249
K
(in zerototal momentum frame]
has changed. The first term on the right— sum of masses of individual constituents — has decreased by 0.93 kilogram: 2
)
~ ( 2
kilogram
The second term — sum of kinetic energies, includ ing “kinetic energy’’ of photons and neutrinos produced — has increased by the same amount: 2) /■= 1
)
“ ( 2
/ after
\ i= 1
kilogram / before
The first term on the right side of this equation — the original heat content of the bomb — is practi cally zero by comparison with 0.93 kilogram. Thus part of the mass of constituents has been con verted into energy; but the mass of the system has not changed. The mass of the products of a nuclear fission explo sion (Section 8.7: fragments of split nuclei of ura nium, for example) — contained in an underground cavity, allowed to cool, collected, and weighed — is this mass less than the mass of the original nuclear device?
Yes! The key point is the waiting period, which allows heat and radiation to flow away until trans muted materials have practically the same heat con tent as that of original bomb. In the expression for the mass of the system M.system =
2
[in zerototal momentum frame]
the second term on the right, the kinetic energy of thermal agitation — whose value rose suddenly at the time of explosion but dropped during the cool ing period — has undergone no net alteration as a consequence of the explosion followed by cooling. In contrast, the sum of masses
has undergone a permanent decrease, and with it the mass M of what one weighs (after the cooling period) has dropped (see the figure).
Em;
sum of masses of individual particles
250
CHAPTER 8
COLLIDE. CREATE. ANNIHILATE.
Does Einstein’s statement that mass and energy are equivalent mean that energy is the same as mass?
No. Value of energy depends on the freefloat frame of reference from which the particle (or isolated system of particles) is regarded. In contrast, value of mass is independent of inertial frame. Energy is only the time component of a momenergy 4vector, whereas mass measures entire magnitude of that 4vector. The time component gives the magnitude of the momenergy 4vector only in the special case in which that 4vector has no space component; that is, in a frame in which the momentum of the particle (or the total momentum of an isolated system of particles) equals zero. Only as measured in this spe cial ze ro m o m en tu m frame does energy have the same value as mass.
Then what is the meaning of Einstein’s statement that mass and energy are equivalenr?
Einstein’s statement refers ro the reference frame in which the particle is at rest, so that it has zero momenmm p and zero kinetic energy K. Then E = m\ K * m { 0 .ln that case the energy is called rhe rest energy of the particle:
In this expression, recall, the energy is measured in units of mass, for example kilograms. Multiply by the conversion factor to express energy in convenrional units, for example joules (Table 71). The result is Einstein’s famous equation: p
= ffKp'
Many treatments of relativity fail to use the sub script “rest’’— needed to remind us that this equiv alence of mass and energy refers only to the rest energy of the particle (for a system, the total energy in the zerototalmomentum frame). W ithout delving into all fine points of legalistic phraseology, how significant is the conversion factor in the equation = mc^}
The conversion factor c^, like the factor of conver sion from seconds to meters or miles to feet (Box 32), today counts as a detail of convention, rather rhan as a deep new principle.
If the factor is not the central feature of the relationship between mass and energy, what is cen tral?
The distinction between mass and energy is this: Mass is the magnitude of the momenergy 4vector and energy is the time component of the same 4vec tor. Any feature of any discussion that emphasizes this contrast is an aid to understanding. Any slurring of terminology thar obscures rhis distinction is a potential source of error or confusion.
Is the mass of a moving object greater than the mass of the same object at rest?
No. It is the same whether the object is at rest or in motion; the same in all frames.
Really? Isn’t the mass, M, of a system of freely moving particles given, not by the sum of the masses W; of the individual constiments, but by the sum of
Ouch! The concept of “relativistic mass” is subject to misunderstanding. That’s why we don’t use it. First, it applies the name mass— belonging to the
REFERENCES energies £,■ (hut only in a frame in which total mo mentum of the system equals zero)? Then why not give a new name and call it “relativistic mass’’ of the individual particle? Why not adopt the notation «, rel
= E; = m; + K,
?
W ith this notation, can’t one then write
25 1
magnimde of a 4vector— to a very different con cept, the time component of a 4vector. Second, it makes increase of energy of an object with velocity or momentum appear to be connected with some change in internal strucmre of the object. In reality, the increase of energy with velocity originates not in the object but in the geometric properties of spacetime itself.
[in zerototal momentum frame] i =
1
In order to make this point clear, should we call invariant mass of a particle its “rest mass’’?
That is what we called it in the first edition of this book. But a thoughtful student pointed out that the phrase “rest mass’’ is also subject to misunderstand ing; W hat happens to the “rest mass’’ of a particle when the particle moves? In reality mass is mass is mass. Mass has the same value in all frames, is invariant, no matter how the particle moves. {Gali leo: “In questions of science the authority of a thou sand is not worth the humble reasoning of a single individual.’’}
Can any simple diagram illustrate this contrast be tween mass and energy?
Yes. The figure shows the momentumenergy 4vector of the same particle as measured in three different frames. Energy differs from frame to frame. Momentum differs from frame to frame. Mass (magnitude of 4vector, represented by the length of handles on the arrows) has the same value, m = 8, in all frames.
particle at rest:
p ' =  6
p =0
8
£=8
mass
LABORATORY FRAME
3 \ \ \£ '= 1 0 mass
ROCKET FRAME
SUPERROCKET FRAME
REFERENCES Quotation from Count Rumford in Section 8.2: Sanborn C. Brown, Benjamin Thompson, Count Rumford (MIT Press, Cambridge, Mass., 1979), page 220. Reference to measurement of very small effects in Section 8.2: Vladimir Bra ginsky and A. B. Manukin, “Quantum nondemolition,” in Measurement of Weak Forces in Physics Experiments, edited by David H. Douglas (University of Chicago Press, 1977).
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CHAPTER 8
COLLIDE. CREATE. ANNIHILATE.
Quotation from biography of A. H. Compton in Section 8.4: Robert S. Shankland, Dictionary of Scientific Biography, edited by Charles Coulston Gillespie, Volume III (Charles Scribner’s Sons, New York, 1 9 7 1). Compton scattering reported in A. H. Compton, Physical Review, Volume 2 2 , pages 4 0 9  4 1 3 (1 9 2 3 ). The poly electron mentioned in Section 8 .5 has been independently generated, through interaction of a slow positron with the electrons of a metal surface, by Alan Mills, Jr., at Bell Telephone Laboratories, as reported in Physical Review Letters, Volume 4 6 , pages 7 1 7  7 2 0 (1 9 8 1 ). Final quotation in Box 81: Timothy Ferris, Coming of Age in the Milky Way (Anchor Books, Doubleday, New York, 1 9 8 8 ), page 3 4 4 . Sample Problem 85 was suggested by Chet Raymo’s science column in the Boston Globe, May 2, 1 9 8 8 , page 35. Galileo quote in final dialog: Galileo Galilei, Dialogo dei due massimi sistemi del mundo, Landini, Florence. Translation by S. Drake, Galileo Galilei — Dialogue Concemingthe Two Chief World Systems— Ptolemaic and Copemican, University of California Press, Berkeley and Los Angeles, 19 5 3.
ACKNOWLEDGMENTS We thank colleagues old and young for the comments that helped us clarify, formulate, and describe the concept of mass in this chapter and in the final dialog, and very specially Academician Lev B. Okun, Institute of Theoretical and Exper imental Physics, Moscow, for correspondence and personal discussions. W e be lieve that our approach agrees with that in two of his articles, both entitled “The Concept of Mass,” which appeared in Physics Today, June 1 9 8 9 , pages 31  36, and Soviet PhysicsUspekhi, Volume 3 2 , pages 6 2 9  6 3 8 (July 1 9 8 9).
EXERCISES
You now have at your disposal the power of special relativity to provide physical insight and accurate pre dictions about an immense range of phenomena, from nucleus to galaxy. The following exercises give only a hint of this range. Even so, there are too many to carry out as a single assignment or even several assignments. For this reason — and to anchor your understanding of relativity— we recommend that
253
you continue to enjoy these exercises as your study moves on to other subjects. The following table of contents is intended to help organize this ongoing attention. R em inder: In these exercises the symbol v (in other texts sometimes called f i ) stands for speed as a fraction of the speed of light c. Let be the speed in conventional units; then v = v ^ ^ / c .
CONTENTS mass and energy 8 1 82
821 822
Examples o f conversion 2 5 4 Relativistic chemistry 2 5 4
823 824
photons 83 84 85 86 87
Pressure o f light 2 5 4 M easurem ent o f photon energy 2 5 4 Einstein’s derivation: Equivalence of energy and mass 2 5 4 Gravitational red shift 2 5 8 Density o f the com panion of Sirius 2 5 8
creations, transformations, annihilation 88 89 810 811 812 813 814 815 816 817
N uclear excitation 2 5 9 Photon braking 2 5 9 Photon integrity 2 5 9 Pair production by a lonely photon? 2 5 9 Photoproduction o f a pair by two photons 2 5 9 Decay o f positronium 2 6 0 Positron  electron annihilation 1 2 6 0 P ositronelectron annihilation II 260 Creation o f p ro to n a n tip ro to n pair by an electron 2 6 1 Colliders 2 6 1
825 826 827 828
Speeding light bulb 2 6 4 D oppler shift at the lim b o f Sun The expanding universe 2 6 4 Twin Paradox using the D oppler shift 2 6 4 D oppler line broadening 2 6 4 ^restconv ~ fitom the D oppler shift 2 6 4 Everything goes forward 2 6 5 Decay o f 7r°meson 2 6 7
Compton scattering 829 830 831 832
C om pton scattering 2 6 7 C om pton scattering examples 2 6 8 Energy o f a photon and frequency of light 2 6 8 Inverse C om pton scattering 2 6 9
tests of relativity 833 834 835 836 837 838 839
Photon energy shift due to recoil of em itter 2 7 0 Recoilless processes 2 7 0 Resonant scattering 2 7 1 M easurem ent o f D oppler shift by resonant scattering 2 7 1 Test o f the gravitational red shift I 272 Test o f the gravitational red shift II 2 7 2 Test o f the Tw in Paradox 2 7 2
Doppler shift 818 819 820
D oppler shift along the xdirection 2 6 3 D oppler equations 2 6 3 The physicist and the traffic light
freeforall! 840 841 263
264
M om entum w ithout mass? 2 7 3 The photon rocket and interstellar travel 2 7 4
25 4
EXERCISE 81
EXAMPLES OF CONVERSION
MASS AND ENERGY 81
examples off conversion
a How much mass does a 100w att bulb dissi pate (in heat and light) in one year? b The total electrical energy generated on Earth during the year 1 9 9 0 was probably between 1 and 2 X 10*^ kilowatthours. To how much mass is this energy equivalent? In the acmal production of this electrical energy is this much mass converted to en ergy? Less mass? More mass? Explain your answer. c Eric Berman, pedaling a bicycle at full throttle, produces onehalf horsepower of u se fu l power (1 horsepower = 7 4 6 watts). The human body is about 2 5 percent efficient; that is, 75 percent of the food burned is converted to heat and only 2 5 percent is converted to useful work. How long a time will Eric have to ride to lose one kilogram by the conversion of mass to energy? How can reducing gymnasiums stay in business?
82
relativistic chemistry
One kilogram of hydrogen combines chemically with 8 kilograms of oxygen to form water; about 10® joules of energy is released. a Ten metric tons (10'' kilograms) of hydrogen combines with oxygen to produce water. Does the resulting water have a greater or less mass than the original hydrogen and oxygen? W hat is the magni tude of this difference in mass? b A smaller amount of hydrogen and oxygen is weighed, then combined to form water, which is weighed again. A very good chemical balance is able to detect a fractional change in mass of 1 part in 10®. By what factor is this sensitivity more than enough — or insufficient— to detect the fractional change in mass in this reaction?
PHOTONS 83
pressure off light
a Shine a onewatt flashlight beam on the palm of your hand. Can you feel it? Calculate the total force this beam exerts on your palm. Should you be able to feel it? A particle of what mass exerts the same force when you hold it at Earth’s surface? b From the solar constant (1 .3 7 2 kilowatts/ square meter. Sample Problem 8 5 ) calculate the pressure of sunlight on an Earth satellite. Consider both reflecting and absorbing surfaces, and also “ real” surfaces (partially absorbing). Why does the color of the light make no difference?
c A spherical Earth satellite has radius r = 1 meter and mass m = 1 0 0 0 kilograms. Assume that the satellite absorbs all the sunlight that falls on it. W hat is the acceleration of the satellite due to the force of sunlight, in units of g, the gravitational accel eration at Earth’s surface? For a way to reduce this “disturbing” acceleration, see Figure 92. d It may be that particles smaller than a certain size are swept out of the solar system by the pressure of sunlight. This certain size is determined by the equal ity of the outward force of sunlight and the inward gravitational attraction of Sun. Estimate this critical particle size, making any assumptions necessary for your estimate. List the assumptions with your answer. Does your estimated size depend on the particle’s distance from Sun? Reference; For pressure of lighr measuremenr in an elementary labo ratory, see Robert Pollock, American Journal of Physics, Volume 31, pages 901904 (1963). Pollock’s method of determining the pres sure of light makes use of resonance to amplify a small effea to an easily measured magnitude. Dr. Pollock developed this experiment in collaboration with the same group of firstyear students at Prince ton University with whom the authors had the privilege to work out the presentation of relativity in the first edition of this book.
84
measurement off photon energy
A given radioactive source emits energetic photons (Xrays) or very energetic photons (gamma rays) with energies characteristic of the particular radioactive nucleus in question. Thus a precise energy measure ment can often be used to determine the composition of even a tiny specimen. In the apparatus dia grammed in the figure on page 2 5 5 , only those events are detected in which a count on detector A (knockedon electron) is accompanied by a count on detector B (scattered photon). W hat is the energy of the incoming photons that are detected in this way, in units of the rest energy of the electron?
85
Einstein's derivation: equivalence off energy and m ass— a worked example
P ro b le m From the fact that light exerts pressure and carries energy, show that this energy is equivalent to mass and hence— by extension — show the equivalence of all energy to mass. C om m entary: The equivalence of energy and mass is such an important consequence that Einstein very early, after his relativistic derivation of this result, sought and found an alternative elementary physical line of reasoning that leads to the same conclusion. He envisaged a closed box of mass M initially at rest, as shown in the first figure. A directed burst of electro
EXERCISE 85
EINSTEIN'S DERIVATION: EQUIVALENCE OF ENERGY AND MASS
Electron detector
Thin foil (loosely bound electrons) \
255
3 : 4 : 5 triangle
''^Collimating slit Shielding Photon detector
Source of photons of characteristic energies
EXERCISE 84. Measurement of photon energy.
magnetic energy is emitted from the left wall. It trav els down the length L of the box and is absorbed at the other end. The radiation carries an energy E. But it also carries momenrum. This one sees from the fol lowing reasoning. The radiation exerts a pressure on the left wall during the emission. In consequence of this pressure the box receives a push to the left, and a momentum,/). But the momentum of the system as a whole was zero initially. Therefore the radiation carries a momentum p opposite to the momentum of the box. How can one use knowledge of the transport of energy and momentum by the radiation to deduce the mass equivalent of the radiation? Einstein got his answer from the argument that the center of mass of the system was not moving before the transport pro cess and therefore cannot be in motion during the transport process. But the box obviously carries mass to the left. Therefore the radiation must carry mass to the right. So much for Einstein’s reasoning in broad outline. Now for the details. From relativity Einstein knew that the momentum /) of a directed beam of radiation is equal to the energy E of that beam (Section 8.4; both p and E measured in units of mass). However, this was known before Ein stein’s relativity theory, both from Maxwell’s theory of electromagnetic radiation and from direct observa
tion of the pressure exerted by light on a mirror suspended in a vacuum. This measurement had first successfully been carried out by E. F. Nichols and G. F. Hull between 1901 and 1903. (By now the exper iment has been so simplified and increased in sensitiv ity that it can be carried out in an elementary labora tory. See the reference for Exercise 83.) Thus the radiation carries momentum and energy to the right while the box carries momenrum and mass to the left. But the center of mass of the system, box plus radiation, cannot move. So the radiation must carry to the right not merely energy but mass. How much mass? To discover the answet is the object of these questions. a W hat is the velocity of the box during the time of ttansit of the radiation? b After the radiation is absorbed in the other end of the box, the system is once again at rest. How far has the box moved during the transit of the radiation? c Now demand that the center of mass of the system be at the same location both before and after the flight of the radiation. From this argument, what is the mass equivalent of the energy that has been transported from one end of the box to the other?
