tayl01-001-045.i

RELATIVITY

Chapter 1 The Space and Time of RelativityChapter 2 Relativistic Mechanics

Two great theories underlie almost all of modern physics, both of them discov-ered during the first 25 years of the twentieth century. The first of these, rela-tivity, was pioneered mainly by one person, Albert Einstein, and is the subjectof Part I of this book (Chapters 1 and 2). The second, quantum theory, was thework of many physicists, including Bohr, Einstein, Heisenberg, Schrödinger,and others; it is the subject of Part II. In Parts III and IV we describe theapplications of these great theories to several areas of modern physics.

Part I contains just two chapters. In Chapter 1 we describe how several ofthe ideas of relativity were already present in the classical physics of Newtonand others. Then we describe how Einstein’s careful analysis of the relation-ship between different reference frames, taking account of the observed in-variance of the speed of light, changed our whole concept of space and time. InChapter 2 we describe how the new ideas about space and time required a rad-ical revision of Newtonian mechanics and a redefinition of the basic ideas —mass, momentum, energy, and force — on which mechanics is built. At the endof Chapter 2, we briefly describe general relativity, which is the generalizationof relativity to include gravity and accelerated reference frames.

PARTI

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1

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2

C h a p t e r 1The Space and Time of Relativity

1.1 Relativity1.2 The Relativity of Orientation and Origin1.3 Moving Reference Frames1.4 Classical Relativity and the Speed of Light1.5 The Michelson–Morley Experiment�

1.6 The Postulates of Relativity1.7 Measurement of Time1.8 The Relativity of Time; Time Dilation1.9 Evidence for Time Dilation1.10 Length Contraction1.11 The Lorentz Transformation1.12 Applications of the Lorentz Transformation1.13 The Velocity-Addition Formula1.14 The Doppler Effect�

Problems for Chapter 1�Sections marked with a star can be omitted without significant loss of continuity.

1.1 Relativity

Most physical measurements are made relative to a chosen reference system. Ifwe measure the time of an event as seconds, this must mean that t is 5seconds relative to a chosen origin of time, If we state that the positionof a projectile is given by a vector we must mean that the posi-tion vector has components relative to a system of coordinates with adefinite orientation and a definite origin, If we wish to know the kineticenergy K of a car speeding along a road, it makes a big difference whether wemeasure K relative to a reference frame fixed on the road or to one fixed onthe car. (In the latter case of course.) A little reflection should con-vince you that almost every measurement requires the specification of a refer-ence system relative to which the measurement is to be made. We refer to thisfact as the relativity of measurements.

The theory of relativity is the study of the consequences of this relativityof measurements. It is perhaps surprising that this could be an important sub-ject of study. Nevertheless, Einstein showed, starting with his first paper on rel-ativity in 1905, that a careful analysis of how measurements depend oncoordinate systems revolutionizes our whole understanding of space and time,and requires a radical revision of classical, Newtonian mechanics.

K = 0,

r = 0.x, y, z

r = 1x, y, z2,t = 0.t = 5

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Section 1.2 • The Relativity of Orientation and Origin 3

In this chapter we discuss briefly some features of relativity as it appliesin the classical theories of Newtonian mechanics and electromagnetism, andthen we describe the Michelson–Morley experiment, which (with the supportof numerous other, less direct experiments) shows that something is wrongwith the classical ideas of space and time. We then state the two postulates ofEinstein’s relativity and show how they lead to a new picture of space and timein which both lengths and time intervals have different values when measuredin any two reference frames that are moving relative to one another. InChapter 2 we show how the revised notions of space and time require a revi-sion of classical mechanics. We will find that the resulting relativistic mechan-ics is usually indistinguishable from Newtonian mechanics when applied tobodies moving with normal terrestrial speeds, but is entirely different whenapplied to bodies with speeds that are a substantial fraction of the speed oflight, c. In particular, we will find that no body can be accelerated to a speedgreater than c, and that mass is a form of energy, in accordance with thefamous relation

Einstein’s theory of relativity is really two theories. The first, called thespecial theory of relativity, is “special” in that its primary focus is restricted tounaccelerated frames of reference and excludes gravity. This is the theory thatwe will be studying in Chapters 1 and 2 and applying to our later discussions ofradiation, nuclear, and particle physics.

The second of Einstein’s theories is the general theory of relativity,which is “general” in that it includes accelerated frames of reference and grav-ity. Einstein found that the study of accelerated reference frames led naturallyto a theory of gravitation, and general relativity turns out to be the relativistictheory of gravity. In practice, general relativity is needed only in areas whereits predictions differ significantly from those of Newtonian gravitational theo-ry. These include the study of the intense gravity near black holes, of the large-scale universe, and of the effect the earth’s gravity has on extremely accuratetime measurements (one part in or so). General relativity is an importantpart of modern physics; nevertheless, it is an advanced topic and, unlike specialrelativity, is not required for the other topics we treat in this book. Therefore,we have given only a brief description of general relativity in an optionalsection at the end of Chapter 2.

1.2 The Relativity of Orientation and Origin

In your studies of classical physics, you probably did not pay much attention tothe relativity of measurements. Nevertheless, the ideas were present, and,whether or not you were aware of it, you probably exploited some aspects ofrelativity theory in solving certain problems. Let us illustrate this claim withtwo examples.

In problems involving blocks sliding on inclined planes, it is well knownthat one can choose coordinates in various ways. One could, for example, use acoordinate system S with origin O at the bottom of the slope and with axes horizontal, vertical, and across the slope, as shown in Fig. 1.1(a). An-other possibility would be a reference frame with origin at the top of theslope and axes parallel to the slope, perpendicular to the slope, and

across it, as in Fig. 1.1(b). The solution of any problem relative to theframe S may look quite different from the solution relative to and it oftenhappens that one choice of axes is much more convenient than the other. (Forsome examples, see Problems 1.1 to 1.3.) On the other hand, the basic laws of

S¿,O¿ z¿

O¿ y¿O¿ x¿O¿S¿

OzOyOx

1012

E = mc2.

(a)

(b)

O x

Frame Sy

Frame S�y�

O�

x�

FIGURE 1.1(a) In studying a block on anincline, one could choose axes horizontal and vertical and putO at the bottom of the slope.(b) Another possibility, which isoften more convenient, is to use anaxis parallel to the slope with

perpendicular to the slope, andto put at the top of the slope.(The axes and point out ofthe page and are not shown.)

O¿ z¿OzO¿

O¿ y¿O¿ x¿

OyOx

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4 Chapter 1 • The Space and Time of Relativity

motion, Newton’s laws, make no reference to the choice of origin and orienta-tion of axes and are equally true in either coordinate system. In the languageof relativity theory, we can say that Newton’s laws are invariant, or unchanged,as we shift our attention from frame S to or vice versa. It is because thelaws of motion are the same in either coordinate system that we are free to usewhichever system is more convenient.

The invariance of the basic laws when we change the origin or orienta-tion of axes is true in all of classical physics — Newtonian mechanics, electro-magnetism, and thermodynamics. It is also true in Einstein’s theory ofrelativity. It means that in any problem in physics, one is free to choose the ori-gin of coordinates and the orientation of axes in whatever way is most expedi-ent. This freedom is very useful, and we often exploit it. However, it is notespecially interesting in our study of relativity, and we will not have muchoccasion to discuss it further.

1.3 Moving Reference Frames

As a more important example of relativity, we consider next a question involv-ing two reference frames that are moving relative to one another. Our discus-sion will raise some interesting questions about classical physics, questions thatwere satisfactorily answered only when Einstein showed that the classicalideas about the relation between moving reference frames needed revision.

Let us imagine a student standing still in a train that is moving with con-stant velocity v along a horizontal track. If the student drops a ball, where willthe ball hit the floor of the train? One way to answer this question is to use areference frame S fixed on the track, as shown in Fig. 1.2(a). In this coordinatesystem the train and student move with constant velocity v to the right. At themoment of release, the ball is traveling with velocity v and it moves, under theinfluence of gravity, in the parabola shown. It therefore lands to the right of itsstarting point (as measured in the ground-based frame S). However, while theball is falling, the train is moving, and a straightforward calculation shows thatthe train moves exactly as far to the right as does the ball.Thus the ball hits thefloor at the student’s feet, vertically below his hand.

Simple as this solution is, one can reach the same conclusion even moresimply by using a reference frame fixed to the train, as in Fig. 1.2(b). In thiscoordinate system the train and student are at rest (while the track moves tothe left with constant velocity ). At the moment of release the ball is at rest(as measured in the train-based frame ). It therefore falls straight down andnaturally hits the floor vertically below the point of release.

The justification of this second, simpler argument is actually quite subtle.We have taken for granted that an observer on the train (using the coordinates

) is entitled to use Newton’s laws of motion and hence to predict thata ball which is dropped from rest will fall straight down. But is this correct?The question we must answer is this: If we accept as an experimental fact thatNewton’s laws of motion hold for an observer on the ground (using coordi-nates ), does it follow that Newton’s laws also hold for an observer in thetrain (using )? Equivalently, are Newton’s laws invariant as we passfrom the ground-based frame S to the train-based frame Within the frame-work of classical physics, the answer to this question is “yes,” as we now show.

Since Newton’s laws refer to velocities and accelerations, let us first con-sider the velocity of the ball. We let u denote the ball’s velocity relative to theground-based frame S, and the ball’s velocity relative to the train-based S¿.uœ

S¿ ?x¿, y¿, z¿

x, y, z

x¿, y¿, z¿

S¿-v

S¿

S¿,

y

x

v

OFrame S fixed to ground

(a)

(b)

�v

y�

x�O�

Frame S� fixed to train

FIGURE 1.2(a) As seen from the ground, thetrain and student move to the right;the ball falls in a parabola and landsat the student’s feet. (b) As seenfrom the train, the ball falls straightdown, again landing at the student’sfeet.

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Section 1.3 • Moving Reference Frames 5

Since the train moves with constant velocity v relative to the ground, wenaturally expect that

(1.1)

We refer to this equation as the classical velocity-addition formula. It re-flects our common-sense ideas about space and time, and asserts that velocitiesobey ordinary vector addition. Although it is one of the central assumptions ofclassical physics, equation (1.1) is one of the first victims of Einstein’s relativity.In Einstein’s relativity the velocities u and do not satisfy (1.1), which is onlyan approximation (although a very good approximation) that is valid when allspeeds are much less than the speed of light, c. Nevertheless, we are for the mo-ment discussing classical physics, and we therefore assume for now that theclassical velocity-addition formula is correct.

Now let us examine Newton’s three laws, starting with the first (the lawof inertia): A body on which no external forces act moves with constant veloc-ity. Let us assume that this law holds in the ground-based frame S. This meansthat if our ball is isolated from all outside forces, its velocity u is constant. Since

and the train’s velocity v is constant, it follows at once that isalso constant, and Newton’s first law also holds in the train-based frame Wewill find that this result is also valid in Einstein’s relativity; that is, in both clas-sical physics and Einstein’s relativity, Newton’s first law is invariant as we passbetween two frames whose relative velocity is constant.

Newton’s second law is a little more complicated. If we assume that itholds in the ground-based frame S, it tells us that

where F is the sum of the forces on the ball, m its mass, and a its acceleration,all measured in the frame S. We now use this assumption to show that

where are the corresponding quantities measured rela-tive to the train-based frame We will do this by arguing that each of

is in fact equal to the corresponding quantity F, m, and a.The proof that depends to some extent on how one has chosen to

define force. Perhaps the simplest procedure is to define forces by their effecton a standard calibrated spring balance. Since observers in the two frames Sand will certainly agree on the reading of the balance, it follows that anyforce will have the same value as measured in S and that is, *

Within the domain of classical physics, it is an experimental fact that anytechnique for measuring mass (for example, an inertial balance) will producethe same result in either reference frame; that is,

Finally, we must look at the acceleration.The acceleration measured in S is

where t is the time as measured by ground-based observers. Similarly, the ac-celeration measured in is

(1.2)aœ =duœ

dt¿

S¿

a =dudt

m = m¿.

F = Fœ.S¿;S¿

F = FœFœ, m¿, aœ

S¿.Fœ, m¿, aœFœ = m¿aœ,

F = ma

S¿.uœuœ = u - v

uœ

u = uœ + v

*Of course, the same result holds whatever our definition of force, but with some defi-nitions the proof is a little more roundabout. For example, many texts define force bythe equation Superficially, at least, this means that Newton’s second law istrue by definition in both frames. Since and (as we will show shortly), itfollows that F = Fœ.

aœ = am = m¿F = ma.

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where is the time measured by observers on the train. Now, it is a central as-sumption of classical physics that time is a single universal quantity, the samefor all observers; that is, the times t and are the same, or Therefore, wecan replace (1.2) by

Since

we can simply differentiate with respect to t and find that

(1.3)

or, since v is constant,We have now argued that and Substituting into

the equation we immediately find that

That is, Newton’s second law is also true for observers using the train-basedcoordinate frame

The third law,

is easily treated. Since any given force has the same value as measured in S orthe truth of Newton’s third law in S immediately implies its truth in

We have now established that if Newton’s laws are valid in one referenceframe, they are also valid in any second frame that moves with constant veloc-ity relative to the first. This shows why we could use the normal rules of pro-jectile motion in a coordinate system fixed to the moving train. Moregenerally, in the context of our newfound interest in relativity, it establishes animportant property of Newton’s laws: If space and time have the usual proper-ties assumed in classical physics, Newton’s laws are invariant as we transferour attention from one coordinate frame to a second one moving withconstant velocity relative to the first.

Newton’s laws would not still hold in a coordinate system that wasaccelerating. Physically, this is easy to understand. If our train were accelerat-ing forward, just to keep the ball at rest (relative to the train) would require aforce; that is, the law of inertia would not hold in the accelerating train. To seethe same thing mathematically, note that if and v is changing, isnot constant even if u is. Further, the acceleration as given by (1.3) is notequal to a, since is not zero; so our proof of the second law for the train’sframe also breaks down. In classical physics the unaccelerated frames inwhich Newton’s laws hold (including the law of inertia) are often calledinertial frames. In fact, one convenient definition (good in both classical and

S¿dv>dt

aœuœuœ = u - v

S¿.S¿,

1action force2 = -1reaction force2

S¿.

Fœ = m¿aœ

F = ma,aœ = a.m¿ = m,Fœ = F,

aœ = a.

aœ = a -dvdt

uœ = u - v

aœ =duœ

dt

t = t¿.t¿

t¿

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Section 1.4 • Classical Relativity and the Speed of Light 7

*More precisely, In fact, the determination of c has become so ac-curate that since 1984, the meter has been defined in terms of c, as the fraction

of the distance traveled by light in 1 second. This means that, by defini-tion, c is exactly.299,792,458 m>s1>299,792,458

c = 299,792,458 m>s.

