1. Cross-rcfercncc Table I w t ween j~roblemnu~nl~cri n J .I). Jackson's Classical Elcctrodynamics and equivalent ~ ~ r o b l c n ~number in t h i s book. n~nltmef, In 2nd r.l~tl.~nII~II~'I'*. 14 10 14.9 14.11 , 14.10 14.11 14.11 14.13 j 14 12 14.14 '14.1) numhcr! [lul. in 2nd va ent ed~tl,,ni ltun~hrr 9.24 9.11 9.25 , 9.15 iIO.1 IO.1 10.2 10.2 "Umber] ~J;I-~In2nd r ~ l ~ t l , ~ nn111rlbrr 7.1 ' 7.11 7.2 2 7.3 . 7.3 'nunlbcr an 2nd / vJrnt ,.J,,~,,~n,,mhrr 1.1 , 1.1 1.2 1 1 . 1 ~ 1.3 ' 1.15 1.4 ' 1.3 1.5 1 1 . 4 . 4J 1.7 1.6 4.7 4.3 7,6 7.1.' 10.5 ; 10.5 1.8 1.7 4.8 4.4 numhrrn f ul in :n, VA,r~ltl,,n nun~hrr J.16 '.IV 3.19 3.20 3.20 : 3 7 3.22 , 3.14 7.7 7 . 3 10.6 10.6 I 15.1 1.9 I.8 1.10 1.9 1.11 1.10 I . l.U 1.15 . 1.12 1.16 ' 1.13 I 4.9 4.5 4.10 4.6 4.11 4.7 4.11 4 4 3 4.9 5.1 1 5J 5.2 j 5.2 2.2 5.3 ! .5.3 7.8 ; 7.14 7.10 7,12 7.h 7.14 7.77 1 5 7.15 1 7.16 'IJ6 j 7.17 7.17 7.18 'IJ8 I 7.1V 10.7 10.7 11.1 'I.' 11.2 11.2 11.3 11.4 11.4 'I" I 1 11.6 2.5 1 2.5 5.6 # .(.I I.J 8.1 12.1 5 15.13 ISJO 2.6 2.6 5.7 6 8.2 1 j 12.6 1.7 2.7 5.8 5.7 '.' 12.4 12.7 2.8 ' 2.8 5.9 5.8 12.5 11.8 16.5 16.5 16.6 16.6 16.7 I 5 4 8JI 13J 13J 6 - 8 16.8 3J 3.2 11-13 8.12 13.3 13.2 16.9 16.9 3.2 3.3 b.1 6.1 13.3 13.3 16.10 16.10 11.14 ' 11.15 11J8 1 UJl.2 3.3 3.5 3.7 3.8 3.9 3J0 3 . 3J2 3.13 3J4 3J5 15J2 15.4 I1S.V 15.5 : 15.2 I . j I5.J 15.7 / 15.4 , I S . S 15.9 3.4 3.5 3 3 3.7 3.8 3.9 JJO 3.p 3 J 1 3 15.6 6.2 6.3 6.4 6.5 6.8 9 6.10 6fi -- 6-12 6J6 7 6.2 6.3 - 6 . 4 6.5 - 6-13 6.15 .. 6.8 6.9 6.10. 7 6J8 9.2 9A3 9-5 9.6 9.8 954 9.15 9J7 9J9 9.20 9.23 9.3 9.4 9.5 9J2 9.13 9 . 1 9.7 9.8 9.9 9JO 13.4 3.5 14J 14.2 14.3 14.4 14.5 14.6 14.7 13.4 13.5 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8 114.8 I6J2 16J3 ''IJ 17.3 17" I - 1 6 u 16J1 17.1 17-2 17.3 I7q4 17.6 17.7 7 . 8 17.8

2. CONTENTS 5 l~rc'iace ", F I I ~I.'u~tl~crRcaJ~np IIII~IIJU~II~II C'l~apterI In~ruductlunto Clactrustrtic~ Cl~apter2 I5oundary - Value Prc~hlc~i~sI~;lectrostaticsI ('l~apterJ Ilourldarv . L'alue i'rohic~iisk.lcclrotalli~11 ('I1aptcr 4 hl ulttpc~les.I:lectrost~t~cs111 XIacrc~>cop~~3laJ1.1 0 1 ~ l r .I ~ I L Chapter 5 M a g ~ ~ e t o s t ~ t ~ c s Chapter 6 T i ~ n eVarying F~elJs.Sl3xwc.ll's ~-.~IIJIIOII~ * Chapter 7 Plane 1:lc.ctrc~1ilapnc.t1~W ~ v c . ,and Vavs I'r~~papatloc~ Cliapter 8 Wavc Guldes and Kesona~lt('av~ttr Cllapter 9 Silnple HaJi~t111gSy'itc~~ialld I ) ~ t ' l ' r ; l i l l ~ l l C11aptc.r 10 hlagnctohylroJyn;,~~liis.IIIJ I'~;,SIIIJIJI1!:>li Chapter II 'Special Thcury u l H e l r t ~ v ~ t y Chapter 12 Dynamics of Relativisttc Particles and I:lectro~~irp,net~c1 IcIL" :Ill ch-i ter 13 ~oeisionsbetween Cll~rgeJ1'rrticlas.l llcrgy his. r l ~ dSi.~ttar~rlg Chapter 14 Radiation by hluvil~gCl1argc.r Chapter I 5 Bremsotral~l~urg.MctIioJ UI' V~rtudl( J I I : I ~ ~ ~ .Krcl~at~bc.1ja1.1 Processers Chapter 16 Multipole Fields Chapter 17 Radiation Dam ing, Self-F~eldsof a Particle Scattering arid Absorption o f ladialion by r Buund Systcni JSJ 4 0 ,' N. B. Number system used in this book can be classified under 1hrc.ecategories: ( I ) All problems from the first edi~ionof J. D. Jackson's Classical Elec~ro- dynamicsare numbered exactly as in the text. ( 2 ) Additional problems taken from the second edition are nirrkcd with two numbers. The first nuniber follows tile scquenqe of th~sbot)k while the second number. which is in brackets and niarke3 with an astcrick, is the number in the second edition. e.g. 6.17(*6.16). 6.17 is the sequential number in this book and the question has been taken froni problem 6.16 of t he second ediiion. (3) For problems which arz taken from other sources problem statements have been included. 3. Thc rlccd f;)r a t h o r o u ~ hrevislnrr and enlarpcn~cntof t l ~ cfirst edition of tlrrs b t ~ ~ 4 . 1111prcswd011 t11c I .1rrtl111r i l y rlllnlcrous sug~estionsand criticrsrrl lrom readers and colleagues near 2nd afar: tlic rcsult 1s this volumc which con- talns rclined sl)lutlons to about two llurldrcd "classical" as well as rllodcrn problcn~son clectro~~~ap~ctisrn., l a r ~ cr~urllbcrof wlutior~sin tl~isbook arc based on tile problcrlls i r ~t111. tirsl edition of "('lassrcal IIlcctrtdynanliss" by J.1). Jackson. N'c llave not In 311 instances succccdcd In arrivln~at tlic results given by Jaikwn. and i r ~s u c l ~cascs cc Ilavc Juely indicated tllc differcnce. A few problenls o f thc second edrtiun of J;rckson's text havc also been incorporated. These arc n~arkcdoft' wrtl~an astcrrsk ( * ) for ~dentil'icatror~.Illcrc are also sonie other problems. Tlrc s o l ~ ~ t r i ~ n sare oftcn worked out in scvcral ways. Tlre bonk was first tlcs~cncdto liclp tllc upper undcrpraduatc students in Tarwan lo hcttcr appreil.ttc ~ I I Cs p l c ~ ~ d o r01 111: first cclition of Jackson's text I)! ~ I V I I I ~t t ~ cstutlcnt, !~ft'r~rsntprdit1i.i thcy 1113Y I~CCOIIICreasonably proliclent In llic u s e I?! tllc 11~a1hcn1at1;alI e i ) ~ n ~ q ~ r o~ n dapproaclles tor x)lvrng clcctro- magnetic problen~s. The purpose of t111swork is twclfold. ( I ) to train students in thinking about electromagnct~iphcnortlcna in mathematical terrrls. and ( 2 ) to Ilelp nudents to develop :I pl~vslcalrntcrprctaticin of rnathcnlatical problem:,. D~rcthanks ; t r ~ t l f r ~ t c r n ~ lcrcctrnc pivcn to Dr. I.cc Clrov. I)r W.F. J u n h ~ n .Dr '.L I cr.. I)r Scr{~oIr l'.P clcl S:rn~oJ I I ~ I I ' I ~ ~ I ~ ~ V ~ II111 . S~)IJIIC)wllo st1~rcJN I I I I 115 tilerr deep u r ~ d c r s t a n d ~ r ~ ~o f tlrc crl!>l:.i~ ~.J.;I I > ~ t r not Icast. we sclltl olrr r-Jrrll 111;rnksto 311. Lo and Sir. S11c.n Ir I & I . .I gr~tlualcstudcnr In astrvc i3lldhl~i)at 1slng Iiua Ilnrvcrsrty. T a ~ w d r ~ .h:lpt.r! I I111 prcparrng s~jlutions11, onie 14 tlrc prohlcnis i r ~i l l ~ p l c r s0 . I I ant! 1: of Jackson's text. first cdrt~ori The contri. butior~of tllc sccontl a ~ ~ t l ~ t > rto tl~rshvoh is dcdicatcd to I)r I I :I (;oul~l. Professor J D . Jackson ar~t!I'rot-essor K K.Pathrw. In spite of all the gouJ counsel u c 11ac.crccelvcd and t l ~ c(arc excrclscd during preparatron. errors undoubtedly rcnirrn; notice of such crrcvs would be rnosl wclcurnc. as woc~ldany ~ ~ r l l c rsugyestrons for furtller Inipro.cnicnt on tllc nldtcr~alin tl11. prcscnt I.,O~~L. Suen. K ~ n sf'o~> lkstcvc~.(;. A. T u n ~ h a iIJntverb~t) f'ljrnlc-r Assistant Professor of l'l~ys~cs P.O. Box 808 Universidad Franc~scodc I'aul~S;lntrr~Jcr Taichung. Tarwan. Cucuta. Colombia. 4. 711c suhjccr o f spcilnl fllncllons of ri~athcn~aticalphysics. Slumi-l tllcury. llic soltilion of ordlnary dilfcrcn~ialequations. hyperpcornctric ftj . '1, . ~ r ccovcrcd in nun! hooks. strcli a r - ( b Arlkcn. htatlicllia~icalMcll~odsfor Physicists, hcadcmic I'rcss. 1970. (I)K) t-.nucry and Krylwicki. Mat hcmnlics for Physicists. ttapcr-Row ( 1 htatlicws and Walkcr. hla~licn~at~calMe~hodsof Pliysics. Bc . 1964. ('') Wlii~~akcrand Wulwn. htodcrii Annlysis. Cambridge Press. (51F) Xtorsc and Fcshbnch. hlclliods of I r~coi:ticd I'hysics, McGr: 1053. S ~ ~ r ~ i i L ~ o i l I c1 1 c r 11. p. 424 4 4 0 ; DK. 280--2XX; .51V, Cli. 0 . . 71') -748. L)~ilcrcntlnI Equat~onsstngular polnls. clc. -- A, p. 387-408; DK. p : JtF. p. 2 ? - 5 7 6 . and p. 667-674: UW, p. 194-2 10. Ilypcrgeomc~ricfunclions - DK. p. 303- 308; MF. p. 541 ff; WW, p. 28 Lcccndre func~ions- A, Ch. 12: MF. p. 593-600 and p. 1325-1328; h! 7 ; V'.p. 355-365. (also Wa~son. Hcsscl Func~inns.Carllhridp I3owm;tn. Bcsscl Func1il)ns. 13cril1anilri. a sllort ~-olnprclicns~vc~ M M ) L e3sy to read.) Conformal mapping and ~ h cusc ol' coniple v.~rlablcs f o r IWO dlriic polcnt131problcrns: Jeans. p. 261-286. hlaxwcll. VoI. I. c h a p ~ c rXII. 15lorsc and Fcshbach. Vol. I . p 44?- 453;'ol. 2 p. 1215 - 1252. A p o d l1111cm a ~ h e m a ~ i chook on the sllhject is L. Blehcrhach, "Cor Irl~pp~ng".Chclsca Publ. Co.. N 1'. 1904. Sr~i)tllc. scct. 4.15 lo 4.31 ( p 7 0 105) pmd dissursions wrth ah . . cai~ipIcs.(Smythe. "Slar~cand I)yll~nilck.lcctrlcl~v.) ('onlorrn;rl ~r~nsl'orriial~oris. I I Kobcr. I)1ctlon3n ol' Coniornlsl r c p r c w n ~ a ~ ~ o n c .1)ovcr. 1957 pages of c1aniplc5. oilen troni ~ c r o d ~ n a ~ r ~ ~ c s .hul w ~ t hrnany f-!ii111. t'ro~ir clcctros~a~~c's.c.p p . 117; S c h w ~ r r C l ~ r i s ~ o f l c l~ransfclrrlialrt~ 1rc;llcd in Par1 IV. p, I41 -- IbS. will1 nulncrous cxarnples. I:. Durand. Eleclroslarlque el Magnetosta~~que.Masslon. Rris. 1953- 1s devo~edto twodirnenslonal polential problems with many ex . . and nice plclures of the c q u ~ p ~ c n t l a l sand stream lines. 5. I4nZqi if q's arc inside V. ./' E . ; da = , s 10 if no ch3rpc is enclosed arid tlic cqulvalcrit diffcrent~rllform is For a siniplc problem, if a Gaussian surface is constructed accordln~ to physical geometry of the problem. then tllc problcni is solved. Otherwise.. Gauss's law pves litplace's equation (V14 =O)or Poisson's equatiori (C' @ = -4 n q). wl~icliservcs as tlle starting p ~ n tor most or tlic problems. The LpLcc opcrator is geometry dependent. Thc special functions associated w ~ t hcoordlnatc systerns are shown as the following:- (i) Cartesian Coord~nates- - - othogonal cxpanslons of sine and cosine funct~ons. ( 1 1 ) Cylindrical Coordinates - - - Bessel functlon ( J p ) (111) Sphcrical ordinates - - - Legendre polynomirll (Py), associated Lcgendrc polynomial (PT).spherical harmonics (Yy,) and sphcrical Bcsrl function up). ( h ) 3fult~ p l ccxparislon - . - 1s the most often used method in elcctro- sr ~ t l i.lr,tl clccirodynrlmic problcms. Hv dlrect Taylor's series expan- ~ I ~ I I Ioi IIIC ;,~IICI.:IJI: . ( I I ~ C ~1110nop(>le( ? ). ~ I I I O I C (dlpi)lcmonient and qusdrupolc (quadrtlpole nioment a~idtlic lilglicr terms iri x' or lilglicr niultipolc mnmrnts. Ic) Green's funcrit~nG(C. 19 - It 1s 31so 3 pcneral mcthnd to obtain x>lut~nnfor pntcrit~~lIt 1s dctluccd frnrn Lrccn's tticnrcnl Green's funct~ondepends only Si' (the displacement vector from the oriejn ro the charge distribution) and it (the displacement vector from the orign to the point of obunat~on).Using the boundary 6. In complex plane. I = x + )) = rc l e drid tlie cquivalrncc of U 1s #' wllcre W = 1; + JV.Botll C: and V are real and arc called Conjup~te Fund lons. Tllc farnilics of curves U (x. y) = conslant and L (x. y) = constant arc ortllo~onal. Thcy can hc i~iterprcteclas cquipotrntlals and tl~elines of forcc in clcc~trc~~tatri problen~s.Thc introduction of transformalion in the complcx planc (cor~lurllial transformation) gives the frcedom of alternation of the elapitudr: o l arlglc, and lcngth to rcducc the problcm to 3 trivially simple form. Conjugatc functions are vcry useful in two dimcnsion~lproblcll~s,(for example infiiite parallel cylinders which can be expanded to inlinitc radius so that they become planes; or shrink to zero radius thus bccorning line charges). The expression W for thwe problems can be written as which b derived from a more gcneral txprcssion where Zi is the position of the linear charge with charge density Oi. The problem b to mnsider the equipotentials generated by two equal and opposite Linear charge U a I at y = a and 6 =-I at y = -a. The conjuate functions U and V become and X' + (y-a coth l!)'