Instructor’s Solution Manuals to Introduction to Electrodynamics

297 Pages·2012·49.46 MB·English

by  David J. Griffiths

Instructor’s Solution Manual
Introduction to Electrodynamics
Fourth Edition
David J. Gri�ths
2014
2
Contents
1 Vector Analysis 4
2 Electrostatics 26
3 Potential 53
4 Electric Fields in Matter 92
5 Magnetostatics 110
6 Magnetic Fields in Matter 133
7 Electrodynamics 145
8 Conservation Laws 168
9 Electromagnetic Waves 185
10 Potentials and Fields 210
11 Radiation 231
12 Electrodynamics and Relativity 262
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3
Preface
Although I wrote these solutions, much of the typesetting was done by Jonah Gollub, Christopher Lee, and
James Terwilliger (any mistakes are, of course, entirely their fault). Chris also did many of the figures, and I
would like to thank him particularly for all his help. If you find errors, please let me know (gri�[email protected]).
David Gri�ths
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CHAPTER 1. VECTOR ANALYSIS 3
CHAPTER 1. VECTOR ANALYSIS 3
4 CHAPTER 1. VECTOR ANALYSIS
CHAPTER1. VECTORANALYSIS 3
Chapter 1
Chapter 1
Chapter 1
Vector Analysis
Chapter 1
Vector Analysis
Vector Analysis
Vector Analysis
Problem 1.1
✒✣
Problem 1.1
(a) FrPomrobthleemdi1a.Pg1rraombl,em|B1+.1C|cosθ3 =|B|cosθ1 +|C|cosθ2. Multiply by |A|. C ✒✣
(b)S||SAIFAAfioom×(nˆ:||ra||iBB)(AltiBas|SS|IAhFAAf·r++oier(tml+nˆ:o||Bhy||gmCCiBB(Ae:lieCBas+||n|·urt++B(tcs)lhe+nBhyCioren=CC(i:e+sataC)+dθ||)|θlu((Bvics)3=C3S||SAIanFacACiAAofeion=n)ˆargim+=s×c|ntˆ:oA||)×rsS||SAI✓||Ft✓smAAiaBBf=((Aev3=iooliCi3rB|A·amms×Bn,enˆ:Aorn||Bˆ·r++t=c|||(t|smAhl,i+BBθ(A)BAthsyplie|=eBCCa3osi|+|e:|·+eBonBCB·Arr++t+B|d||(t|=u|i|hl+B✓csi)BG)hAyBnAp|n|(a+eCiCCo3+|e:n=+tAcogiB+|sC.|t|·+i)d=Bro|i||Cθ|θuCnsBnAaE(B|v×(csi3)=sC3iAnamgtAc|Cei|o|nnˆ..|θn=·i=gBico|csC+sACn×,st1t)soHroisθ=θ(iθogi||na|(n·vsuBnB.1D)3✓=+CA3rBsaCAm.e|1✓oθti✓θnnˆˆ()Aypo=n|c+|3u31A1D)+|×+,|(+otos’tB.A|=t✓+Csn==ˆoi||i+Cof|·BnABnB1|(|pA(rBot|rAVtAc+|Ct|||+|||θr·fi)oAB|opcAhBC|p+CnseCo3×r|+|osst+io|g|Bre|co||ndsA.h|Csθ=||Co|isCC|ts|c1BnθAoieθu|c((podCssnp||3ou1Dt)|Ac+|o||cirua|cp·.isCrsBθtopncn=ˆ|Csntosocgio×1a✓|os(raonitsgAs|tθ|θ+Cegno1dis+.|θfinssBsCp2✓2✓,nedi|r1θuoi+iθt(||sd.2r2.,on|nAh|3Cuiud1D)Cocθ+..stcied.c✓|Mθdttsi2n=tTˆ||oo|CMs|ti21ufc|p(sCspsotnˆteifAoo+Ccuinaˆ|tiθ+so|rfrnun|stslB.prgc1olli.|rlsltθilθids|be|oordioid|i+botAh|Cs22,notCosuiwuspwic.u.pesidrdsscθ✓|||ott|lttClst|sMti2u2cypiysCissrprAfvv.onˆ|tciiottaiiθursnrbebceb|sthbhl.ng1oMlilo)si)udyboθuaθyeadtaisi+iwsu2t2tp,nuttsulθ|i..ty|ltisAlvid2ycθ|iAvrttsvC.eMttii2ie|ibsebph)f|snˆ)|tn)iˆouyaulsn,ˆrcylt.t.lilo|iidC.boAvtbsiwuephsy|θ)tltsnˆ2ayir|v..pAtibebht)u|yae.ttr|i!Av1|eB,|)nˆθ|S|B"1cB.e#oθ|c1cBsθot3θsθi1✯3θBo1✯B$n+!+7!C||CCθθ|(CCB22||dc"θ|co"1coB#os#ostθsθ2✒θ3θ✣p21✯B}}r$$o+||}}CB!d✲C|||||ACBuCssθii✲C2nnc|||A"cθθsts#o12ii)nnsθaθθ212n$d}}||CB✲||Assiinnθθ12