Isaac Newton(1642–1727, English)

Newton was possibly the greatestscientific genius of all time. In addi-tion to his laws of motion and histheory of gravity, his contributionsincluded the invention of calculusand important discoveries in op-tics. Although he believed in “ab-solute space” (what we would callthe ether frame),Newton was wellaware that his laws of motion holdin all unaccelerated frames of ref-erence. From a modern perspec-tive, it is surprising that Newtondevoted much of his time to find-ing ways to manufacture gold byalchemy and to dating the creationof the world (3500 B.C. was hisanswer) using biblical chronology.

v

Speed c seen from S

Speed c � v seen from S� Speed c � v seen from S�

S�S

relativistic mechanics) of an inertial frame is just that it is a frame where thelaw of inertia holds. The result we have just proved can be rephrased to saythat an accelerated frame is noninertial.

1.4 Classical Relativity and the Speed of Light

Although Newton’s laws are invariant as we change from one unacceleratedframe to another (if we accept the classical view of space and time), the sameis not true of the laws of electromagnetism.We can show this by separately ex-amining each law — Gauss’s law, Faraday’s law, and so on — but the requiredcalculations are complicated. A simpler procedure is to recall that the laws ofelectromagnetism demand that in a vacuum, light signals and all other electro-magnetic waves travel in any direction with speed*

where and are the permittivity and permeability of the vacuum. Thus ifthe electromagnetic laws hold in a frame S, light must travel with the samespeed c in all directions, as seen in S.

Let us now consider a second frame traveling relative to S and imag-ine a pulse of light moving in the same direction as as shown on the left ofFig. 1.3.The pulse has speed c relative to S.Therefore, by the classical velocity-addition formula (1.1), it should have speed as seen from Similarly, apulse traveling in the opposite direction would have speed as seen from

and a pulse traveling in any other oblique direction would have a differentspeed, intermediate between and We see that in the frame thespeed of light should vary between and according to its directionof propagation. Since the laws of electromagnetism demand that the speed oflight be exactly c, we conclude that these laws — unlike those of mechanics —could not be valid in the frame

The situation just described was well understood by physicists towardthe end of the nineteenth century. In particular, it was accepted as obvious thatthere could be only one frame, called the ether frame, in which light traveledat the same speed c in all directions. The name “ether frame” derived from thebelief that light waves must propagate through a medium, in much the sameway that sound waves were known to propagate in the air. Since light propa-gates through a vacuum, physicists recognized that this medium, which no one

S¿.

c + vc - vS¿c + v.c - v

S¿,c + v

S¿.c - v

S¿,S¿

m0e0

c =11eo mo

= 3.00 * 108 m>s

FIGURE 1.3Frame travels with velocity vrelative to S. If light travels with thesame speed c in all directionsrelative to S, then (according to theclassical velocity-addition formula) itshould have different speeds as seenfrom S¿.

S¿

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had ever seen or felt, must have unusual properties. Borrowing the ancientname for the substance of the heavens, they called it the “ether.” The uniquereference frame in which light traveled at speed c was assumed to be the framein which the ether was at rest. As we will see, Einstein’s relativity implies thatneither the ether nor the ether frame actually exists.

Our picture of classical relativity can be quickly summarized. In classi-cal physics we take for granted certain ideas about space and time, all basedon our everyday experiences. For example, we assume that relative velocitiesadd like vectors, in accordance with the classical velocity-addition formula;also, that time is a universal quantity, concerning which all observers agree.Accepting these ideas we have seen that Newton’s laws should be valid in awhole family of reference frames, any one of which moves uniformly relativeto any other. On the other hand, we have seen that there could be no morethan one reference frame, called the ether frame, relative to which the elec-tromagnetic laws hold and in which light travels through the vacuum withspeed c in all directions.

It should perhaps be emphasized that although this view of natureturned out to be wrong, it was nevertheless perfectly logical and internallyconsistent. One might argue on philosophical or aesthetic grounds (as Einsteindid) that the difference between classical mechanics and classical electromag-netism is surprising and even unpleasing, but theoretical arguments alonecould not decide whether the classical view is correct. This question could bedecided only by experiment. In particular, since classical physics implied thatthere was a unique ether frame where light travels at speed c in all directions,there had to be some experiment that showed whether this was so.This was ex-actly the experiment that Albert Michelson, later assisted by Edward Morley,performed between the years 1880 and 1887, as we now describe.

If one assumed the existence of a unique ether frame, it seemed clearthat as the earth orbits around the sun, it must be moving relative to the etherframe. In principle, this motion relative to the ether frame should be easy todetect. One would simply have to measure the speed (relative to the earth) oflight traveling in various directions. If one found different speeds in differentdirections, one would conclude that the earth is moving relative to the etherframe, and a simple calculation would give the speed of this motion. If, instead,one found the speed of light to be exactly the same in all directions, one wouldhave to conclude that at the time of the measurements the earth happened tobe at rest relative to the ether frame. In this case one should probably repeatthe experiment a few months later, by which time the earth would be at a dif-ferent point on its orbit and its velocity relative to the ether frame shouldsurely be nonzero.

In practice, this experiment is extremely difficult because of the enor-mous speed of light.

If our speed relative to the ether is the observed speed of light should varybetween and Although the value of is unknown, it should onaverage be of the same order as the earth’s orbital velocity around the sun,

(or possibly more if the sun is also moving relative to the ether frame). Thusthe expected change in the observed speed of light due to the earth’s motion is

v ' 3 * 104 m>s

vc + v.c - vv,

c = 3 * 108 m>s

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Section 1.4 • Classical Relativity and the Speed of Light 9

about 1 part in This was too small a change to be detected by direct mea-surement of the speed of light at that time.

To avoid the need for such direct measurements, Michelson devised aninterferometer in which a beam of light was split into two beams by a partiallyreflecting surface; the two beams traveled along perpendicular paths and werethen reunited to form an interference pattern; this pattern was sensitive to dif-ferences in the speed of light in the two perpendicular directions and so couldbe used to detect any such differences. By 1887, Michelson and Morley hadbuilt an interferometer (described below) that should have been able to detectdifferences in the speed of light much smaller than the part in expected.Totheir surprise and chagrin, they could detect absolutely no difference at all.

The Michelson–Morley and similar experiments have been repeatedmany times, at different times of year and with ever-increasing precision, butalways with the same final result.* With hindsight, it is easy to draw the rightconclusion from their experiment: Contrary to all expectations, light alwaystravels with the same speed in all directions relative to an earth-based refer-ence frame even though the earth has different velocities at different times ofthe year. In other words, light travels at the same speed c in all directions inmany different inertial frames, and the notion of a unique ether frame with thisproperty must be abandoned.

This conclusion is so surprising that it was not taken seriously for nearly20 years. Rather, several ingenious alternative theories were advanced that ex-plained the Michelson–Morley result but managed to preserve the notion of aunique ether frame. For example, in the “ether-drag” theory, it was suggestedthat the ether, the medium through which light was supposed to propagate,was dragged along by the earth as it moved through space (in much the sameway that the earth does drag its atmosphere with it). If this were the case, anearthbound observer would automatically be at rest relative to the ether, andMichelson and Morley would naturally have found that light had the samespeed in all directions at all times of the year. Unfortunately, this neat expla-nation of the Michelson–Morley result requires that light from the stars wouldbe bent as it entered the earth’s envelope of ether. Instead, astronomical ob-servations show that light from any star continues to move in a straight line asit arrives at the earth.†

The ether-drag theory, like all other alternative explanations of theMichelson–Morley result, has been abandoned because it fails to fit all thefacts. Today, nearly all physicists agree that Michelson and Morley’s failure todetect our motion relative to the ether frame was because there is no etherframe. The first person to accept this surprising conclusion and to develop itsconsequences into a complete theory was Einstein, as we describe, starting inSection 1.6.

104

104.

*From time to time experimenters have reported observing a nonzero difference, butcloser examination has shown that these are probably due to spurious effects such asexpansion and contraction of the interferometer arms resulting from temperature vari-ations. For a careful modern analysis of Michelson and Morley’s results and many fur-ther references, see M. Handschy, American Journal of Physics, vol. 50, p. 987 (1982).†Because of the earth’s motion around the sun, the apparent direction of any one starundergoes a slight annual variation–an effect called stellar aberration. This effect isconsistent with the claim that light travels in a straight line from the star to the earth’ssurface, but contradicts the ether-drag theory.

Albert Michelson(1852–1931, American)

Michelson devoted much of his ca-reer to increasingly accurate mea-surements of the speed of light,and in 1907 he won the NobelPrize in physics for his contribu-tions to optics. His failure to de-tect the earth’s motion relative tothe supposed ether is probably themost famous “unsuccessful” ex-periment in the history of science.

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1.5 The Michelson–Morley Experiment�

�More than a hundred years later, the Michelson–Morley experiment remains the sim-plest and cleanest evidence that light travels at the same speed in all directions in all in-ertial frames — what became the second postulate of relativity. Naturally, we think youshould know a little of how this historic experiment worked. Nevertheless, if you arepressed for time, you can omit this section without loss of continuity.

Figure 1.4 is a simplified diagram of Michelson’s interferometer. Light fromthe source hits the half-silvered mirror M and splits, part traveling to the mir-ror and part to The two beams are reflected at and and returnto M, which sends part of each beam on to the observer. In this way the ob-server receives two signals, which can interfere constructively or destructively,depending on their phase difference.

To calculate this phase difference, suppose for a moment that the twoarms of the interferometer, from M to and M to have exactly the samelength l, as shown. In this case any phase difference must be due to the differ-ent speeds of the two beams as they travel along the two arms. For simplicity,let us assume that arm 1 is exactly parallel to the earth’s velocity v. In this casethe light travels from M to with speed (relative to the interferome-ter) and back from to M with speed Thus the total time for theround trip on path 1 is

(1.4)

It is convenient to rewrite this in terms of the ratio

which we have seen is expected to be very small, In terms of (1.4)becomes

(1.5)t1 =2lc

1

1 - b2 L2lc

11 + b22

b,b ' 10-4.

b =vc

t1 =l

c + v+

lc - v

=2lc

c2 - v2

c - v.M1

c + vM1

M2 ,M1

M2M1M2 .M1 ,

c u

v

(a)

(b)

v

l

l2

1M1

M2

M Observer

Lightsource

FIGURE 1.4(a) Schematic diagram of theMichelson interferometer. M is ahalf-silvered mirror, and aremirrors. The vector v indicates theearth’s velocity relative to thesupposed ether frame. (b) Thevector-addition diagram that givesthe light’s velocity u, relative to theearth, as it travels from M to The velocity c relative to the etheris the vector sum of v and u.

M2 .

M2M1

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Section 1.5 • The Michelson–Morley Experiment 11

In the last step we have used the binomial approximation (discussed inAppendix B and in Problems 1.12–1.14),

(1.6)

which holds for any number n and any x much smaller than 1. (In the presentcase and )

The speed of light traveling from M to is given by the velocity-addi-tion diagram in Fig. 1.4(b). (Relative to the earth, the light has velocity u per-pendicular to v; relative to the ether, it travels with speed c in the directionshown.) This speed is

Since the speed is the same on the return journey, the total time for the roundtrip on path 2 is

(1.7)

where we have again used the binomial approximation (1.6), this time with

Comparing (1.5) and (1.7), we see that the waves traveling along the twoarms take slightly different times to return to M, the difference being

(1.8)

If this difference were zero, the two waves would arrive in step and in-terfere constructively, giving a bright resultant signal. Similarly, if were anyinteger multiple of the light’s period, (where is the wavelength),they would interfere constructively. If were equal to half the period,

(or or ), the two waves would be exactly out of stepand would interfere destructively. We can express these ideas more compactlyif we consider the ratio

(1.9)

This is the number of complete cycles by which the two waves arrive out ofstep; in other words, N is the phase difference, expressed in cycles. If N is aninteger, the waves interfere constructively; if N is a half-odd integer

the waves interfere destructively.The phase difference N in (1.9) is the phase difference due to the earth’s

motion relative to the supposed ether frame. In practice, it is impossible to besure that the two interferometer arms have exactly equal lengths, so there willbe an additional phase difference due to the unknown difference in lengths.Tocircumvent this complication, Michelson and Morley rotated their interferom-eter through observing the interference as they did so.This rotation wouldnot change the phase difference due to the different arm lengths, but it shouldreverse the phase difference due to the earth’s motion (since arm 2 would now

90°,

AN = 12 , 32 , 52 , Á B ,

N =¢t

T=

lb2>cl>c =

lb2

l

2.5T, Á 1.5T,¢t = 0.5T¢t

lT = l>c ¢t¢t

¢t = t1 - t2 Llc

b2

n = - 12 .

t2 =2l3c2 - v2

=2l

c31 - b2L

2lc

A1 + 12 b2 B

u = 3c2 - v2

M2

x = b2.n = -1

11 - x2n L 1 - nx

TAYL01-001-045.I 12/10/02 1:50 PM

Albert Einstein(1879–1955,German–Swiss–American)

Like all scientific theories, relativitywas the work of many people.Nevertheless, Einstein’s contribu-tions outweigh those of anyoneelse by so much that the theory isquite properly regarded as his. Aswe will see in Chapter 4, he alsomade fundamental contributionsto quantum theory, and it was forthese that he was awarded the1921 Nobel Prize in physics. Theexotic ideas of relativity and thegentle, unpretentious persona ofits creator excited the imaginationof the press and public, and Ein-stein became the most famous sci-entist who ever lived. Asked whathis profession was, the aged Ein-stein once answered,“photograph-er’s model.”


be along v and arm 1 across it). Thus, as a result of the rotation, the phasedifference N should change by twice the amount (1.9),

(1.10)

This implies that the observed interference should shift from bright to darkand back to bright again times. Observation of this shift would confirmthat the earth is moving relative to the ether frame, and measurement of would give the value of and hence the earth’s velocity

In their experiment of 1887, Michelson and Morley had an arm length(This was accomplished by having the light bounce back and forth

between several mirrors.) The wavelength of their light was andas we have seen, was expected to be of order Thus the shiftshould have been at least

(1.11)

Although they could detect a shift as small as 0.01, Michelson and Morleyobserved no significant shift when they rotated their interferometer.

Michelson and Morley were disappointed and shocked at their result,and it was almost 20 years before anyone drew the right conclusion from it —that light has the same speed c in all directions in all inertial frames, the ideathat Einstein adopted as one of the postulates of his theory of relativity.

1.6 The Postulates of Relativity

We have seen that the classical ideas of space and time had led to twoconclusions:

1. The laws of Newtonian mechanics hold in an entire family of referenceframes, any one of which moves uniformly relative to any other.

2. There can be only one reference frame in which light travels at the samespeed c in all directions (and, more generally, in which all laws of electro-magnetism are valid).

The Michelson–Morley experiment and numerous other experiments in thesucceeding hundred years have shown that the second conclusion is false.Light travels with speed c in all directions in many different reference frames.

Einstein’s special theory of relativity is based on the acceptance of thisfact. Einstein proposed two postulates, or axioms, expressing his convictionthat all physical laws, including mechanics and electromagnetism, should bevalid in an entire family of reference frames. From these two postulates, hedeveloped his special theory of relativity.

Before we state the two postulates of relativity, it is convenient to ex-pand the definition of an inertial frame to be any reference frame in which allthe laws of physics hold.