= a2 csch' U (X -3 cot V)' + y' = a' cosec'~ 7. .!I liluslratron is given in problem 2.13 by niodifyinp an esanlple form S~IIIIJI~. TIIC virtues of cornplcx ~nnsf~>rrnalionb best Jcmonstrared in Schwarlr Iran,. !;,rirl~r~onwhich pcrmits the interior of the polygon in z plane to be trrnblorllud 11) Itlr real anis of Ihe L, phnc and their relation IS given by the con~plcxderivj- I1ve 8. CHAPTER I - ( a ) In static C~UI~I~IIUIII. L: inslde a LI~II~ULI~II IIIIISI 115 , (..,LI,I.~II ~III.ILC 13 ~ U I ~ S I I U C ~ C ~ ~just urldcr tllc rurtaic oI t11c ~ ~ J I I ~ ~ L I C I U I A - E = 0 ; V . l ' = O = J r ; y i.e. 9 = O insidc the Gaussnn surl~cc. (b) Cons~ructa C~ussi~nsurface inside the conductor just under the surt'acc Since there is no electric field inside the conductor. External held induces charges on the outer surface only and the electric field terinlnatcs at the outer surface. d .c- c Einsidc = O / I f there are charges inside t h e [ ~ c o ~ ~ d u c t o rthe chargcr would induce -- equal and opposite charges on thc interia surface of th: hollow conJuctor sucl~ t11at i t encloses no net charges. There is no E field insidc the ~.o~iJuctor,but 111~ cxcess charges of the same sign and quantity distributed 011 the ouler surhcc, w l ~ i c l ~produces electric field. The electric field does not dcpcnd on [he dislll. bution of cllarges inside the hollow conductor. but on thc geometry of the hollow conductor. Froni A A (c) n . ( E 2 - E , ) = 4 no 9. 3.1-- (a) Since the potential is specified at thc 5, surface of the sphere the potential inside the sphere can be expressed in general form I 1 . Since there is no charge at the center of the sphere. therefore Blm = 0. After 3 little modifications, rl~cexpression becorncs CHAPTER 3 The spher~calharmon~cpart 18. v )is g (0.V ) = V(O c 6 < ~111). and 7 - g(O.6) = -V(nln c O ,=:) n Boundary -Value Problems in Electrostatics,I1 1 m C gf0.6) = 1 AcrnYcm(O.@) on the surface ( o;;l=-e For the case n > I : It can be shown that all the coefficient Afo must be zero and all A with P =m=odd integers should also vanish. cm Therefore, the expansion of the potential up to P-3 is:- - i 10. Physici~llythe problem depends very much on 4 and in fact. for every, n/ n the potential changes sign. So,m=O terms must -vanish that is Ap, 4. For the cau n 1. tlic potential is synimetrical with respect to the origin and also with respect Z axis. This implies tlwt the parity should be even. pYP,, (0 .6)= (-1)p-m~ p r n(0 .@I t hcrcforc,Q=m=odd must be zero also. (1)) For tlrc cpcci:~lcase n= 1: - -;VJ;!A1.l = - G I - I A,.: = - A 2 . ? = 0- A,,, E - A ; . . ~ = - iV -= - -- 35 a =A. 7- "-' The potential expansion of the potential becomes:- If wc perform a transformation of rotation with respect to x axis. we obtain Tlx problem exhibits adniuthal synlmetry. m t I t C ( r . 0 ) ~1: [ACr + Bvr I PC(sos 0) ( I t = o 3 i . h 3 1 1 d a < 1 G b . The boundary conrlit~onsArc: for 0 < 0 r 'V -( - I { r $(~,o.Q)=~t.=- I--- ( - ) ~ ~ * ~ P : ~ * I ( ~ O ~ U ) I; 9-" 'V t l T h ~ sqrccs with the case tlul .:(r. U = ;: 2 ) = '(It . r 1, hc~;ltrsc3 conlmon lcrm cosU can he bc~orizcdout in all the add Lczendre polynomrdl. Wlicn 0 = n/?, rlic surnnialion vanlshcs 311 together, 2nd grvcs ?(r. 0 = xi') = V. To show the equivalence of Eq(1) and Eq(?) of the inlcrior problem of s hollow sphere a1 poten1i31 V(U'. 0') on the surface. II IS neccs sary IO de~~lonstratcEq(l) and Eq(2) arc rlle polenlial in the tntenor of thc spherical shell of inrrer radil~sJ . For the exlerior problcnt O= ~ ( 0 . 9 ) from Eq(3.25) we have 14. ; -a - a Wa.0:4')dRA I 4n ( x ~ + a 2 - 2 a x c o ~ ) ~ ' '.. )r the interior problem El is opposlt sign Replace @(a.01 6') byan, := a t ' [O'. 6')and x by r, then #here cos 7 = cosOcos0.+ sin0sinOi-odd -@I. On the other hand a gcneral ~lutionfor a boundary-value problem is gven in Eq(3.61) 501 intcrior problcm r - 0. Din, must be set io zero. AI r=a, tlic potential sknown, so that on F Thercforc, o ( r )= r r A,-,,,(' ~YQ,,,(o.01 (4) t o m - -9 a -$(I ) and tlq(2) are tlic polentills in thc intcrror of the spherical sliell of radius a. 3y subst1tutlon the expllclt expression of /a, in Eq(4), we obtain Compring Eq(3) and Eq(4'). we notice that to show their equtvalcnce. 11 is suffictent to provc Let's consider rhe left hand side of Eq(5) as an arbitrary function expanded in spherical harmonic as given in Eq(3.58) 0. Pa(a2 - r2)On) - .. ..-- FcmMo.Q) r - Vrn=.P a(a' - r2)P nl F~rn= j dn y;rn('V@) (rz + a1 - rcosy,.~/l - ra But, -- aG-- 1 a(rl+ 2' - ?a r c ~ s y ) 3 1 ~- ar. l r ~ a 15. + h 7 it--- h =I-- m -+ ,h P1--tJ --L I h 4- m IN - h e + Icr Y 2- -d 0. 7U --I n m."- 16. - . - Llrlcar 111 u B is a X I a Ivector :., U X J ts only pokr vector pot.i~hir Quadratic In U: (E .B) l a n d (B jj (b) Ulidcr ~ i n ~ ereversal, [his ~eneralizedOhm's law does not rcrnairi invariant. In fact. it changesinto. . . . . - = -(go + p , E . ~ ) ) - i+ R ( ~ x ~ ) - & ( ~ - J ) U T l ~ c11311 errect tcrm docs not change sign. but the others all do. TIIIS refccls the physical facl that Ohm's law describes ~ h rirrcvcrsiblc curl- version of enerFy from e m fields into mechanical motion (I~eai).Clndcr s1rc11 circumst~nccsit is inconect to require time-reversal ~nvari;lnce of the pcnerali-cd Ot,mls law. 17. long as r +0. Now. 2 r - d < C O S ~ ~ S -= = d R r3 I* wllcre 0 is the angle between d s e n d F,and d flisthe solid angle subtended by. . 11 s at r. Thus. bbhcrc 52 IS the solid angle subtended at 2 by the complete circuit. ,111cr11~tit~111et11od Let 6 s' be an infinitesimal displacement o l the loop. Tllis infinitesimal displacement changes 0the solid angle formed by the closed loop and subtended at P. d f ~6 ; d (65 1 ) = x S' where (d f X 6 3 is the equivalent change inarea . d f ~II .'.. 652 0 -6;- f x3 From . I d f ? X z d B = - The mapetic-field due to one loop is (where z is the distance from the center of the loop to the point of observation and R is the radius of the circular loop) 18. p., I b2 Ba = for r' = J ( x - c ) ~ + y 3 < b (4) 2zi (aa - b2) B2 = PO I b' ~ a r '(a' - b2) for r'> b and thc vector induction $ is X X - C 6, = - & i+ B, d(x - c)' + yi J(x- c): + yl ' The actual magentic field B'is sum of El and 5'. In region r >a: d 111 thc rcpon r' < b (and r < a), we f~nd B = - /J.,Ic .,.2n( 32 - h2) ' In I ~ I Cregion r'> b but r < 2. we find ( 3 ) Erwn the symmetry ui the probleln. thc only conipollcril 01. :I 15 :I.:,' where By expanding (x- ;;')-I in Ierm of cylindrial coorJi113tc.uc O ~ I J I I , 2 1 m Of AQ(8. 1 ) = -I I 1 j tiAe1"""- 0 'LO> A/ I,,, IA S . )L,,,(kY.) 1: C 2 111 : - m ,, ~"tir'dQ~us0'6 (30s0' ) 6 ( I ' - 3 ) I We notice that n~nwsl be equal lo unltj. and A( 1.0 r A(p.1J so 1lla1PO 19. J, (kb) i3 Y I I3! I 'llolltral rule wc know that Ptrn = Pimu-----= - k P-+o Y y-o a ( k ~ ) ? Nolice t l ~ cLaplace transform for Rc(v) > - H Puttinf p a z. t -. k, a - a, v = I we get where tn rhe l ~ s rs ~ e pwc have made use of the tact that I'(!i2) = I : & -- ~- - 5.5 This problem is solved hy first cons~tler inp crnc nf the curren't carrying I o ~ pa n field source and,,the other loop ttitcr acts with thc,field wtfh a p3rttiul3r or~ent~tton,i'c.the plane ol' ~ h cloop is - In the xz plane. Thc general solur~onIS ohtained try multiplicatton of Le~endrc polynon~i31to thc result 3s 11w3s donr jn problems In ch~pter3. /' Y ( r > ~f3)dJn . / > .> > . . - = J ( J ' ( u . r ) - I , . I , B : d l7, lotrp I T 1n xr plane. u e haw J = J .0 ' a where Jo = term vantshes h 20. .. . . ' ' . ... _ .-. . . .~. . . ... . .. . . . . - . ,,. ... .. ._ I . ... . - . .: . . . . :. _ .. - . . .. . . . . - . . . . . - ' . . . .. . . . 21. Il~reto the symmetry of the problem we do not have llcrc a z-dependence. The solution of L3pIace's equation in c~rcularcylindrical coordinates is.for S i n ~ cI r C = O. uec IIJVC 1P = ,, + U, III r. Vc Call or1111tile P = O tcrl~lsrr~ 1llc general expansion of the poter~tlal.as tile L . ~ I I S I ~ I I ~I C ~ I I Iadds llothing to 1 1 1 ~liclds 2nd tl~el o g ~ r ~ t l ~ r n ~ stcrln wl~icl~has 3 si~~gularityat zero and at il~fil~ity 1s tile potential ol' a llr~carmonopole distrlhutlo~~.So tile expressloa for the poter~tialcan bc writtcn as Sirlie I'rorrr sylnrlletry 111at4, (r. 0 ) = 111(r. - U ). ho 111atCV = L)Q = O.I-or r > b IIIC 11ow11tialnlust he of tlw li)rri~. . rile lirst tcrrll gives tlu: uniform field. 11 = h = Go at large di5unccs. 111at15 I I satisfies B.C. (iv). Fur the inner regions the potentWI must he w r < a, *i = 1 bprPCOSQO I (5Y = l LZ) .rce appendix Using Eq(J) and Eq(5) in the four conditions'we obtain (all coefficients with P + 1 vanish) pa2& - pyl - a2 6, = 0 ~q.16)can be solved to give From Eq(s) 4, =61 rcosO = 61 x 22. 3 20 0 u 0 0 x g; --. 2, -- 2 0 2 5 -2.0 a -m b --3- 0 II -2-FJ - -3 Y n.' 2 G' p 0--Y w g 2 24 - II 1?-u 0- w & : % 0 - c e.2) 4 23. .- (a) Using analog of image method in electrostatia, the vectbr potential in region (I) contains two terms: , . where j*(%') is the i m g e current density with coordinates (x.y.-2). In regon ( 2 ) there is no current density. but the current density In region 1 is reduced in r~~~gnitudeand denoted by J'. ,pplyir~gthe boundary condilions a1 z = 0 (LC. 81, = pB2,; B1, = pB2y; B I Z= BZz), the following relalionshp can be obtained A Since the aAlY ~ A I X )(--- ax a~ . ,.- . . . refer to hprimcd system while the integration - of the vector potential refer to the primed system. So the differentiation can be taken iruidc the integral. The-following equations are-obtained at -- z = 0 from equation, (3). (4) and (5) respectively. Sice these equations hoid for- arbitrary x'y' and i,so the coeff. must .. vznirh. -- -- -- - - - - . - 24. Tlre D duc to the inlage loop at the cenrer of the real loop is in t l ~ c zdirt 2m' rn' lion anll has a magnitude (froni Eq(S.41) ) B, = -= -(2d)' 4d3 l lcncc rnni a 1 - 3mm' - 3n7 l2 a4 C( -- 1 F,=F,.=O. F , = - -- 4 G a d dl 1 =T= Sc2d4 (n For tlic orienration (b). B of rllc image loop is in same direction as rn b u ~it halt' as large as U, ahove F = Fx y = o , 5.10 I I In a uniformly rnagnetizcd nrcdiuni all tlie inrcrnal currents cancel otlt. Tlic 11 field can bc tlloughr as fornted by a layer of equl v2len1 "polc cli~rgc" localed on tllc rnngnc t ~ cpole faces in tile sanie rilarlncr ns an L I ? ctcc~rosl~liificklwotrld hc fornicd by e l e c l r ~ ~ cli;lrgcs so placct!. Since we arc onlv ~ n ~ c r e c ~ e d , in tlrc field along the z axis. Any arbirrar) point I' alone the L axis is cllosen as sllown in tllc diagram. The polc conliricd iri 11ic rtnc r3d1us r is dq, =--My2m dr licrc . l 1 3 1s JISO I l ~ ts11r1-a~~0 0 1 ~dens11y. - I3c;;rlrsc oi the symmetry of rllc prot)lenr, wc T IIJC t~nl!, rl~c:~si.tl component pvcn by Srnitl.irl> l o r ~ h cI[l~crpolc 1 3 ~ 2a11d !lie ~ u ~ a ln i ~ ~ r i e l i ~ticld insldc of t11c c.! llndcr - - l ~$1. in: C) + L'OS ~7 14," = 11. + 11- = ( I - ------ ------ 25. whch Is pslrrvr slnce i t IS directed upward. In lree spacc I3 = u,II. Thc above pven e*prculon is also lor B i ~ l doutside the cylnder, while inside the cylinder we use the followng equatlon IT= 6 - 4nhi or u ; IS + 4n $;. The B field inside and outside o l the iyllndrical Inaenets arc: inside: = 0 + l n " -sa -3s - 2 5 ' NEGATIVE POSITIVE This problem is not solved properly. Correct answer wns not obtained, but here presents two approaches ior discussion sake. From Eq(S.12) First approach:- It is given th;lt there is no rnacroscop~ccurrcnt. I therefore. ~ ' ( ~ 1= f m ( x ) = c ( i T ~M) and .