⇥ (b) For⇥the genera⇥l case, see G. E. Hay’s Vector and Tensor Analysis!, C"h#apt$er! 1,"S#ect$ion 7 (dot product) and
(b)Forthegeneralcase,seeG.E.Hay’sVector and Tensor Analysis, Chapter1,Section7(dotproduct)and
Section 8 (cross product)
(b) For the geneSreacStlieoccnatsi8eo,n(csr8eoes(scGrp.orsoEsd.upHcrtoa)dy’uscVt)ector and Tensor Analysis, Chapter 1, Section 7 (dot product) and
ProblemSe1ct.i2on 8P(rcPorborlsoesbmpler1om.d2u1c.t2)
Problem 1.2 CC C
The triple cross-TphreTohtdreuipctlertipcirsloesnsc-orpotrsoisdn-upcgrtoedinsuencrotatlisinansgsoeotnceiinraatlgiaevsneseo.rcaiFaltoairvsese.oxcFaioamrtiepvxelae.m,Fpoler, example,✻✻ ✻
suppoTsheeAtrip=leBcsruoapsspnsu-odppsrpeoCoAdsuei=csAtpBise=ranpnoBdetniCadnnicdigsuenCplaeerrripasetlnopadesiAsrcopu,cleainraadtstiiocviuenA.la,rtFhaotseroiendAxitaa,hmgearpsadliemian,g.rtahme. diagram.
Thensu(Bpp×osCe)ApoT=ihneTBtnsh(eaoBnnu×d(tBC-Co)×f-piCtsohi)nept-pesproapoinugetnte-sd,ofioa-cutunhtlde-a-orpfAa-ttgoh×ee,A-(apBn,ad×gaesAC,×ian)n(Bdpth×oAeiCn×dt)isa(pBgodri×onawtmCsn.d),opwoni,nts down, ✲✲AA==BB✲A=B
Aan×d(hBTaanh×sdeCnmha)a(.sBgnm⇥itaCuAag)nnd×dieAapt(nuhoB×Addain×seB(thBCmsAaC×)asoB..guCmnCtiB-)at.ou.gufdn-Btetihut(AuetA-dBp(e×ACagA.⇥BeBB,B)uCa)=tn.=d(AB0A0×,u,⇥tsBos(()oAB(=(A×⇥A0×BC,⇥s))BBo=p))(o⇥×Ai0n×CC,tBss=o=)d×o(0AwC0n6=×≠=,B0)×̸=CB×=B❂C×0❂C̸=❄A❄A×B×((×BB❂C××CC)❄)A×(B×C)
A (B C).
⇥ ⇥
Problem 1.3 Problem 1.3 z z✻
Problem 1.3Problem 1.3 ✻ z
✻
A=+1xˆ+1yˆ 1ˆz;A=√3; B=1xˆ+1yˆ+1ˆz;
A =+A1=xˆ++11xˆyˆ+1yˆ1Aˆz=;1A+ˆz;1=Axˆ√=+3−1p;yˆB3; B=1=ˆz1;xˆ1Axˆ+=+1√1yˆ3yˆ+;+B11ˆ=zˆz;;1BBxˆ=+=1√pyˆ33.+. 1ˆz; ✣B
A·B =A+B1=++11+−A11·BA=�·=B11+===11++A=11B−+AcB11o=sc−θo1s1==✓==A√1Bp3=c√3oAps3θB3c=occsoo√sθs3✓θ√=3cc√cooos3ssθ√θ✓⇒3==ccoo11ss..θθ. cosθ. ✣Bθ ✣✲By
· −θ=�cos−1 31 ≈70−.5288◦ ⇒) 33 ⇒ θ ❲ ✲θ y ✲y
θ = c✓os=−1co%s31�P&1r≈�o13b�7le0⇡θm.5=72108c..458o2%s◦−88&1�%31&≈70.5288◦ x✰ x✰ xA❲✰A A❲
Problem 1.4
Problem 1.4 PTrhoebclreosms-p1ro.4ductofanytwovectorsintheplanewillgiveavectorperpendiculartotheplane. Forexample,
wemightpickthe base(A)and the leftside (B):
Thecross-productofanytwovectorsintheplanewillgiveavectorperpendiculartotheplane. Forexample,
Thecross-productTofhaencyrotsws-oprvoedcutcotrosfinantyhtewpolavencetowrisllingtivheepalavnecetwoirllpgeirvpeeanvdeicctuolrapretropetnhdeicpullaanret.oFthoerpelxaanme.pFleo,rexample,
we might pick the base (A) and the left side (B):
we might pick th⃝ec2bw00ae5sPmeeai(rgsAhont)Epdaiuncckadtitothnh,eIenbcl.a,esUfetpp(seAirdS)eadad(nlBedR)ti:vheer,NleJf.tAsllidrigeht(sBre)s:erved. Thismaterialis
protectedunderallcopyrightlawsastheycurrentlyexist. Noportionofthismaterialmaybe
A= 1xˆ+2yˆ+0ˆz; B= 1xˆ+0yˆ+3ˆz.
