Introduction to Electrodynamics (solutions) - ch12

Prévia do material em texto

Chapter 12
Electrodynamics and Relativity
Problem 12.1
Let u be the velocityof a particlein S, u its velocityin 5, andv the velocityof 5 with respectto S.
Galileo'svelocityadditionrulesaysthat u =u+v. For a freeparticle,u is constant(that'sNewton'sfirst
lawinS).
(a)If v isconstant,thenu =u-v isalsoconstant,soNewton'sfirstlawholdsin5, andhenceS isinertial.
(b)If 5 is inertial,thenu isalsoconstant,sov =u - u isconstant.
Problem12.2
(a)mAUA+mBUB=mcuc +mDuDi Ui=Ui+v.
mA(uA+v) +mB(uB +v) =mc(uc+v) +mD(uD+v),
mAnA+mBUB+ (mA+mB)v=mcuc +mDuD+ (mc+mD)v.
Assumingmassis conserved,(mA+mE)= (mc+mD),it followsthat
mAUA+mBuB =mcuc+mDuD,somomentumisconservedin5.
(b) 1 2 1 2 - 1 2 1 22mAUA+ 2mBUB- 2mcuC+ 2mDuD =>
!mA(U~+2UA' v + V2)+ !mB(U~+ 2UB' v +v2)= !mc(ub+2uc' v +V2)+~mD(ub+2UD'v +V2)
!mAU~+ ~mBU~+2v. (mAuA+mBuB) + !v2(mA+mE)
= ~mcub+ !mDub+2v. (mcuc+mDuD)+~v2(mc+mD)'
But themiddletermsareequalby conservationof momentum,andthelast termsareequalby conservation
f 1 -2 + 1 -2 1 -2 + 1 -2 d0 mass,so 2mAuA 2mBuB= 2mcuC 2mDuD' qe
Problem 12.3
(a) Va=VAB +VBC' VE = VAB+VBC ~ Va (1- ~ ) =>~ = ~., 1+VABVBC/C2 C va C
In mi/h, c = (186,000mi/s) x (3600sec/hr)=6.7x 1O8mi/hr.
. ~ = (5)(60J =6 7 X 10-16=>16.7x 10-14%error I (pretty small!).. va (6.7xlO )2 . ,
~
[10].
(b)(!c+ic) / (1+~.i) = (~c)/ en =l.:!rJ (stllliessthanc).
(c)Tosimplifythenotation,let/3==VAC/C,/31==VAB/e,/32==vBc/e.ThenEq. 12.3says:/3=f~%~;2'or:
/32= /3f+ 2/31/32+/3~ = 1+2/31/32+/3f/3~- (1+/3f/3~- /3f- /3~)=1 - (1 - /3f)(l - /3~)=1- A
(1+2/31/32+/3f/3~) (1+2/31/32+/3f/3~) (1+2/31/32+/3f/3~) (1+/31/32)2 '
219
~
220 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
where~==(1- ,8f)(1 - ,8~)/(1+,81,82)2isclearlyapositivenumber.So,82<1,andhenceIvAGI<c. qed
Problem 12.4
(a) Velocityof bulletrelativeto ground:~c+ lc =~c=~gc.
Velocityofgetawaycar:~c=192C.Since Vb>vB' I bulletdoesreachtarget.!
1c+1c ~c 5 20
(b) Velocityof bulletrelativeto ground::+1\ =T ='fc=28c.2 3 6
Velocityofgetawaycar:~c=¥Bc. Since Vg>Vb, I bulletdoesnot reachtarget.I
Problem 12.5
(a) Light fromthe90thclocktook i~~~8°:'/S=300s =5 min to reachme,so thetimeI seeontheclockis
111:55am.I
(b) I observe112noon. I
Problem 12.6
{
lightsignalleavesa at timet~;arrivesat earthat timeta=t~+dalc,"7'
lightsignalleavesbat timet~;arrivesat earthat timetb=t~+dblc.
:.~t=tb-ta =t~-t~+ (db-da) =~t'+ (-v~t'cosO)=~t' [1- ~cosO].c c c
(Heredaisthedistancefroma toearth,anddbisthedistancefrombtoearth.)
, . vsinO~t
I
vsinO.. .
~s =v~t SIll0 =(1 I 0); u=(1 v 0)11sthetheapparentveloCity.- V ccos - Ccos
du - v[(I- ~cosO)(cosO)- sinO(~sinO)]= O:::}(1- ~cosO)cosO=~sin20dO- (1- ~COSO)2 C
v.2 2 ) _V:::}cosO=-(SIll 0+cos0 --c c
1
0 -1( I)
I
A h. . al 1 vV1-v2/c2 v
max= COS V c. t t 1Smax1m ange,u = 1-v2/c2 = V1-v2/c2'
Asv ~ c,I u ~ 00, I becausethedenominator~ 0,eventhoughv<c.
Problem 12.7
The studenthasnot takenintoaccounttimedilationof themuon's"internalclock". In the laboratory,the
muonlasts'YT=...; T , whereT is the "proper"lifetime,2 x 10-6s. Thus1-v2/c2
V2 1
c2= 1+ (Tcld)2;
d d
v= = -vl-v2/c2, whered= 800m.
TI../l- v2/c2 T
(
T
)2 2 v2 2[(
r
)
2 1
] 2 1d v =1- e2; v d +e2 =1; v =(rld)2 + (1/e)2'
Te- (2x 10-6)(3x 108)- ~-~. V2 - 1 - 16.
d - 8.00 - 8- 4' c2 - 1+9/16- 25' Iv = ~e.1
221
Problem 12.8
(a) Rocket clock runs slow; so earth clock reads 'Yt=V 1 2 .1 hr. Here "1 =V 1 2 =~ =_45.i-v /c2 i-v /C2 1-9/25
.'.Accordingto earthclockssignalwassent11hr and15minIaftertake-off.
(b) By earthobserver,rocketis nowa distanceac) (~)(1hr) = ~chr (three-quartersofa lighthour)away.
Light signalwill thereforetake~hr to returnto earth. Sinceit [eft 1 hr and 15min afterdeparture,light
signalreachesearth12hrs aftertakeoff.!
(c)Earthclocksrunslow:trocket="1' (2hrs)=~.(2hrs)=12.5hrs.1
Problem 12.9
L =2L . h =h. so2. =...L= Vl- (1)2= Ii. 1 =1- V2 =..1..v2 =1-..1. =13.lv =V13c Ic V"Ye 'Y.' 'Ye 'Y. 2 V4';:Y; ~ 16'~ 1616' 4'
Problem 12.10 .
Saylengthofmast(at rest)is 1.To anobserverontheboat,heightof mastis [sin0, horizontalprojection
is [cosO. To observeron dock, the formeris unaffected,but the latter is Lorentzcontractedto l[ cosO.'Y
Therefore:
- [sin 0 .
I
-tan 0
tan0=1 ="1 tan0, or tan0= .
::y[cosO vI - V2/C2
Problem 12.11 .
Naively,circumference/diameter=~(27rR)/(2R)=trh =7rVl- ((;JR/C)2- but this is nonsense.Point
is: anacceleratingobjectcannotremainrigid,in relativity.To decidewhatactuallyhappenshere,youneeda
specificmodelfor theinternalforcesholdingthedisktogether.
Problem 12.12
(iv) =>t =~+~. Put this into (i), andsolvefor x:
(
l VX
) (
v2
) - 1 - x - I Ix = 'YX- 'Yv ::y+ e2 ='YX1- e2 - vt='Yx"12- vt= ::y- vt; x ='Yex+vt). ..(
Similarly,(i) =>x = it+vt. Put thisinto(iv)andsolvefort:'Y
- 'YV(
X
) (
v2
)
V - t v -
t ='Yt- - - +vt ='Yt 1- - - -x =- - -x;e2 "1 c2 C2 "1 C2 It='Y(l+~x).I..(
Problem 12.13 '-~
Let brother'saccidentoccurat origin,timezero,in bothframes.In systemS (Sophie's),thecoordinates
of Sophie'scry are x = 5 X 105m, t = O. In system.5(scientist's),l ='Y(t - ~x) = -'YVX/C2. Since
this is negative,Sophie'scryoccurredbeforetheaccident,in.5. "1 = V 1 = v'16~3144= 15
3. So
1-(12/13)2 -
f=- (l!) (He) (5x 105)/c2=-12 X 105/3 X 108=-4 X 10-3. 14 X 10-3 s earlier. I
Problem 12.14
(a) In S it movesa distancedy in timedt. In.5, meanwhile,it movesa distancedy = dy in time dl =
"((dt- ~dx).
dy dy (dy/dt)
1
- uy - Uz
.'. dl ='Y(dt- ~dx) = "1(1- ~~~)j or Uy= "1(1- 7) ; Uz= "1(1- 7)'
222 CHAPTER 12. ELECTRODYNAMICSAND RELATIVITY
(b) taniJ =- Uy =- Uy/ b(1- ~ )] =.!. (-Uy) .
Ux (ux - v)/ (1-~) 'Y(ux- v)
In this caseU =-ccos{}' U =csin{}=}taniJ= 1.( -esinO ).x , Y '"Y -eeosO-v
- 1
(
sin{}
) -. 0tan {}=- {} / . [Comparetan {}='Y~~n0 in Prob. 12.10.The point is that velocitiesaresensitive'Y cos +vc s
not only to the transformationof distances,but alsoof times. That's why thereis no universalrulefor
translatingangles-you haveto knowwhetherit's ananglemadeby a velocityvectoror a positionvector.]
Problem 12.15
5 ~c- lc (1/4)c 2
Bullet relativeto ground:-c, Outlawsrelativeto police: 4 321 =(5/8) =-c.7 1-4'2 5
2c- ~c -(1/28)c 1
Bullet relativeto outlaws:17- ~~i = (13/28) =- 13c. [Velocityof A relativeto B is minusthevelocity
of B relativeto A, soall entriesbelowthediagonalaretrivial. Notethat in everycaseVbullet< Voutlaws,sono
matterhowyoulookat it, thebadguysgetaway.]