S o lu tio n a During the transit of the radiation the mo mentum of the box must be equal in magnitude and opposite in direction to the momentum p of the radi ation. The box moves with a very low velocity v. Therefore the Newtonian formula Mv suffices to cal culate its momentum: Mv = —p — — E From this relation we deduce the velocity of the box, v = E IM
EXERCISE 85, first figure. Transfer of mass by radiation.
b
The transit time of the photon is very nearly
256
EXERCISE 85
EINSTEIN'S DERIVATION: EQUIVALENCE OF ENERGY AND MASS
/ = L meters of lighttravel time. In this time the box moves a distance A x = pt = — EL/M
c If the radiation transported no mass from one end of the box to the other, and if the box were the sole object endowed with mass, then this displace ment A x would result in a net motion of the center of mass of the system to the left. But, Einstein reasoned, an isolated system with its center of mass originally at rest can never set itself into motion nor experience any shift in its center of mass. Therefore, he argued, there must be some countervailing displacement of a part of the mass of the system. This transport of mass to the right can be understood only as a new feature of the radiation itself. Consequently, during the time the box is moving to the left, the radiation must transport to the right some mass m, as yet of unknown magni tude, but such as to ensure that the center of mass of the system has not moved. The distance of transport is the full length L of the box diminished by the distance A x through which the box has moved to the left in the meantime. But A x is smaller than L in the ratio EjM. This ratio can be made as small as one pleases for any given transport of tadiant energy E by making the mass M. of the box sufficiently great. Therefore it is legitimate to take the distance moved by the radiation as equal to L itself. Thus, with arbitrarily high preci sion, the condition that the center of mass shall not move becomes M A x + mL = 0 Calculate the mass m and find, using A x from part b, m A
x M.i l
= { E L /M ){ M ID
or, finally, m= E In conventional units, we have the famous equation £conv = W e conclude that the process of emission, rransport, and reabsorption of radiation of energy E is equivalent to the transport of a mass m = E from one end of the box to the other end. The simplicity of this derivation and the importance of the result makes this analysis one of the most interesting in all of physics. D iscussion; The mass equivalence of radiant en ergy implies the mass equivalence of thermal energy and — by exrension — of orher forms of energy, ac
cording to the following reasoning. The energy that emerges from the left wall of the box may reside there originally as heat energy. This thermal energy excites a typical atom of the surface from its lowest energy state to a higher energy state. The atom returns from this higher state to a lower state and in the course of this change sends out the surplus energy in the form of radiation. This radiant energy traverses the box, is absorbed, and is ultimately converted back into ther mal energy. Whatever the details of the mechanisms by which light is emitted and absorbed, the net effect is the transfer of heat energy from one end of the box to the other. To say that mass has to pass down the length of the box when radiation goes from one wall to the other therefore implies that mass moves when thermal energy changes location. The thermal energy in turn is derived from chemical energy or the energy of a nuclear transformation or from electrical energy. Moreover, thermal energy deposited at the far end of the tube can be converted back into one or another of these forms of energy. Therefore these forms of energy — and likewise all other forms of energy— are equivalent in their transport to the transport of mass in the amount m = E. How can one possibly uphold the idea that a pulse of radiation transports mass? One already knows that a photon has zero mass, by virtue of the relation (Section 8.4) (mass)^ = (energy)^ — (momentum)^ = 0 Moreover, what is true of the individual photon is true of the pulse of radiation made up of many such photons: The energy and momentum are equal in magnirude, so that the mass of the radiation necessar ily vanishes. Is there nor a fundamental inconsistency in saying in the same breath that the mass of the pulse is zero and that radiation of energy E transports the mass m = E from one place to another? The source of our difficulty is some confusion be tween two quite different concepts; (1) energy, rhe time component of the momentum  energy 4vector, and (2) mass, the magnitude of this 4vector. When the system divides itself into two parts (radiation going to the right and box recoiling to the left) the components of the 4vectors of the radiation and of the recoiling box add up to identity with the compo nents of the original 4vector of the system before emission, as shown in the second figure. However, rhe magnitudes of the 4vectors (magnirude = mass) are not additive. No one dealing with Euclidean geome try would expect the length of one side of a triangle to be equal to the sum of the lengths of the other two sides. Similarly in Lorentz geometry. The mass of the system (M ) is not to be considered as equal to the sum
EXERCISE 85
EINSTEIN'S DERIVATION: EQUIVALENCE OF ENERGY AND MASS
of the mass of the radiation (zero) and the mass of the recoiling box (less than M ). But components of 4vectors are additive; for example, / energy o f\ ^ / energy o f \ __ / energy of \ V system / \ radiation / \recoiling box/ Thus we see that the energy of the recoiling box is M — Not only is the energy of the box reduced by the emission of radiation from the wall; also its mass is reduced (see shortened length of 4vector in dia gram). Thus the radiation takes away mass from the wall of the box even though this radiation has zero mass. The inequality /(mass mass o f\ f\ ^ / mass of \ \radiation[zero}/ \ system /
mass of V^coiling box)
(
is as natural in spacetime geometry as is the inequality 5 # 3 + 4 for a 345 triangle in Euclidean geom etry. W hat about the gravitational attraction exerted by the system on a test object? O f course the redistribu tion of mass as the radiation moves from left to right makes some difference in the attraction. But let the test object be at a distance r so great that any such redistribution has a negligible effect on the attraction. In other words, all that counts for the pull on a unit test object is the total mass M as it appears in New ton’s formula for gravitational force: force per \ _ GM
( unit mass
257
Even so, will not the distant detector momentarily experience a lessthannormal pull while the radiation is in transit down the box? Is not the mass of the radiation zero, and is not the mass of the recoiling box reduced below the original mass M of the system? So is not the total attracting mass less than normal during the process of transport? No! The mass of the system — one has to say again — is not equal to the sum of the masses of its several parts. It is instead equal to the magnimde of the total momentum  energy 4vector of the system. And at no time does either the total momentum (in our case zero!) or the total energy of the system change — it is an isolated system. There fore neither is there any change in the magnitude M of the total momentum  energy 4vectors shown in the second figure. So, finally, there is never any change in the gravitational attraction. There is one minor swindle in the way this problem has been presented: The box cannot in fact move as a rigid body. If it could, then information about the emission of the radiation from one end could be ob tained from the motion of the other end before the arrival of the radiation itself— this information would be transmitted at a speed greater than that of light! Instead, the recoil from the emission of the radiation travels along the sides of the box as a vibra tional wave, that is, with the speed of sound, so that this wave arrives at the other end long after the radia tion does. In the meantime the absorption of the radiation at the second end causes a second vibrational wave which travels back along the sides of the box. The addition of the vibration of the box to the prob
Radiation (zero rest mass)
M
Energy A
Momentum BEFORE
AFTER
EXERCISE 85, second figure. Radiation transfers mass from place to place even though the mass of the radiation is zero!
258
EXERCISE 86
GRAVITATIONAL RED SHIFT
lem requires a more complicated analysis but does not change in any essential way the results of the exercise. References: A. Einstein, Annalen der Physik, Volume 20, pages 6 2 7  6 3 3 (1906). For a more careful treatment of the box, see A. P. French, Special Relativity (W . W . Norton, New York, 1968), pages 1 6  1 8 and 2 7  2 8 .
86 gravitational red shift N ote: Exercises 86 and 87 assume an acquaintance with the following elementary facts of gravitation. (1) A very small object— or a spherically sym metric object of any radius— with mass M attracts an objea of mass m — also small or spherically symmetric— with a force
F=
GMm
Here r is the distance between the centers of the two objects and G is the Newtonian con stant of gravitation, G = 6 .6 7 X 1 0 “ ** (meter)^/(kilogramsecond^). The work required to move a test particle of ( 2) unit mass from r t o r + dr against the gravita tional pull of a fixed mass M is GM{dr/r'^). Translated from conventional units of energy to units of mass this work is GM dr d W coov = ■
dr M* — .2
per unit of mass contained in the test particle. (3) The symbol M* = GMjc'^ in this formula has a simple meaning. It is the mass of the center of attraction translated from units of kilograms to units of meters. For example, the mass of Earth = 5.974 X lO^'* kilo grams) expressed in length units is = 4.44 X 10“ ^ meters, and the mass of Sun (M s„„= 1.989 X 1030 kg) is M*su„ = 1.48 X 1Q3 meters. (4) Start the test particle at a distance r from the center of attraction of mass M and carry it to an infinite distance. The work required is IF = M */r in units of mass per unit of mass con tained in the test particle. So much for the minitutorial. Now to business. a W hat fraction of your rest energy is converted to potential energy when you climb the Eiffel Tower (300 meters high) in Paris? Letg* be the acceleration of gravity in meters/meter^ at the surface of Earth: = .
1 'Earth
g Earth
b W hat fraction of one’s rest energy is converted to potential energy when one climbs a very high lad der that reaches higher than the gravitational influ ence of Earth? Assume that Earth does not rotate and is alone in space. Does the fraction of the energy that is lost in either part a or part b depend on your original mass? c Apply the result of part a to deduce the frac tional energy change of a photon that rises vertically to a height z in a uniform gravitational field g*. Photons have zero mass; one can say formally that they have only kinetic energy E = K. Thus photons have only one purse — the kinetic energy purse— from which to pay the potential energy tax as they rise in the gravitational field. Light of frequency / i s composed of photons of energy E = hf/c^ (see Exercise 831). Show that the fractional energy loss for photons rising in a gravitational field corresponds to the following fractional change in frequency: = —g*z
[uniform gravitational field]
/ N ote: W e use / for frequency instead of the usual Greek nu, V, to avoid confusion with v for speed. d Apply the result of part b to deduce the frac tional energy loss of a photon escaping to infinity. (To apply b for this purpose is an approximation good to one percent when this fractional energy loss itself is less than two percent.) Specifically, let the photon start from a point on the surface of an astronomical object of mass M (kilograms) or AI* (meters) = GM/c^ and radius r. From the fractional energy loss, show that the fractional change of frequency is given by the expression M* [escape field of spherical object]
f This decrease in frequency is called the g rav itatio n al re d shift because, for visible light, the shift is toward the lowerfrequency (red) end of the visible spectmm. e Calculate the fractional gravitational red shifts for light escaping from the surface of Earth and for light escaping from the surface of Sun. D iscussion: The results obtained in this exercise are approximately correct for light moving near Earth, Sun, and white dwarf (Exercise 87). Only general relativity correctly describes the motion of light very close to neutron star or black hole (Box 92).
87 density of the companion of Sirius N ote: This exercise uses a result of Exercise 86. Sirius (the Dog Star) is the brightest star in the heavens. Sirius and a small companion revolve about
EXERCISE 812
PHOTOPRODUCTION OF A PAIR BY TWO PHOTONS
one another. By analyzing this revolution using New tonian mechanics, astronomers have determined that the mass of the companion of Sirius is roughly equal to the mass of our Sun (A1 is about 2 X 10^° kilo grams; M* is about 1.5 X 10^ meters). Light from the companion of Sirius is analyzed in a spectrometer. A spectral line from a certain element, identified from the pattern of lines, is shifted in frequency by a frac tion 7 X lO""* compared to the frequency of the same spectral line from the same element in the laboratory. (These figures are experimentally accurate to only one significant figure.) Assuming that this is a gravita tional red shift (Exercise 86), estimate the average density of the companion of Sirius in grams/centimeter^. This type of star is called a w h ite d w a rf (Box 92).
CREATIONS, TRANSFORMATIONS, ANNIHILATIONS A nucleus of mass m initially at test absorbs a gamma ray (photon) and is excited to a higher energy state such that its mass is now 1.01 m. a Find the energy of the incoming photon needed to carry out this excitation. b Explain why the required energy of the incom ing photon is greater than the change of mass of the nucleus.
Om® (at rest)
AFTER
c O 1.01
^ m
EXERCISE 88. Excitation of a nucleus by a gamma ray.
O
 ^
be found)
A
AFTER
' " OQ
^
(not known)
(at rest)
EXERCISE 89. Stopping a nucleus by emission of a gamma ray.
810 photon integrity Show that an isolated photon cannot split into two photons going in directions other than the original direction. (Hint: Apply the laws of conservation of momentum and energy and the fact that the third side of a triangle is shorter than the sum of the other two sides. W hat triangle?)
811
88 nuclear excitation
BEFORE A / V X A/ V ^
BEFORE ^
259
pair production by a lonely photon?
A gamma ray (highenergy photon, zero mass) can carry an energy greater than the rest energy of an electron  positron pair. (Remember that a positron has the same mass as the electron but opposite chatge.) Nevertheless the process (energetic gamma ray) (electron) + (positron) cannot occur in the absence of other matter or radia tion. a Prove that this process is incompatible with the laws of conservation of momentum and enetgy as employed in the labotatory frame of reference. Ana lyze the alleged creation in the frame in which electron and positron go off at equal but opposite angles ± (f) with the extended path of the incoming gamma ray. b Repeat the demonstration— which then be comes much more impressive— in the centerofmomentum frame of the alleged pair, the frame of refer ence in which the total momentum of the two resulting particles is zero.
89 photon braking
812 photoproduction of a pair by two photons
A moving radioactive nucleus of known mass M emits a gamma ray (photon) in the forward direction and drops to its stable nonradioactive state of known mass m. Find the energy of the incoming nucleus (BEFORE diagtam in the figute) such that the result ing mass m nucleus is at rest (AFTER diagtam). The unknown energy of the outgoing gamma ray should not appear in your answer.
Two gamma rays of different enetgies collide in a vacuum and disappear, bringing into being an electron  positron pair. For what ranges of energies of the two gamma tays, and fot what range of angles between their initial directions of propagation, can this reaction occur? (Hint: Start with an analysis of the reaction at threshold; at threshold the electron and positron are relatively at rest.)
260
EXERCISE 813
DECAY OF POSITRONIUM
mQ
BEFORE
BEFORE
A
Qm
m O
B
A
AFTER
D
A
/ V W C
(at rest)
^
EXERCISE 813. Decay of positronium in flight.
813
decay off positronium
A moving "atom ” called positronium (an electron and positron orbiting one another) of mass m and initial energy £ decays into two gamma rays (highenergy photons) that move in opposite direaions along the line of motion of the initial atom. Find the energy of each gamma ray, Ec and £ o , in terms of the mass m and energy E/^^ of the initial particle. Check that Ec = £ d in the case that the initial particle is at rest.
814
positron—oloctron annihilation I
A positron e"*" of mass m and kinetic energy K is annihilated on a target containing electrons e~ (same mass m) practically at rest in the laboratory frame:
AFTER
EXERCISE 815, first figure. Positron  electron annihilation.
815
positron—oloctron annihilation II
A positron e'*’ o f mass m and kinetic energy K is annihilated on a target containing electrons e~ (same mass m) practically at rest in the laboratory frame; e‘*'(fast) + e~(at rest) * radiation
^■'■(fast) b e (at rest)
' radiation
a By considering the collision in the centerofmomentum frame (the frame of reference in which the total momentum of the initial particles is equal to zero), show that it is necessary for at least two gamma rays (rather than one) to result from the annihilation. b Return to the laboratory frame, shown in the figure. The outgoing photons move on the line along which the positron approaches. Find an expression for the energy of each outgoing photon. Let your deriva tion be free of any reference to velocity. c Using simple approximations, evaluate the an swer to part b in the limiting cases (1) very small K and (2) very large K. (Very small and very large compared with what?)
BEFORE
Qm
mQ.
B
A
The resulting gamma rays go off at different angles with respea to the direaion of the incoming positron, as shown in the first figure. a Derive an expression for the energy of one of the gamma rays in the laboratory frame as a fiinaion of the angle between the direction of emergence of that gamma ray and the direction of travel of the positron before its annihilation. The gamma ray en ergy should be a funaion of only the energy and mass of the incoming positron and the angle of the outgo ing gamma ray. (Hint: Use the law of cosines, as applied to the second figure.) P d ^ = P a ^ + P c ^ ~ 2/>a/»cCos (f)c
b Show that for outgoing gamma rays moving along the positive and negative xdirection, the results of this exercise reduce to the tesults of Exercise 814.
(al rest)
AFTER
D
c
LABORATORY FRAME EXERCISE 814. Positronelectron annihilation.
EXERCISE 815, second figure. Conservation of vector momen tum means that the momentum triangle is closed.