An inertial frame is any reference frame (that is, system of coordinates andtime t) where all the laws of physics hold in their simplest form.

x, y, z

¢N =2lb2

lL

2 * 111 m2 * 110-422590 * 10-9 m

L 0.4

10-4.b = v>c l = 590 nm;l L 11 m.

v = bc.b,¢N

¢N

¢N =2lb2

l

TAYL01-001-045.I 12/10/02 1:50 PM

Section 1.6 • The Postulates of Relativity 13

Notice that we have not yet said what “all the laws of physics” are; to a largeextent, Einstein used his postulates to deduce what the correct laws of physicscould be. It turns out that one of the laws that survives from classical physicsinto relativity is Newton’s first law, the law of inertia. Thus our newly definedinertial frames are in fact the familiar unaccelerated frames where a body onwhich no forces act moves with constant velocity. As before, a reference frameanchored to the earth is an inertial frame (to the extent that we ignore thesmall accelerations due to the earth’s rotation and orbital motion); a referenceframe fixed to a rapidly rotating turntable is not an inertial frame.

Notice also that in defining an inertial frame, we have specified that thelaws of physics must hold “in their simplest form.” This is because one cansometimes modify physical laws so that they hold in noninertial frames as well.For example, by introducing a “fictitious” centrifugal force, one can arrangethat the laws of statics are valid in a rotating frame. It is to exclude this kind ofmodification that we have added the qualification “in their simplest form.”

The first postulate of relativity asserts that there is a whole family ofinertial frames.

FIRST POSTULATE OF RELATIVITY

If S is an inertial frame and if a second frame moves with constant velocity rela-tive to S, then is also an inertial frame.

We can reword this postulate to say that the laws of physics are invariant as wechange from one reference frame to a second frame, moving uniformly rela-tive to the first. This property is familiar from classical mechanics, but inrelativity it is postulated for all the laws of physics.

The first postulate is often paraphrased as follows: “There is no suchthing as absolute motion.” To understand what this means, consider a frame attached to a rocket moving at constant velocity relative to a frame S anchoredto the earth. The question we want to ask is this: Is there any scientific sense inwhich we can say that is really moving and that S is really stationary (or,perhaps, the other way around)? If the answer were “yes,” we could say that Sis absolutely at rest and that anything moving relative to S is in absolute mo-tion. However, the first postulate of relativity guarantees that this is impossi-ble:All laws observable by an earthbound scientist in S are equally observableby a scientist in the rocket any experiment that can be performed in S canbe performed equally in Thus no experiment can possibly show whichframe is “really” moving. Relative to the earth, the rocket is moving; relative tothe rocket, the earth is moving; and this is as much as we can say.

Yet another way to express the first postulate is to say that among thefamily of inertial frames, all moving relative to one another, there is nopreferred frame. That is, physics singles out no particular inertial frame asbeing in any way more special than any other frame.

The second postulate identifies one of the laws that holds in all inertialframes.

SECOND POSTULATE OF RELATIVITY

In all inertial frames, light travels through the vacuum with the same speed,in any direction.c = 299,792,458 m>s

S¿.S¿;

S¿

S¿

S¿S¿

TAYL01-001-045.I 12/10/02 1:50 PM


O

S

FIGURE 1.5The chief observer at O distributesher helpers, each with an identicalclock, throughout S.

This postulate is, of course, the formal expression of the Michelson–Morley re-sult. We can say briefly that it asserts the universality of the speed of light c.

The second postulate flies in the face of our normal experience. Nev-ertheless, it is now a firmly established experimental fact. As we explore theconsequences of the two postulates of relativity, we are going to encounterseveral unexpected effects that may be difficult to accept at first. All ofthese effects (including the second postulate itself) have the subtle proper-ty that they become important only when bodies travel at speeds reasonablyclose to the speed of light. Under ordinary conditions, at normal terrestrialspeeds, these effects simply do not show up. In this sense, none of the sur-prising consequences of Einstein’s relativity really contradicts our everydayexperience.

1.7 Measurement of Time

Before we begin exploring the consequences of the relativity postulates, weneed to say a word about the measurement of time. We are going to find thatthe time of an event may be different when measured from different frames ofreference. This being the case, we must first be quite sure we know what wemean by measurement of time in a single frame.

It is implicit in the second postulate of relativity, with its reference tothe speed of light, that we can measure distances and times. In particular,we take for granted that we have access to several accurate clocks. Theseclocks need not all be the same; but when they are all brought to the samepoint in the same inertial frame and are properly synchronized, they mustof course agree.

Consider now a single inertial frame S, with origin O and axes Weimagine an observer sitting at O and equipped with one of our clocks. Usingher clock, the observer can easily time any event, such as a small explosion, inthe immediate proximity of O since she will see (or hear) the event the mo-ment it occurs. To time an event far away from O is harder, since the light (orsound) from the event has to travel to O before our observer can sense it. Toavoid this complication, we let our observer hire a large number of helpers,each of whom she equips with an accurate clock and assigns to a fixed, knownposition in the coordinate system S, as shown in Fig. 1.5. Once the helpers arein position, she can check that their clocks are still synchronized by havingeach helper send a flash of light at an agreed time (measured on the helper’sclock); since light travels with the known speed c (second postulate), she cancalculate the time for the light to reach her at O and hence check the setting ofthe helper’s clock.

With enough helpers, stationed closely enough together, we can be surethere is a helper sufficiently close to any event to time it effectively instanta-neously. Once he has timed it, he can, at his leisure, inform everyone else of theresult by any convenient means (by telephone, for example). In this way anyevent can be assigned a time t, as measured in the frame S.

When we speak of an inertial frame S, we will always have in mind a sys-tem of axes Oxyz and a team of observers who are stationed at rest through-out S and equipped with synchronized clocks. This allows us to speak of theposition and the time t of any event, relative to the frame S.r = 1x, y, z2

x, y, z.

TAYL01-001-045.I 12/10/02 1:50 PM

Section 1.8 • The Relativity of Time; Time Dilation 15

1.8 The Relativity of Time; Time Dilation

We are now ready to compare measurements of times made by observers intwo different inertial frames, and we are going to find that, as a consequence ofthe relativity postulates, times measured in different frames inevitably dis-agree. To this end, we imagine the familiar two frames, S anchored to theground and anchored to a train moving at constant velocity v relative to theground.We consider a “thought experiment” (or “gedanken experiment” fromthe German) in which an observer at rest on the train sets off a flashbulb onthe floor of the train, vertically below a mirror mounted on the roof, a height habove. As seen in the frame (fixed in the train), a pulse of light travelsstraight up to the mirror, is reflected straight back, and returns to its startingpoint on the floor. We can imagine a photocell arranged to give an audiblebeep as the light returns. Our object is to find the time, as measured in eitherframe, between the two events — the flash as the light leaves the floor and thebeep as it returns.

Our experiment, as seen in the frame is shown in Fig. 1.6(a). Since is an inertial frame, light travels the total distance at speed c. Therefore, thetime for the entire trip is

(1.12)

This is the time that an observer in frame will measure between the flashand the beep, provided of course, that his clock is reliable.

The same experiment, as seen from the inertial frame S, is shown inFig. 1.6(b). In this frame the light travels along the two sides and of thetriangle shown. If we denote by the time for the entire journey, as measuredin S, the time to go from A to B is During this time the train travels a dis-tance and the light, moving with speed c, travels a distance (Note that this is where the postulates of relativity come in; we have taken thespeed of light to be c in both S and ) The dimensions of the right triangleS¿.

c ¢t>2.v ¢t>2,¢t>2.

¢tBCAB

S¿

¢t¿ =2hc

2hS¿S¿,

S¿

S¿

�v

A D

h

BS

Flash Beep

(a) (c)

S�

v�t/2

c�t/2

v

A D C

BS

Flash

(b)

S�

Beep

FIGURE 1.6(a) The thought experiment asseen in the train-based frame (b) The same experiment as seenfrom the ground-based frame S.Notice that two observers areneeded in this frame. (c) Thedimensions of the triangle ABD.

S¿.

TAYL01-001-045.I 12/10/02 1:50 PM


are therefore as shown in Fig. 1.6(c).Applying Pythagoras’s theorem, wesee that*

or, solving for

(1.13)

where we have again used the ratio

of the speed to the speed of light c. The time is the time that observers inS will measure between the flash and the beep (provided, again, that theirclocks are reliable).

The most important and surprising thing about the two answers (1.12)and (1.13) is that they are not the same. The time between the two events,the flash and the beep, is different as measured in the frames S and Specifically,

(1.14)

We have derived this result for an imagined thought experiment involving aflash of light reflected back to a photocell. However, the conclusion applies toany two events that occur at the same place on the train: Suppose, for instance,that we drop a knife on the table and a moment later drop a fork. In principle,at least, we could arrange for a flash of light to occur at the moment the knifelands, and we could position a mirror to reflect the light back to arrive just asthe fork lands. The relation (1.14) must then apply to these two events (thelanding of the knife and the landing of the fork). Now the falling of the knifeand fork cannot be affected by the presence or absence of a flashbulb and pho-tocell; thus neither of the times or can depend on whether we actuallydid the experiment with the light and the photocell. Therefore, the relation(1.14) holds for any two events that occur at the same place on board the train.

The difference between the measured times and is a direct conse-quence of the second postulate of relativity. (In classical physics ofcourse.) You should avoid thinking that the clocks in one of our frames mustsomehow be running wrong; quite the contrary, it was an essential part of ourargument that all the clocks were running right. Moreover, our argumentmade no reference to the kind of clocks used (apart from requiring that theybe correct). Thus the difference (1.14) applies to all clocks. In other words,time itself as measured in the two frames is different. We will discuss theexperimental evidence for this surprising conclusion shortly.

Several properties of the relationship (1.14) deserve comment. First, ifour train is actually at rest then and (1.14) tells us that

That is, there is no difference unless the two frames are in relative¢t = ¢t¿.b = 01v = 02,

¢t = ¢t¿,¢t¿¢t

¢t¿¢t

¢t =¢t¿31 - b2

S¿.

¢tv

b =v

c

¢t =2h3c2 - v2

=2hc

131 - b2

¢t,

a c ¢t

2b2

= h2 + av ¢t

2b2

ABD

*Here we are taking for granted that the height h of the train is the same as measuredin either frame, S or We will prove that this is correct in Section 1.10.S¿.

TAYL01-001-045.I 12/10/02 1:50 PM

Section 1.8 • The Relativity of Time; Time Dilation 17

motion. Further, at normal terrestrial speeds, and thus thedifference between and is very small.

Example 1.1

The pilot of a jet traveling at a steady sets a buzzer in the cockpit togo off at intervals of exactly 1 hour (as measured on the plane). What wouldbe the interval between two successive buzzes as measured by two observerssuitably positioned on the ground? (Ignore effects of the earth’s motion; thatis, consider the ground to be an inertial frame.)

The required interval between two buzzes is given by (1.14), withand Thus

We have to be a bit careful in evaluating this time. The number in thedenominator is so close to 1 that most calculators cannot tell the difference.(It takes 12 significant figures to distinguish from 1.) In this situa-tion the simplest and best course is to use the binomial approximation,

which is an excellent approximation, provided x is small.[This important approximation was already used in (1.6) and is discussed inProblems 1.12–1.14 and in Appendix B.] In the present case, setting and we find

The difference between the two measured times is or 1.8nanoseconds. (A nanosecond, or ns, is ) It is easy to see why classicalphysicists had failed to notice this kind of difference!

The difference between and gets bigger as increases. In modernparticle accelerators it is common to have electrons and other particles withspeeds of and more. If we imagine repeating our thought experimentwith the frame attached to an electron with Eq. (1.14) gives

Differences as large as this are routinely observed by particle physicists, as wediscuss in the next section.

If we were to put (that is, ) in Eq. (1.14), we would get theabsurd result, and if we put (that is, ), we would getan imaginary answer. These ridiculous results suggest (correctly) that mustalways be less than c.

v 6 c

vb 7 1v 7 c¢t = ¢t¿>0;

b = 1v = c

¢t =¢t¿41 - 10.9922 L 7¢t¿

b = 0.99,S¿0.99c

v¢t¿¢t

10-9 s.5 * 10-13 hour,

= 1.0000000000005 hours

= 11 hour2 * A1 + 12 * 10-12 B

¢t = ¢t¿11 - b22-1>2 L ¢t¿ A1 + 12 b2 B

n = -1>2,x = b2

11 - x2n L 1 - nx,

1 - 10-12

¢t =¢t¿31 - b2

=1 hour31 - 10-12

b = v>c = 10-6,¢t¿ = 1 hour

300 m>s

¢t¿¢tb V 1;v V c

TAYL01-001-045.I 12/10/02 1:50 PM


This is one of the most profound results of Einstein’s relativity: The speed ofany inertial frame relative to any other inertial frame must always be less than c.In other words, the speed of light, in addition to being the same in all iner-tial frames, emerges as the universal speed limit for the relative motion ofinertial frames.

The factor that appears in Eq. (1.14) crops up in so manyrelativistic formulas that it is traditionally given its own symbol,

(1.15)

Since is always smaller than c, the denominator in (1.15) is always less thanor equal to 1 and hence

(1.16)

The factor equals 1 only if The larger we make the larger becomes; and as approaches c, the value of increases without limit.

In terms of Eq. (1.14) can be rewritten

(1.17)

That is, is always greater than or equal to This asymmetry may seemsurprising, and even to violate the postulates of relativity since it suggests aspecial role for the frame In fact, however, this is just as it should be. In ourexperiment the frame is special since it is the unique inertial frame wherethe two events — the flash and the beep — occurred at the same place. Thisasymmetry was implicit in Fig. 1.6, which showed one observer measuring (since both events occurred at the same place in ) but two observers mea-suring (since the two events were at different places in ). To emphasizethis asymmetry, the time can be renamed and (1.17) rewritten as

(1.18)

The subscript 0 on indicates that is the time indicated by a clock that isat rest in the special frame where the two events occurred at the same place.This time is often called the proper time between the events. The time ismeasured in any frame and is always greater than or equal to the proper time

For this reason, the effect embodied in (1.18) is often called time dilation.The proper time is the time indicated by the clock on the moving

train (moving relative to S, that is); is the time shown by the clocks at reston the ground in frame S. Since the relation (1.18) can be looselyparaphrased to say that “a moving clock is observed to run slow.”

Finally, we should reemphasize the fundamental symmetry between anytwo inertial frames. We chose to conduct our thought experiment with theflash and beep at one spot on the train (frame ), and we found that

However, we could have done things the other way around: If aground-based observer (at rest in S) had performed the same experiment witha flash of light and a mirror, the flash and beep would have occurred in thesame spot on the ground; and we would have found that The greatmerit of writing the time-dilation formula in the form (1.18), is¢t = g ¢t0 ,

¢t¿ Ú ¢t.

¢t 7 ¢t¿.S¿

¢t0 … ¢t,¢t

¢t0

¢t0 .

¢t

¢t0¢t0

¢t = g ¢t0 Ú ¢t0

¢t0¢t¿S¢t

S¿¢t¿

S¿S¿.