Problem 12.16
(a) Moving clock runs slow, by a factor 'Y = V 1 =23 . Since 18 years elapsedon the moving clock,1-(4/5)2
~X 18=30yearselapsedonthestationaryclock.151yearsold.I
(b) By earthclock,it took15yearsto getthere,at tc, sod = tC x 15years= 112cyearsI (12lightyears).
(c)It =15years,x = 12cyears.I
(d) It=9 years,x=0.1 [Shegot on at the origin in 5, and rode along with 5, so she's still at the origin. If
youdoubtthesevalues,usetheLorentztransformations,with x andt from(c).]
(e)Lorentztransformations:
{
~='Y(x+vt)
}
(notethat v is negative,sinceS is goingto the left).
t ='Y(t+~x)
:.x=~(12cyrs+ tc. 15yrs)=~ .24cyrs=140c years. I
i =~(15yrs+ t2r .12cyrs)=~ (15+ ~8)yrs=(25+ 16)yrs=141years.I
(f) SetherclockI ahead32years,I from9 to 41 (t -+i). Returntrip takes9 years(movingtime),soherclock
will nowread@QJyearsat herarrival.Notethatthis is ~.30years-preciselywhatshewouldcalculateif the
stay-at-homehadbeenthetraveler,for 30yearsof hisowntime.
(g) (i) t=9 yrs,x = O.Whatis t? t = ~x +f =~ . 9 =257=5.4years,andhestartedat age21,sohe's
126.4yearsold.I (Youngerthanthetraveler(!) becauseto thetravelerit's thestay-at-homewho'smoving.)
(ii) i = 41 yrs, x = O. What is t? t = f =~ . 41 = 1;3 =24.6 years, and he started at 21, so he's
!45.6yearsold.I
speedof -+
Ground Police Outlaws Do they escape?relativeto .j. Bullet
Ground 0 e ie ¥e Yes
Police -e 0 e !e Yes '!-
Outlaws -ie -e 0 --he Yes
Bullet -¥e -!e -he 0 Yes
223
(h)It will takeanother15.4yearsI of earthtimefor the return,so whenshegetsback,shewill sayher
twin'sageis 45.6+ 5.4=~ years-whichis whatwe foundin (a). But notethat to makeit workfrom
traveler'spointofviewyoumusttakeintoaccountthejumpinperceivedageofstay-at-homewhenshechanges
coordinatesfromS to S.
Problem12.17
-a,°rp+a,lfjI+a,2fj2+a,3fj3= -"l(aO - {3aI)(bO- {3bI)+-l(aI - {3ao)(bI- {3aO)+a2b2+a3b3
= -"-?(aobo- {3/bI - {3jl'b°+{32aIbI - aIbI +{3jl'b°+{3/bI - {32aobo)+a2b2+a3b3
= -72aobo(1-{32)+72aIbI(1- {32)+a2b2+a~b3
= -aobo+aIbI +a2b2+a3b3. qed [Note: 72(1 - /32)=1.]
Problem 12.18
(
c~
(
1 000
) (
ct
)(a)I g) ~ -{ ~ ! ~ ~ I(usmg the notation of Eq. 12.24, 10' best compadson).
(
7 0 -7{3 0
)
(b)IA = 0 1 0 0-7{3 0 7 O'
0 0 0 1
(
'YO -'Y~ 0
) (
7 -7/3 0 0
) (
7'Y -7'Y{3-'Y~ 0
)
. . 0 1 0 0 -7{3 7 0 0 -7/3 7 0 0
(c)MultIplythematrIces:A = -'Y~0 'Y 0 0 0 1 0 = -'Y7~ 7'Y{3~ 'Y O'
00010001 0 001
I Yes,I theorderdoesmatter.In theotherorder,"bars"and"no-bars"wouldbeswitched,andthiswouldgive
a differentmatrix.
Problem 12.19 ~
(a) Sincetanh0 = ~~~~~,andcosh20- sinh20=1,wehave:
1 1 cosh0 .
7= = = =coshOj7{3=cosh0tanh0=smhO.
y'1- v2je2 y'1- tanh20 y'cosh2O- sinh20
(
cosh0 - sinh0 0 0
)
. A - - sinh0 cosh0 0 0..1 - 0 0 1 0 .
0 0 0 1 (
cas<jJ sin <jJ 0
)
Compare:R = - sin <jJ cas <jJ 0 .
0 0 1
- u-v fi (uje)-(vje) - tanh<jJ-tanhO
(b) u =lUV ::} - = (U)(V) ::} tanh<jJ=1 h<jJ hO' wheretanh<jJ= uje, tanhO= vje;- C2" e 1- C c - tan tan
tanh4)=fije. But a "trig" formulafor hyperbolicfunctions(CRC Handbook,18thEd., p. 204)says:
tanh<jJ- tanh0 - 1- I1- tanh<jJtanhO=tanh«jJ- 0). :. tanh<jJ=tanh«jJ- 0), or: <jJ= <jJ- O.
224 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
1
i
i
Problem 12.20
(a) (i) I =-c2b.t2+b.x2+b.y2+b.z2= -(5 - 15)2+(10- 5)2+(8- 3)2+(0- 0)2=-100 +25+25=1-50.1
(ii) I No.1 (In such a system Sf =0,so I wouldhaveto bepositive,whichit isn't.)
(iii) I Yes. I :1 _B
6
S travelsin the directionfromB towardA,
makingthetrip in timeIO/c.
5
:. v =-5:0/c5Y=I-~x- ~y.1
4
I
2 ," .. .... .... .. .. .. .. ...ie 5
x
N h v2 - 1 1 - 1 - 1 " 1ote t at C2- 4"+ 4"- 2' so v - Vic, saley
less than c.
2 4 6 8 10
(b)(i) 1= -(3 - 1)2+(5- 2)2+0+0= -4 +9=[[]
(ii) I Yes. 1 By Lorentztransformation:b.(d)=, [b.(ct)- ,8(b.x)].Wewantb.£=0,sob.(ct)= ,8(b.x);or
v b.(ct) (3- 1) 2
E}J
. "
- =~ =( . ) =-. So v = -c, III the+xdIrectIOn.c I lX 5- 2 3 3
(iii) ~ (Insucha systemb.x=b.y=b.z=0soI wouldbenegative,whichit isn't.)
Problem 12.21 } ,
UsingEq. 12.18(iv): b.£=,(b.t- ~b.x)=0=}b.t=~b.x,orv=~tc2=I tB - tA c2.
x XB - XA
Problem 12.22
(a) ct.
worldline
of player1
worldlineof
theball
Truth is, you neverdo communicatewith
theotherpersonrightnow-you communicate
with thepersonhe/shewill bewhenthemes-
sagegetsthere;andtheresponsecomesback
to andolderandwiseryou.
(b) I No way.I It is truethata movingobserv-
er mightsayshearrivedat B beforesheleft
A, but forthe roundtripeveryonemustagree
that shearrivesbackaftershesetout.
ct
A x
B
225
Problem 12.23
(a)
x
~'?
c,t '}.'?
c,t \.'?
c,t \J'?
c,t ./\.
'? "
c,t ./i-'
'? <1-
c,t ./J
'?
c,t
(b) % =slope=H
~ v =~:;c =I O.95c.1
(c) v' =tC,sov= ~c:i:l+sos
- S!.fJ!k r35J
- \37725J=@ =O.95c..(
Problem 12.24
(a)(1- ~)ry'=u';.'(1+~)=ry';I u =,/1 +lry2/C' 11°1
(b) 1 = 1 = coshlJ =coshOJ 1/= 1 u=coshOctanhO=lcsinhO.1Vl-u2/c2 Vl-tanh21J Vcosh21J-sinh21J Vl-u2/c2
Problem 12.25
(a)u. =u, =uco,45°=}, Joe=I Ae.1
(b) Vl-~2/C2= R =~ =J5; 'I}=Vl-:2/C2 ~ l1]x=1]y =v'2col
(c)1]0:,c =I J5 col
{
- - u -v - .,fiTsc-fi75c- rQ"lux- 1":~ - 1-- - ~02 5
(d) Eq. I2.45~ - j;)'f., 2- 1 u 2 2/5c ~
Uy =" (,-~)=/1-. ~ = y'iTo°=[lE]
(e)ijx=,(1]x- (31]°)= VI - ~ (V2c - ~J5 c)=[QJ I ijy =1]y =v'2col
(f) 1 - 1 - V3° - - V3ii
{
ijx =V3Ux =O. .(
}Vl-fj,2/c2- ~- ,'1}- ~ ijy=V3uy=V2c..(
Problem 12.26
/-' - ° 2 2 - 1 2 2 - 2(1 - u2jc2) - r:2l
1]1]/-,--(1]) +1] - (I-u2jc2)(-c +u )--c (I-u2jc2)-~
i i I ! ct 'll 'N1;'1 .....I"JI ,,:>1
I L: Iii I! 41!.'/ 4/, 'If,t..'(f,J/,,-'I -Lf--+ -+--tl-+-
I " , I I J J J I I I I i::PL J - 1-- --I--I " ", i I il I / 'i / I It..- :; ! .i- . ,--LJ-----
-;--;
------
'---71/' I I ' ' I ! / I I I , II ' ' / '
-- -T+- ' II I II l--I/ tZ-1'/ 7/7
I I VJ PJ ./ ./ ./ ./ .//,
fT v-: r It ./ ./ ./ / ./
./ VJ ..)V; ...AI/f ..) /} ./ / ./ ./ ./
./ v ./ r r C--r t-r ./ / ././
./
V
./ V 1...-1/} A. VI /1 ....1
/ /
./
i./V I...--:'V:=lL---fT? fT r ./ +9j-.LL I
i././IA'11...J:1Ji/} ./LLLi--1 ii' I :
&fTlld_i itll i i U
226 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
Problem 12.27
(a) From Frob. 11.31we have "( = tvb2+e2t2.:. T = J ~dt =bJ ~ =~ In(et+ Vb2+e2t2)+k; at
t =0 we want T =0: 0=£Inb+k,sok =_£ Inbj IT =~In [
~(et+ Vb2+e2t2)
]
.
c C e b
(b) VX2 - b2 +x = beeT/hiVX2 - b2= beCT/b- Xj X2- b2= b2e2cT/b- 2xbeCT/b+X2;2xbeCT/b= b2(1+e2CT/b);
x =b(ecr/b+2e-cr/b)=Ibcosh(eTjb).1Also from Frob. 11.31:v =e2tlvb2+e2t2.