A = 1xˆ�+2reyˆpro+du0cedˆz,;inBany=form1o�rˆxby+an0ymyˆea+ns,3wˆzit.houtpermissioninwritingfromthepublisher.
− ⃝c2005PearsonE−ducation,Inc.,UpperSaddleRiver,NJ.Allrightsreserved. Thismaterialis
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reproduced,inanyfo�crm20o1r2bPyeaarnsyonmEeadnusc,awtioitnh,oIuntcp.,eUrmppisesrioSnaidndlwerRitiivnegr,frNomJ.tAhlelrpiugbhtlisshreesre.rved. Thismaterialis
⃝c2005 PearsonEducation,Inc.,UpperSapdrdolteecRteidveurn,dNerJ.alAllclopriygrhigthstrleaswesrvaesdt.hTeyhcisurmreanttelyrieaxliisst. Noportionofthismaterialmaybe
protectedunderallcopyrightlaws astheyrecpurrordeunctelyd,eixnisatn.yNfoormpoorrtiboynaonfytmhiesanms,atweirtihaolumtpayermbeissioninwritingfromthepublisher.
reproduced,inanyformorbyany means,withoutpermissioninwritingfromthepublisher.
CHAPTER 1. VECTOR ANALYSIS 5
xˆ yˆ ˆz
A B= 1 2 0 =6xˆ+3yˆ+2ˆz.
⇥ �� �
� 1 0 3�
�� �
This has the�right dir�ection, but the wrong magnitude. To make a unit vector out of it, simply divide by its
� �
length: � �
A B =p36+9+4=7. nˆ = A⇥B = 6xˆ+ 3yˆ+ 2ˆz .
| ⇥ | A B 7 7 7
| ⇥ |
Problem 1.5
xˆ yˆ ˆz
A (B C)= A A A
⇥ ⇥ � x y z �
�(B C B C ) (B C B C ) (B C B C )�
� y z� z y z x� x z x y� y x �
=�xˆ[A (B C B C ) A (B C B C )]+yˆ()+�ˆz()
� y x y y x z z x x z �
(I’�ll just check�the x-com�ponent; the�others go the sa�me way)
=xˆ(A B C A B C A B C +A B C )+yˆ()+ˆz().
y x y y y x z z x z x z
� �
B(A C) C(A B)=[B (A C +A C +A C ) C (A B +A B +A B )]xˆ+()yˆ+()ˆz
x x x y y z z x x x y y z z
· � · �
=xˆ(A B C +A B C A B C A B C )+yˆ()+ˆz(). They agree.
y x y z x z y y x z z x
� �
Problem 1.6
A (B C)+B (C A)+C (A B)=B(A C) C(A B)+C(A B) A(C B)+A(B C) B(C A)=0.
⇥ ⇥ ⇥ ⇥ ⇥ ⇥ · � · · � · · � ·
So: A (B C) (A B) C= B (C A)=A(B C) C(A B).
⇥ ⇥ � ⇥ ⇥ � ⇥ ⇥ · � ·
If this is zero, then either A is parallel to C (including the case in which they point in opposite directions, or
oneiszero), orelseB C=B A=0, inwhichcaseBisperpendiculartoAandC(includingthecaseB=0.)
· ·
Conclusion: A (B C)=(A B) C either A is parallel to C, or B is perpendicular to A and C.