¢t¿.¢t

¢t = g ¢t¿ Ú ¢t¿

g,gv

gv,v = 0.g

g Ú 1

v

g =131 - b2

=141 - 1v>c22

g.1>31 - b2

TAYL01-001-045.I 12/10/02 1:50 PM

Section 1.9 • Evidence for Time Dilation 19

*The test was actually carried out twice — once flying east and once west — with satis-factory agreement in both cases. The results quoted here are from the more decisivewestward flight. For more details, see J. C. Hafele and R. E. Keating, Science, vol. 177,p. 166 (1972). Since the accuracy of this original experiment has been questioned, weshould emphasize that the experiment has been repeated many times, with improvedaccuracy, and there is now no doubt at all that the observations support the predictionsof relativity.†An alternative characterization is the mean life which differs from by a constantfactor. We will define both of these more carefully in Chapter 17.

t1>2t,

that it avoids the problem of remembering which is frame S and which thesubscript 0 always identifies the proper time, as measured in the frame inwhich the two events were at the same spot.

1.9 Evidence for Time Dilation

In his original paper on relativity, Einstein predicted the effect that is nowcalled time dilation. At that time there was no evidence to support the predic-tion, and many years were to pass before any was forthcoming. The first tests,using the unstable particle called the muon as their “clock,” were carried out in1941. (See Problem 1.27.)

It was only with the advent of super-accurate atomic clocks that testsusing man-made clocks became possible. The first such test was carried out in1971. Four portable atomic clocks were synchronized with a reference clock atthe U.S. Naval Observatory in Washington, D.C., and all four clocks were thenflown around the world on a jet plane and returned to the Naval Observatory.The discrepancy between the reference clock and the portable clocks aftertheir journey was predicted (using relativity) to be

(1.19)

while the observed discrepancy (averaged over the four portable clocks) was*

(1.20)

We should mention that the excellent agreement between (1.19) and(1.20) is more than a test of the time difference (1.18), predicted by special rel-ativity. Gravitational effects, which require general relativity, contribute alarge part of the predicted discrepancy (1.19). Thus this beautiful experimentis a confirmation of general, as well as special, relativity.

Much simpler tests of time dilation and tests involving much larger dila-tions are possible if one is prepared to use the natural clocks provided by un-stable subatomic particles. For example, the charged meson, or pion, is aparticle that is formed in collisions between rapidly moving atomic nuclei (aswe discuss in detail in Chapter 18). The pion has a definite average lifetime,after which it “decays” or disintegrates into other subatomic particles, and onecan use this average life as a kind of natural clock.

One way to characterize the life span of an unstable particle is the half-life†

the average time after which half of a large sample of the particles in ques-tion will have decayed. For example, the half-life of the pion is measured to be

(1.21)t1>2 = 1.8 * 10-8 s

t1>2 ,

p

273 ; 7 ns

275 ; 21 ns

S¿;

TAYL01-001-045.I 12/10/02 1:50 PM


This means that if one starts at with pions, then after halfof them will have decayed and only will remain. After a further

half of those will have decayed and only will remain.After another only will remain. And so on. In general, aftern half-lives, the number of particles remaining will be

At particle-physics laboratories, pions are produced in large numbers incollisions between protons (the nuclei of hydrogen atoms) and various othernuclei. It is usually convenient to conduct experiments with the pions at a gooddistance from where they are produced, and the pions are therefore allowed tofly down an evacuated pipe to the experimental area. At the Fermilab nearChicago the pions are produced traveling very close to the speed of light, atypical value being

and the distance they must travel to the experimental area is about Let us consider the flight of these pions, first from the (incorrect) classical viewwith no time dilation and then from the (correct) relativistic view.

As seen in the laboratory, the pions’ time of flight is

(1.22)

A classical physicist, untroubled by any notions of relativity of time, wouldcompare this with the half-life (1.21) and calculate that

That is, the time needed for the pions to reach the experimental area is 183half-lives. Therefore, if is the original number of pions, the number tosurvive the journey would be

and for all practical purposes, no pions would reach the experimental area.This would obviously be an absurd way to do experiments with pions, and it isnot what actually happens.

In relativity, we now know, times depend on the frame in which they aremeasured, and we must consider carefully the frames to which the times T and

refer. The time T in (1.22) is, of course, the time of flight of the pions asmeasured in a frame fixed in the laboratory, the lab frame. To emphasize this,we rewrite (1.22) as

(1.23)

On the other hand, the half-life refers to time as “seen” bythe pions; that is, is the half-life measured in a frame anchored to the pions,the pions’ rest frame. (This is an experimental fact: The half-lives quoted byphysicists are the proper half-lives, measured in the frame where the particlesare at rest.) To emphasize this, we write (temporarily)

(1.24)t1>21p rest frame2 = 1.8 * 10-8 s

t1>2t1>2 = 1.8 * 10-8 s

T1lab frame2 = 3.3 * 10-6 s

t1>2

N =N0

2183 L 18.2 * 10-562N0

N0

T L 183t1>2

T =Lv

L103 m

3 * 108 m>s = 3.3 * 10-6 s

L = 1 km.

v = 0.9999995c

N0>2n.t = n t1>2 ,N0>81.8 * 10-8 s,

N0>4N0>21.8 * 10-8 s,N0>2

1.8 * 10-8 sN0t = 0

TAYL01-001-045.I 12/10/02 1:50 PM

Section 1.10 • Length Contraction 21

We see that the classical argument here used two times, T and mea-sured in different inertial frames. A correct argument must work consistentlyin one frame, for example the lab frame. The half-life measured in the labframe is given by the time-dilation formula as times the half-life (1.24). With

it is easy to see that

and hence that

(1.25)

Comparing (1.23) and (1.25) we see that

That is, the pions’ flight down the pipe lasts only one-fifth of the relevant half-life. In this time very few of the pions decay, and almost all reach the experi-mental area. (The number that survive is ) That this isexactly what actually happens in all particle-physics laboratories is powerfulconfirmation of the relativity of time, as first predicted by Einstein in 1905.

Example 1.2

The particle ( is the Greek capital L and is pronounced “lambda.”) is anunstable subatomic particle that decays into a proton and a pion

with a half-life of If several lambdas arecreated in a nuclear collision, all with speed on average how farwill they travel before half of them decay?

The half-life as measured in the laboratory is (since is the prop-er half-life, as measured in the rest frame). Therefore, the desired distanceis With

and the required distance is

Notice how even with speeds as large as 0.6 c, the factor is not very muchlarger than 1, and the effect of time dilation is not dramatic. Notice also thata distance of a few centimeters is much easier to measure than a time oforder thus measurement of the range of an unstable particle is oftenthe easiest way to find its half-life.

1.10 Length Contraction

The postulates of relativity have led us to conclude that time depends on thereference frame in which it is measured. We can now use this fact to show thatthe same must also apply to distances: The measured distance between two

10-10 s;

g

distance = vgt1>2 = 11.8 * 108 m>s2 * 1.25 * 11.7 * 10-10 s2 = 3.8 cm

g =131 - b2

= 1.25

b = 0.6,vgt1>2 .¶

t1>2gt1>2

v = 0.6 c,t1>2 = 1.7 * 10-10 s.1¶ : p + p2

¶¶

N = N0>20.2 L 0.9N0 .

T1lab frame2 L 0.2t1>21lab frame2

= 1.8 * 10-5 s = 1000 * 11.8 * 10-8 s2

t1>21lab frame2 = gt1>21p rest frame2

g = 1000

b = 0.9999995,g

t1>2

TAYL01-001-045.I 12/10/02 1:50 PM


*We are taking for granted that the speed of S relative to is the same as that of relative to S.This follows from the basic symmetry between S and as required by thepostulates of relativity.

S¿S¿S¿

events depends on the frame relative to which it is measured.We will show thiswith another thought experiment. In the analysis of this thought experiment, itwill be important to recognize that, even in relativity, the familiar kinematicrelation

is valid in any given inertial frame (with all quantities measured in that frame),since it is just the definition of velocity in that frame.

We imagine again our two frames, S fixed to the ground and fixed to atrain traveling at velocity v relative to the ground; and we now imagine ob-servers in S and measuring the length of the train. For an observer in thismeasurement is easy since he sees the train at rest and can take all the time heneeds to measure the length with an accurate ruler. For an observer Q on theground, the measurement is harder since the train is moving. Perhaps the sim-plest procedure is to time the train as it passes Q [Fig. 1.7(a)]. If and arethe times at which the front and back of the train pass Q and if then Q can calculate the length l (measured in S) as

(1.26)

To compare this answer with we note that observers on the train couldhave measured by a similar procedure. As seen from the train, the observerQ on the ground is moving to the left with speed* and observers on the traincan measure the time for Q to move from the front to the back of the train asin Fig. 1.7(b). (This would require two observers on the train, one at the frontand one at the back.) If this time is

(1.27)

Comparing (1.26) and (1.27), we see immediately that since the times and are different, the same must be true of the lengths l and To calculatel¿.¢t¿

¢t

l¿ = v ¢t¿

¢t¿,

v,l¿

l¿,

l = v ¢t

¢t = t2 - t1 ,t2t1

l¿

S¿,S¿

S¿

distance = velocity * time

vS

Observer Q

(a)

(b)

S� l � v�t

l� � v�t�

�v

S S�

Observer Q

FIGURE 1.7(a) As seen in S, the train moves adistance to the right. (b) Asseen in the frame S andobserver Q move a distance to the left.

v ¢t¿S¿,v ¢t

TAYL01-001-045.I 12/10/02 1:50 PM

Section 1.10 • Length Contraction 23

the difference, we need to relate and using the time-dilation formula. Inthe present experiment the two events of interest, “Q opposite the train’sfront” and “Q opposite the train’s back,” occur at the same place in S (where Qis at rest). Therefore, the time-dilation formula implies that

Comparing (1.26) and (1.27), we see that

(1.28)

The length of the train as measured in S is less than (or equal to) thatmeasured in

Like time dilation, this result is asymmetric, reflecting the asymmetry ofour experiment: The frame is special since it is the unique frame where themeasured object (the train) is at rest. [We could, of course, have done the ex-periment the other way around; if we had measured the length of a house atrest in S, the roles of l and in (1.28) would have been reversed.] To empha-size this asymmetry and to avoid confusion as to which frame is which, it is agood idea to rewrite (1.28) as

(1.29)

where the subscript 0 indicates that is the length of an object measured in itsrest frame, while l refers to the length measured in any frame.The length canbe called the object’s proper length. Since the effect implied by (1.29) isoften called length contraction (or Lorentz contraction, or Lorentz–Fitzgeraldcontraction, after the two physicists who first suggested some such effect). Theeffect can be loosely described by saying that a moving object is observed to becontracted.

Evidence for Length Contraction

Like time dilation, length contraction is a real effect that is well establishedexperimentally. Perhaps the simplest evidence comes from the same experi-ment as that discussed in connection with time dilation, in which unstablepions fly down a pipe from the collision that produces them to the experimen-tal area. As viewed from the lab frame, we saw that time dilation increases thepions’ half-life by a factor of from to In the example discussed, itwas this increase that allowed most of the pions to complete the journey to theexperimental area before they decayed.

Suppose, however, that we viewed the same experiment from the pions’rest frame. In this frame the pions are stationary and there is no time dilationto increase their half-life. So how do they reach the experimental area? Theanswer is that in this frame the pipe is moving, and length contraction reducesits length by the same factor from L to Thus observers in this framewould say it is length contraction that allows the pions to reach the experi-mental area. Naturally, the number of pions completing the journey is thesame whichever frame we use for the calculation.

L>g.g,

gt1>2 .t1>2g,

l … l0 ,l0

l0

l =l0g

… l0

l¿

S¿

S¿.

l =l¿g

… l¿

¢t¿ = g ¢t

¢t¿¢t

TAYL01-001-045.I 12/10/02 1:50 PM


Example 1.3

A space explorer of a future era travels to the nearest star,Alpha Centauri, ina rocket with speed The distance from earth to the star, as mea-sured from earth, is light years. What is this distance as measured bythe explorer, and how long will she say the journey to the star lasts? [A lightyear is the distance traveled by light in one year, which is just c multiplied by1 year, or kilometers. In many problems it is better to write it as

since the c often cancels out, as we will see.]The distance is the proper distance between earth and

the star (which we assume are relatively at rest). Thus the distance as seenfrom the rocket is given by the length-contraction formula as

If then so

We can calculate the time T for the journey in two ways: As seen fromthe rocket, the star is initially away and is approaching withspeed Therefore,

(1.30)

(Notice how the factors of c conveniently cancel when we use andmeasure speeds as multiples of c.)

Alternatively, as measured from the earth frame, the journey lasts for atime

But because of time dilation, this is times T(rocket frame), which istherefore

in agreement with (1.30), of course.Notice how time dilation (or length contraction) allows an appreciable

saving to the pilot of the rocket. If she returns promptly to earth, then as a re-sult of the complete round trip she will have aged only 3.8 years, while hertwin who stayed behind will have aged 8.8 years. This surprising result, some-times known as the twin paradox, is amply verified by the experiments dis-cussed in Section 1.9. In principle, time dilation would allow explorers to

T1rocket frame2 =T1earth frame2

g= 1.9 years

g

T1earth frame2 =L1earth frame2

v=

4 c # years

0.9c= 4.4 years

c # years

=1.7 c # years

0.9c= 1.9 years

T1rocket frame2 =L1rocket frame2

v

v = 0.9c.1.7 c # years

L1rocket frame2 =4 c # years

2.3= 1.7 c # years

g = 2.3,b = 0.9,

L1rocket frame2 =L1earth frame2

g

L = 4 c # years1 c # year,

9.46 * 1012

L = 4v = 0.9 c.

TAYL01-001-045.I 12/10/02 1:50 PM

Section 1.11 • The Lorentz Transformation 25

*Note that our previous two thought experiments were asymmetric, requiring two ob-servers in one of the frames, but only one in the other.

make in one lifetime trips that would require hundreds of years as viewedfrom earth. Since this requires rockets that travel very close to the speed oflight, it is not likely to happen soon! See Problem 1.22 for further discussionof this effect.

Lengths Perpendicular to the Relative Motion

We have so far discussed lengths that are parallel to the relative velocity,such as the length of a train in its direction of motion. What happens tolengths perpendicular to the relative velocity, such as the height of the train?It is fairly easy to show that for such lengths, there is no contraction or ex-pansion.To see this, consider two observers, Q at rest in S and at rest in and suppose that Q and are equally tall when at rest. Now, let us assumefor a moment that there is a contraction of heights analogous to the lengthcontraction (1.29). If this is so, then as seen by will be shorter as herushes by. We can test this hypothesis by having hold up a sharp knife ex-actly level with the top of his head; if is shorter, Q will find himselfscalped (or worse) as the knife goes by.

This experiment is completely symmetric between the two frames S andThere is one observer at rest in each frame, and the only difference is the

direction in which each sees the other moving.* Therefore, it must also be truethat as seen by it is Q who is shorter. But this implies that the knife willmiss Q. Since it cannot be true that Q is both scalped and not scalped, we havearrived at a contradiction, and there can be no contraction. By a similar argu-ment, there can be no expansion, and, in fact, the knife held by simplygrazes past Q’s scalp, as seen in either frame. We conclude that lengths per-pendicular to the relative motion are unchanged; and the Lorentz-contractionformula (1.29) applies only to lengths parallel to the relative motion.

1.11 The Lorentz Transformation

We are now ready to answer an important general question: If we know the co-ordinates and time t of an event, as measured in a frame S, how can wefind the coordinates and of the same event as measured in a sec-ond frame Before we derive the correct relativistic answer to this question,we examine briefly the classical answer.