- C V 2 2 - C Vb2 h2( j b) b2- cosh2(cT/b)-I- sinhCTb -I h (
eT
)V - x x - b - bcDsh(CT/b) COS CT - - e cosh(cTb - ecosh(cT/b- etan b'
(c) 1]1'= "((e,V,0,O)j "( = t =coshT' so1]1'= coshT (e,etanhT' 0,0)=Ie(coshT' sinhT' 0,0).
Problem 12.28
(a) mAUA + mBUB =moUe +mDUD; Ui +vUi = . '.
1+ (uivle2)
UA+v UB+v ue +v UD+v
mA
( I 2)
+mB
( I 2 =me ( 2 +mD ( 0-.1+ UAV e 1+ UBV e ) 1+ uevle ) 1+ uDvle4)
This time,becausethedenominatorsareall different,wecannotconcludethat
mAfiA + mBUB =moUe+mDuD.
As anexplicitcounterexample,supposeall themassesareequal,andUA = -UB = Vj Ue = UD = O.This
is a symmetric"completelyinelastic"collisionin 5, andmomentumis clearlyconserved(0 =0). But the
EinsteinvelocityadditionrulegivesUA=0,UB =-2uj(1+u2je2),ue=UD= -U, soinS theOlfcorrectly
defined)momentumis notconserved:
(
-2u
)m 1+u2/e2 #-2mu.
(b) mA1]A+ mB1]B=me1]e+ mD1]Dj 1]i="((ih +!3ij?).(The inverseLorentztransformation.)
mA"((ijA+!3ij~)+mB"((ijB+!3ij~)=me"((ije+!3ij~)+mD"((ijD+!3ijC];).Thegamma'scancel:
mAfiA+mBijB + !3(mAij~+mBij~)=meije +mDijD+ /3(meij~+mDijC];).
But mi1]?=p?=Ede,soif I energyisconservedI inS (EA+EB =Ee+ED),thensotooisthemomentum
(correctlydefined):
mAfiA+mBijB =meije +mDijD. qed
Problem 12.29
"(me2 - me2 =nme2 =>"( =n + 1= 1 =>1- ~= 1 .
Vl-u2/c2 C ~
. u2 - 1 - 1 - n2+2n+l-l- ~.I - vn(n +2). . ~ - (n+l)2- (11,+1)2- Tfi+T)2'U - n + 1 e.
Problem 12.30
Er =E1 +E2 +"'; pr =PI +P2+...; fir ="((pr - /3Erle)=0=>!3=vie=prelEr.
v = e2pr I Er = I e2(Pi + P2+ . . .)I (E1 + E2 + . . . ).1
Problem 12.31
(m; + m~) 2 2 (m; +m~) 1 V2 1
EJL= e = "(mJLe=>"(= = ; 1- 2" =2j
2m71" 2m71"mJL VI - V2j e2 e "(
2 1 4m2m2 m4+2m2m2+m4- 4m2m2 (m2- m2)2
~=1--=1- 71"I' = 71" 71"I' I' 71"1'= 71" I' jv=
e2 "(2 (m~+m~)2 (m~+m~)2 (m~+m~)2 (
m2 - m2
)m~+m~ e.
227
Problem 12.32
Initial momentum: E2 - p2e2 =m2e4=*p2e2= (2me2)2- m2e4 =3m2e4=*p = V3me.
Initial energy:2me2+me2=3me2.
Eachis conserved,sofinalenergyis 3me2,finalmomentumis V3me.
E2 - p2e2=(3me2)2- (V3me)2e2=6m2e4=M2e4 =*I M =V6m I ~ 2.5m.
(In thisprocesssomekineticenergywasconvertedinto restenergy,soM > 2m.)
v = pe2= V3meC2=I e .1E 3me2 V3
Problem 12.33
First calculatep,ioQ.'senergy:E2 = p2e2+m2e4= I96m2e4+m2e4= ~~m2e4=*E = ~me2.
Conservationof energy: 2me2=EA +EB
}
2E - 2 2
Conservationof momentum: tme2 =PA +PB =~- ~=*tme2=EA - EB A - me.
=*I EA =me2j I I EB =~me2.1
Problem 12.34
Classically,E = ~mv2. In a collidingbeamexperiment,the relativevelocity(classically)is twicethe
velocityof eitherone,sotherelativeenergyis 4E.
1 CD-"- ~s rCD ,E ~S
Let.5bethesystemin which<D is at rest.Its
speedv, relativeto S, is just thespeedof <D
in S.
pO =,(pO- (3p1)=*£.=, (§. - (3p), wherep is themomentumof @ in S., c c
C= ,Me2,so, = ~2;P=-,Mv =-,M(3ejE =, (-f+(3,M(3e)e=,(E +,Me2(32).
,,\,2= 1 =*1- (32= 1 =*(32=1- 1 =1'2;1. E=E E+ [(
E
)2-1 ]
Me2.
I I-IF 7 7 1" MC2 Mc2
- E2 E2 2 1- 2E2 21E =Mc2 + MC2 - Me j E =};jii - Me .
ForE =30GeVandMe2=1GeV,wehaveE=~ -1 =1800-1 =11799GeVI =160E.1
Problem 12.35
(before)
~
A
:°0
EB
(after)
Onephotonis impossible,becausein the "centerof mo-
mentum"frame(Prob. 12.30)we'dbe leftwith a photon
at rest,whereasphotonshaveto travelat speede.
0-- 0
m m
{
Cons.of energy:vpoe2+m2e4+me2=EA +EB.
C f
{
horizontal: PO=&COS60o+liILcosO=*EBCoso=poe-~EA'
}
d ddons. 0 mom.: C C Jo square an a :
vertical: 0 =& sin600- liIL sinO =*EB sinO = ~23EAjC C
228 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
22.2
)
2 1232
EB(cos ()+ sm () =Pac - POCEA+ 4"EA+ 4"EA
=>E1=poc2- POCEA+E~ =[VP5C2+m2c4+mc2- EA] 2
=POC2+m2c4+ 2VP5c2+m2c4(mc2- EA) +m2c4- 2EAmc2+E~.Or:
-POCEA=2m2c4+ 2mc2vp5c2+m2c4- 2EAVP5C2+m2c4- 2EAmc2;
EA(mc2+ Vp5C2+m2c4- poc/2) =m2c4 +mc2Jpoc2+m2c4;
E 2 (mc2+ VP5C2+m2c4) (mc2- VP5C2-t-m2c4- poc/2)A =mc .
(mc2+ VP5C2+ m2c4- poc/2) (mc2- VP5C2+m2c4- Poc/2)
(,,?c'- P6e'- ,,?c' - lPome'- ..",/ P6e'+m'c') I me' (me +2Po +vP'o +m'c')
=mc2 2 2 = - 0
(#c4 - pomc3+ p~t- P5c2- #c4) 2 (mc+ !po)
Problem 12.36
d d
{
dU
(
1
)
1 2 dU
}
F - P - mu - dt -<:2"u. dt
- dt - dt Vl- u2/C2- m Vl- u2/C2+U -2 (1- u2/c2)3/2
m
{
u(u .a)
}
d= a + . qe
Vl-u2/c2 (c2-u2)
ay"
ct
Problem 12.37
At constantforceyougo in "hyperbolic"mo-
tion. PhotonA, whichlefttheoriginat t <0,
catchesup with you, but photonB, which
passestheoriginat t > 0, neverdoes.
Problem 12.38
(a)
0 d1]o d1]odt
[
d
(
c
) ]
1
a =dr =ill dr = dt Vl- U2/C2 ~U2/C2
- C (
1
). (-~)2u'a - 1 u.a- VI - U2/C2 -2 (1- u2/c2)3/2- C(1- u2/C2)2'
a=d1]=dtd1]= 1 d
(
u
)
= 1
{
a +u(-~) -~2u'a
}dr dr dt VI - U2/C2dt VI - u2/C2 VI - U2/C2 VI - U2/C2 2 (1- U2/C2)3/2
1
[
u(u .a)
]
=1 a+ .
(1- u2/C2) (c2 - U2)
229
(b) J.L- o 2 - 1 (u .a)2 1 [(
U2
)
1
]
2
aJ.La --(a) +a:.a:--e2(I-u2/e2)4 + (l-u2/e2)4 a 1- e2 + e2u(u.a)
{ (
2
)
2
2
(
2
) }
1 1 2 2 U U 2 1 2 2
= --(u .a) +a 1- - + - 1- - (u. a) + -u (u. a)
(1 - u2/e2)4 e2 e2 e2 e2 e4
1
{2 (
U2
)2 (u.a)2 U2 U2 }= (l-u2/e2)4 a 1- e2 + e2 (,-1+2-2;2+;2)v
(1- ~)
1
[2 (u. a)2 J-I a +- (1 - u2/e2)2 (e2 - u2) .
(c) 'T]J.L'T]J.L= -e2, so iT ('T]J.L'T]J.L)= aJ.L'T]J.L+'T]J.LaJ.L= 2aJ.L'T]J.L= 0,soI aJ.L'T]J.L= 0.1
(d) KJ.L= ¥r = !r(ml}J.L)=~ I KJ.L'T]J.L = maJ.L'T]J.L= 0.1
Problem 12.39
KJ.LKJ.L= _(KO)2 + K. K. From Eq. 12.70,K. K = (1-~:/C2)'From Eq. 12.71:
KO_~dE- 1 d
(
me2
)
- me [-~ (-I/e2) 2u.a J -m (u.a)- e dr - evil - u2/e2dt viI - u2/e2 - viI - u2/e2 2 (1 - u2/e2)3/2 - e (1 - u2/e2)2'
m
[ U2(u .a) ] m(u .a)But (Eq.12.73): u.F=uFcosO= (u.a)+ 2( 2/2) = ( 2/2)3/2'SOviI - u2/e2 e 1 - u e 1 - u e
KO = uFcosO
evil - u2/C2;
F2 u2F2 COS20
[1 - (u2/e2) COS2OJK KJ.L= - = F2. qedJ.L (l-u2/e2) e2(I-u2/e2) (l-u2/e2)
Problem 12.40
m
[
u(u.a)
J
u(u . a) q /
W= a+ 2 2 =q(E+uXB)=}a+(2 2)=-vl-u2/e2(E+uXB).viI - U2/ e2 e - u e - u m
. u2(u.a) u .a q 2 2
Dotmu:(u.a)+ 2( 2/2)=( 2/2)=-vll-u/e[u.E+.u.(uvXB~;e l-u e l-u e m
=0
u(u. a) q u(u.E) q 1
:. =-vll-u2/e2 . Soa=-vll-u2/e2 [ E+uxB--u(u.E) ]. qed
(e2- U2) m e2 m e2
Problem 12.41
Onewayto seeit is to lookbackat thegeneralformulafor E (Eq. 10.29).For a uniforminfiniteplaneof
charge,movingat constantvelocityin theplane,j = 0 andp = 0, whilep (or rather,a) is independentof t
(soretardationdoesnothing).Thereforethefieldis exactlythesameasit wouldbefor aplaneat rest(except
that a itselfis alteredby Lorentzcontraction).