- V X G = 4 n f x i i i 26. anJ re can simplify llle prohlcn~by inserting images. Sincc wc arc more lallliliar it11 thr ~ I C i t r i ~ a Icase. we firs1 rcplacc the actual problem. shown in Fig. I , by tllc equivalc.nt electrc,static situatiot~sshown in ~i~ 2. 3. and 4. In f'lp. 4 wc have replaccdthc polarization by Ihc crfcctive surfacc charge, o = 1'. The llllrror s y n ~ ~ n e t r yof Fig. 3 and 4 c~isuresthat tllc lit~csof rorcc arc 11orn1~1lo thr. plane wllcn tllcy cross it: hence, the right hand half of Fig. 3.and 4 gives IIIC total field (of itltcrest) in the actual problem. Since wc know tllat tllc forits can hc computed in tcrms of thc stresses in tllc ficlds and that these stresse are numerically equal in the electrical and magnetic cascs. we set tllat c ~ c ttllc ~ 3 1 1 1 ~ .forces in all cascs. Thc forces 3Ct across a thin I:~ycrof vacuc~rl~ :II tllr c ~ ~ J01 tile cyl~nJcr:accordingly. we rlecd IIOI Jistinguil~ hctwecn I!. ti. 1'. o r 1) rllerc. It L i~ very long. the liclds due to layer, of cllargi. at ;l J~ktancc1. I'rorl~tllc plane will bc ncpligiblc; we need to consider only tllc two I.l!crs t IJ (cparated hy an ~ntlnitcsimaldistance. and thc lorce pcr unit area is 2n o: = 2n 1': = 2n M'. If thc area is A = n a:. t l ~ crotnl lorcc oI' attr;~clionis Sonr of t t ~ csteps in tllcthod I are in a way not well justified. (a) hletllod I . The force due to the rnapnetic field ant1 magnetization which is unirorrn and t l ~ c7 3x1s is. -. F = J (hi - i ; ) 3 d 3 x ki(;) x a da, where R = t X A and from l:q(G.107 t A f x ) = f I i - - ; I 1 I u l ~ c r cT-----= - --=- -:-- I x -- x' I J7-ict'-= 9 9 . ~ 0 5 ( 7 - 7. )- and 9 = Qt - G1 -, But. h1 is constant and along r ;Ixlr. thercl~~rc. 1 :.; 27. a a kt kl = - J = = T and k = -- - --- 1Jung IIIC cluuhlc iurrnula + L ,/z+ L' ;us 2x = 2 cos2 x - I . we change variable of thc intcgr;l~lon. n . - n cos'? do 'F' = - 2 m ~ h l ',I 2k, ; J~=T----lB - 2k ,-- -- r - - - -- !' inner radius 3 and ourerrad~~lsI I lll.l to. The normal nearly v a n i s h outside thc wave guide. but In p n c t i a the dielectric for low frequency does not satisfy the a h mentioned condition cl S c,. 33. (4) There are scveral spccial forms of Dirac delta function th may be useful in solving problems in these few chapters. (i) a ring of chargc of radius a and total c l ~ a r gQ inside grounded conducting sphere (page 82), (ii) a uniform line charge of length ?b and total charge Q insi, a grounded conducting sphere of radius b (page 83), (iii) a circular loop of radius a carrying a currcnt I (page 141 J~ = I ~,(cos~*) 6(r'-a) a (iv) the Dirac delta function used in time-varying field and retari ed solution can bc found in Chaplcr 6 (pp. 183-188). (v) a center-fed linear antenna along the z axis (p. 278). A ~ ( x )= I sin (K k lzl)b(x) ~ ( Y ) C J . 2 The contents in chapter 10 of the text is not sufficient for solvin the problems. It is definitely necessary to refer to the references and suggeste reading of this chapter. In solving thesc problems, sometimes certain physic: conditions or concepts are ass~imed.The two termi uscd in thc Plasm? problems need a word of explanation; "fluid" means an ensemble o mobilc electrons; and 'ion' means heavier positively chargcd nuclci and ar. assumed to be "stationary". They serve as the b~ckgroundpositive-charg distribution. The electrons are subjected only to the macroscopic fluctuation.. that may be taken as harmonic oscillation. Sbme variables may have initia values, in such cases. separate the variable into two parts. For exlrnplc in~tidlya static magnetic induction has a value of Bo; then, the expredol for mapetic induction as a function of time and position can be written a, A a A A B (x, t) = co+ ei(k. r -0t) READING REFERENCES: (I) Wave guides: Feynman. Vol. 2 ---- typlsd breezy qualitative physical discussior, brrain and Corson. Chapter 13. Many typical cases with simple mathe matical analysis, no resonant cavities. Borgnis and Paps. "Encyclopaedia of Physics", Vol. XVI, Springr , I,1968. 34. The problem oC finite conductivity and degenerate modes is treated in Collin. Sect. 5.3. It is based on the paper by Pap~dopoulos.Quart. J. Mech. and Appl. hfath. -7. 325 (19S4). The carth ~ n d~onospf~creas a giant resonant cavity (Sit~urnannresonances). Schumarln, Z. Naturiorschung c.149, 250 (1951) - proposal Bdscr and Vagner, Karure Ifi8,638 (1960) - firs1 reliable o b w r ~ ~ l ~ o n For 3 survey. sec the paper hy G~lejs,Vat. bur. St~rld3rds(>OD.-- 1043 (1'965) Walt (p. 1057) anJ Hycroit (p.1071) of the wnlc journal cllcd abuvc. wtllsllcrs~ ilcll~wcll.Crary, Popc. Sm~th.J. Geophys. Kcs. -61, 139 (1956). Son-local cffccis in conductors 2nd superconductors: ..I. U. Pippard. Reports on Progrcss In Physics. 33. 176 (1900). ,Ir[iil~sby Pippard 3 r d M. Tinkham in the I961 Lcs Ilouchcs Sunlrr~cr School volumc. "Low-Tcmperaturc Physics", ed. C. DL' Wi[t, U. Drcyfus, P. G. dc Ccnr~cs.P~ppard'scontr~bution,cdlcd Oynarnlcs of Co~~ductionElcclrons, also exists as a separate book. I Efkctive Dipolc tdomcnrs of Small Aperturcs: Lord hyltigh, Phil. Mag. XLIV, 28 (1897), reprinted in lus Scientific P ~ r s .Vol IV, p.305. 11. A. ~ e t h e .Ihys. Rcv. 66, 163 S. B. Cohn, Proc. J . R . E . ~ .- 1416 (1951); 40, 106'9 (1952) ---- experimental measurement of Peff and 1nCff. Usc of Stokes parameters in astophysical studies of pulsars: Wamplcr, Scargle and Miller, Astrophys. J. Lettcrs -157. L 1 (1969) ---- optical Graham, Lyne, Smith, Nature 225, 526 (7 Feb. 1970) ---- radio Campbell, Ifeiles, Rankin, ~ a t u r e 2 5 ,- 527 (7 Feb. 1970) ) Blue sky: honardo Da Vinci (1500), translated from his notebooks, items 300- 302. p237ff. Vol. I of Jean Paul Richter, --The L i t e 9 Works of Lconardo Da Vinci. 3rd ed. Phaidon, London, 1970.--- J. T } l ~ d ~ l l ,PI:ii. Trans. Roy. S O ~ .(London), Vol 36, 343 (1878) Lurd Rayleigh, Phil. hlag. XU, 107. 274 (1871); XLVII. 375 (1899). For critical opalescence (EinsteinSmoluchowski formula) ---- see Rosenfeld, Theory of Electrons. chapter V. Sect 6.-. I____ 35. CHAPTER 6 Time Varying Fields. Maxwell's Equations (3) Show that for a system of current carrying c.lcmcnts ill crnpty space rhc total energy in the magnetic field is svhcrc J(;) is the current density. From Eq (6. 17), W = ('I . -A 2 = I z.1 . 3 11>1 Y - -.. *l 0 I1 1 - < - 3 -. x l P. P.-) l-4- 0 'I - CI X *i 8 11 -- .? -x 2m 7 ' - A x ! x 1 2x wC-.. e - I I' 3 I II x m -V o '< ,-.. 7 - 2.. -..2 - -. I + -la O_ !? s - II 0 I + W l -! (I , - fK 5 ! ' -- -l - Il 4': e -'tn 3 --7 C, -. >f, -? C -5 0"' 11 + , P r. I 01 0 5 P ", =",-r A O C) 3 0 54. N f o r I s l < a . I-- for 1 x l < a f (x) = m d N =c.I U ( X , O ) I ~ ={z' o r o r l x l > a ,/ -a * 0 f o r I x l > a sin2 [ ( A k ) 3 1 !A (k) 1 = -- 1 a z (A L)' . I A (k) l'max - 2 n (Ak)? a and Ax=-. : I m G ' A k = 3 ~ ! C s ~ n . 1 [ f k o - L ~ a ] d k--. For simpltciry. we consider only the case when the incident hcam is normal to the ~nterfacc.Thc I'oynting flux .. c1 S e n = - r Rn w - R e ~ ( ~ - ~ ) ~ l i ' ~ ~ '1 d d Since k. n and 5 are parallel. therefore i t can be written as 55. - c1 I S I = - Re [ l k l I E'o I' 1 k1 - -and - - n2 8n w kl nI where n, and nl are refractive indices for the respective rncdia. We have to calculate the ratiq of mamitudc of the square of electric fir Assuming that an incident wave colnes from the left and a reflected w: . travelling to the right and in the medium number 2 of thickness d the wavc the resultant of superposition of two waves travelling in opposite directions, a in the third medium, only a transmitted wave travelling to the Icft. This assun lion does not lose generality. In fact it is the observed consequ~ces'in f r ' laboratory. Furthermore it is given that thew media are nonpermiable i.e., pl = p a= p, = I . _I k l X Gfi7 Medium 1 7 x Fr a,= - t / E , --- k l I -- A I - - --- , ,onllnuous ---- [- (kt x L, L; X ErI - - -( k l X Ern)] IJ P ? Solvrng these equations. wc ohriancd. n nI ( I + - 1 ( I - -E]' = nl ,ik,d ~ i k ? d F ": C~(I, L,)d -,; E:- = 1 1 - 56. i thereforr. the minimum occurs when the second term is maximum i.e., when (2 ndh, ) = r/2(thikiwcst term) or d _Ir 4 7.3 Two ~ i ~ r l cscrni-infinite slabs of the urne uniform, ~ s o t r o i ~ ~ c .nun- ~wrr~lcablc.Iosslcss diclccrric with ir~dexof refraction n as sl~own111 ::re Fig. wit11 an ; ~ i ra p of width d. (3) Tlic ralru 01' transmirted power and the ratto vol' rcllcctcd powcr :o ihc incidclit powcr: Polarilrrtron is perpendicular to the phne of irlcidcncc. TIIC I)tjuod3ry cor~dilioi~s: I:,, + li, = El* + Ea- . i (1) cos r (1:. - kli ) = (- n cos i )(Ea* - Ea-1 s~k2(xsinr+ dcosr)+ El-e ik2(xsin r + dcos 1)' = l.,c ~k,(nsini+ dcori) I;?- Cik2( X sin r + d cos I) - E, -e-ika(x sin r + d cos r (4) - "'OSi ) E, &kl (x sin i + d cos i) - (T Simllsr to the method used in problem 7.2. we obtairlcd (n' cos2 i - Cos2r) + (COsa - n2cos2 i ) c ? i k 2 ( ~5 1 " ' + J C O S I ) I' . El ~-.-.- -- - - (cusr + ncosi)' - (cosr - n c o s , ) ~ , 2 i k 2 r ~ sin I + ti it) 1 1 57. -- z-5-. -- -. 2 - 2 ,;si rn - - z . ' g i - - I1 6'' , -;r 3 ='-- c - ; I + - < 0 -... . .. O ec % x =, - 3 - . +- 3 Y I, I, k q z -. -. O ' ? c c I - 3 ( - 3 1 - 1 I 2,:: ? % 8- F ,I, r n , w e .,7 5 Z i S'P Z - PJ s.+ 8 o , aI-? - % -" = r x + 1 - 0 a. = V ' I' - ? . ? !- Y -- - . 3 = I -T. 'd, - - E. I-% S i 58. V A A 0 . N O e A' 59. &I- "I-- m m y- I-. 60. as expected. : . . .- -. -- . - For d + a, X ;(I - I) -, + 0 For d + -. the transmission should be zero. the reflection should be les than 1 bewuw of ohmic losses. (c) For d nor too small - 3 2 x lo4 where we have tnadc. usc uf tl,c fact that L fi = rtcp(l - I ) a - ( I - 1) ,-A = , b . * an', ,e- A ,:= e-Zd/6 0 !r II 2n 6 3n Sketch log T. Re @ = lo-'. - 4 d - 2 d- 2d - Sd Log T = log (32 X lo-') - -- log(l + e -7e cos-) 6 6 "Very small thicknessNmeansd a or less. 61. -3(2) ' ) From h d a u and Liftshitz ("Electrodynamics of Continuous Mcdia" chapfer XI) itis stated that in a non-magnetic and transparant medium in the range of interested frequencies, the rclation betwecn thc elcctric. magnetic field and inductance u d Di = eikEk $ = The components of the dielectric tensor e~ are all real and its principal values are positive. From Maxwell equations and the knowledg that there is no net current in the medium, therefore we have (1) t x r i = " 6 (2) = coeikx - iwt C Clsing Eq( 1) and (31. we obaatned -, w1 k X t < X F ' ) t 7 fj = 0 c- W (b) T ~ Fw3vc V C L I O ~i detineci as k = nk or - n. Subst~tul~ni:to l l ~ rr r s ~ ~ l ~ C obtarned from rcstlon (3). the expression becomes Slncr Eij 1% n varinblc q~nntifythercfore. thc coefficient in the ahovc equation must vanish. i.e.. l n261j -- n,n, - c,I = 0 (6) Tllong the principal axes along x,y,z the eigen value problrrn would be 1 Writing the first term as 7 (n,' +n2' + n,')(ele7e3) and ? ~ = c k l w n clwa multiply both sides by , and after some simplifications, it ky (r,ear3) can be reduced ro 62. W1 c1 w1 c2 w1 c1 Divide every term by ( - - - ) ( -- - - ) ( - - k 1 k2 r2 3 kl -).it becomes (c) Equation (8) is quadratic equation with two roots. There are two mod. of oscillation. In jm isotropic n~cdiumthe induction 6 is a transvcrsc wave, that is 6.n = 0 whcre h is the direction of propagation (or the: is no longtudinal co~nponent).From the result obtained in part (a), u can write This rcscmbles the characteris!ic function. By Caylcy-llamiltzn Theorer- Dl and Dl must be linearly indepcndcnt eigcnvcctor i.e. fia D = 0. a 7.' The M~xwellequations in nonpcrn~eahie(fl = I ) dielectric (9 = 0) rncdiur- d .