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Griffith...ual-pdf) - introduction to electrodynamics (solutions) - ch12, Exercícios de Física

exercícios resolvidos, do livro do Griffiths de eletromagnetismo, separados pos capítulos.

Tipologia: Exercícios

2010

Compartilhado em 26/08/2010

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Baixe Griffith...ual-pdf) - introduction to electrodynamics (solutions) - ch12 e outras Exercícios em PDF para Física, somente na Docsity! Chapter 12 Electrodynamics and Relativity Problem 12.1 Let u be the velocity of a particle in S, u its velocity in 5, and v the velocity of 5 with respect to S. Galileo's velocity addition rule says that u = u + v. For a free particle, u is constant (that's Newton's first law in S). (a) If v is constant, then u = u-v is also constant, so Newton's first law holds in 5, and hence S is inertial. (b) If 5 is inertial, then u is also constant, so v = u - u is constant. Problem 12.2 (a) mAUA + mBUB =mcuc + mDuDi Ui =Ui+ v. mA(uA + v) + mB(uB + v) = mc(uc + v) + mD(uD + v), mAnA + mBUB + (mA + mB)v = mcuc + mDuD + (mc + mD)v. Assuming mass is conserved, (mA + mE) = (mc + mD), it followsthat mAUA+ mBuB =mcuc + mDuD, so momentum is conserved in 5. (b) 1 2 1 2 - 1 2 1 22mAUA + 2mBUB - 2mcuC + 2mDuD => !mA(U~ + 2UA' v + V2) + !mB(U~ + 2UB' v + v2) = !mc(ub + 2uc' v + V2)+ ~mD(ub + 2UD' v + V2) !mAU~ + ~mBU~ + 2v. (mAuA + mBuB) + !v2(mA + mE) = ~mcub + !mDub + 2v. (mcuc + mDuD) + ~v2(mc + mD)' But the middle terms are equal by conservation of momentum, and the last terms are equal by conservation f 1 -2 + 1 -2 1 -2 + 1 -2 d0 mass, so 2mAuA 2mBuB = 2mcuC 2mDuD' qe Problem 12.3 (a) Va = VAB + VBC' VE = VAB+VBC ~ Va (1 - ~ ) => ~ = ~., 1+VABVBC/C2 C va C In mi/h, c = (186, 000 mi/s) x (3600sec/hr) = 6.7 x 1O8mi/hr. . ~ = (5)(60J = 6 7 X 10-16 => 16.7 x 10-14% error I (p rett y small!). va (6.7xlO )2 . , ~ [10]. (b) (!c + ic) / (1+ ~ . i) = (~c) / en = l.:!rJ (stllliess than c). (c) To simplify the notation, let /3==VAC/C,/31==VAB/e, /32==vBc/e. Then Eq. 12.3 says: /3= f~%~;2'or: /32= /3f + 2/31/32+ /3~ = 1 + 2/31/32+ /3f/3~ - (1 + /3f/3~- /3f - /3~) = 1 - (1 - /3f)(l - /3~)= 1 - A (1 + 2/31/32+ /3f/3~) (1 + 2/31/32+ /3f/3~) (1 + 2/31/32+ /3f/3~) (1 + /31/32)2 ' 219 ~ 220 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY where ~ ==(1-,8f)(1-,8~)/(1+ ,81,82)2is clearly a positive number. So ,82< 1, and hence IvAGI< c. qed Problem 12.4 (a) Velocity of bullet relative to ground: ~c + lc = ~c= ~gc. Velocityof getawaycar: ~c= 192C.Since Vb > vB' I bullet does reach target.! 