⇥ ⇥ ⇥ ⇥ ()
Problem 1.7
r =(4xˆ+6yˆ+8ˆz) (2xˆ+8yˆ+7ˆz)= 2xˆ 2yˆ+ ˆz
� �
r =p4+4+1= 3
r
rˆ = r = 2xˆ 2yˆ+ 1ˆz
3 � 3 3
Problem 1.8
(a) A¯ B¯ +A¯ B¯ =(cos�A +sin�A )(cos�B +sin�B )+( sin�A +cos�A )( sin�B +cos�B )
y y z z y z y z y z y z
= cos2�A B +sin�cos�(A B +A B )+sin2�A B�+sin2�A B sin��cos�(A B +A B )+
y y y z z y z z y y y z z y
�
cos2�A B
z z
=(cos2�+sin2�)AyBy+(sin2�+cos2�)AzBz =AyBy+AzBz. X
(b) (A )2+(A )2+(A )2 =⌃3 A A =⌃3 ⌃3 R A ⌃3 R A =⌃ (⌃ R R )A A .
x y z i=1 i i i=1 j=1 ij j k=1 ik k j,k i ij ik j k
This equals A2 +A2 +A2 provided ⌃3 R R� = 1 if j�=� k �
x y z i=1 ij ik 0 if j =k
⇢ 6 �
Moreover, if R is to preserve lengths for all vectors A, then this condition is not only su�cient but also
necessary. ForsupposeA=(1,0,0). Then⌃ (⌃ R R )A A =⌃ R R ,andthismustequal1(sincewe
j,k i ij ik j k i i1 i1
wantA2+A2+A2 =1). Likewise,⌃3 R R =⌃3 R R =1. Tocheckthecasej =k,chooseA=(1,1,0).
x y z i=1 i2 i2 i=1 i3 i3 6
Then we want 2 = ⌃ (⌃ R R )A A =⌃ R R +⌃ R R +⌃ R R +⌃ R R . But we already
j,k i ij ik j k i i1 i1 i i2 i2 i i1 i2 i i2 i1
know that the first two sums are both 1; the third and fourth are equal, so ⌃ R R =⌃ R R =0, and so
i i1 i2 i i2 i1
on for other unequal combinations of j, k. X In matrix notation: R˜R=1, where R˜ is the transpose of R.
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6 CHAPTER 1. VECTOR ANALYSIS
CHAPTER 1. VECTOR ANALYSIS 5
CHAPTER 1. VECTOR ANALYSIS 5
Problem 1.9
y ✻ y❃✻ ✿ ❃ z′✻✻y z′✻✻y
✿
✲x ✲LxooLkoinogkidnogLwodnookwtihnnegtadhxoeiwsa:nxitsh:e axis: ✒ ❄ ✒ ❄
✠ z✠ z✰✰ ■z✰✰x′ &&y■′ &&xy′
z x′ x
A 120 rotation carries the z axis into the y (= z) axis, y into x (= y), and x into z (= x). So A = A ,
� x z
A =A , A =A .
y x z y
0 0 1
R= 1 0 0
0 1
0 1 0
@ A
Problem 1.10
(a) No change. (A =A , A =A , A =A )
x x y y z z
(b) A A, in the sense (A = A , A = A , A = A )
x x y y z z
�!� � � �
(c) (A B) ( A) ( B) = (A B). That is, if C = A B, C C . No minus sign, in contrast to
⇥ �! � ⇥ � ⇥ ⇥ �!
behaviorofan“ordinary”vector,asgivenby(b). IfAandBarepseudovectors,then(A B) (A) (B)=
⇥ �! ⇥
(A B). So the cross-product of two pseudovectors is again a pseudovector. In the cross-product of a vector
⇥
and a pseudovector, one changes sign, the other doesn’t, and therefore the cross-product is itself a vector.
Angular momentum (L=r p) and torque (N=r F) are pseudovectors.
⇥ ⇥
(d)A (B C) ( A) (( B) ( C))= A (B C). So,ifa=A (B C),then a a; apseudoscalar
· ⇥ �! � · � ⇥ � � · ⇥ · ⇥ �!�
changes sign under inversion of coordinates.
Problem 1.11
(a) f =2xxˆ+3y2yˆ+4z3ˆz
r
(b) f =2xy3z4xˆ+3x2y2z4yˆ+4x2y3z3ˆz
r
(c) f =exsinylnzxˆ+excosylnzyˆ+exsiny(1/z)ˆz
r
Problem 1.12
(a) h=10[(2y 6x 18)xˆ+(2x 8y+28)yˆ]. h=0 at summit, so
r � � � r
2y 6x 18=0
� � 2y 18 24y+84=0.
2x 8y+28=0= 6x 24y+84=0 � �
� ) � �
22y =66= y =3= 2x 24+28=0= x= 2.
) ) � ) �
Top is 3 miles north, 2 miles west, of South Hadley.
(b) Putting in x= 2, y =3:
�
h=10( 12 12 36+36+84+12)= 720 ft.