EXERCISE 817
816
creation off proton— antiproton pair by an electron
W hat is the threshold kinetic energy K,i, of the inci dent electron for the following process? electron (fast) + proton (at rest) ^ electron + antiproton + two protons
817 colliders How much more violent is a collision of two protons that are moving toward one another from opposite directions than a collision of a moving proton with one at rest? D iscussion: When a moving particle strikes a stationary one, the energy available for the creation of new particles, for heating, and for other interactions — or, in brief, the available interaaion energy— is less than the initial energy (the sum of the rest and kinetic energies of the initial two particles). Reason: The particles that are left over after the reaction have a net forward motion (law of conservation of momen tum), the kinetic energy of which is available neither for giving these particles velocity relative to each other nor for producing more particles. For this reason much of the particle energy produced in accelerators is not available for studying interactions because it is carried away in the kinetic energy of the products of the collision. However, in the centerofmomenmm frame, the frame in which the total momentum of the system is equal to zero, no momentum need be carried away from the interaction. Therefore the energy available for interaction is equal to the total energy of the incoming particles. Is there some way that the laboratory frame can be made also the centerofmomentum frame? One way is to build two particle accelerators and have the two beams collide head on. If the energy and masses of the particles in each beam are respectively the same, then the laboratory frame is the centerofmomentum frame and all the energy in each collision is available interaaion energy. It is easier and cheaper to achieve the same efficiency by arranging to have particles moving in opposite directions in the same accelerator. A magnetic field keeps the particles in a circular path, “storing” them at their maximum energy for re peated tries at interaction. Such a facility is called a co llid er. The figure on page 262 gives some details of a particular collider. a W hat is the total available interaction energy for each encounter in the laboratory frame of the Tevatron shown on page 262?
COLLIDERS
261
b Now transform to a frame in which one of the incoming particles is at rest (transformation given in Exercise 75). This would be the situation if we tried to build an accelerator in which moving antiprotons hit a stationary target of, say, liquid hydrogen (made of protons and electrons). [Simplify: At 0.9 TeV = 9 X 10** eV what is the effective speed v of the proton? W hat is its momentum compared with its energy? W hat is the value of the time stretch factor y = E/mTi If the target protons were at rest, what energy, in TeV, would the incoming antiproton need to have in order to yield the same interaction energy as that achieved in the Tevatron?
Wait a minute! You keep telling us that energy and momentum have different values when measured with respect to different reference frames. Yet here you assume the “interaction energy” is the same in the Te vatron laboratory frame as it is in the rest frame ofa proton that moves with nearly the speed of light in the Tevatron frame. Is the energy of a system different in different frames, or is it the same?
— There is an important distinction between the total energy of a system and the “available interaction energy,” just as there is an important distinction between your money in the bank and “ready cash” in the bank that you can spend. If some of your money in the bank has been p ut in escrow for payment on a house you are buying, then you cannot spend that part of your bank money to buy a new car. Similarly, the total energy of the protonantiproton system is much smaller in the Tevatron laboratory frame than in the frame in which the proton is initially at rest, but all of the Tevatron laboratoryframe energy can be spent— used to create new particles, for example. In con trast, only a minute fraction of the energy in the frame in which the proton is initially at rest can be spent to create new particles, since total momentum must be conserved; most of the total energy is kept “in escrow” for this purpose. The number and kinds of new particles created must be the same for all observers! Therefore the “ available interaction energy” must be the same for all observers. The central point here is that the Tevatron collider design makes all of the energy in the protonantiproton system “available” for use in the laboratory.
262
EXERCISE 817
EXERCISE 817. Top: Aerial view of the Tevatron ring at Fermi National Accelerator Laboratory in Batavia, Illinois. The ring is 6 .3 kilometers in cir cumference. Bottom: View along the tunnel of the Tevatron. Protons (positive charge) and antiprotons (antiparticle of the proton: same mass, negative charge) circulate in separate beams in opposite direc tions in the same vacuum chamber in the lower ring of superconducting magnets shown in the photo. The upper ring of regular magnets accelerates protons from 8 CeV to 1 5 0 GeV. Some of these protons are injected into the lower set of magnets directly, rotating clock wise. Other protons strike a copper target and create antiprotons at a lower energy that are accumulated over approximately 15 hours in a separate ring (not shown) and then reaccelerated to 15 0 GeV and in serted into the lower ring, circulating counterclock wise. (Opposite charge, opposite motion yields same magnetic force toward the center, hence counterrota tion around the same circle.) Then particles in both beams in the lower ring of magnets are accelerated at the same time from 15 0 GeV to a final energy of 0 .9 TeVper particle. (1 teraelectronvolt = 1 0 ‘^ electronvolts, or approximately 1 0 0 0 times the rest energy of the proton or antiproton.) After acceleration, the beams are switched magnetically so that they cross each other at multiple intersection points around the ring, allowing protons and antiprotons to collide in the laboratory centerofmomentum frame. Detectors at the points of intersection monitor products of the colli sions. Protons and antiprotons that do not interact at one intersection are not wasted; they may interact at another intersection point or on subsequent trips around the ring. The particles are allowed to coast around and around at full energy for as long as 2 4 hours as they interact. Question: Approximately how many revolutions around the ring does a given proton or antiproton make in 2 4 hours? Photographs cour tesy of Fermi Laboratory.
COLLIDERS
EXERCISE 820
THE PHYSICIST AND THE TRAFFIC LIGHT
laboratory xdirection, the photon has an energy E' given by the equation
DOPPLER SHIFT 818
Doppler shift along the xdireclion
N ote: Recall Exercise L5 in the Special Topic on Lorentz Transformation, following Chapter 3. Apply the momenergy transformation equations (Exercise 75) to light moving in the positive xdirection for which p ^ = p = E. a Show that the relation between photon energy E' in the rocket frame and photon energy E in the laboratory frame is given by the equation
E = y (l+ p )E ' =
E' = E y (l —
cos (/))
and moves in a direction that makes an angle (f)' with the x'axis given by the equation
cos 0 ' =
COS 0 “ 1cos 0
b Derive the inverse equations for E and cos 0 as functions of E', cos 0 \ and Show that the results are
(1 + v) E' (1
263
E = E 'y (l + v„^ cos (f)')

, COS ' + t', cos 0  —  ^ 1+ cos 0
(1 + v) E'
[(1  t.)(l + [photon moves along positive xdirection]
b Use the Einstein relation between photon en ergy E and classical wave frequency / , namely = h f or E = hf/c^ and E' = h f /c^, to derive the trans formation for frequency
/' =fy{\f = f 'y ( X +
[wove motion along positive xdirection]
/
This is the Doppler shift equation for light waves moving along the positive xdirection. N ote: W e u se/fo r frequency instead of the usual Greek nu, V, to avoid confusion with v for speed. C Show that for a wave moving along the nega tive xdirection, the equation becomes
r1
/'
[wave motion along negative xdirection]
d Derive the corresponding equations that con vert laboratorymeasured frequency/to rocketmea sured frequency f for waves moving along both posi tive and negative xdirections.
819
c If the frequency of the light in the laboratory frame is / what is the frequency f ' of the light in the rocket frame? Use the Einstein relation between pho ton energy E and classical wave frequency / namely Econv “ ^ /o t E = bf/c^ and E' = h f'/c^, to derive the transformations for frequency
Doppler equations
A photon moves in the xy laboratory plane in a direc tion that makes an angle (f) with the xaxis, so that its components of momentum are p^ —p cos (f) and py = p sin (f) and p^ = 0. a Use the Lorentz transformation equations for the momentum  energy 4vector (Exercise 75) and the relation — /»^ = 0 for a photon to show that in the rocket frame, moving with speed along the
cos 0 ) cos 0 ')
This difference in frequency due to relative motion is called the D o p p le r shift. N ote: We use/ for frequency instead of the usual Greek nu, V, ro avoid confusion with v for speed. d For wave motion along the positive and nega tive xdirection, show that the results of this exercise reduce to the results of Exercise 818. e D iscussion question: Do the Doppler equations enable one to determine the rest frame of the source that emits the photons?
820
the physicist and the traffic light
A physicist is arrested for going through a red light. In court he pleads that he approached the intersection at such a speed that the red light looked green to him. The judge, a graduate of a physics class, changes the charge to speeding and fines the defendant one dollar for every kilometer/hour he exceeded the local speed limit of 30 kilometers/hour. W hat is the fine? Take the wavelength of green light to be 5 30 nanometers = 530 X 10“^ meter) and the wavelength of red light to be 650 nanometers. The relation between wavelength A and frequency/for light is/A = c. Notice that the
264
EXERCISE 821
SPEEDING LIGHT BULB
light propagates in the negative xdirection {(f) = (f)' = n).
821
speeding light bulb
A bulb that emits spectrally pure red light uniformly in all directions in its rest frame approaches the ob server from a very great distance moving with nearly the speed of light along a straightline path whose perpendicular distance from the observer is A Both the color and the number of photons that reach the observer per second from the light bulb vary with time. Describe these changes qualitatively at several stages as the light bulb passes the observer. Consider both the Doppler shift and the headlight effect (Exer cises 819 and L9).
822
Doppler shift at the limb off Sun
Sun rotates once in about 25.4 days. The radius of Sun is about 7.0X10® meters. Calculate the Doppler shift that we should observe for light of wavelength 500 nanometers = 500 X 10“ ^ meter) from the edge of Sun’s disk (the lim b) near the equator. Is this shift toward the red end or toward the blue end of the visible spectrum? Compare the magnitude of this Doppler shift with that of the gravitational red shift of light from Sun (Exercise 86).
823
the expanding universe
N ote; Recall Exercise 310. a Light from a distant galaxy is analyzed by a spectrometer. A spectral line of wavelength 730 nano meters = 730 X 10“ ^ meters is identified (from the pattern of other lines) to be one of the lines of hydro gen that, for hydrogen in the laboratory, has the wavelength 487 nanometers. If the shift in wave length is a Doppler shift, how fast is the observed galaxy moving relative to Earth? Notice that the light propagates in a direction opposite to the direction of motion of the galaxy {(f) = •■1 con v = ntc^ ffrom the Doppler shifft
Einstein’s famous equation in conventional units, Eratconv ~ ^F'd the telativistic expression for en ergy can be derived from (1) the relativistic expression for momentum (derived separately, for example in Exercise 712), (2) the conservation laws, and (3) the
EXERCISE 827
EVERYTHING GOES FORWARD
Doppler shift (Exercise 818). In conventional units, a photon has energy — hf, where h is Planck’s constant and / i s the frequency of the corresponding classical wave. (We use/ for frequency instead of the usual Greek nu, V, to avoid confusion with v for speed.) Divide by to convert to units of mass: E = hf/c^. Expressed in units of mass, a photon has equal energy and momentum. Therefore the momentum of a photon is also given by the equation p = hf/c^. Momentum does differ from energy, however, in that it is a 3vector. In one dimensional motion, the sign of the momentum (positive for motion to the right, negative for motion to the left) is important, as in the analysis below. A _ '“'de of mass emits two photons in opposite directions while remaining at rest in the lab oratory frame. Conservation of momentum requires these two photons to have equal and opposite mo menta and therefore to correspond to the same classi cal frequency / In consequence, they also have the same energy. a First resu lt: Energy released = A w . Now view this process from a rocket frame moving at speed V — t'c o n v / ^ along the direction of flight of the two photons. The particle moves in this frame, but does not change velocity on emitting the photons. The photon emitted in the same direction as the rocket motion will be upshifted in energy (and in corre sponding classical frequency) as compared with the energy observed in the laboratory; the other back wardmoving photon will be downshifted. We can calculate this frequency shift using the Doppler for mulas (Exercise 818). Use the expression m jv for momentum of a particle, equation (78), to state the conservation of momentum (notice the minus sign before the second photon term, representing the pho ton moving to the left):
= m ^ v y b
hf C‘
b Second result: = m . Now add the con dition that energy is conserved in the laboratory frame: ^ b efo re 
Eafter +
(
2)
Compare equations (1) and (2). These two equa tions both describe a particle at rest. Show that they are consistent if Etefore = "^before and E ^ = and that therefore in general
or, in conventional units. ^ re stco n v =
c T h ird result: At any speed, E = my. Next add the condition that energy be conserved in the rocket frame. Place primes on expressions for rocketmeasured energy of the particle and use the Doppler equations to transform the classical frequency back to the laboratory value f. Show that the result is ■^before
E'after + K^hf/
(3)
The salient difference between equations (2) and (3) is that in the rocket frame the particle is in motion. Deduce that the general expression for energy of a particle includes the stretch factor gamma: E = my
ficonv =
m yc^
Reference: Fritz Rohrlich, American Journal o f Physics, Volume 58, pages 3 4 8  3 4 9 (April 1990).
Simplify this expression to
827 (
1)
or m\before
energy released (conventional units) = {Am)c^
[^ 1
1+
"^after +
units result by to convert to conventional units and the equation in the wellknown form
or, in conventional units,
1/2
h f\ 1
»«before =
265
= Am = 2hf/c^ = energy released
Conservation of momentum in both frames im plies a change in particle mass equal to the total energy of the emitted photons. Multiply the mass
everything goes forward
“ Everything goes forward’’ is a good mle of thumb for interactions between highly relativistic particles and stationary targets. In the laboratory frame, many particles and gamma rays resulting from collisions continue in essentially the same direction as the in coming particles. The first figure (top) shows schematically the colli sion of two protons in the centerofmomentum frame, the frame in which the system has zero total
266
EXERCISE 828
DECAY OF 7T°MESON
momentum. A great many different particles are cre ated in the collision, including a gamma ray (the fastest possible particle) that by chance moves per pendicular to the line of motion of the incoming particles: (f)' = 7T/2 radians. The first figure (bottom) shows the same interac tion in the laboratory frame, in which one proton is initially at rest. At what angle (f) does the product gamma ray move in this frame? a From the Doppler equations (Exercise 819), show that the outgoing angle k
2 x 1 0 * ^ km
1 km ^
^
1 km
2 0 0 km
2 0 0 km
 > location to right of center line
TRAVELERS HEADED "NORTH" ON EARTH FIGURE 9 5 . C o m p a r is o n o f t h e p a t h s o f n o r t h w a r d t r a v e l e r s o n E a r t h ’s s u r fa c e w i t h th e w o r l d l i n e s o f b a l l b e a r in g s r e le a s e d s i d e b y s i d e f r o m r e s t n e a r E a r t h ’s s u r fa c e . In both cases the " p a th ” o f each “tra v e le r” sta rts p a ra lle l w ith th a t o f the second traveler (zero i n i t i a l relative velocity). In both cases th is “p a t h ” g ra d u a lly inclines to w a rd the centerline ( “re lative acceleration”). In both cases the p a th s can be accounted fo r in term s o f the local cu rva tu re o f geometry (curvature o f E a r th ’s surface f o r the travelers; cu rva tu re o f spacetim e geom etry— g r a v ita tio n ! — f o r the b a ll bearings). In each dia g ra m , ve rtic a l distances are d r a w n — fo r v iv id n e s s— to a d ifferen t scale th a n h o rizo n ta l distances. B oth d ia gram s suffer fro m th is a d d itio n a l im perfection: they a tte m p t to show, on the f l a t E uclidean surface o f th is page, trajectories th a t can be correctly represented only in term s o f a curved geometry.
you try. This means that the ratio of gravitational mass to inertial mass is the same for all sorts of objects. You have made a great discovery abour mass.” All this time we and our spacetraveler friends are looking down from on high. We see the many treks. We watch the many measurements of distance. Through our intercommunication system we hear and approve as our friends on the ground interpret distance shortening as relative acceleration — and relative acceleration as “gravitation.” But then they get into weighty discussions. They start speaking of “gravitation” as action at a distance. We smile. W hat is at issue— we know — is not action at a distance at all, but the geometry of curved space. All this talk about the identity of “gravitational mass” and “inertial mass” completely obscures the ttuth. Curvature and nothing more is all that is required to describe the increasing rate at which A and B approach each other.
Curvature alone accounts for relative acceleration
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9.6 GRAVITATION AS CURVATURE OF SPACETIME spacetime curvature accounts for tidal accelerations of objects Spacetim e curvature dem onstrated by change in separation of two originally parallel worldlines
Acceleration tow ard Earth: Totalized effect of relative accelerations, each particle tow ard its neighbor, in a chain of test particles that girdles globe
Einstein smiles, too, as he hears gravitation described as action at a distance. Curvature of spacetime and nothing more, he tells us, is all that is required to describe the millimeter or two change in separation in 8 seconds of two ball bearings, originally 20 meters apart in space above Earth, and endowed at the start with zero relative velocity. Moreover, this curvature completely accounts for gravitation. “W hat a preposterous claim!” is one’s first reaction. “ How can such minor— and slow — changes in the distance between one tiny ball and another offer any kind of understanding of the enormous velocity with which a falling mass hits Earth?” The answer is simple: Many local reference frames, fitted together, make up the global structure of spacetime. Each local Lorentz frame can be regarded as having one of the ball bearings at its center. The ball bearings all simultaneously approach their neigh bors (curvature). Then the largescale structure of spacetime bends and pulls nearer to Earth (Figure 96). In this way many local manifestations of curvature add up to give the appearance of longrange gravitation originating from Earth as a whole. In brief, the geometry used to describe motion in any local freefloat frame is the flatspacetime geometry of Lorentz (special relativity). Relative to such a local freefloat frame, every nearby electrically neutral test particle moves in a straight line with constant velocity. Slightly more remote particles are detected as slowly changing their velocities, or the directions of their worldlines in spacetime. These changes are de scribed as tidal effects of gravitation. They are understood as originating in the local curvature of spacetime. From the point of view of the student of local physics, gravitation shows itself not at all in the motion of one test particle but only in the change of separation of two or more nearby rest particles. “Rather than have one global frame with gravitational forces we have many local frames without gravitational forces. ’’ However, these local dimension changes add up to an effect on the global spacetime structure that one interprets as “gravitation” in its everyday manifestations. In contrast, Newton supposed the existence of one ideal overall reference frame. For him, “Absolute space, in its own nature, without relation to anything external, remains always similar and immovable.” The ball bearing or spaceship is regarded by Newton as actually accelerated with respect to this ideal frame. The “gravitational force” that accelerates it acts mysteriously across space and is produced by distant objects. That the man in the spaceship finds no evidence either of the acceleration or the force is an accidenr of nature, according ro the Newtonian view. Pundits used to interpret this accident of nature as the fortuitous equality of “gravitational mass” and “inertial mass” or in other “ learned” ways. In conversations with one of the authors of this book at various times over rhe years, Einstein emphasized his great respect for Newton and, in particular, his admiration for Newton’s courage and judgment. He stressed that Newton was even better aware than his seventeenthcentury critics of the difficulties with the ideas of absolute space and time. To postulate those ideas was nevertheless the only practical way to get on with the task of describing motion in Newton’s century. In effect, Newton chopped the problem of motion into two parts: (1) space and time and their meaning: ideas that were puzzling but usable and that were destined to be clarified only 230 years later and (2) the laws of acceleration with respect to that idealized spacetime: laws that Newton gave the world.