A moreelegantargumentexploitsthefactthatE is a vector(whereasB isapseudovector).Thismeansthat
anygivencomponentchangessignif theconfigurationis reflectedin a planeperpendicularto that direction.
But in Fig. 12.35(b),if wereflectin thexy planetheconfigurationis unaltered,sothez componentofE would
230 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
haveto staythesame.Thereforeit mustin factbezero.(By contrast,if youreflectin a planeperpendicular
to the y directionthe chargestradeplaces,so it is perfectlyappropriatethat they componentof E should
reverseits sign.)
Problem 12.42
(a) Field is ao/Eo,andit pointsperpendicularto thepositiveplate,so:
Eo=ao(cos45°x+sin45°y)=
1
;'0 (-x+y).EO v 2EO
(b) FromEq. 12.108,Ex = Exo= -~; Ey = "IEyO= "I)i~o' SoI E =~(-x +"Iy).1
(c) FromFrob. 12.10:tan0 ="I,so10=tan-I "1.1 y
(d) Let ii bea unit vectorperpendicularto theplatesin S-evidently
ii =- sinOx+cosOy;lEI= ;'°'0';1 +"12.
So the angle 4>betweenii and E is:
x
E .ii 1. cosO 2"1_
lEI =cos4>=~(smO+"IcosO) = ~(tanO+"I) =~cosO1+"12 V 1+"12 1+"12
But'" =tan0=sinlJ = v'I-cos21J= I 1 -1 =>...L20= ",2+ 1 =>cosO= 1 .
I coslJ coslJ V cos21J cas I yI+'Y2
Evidentlythefieldis~ perpendiculartotheplatesinS.
Problem 12.43
2 2 ~
( ) E - ~ q(1- v /c) R (E 1292)a - 4 (1 v2. 20)3/2 R2 q. . =>7rEO - ~ sm
SoIcos4>=C :"1"12 ).
IE. da=q(1 - v2/c2) I R2 sin2OdO d4>47rEO R2(1 - ~sin20)3/2
q(1 - V2/C2)
1
" sinOdO. . 2 2
= 27r 2 2 /' Let u =cos0,sodu= - sm0dO,sm 0=1- u .47rEO 0 (1- ~sin 0)3 2
q(l-v2/c2) (I du q(l-v2/c2) (
C
)3 {I du= 2Eo 1-1 [1- ~+~u2J3/2 = 2EO V J-I (C2 -1 +u2)3/2'c c ~
1
+1 2
(
V
)
3 2
The integralis: .ju 2 =(c2- 1) £.= -;; (1- v2/ei).(c2 - 1) c - 1+U2 -1 ~ v~ ~
I q(1 - v2/C2) (C)3 (~)3 2 =. .(So E.da= 2EO v C (l-v2/c2) q
1 -.2..~/Loq2(I-v2/c2)2vsinO Rx(fi);
(b) UsingEq. 12.111andEq. 12.92,S = /Lo(E X B) - /Lo47rEO47r R4(1- ~sin20)3 (~
-(}
231
q2 (l-v2je2)2vsinOA
S =1- 2 2 2 (J.
1611"£0 R4(1- ~sin 0)3
Problem 12.44
(a) Fields of A at B: E = 4;fO~Yj B =O. So force on qB is IF =4:£0 q~;BY.
*-x~
-L y (b) (i) FromEq. 12.68:IF =~Ty.1(Note:heretheparticleis at restin S.)
W 2 2d
(
..
) Fr E 1292
. h 0- 90°. E- - 1 qA(1- v je ) 1 A - 'Y qA A- 11 am q. . , WIt - . - 4
-
( 2j 2)3/2.ny - _4 .n yX 11"£01 - v e u- 11"£0u-
V qA (this also follows from Eq. 12.108).
B i- 0,butsinceVB=0in S, thereisnomagneticforceanyway,andIF = ~TY I (asbefore).
Problem 12.45
Here 0 =90°,~=y, lb=Z,Iz-=r, so(usinge2= IjJ-lo£o):
E=-~ 'Y A
411"£0r2 y,
B = -~ v 'Y A
411"£0C2 r2 z,
1
where'Y= VI - v2je2.
Notethat(E2- B2e2)=(~)2'Y2(1-~) =(~)2 is invariant,becauseit doesn'tdependonv. Wecan
usethisasa check.
SystemA: VA =v,soE =- q 1 A
411"£0r2 y,
1
q . ~ 1 Z,where1= Vl- v2jc2
B --- 22- 411"£0c r
2- 2 2- 2
F =q[E +(-vx) X B]= !Ll [y- ~(xX z)] = !Ll (1+~)y.~~~ ~ ~~~ ~
SystemB:
~
VB= v+v - 2v
l+v2je2 - (1+V2jc2)
'YB = - 14v2/c2 = (1+v2jc2) =(1+v2jc2)=-2 V2.
vI (t+V2/C2)2V1-2~+V4 (l-v2je2) 'Y (1+
C2),Vb'YB=2v12.
c C4
:.E=-~~12 (I+v2 )A.B- q 2v12A411"£0r2 e2 y, - -- 4
--z.
11"£0e2 r2
[
2 2 2 4 4 2 2 I 2 ]Check: E2 - B2C2 =(-L- ) ,:y4(1+ v +v - v ) =(-L- ) 14- =(-L- ) . ./41rfor I ~ C4 ~ 41rfor ¥ 41TfOr
q2 12 v2
F =qE =--4 "2(1 + "2)Y. (+qat rest=>nomagneticforce).[Check:Eq. 12.68=>FA = J-FB. ./]11"£0r c -y
SystemC: ve=O. E=-~~Yj B=O; F=qE=_L~y.
411"£0r2 411"£0r2
[Therelativevelocityof Band C is 2vj(1+ v2jc2),andthecorresponding'Y is 12(1+ v2je2).
=>Fe = -y2(t+~2/c2)FB'./]
So Eq. 12.68
232 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
Summary:
}~~ ";'-~'I"
Problem 12.46
(a)FromEq. 12.108:
-- -- -- -- 2 V v
E. B =ExBx +EyBy +EzBz =ExBx +"/ (Ey- vBz)(By+zEJ +,,/(Ez+vBy)(Bz- ZEy)c c
v ~ v2 V ~ V2
=ExBx+,,/2{EyBy+ ci/'yEz - vP(Bz - C2EzBz+EzBz- ci/'yEz +vP(Bz - c2EyBy}
= ExBx +,,/2[EyBy(1- ~:)+EzBz (1- ~:)] = ExBx +EyBy + EzBz = E .B. qed
(b) P;2 - C2jp =[E; + ,,/2(Ey- VBz)2 +,,/2(Ez +VBy)2] - c2[B; +,,/2(By+;Ez)2 + ,,/2(Bz - ;Ey)2]
=E2 +",2 (E2 - 2E £B +v2B2 +E2 +2E £B +v2B2 - C2B2 - c22vk E
x / Y "')fv, z z z 7v, y y y ;li Uy z
V2 ~ V2-c2-E2-C2B2+C22 E -c2-E2 ) -C2B2c4Z z zy c4Y x
2 2 2 2
[
2(
V2
) 2(
V2
) 2 2(
V2
) 2 2(
v2
)]
=E -cB +"/ E 1-- +E 1-- -cB 1-- -cB I--x x Y c2 z C2 Y c2 z C2
=(E2 +E2 +E2) - c2(B2 +B2 +B2) =E2 - B2C2 qedx Y z x Y z .
(c) I No.1For if B =0 in onesystem,then(E2 - C2B2) is positive.Sinceit is invariant,it mustbepositivein
anysystem.ThereforeE =I- 0 in all systems.
Problem 12.47
(a) Making the appropriate modifications in Eq. 9.48 (and picking 8 =0 for convenience),
E(x, y,z,t) = Eocos(kx- wt)Y, B(x,y,z,t) = Eocos(kx- wt)z, wherek ==~.c c
(b) UsingEq. 12.108to transformthefields:
Ex =Ez=0, Ey=,,/(Ey- vBz)=,,/Eo[cos(kx- wt)- ~cos(kx- Wi)]=o:Eocos(kx- wi),
Bx=By=0,
- v
[
1 v
]
Eo
Bz = ,,/(Bz- ZEy) = ,,/Eo- cos(kx- wt)- z cos(kx- wt) =0:- cos(kx- wi),c c c c
where 0:=="/(1-~)=
1- vlc
l+vlc.
(-),y (-),y2(1 + )y (- 47r;Or2)Y
(- 47rfQor2) "/ Z ( )2V 2A 0- 47rfOr 2"/ Z
(- 47r:r2),(1+ )Y () 2(1 v2) A (- 47r;or2)Y- 47rfOr "/ + 2 Y
233
NowtheinverseLorentztransformations(Eq. 12.19)=*x = 'Y(x+vf) andt = 'Y(f + ;x),so
kx-wt='Y[k(X+Vf)-w(f+ ;x)]='Y[(k-~~)x-(w-kv)t]=kx-wf,
-
(
WV
)where(recallingthatk =wie): k =='Y k - ~ ='Yk(1- vie) =ak andw=='Yw(1- vic)=aw..