We consider our usual two frames, S anchored to the ground and an-chored to a train traveling with velocity v relative to S, as shown in Fig. 1.8. Be-cause the laws of physics are all independent of our choice of origin andorientation, we are free to choose both axes Ox and along the same line,parallel to v, as shown. We can further choose the origins of time so that

at the moment when passes O. We will sometimes refer to thisarrangement of systems S and as the standard configuration.S¿

O¿t = t¿ = 0

O¿ x¿

S¿

S¿ ?t¿x¿, y¿, z¿,

x, y, z

Q¿

Q¿,

S¿:

Q¿Q¿Q, Q¿

Q¿S¿,Q¿

v

S S�

xvt x�

y� � y

P�

O�O

FIGURE 1.8In classical physics the coordinatesof an event are related as shown.

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Galileo Galilei(1564–1642, Italian)

Considered by some the father ofmodern science, Galileo under-stood the importance of experi-ment and theory and was a masterof both. Although he did not dis-cover the telescope, he improvedit and was the first to use it as atool of astronomy, discovering themountains on the moon, phases ofVenus, moons of Jupiter, stars ofthe Milky Way, and sunspots androtation of the sun. Among hismany contributions to mechanics,he established the law of inertiaand proved that gravity acceleratesall bodies equally and that the peri-od of a small-amplitude pendulumis independent of the amplitude.He understood clearly that thelaws of mechanics hold in all unac-celerated frames, arguing that in-side an enclosed cabin it would beimpossible to detect the uniformmotion of a ship. This argumentappeared in his Dialogue on the TwoChief World Systems and was usedto show that the earth could per-fectly well be moving in orbitaround the sun without our beingaware of it in everyday life. Forpublishing this book, he was foundguilty of heresy by the Holy Officeof the Inquisition, and his bookwas placed on the Index of Prohib-ited Books — from which it wasnot removed until 1835.


Now consider an event, such as the explosion of a small firecracker, thatoccurs at position and time t as measured in S. Our problem is to calcu-late, in terms of (and the velocity ) the coordinates of thesame event, as measured in — accepting at first the classical ideas of spaceand time. First, since time is a universal quantity in classical physics, we knowthat Next, from Fig. 1.8 it is easily seen that and (and, similarly, although the z coordinate is not shown in the figure).Thus, according to the ideas of classical physics,

(1.31)

These four equations are often called the Galilean transformation afterGalileo Galilei, who was the first person known to have considered the invari-ance of the laws of motion under this change of coordinates. They transformthe coordinates of any event as observed in S into the correspondingcoordinates as observed in

If we had been given the coordinates and wanted to findwe could solve the equations (1.31) to give

(1.32)

Notice that the equations (1.32) can be obtained directly from (1.31) by ex-changing with and replacing by This is because therelation of S to is the same as that of to S except for a change in the signof the relative velocity.

The Galilean transformation (1.31) cannot be the correct relativistic rela-tion between and (For instance, we know from time dila-tion that the equation cannot possibly be correct.) On the other hand,the Galilean transformation agrees perfectly with our everyday experience andso must be correct (to an excellent approximation) when the speed is smallcompared to c. Thus the correct relation between and willhave to reduce to the Galilean relation (1.31) when is small.

To find the correct relation between and we consid-er the same experiment as before, which is shown again in Fig. 1.9. We havenoted before that distances perpendicular to v are the same whether measuredin S or Thus

(1.33)y¿ = y and z¿ = z

S¿.

x¿, y¿, z¿, t¿,x, y, z, tv>c x¿, y¿, z¿, t¿x, y, z, t

v

t¿ = tx¿, y¿, z¿, t¿.x, y, z, t,

S¿S¿-v.vx¿, y¿, z¿, t¿x, y, z, t

t = t¿ z = z¿ y = y¿ x = x¿ + vt¿

x, y, z, t,x¿, y¿, z¿, t¿

S¿.x¿, y¿, z¿, t¿x, y, z, t

t¿ = t

z¿ = z

y¿ = y

x¿ = x - vt

z¿ = zy¿ = yx¿ = x - vtt¿ = t.

S¿x¿, y¿, z¿, t¿vx, y, z, t,

x, y, z,

v

S

Both measured in S Measured in S�

S�

xvt x�

y� � y

P�

O�O

FIGURE 1.9The coordinate is measured in

The distances x and aremeasured at the same time t in theframe S.

vtS¿.x¿

TAYL01-001-045.I 12/10/02 1:51 PM

Section 1.11 • The Lorentz Transformation 27

exactly as in the Galilean transformation. In finding it is important to keepcareful track of the frames in which the various quantities are measured; in ad-dition, it is helpful to arrange that the explosion whose coordinates we are dis-cussing produces a small burn mark on the wall of the train at the point where it occurs. The horizontal distance from the origin to the mark at as measured in is precisely the desired coordinate Meanwhile, the samedistance, as measured in S, is (since x and are the horizontal dis-tances from O to and O to at the instant t, as measured in S). Thusaccording to the length-contraction formula (1.29),

or(1.34)

This gives in terms of x and t and is the third of our four required equations.Notice that if is small, and the relation (1.34) reduces to the first of theGalilean relations (1.31), as required.

Finally, to find in terms of and t, we use a simple trick. We canrepeat the argument leading to (1.34) but with the roles of S and reversed.That is, we let the explosion burn a mark at the point P on a wall fixed in S, andarguing as before, we find that

(1.35)

[This can be obtained directly from (1.34) by exchanging with and re-placing by ] Equation (1.35) is not yet the desired result, but we can com-bine it with (1.34) to eliminate and find Inserting (1.34) in (1.35), we get

Solving for we find that

or, after some algebra (Problem 1.37),

(1.36)

This is the required expression for in terms of x and t. When is muchsmaller than 1, we can neglect the second term, and since we get in agreement with the Galilean transformation, as required.

Collecting together (1.33), (1.34), and (1.36), we obtain our required fourequations.

(1.37) t¿ = g¢ t -vx

c2 ≤ z¿ = z

y¿ = y

x¿ = g1x - vt2

t¿ L t,g L 1,v>ct¿

t¿ = g¢ t -vx

c2 ≤

t¿ = gt -g2 - 1gv

x

t¿

x = g3g1x - vt2 + vt¿4t¿.x¿

-v.vx¿, t¿x, t

x = g1x¿ + vt2

S¿x, y, z,t¿

g L 1vx¿

x¿ = g1x - vt2

x - vt =x¿g

O¿P¿vtx - vt

x¿.S¿,P¿,O¿P¿

x¿,

TAYL01-001-045.I 12/10/02 1:51 PM

Hendrik Lorentz(1853–1928, Dutch)

Lorentz was the first to writedown the equations we now callthe Lorentz transformation, al-though Einstein was the first to in-terpret them correctly. He alsopreceded Einstein with the lengthcontraction formula (though, again,he did not interpret it correctly).He was one of the first to suggestthat electrons are present inatoms, and his theory of electronsearned him the 1902 Nobel Prizein physics.


These equations are called the Lorentz transformation, or Lorentz–Einsteintransformation, in honor of the Dutch physicist Lorentz, who first proposedthem, and Einstein, who first interpreted them correctly. The Lorentz trans-formation is the correct relativistic modification of the Galilean transforma-tion (1.31).

If one wants to know in terms of one can simply ex-change the primed and unprimed variables and replace by in the nowfamiliar way, to give

(1.38)

These equations are sometimes called the inverse Lorentz transformation.The Lorentz transformation expresses all the properties of space and

time that follow from the postulates of relativity. From it, one can calculate allof the kinematic relations between measurements made in different inertialframes. In the next two sections we give some examples of such calculations.

1.12 Applications of the Lorentz Transformation

In this section we give three examples of problems that can easily be analyzedusing the Lorentz transformation. In the first two we rederive two familiar re-sults; in the third we analyze one of the many apparent paradoxes of relativity.

Example 1.4

Starting with the equations (1.37) of the Lorentz transformation, derive thelength-contraction formula (1.29).

Notice that the length-contraction formula was used in our derivationof the Lorentz transformation.Thus this example will not give a new proof oflength contraction; it will, rather, be a consistency check on the Lorentztransformation, to verify that it gives back the result from which it wasderived.

Let us imagine, as before, measuring the length of a train (frame )traveling at speed relative to the ground (frame S). If the coordinates of theback and front of the train are and as measured in the train’sproper length (its length as measured in its rest frame) is

(1.39)

To find the length l as measured in S, we carefully position two observers onthe ground to observe the coordinates and of the back and front of thetrain at some convenient time t. (These two measurements must, of course,be made at the same time t.) In terms of these coordinates, the length l asmeasured in S is (Fig. 1.10)

l = x2 - x1 .

x2x1

l0 = l¿ = xœ2 - xœ

1

S¿,xœ2 ,xœ

1

vS¿

t = g¢ t¿ +vx¿c2 ≤

z = z¿ y = y¿ x = g1x¿ + vt¿2

-v,vz¿, t¿,x¿, y¿,x, y, z, t

TAYL01-001-045.I 12/10/02 1:51 PM

Now, consider the following two events, with their coordinates as measuredin S.

Event Description Coordinates in S

1 Back of train passes first observer2 Front of train passes second observer

We can use the Lorentz transformation to calculate the coordinates of eachevent as observed in

Event Coordinates in

12

(We have not listed the times and since they don’t concern us here.) Thedifference of these coordinates is

(1.40)

(Notice how the times and cancel out because they are equal.) Since thetwo differences in (1.40) are respectively and l, we conclude that

or

as required.

Example 1.5

Use the Lorentz transformation to rederive the time-dilation formula (1.18).In our discussion of time dilation we considered two events, a flash and

a beep, that occurred at the same place in frame

xœflash = xœ

beep

S¿,

l =l0g

l0 = gll¿ = l0

t2t1

xœ2 - xœ

1 = g1x2 - x12

tœ2tœ

1

xœ2 = g1x2 - vt22

xœ1 = g1x1 - vt12

S¿

S¿.

x2 , t2 = t1

x1 , t1

Section 1.12 • Applications of the Lorentz Transformation 29

v

S

x1 t1 t2 � t1x2

FIGURE 1.10If the two observers measure and at the same time then l = x2 - x1 .

(t1 = t2),x2

x1

TAYL01-001-045.I 12/10/02 1:51 PM


The proper time between the two events was the time as measured in

To relate this to the time

as measured in S, it is convenient to use the inverse Lorentz transformation(1.38), which gives

and

If we take the difference of these two equations, the coordinates anddrop out (since they are equal) and we get the desired result,

Example 1.6

A relativistic snake of proper length 100 cm is moving at speed tothe right across a table. A mischievous boy, wishing to tease the snake, holdstwo hatchets 100 cm apart and plans to bounce them simultaneously on thetable so that the left hatchet lands immediately behind the snake’s tail. Theboy argues as follows: “The snake is moving with Therefore, itslength is contracted by a factor

and its length (as measured in my rest frame) is 80 cm. This implies thatthe right hatchet will fall 20 cm in front of the snake, and the snake will beunharmed.” (The boy’s view of the experiment is shown in Fig. 1.11.) Onthe other hand, the snake argues thus: “The hatchets are approaching mewith and the distance between them is contracted to 80 cm. SinceI am 100 cm long, I will be cut in pieces when they fall.” Use the Lorentztransformation to resolve this apparent paradox.

Let us choose two coordinate frames as follows: The snake is at rest inframe with its tail at the origin and its head at Thetwo hatchets are at rest in frame S, the left one at the origin and theright one at x = 100 cm.

x = 0x¿ = 100 cm.x¿ = 0S¿

b = 0.6,

g =131 - b2

=121 - 0.36

=54

b = 0.6.

v = 0.6c

¢t = tbeep - tflash = g1tœbeep - tœ

flash2 = g ¢t0

xœflash

xœbeep

tflash = g¢ tœflash -

vxœflash

c2 ≤

tbeep = g¢ tœbeep -

vxœbeep

c2 ≤

¢t = tbeep - tflash

¢t0 = ¢t¿ = tœbeep - tœ

flash

S¿,

TAYL01-001-045.I 12/10/02 1:51 PM

Section 1.12 • Applications of the Lorentz Transformation 31

xL � 0

v

x � 80 cm

t � 0

xR � 100 cm

FIGURE 1.11As seen in the boy’s frame S, thetwo hatchets bouncesimultaneously (at ) 100 cmapart. Since the snake is 80 cm long,it escapes injury.

t = 0

As observed in frame S, the two hatchets bounce simultaneously atAt this time the snake’s tail is at and his head must therefore be

at [You can check this using the transformation with and you will find that as required.] Thus,as observed in S, the experiment is as shown in Fig. 1.11. In particular, theboy’s prediction is correct and the snake is unharmed. Therefore, the snake’sargument must be wrong.

To understand what is wrong with the snake’s argument, we must ex-amine the coordinates, especially the times, at which the two hatchetsbounce, as observed in the frame The left hatchet falls at and

According to the Lorentz transformation (1.37), the coordinates ofthis event, as seen in are

and

As expected, the left hatchet falls immediately beside the snake’s tail, at timeas shown in Fig. 1.12(a).

On the other hand, the right hatchet falls at and Thus, as seen in it falls at a time given by the Lorentz transformation as

We see that, as measured in the two hatchets do not fall simultaneously.Since the right hatchet falls before the left one, it does not necessarily haveto hit the snake, even though they were only 80 cm apart (in this frame). Infact, the position at which the right hatchet falls is given by the Lorentztransformation as

and, indeed, the hatchet misses the snake, as shown in Fig. 1.12(b).The resolution of this paradox and many similar paradoxes is seen to be

that two events which are simultaneous as observed in one frame are not neces-sarily simultaneous when observed in a different frame. As soon as one recog-nizes that the two hatchets fall at different times in the snake’s rest frame, thereis no longer any difficulty understanding how they can both miss the snake.

xœR = g1xR - vtR2 = 5

4 1100 cm - 02 = 125 cm

S¿,

tœR = g¢ tR -

vxR

c2 ≤ =54

¢0 -10.6c2 * 1100 cm2

c2 ≤ = -2.5 ns

S¿,xR = 100 cm.tR = 0

tœL = 0,

xœL = g1xL - vtL2 = 0.

tœL = g¢ tL -

vxL

c2 ≤ = 0

S¿,xL = 0.

tL = 0S¿.

x¿ = 100 cm,t = 0,x = 80 cmx¿ = g1x - vt2;x = 80 cm.

x = 0t = 0.

x�L � 0

(a) (b)

t �L � 0 t �R � �2.5 ns

x� � 100 cm x�R � 125 cm

FIGURE 1.12As observed in both hatchetsare moving to the left. The righthatchet falls before the left one, andeven though the hatchets are only80 cm apart, this lets them fall atpositions that are 125 cm apart.

S¿,

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1.13 The Velocity-Addition Formula

In Section 1.3 we discussed the classical velocity-addition formula. This relatesthe velocity u of a body or signal, relative to a frame S, and its value relativeto a second frame or equivalently,

(1.41)

Here v is the velocity of relative to S, and the formula asserts that in classi-cal physics, relative velocities add and subtract like vectors. Notice that here, aselsewhere, we use u and for the velocities of a body or signal relative to thetwo frames, while v denotes the relative velocity of the frames themselves.