> A 1 5 '1 = 0. ,, ( I ) GXE+;,, (3) . fi = O. For one dimtnsional plane wave solution, wirhout loosing generality. we tsk: the direction of propagation lo bc the x axis and fi is the unlt normal vector ol the phnc wave. Then the Maxwell equations becomes a aE I a i i( 5 ) x - + - - = 0. ax c a t They can be reduced to 32 - - c a 4no - ax2 7 all a t -01 - a< 4no and n .dH = 0,fi .( - dt + T ddt + dx) = 0 t h ~latter two c ax 63. , h o w t h t the transversdity of E and fl fields. The transverse :nt a n be obtained from Eq(9) and the solution Tor and H are t, 50 p(x.1) if chosen to represent E and ti. p(x.t) is a function of ' , ,!el and time. If the variables are separable then ring to Eq(9), it becomes .e, the general solution medium which is homogeneous, isotropic and with a refractive ~ndex ' . J), k must satisfy the following conditions: W e + i - and k = - II (w) C C (10) ( d x , ()is real, i.c., ~ ' ( x ,t)* =p(x. t) . W W [,,yw)e-l n*(w)x + ~ * ( ~ ) c'7n*(W)x~ -QO :ard to show n(-w) = n*(w) from the above equality, however if wc 1 00 J d o e' iW' [A (w) + B(w)j1 (0. t) = -~ n d -ic t)-- I * 1 -- j d ~ e - ~ ~ ' [ A ( w )- B(w)l n o a x - fin- burier transform. we have 64. 7.8 (a) A long plane wave train w h ~frequency u exactly w. incident normally at I = 0 on a semi-infmite dielectric with refractive i n d u n(w). rile boundary conditions (i) tangential electric field: u(0. I) = B (t)c - '' sinwet (ii) normal derivative of the E field: and I " "(W)X I cwn(w) - iw2n(w) - ~ w i n ( w , ) e ~ ~ ( t J dwe-IW' .-.-- w wn(w) ((r - iwl- + w; ] (b) Causaliry. I! IS convenient to introduce sepjrate symbols for the real and ilnagnary parts of w and n. We arc 3~I:cd10 prove that .if: (i) n(w) is analyric in w, regular for rl Z 0 (ii) 1 + 0 if q 2 0(we do not seem to need this 11' we have u > 0). (iii) n -- I as I n 1 -.= for q >0 (We seem to need (Rew, Im n) bounded as In I-. +- for q > 0) then the transmitted wave ET(x. I) = 0 if x-ct > O wirh [he full expression for the transmitted wave is: WO -teiw[n(w)x-c~IIc %(x. 1) = -I2 a -00 [n(w) + I I (wf - ( w + ie)' ] d w (2) It may not he necessary to take T) >O in these condtions; 17 > 0 m3y be adcquatc. It' we consider the integral of Eq(l), the transmitted m v e whcn CJ is complex, we see that [wo2 - (w + i r)'] -' 65. - has polu at: w = fwo - ic. c >0. but is regular for r)> 0. Since v >0 or M # 0 and since n(w) is regular for rl >0, the quantity 2/(n + 1) is also regular there. And the exponential is regular for all finite values of the real parameters, x and 1. Thus. we can apply Cauchy's theorem to the the contour C: r, = 0,-R d E 0 and v > 0. w PLANE 1" Therefore, what we must prove is t h ~ tif then the integral along the semicircular arc in Eq(3) goes to zero. Let us write n = I + a + i p . + H where both a and p tend to zero as I w I OD. Then the integrand along the arc becomes, (since 1 I= -- a' - (W + ic)' R ? I1 + O(I1R)I) I + ( a + & ) 1 I i [ [ x ( ~ + a ~ c t ) / c ~ - ~ ~ ~ ( ~ + ~ ) - c t ] ~ c , - ~ ~+ir))ps/c 1 R"I+O(K)I c I + 7 (a + ip) & Wc now see that if the exponential factors remain fmite, since the Icrigth of the arc is nR, thc l/R2 factor dominates and the integal along ~lle axis. tends to zero as R goes to is bounded and can cause no trouble. If x - ct >0.we can always find an R large enough to make a small enough that x(l + a ) - ct >0 (if a > 0. there is, of cout's?. no problem). Hence, Since r) >0on the arc. Finally e-(t + ir))pxlc = e-CpxI~,-lrlpxlc must be considered. The second factor has unit absolute value but the factor ,-Evxfc could uuse trouble for [ - R. Hence, we must impose the further condit~onthat fl goes to zero fast enough as I w I -.-that 1 1 b bounded on the arc. With this condi. tion we have the required rault. It is to be noticed that this problem has been omitted in the second edition. 66. From Eq (7.93) in the tcxt 0 P 2n 2 = I - - z shows that n is only real when w > up. but it is purely imafinary when up > w. It is desirable to look for the - 00 m relationship between the rwl and the imaginary parts of the refractive index. The refractive index can be represented as where F ( o ) vanislics as w - -. which utisfics the given condition t h ~ ~ n(w) = I, whcn o --and F'(o) is analytic everywhere. Assume tl~itthere i? is a singularity o ~ ithe real axis w, > 0.Let us study the expression AI ~tifinityn ( o ) - 1, rhcrcforc thc Intcpatton approaches to 71x0. It is ana1yt1 cvcrywl~crc.Thc integration ltas to be broken into ,3 parts ( I ) from --to wo - 9; (2) 31 wo and ' (3) from w. + 9 tow. hlaking usc of Caushy Principal value notation we have Writc the rcfractivc ~ndex In tllc iornl oc' real and ima~inary parts :! n ( c ) = n'(w) + in"(o) and since n(o) = Ick(o)] l o which is an odd funct~nl with respect to w (k (w) is an even iunctioti) " n'(w') + tn"(w7 - I .- ir (n'(w)+in"(w) - I ] = I' J - dw' -00 w' - w I " nW(w') .n'(w) - I = - P J , d w' n - W w - w -I 00 w n"(w' ) . n'(w) = I + 'I' J r-. dw' " o - u - S~n~~larly. I " n'(w') - 1 n"(w) = -1 d w ' t K n-" a ' - - w (c) It is more convenient to discuss part (c) first. In cl~ssiwlmodel II1; . , , indcx of refraction is based on a collection of damped electronic 0~~5cill~lofs " 67. To verify that this a p e a the result obtained from (a) it is necessary to expreu n(w) in terms of real and imaginary parts separately. 2 n ~ e ' imaginary part:- nS*(w)= ----C fkvk w = nW(wk) m (ak' - a')' + Uk2w' k nien. at the pole i.e., w2uk' = -( a t 2 - 0')' aka- a' n' (w) = I + Z n"(wk) = l + C n ' b k ) U k U k YkW -(wk2 - a') d ~ k I, - = - 7iwk Change wk to o' and thc sunirmtion to integration. dfdk It is more convenient to use the summation form of n(&) Differentiate both sideswith respect to In the frequency m g c where the resonant absorption occurs i.e., dn(w) -2nNe2 ?wk + iuk w = wk, then -= dw I: fk m Y'W' If we choose N big enough so b t the whole expression is much bigger dn(u). n unity, -1s a large negative value. Thisis only possible in an anormalous dw persivc region where the group velocity is grcata than the velocity of light vacuo. ere Kt is characterirtd by the resonant absorption frequency. 68. (Altanatlvcmcchod) - . - . . ' , - y 1 .Smcc n(u') is d y t k and repub ta th u ~ p ahif of the a' phnc, then the function . $ 6 is analytic and regular there, except for a pole at w* = o.Let US take w to be real. Let us integrate next the function deried by Eq(1) around the semi- circular closed contour C shown below. I Witlr (Ire lndcntat~onof radius e around w' = w. the ir~rcgrandIS rcgul~rin and on the contour C, and Cauchy's theorem gives Since n(w') + i as Io'I +-, the contribution of the arc of radius R Therefore If we also pass to the limit as e + 0.the first two integrals arc jus: ~llcC~uchy principal value and the integral around the indentation is half the integral (clockwise) around the circle around the pole o'= w, and the integral is given dlrcctly by the Cauchy res~duetheorem. We observe that on the left hand s~dcof Eq(?), w' is now real. Taking the real and imaginary p r t s of both sides of Eq('-) we obtain 1 " Im(n(w0)) Ke(n(w)l = I + - P I , dw' n -aa w - w 69. , ,%?II; - r c ~ l ~ l s ~ ~ ~('.71 ill tlic ten!. p r l s (a) and (b), we see l l i a l in order l o hc .: ~ ~ ~ ~ r l ~ i ~ l ~ l clo rcal ficlds. E and 8, we must have - 11131 is. Kcn(-w) = Re n(w) : Irn n(--w) = - In1n(w) Tlirrcforc. upori usirig as tlic variable of integration w' = x wlicn w' < 0 and w' = y wlien a'> 0.was obtain . - . -4 I rn Irn n(w') 0 Im n(x) dx rn Irn n(v) d d = P $ + P I - - m a' - W - m X - W 0 Y - a dy . - I I -'Y Let us sct x = -y and use -- -= - y - w y - w y = - w 2 ' W' 1111n(w' ) Let ussct next y = w' toget Rcn(w)= I + P J d d n (1 w'? - w2 m w l R c n ( w ) - l ) Likewise. In1ri(w) = - -:-P -- --- dw' n , w' -- W 7.10 11 pl~ncelccrrcn~~gnct~cwave 1s incidc~itnormally on a cnndrlctor whose d~clccrric cons~ant and pcrrncahilitv arc t l i ~of free sp~cc. Tlic 1freq~~cncynntl conttrrcr~v~ryarc such 11131, wltluri tlic conductor, tl~ccon- . duct~oncurrent and d~splacementcunent are equal in rnagnltudc. What 1s thc 4 rcflcct~oncocllir~en~.i.c.. tlic ratlo of reflcctcd to inciderit energy? ... -Our prohlcr~i1s that of the reflect~ono i J planc clcc~rornacnct~~u.~c - from a conduit~npplane surface: - J = a F and a t ~ ' d ~ ~ ~ l a a m e n ~= 0 - = -I w c o F a t .. the iJoundary condtions on and 11 arc: P , and H~~~~continuous. i.e., E tanc El + Eel = E; H, = H, - H; P3 In lbuser's textbook appear on page 460, the continuity conditions on the ' tangential components of !!.nd !f for the case of an incoming wave whose 3electric vcctor iz perpendicujsr to thc planc ol lnc~denie;these condit~onsare - expressed by El + E; = E, KI cos8, ~1 case, + it [E; - 11 = - El U P 1 W P 1 !03 i 70. -- The solution oT these equations.~ .- and ZK, and E2= El K2 + K , + iy Ylcldll~pTor the arnpl~ludcreflection coefficient Tile energy reflection cocfLcient is 1 R I' :the values of K l , K, and y are . . bY where we h v e used the fact tht cI = E,. p, = P O : wc also have ( c i 1 ' ( I 5.01) and ( 1 5.62) of Hauser's text) wtli 6, = 0.o =- w co. 71. In circular polarization basis: 2 6 + a &- 1e- iwt E = ( a + e e+ - 1 b = a+' + a- , s1 = 23+a cos(6- - 6+), -.- --(a) Stokes paramelm: b = 3. Sl = -- 1. s, = 2--- 5, = -2. n Linear baas = 1, a2 = fl, 6z -61 = - -4 s3 < 0 helicity t, semimajor axis = 1 +d-T 2 -- I - -6- - '+= 58". ~mi-minoraxis = 2 - - 2 - -- 72. - (b) Stoker parameters: a .r, 25, st = 0.- sa - 27,- S, = 7. 5 5 I , = -,E' 24 7 cosQ2 - & I ) = - sin(6, - 6,) = - 25 ' 25 I i 7 'XU 1 1 J i a + = 4, a _ = 3, cos(6--b*) = 0 A plane wave of frequency w is incident normally from vacuur~lon 3 sc1111- inlinite slab of material with a complex index of refraction n(w) ( I I ' = c(wj] Fro111 Eq(7.58) with due attention to the convcntlon about tlie direction of E" in Fig 7.10. we find E: - 2,3 Transmitted amplitude: - - .-!.+ c' 1 - n Reflected amplitude: - u Eo I + n Examination of the derivation shows that n a n be couple. Sinularly 1 fi' = 6(k'x k?)= n(k x E') holds n complex. Transmitted power 2 2 4Re n = R e [ n * - -1 =-2zT lnc~dentpower - . l + n * 1 + n I 1 +nl - w H'olHo - f 6 / ~ o (b) Lookat ----- . ---, f i w - - A - Iw 4(n2)* 4 1 nl' )Eo2Je . 