E(x,y,z,f) =Eocos(kx- wf)y, B(x,y,z,f) = Eocos(kx- wf)z,c
- - - - /1- vlcwhereEo=aEo, k=ak, w=aw, anda =1 I
.
1+v c
Conclusion:
(c)Iw=w 1- vlc
l+vlc' ThisistheI DopplershiftI for light.
- 211" 211"- ~ The velocityof the\ - - - .
A = k - ak a
- w- w
wavein S is v=-A =\" =@]
211" A
samein any inertial system).
I Yup, I this is exactlywhatI expected(thevelocityof a lightwaveis the
(d) SinceintensitygoeslikeE2, theratiois i=~~=a2 =I ~~~~~.
DearAI,
The amplitude,frequency,and intensityof the light wavewill all I decreaseto zeroI as you
run fasterandfaster. It'll getso faintyou won'tbe ableto seeit, andso red-shiftedevenyour
night-visiongoggleswon'thelp. But it'll still be going3 x 108mls relativeto you. Sorry about
that.
Sincerely,
David
Problem 12.48
[02 = A~A;tA<7= A8A~tO2+A~A~t12= 'YtO2+ (-'Y{3)t12= 'Y(tO2- {3t12).
~3 = A~A~tA<7= A8A~tO3+ A~A~t13= 'YtO3+ (-'Y{3)t13= 'Y(tO3- {3t13)= 'Y(tO3+ {3t31).
~f23= A~A~tA<7= A~A~t23= t23.
fH = A1A~tA<7= A~AAt30+ A~Att31= (-'Y{3)t30+ 'Yt31= 'Y(t31+ {3tO3).
f12 = AiA;tA<7= AAA~tO2+AtA~t12= (-'Y{3)tO2+'Yt12= 'Y(t12- {3tO2).
Problem 12.49
Supposet"l-'= :HI-'"(+ for symmetric,- for antisymmetric).
fl<A = A I<A Atl-'"I-' "
fAI<= A;A~tl-'"= A;A~t"l-' [BecauseJ.landv arebothsummedfrom0 -+3,
it doesn'tmatterwhichwecall J.landandwhichcallv.]
[I usedthesymmetryof tl-"',andwrotetheA's in theotherorder.]= A~A;(:f:tl-"')
= :f:fI<A.qed
234 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
Problem 12.50
p"VP"V = pO°pOO- pOl pOl - pO2 pO2 - pO3 pO3 - plO plO - p20 p20 - p30 p30
+pH pH +pl2 p12 +p13 p13 +p21 p21 +p22p22 +p23 p23 +p31 p31 +p32 p32 +p33 p33
=-(Ex/C)2- (Ey/C)2- (Ez/C)2- (Ex/C)2- (Ey/C)2- (Ez/C)2+B; +B; +B; +B; +B; +B;
=2B2 - 2E2/C2 =12(B2 - ~:),
which,apartfromtheconstantfactor-~, istheinvariantwefoundinProb.12.46(b).
I G"vG""=2(E2/c2- B2) I (thesameinvariant).
P"VG"v = -2 (pOlGOl+ pO2GO2+pO3GO3)+ 2 (P12G12+p13G13+ p23G23)
(
1 1 1
)-2 -ExBx + -EyBy + -EzBz 2[Bz(-Ez/c) +(-By)(Ey/c)+Bx(-Ex/c)]C C C
2 2 I 4 I-~(E. B) - ~(E.B) =-~(E. B),
=
=
which,apartfromthefactor-4/ c, is theinvariantof Prob.
fundamentalinvariantsyoucanconstructfromE andB.]
Problem 12.51
2
} (
0 COO
)
E - L..2,\:x:- l!2.~:X:
- 4'/1"£0X - 2'/1"x p"v = JLoA -c 0 0 -v .
B - 1&2Av~ - l!2.'\v~ 21TX 0 0 0 0- 4'/1"--xy - 2'/1"-xy 0 V 0 0
12.46(a). [Theseare, incidentally,the only
Problem 12.52
8vP"v = JLoJ". Differentiate:8,,8vP"" =JLo8"J".
But8"8,,=8v8"(thecombinationissymmetric)whilepv" = -P"v (antisymmetric).
.'.8"8,,P""= O. [Why?Well,theseindicesarebothsummedfrom0~ 3,soit doesn'tmatterwhichwe
callJL,whichv: 8"8,,P"v=8v8"pv"=8,,8,,(- P"") = -8,,8vP"v.But if a quantityisequaltominusitself,
it mustbezero.]Conclusion:8"J" =O. qed
Problem 12.53
Weknowthat8"G""=0isequivalenttothetwohomogeneousMaxwellequations,V.B =0andVxE =
- ~~.All wehaveto show,then,is that8,\P"v+8"Pv'\+8"P,\"= 0is alsoequivalentto them.Nowthis
equationstandsfor 64separateequations(JL=0 ~ 3,v=0 ~ 3,A=0~ 3, and4x 4x 4=64).Butmany
ofthemareredundant,ortrivial.
Supposetwoindicesarethesame(say,JL= v). Then8,\P""+8"P,,'\+8"P,\"= O. But P"" = 0 and
P,,'\= - P'\",sothisis trivial:0= O. Togetanythingsignificant,then,JL,v, A mustallbedifferent.They
couldbeallspatial(JL,v,A= 1,2,3=x,y,z - orsomepermutationthereof),oronetemporalandtwospatial
(JL=0,v,A= 1,2or2,3,or 1,3- orsomepermutation).Let'sexaminethesetwocasesseparately.
All spatial:say,JL=1,v =2,A=3 (otherpermutationsyieldthesameequation,orminusit).
8 8 8
83Pl2+81P23+82P31=0 =>8z(Bz)+8x(B:zJ+8y(By)=0=>V .B =O.
235
One temporal:say, J1.=0,v =1,A=2 (otherpermutationsof theseindicesyieldthesameresult,or minus
it).
~FOl +8oFl2+8lF20=0 ~ :y(- ~:r:)+8(~)(B%)+:x (~y)=0,
or-!!..fft+(§j: - 8ff:r:.)=0,whichisthez componentof-~~=VxE. (If J1.=0,v =1,A=2,wegetthey
component;for v =2,A=3 wegetthex component.)
Conclusion:8>.Fp.v+8p.Fv>.+8vF>.p.= 0 is equivalentto V.B = 0 and~~= - V X E, and henceto
8vGp.v= O. qed
Problem 12.54
Ko = q'f/vFov= q('f/lFo1+'f/2Fo2+'f/3FO3)=q(TJ.E)je =I ~1'u, E.! NowfromEq. 12.71weknowthat
Ko =~dd';,.where W is the energyof the particle. Since dr =~dt,we have:
1 dW q IdW I
-1'- =-1'(u.E) ~ - =q(u.E).edt e dt
This saysthepowerdeliveredto theparticleis force(qE) timesvelocity(u) - whichis asit shouldbe.
Problem 12.55
8°1jJ=~1jJ =_!8_1jJ=_!(81jJ8~+81jJ8~+81jJ8~+ 81jJ8:).~ em emm &m ~m &m
8t 8x 8y 8z
From Eq. 12.19,we have: 8t=1', 8t =1'V, 8t = 8t =O.
- 1 81jJ 81jJ. - 81jJ v81jJ
So 8°1jJ=--1'(_8 +v-8 ) or (smceet=XO =-xo): 8°1jJ=1'(_8 - --8 l) =l' [(8oljJ) - ~(811jJ)].e t x Xo e x
8lIjJ =~1>=81jJ8t + 81>8x + 81jJ8y + 81>8z =1'3!..8cp+ 81>=1'( 81jJ - ~8cp)= [(8lljJ)- ~(8°1jJ)].8x 8t8Xl 8x8x 8y8x 8z8x e28t l'8x 8Xl e8xo l'
~1jJ=~=~m+~&+~~+~&=~=~~
8y 8t 8y 8x8y 8y8y 8z8y 8y
~1jJ=~=~m+~&+~~+~&=~=~~
8z 8t8z 8x8z 8y8z 8z8z 8z
1-(:onclusion:8p.1>transformsin thesamewayasap.(Eq. 12.27)-andhenceis a contravariant4-vector.qed
Problem 12.56
Accordingto Prob. 12.53,8ff;"~= 0 is equivalentto Eq. 12.129.UsingEq. 12.132,wefind (in thenotation
of Prob. 12.55):
8Fp.v 8Fv>. 8F>.p. 8 8 8-8 \ +-8 +_8 = >.Fp.v+ p.Fv>.+ vF>.p.x" xp. XV
= 8>.(8p.Av- 8vAp.)+ 8p.(8vA>.- 8>.Av)+ 8v(8>.Ap.- 8p.A>.)
=(8>.8p.Av- 8p.8>.Av)+ (8p.8vA>.- 8v8p.A>.)+ (8v8>.Ap.- 8>.8vAp.)=O. qed
[Notethat 8>.8p.Av= 88~v = 88~v>.= 8p.8>.Av,byequalityof cross-derivatives.]x xp. x x
2:~6 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
Problem 12.57
y
Step1: rotatefromxy to XY, usingEq. 1.29:
x = cosifJx +sinifJy
Y = - sinifJx +cosifJy
<{
~
y
Step2: Lorentz-transformfromXY to XY, using
Eq. 12.18:
X =1'(X - vt) =1'[cosifJx+ sinifJy- .Bet]
Y =Y =- sinifJ x + cosifJY
Z=Z=z
et=1'(et- .BX)=1'[et- .B(cosifJx+sinifJy)]
Step3: RotatefromXY to xii, usingEq. 1.29with negativeifJ:
x
(p x
x =cosifJX - sin ifJY =1'cosifJ[cosifJx+sinifJy- .Bet]- sinifJ[-sin4>x+cos4>Y]
=C'YCOS2ifJ+ sin2 ifJ)x + C'Y..:... 1) sin ifJcos ifJy - ,.B cosifJ (et)
ii=sin ifJX + cosifJY =l' sin ifJ(cos ifJx + sin ifJy - .Bet)+ cosifJ(- sin ifJe +cosifJ y)
=C'Y- 1)sinifJcosifJx + C'Ysin2ifJ+ COS24»y - ,.Bsin 4>(ct)
(
C
) (
l' -1'.BcosifJ -1'.BsinifJ
In matrix form: ~ = -1'.Bc~sifJ C'YCOS2ifJ+ sin2 ifJ) C'Y- 1) sin 4>cos 4>
~ -1'.BsmifJ C'Y-1)sin4>cos4>C'Ysin24>+cos2ifJ)
zOO 0 ~)m.