� � �
(c) Putting in x=1, y =1: h=10[(2 6 18)xˆ+(2 8+28)yˆ]=10( 22xˆ+22yˆ)=220( xˆ+yˆ).
r � � � � �
h =220p2 311 ft/mile; direction: northwest.
|r | ⇡
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⃝c2005PearsponroEtedcutecdatuionnd,eIrnacl.l,cUopppyerirgShatdladwlesRaisvethr,eyNJc.urArellnrtilgyhetxsirset.seNrvoedp.orTtihoinsmofattheirsiamliasterialmaybe
protectedundreeprraoldluccoepdy,riignhatnlyawfosramstohrebyycaunrryenmtelyanesx,iswti.thNooutpopretrimonisosfiotnhiisnmwaritteirnigalfmroamytbheepublisher.
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CHAPTER 1. VECTOR ANALYSIS 7
Problem 1.13
r =(x x)xˆ+(y y )yˆ+(z z )ˆz; r = (x x)2+(y y )2+(z z )2.
� 0 � 0 � 0 � 0 � 0 � 0
(a)r(r 2)= @@x[(x�x0)2+(y�y0)2+(z�z0)2]xˆ+@@py()yˆ+@@z()ˆz=2(x�x0)xˆ+2(y�y0)yˆ+2(z�z0)ˆz=2 r .
(b) r(r1 )= @@x[(x�x0)2+(y�y0)2+(z�z0)2]�12 xˆ+ @@y()�12 yˆ+ @@z()�12 ˆz
=�12()�322(x�x0)xˆ� 12()�322(y�y0)yˆ� 12()�322(z�z0)ˆz
= ()�32[(x x0)xˆ+(y y0)yˆ+(z z0)ˆz]= (1/r 3)r = (1/r 2)rˆ .
� � � � � �
r
(c) @@x(r n)=n r n�1@@x =n r n�1(12 r1 2 r x)=n r n�1rˆ x, so r(r n)=n r n�1 rˆ
Problem 1.14
y =+y cos�+z sin�; multiply by sin�: ysin�=+y sin�cos�+z sin2�.
z = y sin�+z cos�; multiply by cos�: zcos�= y sin�cos�+z cos2�.
� �
Add: ysin�+zcos�=z(sin2�+cos2�)=z. Likewise, ycos� zsin�=y.
�
So @y =cos�; @y = sin�; @z =sin�; @z =cos�. Therefore
@y @z � @y @z
( f) = @f = @f @y + @f @z =+cos�( f) +sin�( f)
r y @y @y @y @z @y r y r z So f transforms as a vector. qed
(rf)z = @@fz = @@fy @@yz + @@fz @@zz =�sin�(rf)y+cos�(rf)z ) r
Problem 1.15
(a) v = @ (x2)+ @ (3xz2)+ @ ( 2xz)=2x+0 2x=0.
r· a @x @y @z � �
(b) v = @ (xy)+ @ (2yz)+ @ (3xz)=y+2z+3x.
r· b @x @y @z
(c) v = @ (y2)+ @ (2xy+z2)+ @ (2yz)=0+(2x)+(2y)=2(x+y)
r· c @x @y @z
Problem 1.16
r·v= @@x(rx3)+ @@y(ry3)+ @@z(rz3)= @@x x(x2+y2+z2)�32
+ @@y y(x2+y2+z2)�32 + @@z z(x2+hy2+z2)�32 i
=()�h32 +x( 3/2)()�522xi+()�32h+y( 3/2)()�522yi+()�32
� �
+ z( 3/2)()�522z =3r�3 3r�5(x2+y2+z2)=3r�3 3r�3 =0.
� � �
This conclusion is surprising, because, from the diagram, this vector field is obviously diverging away from the
origin. How, then, can v = 0? The answer is that v = 0 everywhere except at the origin, but at the
r· r·
origin our calculation is no good, since r = 0, and the expression for v blows up. In fact, v is infinite at
r·
that one point, and zero elsewhere, as we shall see in Sect. 1.5.