2 1 r n n ( ~ ) T 1w RC [ -(E D*-B . H*) ]=Re [-(--- t) n 8n , I 1 +nla I I + n l a E.' o[Re n(w)l [lm n(w)l - I I,, y- zm - - a - - I I + n 1 2~. . 73. LL 1 Compare with the Poyntiiig's vector at a depth z: . . C . . . ....-... . s, = -Re ( x 8 n :.... ..:. 5 . .... . . . - - 4 e- 2 i m n(w)9,-- i' . ..;. I 1 t-nl?. 8 a - . . ... . . . : . :,:...:a+The complex form of Poynting's theorem reads... . With no currents present (J* = 0) the two remaining terms give the continuity i W ~ 2A & equation, .S' + -(E D*- B ti*) = 0. Hencc we should have 8 n as, w Re n(w) Im n(w) NOH'- = - - ~2 ,-2lm n(w) z and this is in az n I l + n 1 2 iw dced the negative of -(Zi 6*- 5 .fie),as required by the continuity Rn equation for energy flow. 4no c (c) For a conductor. n2 = I + i - and the penetration depth b = w ~ u 4nu If w I, we shall prove that an initial disturbace will oscillate at the r ' frequency and will decay in amplitude with a decay constant X = I/(' Proof: On page 223 Jackson shows that disturbaces inside a conductor away in time as e-ht, X = 4n00, where oa the d-c tonduc ' This is not conect. Since oo --lot6rec-', one is not ~ f p n ~ ' 75. 'I use of tile d-i conductivity to find lime variations o f the order o r sec: might be suspect. One must keep the frequency dependence or o(w). P c r l ~ ~ p sthe dutl~or'sIdea ill pluaslng the prohlcm war to g v e some physical scnsc to the idea of initialing 2 disturbancc. hut 11 turns out tlut u-itllout specllylng thc problem in detul i t is better not to distinguish bctween 9 intcrnal and !? applied Thus we now set !Ic,,= 0 and consider ullc clurgc density, one field and one current. -. - .> -- 3 9 Tllc rcleunt equatrolls crrc V - E = 4x6. V I + -= 0 and Ohm's at law. Wc apply Fourlcr transform m tune: 9 (x. t ) = / d ~ c - ' ~ ' 9(x, w); J(x, I) = / dwc ^ i w ' ~ ( x ,,); E (x, 1) = j dwclw1 E (x, w) .- d - 2 Uo W p 2 T O~IIII'SIJWreads 'J (x, U ) = O ( w ) E(x. w) wllcrco(w)=-=-- ( 1 - 1 ~ 7 ) . l ( l - 1 ~ 7 ) The continuity equation pves Combining this with Coulomb's law givcs . Tlierc are non-[rival wlutions for 9 6,w ) only for certain w valuer, namcly the roots of iw - 1 = o 4 no(w) we have 1 1 - iwr--- = 4no(w) wp2 7 i w (1 - i w t ) - 1 = 0 upar Thus iw w2 + - - up1 = 0 7 1 Iknce ,= -- +_ 2 r Jyp' -I For wpr >> 1 . w l - f up. 27 Shxthesolutiomgoas e-iw' we get e-L1(2r)'i w ~ t ,showing the damping 76. Iconstant to be A - and the o d h t i ~ nf r e q ~ qto bC 9.Fimlly, letas21 . - Inbte that if we used a. in I wc would, of couru. get the e n ~ l uresult. At the frequency of the plasma oscillation LJT is always sufficientty large that this step is appropriale. II i F o r w p r > > l . w 2 - - + I 21 - t- -2r + lWPl i Since llle ,elutions go as e-iw' we get e , lhowrng the damping / 1 constant to tie A =- and the oscillation frequency to be up. 27 The collision time 7 is s much more reasonable damping time than -. . (4n0,)-' [Note t h t if we uredgio i/4ro(w)= I. we would, of course, get the erroneous rcsult.] 7.14('7.8) 1 iI I I II .A sryli~sdnlodel oitlle ionosphere is a medium descr~brdby il~cdrelcctrrc i con:t;Lnt C(W) = I - (wp'lwa). Consider the earth is flat and the ionosphere begins suddenly at a height h and extending to infmity. 4 E IIi 'II h e of stght c o m r n ~ c a t i o nb prevented over distances greater than - where 11 IS the height of the antenna above the ground. For example h ;0.1 Krn, the rahus of the earth R=6400 Km.the distance IS approximately 36 h. Usually topograplud features interfere sooner. For our purposes we can neglect ihc curvatlon of the arth from now on. Radiated wave luts the lonospl~crc with angle of inadtrice i greater than I. (Brewstrr'r ~nfle). then total ~ntcmalreflection occurs. Snell's law gves W sin' l = e(w)= 1 -& :., msi.,. 2 wa W I ' II 77. . , .... ......,.... .... .- .. ..._._,...:._ :.-. ': . , , I1---- - n T . j -, 1 1 2 - LC- --.. . - ! L d - d = 2:) ;-- - I . u p no = 3 >. 19' cnl- ' If h = 100 Km, 1l:c.n d = 200 2.86 ~ 5 7 2Km = 360 miles fa) For anglcs i < L there n partial reflection, partial transmimon. The detail; depcnd on the state of plarization of the radiation (see Eq(7.58) ar?d (7.60)). For i > b. total internal reflection. If ionosphere is finite layer, the energy tends to be transmitted outwards from the earth's surface fot FL but foi i >i,, reflected back towards the earth (provided the layer-isnot too thin. The reception pattern is shown in the diagram. The electron density no varies from day to night because it is produced by the action of runlight in the upper atmosphere. During the day the electron density in the lower layers (-100 Km)'is --10"-&" Le, 30 timu the nlght time value .'.,wp(day) - 5 X w,(night). The differenm between daytime and night time behavior of the ionosphere b why certain radio stations in the AM band are required to stop broadcasting at mm. They have closely the same frequency as some pmiously established ststton 500 or 1000 miles away and their skip transmission at night would interfere with the other station. 78. Usc second form. w:Xw'dw' Itc c ( a ) = 1 + --P / n w , - LZz .--. Afi"I '+ w = I + - Jwa2 ----d x n w 1 2 x - u2 If w2 is outside [he rant? of inteeration. tllcn lhcrc arc no prohlcnis: X '+:- a --Itcc(w) = I + - P n I-, .I n 'JI - w - If w,' -< w3 < w Z 2 , Illen wc had bciler he c;lreful. n 0 - -- w, - But note that this ureful result isjust whjt we got beforc when a' I. we wish to locate the n~inirnurnin -. Delb~e 1d ~ i A .; = u / w l ~ .X = wp2/wg2.(A i I). Then c(x) = 1 + -.Put f ( x j = x(I - x) X : C (I = k' (in unitsol wg'/c2) I d2k .1 I I I I I I I I I I I I I ~ . --- 0. At minimum we have 2ff" = (13' dw' -- 2x + ----- . (,lL-.. .2 1 2 (1 - x)' *I% Z Xr e =: + - (I - x)' The11 lll" = (1'): becomes TINSreduces to x3(l - X) + X(X - %) = 0 Evldcnrly for X >> 1. x = !4 u a good approx~mation.This suggests an 1lcra11vesolurion: x = I; - ?(I - x ) / x X o = s, Xn= % - X , _ (I - " " - I )/A For X > I , the rninunum occurs very dose to w = %q. (c) General form of solution (Rob. 7.7(a)) is 85. U(X. I) = .j L A ( ~ )ei"':'/,r;~.< (w) = I.(G)x-- Wt. > l c r l ~~ dof stationary pll3SC (we the sccond cditior~s:.i[lnll 7.1 1 (d) p.316) that doininant contributions to tl~eintecral co~iicCro~iitllosc w d (ck) ct0 or --dues ulicrc - -- d(w) dw x Tlie minimum of the curve (denoted as (A) in the diaprnm) 113sy = t i It.. For to < t < 1,. the horizontal Line intersects the c u r c a[ !~rgcw. The sipal is small in amplitude and of very high frequency. ---- --.. But 31 t > t1 there is a qualitative change. The Fourier intqral now receives contributions from relatively low frequencics. Tlie signal be- comes large and visible. For a time t, >t1 (shown by the linc with points B. B') the sipal has two main frequency components and appears as a beat signalof the stand~rdsort. The high frequency component approaches w ~ .but the low frequency modulation has a lower and lower freqircncy as tlme goes on: For wp = I O L ~= 6 X 10' KC". a b w t. 2 2 X lo-' sec(x = 6 X 10' Km). v h = -- 104t-' HZ,2 n where t is in seconds. Thus for t -0.02 -+ 10 KC, rjo, fal!, throuph the audible range. .:. Name of whistler. 7.19 (*7.18) . Energ Ion of r c d partick passing through a dispenare medium, iw devnbed by r 6 u)a kc-&) 7;(1 - 6,a)). We imagine all - 86. 2 quantltics wiiich are in SPZU: and time through Fourier tra11sror111to q a - a 4 i.e., AC;,t) = J d' k - ~ d wAG,w)eiq ' + . - Similarly, d -.a -L 2 -L iq D (q, w )= 4n9({. w). and E (q. o ) = -< (q. w) (a) Charge in motion: 2 J A 9 (z,t) = &8(x - K(t) ), wit11 (t) = vt. l~~tegrateover d3x first: .A 2 -I 2 2. A A A (b) FIUTI D = cE and E = - iqlp and iq D = 4n9, we get D = -iqc< q2w= 4 n9. = - - 4 n 9 6 , 3 - .- 2 i.e., lp (q, w) = 7 q cG. w) (c) Energy loss: dW A d - J J .E d3x as the ratc at wllicl~work is don: 'Begin with - -dt the fields. We insert the Fourier Tnnsforrns: A -4 . - # .. dW - ~ q.x + iwc . A 2 -=J[JdJq J dw J d3q'J dwaJ*(q'. w') e E({, w)eiq ' x-iwl 1 d t A . (_Note that for convenience we have used complex conjugates for J. , J(x, I) is actually real. This is permitted). Integrate over d3x: = J pqJ dw ldol (2n)3ei(w'- o ) 1 2~ y q ,-' w') E(i,w) --. A i o -# Now we use J 1 a E = -(1 -c) E to write 4 R 87. A 4 - 4 4 iw' _I . d d J'(9.w'). Qq,w) a (Y~.w')-~)E*(~,u'-)E ( { , ~ ) From Eq(l ) the expresion E and Eq(3). we have - . . Subui~umgEq(2) for the 9 in Eq(5), Eq(4) becomes dW 4n1(Ze)' i(w' - w)t--- = --- e d t ( 271)~ I d3q 1 dw I dw' 4' b e can integrate over w' (using 6(w' - - 9)and then note t h t A 6 (w -q v) will make the f~ctore i ( ~ '- = I. Thus 00 To see tha~the result is real we must convert the integrai'6~crj d o to - w OD / d o . We use the r a t y of'@, t). ctc. to *ow ttut +&--) -0 .. . . -.. . 00 w 0 . c*(q, w). Then j dw = I dw + J dw where the second integral can be -m 0 -00 evaluated by putting q + -q and rclabel w' -+ w. Thus, 'I - a3 -. m = - - Bo. 3 2 (Note t h t Be (r= R) = -- 2 B o sine Absorption u o n section: For 6 c obranc'd by Lo~is~der~ngthat the quadrupolc I ~ ~ O I I ~ L ' I I IICIIOI is J ~ r ~ ~ c ' l c ' hrcnsc~rI.c.. Q , , + + Q,, = 0. The lnagrietic ~riducrt~)~i1.. ( I I I ( . I I I CI . I I I c.clrll~'~'"cllls) Fr(:ln Eq (0.43) Vcctor 6( n'): Qa = Q 0 - whcre n = sln Oco*; n2= sinDzi3;and n3= c o d 117. ,nil. tl?r rilnc 3rcr;lFc o r tlie to131 power radiatcd by thc rl~ladrupnlcis . dl' C 1 % - , - dl! = -- q 2 k 6 3 4 X 43 jO1,jP(I -,,a 'IL 4 J!! 32n - (1-=s The peneral expansion o l the potentia! ( :.O.o.t ) I S a(r.0, t ) = c-iw12~ ~ ~ ( c o s ~ ) l ~ / ' ) ( k r ) a ~ wherc h p ( l ) = jF (kr) + Inf (kr) spher~ I li3nkcl funs1 Solving for a&?.wc have, - The asv~~lptoticform of the 113nkcl (unction ( k r ?? 1 ) 119. For t l ~ cncar zone A T l ~ er;lrll;ltlon lields ( 1: and U Ire In lcrms o f 0 scalar po1en11;lland x, the vcclor potcn~ial.In the form clt' a -.- -l~,rcnt/gsugc gives I dtl. I ., + A :.- = 0 c 01 - - I a+J .A i-- -- = l k V c - l d l l I ( ( ) '1!:!kIBB(Los0 ) a1 ( 5 ) L,JJ t i v ( ''( l a 1 TI115 problem w ~ l lbe slmplrticd J g r c ~ tJcal I( we consider t l ~ ed~poler ~ d l a t r i , ~ only The ~ t ~ u l t ~ p o l sc x p a ~ d s ~ o nol' rllc p o t e n t ~ ~ l(as 11 1s In Clupter 4 ) cuso I = 3ve-lwl cosoy i p , r = a 2 ;; 3The radiation fields are: -ij= kz ( -,z Ve-iut )1 -- .- r In a more general manner. we c m make use of the fact that Er mrushcs at I;lr zone: Due to the symmetry of the problem. Ad does 11o1mist 120. I aTllc cxprcsslon --- ( sinOA ) vanishes for all Y. this inipilcs 1h21~ I I I ~;I,, rs1n0 a0 o IS 1101 3 1~11illonufO or A, is den tic ally zero. A poss~bles u l u t ~ o ~ ~lor ,lo I, -liowcver ,wlicn 11 IS checked wttl~dipole expression for A . -. -- i k r 3 cilrr ,I = .,r, '- ) m d A = -ik( -a ' ~ c - ' ~ ') (-; ) c o d r 2 11 ~upgclts111;rt ,l,.;liould ]lave (Iic following exprcssiun ln,rc.;rJ 01 r1l.11 I I I I.i;( h ) d l , , -v c - i u t r l~p(''(kr) [ I - -I -- JllnU Gdd h $ l ) ( h ) lkl I aTl~iscxprcssron does not give --(sinBA,) = 0. rsinO ad With t h ~ scxprcsslun. the B field is 4 4 and thc elcctr~clicld is E = n x B llic avcrag power r~diatcdper u ~ tsolid angle is Iri r3d1311onrtJlie, thc crossed trrrns can be neglected 3nrl lor J~llolc raJi:~lionwe I~avc: 121. r.h:ct; ;I.:rccL it11 lIq(9.19) in the text, for radiation zoric ( I - I/(ikr)k I 2 - > ikr C i k r 1) =.k- ( ~ . ~ . ~ . - l u l ~ : ) (L)EunO = k2(" x p ) ( - ) r @ r dl' 9c= ---- k'(v2j4 )sin20 d!Z ??R Thrs rrsr~ltIjves tlie s m e expression which wc oblarncd I)rclcnrsl) -- 1) .4 --- (3) Thc current dcrisity Ii u ~ ~ l ~ i ~ c ~ l111the tl1111 antenna. / 4 .. * 'n --- J ( x ) = l s i n (z-7 ):, () h ( ; : ) c 3 d Frum t.qfc) 7). 