9.6
GRAVITATION AS CURVATURE OF SPACETIME
FIGURE 9 6 . L o c a l c u r v a t u r e a d d i n g u p to t h e a p p e a r a n c e o f lo n g r a n g e g r a v i ta t i o n . The shortening o f d ista n ce between a n y one p a ir , A a n d B, o f h a ll hearings is sm a ll w hen the d istance its e lf is sm all. H owever, sm a ll separation between each h a ll hearing a n d its p a rtn e r dem ands m any p a ir s to encompass E arth. T he to ta lize d shortening o f the circumference in a n y g iv e n tim e — the shortening o f one separation tim es the num ber o fse p a ra tio n s— is in dependent o f the fineness o f the subdiv ision. T h a t to ta lize d p u llin g in o f the circumference carries the whole necklace o f masses in w a rd . T h is is fre e fa l l , th is is g ra v ity , th is is a large scale m otion interpreted a s a consequence o f local curvature. Exam ple: O rig in a l separation between A a n d B — a n d every other p a ir: 2 0 meters T im e o f observation: 8 seconds Shortening o f separation in th a t tim e: 1 m illim eter F ractio n a l shortening: 1 m illim e te rf 2 0 meters = 1 1 2 0 ,0 0 0 Circumference o f E a rth {length o f a iry necklace o f h a ll hearings): 4 .0 0 3 0 X 10^ meters Shrinkage o f th is circumference in 8 seconds: 1 /2 0 ,0 0 0 X 4 .0 0 3 0 X 10^ meters = 2 0 0 1 .5 meters Decrease in the d ista n ce fro m the center o f E a rth (drops by the sam e fa c to r 1 /2 0 ,0 0 0 ) : 1 /2 0 ,0 0 0 X 6 .3 7 1 X 10^ meters = 3 1 5 meters. T h is app a ren tly largescale effect is ca u sed — in E in s te in ’s p ic tu r e — by the a d d itio n o f a m u ltitu d e o f sm allscale effects: the changes in the local dim ensions associated w ith the curvature o f geometry (fa ilu r e of& to rem ain a t rest a s observed in the fre eflo a t fra m e associated w ith A).
W hat is the source of the curvature of spacetime? Momenergy is the source. In Chaprer 8 we saw the primacy of momenergy in governing interactions between particles. Crash of mass on mass, no matter how elastic or how destructive, leaves the total momenergy of the system quite unaltered. By what miracle does this come about? Education of momenergy from birth onward to good behavior? Goodness of heart?
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HOW SPACETIME CURVATURE CARRIES INFLUENCE FROM ONE MASS TO ANOTHER The necklace of ball bearings (Figure 96) as they ap proach Earth, examined more closely, reveals a re markable feature of spacetime curvature outside a
great, essentially uniform, essentially isolated sphere of mass. The curvature in its character is totally “ tideproducing,” totally “ noncontractile.”
A n arra y o f test masses covering the surface o f a hollow sphere fre ely flo a tin g above the E a r th ’s surface w i l l sh rin k in tw o dim ensions a n d lengthen in one. T he volum e rem ains constant; only the shape changes. T h is change is evidence o f the noncon tra c tile, tid e d r iv in g spacetim e curvature outside E arth.
What do these descriptive terms mean, and how do we verify that they apply? W e look at a cluster of ball bearings dotted here and there over the surface of an imaginary small sphere, all momentarily at rest relative to each other and relative to Earth. That shape, how ever, as the seconds tick by, changes from sphere to ellipsoid. How come? First let’s look at the two dimen sions of the sphere that lie perpendicular to each other but parallel to Earth's surface. Both these dimensions of the sphere shrink os the boll bearings converge toward Earth’s center. The updown dimension of the pattern, however, lengthens, and twice as much. Why? Newton says because of the greater gravitational acceleration of the one nearer Earth. Einstein says because twoper cent stretch in that dimension compensates onepercent shrinkage in the other two dimensions and keeps the volume of the pattern unchanged. Spacetime curvature.
Spacetim e controls momenergy
yes; but a totally noncontractile curvature. Einstein’s famous equation, stated in simple terms, tells us how spacetime curvature responds to mass: ap p ro p riate m easure of spacetim e contractile curvature at an y p lace, an y time, in any Lorentz fram e
(
density of e n e rg y \ at that lo cale j perceived in that I Lorentz fram e /
Outside, no mass, no energy, a spacetime curvature that is totally noncontractile. Inside Earth, however, there is mass, therefore there is energy — or in a mov
Obedience to the eyes of a corps of bookkeepers? No, Einstein taught us. The enforcing agency does not lie far away. It’s close at hand. It’s the geometry of spacetime, right where the crash takes place. N ot only does spacetime grip isolated mass, telling it how to move. In addition, in a crash it sees to it that the participants neither gain nor lose momenergy. But there is more! Spacetime, in so acting, cannot
9.6
GRAVITATION AS CURVATURE OF SPACETIME
ing frame, energy plus energy flow — and therefore spacetime curvature there has a contractile character. The ball bearings — when shafts are drilled for them so that not one of them encounters any obstacle to freefloat motion — start to converge vertically as well as horizontally. The volume shrinks. That, overlooking de tails, is what we mean when we say that “ mass grips spacetime, telling it how to curve.” There is no Earth mass out at Moon's orbit. How then does Einstein’s spacetime geometry account for Moon’s motion? Answer: Earth’s mass imposes on spacetime a contractile curvature throughout Earth’s in terior, as a jumper’s feet impose a contractile curvature on a trampoline. That contractile curvature, where the feet push, forces on the surrounding nontear fabric a corresponding lateral stretch. That effect transmits itself in ever more dilute measure to the ever more remote regions of the trampoline.
287
Likewise spacetime does not tear. Its fabric just above Earth’s surface experiences the same lateral contractil ity as it does just below the surface. Not so with the curvature in the twodimensional domain defined by time and by direction perpendicular to Earth’s surface. In that one plane, curvature within Earth is contractile but suddenly jumps just above Earth’s surface to the opposite character. Hence the tideproducing charac ter of spacetime curvature outside Earth. A point twice as far from Earth’s center lies on an imaginary Earthcentered sphere that encompasses eight times the vol ume. There the tideproducing curvature experiences eight times the dilution and has one eighth the strength. Despite this rapid dilution of tideproducing power with distance, it has strength enough at Moon, 60 Earth radii away from Earth’s center, to deform Moon from sphere to ellipsoid, 1738.35 kilometers in radius along the EarthM oon direction, 1738.15 kilometers in radius for each of the other two perpendicular directions. Easy as it is to regard Earth as running the whole show. Moon too has its part. Like an infant standing on the trampoline some distance from its mother, it imposes its own small curvature on top of the curvature evoked by Earth. That additional curvature, contractile in Moon’s interior, has tidedriving character outside. W ere the Earth an ideal sphere covered by an ideal ocean of uniform depth, then Moon would draw that ocean’s surface 35.6 centimeters higher than the average in two domains, one directly facing Moon, one directly oppo site to it — simultaneously lowering those waters 17.8 centimeters below the average on the circle of points midway between the two. (These low figures show how important are funneling and resonant sloshing in deter mining heights of actual ocean tides on Earth.)
T he deform ation o f the nontear tram poline fa b r ic u n d er the ju m p e r ’s fe e t a n d elsewhere is analogous to the nontear cu rva tu re o f spacetim e geometry inside E a rth a n d elsewhere.
The local contractile curvature of spacetime at Moon’s location added up along Moon’s path yields the ap pearance of longrange gravitation, similar to that il lustrated in Figure 96. Box 21 tells a little of the many influences that have to be taken into account in any fuller treatment of the tides.
maintain the perfection assumed in textbooks of old. To every action there is a corresponding reaction. Spacetime acts on momenergy, telling it how to move; momenergy reacts hack on spacetime, telling it how to curve. This “handshake” between momen ergy and spacetime is the origin of momenergy conservation — and the source of spacetime curvature that leads to gravitation (Box 91).
M om energy tells spacetim e how to curve
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9.7 GRAVITY WAVES gravitational energy moving at light speed Gravity waves from collapsing matter
In the depths of an illfated, collapsing star, billions upon billions of tons of mass cave in and crash together. The crashing mass generates a wave in the geometry of space— a wave that rolls across a hundred thousand lightyears of space to “jiggle” the distance between two mirrors in our Earthbound gravitywave laboratory. A cork floating all alone on the Pacific Ocean may not reveal the passage of a wave. But when a second cork is floating near it, then the passing of the wave is revealed by the fluctuating separation between the two corks. So too for the separation of the two mirrors. There is, however, this great difference. The corktocork distance reveals a momentary change in the twodimensional geometry of the surface of the ocean. The
B O X 9 2
COMPACT STELLAR OBJECTS Three kinds of astronomical objects exist comparable in mass to Sun but very much smaller. Two of these have been observed; the third seems an inevita ble result of Einstein’s theory. A white dwarf star is a star of about one solar mass, with radius about 5000 kilometers. (The radius of Earth is 6371 kilometers.)Thisgives the white dwarf a density of approximately 10’ kilograms/meter^ (or one metric ton per cubic centimeter). As of 1990, approximately 1500 white dwarfs have been iden tified. White dwarfs were observed and studied astronomically long before they were understood theoretically. Today we have come to recognize that a white dwarf is a star that quietly used up its fuel and settled gently into this compact state. The electrons and nuclei that make up the body of a white dwarf are not separated into atoms. Instead, the electrons form a gas in which the nuclei swim. The pressure of this “ cold” electron gas keeps the white dwarf from collapsing further. S. Chandrasekhar calculated in 1930 that no white dwarf can be more mas sive than approximately 1.4 solar masses (“ Chandrasekhar limit” ) without collapsing under its own gravitational attraction. His analysis assumed the mix of electrons and nuclei to be unaltered under compression by a load so heavy, an assumption that had to be modified in later years. Today we recognize that enormous compressions squeeze electrons into combining with protons to make neutrons. At compressions near the Chandrasekhar limit, the electron gas transforms into a neutron gas, the interior of the star becomes a giant nucleus, and the whole nature of the compact object changes to that of a neutron star. A neutron star has roughly the same density as an atomic nucleus, of the order of 10'^ kilograms/meter^, or one Earth mass per cube of edge length 400 meters. The radius of a neutron star is approximately 10 kilometers.
9.7
GRAVITY WAVES
mirrortomirror distance reveals a momentary change in the threedimensional ge ometry of space itself. The idea of extracting energy from ocean waves is old. After all, the ability of a water wave to change a distance lets itself be translated into the ability to do work. The same reasoning applies ro a gravity wave. Because it can change distance, it can do work. Ir carries energy. Energy once resident as mass in the interior of a star has radiated out to us and to all the universe. O f all the workings of rhe grip of gravity, none is more fascinating or opens up for exploration a wider realm of ideas rhan a gravity wave. None pushes to a higher pirch the art of detecting a small effect, and none gives more promise of providing an unsurpassable window on cataclysmic events deep inside troubled stars. Nevertheless, no other great prediction of Einstein’s geometric theory of graviry srands today so far from triumphant exploitation. As of this writing, not one of the nine ingenious
How often is a neutron star formed? Towards answering this still open ques tion we have one important lead: In our own galaxy we see one supernova explosion on average about every 300 years [most recent supernova in the Large Magellanic Cloud, a satellite structure near our galaxy, on February 23, 1987; one seen by Kepler, October 13, 1604; one seen by Tycho Brahe, November 6, 1572; earlier ones: 1181 a . d . ; July 4, 1054 a . d . ; 1006 a . d . (the brightest); 185 a . d . ; and two possibles in 386 a . d . and 393 a . d .] . In such an event a star teetering on the edge of instability finally collapses. The Niagara Falls of infalling mass in some cases go too far and overcompress the inner region of the star. That region thereupon acts like a spring, or explosive charge, and drives off the outer portions of the star. This explains the spec tacular luminosity that is such a prominent feature of a supernova. The core that remains becomes a neutron star in some events, it is believed, in others a black hole. Neutron stars were predicted in 1934 but not observed until 1968. Many neutron stars spin rapidly — with a period as short as a few milliseconds. A neutron star typically has an immense magnetic field. When that field is aligned at an angle relative to the axis of spin of the star (as in Earth, for example), it sweeps around like a giant whisk brush through the plasma in the space around the star. The periodic shock to the electrons of the plasma from the periodic arrival of this field excites those electrons to radiate periodic pulses of radio waves and visible light — both observed on Earth. Because of this behavior, such neutron stars are called pulsars. As of 1990, nearly 500 pulsars have been identified. A black hole is an object created when a star collapses to a size so small that strong spacetime curvature prevents it from communicating outward with the external universe. Even light cannot escape from a black hole, whence its name. No one who accepts general relativity has found any way to escape the prediction that black holes must exist in our galaxy. Strong evidence for the existence of black holes has been found, but it is not yet convincing to all astrophysicists. A black hole can have a mass as small as a few times the mass of our Sun. A black hole of three solar masses would have a “ radius” of about 9 kilometers. There is no theoretical upper limit to its mass.
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How to detect gravity w aves
290
Gravity waves result from time delay
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detectors built to this day has proved sensitive enough to secure any generally agreed detection of an arriving gravity wave. Does any truly simple line of reasoning assure us that gravity will inescapably carry energy away from two masses that undergo rapid change in relative position? Yes is the conclusion of a little story that savors of mythology. The Atlas of our day, zooming through space in free float, insists as much as ever on maintaining physical fitness. He pumps iron, not by raising iron against the pull of Earth’s gravity, but by throwing apart two identical great iron spheres. Alpha and Beta. He floats between those minor moons and plays catch with them. Each time they fall together under the influence of their mutual gtavity, he catches them, absorbs their energy of infall in his springlike muscles, and flings them apart so that they always travel the same distance before returning. It’s an enchanting game, but Atlas finds that it’s a losing game. When the masses fly back together, they never yield up to him as much energy as he must supply to throw them apart again. Why not? Say the central point in two words: time delay. Like any force that makes itself felt through the emptiness of space, the force of gravity cannot propagate faster than the speed of light. This limitation imposes a delay on the attraction between the two iron spheres. Alpha, on each little stretch of its outbound path, feels a pull that originated from Beta when the two were a tiny bit closer than they are now. The actual force that’s slowing Alpha is therefore a tiny bit bigger than we would judge from thinking of them as stationary at their momentary separation. On its return trip inbound along the same little stretch of path. Alpha experiences a helping pull that originated from Beta when the two had a separation slightly greater than its present value. The actual force that’s speeding Alpha inward is therefore a tiny bit less than we would judge from thinking of them as stationary at theit momentary separation. In each stretch of their outbound trip, the two masses have to do more work against the pull of gravity than they get back — in the form of work done on them by gravity— on the same stretch of path inbound. A calculable amount of energy disappears from the local scene on each outin cycle of Atlas’s exercise. Yet the total energy must somehow be conserved. Therefore the very gravity that steals energy from Atlas and his iron, or from any two
Year FIGURE 9 7 . T w o w h irlin g neutron sta rs fu r n is h a g ia n t do ck, whose tim ekeep in g h a n d is the line, evertu rn in g, th a t separates the centers o f those tw o stars. T h a t h a n d does not today keep the “slo w ” schedule (stra ig h t h o rizo n ta l line) one m ig h t have expected fro m its tim in g a s m easured in 1 9 7 4 . T he d o w n w a rd sloping curve show s g ra v ity w a v e theory's p rediction o f the shortening in the tim e required to accum ulate a ny specified num ber o f revolutions. T he dots show the a c tu a l observed shortening in th a t tim e.