Problem 1.17
v =cos�v +sin�v ; v = sin�v +cos�v .
y y z z y z
�
@vy = @vy cos�+ @vz sin�= @vy @y + @vy @z cos�+ @vz @y + @vz @z sin�. Use result in Prob. 1.14:
@y @y @y @y @y @z @y @y @y @z @y
= @vy cos�+ @vy sin� ⇣cos�+ @vz cos⌘�+ @vz s⇣in� sin�. ⌘
@y @z @y @z
@vz = ⇣@vy sin�+ @vz cos�⌘= @vy⇣@y + @vy @z sin�+ ⌘@vz @y + @vz @z cos�
@z � @z @z � @y @z @z @z @y @z @z @z
= @vy sin�+ @vy cos� ⇣sin�+ @vz si⌘n�+ @vz⇣cos� cos�. So⌘
� � @y @z � @y @z
⇣ ⌘ ⇣ ⌘
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8 CHAPTER 1. VECTOR ANALYSIS
@vy + @vz = @vy cos2�+ @vy sin�cos�+ @vz sin�cos�+ @vz sin2�+ @vy sin2� @vy sin�cos�
@y @z @y @z @y @z @y � @z
@vz sin�cos�+ @vz cos2�
� @y @z
= @vy cos2�+sin2� + @vz sin2�+cos2� = @vy + @vz. X
@y @z @y @z
Problem 1.18 � � � �
xˆ yˆ ˆz
(a) v = @ @ @ =xˆ(0 6xz)+yˆ(0+2z)+ˆz(3z2 0)= 6xzxˆ+2zyˆ+3z2ˆz.
r⇥ a � @x @y @z � � � �
�x2 3xz2 2xz�
� � �
� �
� xˆ yˆ ˆz �
� �
(b) v = @ @ @ =xˆ(0 2y)+yˆ(0 3z)+ˆz(0 x)= 2yxˆ 3zyˆ xˆz.
r⇥ b � @x @y @z � � � � � � �
�xy 2yz 3xz�
� �
� �
� xˆ yˆ � ˆz
� �
(c) v = @ @ @ =xˆ(2z 2z)+yˆ(0 0)+ˆz(2y 2y)= 0.
r⇥ c � @x @y @z � � � �
�y2 (2xy+z2) 2yz�
� �
� �
Problem 1.1�9 �
� �
y As we go from point A to point B (9 o’clock to 10 o’clock), x
increases, y increases, v increases, and v decreases, so @v /@y >
x y x
v 0, while @vy/@y < 0. On the circle, vz = 0, and there is no
dependence on z, so Eq. 1.41 says
v B v
@v @v
y x
v=ˆz
r⇥ @x � @y
A x ✓ ◆
points in the negative z direction (into the page), as the right
z
hand rule would suggest. (Pick any other nearby points on the
v circle and you will come to the same conclusion.) [I’m sorry, but I
cannot remember who suggested this cute illustration.]
Problem 1.20
v=yxˆ+xyˆ; or v=yzxˆ+xzyˆ+xyˆz; or v=(3x2z z3)xˆ+3yˆ+(x3 3xz2)ˆz;
� �
or v=(sinx)(coshy)xˆ (cosx)(sinhy)yˆ; etc.
�
Problem 1.21
(i) (fg)= @(fg)xˆ+ @(fg)yˆ+ @(fg)ˆz= f@g +g@f xˆ+ f@g +g@f yˆ+ f@g +g@f ˆz
r @x @y @z @x @x @y @y @z @z
=f @g xˆ+ @g yˆ+ @g ˆz +g ⇣@f xˆ+ @f yˆ⌘+ @f ˆz⇣ =f( g)+⌘g( f⇣). qed ⌘
@x @y @z @x @y @z r r
⇣ ⌘ ⇣ ⌘
(iv) (A B)= @ (A B A B )+ @ (A B A B )+ @ (A B A B )
r· ⇥ @x y z� z y @y z x� x z @z x y� y x
=A @Bz +B @Ay A @By B @Az +A @Bx +B @Az A @Bz B @Ax
y @x z @x � z @x � y @x z @y x @y � x @y � z @y
+A @By +B @Ax A @Bx B @Ay
x @z y @z � y @z � x @z
=B @Az @Ay +B @Ax @Az +B @Ay @Ax A @Bz @By
x @y � @z y @z � @x z @x � @y � x @y � @z
⇣ A @Bx⌘ @Bz � A @By� @Bx ⇣=B ( A⌘) A⇣( B). ⌘qed
� y @z � @x � z @x � @y · r⇥ � · r⇥
� � ⇣ ⌘
(v) (fA)= @(fAz) @(fAy) xˆ+ @(fAx) @(fAz) yˆ+ @(fAy) @(fAx) ˆz
r⇥ @y � @z @z � @x @x � @y
⇣ ⌘ ⇣ ⌘ ⇣ ⌘
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CHAPTER 1. VECTOR ANALYSIS 9
= f@Az +A @f f@Ay A @f xˆ+ f@Ax +A @f f@Az A @f yˆ
@y z@y � @z � y@z @z x@z � @x � z@x
⇣ + f@Ay +A @f f@Ax⌘ A ⇣@f ˆz ⌘
@x y@x � @y � x@y
=f @Az ⇣ @Ay xˆ+ @Ax @Az yˆ+ @⌘Ay @Ax ˆz
@y � @z @z � @x @x � @y
h⇣ A @⌘f A� @f xˆ+ A� @f ⇣ A @f yˆ+⌘ iA @f A @f ˆz
� y@z � z@y z@x � x@z x@y � y@x
=f( Ah)⇣ A ( f). ⌘qed ⇣ ⌘ ⇣ ⌘ i
r⇥ � ⇥ r
Problem 1.22
(a) (A )B= A @Bx +A @Bx +A @Bx xˆ+ A @By +A @By +A @By yˆ
·r x @x y @y z @z x @x y @y z @z
⇣ + A @Bz +A @Bz⌘+A ⇣@Bz ˆz. ⌘
x @x y @y z @z
⇣ ⌘
(b)ˆr= r = xxˆ+yyˆ+zˆz. Let’s just do the x component.