1c+1c ~c 5 20 (b) Velocity of bullet relative to ground: :+ 1\ = T = 'fc= 28c.2 3 6 Velocity of getaway car: ~c = ¥Bc. Since Vg > Vb, I bullet does not reach target. I Problem 12.5 (a) Light from the 90th clock took i~~~8°:'/S= 300 s = 5 min to reach me, so the time I see on the clock is 111:55 am. I (b) I observe 112 noon. I Problem 12.6 { light signal leaves a at time t~; arrives at earth at time ta = t~+ dalc,"7' light signal leaves b at time t~; arrives at earth at time tb = t~+ dblc. :.~t=tb-ta =t~-t~+ (db-da) =~t'+ (-v~t'cosO) =~t' [1- ~cosO].c c c (Here da is the distance from a to earth, and db is the distance from b to earth.) , . vsinO ~t I vsinO.. . ~s = v~t SIll 0 = (1 I 0); u = (1 v 0) 11sthe the apparent veloCity.- V c cos - Ccos du - v[(I- ~cosO)(cosO) - sinO(~ sinO)] = O:::}(1- ~cosO) cosO = ~ sin2 0dO - (1 - ~COSO)2 C v.2 2 ) _V:::} cosO= -(SIll 0 + cos 0 --c c 1 0 -1 ( I) I A h. . al 1 vV1-v2/c2 v max = COS V c. t t 1Smax1m ang e, u = 1-v2/c2 = V1-v2/c2' As v ~ c, I u ~ 00, I because the denominator ~ 0, eventhoughv < c. Problem 12.7 The student has not taken into account time dilation of the muon's "internal clock". In the laboratory, the muon lasts 'YT =...; T , where T is the "proper" lifetime, 2 x 10-6 s. Thus1-v2/c2 V2 1 c2 = 1 + (Tcld)2; d d v= = -vl-v2/c2, whered= 800m. TI../l- v2/c2 T ( T ) 2 2 v2 2 [( r ) 2 1 ] 2 1d v = 1- e2; v d + e2 = 1; v = (rld)2 + (1/e)2' Te - (2 x 10-6)(3 x 108) - ~ -~. V2 - 1 - 16. d - 8.00 - 8 - 4' c2 - 1+ 9/16 - 25' Iv = ~e.1 223 (h) It will take another 15.4 years I of earth time for the return, so when she gets back, she will say her twin's age is 45.6 + 5.4 = ~ years-which is what we found in (a). But note that to make it work from traveler's point of viewyou must take into account the jump in perceivedage of stay-at-home when she changes coordinates from S to S. Problem 12.17 -a,°rp+ a,lfjI+ a,2fj2+ a,3fj3= -"l(aO - {3aI)(bO- {3bI)+ -l(aI - {3ao)(bI - {3aO)+ a2b2 + a3b3 = -"-?(aobo - {3/bI - {3jl'b° + {32aIbI - aIbI + {3jl'b°+ {3/bI - {32aobo)+ a2b2 + a3b3 = -72aobo(1- {32)+ 72aIbI(1- {32)+ a2b2+ a~b3 = -aobo + aIbI + a2b2+ a3b3. qed [Note: 72(1 - /32) = 1.] Problem 12.18 ( c~ ( 1 000 ) ( ct )(a) I g) ~ -{ ~ ! ~ ~ I(usmg the notation of Eq. 12.24, 10' best compadson). ( 7 0 -7{3 0 ) (b) IA = 0 1 0 0-7{3 0 7 O' 0 0 0 1 ( 'YO -'Y~ 0 ) ( 7 -7/3 0 0 ) ( 7'Y -7'Y{3 -'Y~ 0 ) . . 0 1 0 0 -7{3 7 0 0 -7/3 7 0 0 (c) MultIply the matrIces: A = -'Y~ 0 'Y 0 0 0 1 0 = -'Y7~ 7'Y{3~ 'Y O' 00010001 0 001 I Yes, I the order does matter. In the other order, "bars" and "no-bars" would be switched, and this would give a different matrix. Problem 12.19 ~ (a) Since tanh 0 = ~~~~~,and cosh20 - sinh20 = 1, we have: 1 1 cosh 0 . 