Problem 12.58
In center-of-momentumsystem,thresholdoccurswhenincidentener-
gy is just sufficientto coverthe restenergyof theresultingparticles,
with none"wasted"askineticenergy.Thus, in labsystem,wewant
theoutgoingK andE to havethesamevelocity,at threshold:
1f P
0-- --0
00
KE
before(CM)
after(CM)
0-- 0
1f P
Before
(X)---+-
KE
After
Initial momentum:P7T;initial energyof 1f: E2 - p2e2=m2e4:::} E; =m~c4+p~c2.
Total initial energy:mpe2+ vm;c4 +p;e2. Thesearealsothefinalenergyandmomentum:E2- p2e2=
(mK +m~:Ye4.
(mpe2+vm;c4+p;c2)2- p;C2=(mK +m!Yc4
2m e2
m;!, + -;f-vm;e2 +p; e+m;!' +rJ!c2- rJ!c2 =(mK +mE)2!,
2:p vm;e2 +p; =(mK +mE)2- m;-m;
237
(
2 2 2
)
4m; ( )4 2( 2 2)( )
2 4 4 2 2 2m".c +p". ~ = mK +mI; - mp+m". mK +mI; +mp+m". + mpm".c
4m2
-TP; = (mK +mI;)4- 2(m;+m;)(mK +mI;)2+ (m; - m;)2c
P".= 2~ v(mK +mI;)4- 2(m~+m~)(mK+mI;)2+ (m~- m~)2p
= (2m~c2)cv(mKc2 +mI;c2)4- 2[(mpc2)2+ (m".c2)2](mKc2+mI;c2)2+ [(mpc2)2- (m".c2)2]2
= 2C(~OO)V(1700)4 - 2[(900)2+(150)2](1700)2+[(900)2- (150)2]2
= ~V(8.35 X 1012)- (4.81X 1012)+ (0.62x 1012)= ~(2.04 X 106)= 11133MeVIc.!
Problem 12.59
In CM:
P P0-- --0
Before
rP
u;;;~u
sp)
After
Lx (p=magnitudeof 3-momentum
in CM, 4> =CM scatteringangle)
Outgoing4-momenta:rP =(~,pcos4>,psin4>,O)jsP=(~,-pcos4>,-psin4>,O)o
In Lab: 0---+- 0
TP
~-"S
Problem:calculate0, in termsof p, 4>.
Before
Lorentztransformation:Tx= -y(rx- {3rO)jTy =rvj Bx= -y(sx- {3s0)jBy= By.
NowE = -ymc2jP = --ymv (v hereis to the left);E2 - p2c2= m2c4,so{3= -1jfo
:. Tx=-y(pcos4>+ 1jf~)=-yp(1+COg4»j Tv =psin4>jBx=-yp(1- COg4»; By=-psin 4>.
r .S -y2p2(1- COS2 4» - p2sin24>cosO=- =
TB Vh2p2(1 + COg4»2+ p2sin24>][-y2p2(1- cos4»2+p2sin24>]
- (-y2- I) sin24>
- Vh2(1+cos4»2+sin24>]h2(1- cos4»2+ sin24>]
- (-y2- I) - (-y2- I)
- [-y2e;i~o:<p)2+I] [-y2e~i~o:<p)2+I] - V (-y2cot2~+I) (-y2tan2~+I)
238 CHAPTER 12, ELECTRODYNAMICS AND RELATIVITY
w
cos()=
V(1 +cot2~+wcot2~)(1+tan2~+wtan2~)
(wherew ==,.,?- 1)
,
() - r2 - 4 (1 + ) - 4 2 t () - -2r..sm - sin"" r - i? w - ~"( , SO an - (-y2-1)sin""
( ) . 2 I 2c2Or,since("(2- 1)="(2 1-? ="(2~, tan() ="(V2 sin </>.
w sin 1!.cos 1!.2 2
- V(CSc2~+wcot2~)(sec2~+wtan2~)- V(1+WCOS2~)(1+wsin2~)
- ~wsin</> - sin</>
- V[1 +~w(1+cos</»][1+~w(l- cos</»]- V[(~ +1)+cos</>][(~+1)- cos</>]
- sin</> - sin</> - 1 h 2- 4 4- - - , were r - - + -,
V(~+1)2-cos2</> V~+t+sin2</> VI + (rj sin</»2 W2 w
/$." .
\<0'.
0/
,x/
fL
1
w
r j sin</>
Problem 12.60 d dt dt - 1 '= mu ,* =K (a constant) =>'!lfdT=K, But dr - Vl-u2/c2' P Vl-u2/c2
'J!.. ( u )=Kyl-u2jc2, Multiplyby ~;=~:' 'dt Vl-u2/c2 m
dt ~(
u
)=!£(
u
)=K -/1- u2jc2 Let w =~dxdt VI - U2j C2 dx VI - U2j c2 m u' VI - U2j C2.
dw =K ~; wdw =~!£W2=!:.-; d(W2)=2K =>d(w2)=2K (dx),dx mw dx 2dx m dx m m
:. w2=2Kx+ constant.But at t =0,x =0andu=0(sow=0),andhencetheconstantis O.m
2K U2
W2 =-x =1 2j 2 jm -u c
2 2Kxjm - c2 ,
U =1+ 2Kx - 1+(mc2)
,
mc2 2Kx
2 2Kx 2Kx 2u =---u ;m mc2
dx c ;
dt = VI +(~;~)
u2(1+ 2Kx) =2Kx,mc2 m
J
mc2
ct= 1+ (2Kx) dx.
Let mc2 =a2,2K - , ct= J~ d..;x x. Let x ==y2jdx=2ydyj ..;x=y. o~
ct=J -/y2y+a22ydy=2J Vy2 + a2dy=[YVy2+a2+a2ln(y+Vy2+a2)]+ constant,
At t =0, x =0 =>y =0,so0=a2Ina+constant=>constant= -a2 Ina.
:. ct = YVy2 + a2+ a2ln(yja + v(yja)2 + 1) =a2 [ (~)V (~)2 + 1+In(~+/ (~)2 +1)],
Let:z ==yja =..;xv-?!b=V~'!j,ThenI ¥!d=z~ +In(z+~),I
239
Problem 12.61
......--........
(a) x(t) = f; [VI +(at)2-1], wherea = ::c' Theforceof +qon
-q will bethemirrorimageof theforceof -q on+q (in thex axis),
sothenetforceis in thex direction(thenetmagneticforceis zero).
All weneedis thex componentof E.
The fieldat +qdueto -q is: (Eq. 10.65)
q ,z
E=--( )3 [u(c2_v2)+u(~'a)-a(~'u)].47f/Oo~. U
U =Cot- v ~ Ux =ci - v = t(cl- v,z)j~.U=0'1-- ~.v =(0'1--Iv); ~.a = la. So:
Ex = -~ ,z
[~(cl- m)(c2 - V2)+~(cl - )4)la -"a(O'1-- jAJ)]47f/Oo(0'1-- vl)3 ,z "z ,
~
tca(12 - ~2)=-carP/~
=--
4q ( 1 1)3[(cl-m)(c2-v2)-cacP.]7f/Oo0'1-- V
x
............-..-........
-d/2""".""'.:'q
The forceon +q is qEx, andthereis anequalforceon -q, sothenetforceon thedipoleis:
2q2 1
F =- 47f/Oo(c,z-lv)3 [(cl - V,z)(C2- V2)- cad2]x.
It remainsto determine,z,1,
v, anda, andplugthesein.
dx c 1 1 2 cat catr /v(t) =-d
=--
2V 2a t =V ; v =v(tr)=-T ' whereT ==vI +(atr)2.t a 1+(at)2 1- (at)2
dv ca
(
1
)2a2tr ca [ 2 2]
ca
a(tr)=dtr=T +catr-2 ~ =T31+(atr)- (atr) =f3'
Nowcalculatetr: c2(t- tr)2=,z2=12+rP;1=x(t)- x(tr)= f;[VI + (at)2- VI +(atr)2], so
y - 2ttr+Yr= ~ [1+(~2 +1+(a}{)2- 2Vl +(at)2vl +(atr)2]+ (d/C)2
(*) VI +(at)2vl +(atr)2=1+a2ttr+~(':,d)2.Squarebothsides:
.4.h.4.h 1(
ad
)
4
(
ad
)
2
(
ad
)
2
,X +(at)2+(atr)2+a/ t; =,X+a/ t; +:1~ +2a2ttr+ ~ +a2ttr~
2 2 (
ad
)
2
(
d
)
2 a2
(
d
)
4
t +tr - 2ttr- ttr ~ - ~ - 4 ~ =O.
At this pointwe couldsolvefor tr in termsof t, but sincev anda arealreadyexpressedin termsof tr it is
simplerto solvefor t (in termsof tr), andexpresseverythingin termsof tr:
t2-ttr[2+(:d)2]+[t;_(~)2 _~2(~)4]=O~
t=~{tr[2+(:dr] ~ t~r +4(acdr+(:df] - ~ +4(~r+a2(~f}
=tr[1+~(:dr] ~ [1+(atr)2](~r [1+(~:r]
240 CHAPTER12. ELECTRODYNAMICS AND RELATIVITY
Whichsign?For smalla wewantt ~ tr +die, soweneedthe+ sign:
[
1
(
ad
)
2
]
d I
(
ad
)
2
t =tr 1+"2 7 + ~TD, whereD ==VI + 2e .
So1-=e(t- tr) =?1-= ~(Q:)2 +dTD. NowgobacktoEq.(*) andsolvefor0+(at)2:
VI + (at)2=~{ 1+ ~(aedf +a2tr[tr (1+~(:d f) +~TD]}
=~{II +~atr)2J[1+~(aedf] + a2~rdTD} = [1+~(aedf]T + a2~rdD.
T2
e
[ ]
e
{
rv 1
(
ad
)
2
]
a2trd
} (
d.