9.7
GRAVITY WAVES
masses that rapidly change their relative position, must somehow all the time be transporting the stolen energy to the faraway. That inescapable theft of energy is in its quality, its directional distribution, and its magnitude none other than what Einstein had treated long before under the head of gravity radiation and what we now call gravity waves. Atlas couldn’t “see” those gravity waves. Neither have we today yet succeeded in detecting directly the gravity waves we feel sure must be radiating from sources dotted here and there in the galaxy and in the universe. However, we have an exciring indirecr confirmation that gravity waves exist— not through their action on any receptor, bur through the energy they carry away from a whirling pair of neutron srars. That particular “binary pulsar” first revealed itself to Joseph H. Taylor, Jr., and Russell A. Hulse by periodic pulses of radio waves picked up on the huge disklike antenna at Arecibo in Puerto Rico. As one of these neutron stars spins on its axis, its magnetic field spins with it, giving timing comparable in accuracy to the best atomic clock ever built (Box 92). Thanks to this happy circumstance, Taylor and his colleagues have been able to follow the evershortening separation of the two stars and the everhigher speed they attain as they slowly spiral in toward an ultimate catastrophe some 400 million years from now. The timing of the orbits gives us a measure of energy lost as the stars spiral in. No reasonable way has ever been found to account for the thus observed loss of energy except gtavitational radiation. As of September 1989, 14 years after first observation, this loss of enetgy agrees with the rate predicted by theory to bettet than one percent (Figure 97). Gravity waves and pulses of gravity radiation are sweeping over us all the time from sources of many kinds out in space. Detecting them, however, we are no better than the primitive jungle dweller unable to detect and even totally unaware of the tadio waves that carry past her every minure of the day music, words, and messages. However, experimentalists are working out ingenious technology and building detec tor instrumentation of evetgrowing sensitivity (Figures 98 and 99). Few among them have any doubt of their ability to detect pulses of gravity radiation from one or another star catastrophe by sometime in the first decade of the twentyfirst century.
FIGURE 9 8 . The proposed M I T  Caltech g ra v ity w a v e detector w ill ( I ) use the beam fro m a laser (left), (2 ) sp lit i t by a device (center) analogous to a h a lfsilvere d mirror, (3 ) send one h a lfstren g th beam to one fa r a w a y m irror (top) a n d the other to the other fa r a w a y m irror (right), (5 ) allow these beams to undergo m a n y m a n y reflections (not show n), a n d (6 ) recombine them a t the detector (bottom ). A g r a v ity w a v e in c id e n t on E a rth w ill slig h tly shorten the 4 kilo m eter d ista n ce to the one m irror a n d s lig h tly lengthen the 4kilo m eter d ista n ce to the other mirror. T h is rela tive a ltera tio n in the p a th length o f the laser beams, i f big enough, a m p lified enough, a n d p ic k e d u p by detectors sensitive enough, w ill reveal the passage o f the g r a v ity
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G ravity w aves steal en ergy from orbiting neutron stars
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FIGURE 99. P r o to ty p e g r a v i t y  w a v e d e te c to r , C a l i f o r n ia I n s t i t u t e o f T e c h n o lo g y , P a s a d e n a . The laser beam is ta ilo red (lower right} f o r entry in to the beam sp litte r (located where the tw o long lig h t p ip es meet, ju s t to the le ft o f center in the photograph). The m irrors a t the ends o f these tw o eva cu a ted lig h t pipes lie outside the boundary o f the photograph.
Astronomy uses signals of many kinds — light, radio waves, and Xrays among them — to reveal the secrets of the stars. O f all signals from a star, none comes out from deeper in the interior than a gravity wave. Among all violent events to be probed deeply by a gravity wave, none is more fascinating than the dance of death of two compact stars as they whirl around each other and undergo total collapse into . . . a black hole!
9.8 BLACK HOLE over the edge with a scream of radiation
‘Escape velocity c” implies black hole
A black hole is a domain whose mass is so tightly compacted that nothing can escape from it, not even light. Everything that falls in is caught without hope of escape (Figure 910). To fire a missile from Moon’s surface so that it escapes that satellite’s attraction demands a speed of 2.3 8 kilometers per second or greater. The critical speed for escape from Earth — in the absence of drag from the atmosphere— is 11.2 kilometers per second. When the object does not rotate and is so compact that even light cannot escape, the “effective radius’’ or socalled “horizon radius’’ is
(effective radius)
'circumference of region' out of which ^ light cannot escape , 27T = 2 X (1.47 kilometers) X
Black hole still exerts “ pull” of gravity
mass of black hole ' expressed in , number of Sun masses j
When a star or cloud of matter collapses to a black hole it disappears from view as totally as the Cheshire cat did in Alice in Wonderland. The cat, however, left its grin behind; and the black hole — via the effect of spacetime curvature that we call gravity — exerts as much “pull” as ever on normal stars in orbit around it. They are like participants in a formal dance with lights turned low. Only the white dress of the girl is visible as she whirls around in the arms of her blacksuited companion. From the
9.8
BLACK HOLE
293
p articles a s yet undetected angular
known particles
momentum
gravitational and electrom agnetic
mass ch arg e an gu lar momentum FIGURE 9 1 0 . W h a tever objects f a l l in to a black hole, they possess a t the e n d — a s seen fro m o u tsid e — only mass, a n g u la r m om entum , a n d electric charge. N o t one other characteristic o f a n y in fa llin g object rem ains to betray its p a s t — not a h a ir. T h is leads to the saying, “A black hole has no h a ir .”
speed of the girl and the size of the circle in which she swirls, we know something of the mass of the invisible companion. By such reasoning it was possible to conclude by 1972 that the optically invisible companion of one longknown star has a mass of the order of 9.5 solar masses. This remarkable object came first to attention because in December 1971 the Uhuru orbiting Xray observatory detected Xray pulsations with time scales from one tenth to tens of seconds from an object located in the Cygnus region close to the known star. Why does it give off Xrays? And why does the intensity of the Xrays vary rapidly from instant to instant? The gas wind from the visible companion varies from instant to instant like the smoke from a factory chimney. This gas, falling on a compact object, gets squeezed. To picture the how and why of this squeeze, look from a lowflying plane at the streams of automobiles converging from many directions on a football stadium for a Saturday afternoon game. The particles and the gas are pushed together as surely as the cars in the traffic. The compression of the traffic raises the temper of the driver, and the compression of the gas raises its temperature as air is heated when pumped in a bicycle pump. However, because the gas falls from an object of millions of kilometers in size to one a few kilometers across, the compression is so stupendous that the temperature rises far above any normal star temperature, and Xrays come off. The time scale of the fluctuations in Xray intensity depends on the size of the object that is picking up the star smoke, a size less by a fantastic factor than that of any normal star. Could the object be a white dwarf (Box 92)? No, because such a star would be
Cygnus X 1 : A black hole?
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GRAVITY; CURVED SPACETIME IN ACTION C ^ ^ B L E 92 T ]> 
BLACK HOLES FOR WHICH THERE WAS SUBSTANTIAL EVIDENCE AS OF SEPTEMBER 1989 (Uncertainties in masses are of the order of 20 to 50 percent.) Astronomical designation of black hole Cygnus X1 LMC X1 AO 62000 LMC X3 SS433 Black hole at center of our galaxy
Black hole at center of our g a la x y?
Q u a sa r energy output from matter swirling into black hole?
Highefflciency conversion of gravitational en ergy to radiation
Mass (in solar masses) 9.5 2.6 3.2 7.0 4.3 3.5 X 10*
visible. A neutron star? No, because even matter compressed so tightly that it is transformed to neutrons cannot support itself against gravity if it has a mass much over two solar masses. No escape has been found from concluding that Cygnus X1 is a black hole. This great discovery transformed black holes from pencilandpaper objects into a lively and evergrowing part of modern astrophysics (Table 91). Much attention went in the 1980s to a presumptive black hole with a mass of about three and a half million times the solar mass and a horizon radius of about ten million kilometers. It floats at the center of our galaxy, rhe Milky Way. Around it buzz visible stars of the everyday kind, most of them fated to fall eventually into that black hole and increase its mass and size. That stars close to the center of our galaxy go around as fast as they do is one of the best indicators we have for the presence, and one of the best measures we have for the mass, of the central black hole, which is itself invisible. In contrast to dead solitary black holes, the most powerful source of energy we know or conceive or see in all rhe universe is a black hole of many millions of solar masses, gulping down enormous amounts of matter swirling around it. Maarten Schmidt, working at the Mount Palomar Observatory in 1956, was the first to uncover evidence for these quasistellar objects, or quasars, starlike sources of light located not billions of kilometers but billions of lightyears away. Despite being far smaller than any galaxy, the typical quasar manages to put out more than a hundred times as much energy as our own Milky Way, with its hundred billion stars. Quasars, unsurpassed in brilliance and remoteness, we call lighthouses of the heavens. Observation and theory have come together to explain in broad outline how a quasar operates. A black hole of some hundreds of millions of solar masses, itself built by accretion, accretes more mass from its surroundings. The incoming gas, and starsconvertedtogas, does not fall in directly, any more than the water rushes directly down the bathtub drain when the plug is pulled. Which way the gas swirls is a matter of chance or past history or both, but it does swirl. This gas, as it goes round and round, slowly makes its way inward to regions of everstronger gravity. Thus com pressed, and by this compression heated, the gas breaks up into electrons — that is negative ions — and positive ions, linked by magnetic fields of force into a gigantic accretion disk. Matter little by little makes its way to the inner boundary of this accretion disk and then, in a great swoop, falls into the black hole, on its way crossing the horizon, the surface of no return. During that last swoop, hold on the particle is relinquished. Therefore, the chance is lost to extract as energy the full 100 percent of the mass of each infalling bit of matter. However, magnetic fields do hold onto the ions effectively enough for long enough to extract, as energy, several percent of the
9.8
BLACK HOLE
I Co
g 3O
u
o w c
X A L B E R T E IN S T E IN Ulm, G erm any, M a rch 14, 1 8 7 9 — Princeton, N e w Jersey, A p r il 18, 1 9 5 5
“Newton himself was better aware of the weaknesses inherent in his intellectual edifice than the generations which followed him. This fact has always roused my admiration.’’ * ★ ★ “Only the genius of Riemann, solitary and uncomprehended, had already won its way by the middle of the last century to a new conception of space, in which space was deprived of its rigidity, and in which its power to take part in physical events was recognized as possible.’’ ★ ★ ★ “ All of these endeavors are based on the belief that existence should have a completely harmonious structure. Today we have less ground than ever before for allowing ourselves to be forced away from this wonderful belief.’’ mass. In contrast, neither nuclear fission nor nuclear fusion is able to obtain a conversion efficiency of more than a fraction of a percent. O f all methods to convert bulk matter into energy, no one has ever seen evidence for a more effective process than accretion into a black hole, and no one has even been able to come up with a more feasible scheme for one. O f all the features of black hole physics in action, none is more spectacular than a quasar. And no lighthouse of the skies gives more dramatic evidence of the scale of the universe.
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9.9 THE COSMOS a final crunch? Expanding universe; Evidence for big bang beginning
“ O p e n ” universe expanding forever? O r “ clo sed ” universe that recontracts to crunch? An open question!
The more distant quasars and galaxies are, the greater the speed with which they are observed to be receding from us. This expansion argues that somewhere between ten and twenty billion years ago the universe began with a big bang, a time before which there was no time. We see around us relics of the big bang, not only today’s rapidly receding galaxies but also today’s abundance of the chemical elements — some among them still radioactive, the “still warm ashes of creation” (V. F. Weisskopf) — and today’s greatly cooled but still allpervasive “primordial cosmic fireball radiation.” We now believe that in the first instants of its life, the entire universe filled an infinitesimally small space of enormous density and temperature where matter and energy fused in a homogeneous soup. Immediately the universe began expanding. After about 10“ ^ seconds it had cooled enough that subatomic particles condensed from the matterenergy soup. In the first three minutes after the big bang, neutrons and protons combined to make heavier elements. Eons later stars and galaxies formed. Never since has the universe paused in its continual spread outward. Will the universe continue expanding forever? Or will its expansion slow, halt, and turn to contraction and crunch (Table 92), a crunch similar in character but on a far larger scale than what happens in the formation of a black hole? Great question! No one who cares deeply about this question can fail to celebrate each week that week’s astrophysical advances: instruments, observations, conclusions. We have come to the end of our journey. We have seen gravity turned to float, space and time meld into spacetime, and spacetime transformed from stage to actor. We have examined how spacetime grips mass, telling it how to move, and how mass grips spacetime, telling it how to curve. O f all the indications that existence at bottom has a simplicity beyond anything we imagine today, there is none more inspiring than the unsutpassed simplicity of gravity as we now see it.
REFERENCES Extended portions of this chapter were copied (and sometimes modified) from John Archibald W h e e l e r , Into Gravity and Spacetime (Scientific Ameri can Library, a division of HPHLP, New York, 1990). For details of Galileo’s views on motion, see Galileo Galilei, Dialogues Concerning Two New Sciences, originally published March 1638; one modern translation is by Henry Crew and Alfonso de Salvio (Northwestern University Press, Evanston, 111., 1950). How Newton came only in stages to the solution of the problem of fall is told nowhere with such care for the fascinating documentation as in Alexander Koyre, “A Documentary History of the Problem of Fall from Kepler to Newton,” Transactions of the American Philosophical Society, Volume 45, Part 4 (1955). Keynes quotation under Newton portrait: Reprinted by permission of the pub lisher, Horizon Press, from Essays in Biography by John Maynard Keynes, copyright 1951.
THE COSMOS
A CLOSEDMODEL UNIVERSE COMPATIBLE WITH OBSERVATION Radius at phase of maximum expansion
18.9 X 10^ lightyears or 1.79 X 10^^ meters
Time from start to maximum size
29.8 X 10^ years or 2.82 X 10^^ meters
Radius today
13.2 X 10^ lightyears
Time from start to today’s size
10.0 X 109 years
Time it would have taken from start to today’s size if the entire expansion had occurred at today’s slowed rate of expansion
20.0 X 109 years
Ptesent expansion rate
An extra increment of recession velocity of 15.0 kilometers/second for every extra million lightyears of remoteness of the galactic cluster
Fraction of the way around the 3sphere universe from which we can in principle receive light today
113.2 degrees   = 6 2 .9 % 180 degrees
Fraction of the matter in the 3sphere universe that has been able to communicate with us so far
74.4%
Number of new galaxies that come into view on average every three days
One!
Average mass density today
14.8 X 10“ ^^ kilogram/meter^
Average mass density at phase of maximum expansion
5.0 X 10“ ^^ kilogram/meter^
Rate of increase of volume today
1.82 X 10®^ meters^/second
Amount of mass
Alconv “ 5.68 X 10’^ kilograms In geometric units: M= = 4.21 X 1026 njeters
Equivalent number of suns like ours
2.86 X 1023
Equivalent number of galaxies like ours
1.6 X 1012
Equivalent number of baryons (neutrons and protons)
3.39 X 10«9
Total time, big bang to big crunch
59.52 X 109 years
/ / /
2 9 7
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Figure 92: Figure and data from Journal of Spacecraft, Volume 11 (September 1974), pages 6 3 7 6 4 4 , published by the American Institure of Aeronautics and Astronautics. Data also from D. B. De Bra, APL Technical Digest, Volume 12: pages 1 4 2 6 . Figure 93 from Philosophiae Naturalis Principia Mathematica (Joseph Streater, London, July 5, 1686); Morre translation into English revised and edited by Florian Cajori and published in two paperback volumes (University of California Press, Berkeley, 1962). This is also the source of rhe quote in Section 9.6: “Absolute space, in its own namre, without relation to anything external, remains always similar and immobile.” Three quotations under the Einstein picture come from Albert Einstein, Essays in Science (Philosophical Libraty, New York, 1934). Quotation in Section 9.6: “Rather than have one global frame with gravitational forces we have many local frames without gravitational forces.” Steven Schutz, in January 1966 final examination in course in relativity, Princeton University. For an exciting and readable overview of the experimental proofs of Einstein’s general relarivity theory, see Clifford M. Will, Was Einstein Right? Putting General Relativity to the Test (Basic Books, New York, 1986). In particular (Chapter 10, pages 1 8 1 2 0 6 ), he describes at some length the emission of gravity waves by the binary pulsar system studied by Joseph H. Taylor, Jr., and Russell A. Hulse.
