PHYS4102 HOMEWORK

CHAPTER 15

Assigned Tuesday, April 19, Thursday, April 28

These problems are taken from Griffiths.

Problem 1. This problem deals with the relativistic velocity addition for-mula. It is composed of two problems from Griffiths.

450 Chapter 10 Electrodynamics and

second postulate. In special relativity, as we shall see, it is replaced byvelocity addition rule:

V4g -f VsgvAc:

t I vlsvsct -

" '

For "ordinary" speeds (vap 11 c, vsg 11 c), the denominator is so close to I thrdiscrepancy between Galileo's formula and Einstein's is negligible. On thehand, (10.7) has the desired property thatif vnpconsistent with (10.6):

c*vsg

| + cvs-cc'

But how can Galileo's rule, which relies on nothing but the most primitiwmetrical inference, possibly be wrong? And if it rs wrong, what does this do toclassical physics? (As a matter of fact, I used this very rule in equation (10-llsupport the principle of relativity itself!) The answer is that special relativityus to alter our notions of space and time themselves, and therefore also of suclrived quantities as velocity, momentum, and energy. Although it developedcally out of Einstein's contemplation of electrodynamics, the special theorl'-limited to any particular class of phenomena-rather, it is a description of thetime "arena" in which c// physical phenomena take place. And in spite of theence to the speed of light in the second postulate, relativity has nothing to dolight: c is evidently a fundamental velocity, and it happens that light (alsoand, presumably, gravitons) travel at that speed, but it is possible to conceircuniverse in which there are no electric charges, and hence no electromagnetkor waves, and yet relativity would still prevail. Because relativity defines theof space and time, it claims authority not merely over all presently knownena, but over those yet to be discovered. It is, as Kant would say, aany future physics."

Problem l0.l(a) What's the percent error introduced when you use Galileo's rule, instead dstein's, with r,a3 : 5 milh and vss : 60 mi/h?(b) Suppose you could run at half the speed of light down the corridor of a tnithree-quarters the speed of light. What would your speed be relative to ground!(c) Prove, using equation (10.7), thatif uts < c and vBs 1 c then us ( c. Inresult.

Problen 10.2 As the outlaws escape in their getaway car, which goes f;c, the poliafires a bullet from the pursuit car, which only goes ic. (Fig. 10.4). The muzzlc(speed relative to gun) of the bullet is |c. Does the bullet reach its target (a)Galileo, (b) according to Einstein?

: c, then automatically vrc =

:c

l0-1 The Special Theo

10.1.2 The Ge

In this section lintroduce the tltime dilation. I10.1.3 the samemations.

(i) The relspeed along a srhangs a light bttions at the speron the train wilreaches the bacl(b) light reacherthe ground thesfrom the bulb. tlshorter distanceobserver, therefcan express train,

Ttyo evenksimultaneo

This is a prfrom Einstein's rdirections for earsidered preposte:crepancy becomr

Of course, ineity: you hear t,the source of thetrivial error. havimust correct forto reach you. Wlmake this correcr

vAc:

f l+t

Flguro 105

S to S'. Derive the analogous formulas for velocities in the 1'- and :dtrucndirections perpendicular to the relative motion of S and S').(b) A spotlight is mounted on a boat so that its beam makes an angk I rd(Fig. 10.21). If this boat is then set in motion at speed v. what angle 0' dm ron the docfr say the beam makes with the deck? Compare Problem lo-E' uddifference.

Problem 10.13 You probably did Problem 10.2 from the point of view of an obcground. Now do it from the point of view of the police car, the outlas's. end ttThat is, fill in the gaps in the following table:

RelativetoI Gnouno Por,rcn Ourllws Burrpr Do rssr

Gnoulrp

Por,rcB

Ourr'nws

Bur,r.pr

Problcm 10.14 (Twin paradox reconsidered). On their 21st birthday, one ts-ina moving sidewalk, which carries her out to star X at a speed f c; the other tsbhome. When the traveling twin gets to star X, she immediately jumps onto themoving sidewalk and comes back to earth, again at speed $c. She arrives on lrcrbirthday (as determinedby her watch).(a) How old is her twin brother (who stayed at homeX(b) How far away is star X? (Give your answer in light years.)Call the outbound sidewalk system ,S' and the inbound one S " (the earth system bAll three systems set their master clocks, and choose their origins, so that x = rxo : 0, t : t' : t' : 0 at the moment of departure.(c) What are the coordinates (x, r) of the jump (from outbound to inbound.s?(d) What are the coordinates (x' , t') of the jump in S'?(e) What are the coordinates (xo , t') of the jump in J" ?(f) If the traveling twin wants her watch to agree with the clocks in J " , what mustto her watch immediately after the jump? If she does this, what will her watch readshe gets home? (This won't change her age, of course-she's still 39-it'll just makewatch agree with the standard synchronization in S".)