1 1 1 ~vector potential can be obtaincd for radlatrcln ronc 4 -L If tl~cantenna is just onc w ~ v clength in length. i.c., kd = 2n - -. 1 1 1 ~r:rd~.l~rc)nficlds can bc obtained by D = ikn x A and the clcctric field 4 4 IS pt,cn 3s = U x i1. The radiated power per unit solid anglc is dP c -.4 c - - - ? - - R e I r ' j l . [ : x B * l = - r 2 1 ~ 1 1 d!! 8 3 8 1 ( 1 ) consider I hc Ions w:lvelength limit (kdtl ) ( 2 ) If the anvnna is exactly one wavckn@h long(kd = 21) 122. 1 1 ('i~:lrlgc' i.~rl.~t>lc. I ~ ; I ~ I I( 1 + L ) = -- 311d( 1 - 7 ) =-- Inr t l ~ csccor~tlrcrnl. 2n Zx l l i ~ . r i,c I~trve 1' 4 7 I -COSI ' -- f (-- - 2 0 I )dt I - cost I, i~!~ivc11;I, I C I I IIIJ! 4n ( ) dt = 3.06 thcrcforc, the total pnu.-0 t 123. . . ;rs ran CIICUI~IC the vector potcntral due to the multipolc cxpanslon: The elcctni &pole: l e i k r 'n A ( x ) = - ? , j j l s 1 n c = - ~ ) 6 ( n ) I , ( y ) d x d y d ~ = 0. cr d 11 v~nisl~esbrc~usche inttgrand is and odd function. The rrilgnelrc dipole: - e ~ k r I I - -~ ( x j =-(---1k) I j ( u ' x J ) x n d 3 x = 0 . cr r -c -(:.. u ' x I vantshcs). Tlie electrrc qu;lJrupole: 1l1clowcst IIUII-varushir~g~nultll~olcrnorncnt is clci~ricquulrupolc rnorilcnt 1.11~~1131;~'dc11~1tyrnuht bc i~lcul~ltcd.111 order to ~ ~ ~ C U I J I Cthe q ~ ~ d ~ i : j ~ ~ l l t inorncril cxplrc~ty. i o q = q . 7 2nl 2n :., q = -ws( - z ) 6 ( x ) 6 ( y j iwd d Q4= J ( 3 ~ 2 ~ - " 6 ~ )Y( ;)d3x -- 1 All the offdiagonal terms vanish due to L( x ) and 6( y ).The diapnal clernenp .z- are not all independent, but they are rclatcd as (b) The angular distribution of the power radiated is d' For radiation zone - O) 27: aprrtufrs -. I ePu if-;fda.Thus. E ( 7 > O ) z - V x - j 2 x Ee- 2n Aperture lhf a A A I i.. Ii~ XLj t r t ( l -ik. r-)&;(toorderof--) ?n r r ( 2 ) aperture Cornpanng Eq (2)and (I). we recognize fmm the first term of kq (2) that & m = - I j i x Eda' aperture ' ~ I Ccfrcclivc maenctic dipole moment is thus 138. - - l h c sccond t c m of Eq (2) can be rewritten as follows Wllerc wc liavc uscd the vcctor idcntity A On thc first tcrrn of Ip .which now beconlcs lkr _ A -. eikr A A A k da' [ i (r" E') x k I,, =- ( k x [ jd3'%( r. - E )'trn n rn Apenure 471r Comparing this tcml with the dipole term in Lq (I ). wc find The 'second termof Igcan be wcn to bc the magnetic quadrupoli by the following arguments. Using thc fact that thc L' ficld is to thc surfacc of th: perfectly conducting wall, we have u o u n d edge of rperlure A & E has only i component therc and dP'is on the x - y plane A 0 = j V'x(x,'x;l E ) - i d a ' (by Stokcs' thcorcm) Aperture A A A = j d a ' { x ~ i , xE - i t x ' i n X E . i + x . x ' ( v ' x E ) . i ) J n J aperture A A A = jda'(x;I(F x i ) , + x ; ( E x i ) , t x:x;(D'x C ) * i } aperture llierefore. - i d a ' ( ~ , ( E x i ) + x i ) n 1'- j d a ' s ' x * i k U - i I I n I Apcrturr aperture A In view of this, the second terrh of Ip can be rewritten as: 139. Finally, A 111 = -/ ? ( i . u ) d ~ ' Inu apcrlurc 1 :L IF:. Sptlelc ~ 1 -rxilus K with good c o r l d u c ~ i ~ ~ ~ ~0 S U L ~ I~t131 = C - M R ~ n dkK .n]. d GO @ c ~ I I Cproblen o r E and I3around the spherr. ---Y 0= constant on sphere; EQ along z u i s a' Q = - E,( r --1 cog + const. /' r2 I A :.. ,I = a3EL 2 2.1' - 1 1 - -) I kr = L., ( l + -- ) ins0 I =o. 1' r 3 3 hlagnctostatic part: Boundary cocdition is B, = 0 ;Ir r = R. 140. ,h s o r p l ~ o ~ ~cross bcctron: Fnr 6 .:-K we can use Sect 8.1 to coniputc the powcr I61sst ~ t111, , , I ( ! ; . , , ;,,; c . - c 3 t x c CLIIICIII IS Kc,, = --c*, x lib,, = - - x 7 R, s11102 4n 4n - Q ~'cltcr 1lcr U I I I I .UCJ 01 ~ u r l i c c01 spl~ereI j FIVCII l)y I-(l(SI I 111.:1:~11! 1'&1tcr111111 .1rc3. c I' :,Arc3 = -u: I I . . - I: . ,Ill. 8 R S c ~ ~ r e ~ ~ l i g c r U s sscctlon I1 >, -.K rllcrc wll be just mall corrections of 0rJc.r :. K I,, I ~ I L . . I I ~ ~ ,,,. I I O I I ~ - l ~ u l ~ ~ c d ~ ,hr a pcrfcctly conducl~rigspllcrc. 1IIC II,I.I!X . I II . '11 . I On CIU,) >cillon !ur s t ~ i l ~3 sphere is o = -0 ; ~ ) ~K: kc !13vc 2 I I ' = 3 I r I< 1-- -.. b ;s: -- :I 311dkK c:I one can say o,~;? or. ( kK 1' I< 141. h l a y n ~ t o l i y d r o d v n a m i c sand Plasma Physics I 10.1 i Vc cohsidcr part (a) of the probleni as static. From Iiq(lO.S),lwc Iuvc. i 4 n , 4 n 'C >.,11 = -J ; o r . f 'B'. d < = -,I T - G d aC c , 'tlc~lccthe crlrrerl t.ilcnsit ics and the magnctic i~iductioriirisiclc rllr cylinder arc:R ( I ) 0 < r 6' 7 - - 41 X l r J, = - f c ; B = - c; n R2 c H 2 R f2) - < r < R ' / 2 4 1 1 ' / 8 1 ~ ~ = - - - - - k ; B = - [R2-r2] $3 n ~ ' 3 c r R 2 The magnctic induction B so obtaiqed is time independent or stc;~dysl2te.#, (h) Ncglcc~ing the displaccmcnt',c_urrcnt due to the po~ential- diffcrcncc bctwccn tlir cylinders. and assumiag,~B(r. q, z, t) = B, (1) Us ( r , q. Z) i.c.. the mapclic induction is separabe h t o a time and a spatial parts. From t q ( 10.1 1). the tlmc dcpendent 'magnetic induction bccornes -- c - I Furthermore, In both caws V' Bq g i m 1/ra ca-81r ,no,i1 - Solution: B = - e c R 142. lnlti3l condition: Thc axial rnagnctic induction B,,, the rat . . . inifinitc tuhc 1s KO.Fro111 Il~cforce cquation (10.23). we have Thc quasi-cquilbricm = 3, and ntglcctir~gtl~cgravitational intcr . 113vc If we usr Iiq(l). tf1c11wc have Bq is produced due to the axial cuncnt and is 3 funclion of r only ('. --a A I a an s x i r x i i ) = (n;, + B,LI x [; - ( r b ) L - - a r a 7 Sinsc tllc problcrn is only r dcpcndent. lhercfore 143. , , l a 1 5 ,~cordi~igro scction 10.6. the inlcgr~lionijf Far3d~y.sIJU' (T X E + - -c at = 0) gives.- I d j t ? X :) .;,la ' - ---/.B- l l d ~= 0 c dt - . I , C-I:, I I>s. . .- Ilcncc, li+' = vA' (1 + x' sin' 0) = vA2 . . . . . . . . . . . . . . 112 1 . 2 B( ix + i, xa cos 0 sin 0 A v,'" is perprndicuhr to the field and in the plane containir~gk and B, Thus i',(" is paral!el to the mapletic field B,. I Alterna~ivemethod I 1 From Eq (10.66) . .- 146. Thc pressure p is a ~uriclionof thc density (9) and the surface a pr p (C. a) = 5' p a d 9 + ( -) p a d s a s Assunling the surfacc area does not cliange. thcn P p (9,a) = s' grad 9 . 2 d 11, makesan angle 0 with k. k t k be along the x axis, so the r !I_l~ctionB along !hc x dircction is constant ant1 taking the gradic k direction ( i t . the dircctior~of thc x axis). I;.q(I) can bc.lsirnplilic av, r' ;I? I an, Iavx - - - -- ari + vx- - - - 11, - - - - at ax 5'a 3 ~ 3 O X 4 ~ 9 a . r) v, av, - o, an,+ vx-- at a x 4 n ~ ax IS thr vrloc~rvof thc wnvc iront rrjcc 6 ( u, I ) = 0 3111 ir define( a - 3 0 a .-- - a d o 3 at -at ao arid - E - - .rlicn the I'lrst equatrl. ax a x a0 147. Sill1113rly.wc convcrl 311 I ~ I Tccludlrorls ill El ( 2 )~nlcrrn of [lie p3r3rnctcr 0,3 r d 111c.ybccornc civz Bx a 4u --- - A - =o 20 J n p a@ Tlie condition for thcse simultmeous equations that givcs non-triv~lsolutions 5 9 a v , is in thc form of dctcrminant (the variable arrangement as - -, ao ' d o 3 avz --a 0, aid dl), 30 ' s - ' 36 a (., 1 148. Bx 0 0 u 0 - 0 4 n9 -B, o o o u o - 4n 9 A u J.'run~I-q( 10.70) the tilfvsn velocity is dslir~cd V A = ,/-In7 nls 1~ycrohydrodynamiccharacteristic relation in Eql3) bccon~cs it1' - V , , ? ) ( ( u 2 - s2)(u2 - V A X 2 ) - CJ?c wit11 rcspect to the rvector (wavcfront norr~lal)and ( c ) l'n! L',, L)'.s. the solutions become U,: = (VAcostlj1 ; UZ2 2: v ~ ' J ~ ~ u , '= 0 tVc tiavc proval in part (b) that the solution UI' ~~ ~ _ c o s O ) 'leads to a trJnsvcrsr componcrlt normal to the rnagnc2c field B and I:. From Jiq(10.75) rc ncrllic II,,II t11r rri;!gnetic field is also 1 B,. Thcrcfore. tl~ccomponent ic ind ti!;. pl3!1~'iorlt;iln~~>gII and k. For the other soutions wc cnn sllor. wlth thc wlne n~rtllcldas i:? p:lrr (b)that the apprcciablc component of .clocitl; ic panllcl t f l rllr nlaplc't:c field. 10.4 Tabng LqjlO:23) for nonviscuous, perfectly conductinp fluid and constant density 9 , is a c ~ e dupon by a gravitational porcr~rial3 as A d 2 Let n = B, + n, f... I). Thr problem is static in thc bcginning. thcrcl'ore .'., T ( p + p> + l.Q,+)= 0 Eq ( I ) is reduced to d~ I a - 4 9 0 -P ( 9 -V) B d t 4 n . -. 150. BI For inconiprcssiblc psrallcl flow, = (dism~scdin Mapnet( , rs - - . - dynamics by Alan Jeffrcy. 1066 p a F 83). - n,v = - f i n s The remaining rehtionsi~ipis By substitution of thc rcsult in IJq(4) to Eq(S),we have 2 U after the operator del (V)can bc rcphccd by k', since Ei is constan (b) The initial disturbance E,(x. 0) cvolves according to Eq(6) - a ii, d (t,, 3 ) B, = - whcre VA i~the Alfvel~wl. a t A Let f l ~ cinitial mspnetic indt~ctionI), bc along thc z axis (without ,. gcnera1i:y). it hecomes i.e. the disturbance in later times is 1 to IUfvcn velocity. 151. 1i I* I,X'!.i::l~CCCSno1 q r c :r l ~ hI I I ~ )I,I[ tllc hooks U I hi^ iublecl. Tlir: carnpl~x condtrc.hvil? occurs In X.C. elcctrlc field nor In1 3 1 l i citlnl~n -> v = a o E t 3,1> V, therefore -. . velocity of the phuna, the oscillatjon becomes very large. This gives a sing, . ' i n Eq(7). but il can be integatqd by introducing complex w. landau , ' that under certain conditions the kciltations became severely damped. 155. 10.7 (3) Vi~hexlernal connanl rnapcric lield 0, and auuminp all mriablcs - d IS cxp (ili - r -- i d ). 111osearc: 2 - E = i + iec ~ ~ k- r - U I ) 2 n ( o r 9 ) = n, (or 1 (c) (i) k 9 b ; can be reduced to w2 (u2 - wp2)(w2- up2-k 2 ~ 2 ) 2= G)B2 (u2- k2c2)' (u2 - u p 2 ) (a2- u B 2 ) ( w 2 - k 2 ~ 2 ) 2+ c.+'u2 - 2u2wp2(a2- k2c2) = o - If w == 9.then we can assume also a >> q.The cqwtlon beconles - - w2 - lw2(k2c2 - op)2 + (k2e2 - wp2)' = 0 and the solution is w2= k2c2- c.+2 + 2 9 k c (ii) k 1 b: Eq(3) becomes 156. m d the approximaled soiu~ionu simply (for up>> yl ) - a=== UP' I I:q(10.1GO) are applicable. but with mod~fi~arron111 rlir Iurcc ~ q ~ ~ ~ ~ - - I -. -. ullcrc. lor our C ~ Y : we drop the pressure term 2nd let E - E + -v X B, 7'11~ i 157. -* -1 .,. II In terms of k E, from Eq(51: . 4 d 2 c' (k .E) (k .6) llur fro111tq(5 ) wc also Ilavc I)i.idin? I:qf7I I)!. I:q(H) we obtain thc sot~chtfor rcsuir. n;~~l.rciy. 3 . 2 1 1 W-(w- - L I ~ ' ) ( W ~- wP2- k c ) -. - = Y 3 1 ( ~ ' - k ' c 2 ) [ w ' ( ~ ' - ~ p ' - k ' c ' ) + ~ ~ ~ c ~ ( k . h ) ' ](9) _ L 2 d (c) (i) k E U, implies (k . 6)' = k2 From Eq(9) u2(u?