ANSW ERS TO O D D N U M B ER ED EXERCISES
chapter 1 1  1 a 10.2 meters b 270 meters c 10^ meters d 10"* kilometers ~ 2 times BostonSan Francisco distance l3 a 2.6 X 10*^ meters b 5.3 X 10“ ^ sec ond c 1.85 X 10“ *° hours d 52 weeks e 5.4 X 10° furlongs l5 a 4 years b 4 /5 the speed of light = 2.4 X 10® meters/second l7 a 4 meters b V7 meters = 2.65 meters c V l5 meters = 3.87 meters d 2 meters e 4 meters (same as part a) l9 a 2 X 10’ years h v — 0.995 c 6.33 X 10"* years, v = 0.9995 d r /= 1  5 X 10“ ** = 0.99999999995 11 la 2 X 10“ "* second b 133 halflives; (1/2)*’’ ~ lO” "*** c 3 halflives d zero space separation (creation and decay occur at the same place in rocket frame) e 3 halflives = 4.5 X 10“ ° second
chapter 2 2  la hit the ceiling b same answer c Rider cannot tell when elevator reaches top. 23 Set clock to 10 seconds, start when reference flash arrives. 2 5 a Experiment in progress for 1/0.96 = 1.04 meters of time. In this time, test particle falls 6 X 10“ *^ meters, about 10“ ^ diameter of a nucleus, b 3 X 10“ "* second, 10’ meters 27 3.6 millimeters; 19.7 seconds. Spacetime region: 20 meters X 20 meters X 20 meters of space X 59 X 10® meters of time 2 9a de crease (think of each ball bearing in an elliptical orbit around the center of Earth) b apart c No, you cannot distinguish rising from falling. At the top you notice nothing inside the coach. 211 = 0.735 thespeedoflight. 2  13a Effec tive time of fall: 4.67 seconds. Net velocity of fall: 1284 meters/second. b Angle of deflection: 4.3 X 10“ ° radian = 2.5 X 10“ "* degree = 0.88 second of arc
chapter 3 3  la 6 0 seconds b 4 5 seconds against the current, 2 2 .5 seconds with the current, 6 7 .5 seconds round trip c No 33 If different kinds of clocks ran at different rates in a freefloat rocket frame, then this difference could be used to detect the relative velocity of the laboratory from inside the rocket, which violates the Principle of Relativity. This does not mean that the common rate of rocket clocks will be the same as measured in rocket and laboratory frames. 3 5 a 11.5 lightyears b 9 .4 3 years c v = 0 .6 d 8 years = the interval between the two events. 37 The bullet misses. Coincidence of A and A ' (event 1) and firing of the bullet at the other end of spaceship 0 (event 2) cannot be simultaneous in both rocket reference frames. The right panel of the figure is wrong. Consistent with the Train Paradox (Section 3 .4 ), spaceship O' (standing in for the train frame) will observe the bullet to be fired before coincidence of A and A ', thus accounting for the fact that bullet misses. 39a sin yy = (in meters/meter) b siny/ ~ \j/ ~ 10“ "*radian = 21 seconds of arc c sin y / and tan y / are both approximately equal to y / for small y/. Therefore the difference between the two predictions cannot be used to distinguish between relativ istic and nonrelativistic ptedictions. d in a direction 0 .5 2 4 radians = 3 0 degrees ahead of transverse 3 llg (l) = 10“ ^, r'buUet “ 2 X 10“ °. Their product is 2 X 10“ *’, very small compared with 1; thetefore we expect t'buiiet to be the sum of t'buUct ^tid the form verified in everyday experience at low speeds. (2) Vb,ji„ = 2 4 / 2 5 = 0 .9 6 (3) I'buiiet = t'ught = + 1 (" 4 Wg = 9 units 71 la proton: 938 MeV; electron: 0.511 MeV b P i ^ ^ 0.12. Proton kinetic energy at limit ~ 6 MeV. Electron kinetic energy at limit ~ 3.4 X 10~^ MeV = 3.4 keV. Yes, designer of color TV tubes (electron kinetic energy — 25 keV) must use special relativity.
chapter 8 8 la approximately 35 X 10“ ® kilograms = 35 micrograms b approximately 600 kilograms. More, c approximately 6 X 10^^ seconds or about 2 million years! Chemical burning in Eric’s body produces large quantities of waste products. Elimi nation of these products carries away mass enormously faster than mass is carried away as energy. 83a Force is approximately 3 X 10“ ^ newtons, or the weight of 3 X 10“ kilograms. You should not be able to feel it. b pressure on a perfectly absotbing satellite == 5 X 10“ ^ newton/meter^; on a perfectly reflecting satellite == 9 X 10“ ^ newton/meter^; somewhere in between for a partially absorbing surface. Total enetgy absorbed/meter^, not color of the incident light, determines pres sure. c acceleration approximately 10“ ^ g d particle radius approximately 10~® meter, independent of the distance from Sun 87 density approximately 5 X 10^® kilograms/meter^ = 5 X 1 0 ’ grams/centimeter^, or 50 million times the den
301
302
ANSWERS TO ODDNUMBERED EXERCISES sity of water! 89 m^)/{2m) 81 la From conservation equa tions, show that cos (/) > 1, which is impossible, b If the total momentum is zero after the collision, it must be zero before the collision. But the alleged single photon before the collision cannot have zero momentum. Therefore the reaction is impossible. 813 2Ec = E^\ {Ey^ — and 2Ep = E ^ ~ ( E ^ — If the particle is at rest, then Ej^ = m and E ^ = Ep = m /2. 815a = w(£ b m)l\E + m  (E^ cos 0c] 8 17a 1.8 TeV b £ « 1700 TeV 819e No 821 When the bulb is seen way ahead, its light is very intense and radically blueshifted. While still seen ahead, there is an angle of observation (de pending on the speed) at which the light is red, but dim. As the bulb is seen to pass the observer, its light is infrared and very dim. As the bulb is seen to retreat into the distance, its light is extremely dim and radically redshifted. 823a v = 0.38 b 13 X 10^ years c Allowance for gravitational slowing will decrease the estimated time back to the start of the expansion. 825 l ^ f / f ~ The observed frequency will increase for molecules approaching the observer and decrease for molecules receding from the observer. The overall effect— at temperatures for which Newtonian expressions are valid — is to produce a spread of frequencies approximated by the expression above (“ Doppler line broadening”). 821 E ' = m /2, E = m, (f) = 30 degrees. 835a The incident gamma ray (with excitation energy £ ) imparts a small kinetic energy K to the iron atom, for which Newtonian expression is valid: K = p'^/2m = E'^/2m, since p = E {os the gamma ray. Then (energy of recoil)/(energy for excitation) = K /£ ~ £/(2w ) = 1.4 X 10“ ^. But fractional resonance width (6 X 10“ *^) is smaller than this by a factor of almost a million, so the iron nucleus cannot accept the gamma ray and conserve energy, b One gram is about 10^^ atoms. If the m in the above equation increases by the factor 10^^, then the energy of recoil is a factor 10^^ smaller, and the nucleus will not notice the residual mismatch in energy. 837 A / / / = — gz/eY, v = 0.7 X 10“ ® meter/second towards emitter 839 A //( /,A T ) = (3/2)/^/(wc^) ~ 1.2 X 10“ '^ per degree.
INDEX
Abbott, James, 135 (Exercise 41) aberration of starlight, 81 (Exercise 39) absolute elsewhere, 181 “absolute” space and time (Newton), 160, 284 abuse of the concept of mass, 244251 (Section 8.8) acceleration, relative, as witness to gravity, 3036 (Sec tions 2.3, 2.4), 280287 (Sections 9.4, 9.5, 9.6) accelerationproof clocks, 152 active future, 182 addition of velocities, 8 2 84 (Exercise 311), 103110 (Secrion L.7) Aging, Principle of Maximal, 150 Akihito, Emperor of Japan, 138 American Civil War, 25 Andromeda galaxy Enterprise in, 106107 (Box L2) trip to by rocket, 2 223 (Exercise 19) trip to by Transporter, 23 (Exercise 110) angles, transformation of, 114115 (Exercise L6) annihilation, positronelectron, 237238, 242243 (Sample Problem 84), 260 (Exercises 814, 815) appearance, visual, of relarivistic objects, 64, 9 293 (Ex ercise 317) Arecibo radio antenna (Puerto Rico), 291 arrow of momenergy, 191195 (Section 7.2) autobiography of a phoron, 184185 (Exercise 64) available interaction energy, 261 (Exercise 817) backyard zoo of particles, 235 (Box 81) bad clock, 112113 (Exercise L2) barn and pole paradox, 166 (Exercise 54) Barrlett, Sreven, 19 Bay of Fundy, tides in, 3233 (Box 21) Berman, Eric, 254 (Exercise 81) beta (Greek p), symbol for speed, 41, 253 Betrayal, Great, 108109 (Box L1) black hole, 289 (Box 92), 292295 (Section 9.8) as source of neutrinos, 80 (Exercise 38) bomb fission, 249 hydrogen (fusion), 248249 Super, 108109 (Box L1) bounce, freefloat, 45 (Exercise 22) Braginsky, Vladimir, 36, 223 broadening of spectral lines, Doppler, 264 (Exercise 825) bulb flickering, paradox of, 186187 (Exercise 67) speeding, 264 (Exercise 821) c (speed of lighr), see light speed Caesar, Julius, 106107 (Sample Problem L2)
cannonball, human, 45 (Exercise 21) Canopus, trip to, 121134 (Chapter 4) cat, Cheshire, 292 causaliry, light speed limit on, 171 (Section 6.1), 180183 center of momentum frame, 246251 Cerenkov radiation, 8081 (Exercise 38) Chandrasekhar, S., 288 Chandrasekhar limir, 288289 (Box 92) chemistry, relativistic, 254 (Exercise 82) Civil War, American, 25 Cleoparra, 228 clock accelerationproof, 152 atomic, test of twin effect, 131 bad, 112113 (Exercise L2) construction of, 78 (Exercise 33) lightflash, 12 reference, 37 clock paradox, see Twin Paradox clocks latricework of, 3739 (Secrion 2.6), 4 5 4 6 (Exer cises 23, 24) plane of agreement of, 120 (Exercise L15) run at different rates in gravitational field, 118 (Ex ercise L13) “run slow?”, 7677 (Box 34) collapse, gravitational, 288, 292295 (Section 9.8) colliders, 261262 (Exercise 817) collision, 221252 (Chapter 8) analyzing, 239 (Box 82) elastic, 222, 240241 (Sample Problem 83) inelastic, 222223 solving problems, 239 (Box 82) comet, 35 communication, time delay in, 3940 communications storm, 48 (Exercise 211) compact stellar objects, 288289 (Box 92) components of momenergy, 195 199 (Section 7.3) energy, 201206 (Section 7.5) momentum, 199200 (Section 7.4) Compton, Arthur Holly, 229 Compton scattering, 229, 231, 267268 (Exercise 829) examples of, 268 (Exercise 830) inverse, 269270 (Exercise 832) computer size, 22 (Exercise 18) cone, light, partition in spacetime, 177 183 (Section 6.3) conscienceguided satellite, 277279 conservation laws, see energy; momentum; momenergy conserved, defined, 208209 (Box 73) constant, defined, 208209 (Box 73)
303
304
INDEX
contracting train paradox, 187188 (Exercise 68) contraction, Lorentz, 6365 (Seaion 3.5), 126127 (Section 4.7) for cosmic rays, 215216 (Exercise 77) described by stretch factor, 157 how it occurs, 119120 (Exercise L14) or rotation?, 9 293 (Exercise 317) conversion factors for energy, 203, 250 miles to meters, 2, 16, 5859 (Box 32) for momentum, 200 seconds to meters, 6, 12, 16, 5859 (Box 32) conversion of mass to energy, 237244 (Section 8.7), 254 (Exercise 8.1) cosmic rays, 160, 215216 (Exercise 77) cosmos, 296297 (Section 9.9) creation of protonantiproton pair by an electron, 261 (Exercise 816) curvature of Earth, 281283 (Section 9.5) equation, Einstein’s, 286 of spacetime, 280287 (Sections 9.4, 9.5, 9.6) Daytime surveyor, 1 4 (Section 1.1), 1617 (Box 11) decay mu meson, 2 324 (Exercise 111) pinaught meson, 267 (Exercise 828) piplus meson, 24 (Exercise 112) positronium, 260 (Exercise 813) defleaion of starlight by Sun, 5051 (Exercise 213) density of companion of Sirius, 258259 (Exercise 87) detonator paradox, 185186 (Exercise 65) deuterium, combined with helium, 237 Dicke experiment, 36, 4 8 5 0 (Exercise 212) dimension, transverse, invariance of, 6567 (Section 3.6) distance invariance of, 4, 17 proper, 174 dog and passenger paradox, 2526 Dog Star (Sirius), 135 (Exercise 41), 258259 (Exercise 87) Doppler shift along xdirection, 114 (Exercise L5), 263 (Exercise 818) at limb of Sun, 264 (Exercise 822) E = mi? from, 264265 (Exercise 826) equations, 263 (Exercise 819) line broadening, 264 (Exercise 825) measurement of by resonant scattering, 271272 (Exercise 836) Twin Paradox using, 264 (Exercise 824) down with relativity!, 79 (Exercise 36) DUMAND experiment, 80 (Exercise 38) dwarf, white, 258259 (Exercise 87), 288 (Box 92)
E = m?, 203, 206, 250 from Doppler shift, 264265 (Exercise 826) Earth curved, 281283 (Section 9.5) mass in units of meters, 258 surface of as a freefloat frame, 46 (Exercise 25) Eigenzeit, 11; ree also proper time Einstein, Albert admiration for Newton, 284, 295 curvature equation, 286 eliminate gravity, 28 epigram, iii equivalence of energy and mass, 250, 254258 (Exercise 85) and Galileo and Newton, 275276 (Section 9.2) and gravity, 275298 (Chapter 9) happiest thought of life, 25, 44 picture and quotes, 295 special relativity, 5 Train Paradox, 6263 Einstein puzzler, 78 (Exercise 32) elastic collision, 222, 240241 (Sample Problem 83) electrodynamics, quantum, 185 (Exercise 64) electron, 235 (Box 81) creation of proton  antiproton pair by, 261 (Exercise 816) electronpositron annihilation, 237238, 242243 (Sample Problem 84), 260 (Exercises 814, 815) electronpositron pair production, see photon electrons, fast, 215 (Exercise 76) elsewhere, absolute, 181 Emperor Akihito, 138 Emperor Hirohito, 137 emptiness of spacetime, 5657 (Box 31) encounter, particle, 239 (Box 82) energy, 196, 213 (Table 7.1) conserved in a collision, 1 8 9 1 9 0 (Section 7.1), 207, 2 2 2 2 2 3 (Section 8.2), 239 (Box 8.2)
conversion of mass to, 2 3 7 2 4 4 , (Section 8.7), 254 (Exercise 81) interaction, 261 (Exercise 817) kinetic, 201, 203, 206 Newtonian, lowvelocity limit, 190, 203, 205 (Box 7 2) and mass, 201, 203, 206, 250251, 254258 (Exercise 85) production of, in Sun, 242245 (Sample Problem 8 5) quantities related to, 213 (Table 71) rest, 201, 203, 250 shift of, due to recoil of emitter, 270 (Exercise 833) threshold, 236, 259 (Exercise 812), 261 (Exercise 816) as “time” part of momenergy, 201206 (Section 7.5)
INDEX transformation of, 215 (Exercise 75) in unit of mass, 190, 203 without mass (photon), 228233 (Section 8.4), 273274 (Exercise 840) energy of light, 230 energy of photon and frequency of light, 268269 (Exer cise 831) Engelsberg, Stanley, 4 546 (Exercise 24) Enterprise, Starship, 106107 Eotvos, Baron Roland von, 36 equivalence of energy and mass, 250, 254258 (Exercise 85) ether theory of light propagation, 84, 88 Euclidean 3vector, 192 (Box 71) Euclidean geometry, 8, 11, 126, 143, 151, 172, 177, 192 (Box 71), 198, 279 event, 10, 16 and interval, 9 1 1 (Section 1.3) locating, with latticework of clocks, 3739 (Section 2 . 6)
not owned by any frame, 43 reference, 38 events relation between, 11, 172 —177 (Section 6.2)
time of, 38, 137139 (Section 51) evidence, experimental, for Twin Paradox 131134 (Sec tion 4.10) expanding universe, 82 (Exercise 310), 264 (Exercise 823), 296297 (Section 9.9) experimental evidence for Twin Paradox, 131134 (Sec tion 4.10), 272273 (Exercise 839) fast electrons, 215 (Exercise 76) fast protons, 214215 (Exercise 74) faster than light?, see light, faster than? Federation, 108109 (Box L1) Feynman, Richard, 1 firing meson, 110 (Sample Problem L3) fission, 237238 bomb, 249 Fizeau experiment, 120 (Exercise L16) flash, reference, 38 flickering bulb paradox, 186 187 (Exercise 67) floating to Moon, 2 526 (Section 2.1) force of gravity, eliminate, 2629 (Section 2.2) four times light speed, 89 (Exercise 315) fourvector, momenergy as, 191 195 (Section 7.2) frame center of momentum, 246251 Earth, 46 (Exercise 25) freefloat, see freefloar frame inertial, see freefloat frame laboratory, 5, 41 local, see freefloat frame