i.L-

EAl,:tl

t - l '

ra,

- l {a-qbra

-

ro.t-a Tb t

(If Frrhea erptesr

Usingro (instethe second tolI meter (in vac

then the Loren

Or, in matrix f(

Letting the Gree

where A is the Lrlabels the row, thabstract mannertion, in which thwould be more ,changed.

If this remir

Speed

-:Of-

1

2 CHAPTER 15

Problem 2. This problem deals with the Lorentz transformation equations.

f l+t

Flguro 105

S to S'. Derive the analogous formulas for velocities in the 1'- and :dtrucndirections perpendicular to the relative motion of S and S').(b) A spotlight is mounted on a boat so that its beam makes an angk I rd(Fig. 10.21). If this boat is then set in motion at speed v. what angle 0' dm ron the docfr say the beam makes with the deck? Compare Problem lo-E' uddifference.

Problem 10.13 You probably did Problem 10.2 from the point of view of an obcground. Now do it from the point of view of the police car, the outlas's. end ttThat is, fill in the gaps in the following table:

RelativetoI Gnouno Por,rcn Ourllws Burrpr Do rssr

Gnoulrp

Por,rcB

Ourr'nws

Bur,r.pr

Problcm 10.14 (Twin paradox reconsidered). On their 21st birthday, one ts-ina moving sidewalk, which carries her out to star X at a speed f c; the other tsbhome. When the traveling twin gets to star X, she immediately jumps onto themoving sidewalk and comes back to earth, again at speed $c. She arrives on lrcrbirthday (as determinedby her watch).(a) How old is her twin brother (who stayed at homeX(b) How far away is star X? (Give your answer in light years.)Call the outbound sidewalk system ,S' and the inbound one S " (the earth system bAll three systems set their master clocks, and choose their origins, so that x = rxo : 0, t : t' : t' : 0 at the moment of departure.(c) What are the coordinates (x, r) of the jump (from outbound to inbound.s?(d) What are the coordinates (x' , t') of the jump in S'?(e) What are the coordinates (xo , t') of the jump in J" ?(f) If the traveling twin wants her watch to agree with the clocks in J " , what mustto her watch immediately after the jump? If she does this, what will her watch readshe gets home? (This won't change her age, of course-she's still 39-it'll just makewatch agree with the standard synchronization in S".)

i.L-

EAl,:tl

t - l '

ra,

- l {a-qbra

-

ro.t-a Tb t

(If Frrhea erptesr

Usingro (instethe second tolI meter (in vac

then the Loren

Or, in matrix f(

Letting the Gree

where A is the Lrlabels the row, thabstract mannertion, in which thwould be more ,changed.

If this remir

Speed

-:Of-

ics and Relat iv i tY

z-directions (that is.

ngle 0 with the deck: 0' does an obsen'e;10.8, and exPlain thc

of an observer on th.:laws. and the bulla

rday, one twin gers :r:; the other twin sr'-rnnps onto the return-qhe arrives on her -r+:t

the earth sYstem rs 5ns, so that r : .r

r inbound sidewdl r

rS",whatmu$s.h-ill herwatch read rl

39-it'll just malr I

10.1 The Special Theory of Relativity 467

(g) If ttre traveling twin is asked the question, "How old is your brother right now, andwhich of you is younger?", what is the correct reply (i) just before she makes the jump,(ii) just after she makes the jump? (Nothing dramatic happens to her brother during thesplit second between (i) and (ii), of course-what does change radically is his sister'snotion of what "right now" means.)