- G;;) (w' - c'k2 - w;)'I= ~6 (w' - c3k2)(u2--w ~ ) ( w '- k2c') The roots arc thcrelbre w' - ',; = 0 and w' (m2 - c?k2- Wi)= = (w' - CT )? . ( 1 0) We observc that for propaption parallel to 6 thc diclcctric constsnt (or square of thc indcn of refraction) (7.102) is rccovcrcd. rllar is. 158. -If wit = 0.tl~rroots of' w could br 0md 2 d c 2 k 2 + w:, .For 6 . -- . we cau assurnc 011: roots to bc of thc form Mcrc (I, fi 2nd -y arc to bc dctcr~nincd.Asumulg our roots to . . above form. the prodr~ctof thc threc roots upon cxpanion. bcconlcs to first r~rdcrin ( - ). Comparing the coefficients of various : 1 LJ in Lq(l2) and Lq(l3) wc find: 159. u l ~ e r ewe llave assumed w > 0 A 3 [ii) k 1 Bo thus k - b = 0 ? - # - , Eq( 14) can be expanded to give w4 - w' (2u6 + k2c' + wf,) + (wi + w;k- i- t wik2c') = 0. Using the quadratic iormul, one finds where we have neglecred - terns. -. a; 160. chapter I I m d I t are mainly concerned with the c6nslanlreallve transformation (Lorent2 tnrufonnation). Only a few problems tackle the subject of ~cceleration.The discussion of accelention in Lorentr tram. formaiion can be found in problem 11.4. 11.s and 12.4. The events are not observed in the same manner in an accelented frame as in a constant relalive. velocity frame. In doing thest problenu. we consider that the mowng frame is at constant velocity at that instant. As in the case of problem 12.4. the instmtaneous velocity of the object is taken as the constant relative velocity of the moving frame. The coricept of "proper time". that is the time registered by tlie accclatcd clock. may be used. The proper time interval is: A T .= / dl-v:(t),c: dl Physical problems require a clear analysis to determine tlie " I I I ~ V I I ~ ~ "and the "stationary" reference frame before the problem can be solved nl3thcnlat1- cally (Prob. 11.4 is obvious, but not so in prob. 12.4). It IS ~nicrcitlng10 note that the rectilinear translat~ur~Lurc~it/tr111~- formation can bc thought of a rotation transformation (x; = x : K; = , . x i = x3cosJI - x,sinJI; x', =x,sinJI + x,cos$). It is exactly like I rotd- ional transformation with respect tolone of the axes in Cartesian Coordi~iatci. but the difference is that JI is imaginary and has the following relatto~isliip with the Lorcntz transformation v2 1 tan i, = v/c; .sin$ = -ivlc and c O s ~' ( 1 - -T)-l ,/- C It IS obv~ous that two ficiite rotation transformations do not comniutc; tlirrcfure ne1r11c.rdo the two Lorentz transformations (Prob. 11.?). Prob. 12.1. 12.2 and 12.3 are typical problems in part~clcphysics. One of thc better references on this subject is R. Hagedorn, "Relat~v~stic ~incmaticsf)Angular dependent threshold energy for a production reaction 1s demonstrated in prob. 12.3. For Van Allen radiation prob., it is helpful to go over the last few sections of chapter 7. 161. Ekrrz rcfcrences and comments ( 1) I(el3tivit!. including Lagrangian formulation: Itc~nl;i;t,"Tl~cory of Elementary Particlcs", Chapter 1 on the b r e n t z group. and p.160- 174 on various Lagran~ians. 2) Electromgnetism and Relativity: Lorrain and Corson (elementary rcferencc)Chaptcr 5 and 6 discuss special relativity. Chapter 5 is for general concepts of spacc, time, general transformations of various quantities of nicchan~cs(e.g. velocity, nicnientum, acceleration, force, energy :~ndctc.) and mentions scvcral important experiments. In chapter 6. thc derivation of h1:!xwcll's equations b u d on spccirl rc1311i11yis sliown. E. 1'. Ncy , "Elcctromagretism and Relativity ". f 3) Hcl~tivisticKinematics: I?. I l:tycclorn, "Relativistic Kinematics" (Benjamtn. N.1'. 1063) It covers h103d interest in this subject wit11 exnnij~le~. !I:,(, 1 ~ ~ ~ ~ h sor1 clcnientary particlt physics. ( 4 ) 5tandclstam d h p m s , Kibble function. st-u, ctc.: Ii~gcdorn."Kclativistic Kinematics", p.70-78. Omncr and Froissart, "Mandelstam Theory and R e g c Poles", p.54-71. csp. p.66-70 for several examples. T. IY. R. Kibhlc.Phys. Rev. 117, 1159 (1960) - oripn of the name. 15) 1)aru-in-Ureit Lagrangian: ( a ) I'lasma pliysics application -- A. N. Kaufnian and P. S. Rostlcr. I'hys of Fluids -14,446 (1971). (b) Quantum-mechanical uses -- H. A. Bcthc and E. Salpeter. "Ouantum Mechanics of One and Two-electron Atoms", p.170 ff. csp. p.180-1 and p.192-5. 162. Special Theory of Relativity 11.1 The postulate: The velocity of a dependent of motion of the sol (a) The clock moves perpendic light path: Let At be the tin - . : between emission of the pulsc . .. reception of the signal. A t = L J m L 2 -C M - c d : (b) The clyck moves parallcl 11 * path. Iiere we have to assurn, transformation is known. We 1 that the contraction of the d . bctwccn Ihc mirror and thc pl:. The time interval for the light travel from thc flash to the mirr~. and that from the mirror to the . is At1 At = 2d c ,./- So, hoth cases have the same time dilatation factor. - ~ Alterngtive method 163. The dock is moving LI Kgwith velocity v, with respect to an observer in K. The length d is perpendicular to the dircction of motion. The electro- magnetic wave is radiated from F and after being reflected from mirror M. it is received at P and one "tick" is observed. When the clock is stationary.the clicks are observed every (2dIc)sec.. where c u the velocity of light. Problem (1 1.3b) states that !he wave equation is invariant under Lorentz transformation. Therefore. a plane electromagnetic wave in one coordinate frame K will also appear as a plane wave in another coordinilte frame K' moving with constant vclocity with respect to K. Lct ~ d . 1 )be t h e + e l e c t ~ m a ~ e t r cfield tensor which contams the space time information of E and It. In t'rune K a plane wave is represented by where fW arc appropnate constant coefficients, and k and w are the wave vector and Ylgular frequency of the wave. In the coordinate system K' the plane will be -b whcm the fW are again constant coefficients, and k' and w' are the wave wctor and frequency as seen in K'. We also know that FW and Few are related by the Lorenu transformation matrix, -b -. F'ru (x', t') = L-'F1: = l:pA1:,,FXo(x. t) 1' (3) where summation convention is applied and I:h given by From equations (1) and (3) we get, 164. Companng w ~ t hequation (2) we p t , - -. .. - . re,& - -:(5) Let us inlroduce the concept of wave equation invariance; for equation ( 5 ) 11, be ialid 31 a11 points in spacc.timc, the p h factors on both sida be equal. thal is + From ( 6 ) I I is clear that k and w must form the space and time parts of 3 4 vector ku: Thcrcfore. the Lure~llztransormalion of this vector will be. W k', = 7(kK - O F ) k = y(kx - ~ w l c ' ) k', = k, k;= k, or The inverse transformation corresponding to the set of equations (8) IS: k, = k', w = 7 ( ~ '+ VL',) 1This will be seen by an o b ~ m r K while clock is mown8 wilh K'. 165. 4 -.If thr wave vector k' and v are not in the same dire~tionbut k' makes an 4 --. W * W an~Jco', while k, P and 1 k 1 = - while k I = -, from cquatioll(9) we C C W' v ,coso' + w' ~ ' ( C O S O '+ 4) 3- or wcod =coso = - I - V:/C~ (1 0) 4I - v2/c= hzrting 111e last eqution in-Eq(9) we *ti- , w e . W"-. . w = r ( w + - c o d ' ) = C d-; (1 + 5C ~ 0 ~ 0 ' ) (1 1) From Eq(10) and (11) we obtain. v c o d ' + - coso = C 1 + Y c d ' C From (12) it can be shown that, I -v1/ c? sin0' tan0 = 4 - cosOr + v/c Thus. we get the following two transformed equations. W = --- v - ,in(). (I,) (I + cod'); tan0 - -- - coso' + v/c wl~ichdescribe the frquency and angle relation of elec~onlagneticwavcs in two coordinate systems. From the figure, it is seen that k is perpendicular to v, therefore 0'= H. From (13) we get Ule two relationshjps Thus the frequency of the electromagnetic wave in K which is I th J i S F of the wave in system K' which in turn is the time dilatation as observed by an observer in system K. (b) In this case the system is moving in such a way that d is parallel to the direction of v and a tick is recorded every 2dlc KC in the system K' in which clock b stationary. Thc problem reduces to find out the contraction in length d. The time and position of each point in K' system is related bv the corrcs- ponding space-time coordinates as follows, = 7 (x' + 1'1') y = Y' 2 = 2' 166. Call X I ' and xl' the end points representing the length d'; call x l an the cnd polnts obscrvcd by an stationary obscrvcr in thc systcm K, the, and one tick in I;' svstcm will bc related by corresponding observatiorl system by the following equation: Thus, the time dilatation between the two systcms is characterized h! factor 1 , obtained previously in (a). 4- 11.2 Using the notation of Eq(1 1.75) and Eq(11.76), the Lorcntz transo lions Ll and L, along the same direction along x axis arc: and L, is It can be shown simil3rily that L, LI equals to the =me matrix. LJ = LI 1 L: L , , thal is L, and I., commute. y3 = 7, y?(1+DIPl ) and U3y3 = i y l 71(PIPl) notice they are sSn' V I - v1 c31 with respect ot I and 2 exchange, i.c.. v, = - I + v, v, /cl (b) In the caw when one Lorentz transformation is along the x axis ' the other Lorentz transformation is along the y axis, then 167. -11 is obvious that LxLy f $4. If we consider a transformation of v = iv, + jv2, it a n easily be converted to a simple Lorentz transformation by rotating both axes of the fixed coordinate frame and moving frame such tlut dne of the axes align with the direction of the velocity. 9r 0 ,/- 7 ,, By comparing the matrix elements in Lx+ or that in Lx+ none of rhcm resemble the matrix element in Eq(l). Notiuc the angle of rotation of [he (1) stationary reference frame is 0 where tan0 = v2/vI and ( 2 ) lnovlng frame is 8' where tan8' = van8. Alternative merhod Xo coshq 0 0 -sinhq 1 0 (4 Forboostinz-direction ~ ~ = A x w h c r e x = ( ~ ~ ~ , ~ = ( ~: x3 -sing 0 0 C O S ~ ? Obvious from the fact that A = e4K3 tanh g = fl that succesive boosts in z-direction commute. To find equivalent single boost, the simplest way is to do the matrix multiplication: coshq, 0 0 -sinhpl -. - 0 0 co . - 168. tanhlpl + tanhlp, Tl~cvelocity of A is fl = tanh(lp, +ql) = or = -----_01 +PI - I + tanhql t a n h ~ , I + P I ~ I (b) Since [K .K,] # 0 (in fact. [K, $1 ] = -S,) sucessive boosts in x, ~ h c , ~y dircc~ionsdo not commute. Conslcler [Al.Al] = [(-Klsinhgl + K:cosllql + ( I - K , ' ) ) . (-Klsinhlpa + ~ ~ ' c o s h l p ~+ (I - K]' ) ) Note [ A, ll J = AB - BA: A. B = AB + BA ' [ A l ,Al ] = s l n h ~ ,s i n h ~ ~[KlJ l ] + smhlpl (I-coshQ~)[KI.K2:1 1" ( S K ) = -gkz = +KII Hence [Al ,A2] = -sinhql sinhql S3+ sinhg, (I -coshql ) K g -SIII~$:(l -c.oshqI ) K2g # 0. The c2rnmutator is nothing simple because the matrix g is present. Transfornlation with v = vl i + vlj : Then, A = e ~ ( $ ~ I+ K ~ ) "I = e q , K ~ ,-lp?K2 11.3 I a 2 The wave equation: O"lr = 0'9 - --* = 0. c n a ata2 (a) Galilean transformation: x = x'+ vt'; y = ye; z = 2 ' ; I = I ' 2nd c' =c-v 169. a5 a P a, P a' 7 '-= (-+2v-+v7+ az az ' at= ata axat ax 1 a=v'. q - 7 -,-* = 1. ( - - + ~ V - - + V ' ~8 * e l at' - at. axata' therefore, the wave equation is not invariant under Galilean transformation. t '. ad ad- E apu-(b) Lorentz Transformation: xp aWxv; -v= ,, P 3% 4 a a0'' 6 = ( +A%- -) # when avA is given in V - I x p axA axp Eq. (11.75) Hence, the D'Alembertian gives 11.4 The coordinate system K' moves with a velocity V rclative to system K .Lorents transformatio o f the differential are: vdx-ddx '+Vdt '); dy-dy ': dz=& ' and d t v d t '+ - dx ') and of the velocities are: c2 I $= u;+v Gl v.0 (along the x axis) and U i = . I + - V.U.