r px2+y2+z2
[(ˆr )ˆr] = 1 x @ +y @ +z @ x
·r x p @x @y @z px2+y2+z2
⇣ ⌘
= 1 x 1 +x( 1) 1 2x +yx 1 1 2y +zx 1 1 2z
r p �2 (p )3 �2(p )3 �2(p )3
= 1nxh 1 x3+xy2+xz2 i = 1 hx x x2+iy2+zh2 = 1 x iox =0.
r r � r3 r r � r3 r r � r
Same goes for�the othe�r components. H� ence: �(ˆr )ˆr=� 0 . � � �
·r
(c) (v )v = x2 @ +3xz2 @ 2xz @ (xyxˆ+2yzyˆ+3xzˆz)
a·r b @x @y � @z
=⇣x2(yxˆ+0yˆ+3zˆz)+3xz⌘2(xxˆ+2zyˆ+0ˆz) 2xz(0xˆ+2yyˆ+3xˆz)
�
= x2y+3x2z2 xˆ+ 6xz3 4xyz yˆ+ 3x2z 6x2z ˆz
� �
=�x2 y+3z2 �xˆ+2�xz 3z2 2y �yˆ 3�x2zˆz �
� �
Problem 1.23 � � � �
(ii) [ (A B)] = @ (A B +A B +A B )= @AxB +A @Bx + @AyB +A @By + @AzB +A @Bz
r · x @x x x y y z z @x x x @x @x y y @x @x z z @x
[A ( B)] =A ( B) A ( B) =A @By @Bx A @Bx @Bz
⇥ r⇥ x y r⇥ z� z r⇥ y y @x � @y � z @z � @x
[B⇥(r⇥A)]x =By @@Axy � @@Ayx �Bz @@Azx � @@Axz� � � �
[(A )B] = A @ +A @ +A @ B =A @Bx +A @Bx +A @Bx
·r x x@x� y@y �z@z �x x @x � y @y z @z
[(B )A] =B @Ax +B @Ax +B @Ax
·r x �x @x y @y z �@z
So [A ( B)+B ( A)+(A )B+(B )A]
⇥ r⇥ ⇥ r⇥ ·r ·r x
= A @By A @Bx A @Bx +A @Bz +B @Ay B @Ax B @Ax +B @Az
y @x � y @y � z @z z @x y @x � y @y � z @z z @x
+A @Bx +A @Bx +A @Bx +B @Ax +B @Ax +B @Ax
x @x y @y z @z x @x y @y z @z
= B @Ax +A @Bx +B @Ay @A/x +@A/x +A @By @B/x +@B/x
x @x x @x y @x � @y @y y @x � @y @y
+Bz �@@A/zx +@@Axz + @@A/�zx +Az �@@B/zx +�@@Bxz +�@@B/zx �
= [ (A B)] (same for y and z)
r �· x � � �
(vi) [ (A B)] = @ (A B) @ (A B) = @ (A B A B ) @ (A B A B )
r⇥ ⇥ x @y ⇥ z� @z ⇥ y @y x y� y x � @z z x� x z
= @AxB +A @By @AyB A @Bx @AzB A @Bx + @AxB +A @Bz
@y y x @y � @y x� y @y � @z x� z @z @z z x @z
[(B )A (A )B+A( B) B( A)]
x
·r � ·r r· � r·
=B @Ax +B @Ax +B @Ax A @Bx A @Bx A @Bx +A @Bx +@By +@Bz B @Ax +@Ay +@Az
x @x y @y z @z � x @x � y @y � z @z x @x @y @z � x @x @y @z
� � � �
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10 CHAPTER 1. VECTOR ANALYSIS
=B @Ax +A @B/x +@B/x +@By + @Bz +B @A/x @A/x @Ay @Az