10.1.4 The Structure of Spacetime

(i) Four-vectors. The Lorentz transformations take on a simpler appearancewhen expressed in terms of the new quantities

xo:ct , A: ! c

Using x0 (instead of r) and B (instead of y) amounts to changing the unit of time fromthe second to the meter-l meter of x0 corresponds to the time it takes light to travel1 meter (in vacuum). If, at the same time, we number the r, y, z coordinates, so that

yl :y, x2:y, x3:z

then the Lorentz transformations read

(r0.27)

Or. in matrix form:

(10.28)

Letting the Greek indices run from 0 to 3, this can be distilled into a single equation:J

(xr) ' - D (l t)x'v--0

where A is the Iorentz hansformation matrix in equation (10.28) (the superscript trrlabels the row, the subscript y labels the column). One virtue of writing things in thisabstract manner is that we can handle in the same format a more general transforma-tion, in which the relative motion isnot along a commonx-x' axis; the matrix Awould be more complicated, but the structure of equation (10.29) remains un-changed.

If this reminds you of therotstions we studied in Chapter 1, it's no accident.

(xo) ' : f (xo-pxt))

(xr) ' : y(r t - 0ro) I

(x2\ ' : x2 \

(x3) ' - x3 )

(10.2s)

(10.26)

(r0.29)

[]l :(1' I'l ilfi)

Do rnnv EscAPE'

Problem 3. This problem deals with the Lorentz transformation equationsand invariant intervals.

470 Chapter 10 Electrodynamics arrd Rela: . *-

When you transform to a moving system, the time between A and B is alt<-=:(t' + t), and so isthe spatial separation (d' + d), but the interval l remains'.-same.

Notice that, depending on the two events in question, the interval can be F's'tive, negative, or zero:

1. If 1 ( 0 we call the interval timelike, for this is the sign we get when the :."occur at the same place (d : 0), and are separated only temporally.

2. lt I > 0 we call the interval spacelike, for this is the sign we get when the -'roccur at the same time (t : 0) and are separated only spatially.

3. If / : 0 we call the interval lightlike, for this is the relation that holds u'her-. : :s

two events are connected by a signal traveling at the speed of light.

In fact, if the interval between two events is timelike, there exists an in.':- "esystem (accessible by Lorentz transformation) in which they occur at the same p rFor if I hop on a train going from (.4) to (B) at the speed v : d/t,leaving eve:' .

when it occurs, I shall be just in time to pass ,B when i/ occurs. In the train syste;:: '-and B take place at the same point. You cannot do this Ior aspacelike intena- rcourse, because vwould have to be greater than c, and no observer can exceec'?Espeed of light (if he did,7 would be imaginary and the Lorcntz transforma:i:'trwould be nonsense). On the other hand, if the interval is spacelike then there e\:!i! rsystem in which the two events occur at the same time (see Problem 10.19).

Problem l0.lE(a) Event .4 happens at point (r : 5, ! : 3, z : 0) and at time t given by ct : I 5 :

-':rrB occurs at (10, 8, 0) and cr : 5, both in system S.(i) What is the invariant interval between A and B?(ii) Is there an inertial system in which they occur simultaneouslyZ If so. fi:: tt

velocity (magnitude and direction) relative to S.(iii) Is there an inertial system in which they occur at the same point? If so. fi:-,: tt

velocity relative to ,S.(b) Repeatpartafor A : (2,0, 0) , cr : l ; and.B : (5, 0, 0) , cr : 3.

Problem 10.19 The coordinates of event A are (x1, t), and the coordinates of event I rr(x s, t il. Assuming the interval between them is spacelike, find the velocity of the s. s:*in which they occur at the same time.

(iii) Space-time diagrams. If you wish to represent the motion of a par:u:rgraphically, the normal practice is to plot position versus time (that is, x runs rt'cally and / horizontally). On such a graph, the velocity can be read off as the slo- cthe curve. For some reason the convention is reversed in relativity: Everyone i':6position horizontally and time (or, better, x0 : ct) vertically. Velocity is then Er:lby the reciprocal of the slope. A particle at rest is represented by a vertical linr rphoton, traveling at the speed of light, is described by a 45" line; and a rocket g.'r1at some intermediate speed follows a line of slope c/v : l/P. The trajecton { rparticle on such a space-time diagram is called its world line (Fig. 10.22).

10.1 The Special Thec

Suppose,no material objless than 1. Abounded by thelocus of all poiron, and you monarrow: Your "ever point you fiyour "past," income. As for ththe generalizedfact, there's nohave to travel fainaccessible to 1

I've been i1the page, the "wBecause their b<light cone and tlforward light co

Notice thaltells you at a glz

Problem 4. Taylor 15.10