The Michelson–Morley experiment shows that the classical formula(1.41) cannot be correct, since it would contradict the universality of the speedof light. In this section we use the Lorentz transformation to derive the correctrelativistic velocity-addition formula.

Let us imagine some moving object whose velocity we wish to discuss.(For example, this object could be a space rocket, a subatomic particle, or apulse of light.) We consider two neighboring points on its path, as in Fig. 1.13.We denote by and the coordinates of thesetwo points, as measured in S, and by and the times at which the objectpasses them. The velocity as measured in S, is then given by

(1.42)

where and so on (and these equations may, strictly speaking,be valid only in the limit that the two points are close together, ). Thevelocity relative to is defined in the same way, using the coordinates andtimes measured in That is, and so on.

We can now use the Lorentz transformation to relate the coordinatesand times of S to those of and then, using definition (1.42), relate the corre-sponding velocities. First, according to the Lorentz transformation (1.37)

and

Subtracting these equations, we find that

From these we can calculate the components of First

uœx =

¢x¿¢t¿

=g1¢x - v ¢t2g1¢t - v ¢x>c22

uœ.

¢t¿ = g¢¢t -v ¢x

c2 ≤¢x¿ = g1¢x - v ¢t2, ¢y¿ = ¢y, ¢z¿ = ¢z,

xœ1 = g1x1 - vt12, yœ

1 = y1 , zœ1 = z1 , tœ

1 = g¢ t1 -vx1

c2 ≤

xœ2 = g1x2 - vt22, yœ

2 = y2 , zœ2 = z2 , tœ

2 = g¢ t2 -vx2

c2 ≤

S¿,

uœx = ¢x¿>¢t¿,S¿.

S¿uœ¢t : 0

¢x = x2 - x1 ,

ux =¢x

¢t, uy =

¢y

¢t, uz =

¢z

¢t

u = 1ux , uy , uz2,t2t1

r2 = 1x2 , y2 , z22r1 = 1x1 , y1 , z12

uœ

S¿

uœ = u - v

S¿, u = uœ + v,uœ

r2, t2

r1, t1

�r

FIGURE 1.13The velocity of an object isu = ¢r>¢t.

TAYL01-001-045.I 12/10/02 1:51 PM

Section 1.13 • The Velocity-Addition Formula 33

or, canceling the factors of and dividing top and bottom by

(1.43)

Similarly,

or, dividing the top and bottom by

(1.44)

Notice that is not equal to even though this is because thetimes and are unequal.

Equations (1.43) and (1.44) are the relativistic velocity-addition formulas,or velocity transformation. Notice that if both u and are much less than c, wecan ignore the term in the denominators and put to give

and

These are, of course, the components of the classical addition formula

The inverse velocity transformation, giving u in terms of can be ob-tained from (1.43) and (1.44) by exchanging primed and unprimed variablesand replacing by in the familiar way.

Example 1.7

A rocket traveling at speed relative to the earth shoots forward a beamof particles with speed relative to the rocket.What is the particles’ speedrelative to the earth?

Let S be the rest frame of the earth and that of the rocket, with x andaxes both aligned along the rocket’s velocity. The relative speed of the two

frames is We are given that the particles are traveling along the axis with speed (relative to ) and we want to find their speed u rel-ative to S.The classical answer is, of course, that that is, be-cause the two velocities are collinear, and simply add in classical physics.

The correct relativistic answer is given by the inverse of (1.43) (fromwhich we omit the subscripts x, since all velocities are along the x axis).

(1.45) =0.9c + 0.8c

1 + 10.9 * 0.82 =1.7

1.72 c L 0.99c

u =u¿ + v

1 + u¿ v>c2

vu¿u = u¿ + v = 1.7c,

S¿u¿ = 0.9cx¿v = 0.8c.

x¿S¿

0.9c0.8c

-v,v

uœ,uœ = u - v.

uœy L uy and uœ

z L uz

uœx L ux - v

g L 1ux v>c2v

¢t¢t¿¢y¿ = ¢y;uy ,uœ

y

uœy =

uy

g11 - ux v>c22 and, similarly, uœz =

uz

g11 - ux v>c22

¢t,

uœy =

¢y¿¢t¿

=¢y

g1¢t - v ¢x>c22

uœx =

ux - v

1 - ux v>c2

¢t,g

TAYL01-001-045.I 12/10/02 1:51 PM


The striking feature of this answer is that when we “add” torelativistically, we get an answer that is less than c. In fact, it is fairly

easy to show that for any value of that is less than c, the speed u is also lessthan c (see Problem 1.47); that is, a particle whose speed is less than c in oneframe has speed less than c in any other frame.

Example 1.8

The rocket of Example 1.7 shoots forward a signal (for example, a pulse oflight) with speed c relative to the rocket.What is the signal’s speed relative tothe earth?

In this case Thus according to (1.45)

(1.46)

That is, anything that travels at the speed of light in one frame does the sameas observed from any other frame. (We have proved this here only for thecase that u is in the same direction as However, the result is true for any di-rection; for another example, see Problem 1.48.) We can paraphrase this tosay that the speed of light is invariant as we pass from one inertial frame toanother. This is, of course, just the second postulate of relativity, which led usto the Lorentz transformation in the first place.

1.14 The Doppler Effect�

�The Doppler effect is an important phenomenon, with applications that range fromatomic physics, through police radar traps, to the expanding universe. Nevertheless, wewill not be using it again, so, if you are pressed for time, you could omit this sectionwithout loss of continuity.

It is well known in classical physics that the frequency (and hence pitch) ofsound changes if either the source or receiver is put into motion, a phenome-non known as the Doppler effect, after the Austrian physicist ChristianDoppler, 1803–1853, who investigated the effect. There is a similar Dopplereffect for light: The frequency (and hence color) of light is changed if thesource and receiver are put into relative motion. However, the Doppler effectfor light differs from that for sound in two important ways. First, since lighttravels at speed c, one should treat the phenomenon relativistically, and therelativistic Doppler formula differs from its nonrelativistic form because oftime dilation. Second, you will probably recall that the nonrelativisticDoppler formula for sound has different forms according to whether thesource or receiver is moving. This difference is perfectly reasonable for soundthat propagates in air, so that the Doppler shift can legitimately depend onwhether it is the source or receiver that moves through the air. On the otherhand, we know that light (in the vacuum) has no medium in which it propa-gates; and, in fact, relativity has shown us that it can make no differencewhether it is the source or receiver that is “really” moving. Thus the Dopplerformula for light must be the same for a moving source as for a moving re-ceiver. In this respect, the Doppler formula for light is simpler than itsnonrelativistic counterpart for sound.

v.

u =u¿ + v

1 + u¿ v>c2 =c + v

1 + v>c = c

u¿ = c.

u¿v = 0.8c

u¿ = 0.9c

TAYL01-001-045.I 12/10/02 1:51 PM

Section 1.14 • The Doppler Effect 35

The derivation of the relativistic Doppler formula is very similar to thenonrelativistic argument. Let us consider a train (frame ) that is traveling atspeed relative to the ground (frame S) and whose headlamp is a source oflight with frequency as measured in the rest frame of the source. We nowimagine an observer Q at rest on the ground in front of the train and wish tofind the frequency of the light that Q observes.

We first consider the experiment as seen in frame S, in which it is thesource that is moving. In Fig. 1.14(a) and (b) we have shown two successivewave crests (numbered 1 and 2) as they leave the train’s headlight. If the time(as measured in S) between the emission of these crests is then during thistime the first crest will move a distance to the right. During this same timethe train will advance a distance and the distance between successivecrests is therefore

(1.47)

As seen from S, the distance between successive crests is shortened by asa result of the train’s motion. Since the crests are all approaching with speed cand are a distance apart, the frequency with which the observer Qreceives them is

(1.48)

where, as usual,Now, is the time between emission of successive crests as measured in

S. The corresponding time as measured in is the proper time betweenthe two events, since they occur at the same place in Therefore,

(1.49)

and, from (1.48),

(1.50)fobs =1

11 - b2g ¢t¿

¢t = g ¢t¿

S¿.S¿¢t¿

¢tb = v>c.

fobs =c

l=

c

1c - v2¢t=

111 - b2¢t

fobsl

v ¢t

l = c ¢t - v ¢t.

lv ¢t,c ¢t

¢t,

fobs ,

fsce ,v

S¿

S

Q

(a)

(b)

S�Crest 1

S S� Crest 2 Crest 1�

c�t

v�t FIGURE 1.14As seen in frame S, the sourcemoves with speed and thereceiver Q is at rest.

v

TAYL01-001-045.I 12/10/02 1:51 PM


Finally, the frequency measured at the source is just Thus(1.50) implies that

(1.51)

where we have added the parenthesis “(approaching)” to emphasize that thisformula applies when the source is approaching the observer.

The relativistic formula (1.51) differs from its nonrelativistic counterpartonly by the factor of which arose from the time dilation (1.49). In particu-lar, for slow speeds, with we can use the nonrelativistic formula

as one might expect.It is often convenient to rewrite (1.51), replacing the factor of by

(1.52)

to give

(1.53)

We will see shortly that this formula, which we have so far derived only for amoving source, in fact holds whether it is the source or observer that is moving.

The formula (1.53) applies to a source that is approaching the observer.If the source is moving away from the observer, we have only to change thesign of to give

(1.54)

The formulas (1.53) and (1.54) are easy to memorize. In particular, it is easy toremember which is which, since both numerator and denominator lead to theexpected rise in frequency when source and observer are approaching eachother, and the expected drop when they are receding.

We have here analyzed only the cases that the source moves directly to-ward or away from the observer. The case that the source moves obliquely tothe observer is more complicated and is discussed in Problem 1.53.

An important example of the Doppler effect is the famous “redshift” ofthe light from distant stars. A star emits and absorbs light at certain frequen-cies that are characteristic of the elements in the star. Thus, by analyzing thespectrum of light from a star, one can identify which elements it contains. Oncethese elements are identified, one can go further. By seeing whether the char-acteristic frequencies are shifted up or down (as compared to those from asource at rest in the observatory) one can tell whether the star is moving to-ward or away from us. The American astronomer Edwin Hubble found thatthe light from distant galaxies is shifted down in frequency, or redshifted (sincered is at the low frequency end of the visible spectrum), indicating that mostgalaxies are moving away from us. Hubble also found that the speeds of reces-sion of galaxies are roughly proportional to their distances from us, a discoverynow called Hubble’s law. This implied that the universe is expanding uniform-

fobs = A1 - b1 + b

fsce 1receding2

v

fobs = A1 + b1 - b

fsce 1approaching2

1g

= 31 - b2 = 411 - b211 + b2

1>gfobs = fsce>11 - b2, g L 1,1>g,

fobs =fsce

11 - b2g 1approaching2

fsce = 1>¢t¿.fsce

TAYL01-001-045.I 12/10/02 1:51 PM

Edwin Hubble(1889–1953, American)

Raised in midwest America, theyoung Hubble wanted to study as-tronomy, but, at the insistence ofhis father, obtained a law degree atOxford University. After his fatherdied, Hubble returned to collegeto study astronomy and quicklybecame one of the world’s fore-most observational astronomers.Working with what was then thelargest telescope in the world,the 100-inch-diameter reflectingtelescope at Mt. Wilson Observa-tories in California, Hubble estab-lished the two most importantresults of twentieth-century as-tronomy. He demonstrated thatour Milky Way galaxy is only oneof myriads of galaxies in the uni-verse. And he showed that distantgalaxies are receding from us witha speed proportional to their dis-tance, thus establishing that theuniverse is expanding.

Section 1.14 • The Doppler Effect 37

ly; it also provided a convenient way to find the distance of many galaxies,since measurement of a Doppler shift is usually much easier than the directmeasurement of a distance.

Example 1.9

It is found that light from a distant galaxy is shifted down in frequency (red-shifted) by a factor of 3; that is, Is the galaxy approaching usor receding? And what is its speed?

Since the observed frequency is less than the galaxy isreceding, and we use (1.54) to give

or, solving for

That is, the galaxy is receding from us at

The Doppler effect for light has many uses here on the earth. One appli-cation that is all too familiar to many of us is the Doppler radar used by policeto measure our car’s speed. This actually involves two Doppler shifts: As“seen” by our car, the radar is shifted up in frequency as we approach the gun,and the tiny electric currents induced in the car’s body are of slightly higherfrequency than that of the gun. This means that the reflected signal sent backby our car has the same higher frequency, and this signal (from our movingcar) is then raised again as seen by the radar receiver. The total shift is verysmall (since ), but is easily measured and immediately converted togive the car’s speed. In atomic physics a process called laser cooling exploitsthe Doppler effect to slow the thermal motions of the atoms or molecules in agas. The same Doppler effect is also a significant nuisance to atomic spectro-scopists, since it broadens the spectral lines that they wish to measure, as thefollowing example illustrates.

Example 1.10

The atoms in hot sodium vapor give out light of wavelength (measured in the atoms’ rest frame). Since atoms in a vapor move randomlywith speeds up to and even higher, this light is observed with variousdifferent Doppler shifts, depending on the atoms’ speeds and directions. Tak-ing as the atoms’ maximum speed, find the range of wavelengthsobserved.

The minimum and maximum frequencies observed come from atomsmoving directly away from and toward the observer with speed or Since is so small, we can ignore the factor of in (1.51) andthe extreme frequencies are given by

fobs =fsce

1 ; b

gbb = 10-6.v = 300 m>s

300 m>s300 m>s

lsce = 589 nm

b V 1

0.8c.

b = 0.8

b,

A1 - b1 + b

=13

fsce ,fobs

fobs>fsce = 1>3.

TAYL01-001-045.I 12/10/02 1:51 PM

S�

S�

Crest 2

Crest 2 Crest 1

Crest 1

Q

v�t�c�t�

(a)

(b)

� �

FIGURE 1.15As seen in the source isstationary with frequency andthe observer Q is moving at speed to the left.

vfsource

S¿,


Since (both for source and observer), this implies maximum andminimum wavelengths given by

We can write this as

where the maximum shift in the wavelength is

This is a very small shift of wavelength as we should have expected,since is so small compared to c. Nevertheless, such a shift is easily observedwith a good spectrometer. It means that what would otherwise be observedas a sharp spectral line with wavelength is smeared out between

This phenomenon is called Doppler broadening and is one of theproblems that has to be overcome in precise measurement of wavelengths.

We mentioned earlier that the relativistic Doppler shift for light must bethe same whether we view the source as moving and the observer at rest, orvice versa. We check this in our final example.

Example 1.11

Rederive the Doppler formula (1.53) working in the rest frame of thesource (that is, taking the view that the observer is moving).