)l =aVI +(at2)- VI +(atr)2.=a r+"27 T+--;;-D -;t =ad 2eT+trD .
Putting all this in, thenumeratorin squarebracketsin F becomes:
[ ]= {cad(:eT+trD)- e~r[e~r(:df +dTD]}(e2- e2;:t~)-e~~d2
[
d d(atr)2
]
e2 2 2 e2a~
=cad 2eT +vb - 2eT - vb T2[1+(¥'r) - (¥'r) ] - 'f3
e2a~
[
1 1
]
e2ad2
[ ]
c2a~
='f3 "2T2- "2(atr)2- 1 = 2T3 1+(atr)2- (atr)2- 2 =- 2T3 .
q2 c2ad2 ~
. F = 3x.
. . 47r€o[(e1--lv)T]
It remainsto computethedenominator:
{ [
ctr
(
ad
)
2
] (
d .
)eatr}(c1--lv)T= e""2 7 +dTD -ad 2eT+trD T T
=[~¥d2 +cdTD- ~¥d2 - Cd(~r)2D]T=edD[~ ]=dcD.
1+(%)2_(%)2
q2 e2d2a ~
I
q2 a X
:. F =47r€oc3d3D3X = 47"€ocd[I+ (adI2e)2]3/2 (a =:J.
Energymustcomefromthe "reservoir"ofenergystoredin theelectromagneticfields.
1 q2 a [ (
ad
)
2
]3/2 q2 Jloq2(b) F =mea=- - =?1+ - = =-.
247r€ocd[I+ (adI2e)2]3/2 2e 81f€omc2d87rmdt
(forceononeendonly)
. - 2e (Jloq2 )2/3 - - 2mc2 ( JlOq2 )2/3 -..a - d 87rmd 1, so F - d 87rmd 1.
241
Problem 12.62
(a)AI' = (Vie,Ax,Ay,Az) is a 4-vector(likexl' = (et,x,y,z)),so(usingEq. 12.19):V = ,(if +vAx). But
if =0,and
A - JLo(mX f)x
x - 411" 1'3 .
Now (m X f)x =myz - mzij =myz- mzY. So
V =,vJLo(myz- mzy)411" 1'3 .
Nowx=,(x - vt) =,Rx,fj =y =Ry,Z=z = Rz, whereR is thevector(inS) fromthe(instantaneous)
locationofthedipoletothepointofobservation.Thus
V2
1'2=,2R; +R~+R~=,2(R; +R~+R~)+(1- ,2)(R~+R~)=,2 (R2- c2R2sin2B)
(whereB is theanglebetweenR andthex axis,sothat R~+R~=R2sin2B).
:. V =J.Lo v,(myRz - mzRy) .
411",3R3(1-~sin2B)3/2'
butv. (m X R) =v(mX R)x =v(myRz- mzRy), so
V = J.Lov .(m X R)(1 - ~)
411"R3(1
112. 2 n)
3/2 '
- ~ sm u
or,usingJ.Lo=~ andv .(m X R) =R.(v X m): V = 1 Ii. (vX m)(1- ~)
411"fOC2R2 (1 - ~sin2B)3/2.
(b) In thenonrelativisticlimit (V2« e2):
V= ~Ii.(vx m)=~Ii.p,
411"fo e2R2 411"fOR2
with p =v X m
e2 '
whichis thepotentialof an electricdipole.
Problem 12.63
(a) B =-EfKy (Eq.5.56);N =m X B (Eq. 6.1),soN =-EfmK(z X y).
IN =TmKxI =~(Avl2)(UV)X=EfAUV2l2X.
Chargedensityonthefrontside:AO(A=,AO);
Chargedensityonthebackside:X=7AO,whereii =1+;~/C2=>
(b) v/T7
LJv
- 1 (l+v2/c2) l+v2/c2 (l+v2/c2) (
V2
)
,= VI - 4v2/c2= VI +2112+114- 4112= VI - 2v2+114= (1- v2/e2)=,2 1+e2 .(1+v2C2)2 ~ C4 ~ ~ C4
Lengthof frontandbacksidesin this frame:lIT. Sothenetchargeon thebacksideis:
- l 2(
V2
)
A l
(
V2
)q+=A- =, 1+- -- = 1+- Al., e2 " e2
242
Net chargeon frontsideis:
CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
l >..l 1
q- =>"0-=-- =->"l."( "( "( "(2
So thedipolemoment(note:chargeson sidesareequal):
l l
[(
V2
)
l 1 l
]
>"l2
(
V2 V2
) ~l2V2P =(q+)-Y- (q-)- y= 1+- >"l-- ->"l- y =- 1+- -1 +- Y= -y.2 2 C2 2 "(2 2 2 C2 C2 C2.
u h N E
>"l2V2a
(
~ ~
)
l{Lo\ l 22~E =.£.JL2 Z,were a ="(ao,so =p X =- 2 _2 Y X z =-_2 Aa v x.<0 C €o"( "(
So apartfromtherelativisticfactorof "( thetorqueis thesamein bothsystems-but in S it is thetorque
exertedby a magneticfieldon a magneticdipole,whereasin S it is thetorqueexerte<;lby an electricfieldon
anelectricdipole.
Problem 12.64
ChooseaxessothatE pointsin thez directionandB in theyzplane:E = (0,0,E); B = (0,B cos</J,B sin</J).
Go to a framemovingat speedv in thex direction:
:it= (0,-"(vB sin</J,"((E +vBcos</J)); B = (0,"((B cos</J+ ~E), "(B sin </J).c
(1usedEq. 12.108.)Parallelprovided -"(vB sin</J - "((E+vBcos</J)
"((Bcos</J+ ?E) - "(B sin </J ' or
v v ~
-vB2 sin2</J= (B cos</J+ 2E) (E +vB cos</J)=EB cos</J+vB2 cos2</J+2E2 + 2EB cos</J,c c C
2 V 2 (
V2
) v EBcos</JO=vB +2E +EBcos</J 1+2 ; 212 =-B2 E2/ 2'C c l+v c + c
NowE X B =
x y Z
0 0 E
1
=-EBcos</Jx. So v = Ex B
0 B cos</JB sin</J 1+v21C2 B2 + E2/C2. qed
I No, I therecanbeno framein whichE ..LB, for (E. B) is invariant,andsinceit is notzeroin S it can't
bezeroin S.
Problem 12.65
x
Just before:
Field linesemanate
frompresentposition
of particle.
243
x
Just after:Fieldlinesoutsidesphereofradiusctemanatefrom
positionparticlewouldhavereached,had it keptgoingon its
original"flightplan". Insidethe sphereE = O. On the sur-
facethelinesconnectup (sincetheycannotsimplyterminate
in emptyspace),assuggestedin thefigure.
This producesa denseclusterof tangentially-directedfield
lines,whichexpandwith the sphericalshell. This is a pic-
torialwayof understandingthegenerationof electromagnetic
radiation.
Problem 12.66
Equation12.68assumestheparticleis (instantaneously)at restin S. Heretheparticleis at restin 5. So
1- - --
Fl..=-Fl.., Fli=Fli. UsingF=qE,then,I
Fx = Fx = qE:z:,
1 - 1-
Fy =-Fy =-qEy,I I
1 - 1-
Fz = -Fz = -qEz.
I I
InvokingEq. 12.108:
Fx =qEx,
1
Fy= -ql(Ey - vBz)=q(Ey- vBz),I
1
Fz = -ql(Ez + vBy) = q(Ez +vBy).I
But v x B =-vBzx +vByz, so F =q(E+v x B). Qed
Problem12.67
Z.I. .z RewriteEq.12.108withx -+y,y-+Z,Z-+x:
y fi
Ey =Ey
By=By
Ez =I(Ez - vEx) .
Bz=I(Bz+ ;Ex)
Ex=I(Ex +vBz)
Bx=I (Bx- ; Ez)
1E
~- x
v
x This givesthefieldsin system5movingin they directionat speedv.
Now E = (0,0,Eo);B = (Bo,0,0),soEy=0,Ez =I(Eo - vBo),Ex =O.
If wewantE = 0,wemustpickv sothatEo - vBo=0;i.e.Iv =Eo/Bo.1
(TheconditionEo/Bo<cguaranteesthatthereis noproblemgettingtosuchasystem.)
Withthis,By=0,Bz=0,Bx=leBo- .z.Eo)=IBo(I-~) =IBo~=~Bo;113=~Box.1
z
Thetrajectoryin 5: Sincetheparticlestartedoutat restat theorigin
in S, it startedout with velocity-vy in 5. Accordingto Eq. 12.72
it will movein a circleof radiusR, givenby
(
1
) ~p=qBR, orlmv=q -:rBo R=}~ fi
x
The actualtrajectoryis givenby I x=0; fi =-Rsinwl;z=R(1 - coswt);I whereIw = ~.1
244 CHAPTER12. ELECTRODYNAMICS AND RELATIVITY
The trajectoryin S: The LorentztransformationsEqs. 12.18and 12.19,for thecaseof relativemotionin
they direction,read:
x=x x=x
y=,(y - vt) Y=,('f)+vi)
z=z z=z
t=,(t-~y) t=,(t+~y)
So thetrajectoryin S is givenby:
x=o; y=,(-RSinwt+vi)=,{-Rsin[w,(t-;y)] +v,(t- ;y)}, or
(
V2
) [ (
V
)]Y 1+,2- = "'?vt-,Rsinw, t--y~ c2 } (y-vth=-Rsin[w,(t-~y)];
'Y2y(l-~+~)='Y2y
z =R( 1 - COS2wi) =R [1 - cosw, (t - ;y)].
So:Ix=o; y=vt-~sin[w,(t-~y)]; z=R-Rcos[w,(t- ;y)].
We cangetrid of thetrigonometrictermsby theusualtrick:
,(y - vt) =-Rsin [w,(t- ~y)]
} ::} 1,2(y- vt)2+(2'- R)2=R2.1z - R =-Rcos[w,(t- ~y)]
Absentthe ,2, this wouldbe thecycloidwefoundbackin Ch. 5 (Eq. 5.9). The,2 makesit, asit were,an
ellipticalcycloid- samepictureasp. 206,but with thehorizontalaxisstretchedout.