305
Lorentz, see freefloat frame reference, 5; see also freefloat frame rocket, 4143 (Section 2.9) superrocket, 69, 71, 140142 free float, 2545 (Chapter 2) freefloat bounce, 45 (Exercise 22) freefloat (inertial) frame, 2629 (Section 2.2) defined, 31 Earth surface as, 46 (Exercise 25) extent of, near Earth, 3034 (Section 2.3), 46 (Ex ercise 26), 47 (Exercise 28), 285 extent of, near Moon, 4 647 (Exercise 27) local, 3034 (Section 2.3), 284 rocket, 4143 (Section 2.9) stripped down, 121122 (Section 4.2) superrocket, 69, 71, 140142 and test of twin effect, 133 touring spacetime without, 160162 (Section 5.9) verifying, 41, 279 what is same in different, 6062 (Section 3.3) what is not same in different, 5660 (Section 3.2) frequency of light and energy of a photon, 268269 (Exercise 831) Fundy, Bay of, 32 —33 (Box 21)
fusion, 237238 fusion bomb, 248249 future, active, 182 Galilean principle of relativity, 5355 Galilean transformation, 113 (Exercise L3) Galilei, Galileo and gravitational acceleration, 36 and Newton and Einstein, 275276 (Seaion 9.2) picture and quotes, 54 and Leaning Tower of Pisa, 36 and Principle of Relativity, 5355 and tides, 32 gamma (Greek y), stretch factor, 99, 155160 (Section 5.8) gamma rays, 237; see also photon General Conference on Weights and Measures, 12, 58 general relativity, 275298 (Chapter 9) needed for Twin Paradox?, 132 (Box 41) when required, 34, 35, 133, 276, 281 geometry Euclidean, 8, 11, 126, 143, 151, 172, 177, 192 (Box 71), 198, 279 curved space, 280281 (Section 9.4), curved spacetime 284287 (Section 9.6) Lorentz, 8, 11, 126, 143, 151, 172, 177, 192 (Box 71), 198, 284 gigaflop, 22 (Exercise 18) gravitation effect of on clocks, 118 (Exercise L13)
306
INDEX
as curvature of spacetime, 284287 (Section 9.6) tutorial in Newtonian, 258 (Exercise 86) gravitational attraction of system containing photons, 257 gravitational collapse, 288, 292295 (Section 9.8) gravitational radiation, 288292 (Section 9.7) gravitational red shift, 258 (Exercise 86) test of, 272 (Exercises 837, 838) graviton, 153, 176 gravity as curved spacetime, 284287 (Section 9.6) in brief, 275 (Section 9.1) eliminate, 2829 (Section 2.2) radiation, 288292 (Section 9.7) relative acceleration as witness to, 3036 (Sections 2.3, 2.4), 280287 (Sections 9.4, 9.5, 9.6) waves, 288292 (Section 9.7) Great Betrayal, 108109 (Box L1) Great Pyramid, 209 grid, paradox of skateboard and, 116117 (Exercise L12) h, Planck’s constant, 265, 268269 (Exercise 831) handle showing invariant magnitude of momenergy vec tor, 198 headlight effect, 115 (Exercise L9) heat as system property, 224 weighing, 223 helium in Sun, 242245 (Sample Problem 85) Himalaya Mountains, 4849 Hirohito, Emperor of Japan, 137 hole, black, 289 (Box 92), 292295 (Section 9.8) as source of neutrinos, 80 (Exercise 38) Horwitz, Paul, 186 (Exercise 66) Hubble, Edwin, 264 Hubble constant, 264 Hubble time, 264 Hull, Penny, 19, 264, 272 Hulse, Russell A., 291 human cannonball, 45 (Exercise 21)
hydrogen bomb, 248249 hydrogen burning in Sun, 242  245 (Sample Problem 85) hydrogen molecule ion, 233 hyperbola invariant, 143 (Section 5.3), 173174 momenergy, 198 identically accelerated twins paradox, 117118 (Exercise L13) index of refraction and speed of light, 185 (Exercise 64) inelastic collision, 222223 inertia, 31, 189 inettial frame, see freefloat frame integrity of photon, 259 (Exercise 811) interaction energy, available, 261 (Exercise 817) interferometer Fizeau, 120 (Exercise L16)
KennedyThorndike, 8688 (Exercise 313) MichelsonMorley, 8 486 (Exercise 312) verifying freefloat frame using, 46 (Exercise 25) interstellar travel, 274 (Exercise 841) interval, 6 and event, 911 (Section 1.3) invariance of, see invariance of interval as lightlike relation between events, 175177 as spacelike relation between events, 11, 173174 as timelike relation between events, 11, 172173 invariance of distance, 4,17 invariance of interval, 6 7 , 17, 18 for all freefloat frames, 7173 (Section 3.8) preserves cause and effect, 180183 proved, 6 770 (Section 3.7) and spacetime hyperbola, 143 (Section 5.3), 173, 174 and spacetime map, 143 (Section 5.3) used in derivation of the Lorentz transformation, 102 invariance of mass, 197, 246 invariance of momenergy, 194, 198, 210 invariance of speed of light, 60; 8688 (Exercise 313) invariance of transverse dimension, 6 567 (Section 3.6) invariant, defined, 208209 (Box 73) invariant hyperbola, 143 (Section 5.3), 173, 174 inverse Compton scattering, 269270 (Exercise 832) inverse Lorentz transformation, 102103 (Section L.6) Japan, 27, 9697, l6 l Japan Microgravity Center (JAMIC), 27 (Figure 23) Julius Caesar, 106 107 (Sample Problem L2) K"''meson, 72 (Sample Problem 32) Kamisunagawa, 27 KennedyThorndike experiment, 8688 (Exercise 313) Kepler, Johannes, 32 kinetic energy, 201, 203, 206 kinked worldline, 152 155 (Section 5.7) Klingons, 108109 (Box L1) Krotkov, Robert V., 36
laboratory frame, 5, 41 lattice clocks, synchronizing, 3738, 4546, (Exercises 23, 24) latticework of clocks, 3739 (Section 2.6) Law of Addition of Velocities, 8 284 (Exercise 311), 103110 (Section L.7) laws, conservation, see energy, momentum; momenergy Laws, Kenneth L., 77 Leaning Tower of Pisa, 36 length mass in units of, 258 (Exercise 86) time in units of, 1113 (Section 1.4) length along a path, 147 148 (Section 5.5) length contraction, see Lorentz contraction less is more, 154155 (Sample Problem 51), 163164 (Exercise 51)
INDEX light deflection of by Sun, 5051 (Exercise 213) frequency of and energy of a photon, 268269 (Exercise 831) gravitarional red shift of, 258259 (Exercises 86, 87) pressure of, 254 (Exercise 83), 255 rocket propelled by, 274 (Exercise 841) speed of, see lighr speed See also photon light, faster rhan?, 7475 (Box 33), 9 699 (Section L.2), 108109 (Box L1), 122123 (Section 4.3) four times the speed of light?, 8 9 9 0 (Exercise 315) superluminal expansion of quasar 3C273?, 9092 (Exercise 316) things that move faster than light, 8 889 (Exercise 314) light bulb flickering, 186187 (Exercise 67) speeding, 264 (Exercise 821) light cone as partition in spacetime, 177 183 (Section 6.3) lightflash clock, 12 lightlike relation between events, 172177 (Section 6.2) light propagation, ether theory of, 84, 88 light speed as conversion factor, 6, 12, 16, 5859 (Box 32), 200, 203, 250 index of refraction and, 185 (Exercise 64) invarianr magnitude of, 60 (KennedyThorndike experiment), 8 688 (Exercise 313) isotropic (MichelsonMorley experiment), 8486 (Exercise 312) as limit on causality, 171 (Section 6.1), 180183 as limit on observation, 3940 See also light, faster than? lightsecond, 1113 (Section 1.4) lightyear, 12 limb of Sun, Doppler shift at, 264 (Exercise 822) limits of Newtonian mechanics, 34, 113114 (Exercise L4), 217 (Exercise 711) line, world, see worldline line broadening, Doppler, 264 (Exercise 825) linear acceleraror, Stanford, 215 (Exercise 76) local inertial frame, see freefloat frame local moving orders for mass, 277280 (Section 9.3) local time, see proper rime; interval locating events with latticework, 3739 (Section 2.6) Lorentz, Hendrik, 5 Lorentz contraction, 6365 (Section 3.5), 126 127 (Section 4.7) for cosmic rays, 216 (Exercise 77) described by strerch factor, 157 how ir occurs, 119120 (Exercise L14) or rotarion, 9 293 (Exercise 317) Lorentz frame, see freefloat frame
307
LorentzFitzGerald contraction hypothesis, 88 Lorentz geometry, 8, 11, 126, 143, 151, 172, 177, 192 (Box 71), 198, 284 Lorentz interval, 6; see also interval; invariance of interval Lorentz transformation, 95111 (Special Topic) equations, 102 form of, 100 (Section L.4) inverse equations, 102103 (Section L.6) for momenergy componenrs, 215 (Exercise 75) usefulness of, 95 (Secrion L.l) • manhole, paradox of rising, 116 (Exercise L11) map, spacetime, see spacetime map mapmaking in space, 10, 2122 (Exercise 16) in spacetime, 164166 (Exercise 53) mass abuse of the concept of, 244251 (Section 8.8) change in nuclear, 237238 conversion of to enetgy, 237244 (Section 8.7), 254 (Exercise 81) created by material parricle, 234236 (Section 8.6) created by photon, 233234 (Section 8.5) and energy, 201, 203, 206, 250251, 254258 (Exercise 85) energy in unit of, 190, 203 energy without (photon), 228233 (Section 8.4) invariance of, 197, 246 local moving orders for, 277280 (Section 93) loss by Sun of, 242245 (Sample Problem 85) as magnitude of momenergy 4vector, 195, 197 momentum in unit of, 190, 200 momentum without?, 273274 (Exercise 840) photon used to create, 233234 (Section 8.5) proof, 277, 279 “relativistic,” 250251 “rest,” 251 as unit of length, 258 (Exercise 86) use and abuse of the concept of, 244251 (Section 8 .8)
mass of photon, 230 mass of system of particles, 214 (Exercise 72), 224228 (Section 8.3), 247 Maximal Aging, Principle of, 150 maximum speed of walking, 186 (Exercise 66) mechanics Newtonian, 113114 (Exercise L4), 192 (Box 71), 217 (Exercise 711) relarivistic, 192 (Box 71) megaflop, 22 (Exercise 18) meson decay of pinaught, 267 (Exercise 828) firing, 110 (Sample Problem L3) time stretching with, 2324 (Exercise 111), 24 (Exercise 112), 7273 (Sample Problem 32)
308
INDEX
meter defined, 5859 (Box 32) of rime, 1113 (Section 1.4) as unit of mass, 258 (Exercise 86) meter stick, tilted, 115116 (Exercise L10) Michelson  Morley experiment, 8 4 8 6 (Exercise 312) microgravity, 27 (Figure 23), 277 (Figure 92) Minkowski, Hermann, 15 mile defined, 5859 (Box 32) as sacred unit, 1 4 minus sign, 6 8 , 26, 190, 197 minute, unit of distance and time, 1113 (Section 1.4) momenergy as 4vector, 191, 192 (Box 71) analogy of to tree, 210 arrow, 191195 (Section 7.2) components of, 195199 (Section 7.3), 204 (Sample Problem 73) conservation of, 189190 (Seaion 7.1), 207210 (Section 7.6), 247 defined, 191195 (Section 7.2) energy as “time” part of, 201206 (Section 7.5) handle showing invariant magnitude, 198 invariance of, 194, 198, 210 magnitude of is mass, 195, 197 momentum as “space” part of, 199200 (Section 7.4) quantities related to, 213 (Table 71) tree, analogy of, 210 transformation of components of, 215 (Exercise 75) units of, 194, 195, 200, 203 momentum, 196, 213 (Table 7.1) components of, 196 conserved in a collision, 189 190 (Section 7.1), 207, 222223 (Section 8.2), 239 (Box 8.2) derived from conservation law, 217219 (Exercise 712)
of light, 230 Newtonian expression for, 190, 200 as “space” part of momenergy, 199200 (Section 7.4) transformation of, 215 (Exercise 75) in unit of mass, 190, 200 without mass?, 273274 (Exercise 840) momentumenergy 4vector, see momenergy Moon, 2 5 2 6 (Section 2.1), 3233 (Box 21) Moral Principle, Wheeler’s First, 20 Mossbauer effect, 270 Minkowski, Hermann, 15 more is less, 154155 (Sample Problem 51), 163164 (Exercise 51) moving orders for mass, local, 277280 (Seaion 9.3) muons, time stretching with, 23 (Exercise 111)
nanosecond, 5 Neptune, images from, 20 (Exercise 12) neutral or unreachable region, 182 neutrino described, 235 (Box 81) detection of, 80 (Exercise 38) neutron, described, 235 (Box 81) neutron star, 288289 (Box 92) and gravity waves, 290291 Newton, Isaac, 275280 absolute space and time, 160, 284 Einstein’s admiration for, 284, 295 First Law of Motion, 31 and Galileo and Einstein, 275276 (Section 9.2) picture and quotes, 278 Newtonian mechanics, 192 (Box 71) First Law of Motion, 31 gravitational theory, tutorial, 258 (Exercise 86) limits of, 34, 113114 (Exercise L4), 217 (Exer cise 711) Nighttime surveyor, 1 4 (Section 1.1), 1617 (Box 11) nuclear excitation, 259 (Exercise 88) observer, 3940 (Section 2.7) oozing!, 12 oscillator, relativistic, 135136 (Exercise 43) oscilloscope writing speed, 89 (Exercise 314) pair production by photon(s), 233234 (Section 8.5), 259 (Exercises 811, 812) Parable of the Surveyors, 1 4 (Section 1.1), 1617 (Box 11) Parable of the Two Travelers, 281283 (Seaion 9.5) paradoxes contracting train, 187188 (Exercise 68) detonator, 185186 (Exercise 65) Einstein’s train, 6263 flickering bulb, 186187 (Exercise 67) four times light speed, 89 (Exercise 315) identically accelerated twins, 117118 (Exercise L13) passenger and dog, 2526 pole and barn, 166 (Exercise 54) rising manhole, 116 (Exercise L11) mnner on the train, 168 (Exercise 57) scissors, 88 (Exercise 314) skateboard and grid, 116117 (Exercise L12) space war, 7980 (Exercise 37) tilted meter stick, 115116 (Exercise L10) See also Twin Paradox particle, test, 36 (Section 2.5), 4 748 (Exercise 210) particles backyard zoo of, 235 (Box 81) creation of, 234236 (Section 8.6), 261262 (Ex ercises 816, 817)
INDEX creation of by photons, 233234 (Section 8.5), 259260 (Exercises 811 and 812) encounter, 239 (Box 82) measuring speed of, 4041 (Section 2.8) system of, 214 (Exercise 72), 221 (Section 8.1), 224228 (Section 8.3), 244251 (Section 88) timelike wotldline of, 172 used to create mass, 234236 (Section 8.6) virmal, 5657 wotldline of, 143 147 (Section 5.4) partition in spacetime, light cone as, 177 183 (Section 6.3) passenger and dog paradox, 2526 passive past, 182 path, length along, 147—148 (Section 5.5) Peace Treaty of Shalimar, 108 109 (Box L1) Philoponus, John, of Alexandria, 36 photon, 228233 (Section 8.4), 246 from annihilation, 237238 (Section 8.7) autobiography of, 184185 (Exercise 64) braking, 259 (Exercise 89) Compton scattering of, 229, 231, 267270 (Exer cises 829, 830, 832) creation of particleantiparticle pair using, 233234 (Section 8.5) energy of, 228233 (Section 8.4), 268269 (Exer cise 831) energy measurement of, 254 (Exercise 84) energy shift of due to recoil of emitter, 270 (Exercise 833) gravitational red shift of, 258259 (Exercises 86 and 87) integrity of, 259 (Exercise 811) mass of, 228231 (Section 8.4) momentum of, 230 pair production by, 233234 (Section 8.5), 259260 (Exercises 811, 812) resonant scattering of, 271272 (Exercises 835, 836) rocket and interstellar travel, 274 (Exercise 841) used to create mass, 233234 (Section 8.5) physicist and the traffic light, 263264 (Exercise 820) pinaught meson, decay of, 267 (Exercise 828) pipes, speeding (thought experiment), 66 piplus mesons, time stretching with, 24 (Exercise 112) Pisa, Leaning Tower of, 36 place, fundamental to surveying, 9, 16 plane of agreement of clocks, 120 (Exercise L15) Planck, Max, 229 Planck’s constant, 265, 268269 (Exercise 831) plumb bob, deflection of by Himalaya Mountains, 4 849 Poincare, Henri, 5 6 pole and barn paradox, 166 (Exercise 54) polyelectron, 233 positron, 233235
309