We consider again two successive wave crests, but examine theirreception by the observer Q, as shown in Fig. 1.15.As measured in the dis-tance between the two crests is the wavelength (since is thefrequency measured in ), and the time between Q’s meeting the crests wedenote by During the time crest 2 moves a distance to the rightand the observer Q moves a distance to the left. The sum of these twodistances is just

c ¢t¿ + v ¢t¿ = l¿ =c

fsce

l¿,v ¢t¿

c ¢t¿¢t¿,¢t¿.S¿

fscel¿ = c>fsce

S¿,

S¿

lsce ; ¢l.lsce

v

¢l = blsce = 110-62 * 1589 nm2 L 6 * 10-4 nm

¢l

lobs = lsce ; ¢l

lobs = lsce11 ; b2

l = c>f

TAYL01-001-045.I 12/10/02 1:51 PM

Checklist for Chapter 1 39

from which we find that

(1.55)

Now, the frequency with which Q observes wave crests is

(1.56)

where is the time between arrival of the two successive crests as measuredby the observer Q. Since these two events occur at the same place in the ob-server’s frame S, is the proper time and Substituting into(1.56) and then using (1.55), we find that

(1.57)

Apart from the factor this is the nonrelativistic formula for a moving re-ceiver. For our present purposes, the important point is that we can replace using (1.52), and after a little algebra, we obtain exactly our previous answer(1.53), as you should check for yourself. As anticipated, the relativisticDoppler shift for light is the same for a moving observer as for a movingsource, and depends only on their relative velocity v.

g

g,

fobs =g

¢t¿= g11 + b2fsce

¢t¿ = g ¢t.¢t

¢t

fobs =1¢t

¢t¿ =c

1c + v2fsce=

111 + b2fsce

CHECKLIST FOR CHAPTER 1CONCEPT DETAILS

Relativity of measurements Sec. 1.1

The classical velocity-addition formula (1.1)

Invariance of Newton’s laws in Sec. 1.3classical physics

Noninvariance of the speed of light in Sec. 1.4classical physics

Michelson–Morley experiment� (Sec. 1.5)

The postulates of relativity Definition of inertial framesThe two postulates (Sec. 1.6)

Time dilation (1.18)

The speed limit for the relative motion Sec. 1.8of inertial frames is c

Proper time Sec. 1.8

Lorentz–Fitzgerald contraction (1.29)

Proper length Sec. 1.10

The Galilean transformation (1.31)

The Lorentz transformation (1.37)

Relativistic velocity addition (1.43) & (1.44)

The Doppler effect� (1.53)fobs = fsce411 + b2>11 - b2uœ

x =ux - v

1 - ux v>c2 , uœy =

uy

g11 - ux v>c22 , uœz =

uz

g11 - ux v>c22

x¿ = g1x - vt2, y¿ = y, z¿ = z, t¿ = g1t - vx>c22x¿ = x - vt, y¿ = y, z¿ = z, t¿ = t

l = l0>g

¢t = g ¢t0

u = u¿ + v

TAYL01-001-045.I 12/10/02 1:51 PM


PROBLEMS FOR CHAPTER 1

The problems for each chapter are arranged according tosection number. A problem listed for a given section re-quires an understanding of that section and earlier sections,but not of later sections. Within each section, problems arelisted in approximate order of difficulty. A single dot (•) in-dicates straightforward problems involving just one mainconcept and sometimes requiring no more than substitutionof numbers in the appropriate formula. Two dots (••) iden-tify problems that are slightly more challenging and usuallyinvolve more than one concept. Three dots (•••) indicateproblems that are distinctly more challenging, either be-cause they are intrinsically difficult or involve lengthy cal-culations. Needless to say, these distinctions are hard todraw and are only approximate.

Answers to odd-numbered problems are given at theback of the book.

SECTION 1.2 (The Relativity of Orientation and Origin)

1.1 •• At time a block is released from the point Oon the slope shown in Fig. 1.16. The block acceleratesdown the slope, overcoming the sliding friction (coef-ficient ). (a) Choose axes as shown, and resolvethe equation into its x and y components.Hence find the block’s position as a function oftime, and the time it takes to reach the bottom.(b) Carry out the solution using axes with horizontal and vertical, and show that you get thesame final answer. Explain why the solution usingthese axes is less convenient.

Oy¿Ox¿Ox¿ y¿,

1x, y2©F = maOxym

t = 0,

1.2 •• A block slides down the slope of Fig. 1.17 from Owith initial speed The sliding friction (coefficient

) brings the block to rest in a time T. (a) Using theaxes shown, find T. (b) Solve this problem using axes

with horizontal and vertical, and ex-plain why the solution in this frame is less convenient(although it produces the same final answer, ofcourse).

Oy¿Ox¿Ox¿ y¿

mv0 .

above the horizontal (Fig. 1.18). (a) Choosing axesas shown, write down the components of the ball’sinitial velocity and its acceleration g. Hence findthe ball’s position as a function of time andshow that its range up the slope is

Note that when the ball lands, (b) Show thatyou get the same final answer if you use axes with horizontal and vertical. Discuss brieflythe merits of the two choices of axes. (You can findseveral useful trig identities in Appendix B.)

SECTION 1.3 (Moving Reference Frames)

1.4 • A physics lecture demonstration uses a small can-non mounted on a cart that moves at constant veloci-ty v across the floor. At what angle should thecannon point (measured from the horizontal floor ofthe cart) if the cannonball is to land back in themouth of the cannon? Explain clearly your choice offrame of reference.

1.5 • Two students are riding on a flatcar, traveling atspeed u to the right. One is standing at the right endof the car, holding a small vertical hoop at a height habove the car’s floor. The other has a catapult withwhich he plans to shoot a pellet from the car’s floorand through the hoop. He is aiming the catapult atan angle above the floor of the car. If he wants thepellet to be traveling horizontally when it goesthrough the hoop, with what speed (relative to thecar) should he fire the pellet? How far horizontallyshould he be from his companion? Explain yourchoice of reference frame.

1.6 • Consider a head-on, elastic collision between twobodies whose masses are m and M, with Itis well known that if m has speed and M is initiallyat rest, m will bounce straight back with its speedunchanged, while M will remain at rest (to anexcellent approximation). Use this fact to predict thefinal velocities if M approaches with speed and mis initially at rest.

[HINT: Consider the reference frame attached to M.]

1.7 • Use the method of Problem 1.6 to predict the finalvelocities if two bodies of masses m and M, with

approach one another both traveling atspeed (relative to the lab) and undergo a head on,elastic collision.

1.8 •• A policeman is chasing a robber. Both are in carstraveling at speed and the distance between themv

v0

m V M,

v0

v0

m V M.

u

u

Oy¿Ox¿Ox¿ y¿

y = 0.

R =2v0

2 sin u cos1u + f2g cos2 f

1x, y2v0

f

l

y

xO

�

FIGURE 1.16(Problem 1.1)

x

y

�

Ov

0


R

�

O

x

y


1.3 ••• At time a ball is thrown with speed at anangle above a slope that is itself inclined at an angleu

v0t = 0

TAYL01-001-045.I 12/10/02 1:51 PM

Problems for Chapter 1 41

A

D

B C

v


is l. The policeman wishes to shoot the robber with agun whose muzzle velocity is At what angle above the horizontal should he point his gun? Firstsolve this problem using coordinates traveling withthe policeman, as shown in Fig. 1.19. Then sketch thesolution using coordinates fixed to the ground; is theangle of the gun the same as the angle of the bullet’sinitial velocity in this frame? (The advantages of thefirst frame are not overwhelming; nevertheless, it isclearly the natural choice for the problem.)

SECTION 1.4 (Classical Relativity and the Speed of Light)

1.9 • Let us assume the classical ideas of space and timeare correct, so that there could only be one frame, the“ether frame,” in which light traveled at the samespeed c in all directions. It seemed unlikely that theearth would be exactly at rest in this frame and onemight reasonably have guessed that the earth’s speed

relative to the ether frame would be at least of theorder of our orbital speed around the sun

(a) What would be the observedspeed (on earth) of a light wave traveling parallel tov? (Give your answer in terms of c and and thensubstitute numerical values.) (b) What if it were trav-eling antiparallel to v? (c) What if it were travelingperpendicular to v (as measured on earth)? The ac-cepted value of c is (to five signifi-cant figures).

1.10 •• At standard temperature and pressure sound trav-els at speed relative to the air throughwhich it propagates. Four students, A, B, C, D, posi-tion themselves as shown in Fig. 1.20, with A, B, C in astraight line and D vertically above B. A steady windis blowing with speed along the line ABC.If B fires a revolver, what are the speeds with whichthe sound will travel to A, C, and D (in the referenceframe of the observers)? Discuss whether the differ-ences in your answers could be detected.

v = 30 m>s

u = 330 m>s

2.9979 * 108 m>s

v,

1v L 3 * 104 m>s2.v

uu0 .

1.11 ••• It is well known that the speed of sound in air isat standard temperature and pressure.

What this means is that sound travels at speed u in alldirections in the frame S where the air is at rest. In any

u = 330 m>s

other frame moving relative to S, its speed is not uin all directions. To verify this, some students set up aloudspeaker L and receiver R on an open flatcar, asin Fig. 1.21; by connecting the electrical signals fromL and R to an oscilloscope, they can measure the timefor a sound to travel from L to R and hence find itsspeed (relative to the car). (a) Derive an expres-sion for in terms of and where is the car’sspeed through the air and is the angle between vand LR. (We call this since it is the angle between vand the velocity of the sound measured in theframe of the car.)

[HINT: Draw a velocity-addition triangle to represent therelation The law of cosines should give you aquadratic equation for ]

(b) If the students vary the angle from 0 to what are the largest and smallest values of (c) Ifis about (roughly ), what will be theapproximate percent variation in Would this bedetectable?

SECTION 1.5 (The Michelson–MorleyExperiment�)

1.12 • In the discussion of the Michelson–Morley experi-ment, we twice used the binomial approximation

(1.58)

which holds for any number n and any x much small-er than 1 (that is, ). (In the examples, nwas and and was of order )The binomial approximation is frequently useful inrelativity, where one often encounters expressions ofthe form with x small. Make a table show-ing and its approximation for

and 0.1, 0.01, and 0.001. In each casefind the percentage by which the approximationdiffers from the exact result.

1.13 • Do the same tasks as in Problem 1.12, but for thecase In this case give an exact expression forthe difference between the exact and approximateforms. Explain why the approximation gets betterand better as

1.14 • Use the binomial approximation (1.58) (Problem1.12) to evaluate Can you evalu-ate this directly on your calculator?

1.15 • Tom Sawyer and Huck Finn can each row a boat atin still water. Tom challenges Huck to a race in

which Tom is to row the 2000 ft across the Mississippito a point exactly opposite their starting point andback again, while Huck rows to a point 2000 ft direct-ly downstream and back up again. If the Mississippiflows at which boy wins and by how long?

1.16 • An airline, all of whose planes fly with an airspeedof serves three cities, A, B, and C, where B200 mi>h,

3 ft>s,

5 ft>s

11 - 10-202-1 - 1.

x : 0.

n = 2.

x = 0.5,n = - 12

1 - nx11 - x2n11 - x2n10-8.x = b2-

12 ,-1

ƒx ƒ V 1

11 - x2n L 1 - nx

u¿?30 mi>h15 m>s vu¿ ?

180°,u¿u¿.

u = uœ + v.

uœ,u¿

u¿vu¿,u, v,u¿

u¿

S¿,

y

x

l


R

L v� � FIGURE 1.21(Problem 1.11)

TAYL01-001-045.I 12/10/02 1:51 PM


is 320 mi due east of A, and C is the same distancedue north of A. On a certain day there is a steadywind of from the east. (a) What is the timeneeded for a round trip from A to B and back?(b) What is it from A to C and back?

1.17 • In one of the early (1881) versions of Michelson’sinterferometer, the arms were about 50 cm long.What would be the expected shift when he rotat-ed the apparatus through assuming that

and that the expected speed of the earthrelative to the ether was (In this casethe expected shift was so small that no one regardedhis failure to observe a shift as conclusive.)

1.18 •• One of the difficulties with the Michelson–Morleyexperiment is that several extraneous effects (me-chanical vibrations, variations in temperature, etc.)can produce unwanted shifts in the interference pat-tern, masking the shift of interest. Suppose, for exam-ple, that during the experiment the temperature ofone arm of the interferometer were to rise by This would increase the arm’s length by where is the arm’s coefficient of expansion. Provethat this temperature change would, by itself, cause ashift For the dimensions given inProblem 1.17 and taking (the coefficientfor steel) and show that the resultingshift is much larger than the expected shiftdue to the earth’s motion. Obviously a successful ex-periment requires careful temperature control!

SECTIONS 1.8 and 1.9 (The Relativity of Time andEvidence for Time Dilation)

1.19 • An athlete runs the 100-meter dash at 10 m/s. Howmuch will her watch gain or lose, as compared toground-based clocks, during the race?

[HINT: You will need to use the binomial approximation(1.6).]

1.20 • A space vehicle travels at (about) relative to the earth. How much time

will its clocks gain or lose, as compared to earth-based clocks, in a day?

1.21 • (a) What must be one’s speed, relative to a frame S,in order that one’s clocks will lose 1 second per day asobserved from S? (b) What if they are to lose 1minute per day?

1.22 • A space explorer sets off at a steady to adistant star. After exploring the star for a short timehe returns at the same speed and gets home after atotal absence of 80 years (as measured by earthboundobservers). How long do his clocks say that he wasgone, and by how much has he aged? Note: This isthe twin paradox discussed in Example 1.3. It is easyto get the righ t answer by judicious insertion of a fac-tor in the right place, but to understand the resultyou need to recognize that it involves three inertialframes: the earthbound frame S, the frame of theoutward-bound rocket, and the frame of the re-turning rocket. You can write down the time-dilation

S–S¿

g

v = 0.95c

200,000 mi>h 100,000 m>s

¢N L 0.2,¢T L 0.01°C,

a L 10-5>°C¢N = 2al ¢T>l.a

¢l = al ¢T,¢T.

3 * 104 m>s?l = 590 nm

90°,¢N

120 mi>hformula (1.14) for each of the two halves of the jour-ney, and add these to give the desired relation. (No-tice that the experiment is not symmetrical betweenthe explorer and his friends who stay behind on earth— the earthbound clocks stay at rest in a single iner-tial frame, but the rocket’s clock and crew occupy atleast two different frames. This is what allows the re-sult to be unsymmetrical.)

1.23 • (When he returns his Hertz rent-a-rocket after oneweek’s cruising in the galaxy, Mr. Spock is shocked tobe billed for three weeks’ rental. Assuming that hetraveled straight out and then straight back, always atthe same speed, how fast was he traveling? (See thenote in Problem 1.22.)

1.24 •• (a) Use the binomial approximation (1.6) to provethe following useful approximation:

valid when (b) Derive a corresponding ap-proximation for (c) When is close to 1 ( closeto c) these approximations are, of course, useless; inthis case show that if with then

1.25 •• Two perfectly synchronized clocks A and B are atrest in S, a distance d apart. If we wanted to verifythat they really are synchronized, we might try using athird clock, C. We could bring C close to A and checkthat A and C agree, then move C over to B and checkthe agreement of B and C. Unfortunately, this proce-dure is suspect since clock C will run differently whileit is being moved. (a) Suppose that A and C are foundto be in perfect agreement and that C is then movedat constant speed to B. Derive an expression for thedisagreement between B and C, in terms of and d.What is if and (b) Showthat the method can nevertheless be made satisfacto-ry to any desired accuracy by moving clock C slowlyenough; that is, we can make as small as we pleaseby choosing sufficiently small.