(I + - c2 c2 17 (1+~)dU;7u;(!d$) ( 1 - 1 2 ) .dl= cz c2 = c2 -. J dt fi [ 3L--2 (Txf l ) l y2(l+ - C ) ( d t ' + x d x ) (I +- c2 c2 c2 170. 4 di;,, a i ( I - 3 5 3 ' 2 - c 2 + 4 4 . a r - -b -b where U '= 0 in this case. V =U V.U ' 2 4 ( I + -) and a j = g. c2 7 is 5 years measured in the space ship. t=85 years when it i s nlr on cart11 or tile total tirne it takcs i s - 4 X65 ~ 3 4 0years. Thc ! cdrrh is 2 3 4 0 . (h) Tlre distance that thc space s h ~ pcovers in the process o f a c c is approximately 84.5 light years. The distance from carth is I * , Added Consideration: Proof that the second scgnlcnt o f the trl; same earth tinie as the first. Let earth time for first segment he TI , dp dece l3 = - ' g dt where 0 = : b ( ~)( 1 -b2 )=I2 T I 2 I C C P (1) ------. I = I ( T 1 - t 1 JI-P' 'AP(I) P U , ) g (T, - 1) G ( t 1 - 4-, - C P(T,) gf, - - But I -f12(~,) c 171. P (1) lTl + !(T, - t, = 3:I-, - t ) c-== - ,. procedure o f finding ( t ) and substttut~ngi t into the expression I - ,777 is the same as the followed in part (a) tjencc ,.refore. the proper time interval f o r the sccond segmcnt 1s can be argued, and indeed very oftcn is. that a proper solution to : problem lies outside o f the special theory of rclativity since :clcrations are involved in the problem and the special theory dcals t h uniform relative motion only. T h e reader is referred to articles C.D. Scott and J Bronowaki for further rtading. )te: .D.Scott. 'On Solutions o f the Clock Paradox'. A m , J. of Physics (Nov.1959),580. nronowski, 'The Clock Paradox', Scie. A m , (Fcb, 1963). :ssume that (1) u , and u2are nearly equal. S o the K frame is moving ).ith a velocity u with respect t o K frame (u is either u, or ul) and -1) the time interval T o f the two clocks of the sprinters have been ~nchronized. 3 ) Due to the ambiguity o f sirnultaniety of the sprinter 1and Printer 2. When the light signal reaches sprinter l,the starter triggers ne gun. With respect tosprinter 1, the events are simultaneous. but -. 172. for sprinter 2. w h e n he sees the light signal comer back 10 h i m IF^," , i t takes a time interval o f 1. - ;II. , u.licrefl= v/c.jnd 7 = ( 1-pl )." If w e consider b o t l l o b s c r ~ z r s . rllc t l n l r range that gives a n a n l h ~ g ~ l ~ l u iSpparcnl Il;1nJ1 cap i n K frarnc is d - 4 7 + l ) < T k ~ ( ~ * l ~ C C W i t h tile arsurllptlon ( 2 ) T h e [ r u e Ilandicap is r F U . ( h ) T i l e [lllle 2nd p o s ~ t i o no f each sprinter in li' ill 1cru1 o i [llosc 01' K Iralllc. u l ~ d e rtllz L.orcllts lrarlsforrnatiorl 61' C; alld K ' . wr I ~ ac Spr1111cr1 In tisira111~x 1 2 ) = x i " ( c o n d i ~ o nfor no true lland~capIII I;') i.c. ti' hl~ouldIlavc vcloclty 0= cT /d along the x dirccrlon 11.7 Grccn's ttlcorcnl in 4 dimensional analogue IWe can prove th3t O2 - = O where R =(xi, ~ 2 3 3 ,ix4 ) R2 -+ = 1/It2 ~ n dU = A,,= ( A ,igJ, and Eq ( 1) bcco~llcr 173. ( 3 ) Since 111cc l i n ~ g ccurrent is localized. so the suriacc i d 1 1 i.c ~ a k c r ~ a r b i ~ r a r i l vI ~ : c c .T h e cquivale~itsurfilcc c h a r ~ cterrll. J I I ~I I I C clil)oIc I;r.cr tcrrn v ~ n i s l ias S, + m. J , ( f ) docs not dcpcnd on n. t I I ~ c r n ~ t i v chIct11oJ llslng l h c 4 dincnsional I'clrni o f G r c c ~ i ' s r h c o r c ~ l ~ .solvc t l ~ cI I I Ilornogcncous wave cquallolis I n0:A = - J L' ( I ) i p ( a ) S l ~ o u .rliar for a Io.[I s i s lying bclow ~4 .The contribution from the circle at iniinllv g w s to ~ e r obecause R-2 is so small there. With this understanding of th; rlle-lllflg al' tl~cirltcgration over E4 . Eq(6) is valid for coniplex t4 and xJ. arid we need conside: the integral only where JU itl ,t2,t3, tq) is J e f i i ~ c ~ . 1f ilad 13Lr.n a loop running along the imaginary axis abovcxj= icr W, w u u l ~ II3.;c I,--J advanced potentials. A contour running to the Iclt 01' ~ i l r 1111;1;1112ry JXIS above' x j ;rnd to tile r~ghtbeluw. crossing at tllc '110lc.' III Jp , woulJ FIVC 3 I I I I X ~ U I Cofadva~~ccdand retarded poter~t~;rls. (h) F I I I ~lzpu. Bv Eq(11.107) ;:Iv i) ,I I 3I < - ~ (-1I IB I. m e n frame K moving in +z direction with -# 4 E x B - 1 := d mvc~ocity7= c (-) has 2"= 0 and E" = -7 Ea 7r Ez -, In frame motion Y a purely eIectrosta~icuniform field E *'alon~tlle X- direction. e . d -(Tmv;') .=qE'.d -(ymv'3 =qEd dt" dl'. ... . --- -- - -. - .d. .- -0.If 1-0% >0,X' >0and hence(k: +A') >O for all k2.Then 1 r;: +(k; + Xi) has poles at k , = ? i d m .and from Magnusand Obcrhcttinger. p.116, tllc Fourier ~ransformsreduces to 2 re ae ' e i k 9 N w . x ) = v E(W) fi'mdk d-2 [ This rmult holds for w > 0. I lwlq I (b) For w < 0.the argument of K.in Eq(2)isT dm. With ( I -PC)> I Owehave Xq =@d-9 206. Thc elcctric and magnetic fields: &fine y=(I -0'2)-"2 I I For 0 I 111ci1 I ZC f-(2-vt) E7 =T ~+r?(~-~1)2j% T h c results on the right hmd column agree with those from Eq( l l . I 1':) ! z = O a n d e = I . w =(c) For PC>1 wemodify Eq(l) by defining p2 =-- ( ~ ' c - I). T l ~ c n v2 For Ik l I < p. we must be careful of the c3usallty and Krarncrs-KrofllC:L that ~ ( w )= ca(w).This means that for w > 0. Im c > 0. but for w fl 1" The poles in the integral in kl thus appear as 207. Thus. r- . w IwI --- < Q ( ~ , < )= 3 J-!!e ' ~ LI-NJ- J ~ E- 1 )] forw > 0 E 2 v sin((z -VI )ElI I-.ro~llT.I.T.vol. I p. 47,(58) and p.99,( I ) we Ilave, for y > 0: &se (i) (z -vt) > 0: Evidently @(T;.t) = 0,indepcndent of the rclalivc magnitude o f Y = ( z - v l ) a n d a = 9 d m . Case (ii) (z-vt) ,,,,,, . I!n!i%c the wsc of electrically charged particles, tllcrc ISnr) cl~hunccmcnt (11. i r ~ f i l r I I I J ~ I I ~ : ~ ~ ~nlor~upolcsin low vclocity region. d 210. Radiation by Moving Charger Usinp Wiechcit-Lienard po~cntialswe obtained tllc electric and r113gncti. 3s slated In Eqf 13.12) and Eq(l4.14) In this problem wc arc concerned with only the uniform motiori. th: ,? = 0 and t!lc second terrn 11, thc clcctric field vanishes. c ( " - 6)I< cr ~ i ; , ~ ) = - - , / 7' (KR)> j2(KR /+ j t he transverse conlponcnt / r , / . / ' eb t,(x, 1) = /" , 1 and j' (KR)' I / 1 the lonp.ltud~nalcomponenl /. ev1 4.d'. , E, fx, 1') = I -d- jt (RR)' '-'L.borl~oodcf frequencics, the Fourier components vanish. To study thc total powcr ndiltcd, w: usc the fact that the instantaneous energy flux is pvcn by the Poynting's vcctor Then the power radiated per unit solid angle in the direction i. k where R is thc distance to b e source (assumed to be much larger than the extent of thc source). Ilowever, for radiation of frequency o propagting in the direction fi, 222. So. W -- - ~2 - m;, 4n c But because of the decomposition studied earlier, ficncc thc power radiatcd per unit solid anglc in the direction a in thc m r frcqucncy mode is Using thc form already determined for the current. a A e T c(t) - iw,[t+ 1 A, ( r ) = -j dl t C T . 17-Yo(t)l a A a 2 - a and for I r 1 = R >> l ro (I)!, I r - ro(t) l 2 R - r - ro (I) so, for ~e purpose of studving radiation Thus. This is exactly the radiation that one chargc. e, would produce in the sarc motion. tiowever, there are now I/N as many frequency modes in thC radiation and it is necessary to sum over dl the modes to find the total power emitted: 223. OD OD Now, C e ' m 2 1 r ~ =Z m = - w p=-OD 6 (X + P) I icnce. -b d P e2 T T N i~- d (t) n .ro(.ll rn T-=- Z / / d t d t ' b [ ( 1 - -1. t- T + -,I dn;; 4n c3T3 m 0 c c N -. A n ~ ( 1 ) i- ([h x 4t)]. [" - v(tS)][1 - -I-' [ I - d t dt' c *cv(l') I-; J In the limit N -b -, q + 0.Nq = c, lhe sum over m becomes an integral over 1" and d P - - - e T T T n - ro (I) "-70(t,) / / / d t dt'dt" 6 It - - I' dR" 4 n c 3 TZ o o -f + 1'.1 C C i~- ;(I) n .Vtt') L L { / i , x ; ( t ) j .["xY(te)j [ I -- ,- I [ 1 - -1 - 1 ) d t dt' c C n-f.,(t) , The above statement it exact because thc finite r v l p of (t - -- t .. C *-.at)+ -) dew the sum in the previous expresnon to be changed from o. C~ - - -- C to 2 The integration is not dificult and wc have m=-m m'-N BCUW of the periodiaty. Also, the fields arc - and N d d .---- 2 6 [y(t8+ lr - rO(te)l mT-. e .r-. , - t + -11 (r, t) -/ dt' c N - - - cT , 17- = --3n 9 Ymax X 0 1 1 tho other I~acd, 70-> I . Accord~ngt o the discussion preceGng Eq(i-i.Q6),w e must rl!ultlply I([ . G ) tl) rn convcrt 10 powcr rndiated per unit frequency intern!. Thas. a I' (E. w ) = con st ( -)'I3 f (wlo,) E' - If electrons are distributed in energy according to the spectrum N(E)dE -E-'" Jt.Thr power radia~cd1s~pproximately rn3 c5 o how. w < w, means E > (- )'I2. ~ h u s , 3 e B OD 00 --,,-- < P ( u ) > d w = ( p(E.~)h'(E)dEdw a''' E 3 dEdc: 0 i (d) At lower frequencies (a-- 10' KC-' t o -- 6 X 10" rec-I and a 2 0.35 . the Cnb nebula gvtr coniinuola rpectnm~The ndirtion drops off sharply rt a 1.5. Examine n and EcUtorf at the cutoff. For low freq. a 0.35 - 360 - 9 ~ 6 ' 230. li~erciorcII = 1.7 and wc,,,ff = 6 X 10" sec-'. Hence, The above Equation is cons:stcnt with Ihc cutoff of part (a), namcly ECut,ff 2 IOI3 e V (t) Froni problem (14.4) part (b), we haw 3 1 c I t -- = 2.6 X 10" sec = 825 years 2 c4 B2 y The Crah supernova was ~Sservcdin 1054, so the Cr3!~ncbula was ah, 900 years old when Ulc observations cited in thc t c x ~wcre made. 14.16 The incident parttclc e n e r p E (in M c V) I I 2 J E~ - mO2c4 - - -COSO, = -- ( n = l . ' p c ' I 2 p n 3 E o,= cos-' 1 McV electrorl (K.E.) I500 MeV proton 5 GeV p:oll '=c ,+. *e.:': - ---- - - - - - - - . i 4 3 ' ~ 30'. I lo-; I I f : 0 - - 2. - 1 ' I 1 3 ( - 1 . f ) An expression due to Frank and Tam for the energy loss in Cherenkc radiation is given as 231. a narrow spectrum. the refnctive index n does not vary for appreciable ount, then d E 4 2 Z' e' I YI "I --(PIC = (I - - ) ( - --) .- dx h c B'n' c c ..energy radiated per unit distance per unit ircquency rnteral is. . ;efore, t l ~ enSunberof photons radiated per umt distance per urul frequency rval is a frequency interval Aw the nurnber of pholuns radated per unit distance the above equation, we have lev electron: 8 = 0.941 0' = 0.885 U2c)-'= 0.501 I MeV proton: fl a 0.759 8' = .576 @'E)-' = .774 N = 86.1 photons/cm 2V proton: 8 = -9875 8' = .9851 (P2e)-'= .445 232. 14.17 (*14.14) The gpproximation to synchrotron spectrum u g u n as 2w 2 ea - -W P(u,t) = -- Y t ( - I 3 e WC r 9 wc where wc = _~G.,~'(I).It is assumed that the lime swle for change of ~ ( t )is Iong compared to a,-'m d so is adiabetic as far as [lie synchrontron emission is concerned. The validity of the equation is checked 2 which agrccs with the result in Eq(11.3 1) wherc thc coclficicnt is - -vl~crc[he itcliiiicn~--- 0.4114 in coinp3risot1 to rhc exact coct'ficicr~t ; I 1 3 .7i6 01. - l-( ) = 0.1176 deduced from Eq(l4.95) 1 ..- I 4 3 (a) Assuming 7(t) rises linearly and me is neghgbly sm311, we llavc t Y (1) = Ymax ( -T I T t w -2w Thcn < P ( ~ , I ) >= - - - j d l T ( -)t'3exp(-) T n 9 a wc(1) wc(t) t where a,([) = 3 w. ynaX ( y)' 233. 22/3 SD e-y .: 1) (c.1) -2 = - - X Z / 3 ~ J -7mmx dy. - 3n 9 x y4" 00 (11 ?ht-I~nlllinpCorms of the spectrum: (Define 1(x) = J e- y-'I3 dy ) X -(a) For x >> 1 lntcgratc bv par&successively, we have .. ___ .. (h) For > ,'.< I 2 ~ ( ~ 1