7 = = = = coshOj7{3= cosh0tanh0 = smhO. y'1- v2je2 y'1- tanh20 y'cosh2O- sinh20 ( cosh 0 - sinh 0 0 0 ) . A- - sinh0 cosh0 0 0 ..1 - 0 0 1 0 . 0 0 0 1 ( cas <jJ sin <jJ 0 ) Compare: R = - sin <jJ cas <jJ 0 . 0 0 1 - u-v fi (uje)-(vje) - tanh<jJ-tanhO (b) u = lUV ::} - = (U) (V) ::} tanh<jJ = 1 h<jJ hO' where tanh<jJ= uje, tanhO = vje;- C2" e 1 - C c - tan tan tanh 4)= fije. But a "trig" formula for hyperbolic functions (CRC Handbook, 18th Ed., p. 204) says: tanh<jJ - tanh 0 - 1- I1- tanh <jJtanhO= tanh«jJ- 0). :. tanh<jJ= tanh«jJ - 0), or: <jJ= <jJ- O. 224 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 1 i i Problem 12.20 (a) (i) I = -c2 b.t2 + b.x2 + b.y2 + b.z2 = -(5 - 15)2+ (10 - 5)2 + (8 - 3)2 + (0 - 0)2 = -100 + 25 + 25 = 1-50.1 (ii) I No.1 (In such a system Sf = 0, so I would have to be positive, which it isn't.) (iii) I Yes. I :1 _B 6 S travels in the direction from B toward A, making the trip in time IO/c. 5 :. v = -5:0/c5Y= I-~x - ~y.1 4 I 2 ," .. . . .. . . .. .. .. .. .. . ..ie 5 x N h v2 - 1 1 - 1 - 1 " 1ote t at C2 - 4"+ 4"- 2' so v - Vic, sale y less than c. 2 4 6 8 10 (b) (i) 1= -(3 - 1)2+ (5 - 2)2+ 0 + 0 = -4 + 9 = [[] (ii) I Yes. 1 By Lorentz transformation: b.(d) = , [b.(ct) - ,8(b.x)]. We want b.£ = 0, so b.(ct) = ,8(b.x);or v b.(ct) (3 - 1) 2 E}J . " - = ~ = ( . ) = -. So v = -c, IIIthe +x dIrectIOn.c I lX 5 - 2 3 3 (iii) ~ (In such a system b.x = b.y = b.z = 0 so I would be negative, which it isn't.) Problem 12.21 } , Using Eq. 12.18 (iv): b.£ = ,(b.t - ~b.x) = 0 =}b.t = ~b.x, or v = ~t c2= I tB - tA c2. x XB - XA Problem 12.22 (a) ct. world line of player 1 world line of the ball Truth is, you never do communicate with the other person right now-you communicate with the person he/she will be when the mes- sage gets there; and the response comes back to and older and wiser you. (b) I No way. I It is true that a moving observ- er might say she arrived at B before she left A, but for the round trip everyone must agree that she arrives back after she set out. ct A x B 225 Problem 12.23 (a) x ~'? c,t '}.'? c,t \.'? c,t \J'? c,t ./\. '? " c,t ./i-' '? <1- c,t ./J '? c,t (b) % = slope = H ~ v = ~:;c = I O.95c.1 (c) v' = tC, so v = ~c:i:l+sos - S!.fJ!k r35J - \37725J =@ =O.95c. .( Problem 12.24 (a) (1- ~)ry' = u'; .'(1 +~) = ry';I u = ,/1 +lry2/C' 11°1 (b) 1 = 1 = coshlJ = coshOJ 1/= 1 u=coshOctanhO=lcsinhO.1Vl-u2/c2 Vl-tanh21J Vcosh21J-sinh21J Vl-u2/c2 Problem 12.25 (a) u. = u, = uco,45° = }, Joe = I Ae.1 (b) Vl-~2/C2 = R = ~ = J5; 'I} = Vl-:2/C2 ~ l1]x = 1]y = v'2col (c) 1]0: ,c = I J5 col { - - u -v - .,fiTsc-fi75c - rQ"lux - 1":~ - 1-- - ~02 5 (d) Eq. I2.45~ - j;)'f., 2- 1 u 2 2/5c ~ Uy = " (,-~) =/1-. ~ = y'iTo°=[lE] (e) ijx = ,(1]x - (31]°)= VI - ~ (V2c - ~J5 c) = [QJ I ijy = 1]y = v'2col (f) 1 - 1 - V3° - - V3 ii { ijx = V3Ux = O. .