1.26 •• A group of mesons (pions) is observed travelingat speed in a particle-physics laboratory.(a) What is the factor for the pions? (b) If the pions’proper half-life is what is their half-lifeas observed in the lab frame? (c) If there were initial-ly 32,000 pions, how many will be left after they havetraveled 36 m? (d) What would be the answer to (c) ifone ignored time dilation?

1.27 •• Muons are subatomic particles that are producedseveral miles above the earth’s surface as a result ofcollisions of cosmic rays (charged particles, such asprotons, that enter the earth’s atmosphere fromspace) with atoms in the atmosphere. These muonsrain down more-or-less uniformly on the ground, al-though some of them decay on the way since themuon is unstable with a proper half-life of about

In a certain experiment amuon detector is carried in a balloon to an altitude of2000 m, and in the course of 1 hour it registers 650muons traveling at toward the earth. If an iden-tical detector remains at sea level, how many muons

0.99c

11 ms = 10-6 s.21.5 ms.

1.8 * 10-8 s,g

0.8cp

vt

d = 1000 km?v = 300 m>stvt

v

g L 1>12e .e V 1,b = 1 - e,

vb1>g.b V 1.

g L 1 + 12 b2

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would you expect it to register in 1 hour? (Rememberthat after n half-lives the number of muons survivingfrom an initial sample of is and don’t forgetabout time dilation). This was essentially the methodused in the first tests of time dilation, starting in the1940’s.

1.28 ••• Time dilation implies that if a clock moves rela-tive to a frame S, careful measurements made by ob-servers in S [as in Fig. 1.22(a), for example] will findthat the clock runs slow. This is not at all the samething as saying that a single observer in S will see theclock running slow; and the latter statement is, in fact,not always true. To understand this, remember thatwhat we see is determined by the light as it arrives atour eyes. Consider the observer Q in Fig. 1.22(b) andsuppose that as the clock moves from A to B, it regis-ters the passage of a time As measured in S, thetime between these two events (“clock at A” and“clock at B”) is of course However, B iscloser to Q than A is; thus light from the clock whenat B will reach Q in a shorter time than will light fromthe clock when at A. Therefore, the time betweenQ’s seeing the clock at A and seeing it at B is less than

(a) Prove that in fact

(Prove both equalities.) Since is less than theobserver Q actually sees the clock running fast.(b) What will Q see once the clock has passed her?That is, find the new value of when the clock ismoving away from Q.

Your answers here are closely related to theDoppler effect discussed in Section 1.14.The moral ofthis problem is that one must be very careful how onestates (and thinks about) time dilation. It is safe to say“moving clocks are observed to run slow” [where to“observe” means to “measure carefully” as inFig. 1.22(a)], but it is certainly wrong to say “movingclocks are seen to run slow.”

tsee

t0 ,tsee

tsee = t11 - b2 = t0 A1 - b1 + b

t.

tsee

t = gt0 .

t0 .

N0>2n,N0

SECTION 1.10 (Length Contraction)

1.29 • A rocket of proper length 40 m is observed to be32 m long as it rushes past the earth. What is its speedrelative to the earth?

1.30 • A relativistic conveyor belt is moving at speed relative to frame S.Two observers standing beside thebelt, 10 ft apart as measured in S, arrange that eachwill paint a mark on the belt at exactly the same in-stant (as measured in S). How far apart will the marksbe as measured by observers on the belt?

1.31 • A rigid spherical ball (rest frame S) is observedfrom a frame that travels with speed relativeto frame S. Describe the ball’s shape as measured byobservers in

1.32 • Consider the experiment of Problem 1.26 from thepoint of view of the pions’ rest frame. In part (c) howfar (as “seen” by the pions) does the laboratory move,and how long does this take? How many pions re-main at the end of this time?

1.33 •• A meter stick is moving with speed relative toa frame S. (a) What is the stick’s length, as measuredby observers in S, if the stick is parallel to its velocityv? (b) What if the stick is perpendicular to v?(c) What if the stick is at to v, as seen in the stick’srest frame?

[HINT: You can imagine that the meterstick is the hy-potenuse of a 30–60–90 triangle of plywood.]

(d) What if the stick is at 60° to v, as measured in S?

1.34 •• Like time dilation, the Lorentz contraction cannotbe seen directly (that is, perceived by the normalprocess of vision). To understand this claim, considera rod of proper length moving relative to S. Carefulmeasurements made by observers in S [as inFig. 1.23(a), for example] will show that the rod has

l0

60°

0.8c

S¿.

0.5cS¿

0.5c

S

Q

(b)

A

t � 0 t �

0

B

S

(a)

A 0

B

FIGURE 1.22(Problem 1.28) (a) Two observers at rest in frame S at A and Btime the moving clock as it passes them; they find the dilatedtime (b) The single observer Q sees the movingclock at A and B by means of light that has traveled differentdistances, and BQ.AQ

t = gt0 .

S

Q

BA

(b)

(a)

S

l � l0/ �

v

FIGURE 1.23(Problem 1.34) (a) One can measure the Lorentz-contractedlength using two observers to record the positions ofthe front and back at the same instant. (b) What a singleobserver sees is determined by light that left the rod atdifferent times.

l = l0>g

TAYL01-001-045.I 12/10/02 1:51 PM


the contracted length But now considerwhat is seen by observer Q in Fig. 1.23(b) (with Q tothe right of points A and B). What Q sees at any oneinstant is determined by the light entering her eyes atthat instant. Now, consider the light reaching Q at oneinstant from the front and back of the rod.(a) Explain why these two rays must have left the rod(from points A and B) at different times. If the x axishas a graduated scale as shown, Q sees (and a photo-graph would record) a rod extending from A to B;that is, Q sees a rod of length (b) Prove that Qsees a rod that is longer than l. (In fact, at certainspeeds it is even seen to be longer than and theLorentz contraction is distorted into an expansion.)(c) Prove that once it has passed her, Q will see therod to be shorter than l.

SECTIONS 1.11 and 1.12 (The LorentzTransformation andApplications)

1.35 • The two frames S and are in the standard config-uration (origins coincident at and axes parallel, and relative velocity along ). Theirrelative speed is An event occurs on the x axis at

light-seconds (a light-second is the distancetraveled by light in one second, )at time in the frame S. What are its coordi-nates as measured in

1.36 • The Lorentz transformation (1.37) consists of fourequations giving in terms of Solve these equations to give in terms of

Show that you get the same result by in-terchanging primed and unprimed variables and re-placing by

1.37 • Give in detail the derivation of the Lorentz trans-formation (1.36) for starting from equations (1.34)and (1.35).

1.38 • Two inertial frames S and are in the standardconfiguration, with relative velocity v along the lineof the x and axes. Consider any two events, 1 and 2.(a) From the Lorentz transformation (1.37), deriveexpressions for the separations (where etc.) in terms of

(Notice how the transformation of is identical to that of ) (b) If

and whereas what is the relativespeed and what is

1.39 •• The frames S and are in the standard configura-tion with relative velocity along (a) What arethe coordinates in S of an event thatoccurs on the axis with (b) Answer the same for a second event on the axiswith (c) What are the timeintervals and between the two events, asmeasured in S and

1.40 •• In a frame S, two events have spatial separationand temporal separation

A second frame is moving along with nonzero speed and parallel to In itis found that the spatial separation is also 600 m.What are and ¢t¿ ?v

¢x¿S¿Ox.O¿ x¿vOxS¿¢t = 1 ms.

¢y = ¢z = 0,¢x = 600 m,

S¿ ?¢t¿)(¢t

tœ2 = 10 ms.xœ

2 = -1500 m,x¿

tœ1 = 5 ms?xœ

1 = 1500 m,x¿1x1 , y1 , z1 , t12

Ox.0.8cS¿¢x¿ ?v,

¢t¿ = 5 s,¢t = 4 s,¢x = 0x, y, z, t.¢z, ¢t¢x, ¢y,¢z, ¢t.¢x, ¢y,¢x¿ = xœ

2 - xœ1 ,

¢z¿, ¢t¿¢x¿, ¢y¿,

x¿

S¿

t¿,

-v.v

x¿, y¿, z¿, t¿.x, y, z, t

x, y, z, t.x¿, y¿, z¿, t¿

S¿ ?x¿, y¿, z¿, t¿t = 4 s

1 c # sec = 3 * 108 mx = 10

0.5c.Ox

x¿t = t¿ = 0, xS¿

l0 ,

AB.

l = l0>g. 1.41 •• Observers in a frame S arrange for two simultane-ous explosions at time The first explosion is atthe origin while the second is onthe positive x axis 4 light years away

(a) Use the Lorentztransformation to find the coordinates of these twoevents as observed in a frame traveling in the stan-dard configuration at speed relative to S.(b) How far apart are the two events as measured in

(c) Are the events simultaneous as observedin

1.42 •• A traveler in a rocket of length sets up a coor-dinate system with origin anchored at the exactmiddle of the rocket and the axis along the rocket’slength. At she ignites a flashbulb at (a) Write down the coordinates and forthe arrival of the light at the front and back of therocket. (b) Now consider the same experiment as ob-served in a frame S relative to which the rocket istraveling at speed (with S and arranged in thestandard configuration). Use the Lorentz transforma-tion to find the coordinates and of the ar-rival of the two signals. Explain clearly why the twotimes are not equal in frame S, although they were in

(This illustrates how two events that are simulta-neous in are not necessarily simultaneous in S.)

1.43 •• Consider the relativistic snake of Example 1.6, butlet the numbers be as follows: The snake has speed

and proper length of 100 cm (as before), but theboy holds the two hatchets 80 cm apart. (a) Show thatwith these lengths the experiment can be seen as atest of relativity, since the snake will be unhurt if rela-tivity is right (and the boy times things correctly),whereas the snake will definitely be hurt if the classi-cal ideas of space and time are correct. (Naturally, rel-ativity is correct and the snake is unharmed.) (b) Usethe Lorentz transformation to find the positions andtimes of the falling of the two hatchets as measuredby the snake, and use these to verify that it is un-harmed. (Assume the boy bounces the hatchets at

at which time the snake’s tail is at the commonorigin.)

1.44 ••• (a) Consider two frames S and that differ onlyby a rotation in which the x and y axes were rotatedclockwise through an angle to become and Prove that

(and and ). (b) Prove that the standardLorentz transformation can be written as

and

(and and ) where Ex-cept that the trig functions cos and sin are replaced bythe hyperbolic functions cosh and sinh (and that oneterm has changed sign), the Lorentz transformationdoes to x and just what a rotation does to x and y.This is our first indication that should be re-x, y, z, ct

ct

f = tanh-11v>c2.z¿ = zy¿ = y

ct¿ = ct cosh f - x sinh f

x¿ = x cosh f - ct sinh f

t¿ = tz¿ = z

x¿ = x cos u - y sin u and y¿ = y cos u + x sin u

y¿.x¿u

S¿

t = 0,

0.6c

S¿S¿.

xB, tBxF, tF

S¿v

xœB, tœ

BxœF, tœ

F

O¿.t¿ = 0x¿O¿S¿

2d

S¿ ?S¿ ?

0.6cS¿

1x2 = 4 c # years, y2 = z2 = 02.1x1 = y1 = z1 = 02t = 0.

TAYL01-001-045.I 12/10/02 1:51 PM


garded as the four coordinates in some kind of four-dimensional space-time.

SECTION 1.13 (The Velocity-Addition Formula)

1.45 • A rocket (rest frame ) traveling at speed relative to the earth (rest frame S) shoots forwardbullets traveling at speed relative to therocket. What is the bullets’ speed u relative to theearth?

1.46 • As seen from earth (rest frame S) two rockets Aand B are approaching in opposite directions, eachwith speed relative to S. Find the velocity ofrocket B as measured by the pilot of rocket A.

[HINT: Consider a coordinate system traveling withrocket A; your problem is then to find the velocity ofrocket B relative to knowing its velocity u relative to S.]

1.47 •• Using the velocity-addition formula, one canprove the following important theorem: If a body’sspeed u relative to an inertial frame S is less than c, itsspeed relative to any other inertial frame is alsoless than c. In this problem you will prove this resultfor the case that all velocities are in the x direction.

Suppose that is moving along the x axis offrame S with speed Suppose that a body is travelingalong the x axis with velocity u relative to S. (We canlet u be positive or negative, so that the body can betraveling either way.) (a) Write down the body’s ve-locity relative to For a fixed positive (less thanc, of course), sketch a graph of as a function of u inthe range (b) Hence prove that for anyu with it is necessarily true that

1.48 •• Suppose that as seen in a frame S, a signal (a pulseof light, for example) has velocity c along the y axis(that is, ). (a) Write down thecomponents of its velocity relative to a frame traveling in the standard configuration with speed along the x axis of frame S. (b) In what direction isthe signal traveling relative to (c) Using your an-swer to part (a), calculate the magnitude of

SECTION 1.14 (The Doppler Effect�)

1.49 • It is found that the light from a nearby star isblueshifted by 1%; that is, Is the starreceding or approaching, and how fast is it traveling?(Assume that it is moving directly toward or awayfrom us.)

1.50 • A star is receding from us at What is the per-cent shift in the frequency of light received fromthe star?

1.51 • Consider the tale of the physicist who is ticketed forrunning a red light and argues that, because he was ap-proaching the intersection, the red light was Doppler

0.5c.

fobs = 1.01 fsce .

uœ.S¿ ?

vS¿uœ

uy = cux = uz = 0,

-c 6 u¿ 6 c.-c 6 u 6 c,-c 6 u 6 c.

u¿vS¿.u¿

v.S¿

S¿u¿

S¿,uœ

S¿

0.9c

u¿ = 0.6 c

v = 0.5cS¿

shifted and appeared green. How fast would he havebeen going?

1.52 • In our discussion of the Doppler shift, we foundthree superficially different expressions for the re-ceived frequency namely (1.51), (1.53), and(1.57). Show in detail that all three are equal.

1.53 •• Consider a source of light of frequency movingobliquely to an observer Q as in Fig. 1.24(a). (a) Provethat Q receives the light with frequency given bythe general Doppler formula

(b) Check that this formula reduces to our previousresult (1.51) when the source is approaching Qhead-on.

The analysis in part (a) is quite similar to thatleading to (1.51) but the geometry is more complicat-ed. Consider two successive wave crests emitted atpoints A and B as in Fig. 1.24(b). Since A and B are inpractice very close together, the rays and areeffectively parallel. Show that the difference betweenthe lengths and is approximately and hence that the distance between successive crestsas they approach Q is This is the ap-propriate generalization of (1.47), and from here thediscussion is closely parallel.

1c - v cos u2¢t.

v ¢t cos uBQAQ

BQAQ

fobs =fsce

11 - b cos u2g

fobs

fsce

fobs ,

lgreen L 530 nm.21lred L 650 nm,

(b)

v�t

Q

A B

(a)

v

Q

�

FIGURE 1.24(Problem 1.53) (a) Light from the moving source to theobserver Q makes an angle with the velocity v. (b) If twosuccessive wave crests are emitted at A and B, a time apart,then is v ¢t.AB

¢tu

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tayl01-001-045.i

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