positronelectron annihilation, 237238, 242243 (Sample Problem 84), 260 (Exercises 814, 815) positronelectron pair production, 233234 (Section 8.5), 259 (Exetcises 811, 812) positronium, decay of, 260 (Exercise 813) practical synchronization of clocks, 4 5 4 6 (Exercises 23, 24) pressure of light, 254 (Exercise 83), 255 principle of invariance of distance, 4, 17 Principle of Maximal Aging, 150 Principle of Relativity, 5360 (Sections 3.1, 3.2, 3.3) examples of, 6162 (Sample Problem 31), 78 (Exercise 34) Galilean, 5 3 5 5
used in proof of invariance of interval, 73 proof mass (conscience), 277, 279 proper clock, 10 proper distance, 174, 184 (Exercise 63) proper time, 10, 184 (Exercise 63) along a worldline, 148 152 (Section 5.6) tau as symbol of, 155 proton, described, 235 (Box 81) proton  antiproton pair, creation of, 236 protons, fast, 214215 (Exercise 74) pulsar, 289 puppy, 224 puzzler, Einstein, 78 (Exercise 32) Pyramid, Great, 209 Pythagorean theorem, 2, 7 quantum electrodynamics, 185 (Exercise 64) quasar, 9092 (Exercise 316), 114 (Exercise L5), 294295 radar speed trap, 166 168 (Exercise 55) radiation, Cerenkov, 8081 (Exercise 38) radiation, gravitational, 288292 (Section 9.7) radius of a black hole, 292 railway coach rising, 47 (Exercise 29) and tidal accelerations, 3034 (Section 2.3), 281 ray, gamma, see photon ray, X, see photon rays, cosmic, 160, 215216 (Exercise 77) recoilless processes, 270271 (Exercise 834) recoil of emitter, energy shift due to, 270 (Exercise 833) red shift, gravitational, 258 (Exercise 86), 272 (Exercises 837, 838) reference clock, 37 reference event, 38 reference flash, 38 reference frame, 5; see also freefloat frame refraction, index of, and speed of light, 185 (Exercise 64) regions of spacetime, 3436 (Section 2.4), 171183 (Chapter 6) relations between events, 172177 (Section 6.2)
310
INDEX
relative acceleration as witness to gravity, 3036 (Sec tions 2.3, 2.4), 280287 (Sections 9.4, 9.5, 9.6) relative synchronization of clocks, 130 relativistic chemistry, 254 (Exercise 82) “relativistic” mass, 250251 relativistic mechanics, 192 (Box 71) relativistic momentum, 217219 (Exercise 712) relativistic oscillator, 135136 (Exercise 43) relativity general, 34, 35, 132 (Box 41), 133, 276, 281 principle of, 5362 (Sections 3.1, 3.2, 3.3), 78 special, 5, 18, 73, 78 (Exercise 31), 79 (Exercise 36), 131134 (Section 4.10), 270273 (Exer cises 833 to 839) relativity of simultaneity, 6263 (Section 3.4), 128131 (Section 4.9) and contraction of length, 64 See also paradoxes resonant scattering, 271 (Exercise 835) measurement of Doppler shift by, 271272 (Exer cise 836) rest energy, 201, 203, 250 “rest mass,” 251 Riemann, G. F. B., 295 “rigid body” not an invariant concept, 116117 (Exer cise L12), 119120 (Exercise L14) rising manhole paradox, 116 (Exercise L11) rising railway coach, 47 (Exercise 29) rocket frame, 4 143 (Section 2.9) rocket, photon, and interstellar travel, 274 (Exercise 841) rods, latticework of, 3739 (Section 2.6) Roll, Peter G., 36 rotation or contraction?, 9 293 (Exercise 317) Rumford, Count (Benjamin Thompson), 223 Ruml, Frances Towne, 29 runner on the train paradox, 168 (Exercise 57) sacred unit
mile, 1 4 second, 5 7 Satellite (dog), 26 satellite conscienceguided, 277279 pressure of light on, 254 (Exercise 83) scattering Compton, 229, 231, 267270 (Exercises 829, 830, 832) resonant, 271272 (Exercises 835, 836) scissors paradox, 88 (Exercise 314) Schmidt, Maarten, 294 second defined, 5859 (Box 32) as sacred unit, 5 7 as unit of distance and time, 1113 (Section 1.4) Shalimar, Peace Treaty of 108109 (Box L1)
Sheldon, Eric, 19 shift, see Doppler shift; red shift Shurdiff, William A., 19, 77, 198, 213 simultaneity, relativity of, 6263 (Section 34), 64, 128131 (Section 4.9) and transverse plane, 6667 See also paradoxes Sirius, density of companion of, 258259 (Exercise 87) skateboard and grid paradox, 116117 (Exercise L12) Smith, Richard C., 19 Sommerfeld, Arthur, 53 solar constant, 242, 254 (Exercise 83) solar wind, 245 space “absolute” (Newton), 284 as different from time, 18 is ours!, 123124 (Section 4.4) spacelike relation between events, 11, 172177 (Section 6 .2)
space travel, practical, 135 (Exercise 41) space war, 7981 (Exercise 37) spacetime as absolute elsewhere, 181 active future of, 182 emptiness of, 5657 (Box 31) exploded view of regions of, 182 (Figure 65) “Et m . . . ?”, 106107 (Sample Problem L2) light cone as partition of, 177 183 (Seaion 6.3) Lorentz geometry of, 8, 192 (Box 71) mapmaking in, 164166 (Exercise 53) neutral region of, 182 overview of, 119 (Chapter 1) passive past of, 182 regions of, 3436 (Section 2.4), 171 183 (Chap ter 6) surveying, 5 8 (Section 1.2) touring without tefetence ftame, 1 6 0 1 6 2 (Section
5.9) trekking through, 137163 (Chapter 5) units of, 2021 (Exercises 12 and 13) unity of, 7, 1518 (Section 1.5) unreachable region of, 182 spacetime curvamre, 280287 (Sections 9.4, 9.5, 9.6) contractile, 286287 (Box 91) equation (Einstein), 286 gravitation as, 284287 (Section 9.6) noncontractile, 286287 (Box 91) spacetime diagram, see spacetime map spacetime displacement as 4vector, 191194 spacetime geometry, see spacetime; spacetime curvature spacetime interval, see interval; invariance of interval spacetime map, 22 (Exercise 17), 137139 (Section 5.1) constructing, 164166 (Exercise 53) special relativity, 5, 18
INDEX down with, 79 (Exercise 36) four ideas of, 7 3 and swimming, 78 (Exercise 31) tests of, 131134 (Section 4.10), 270273 (Exer cises 833 through 839) spectral lines, Doppler broadening of, 264 (Exercise 825) speed, measuring, 4041 (Section 2.8) speeding light bulb, 264 (Exercise 821) speeding pipes thought experiment, 66 speeding train thought experiment, 6566 speed of light, see light speed speed of walking, maximum, 186 (Exercise 66) speed trap, radar, 166168 (Exercise 55) speeds, comparing, 20 (Exercise 11) Stanford linear accelerator, 215 (Exercise 76) starlight aberration of, 81 (Exercise 39) deflection of by Sun, 5051 (Exercise 213) Starship Enterprise, 106107 stellar aberration, 81 (Exercise 39) stellar objects, compact, 288289 (Box 92) storm, communicarions, 48 (Exercise 211) stretch factor, 99, 155160 (Section 5.8) and Lorentz contraction, 157
as measure of speed, 157 stripped down freefloat frame, 121122 (Section 4.2) Sun conversion of mass to energy in, 242245 (Sample Problem 85) deflection of starlight by, 5051 (Exercise 213) Doppler shift at limb of, 264 (Exercise 822) explosion of, 171 gravitational red shift of light from, 258 (Exercise 86) helium in, 242245 (Sample Problem 85) mass of in units of meters, 258 tidedriving power of, 3233 (Box 21) sunspot, 179180 (Sample Problem 63) Super (superluminal bomb), 108109 (Box L1) super cosmic rays, 215216 (Exercise 77) superluminal expansion of quasar 3C273?, 9092 (Exer cise 316) supernova, 177, 289 superrocket frame, 69, 71, 140142 superspeed Super, 112 (Exercise L1) surveying spacetime, 5 8 (Section 1.2) Surveyors, Parable of, 1 4 (Section 1.1), 1617 (Box 11) swimming and relativity, 78 (Exercise 31) symmetric elastic collision, 240  241 (Sample Problem 83) synchronization of clocks, relative, 130 synchronizing lattice clocks, 3738, 4 5 4 6 (Exercises 23, 24) system of particles, 221 (Section 8.1), 244251 (Section 8  8)
mass of, 214 (Exercise 72), 224228 (Section 8.3), 247248
311
not isolated, 228 system property, heat as, 224
tangent vector to worldline, 194195 tau (Greek T), symbol for proper time, 155 Taylot Bradley James, 179 Katherine Rose, 311 Joseph H., 291 Meredith Christine, 171 Samantha Marie, 23 (Exercise 110) teraflop, 22 (Exercise 18) test particle, 36 (Section 2.5), 4 7 4 8 (Exercise 210) tests of relativity, 131134 (Seaion 4.10), 270273 (Exercises 833 through 839) Thompson, Benjamin (Count Rumford), 223 thought experiments speeding pipes, 66 speeding train, 6 566 threevectors. Euclidean, 192 (Box 71) threshold energy, 236, 259 (Exercise 812), 261 (Exercise 816) tidal effects of large frame, 3 0  3 4 (Section 2.3), 2 8 0 281 (Section 9.4) tidedriving power of Moon and Sun, 3233 (Box 21) tides, 3233 (Box 21), 281, 286287 (Box 91) tilted meter stick paradox 115116 (Exercise L10) time “absolute” (Newton), I 6 O as different from space, 18 of an event, 38, 137139 (Section 5.1) Hubble, 264 and length, 1113 (Section 1.4) and Lorentz transformation, 102 meter of, 12 proper, 10, 148152 (Section 5.6), 155, 184 wrisrwatch 10, 148152 (Section 5.6) time delay in communication, 3940 timelike relation between events, 11, 172177 (Section 6 . 2)
timelike worldline of a particle, 172 time stretching experimental evidence of, 131134 (Section 4.10), 272273 (Exercise 839) with K+ mesons, 7273 (Sample Problem 32) with mumesons, 2 324 (Exercise 111) with piplus mesons, 24 (Exercise (112) and spacetime interval, 21 (Exercise 14) See also Twin Paradox time traveler, 127 128 (Section 4.8) touring spacetime without a reference frame, 160162 (Section 5.9) traffic light, physicist and, 263264 (Exercise 820) train, mass effeas of in collision, 214 (Exercise 73)
312
INDEX
train paradoxes, 6263, 168 (Exercise 57), 187188 (Exercise 68) train rhought experiment, 6566 transformation Galilean, 113 (Exercise L3) Lorentz, 95111 (Special Topic) transformation of angles, 114115 (Exercise L6) transformation of velocity direction, 115 (Exercises L7, L8) transforming worldlines, 164 (Exercise 52) transverse dimension, invariance of, 6 567 (Section 3.6) travel, inrerstellar, 274 (Exercise 841) traveler, time, 127128 (Section 4,8) Travelers, Parable of the Two, 281283 (Section 9.5) traveling clock, synchronization using, 4 546 (Exercise 24) Treaty of Shalimar, 108109 (Box L1) tree analogy to momenergy, 210 Twin Paradox, 125126 (Section 4.6) atomic clocks (“airliner”) test of, 131 circling airplane test of, 133 general relativity needed for?, 132 (Box 41) oneway, 135 (Exercise 42) oscillating iron atom test of, 134, 272273 (Exer cise 839) put to rest, 169170 (Exercise 58) radioactive particle test of, 133 using Doppler shift, 264 (Exercise 824) twins, paradox of identically accelerated, 117118 (Exer cise L13) Two Travelers, Parable of, 281283 (Section 9.5) unit, same for space and time, 1113 (Section 1.4) units, 213 (Table 71) units of energy, 203 units of momenergy, 194 units of momentum, 200 units of spacetime, 1113 (Section 1.4), 2021 (Exer cises 12, 13) unit tangent vector to worldline, 194195 unity of spacetime, 1518 (Section 1.5) universe expanding, 82 (Exercise 310), 264 (Exercise 823), 297 (Table 92) models of, 296297 (Seaion 9.9) unreachable region, 182 uranium bomb, 249 uranium fission, 237
use and abuse of the concept of mass, 244251 (Section 8 . 8)
Van Dam, Hendrik, 79 (Exercise 36) vector, defined, 192 (Box 71) velocities, addition of, 8284 (Exercise 311), 103110 (Section L.7) velocity measuring, 4041 (Section 2.8) velocity of recession from Doppler shift, 114 (Exercise L5), 264 (Exercise 823) velocity of recession from period of light, 82 (Exercise 310) velocity direction, transformation of, 115 (Exercises L7, L8) Verne, Jules, 2526 virtual particles, 5657 (Box 31) visual appearance of relativistic objects, 64, 9293 (Exer cise 317) von Jagow, Peter, 44 walking, maximum speed of, 186 (Exercise 66) war American Civil, 25 space, 7981 (Exercise 37) waves, gravity, 288292 (Section 9.7) weighing heat, 223 Weights and Measures, General Conference on, 12, 58 Weisskopf, V. W., 296 Weyl, Herman, quote, 189 Wheeler's First Moral Principle, 20 white dwarf star, 258259 (Exercise 87), 288 (Box 92) wind, solar 245 worldline, 143147 (Section 5.4) kinked, 152155 (Section 5.7) timelike, of a particle, 172 ttansforming, 164 (Exercise 52) unit tangent vector to, 194195 wristwatch (proper) time along, 148152 (Section 5.6) wristwatch time, 1011 along a worldline, 148152 (Section 5.6) Xray, see photon yvelocity, transformation of, 115 (Exercise L7) year as unit of distance and time, 1113 (Section 1.4) zero mass for photon, 230 zerototalmomentum frame, 246251 zoo of particles, backyard, 235 (Box 81)
THE AUTHORS J ohn A rchibald W heeler (Ph.D., Johns Hopkins University) is one of the world s foremost relativists. He is Joseph Henry Professor Emeritus at Princeton University and, until his retirement in 1986, Blumberg Professor of Physics and Director, Center for Theoretical Physics, at the University of Texas at Austin. A past president of the American Physical Society, he is a recipient of the Enrico Fermi Award (1968), the National Medal of Science (1971), and the Niels Bohr International Gold Medal (1982). Since the appearance of the First Edition of Spacetime Physics, John Wheeler has published a graduate text in general relativity, GRAVITATION, with Kip S. Thorne and Charles W Misner (W H. Freeman, 1970), and a popular treatment of gravity, A Journey into Gravity and Spacetime (Scientific American Library, 1990; distributed by W H. Freeman). E dw in F. T aylor (Ph.D., Harvard University) taught physics for 26 years at the Massachusetts Institute of Technology. He is currently Research Professor in the Department of Physics at Boston University. He is the author of a textbook on in troductory mechanics and An Introduction to Quantum Physics with A. P French (W W Norton, 1978). He has served as Editor of the American Journal of Physics. With MIT undergraduates, Edwin Taylor produced interactive computer pro grams to help students visualize and solve problems in special relativity. These won the 1988 EDUCOM/NCRIPTAL Higher Education Software Awards for Best Physics Software and Best Tool Software. THE BOOK Collaboration on the First Edition of Spacetime Physics began in the mid1960s when Edwin Taylor took a junior faculty sabbatical at Princeton University where John Wheeler was a professor. The resulting text emphasized the unity of spacetime and those quantities (such as proper time, proper distance, mass) that are in variant, the same for all observers, rather than those quantities (such as space and time separations) that are relative, different for different observers. The text has become a standard for modern physics and relativity courses, as well as introduc tory physics. The Second Edition of Spacetime Physics embodies what the authors have learned during an additional quarter century of teaching and research. They have updated the text to reflect the immense strides in physics during the same perkxl and modernized and increased the number of exercises, for which the First Edition was famous. Enrichment boxes provide expanded coverage of intriguing topics. Sample pioblems encourage students to exercise their newfound power. An enlarged final chapter on general relativity includes new material on gravity waves, black holes, and cosmology. The Second Edition of Spacetime Physics provides a new generation of students with a deep and simple overview of the principles of relativity.
ISBN13: 9780716723271 ISBN10: 0716723271
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