y @y x � @x @x @y @z x @x � @x � @y � @z
+Ay �@@Byx� +Az �@@Bzx +Bz @@Azx � � �
=[ (A B)] (same for y and z)
r⇥ � ⇥ �x � � � �
Problem 1.24
(f/g) = @ (f/g)xˆ+ @ (f/g)yˆ+ @ (f/g)ˆz
r @x @y @z
= g@@fx�f@@xg xˆ+ g@@fy�f@@yg yˆ+ g@@fz�f@@gz ˆz
g2 g2 g2
= g12 g @@fxxˆ+ @@fyyˆ+ @@fzˆz �f @@xgxˆ+ @@ygyˆ+ @@gzˆz = grfg�2frg. qed
h ⇣ ⌘ ⇣ ⌘i
(A/g) = @ (A /g)+ @ (A /g)+ @ (A /g)
r· @x x @y y @z z
= g@@Axx�Ax@@xg + g@@Ayy�Ay@@yg + g@@Azz�Az@@xg
g2 g2 g2
= g12 g @@Axx + @@Ayy + @@Azz � Ax@@xg +Ay@@yg +Az@@gz = gr·Ag�2A·rg. qed
h ⇣ ⌘ ⇣ ⌘i
[ (A/g)] = @ (A /g) @ (A /g)
r⇥ x @y z � @z y
= g@@Ayz�Az@@yg g@@Azy�Ay@@gz
g2 � g2
= 1 g @Az @Ay A @g A @g
g2 @y � @z � z@y � y@z
= g(rh⇥⇣A)xg+2(A⇥rg)x⌘(sa⇣me for y and z⌘)i. qed
Problem 1.25
xˆ yˆ ˆz
(a) A B= x 2y 3z =xˆ(6xz)+yˆ(9zy)+ˆz( 2x2 6y2)
⇥ � � � �
�3y 2x 0 �
� � �
� �
(A B�)= @ (6xz)�+ @ (9zy)+ @ ( 2x2 6y2)=6z+9z+0=15z
r· ⇥ � @x � @y @z � �
A=xˆ @ (3z) @ (2y) +yˆ @ (x) @ (3z) +ˆz @ (2y) @ (x) =0; B ( A)=0
r⇥ @y � @z @z � @x @x � @y · r⇥
B=xˆ⇣@ (0) @ ( 2x)⌘ +yˆ� @ (3y) @ (0)� +ˆz⇣ @ ( 2x) @ (⌘3y) = 5ˆz; A ( B)= 15z
r⇥ @y � @z � @z � @x @x � � @y � · r⇥ �
⇣ ⌘ � � ⇣ ⌘
?
(A B)=B ( A) A ( B)=0 ( 15z)=15z. X
r· ⇥ · r⇥ � · r⇥ � �
(b) A B=3xy 4xy = xy ; (A B)= ( xy)=xˆ @ ( xy)+yˆ @ ( xy)= yxˆ xyˆ
· � � r · r � @x � @y � � �
xˆ yˆ ˆz
A ( B)= x 2y 3z =xˆ( 10y)+yˆ(5x); B ( A)=0
⇥ r⇥ � � � ⇥ r⇥
�0 0 5�
� � �
� �
(A )B= x @�� +2y @ +�� 3z @ (3yxˆ 2xyˆ)=xˆ(6y)+yˆ( 2x)
·r @x @y @z � �
(B )A=⇣3y @ 2x @ (xxˆ+⌘2yyˆ+3zˆz)=xˆ(3y)+yˆ( 4x)
·r @x � @y �
⇣ ⌘
A ( B)+B ( A)+(A )B+(B )A
⇥ r⇥ ⇥ r⇥ ·r ·r
= 10yxˆ+5xyˆ+6yxˆ 2xyˆ+3yxˆ 4xyˆ = yxˆ xyˆ = (A B). X
� � � � � r· ·
(c) (A B)=xˆ @ ( 2x2 6y2) @ (9zy) +yˆ @ (6xz) @ ( 2x2 6y2) +ˆz @ (9zy) @ (6xz)
r⇥ ⇥ @y � � � @z @z � @x � � @x � @y
=xˆ( 12y 9⇣y)+yˆ(6x+4x)+ˆz(0)= ⌘21yxˆ�+10xyˆ � ⇣ ⌘
� � �
A= @ (x)+ @ (2y)+ @ (3z)=1+2+3=6; B= @ (3y)+ @ ( 2x)=0
r· @x @y @z r· @x @y �
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