Problem 12.68
(a) D =foE + P suggestsE -+ l..D
}
.. .. . .
H 1B M t B fO H but 1t'Sa httlecleaner1fwed1v1debyPo whilewe'reat 1t,sothat= JloO - suggess -+Po
E -+ ~D = c2D B -+H. Then:Jloofo '
Then (followingthederivationonp. 539):
d
8 8 18
1
8DJloV
~Dov =cV.D =CPt= J fo; ~D1v = -!5""(-cD.,)+(VxH)x = (Jf)x j so ~ = J fJlo,uxV uxV c ut uxV
where1 Jj =(cPt,J f ).1Meanwhile,thehomogeneousMaxwellequations(V.B =0,E =- 0::)areunchanged,
and henceI ~ =0.1
(b)IHJloV =
{
=~:
-Hz
H., Hy
0 -cDz
cDz 0
-cDy cD.,
Hz
}
cDy
-cOD.,
D"" = {-D'
cD., cDy
CD,}
0 Hz -Hy
-cDy -Hz 0 H.,
-cDz Hy -Hz 0
245
(c)If thematerialisatrest,TJv= (-c, 0,0,0),andthesumoverv collapsesto asingleterm:
E
D/JoTJO= C2€p/JOTJO~ D/Jo= C2€p/JO~ -cD = -C2€- ~ D = €E (Eq. 4.32) .(c '
1 1 1 1
H/JoTJO= -G/JoTJo~ H/Jo= -G/Jo~ -H =--B ~ H = -B (Eq.6.31)..(
JL JL JL JL
(d) In general,TJv= ')'(-c, u), so,for JL= 0:
DOvTJv = DOlTJl +DO2TJ2 + DO3TJ3 =cDx(')'ux)+ cDy(')'uy)+ cDz(')'uz) = ')'c(D . u),
povTJv= pOlTJl+pO21J2+ pO3TJ3=Ex (')'ux)+Ey (')'Uy) +Ez(')'uz)= 1:(E.u), soc c c c
DovTJv=~€povTJv~ ')'c(D.u) =C2€(~)(E. u)~ Do u =€(E.u). [1]
y
HOvTJv= HOlTJl+ HO2TJ2+ HO3TJ3=Hx(')'ux)+ Hy(')'uy)+ Hz(')'uz)=')'(Hon),
GovTJv= GOlTJl+GO2TJ2+GO3TJ3=Bx(')'ux)+By(')'Uy)+Bz(')'uz)=')'(B. u), so
1 11
HovTJv= -GovTJv~ ')'(H .u) =-(')')(B 0u) ~ H. u =-(B. u).
JL JL JL
Similarly,forJL=1:
Dlv1Jv = DlO1JO+Dl2TJ2+Dl31J3= (-cDx)(-')'c)+Hz(')'uy)+ (-Hy)(')'uz)=')'(c2Dx +uyHz - uzHy)
= ')'[c2D +(ux H)Jz,
-E
plvTJv = plOTJO+ pl2TJ2+ pl3TJ3= =.(-')'c)+Bz(')'uy)+ (-By)(')'uz)=')'(Ex+uyBz- uzBy)c
= ')' [E+(u x B)]x' soDlvTJv=C2€plV1Jv~
[2]
1
')'[c2D+ (u x H)Jz =C2€(')')[E +(u x B)]x~ D +C2(u x H) =€ [E+(u x B)]. [3]
HlvTJv = HlOTJO+Hl2TJ2+Hl3TJ3=(-Hx)( -')'c)+(-cDzH')'uy) +(cDy)(')'uz)
= ')'c(Hx- uyDz+uzDy)=')'C[H- (uxD)]x,
GlvTJv = GlOTJo+ Gl21J2+ Gl3TJ3=(-BxH -')'c) + (- ~z)(')'Uy) + (~y) (')'uz)
= 1:(C2Bx- uyEz +uzEy) =1: [c2B - (ux E)] , soHlvTJv= ..!..GlV1Jv~
c c x JL
')'C[H- (u x D)]x= ..!..1:[c2B- (ux E)] ~ H - (u x D) =..!..
[
B - 12(u x E)]
0
JLc x JL c
UseEq. [4]asanexpressionfor H, plugthis intoEq. [3],andsolvefor D:
D+ :2UX {(UXD)+~ [B- c~(UXE)]} =€[E+(uxB)];
1 1 1
D +2" [(u.D)u - u2D]=€ [E+(ux B)]- ~(u x B) +-du x (u x E)].c ~ ~
[4]
246 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY
UsingEq. [1]to rewriteu .D:
D(I-~:)
€ 1 1
= --(E. u)u +€[E + (u x B)] - -(u x B) + - [(E. u)u- u2E]
c2 f.LC2 f.LC4
= €
{[
1- ~
]
E - 2-
[
1 - ~
]
(E. u)u +(ux B)
[
1- ~
] }
.
€f.LC4 c2 €f.LC2 €f.LC2
v= 1
- .j€ii" Then
1
~ . ,
Let 'Y= VI - u2JC2
D = 'Y2€{(1 - u:~2)E + (1 - ~:) [(u x B) - :2(E. u)u]} .
Now useEq. [3]asanexpressionfor D, plugthis intoEq. [4],andsolvefor H:
H-ux {- c~(UXH)+€[E+(UXB)]}=t[B- :2(U x E)] j
1 1
[
1
]H + 2" [(u' H)u - u2H] =- B - 2"(ux E) +€(ux E) +€[ux (ux B)].c f.L c
UsingEq. [2]torewriteu .H:
H(I-~:)
1 1
[
1
]
= -"2(B.u)u+- B-2"(uxE) +€(uxE)+€[(B.u)u-U2B]
f.Lc f.L C
= t{ [1- f.LW2]B + (€f.L- C~) [(ux E) +(B. u)~]}.
H =~{(1- ::) B+(:2 - ~) [(uxE)+(B.u)U]}.
Problem 12.69
We knowthat (proper)powertransformsasthezerothcomponentof a 4-vector:Ko =~dd~' The Larmor
2 2
formulasaysthat for v =0,dd'i = IJO6~:(Eq. 11.70).Can wethink of a 4-vectorwhosezerothcomponent
reducesto this whenthevelocityis zero? (-i,
Well,a2smellslike (aVav),but howdowegeta 4-vectorin here?Howaboutr/",whosezerothcomponent
is just c, whenv = O?Try, then:
2
KIJ = f.L
60Q3(aVall)1]IJ.7rC
This has the right transformation properties, but we must checkthat it does reduceto the Larmor formula
whenv -t 0:
dW 1dW 1 ° 1 f.LOQ2 dW f.LOq2-
d
=- -
d
=-cK =-C-
6 3(aVav)1]°,but 1]°= C'Y,so -d
=_
6 (aVav). [Incidentally,this tellst 'Y T'Y 'Y 7rC t 7rC
usthat thepoweritself(asopposedto properpower)is a scalar.If thishadbeenobviousfromthestart,we
couldsimplyhavelookedfor a Lorentzscalarthat generalizestheLarmorformula.]
247
In Frob. 12.38(b)wecalculated(oYO:v)in termsof the ordinaryvelocityandacceleration:
o:vO:v= ')'4
[
a2 + (v. a)2]=')'6 [a2')'-2+ 2..(v. a)2](C2- V2) C2
=')'6[a2(1- ~:) + c~(v.a)2]=')'6{a2- cI2[v2a2-(v.a)2J}.
Nowv. a = vacosB,whereB is theanglebetweenv anda, so:
v2a2- (v .a)2=v2a2(1- COS2B) =v2a2sin2B=IvX a12.
o:vO:v=')'6(a2-I~n.
d:; =~:q:')'6(a2-I v~an, Iwhichis Lienard'sformula(Eq. 11.73).
Problem 12.70
(a) It's inconsistentwith theconstraint"l"K" =0 (Prob.12.38(d)).
(b[We wantto finda4-vectorb"withthepropertythat (d:::+b")"l"=O.Howaboutb" ="'(d:::"lv)"l"? Then
(d:::+b")"l"=d:::"l" +'"d:::"lv("l""l,,).But "l""l"= -C2,sothisbecomes(d:::"l,,)- C2",(d:::"lv),whichis zero,
" . I /10q2
(
do:" 1 do:v
)
I .If wepIck", =l/c2. ThIs suggestsK~ad=~ ~ +C2~"lv"l". Notethat"l"=(c,vb, sothespatIal
componentsofb"vanishin thenonrelativisticlimitv « c,andhencethisstill reducesto theAbraham-Lorentz
£ I [I
.d t II v - 0 -'- d ( V ) - 0 -'- daV + v ~ - 0 daV - v d h b"lormua. ncI ena y,o: "lv- -r dr 0: "lv - -r (fT"lv 0: dr - , so (fT"lv - -0: o:v,an ence can
just aswellbewritten-c\(o:vO:v)"l"']
Problem 12.71
Definethe electriccurrent4-vectoras before:Jf = (cPe,J e), and the magneticcurrentthe sameway:
Jf::,= (cPm,Jm).The fundamentallawsarethen
OvF"v = /10Jf:, a G"V= /10J"v c m' K" = (qeF"V+ q; G"v)"lv.
ThefirstofthesereproducesV.E =(1/€o)PeandV xB = /1oJe+/1o€ooE/ot,justasbefore(p. 539);thesecond
yieldsV . B = (/10/c)(cPm)= /10Pmand-(I/c)(oB/ot +V x E) = (/10/c)Jm,or V x E = -/1oJm- oB/ot
(generalizingpage540).TheseareMaxwell'sequationswithmagneticcharge(Eq. 7.43).The third (following
theargumenton p. 540)says
K -
qe
[E ( B)]
qm
[
-c
( B )
Uy ( Ez ) Uz (Ey )]- + u x +- - + -- + -VI - U2/c2 x c VI - U2/C2 x VI - U2 /C2 c VI - u2/c2 C '
1
{qe[E+(ux B)]+qm[B - ~(u x E)] }
, or
VI - u2/C2 c
qe[E+(uxB)]+qm [B- c~(UXE)],
K1
F =
whichis thegeneralizedLorentzforcelaw (Eq. 7.69).