( }Vl-fj,2/c2 - ~- ,'1}- ~ ijy=V3uy=V2c..( Problem 12.26 /-' - ° 2 2 - 1 2 2 - 2 (1 - u2jc2) - r:2l 1] 1]/-,--(1]) +1] - (I-u2jc2)(-c +u )--c (I-u2jc2)-~ i i I ! ct 'll 'N1;'1 .....I"JI ,,:>1 I L: Iii I! 41!. '/ 4/, 'If,t..'(f,J/,,-'I -Lf--+ -+--tl-+- I " , I I J J J I I I I i::PL J - 1-- --I--I " ", i I il I / 'i / I It..- :; ! .i- . ,--LJ----- -;--; ------ '---71/' I I ' ' I ! / I I I , II ' ' / ' -- -T+- ' II I II l--I/ tZ -1'/ 7/7 I I VJ PJ ./ ./ ./ ./ .//, fT v-: r It ./ ./ ./ / ./ ./ VJ ..) V; ...A I/f ..) /} ./ / ./ ./ ./ ./ v ./ r r C--r t-r ./ / ././ ./ V ./ V 1...-1/} A. VI /1 ....1 / / ./ i./V I...--:'V:=lL---fT? fT r ./ +9j-.LL I i././IA'11...J:1Ji/} ./LLLi--1 ii' I : &fTlld_i itll i i U 228 CHAPTER 12.ELECTRODYNAMICS AND RELATIVITY 22.2 ) 2 1232 EB(cos () + sm () =Pac - POCEA+ 4"EA + 4"EA => E1 =poc2 - POCEA+ E~ = [VP5C2 + m2c4 + mc2 - EA] 2 = POC2+ m2c4 + 2VP5c2 + m2c4(mc2 - EA) + m2c4 - 2EAmc2 + E~. Or: -POCEA = 2m2c4 + 2mc2vp5c2 + m2c4 - 2EAVP5C2 + m2c4 - 2EAmc2; EA(mc2 + Vp5C2 + m2c4 - poc/2) = m2c4 + mc2Jpoc2 + m2c4; E 2 (mc2 + VP5C2 + m2c4) (mc2 - VP5C2-t-m2c4 - poc/2)A =mc . (mc2 + VP5C2 + m2c4 - poc/2) (mc2 - VP5C2 + m2c4 - Poc/2) (,,?c' - P6e' - ,,?c' - lPome' - ..",/ P6e'+ m'c') I me' (me + 2Po + vP'o + m'c') = mc2 2 2 = - 0 (#c4 - pomc3 + p~t - P5c2 - #c4) 2 (mc + !po) Problem 12.36 d d { dU ( 1 ) 1 2 dU } F - P - mu - dt -<:2" u. dt - dt - dt Vl- u2/C2 - m Vl- u2/C2 + U -2 (1- u2/c2)3/2 m { u(u .a) } d= a + . qe Vl-u2/c2 (c2-u2) ay" ct Problem 12.37 At constant force you go in "hyperbolic" mo- tion. Photon A, which left the origin at t < 0, catches up with you, but photon B, which passes the origin at t > 0, never does. Problem 12.38 (a) 0 d1]o d1]o dt [ d ( c ) ] 1 a = dr = ill dr = dt Vl- U2/C2 ~U2/C2 - C ( 1 ). (-~)2u'a - 1 u.a- VI - U2/C2 -2 (1 - u2/c2)3/2 - C(1 - u2/C2)2' a=d1]=dtd1]= 1 d ( u ) = 1 { a +u(-~) -~2u'a }dr dr dt VI - U2/C2dt VI - u2/C2 VI - U2/C2 VI - U2/C2 2 (1 - U2/C2)3/2 1 [ u(u .a) ] =1 a+ . (1 - u2/C2) (c2 - U2) 229 (b) J.L- o 2 - 1 (u .a)2 1 [ ( U2 ) 1 ] 2 aJ.La --(a) +a:.a:--e2(I-u2/e2)4 + (l-u2/e2)4 a 1- e2 + e2u(u.a) { ( 2 ) 2 2 ( 2 ) } 1 1 2 2 U U 2 1 2 2 = --(u .a) + a 1 - - + - 1 - - (u. a) + -u (u. a) (1 - u2/e2)4 e2 e2 e2 e2 e4 1 { 2 ( U2 )2 (u.a)2 U2 U2 }= (l-u2/e2)4 a 1- e2 + e2 (,-1+2-2;2+;2)v (1 - ~) 1 [2 (u. a)2 J-I a +- (1 - u2/e2)2 (e2 - u2) . (c) 'T]J.L'T]J.L= -e2, so iT ('T]J.L'T]J.L)= aJ.L'T]J.L+ 'T]J.LaJ.L= 2aJ.L'T]J.L= 0, so I aJ.L'T]J.L= 0.1 (d) KJ.L= ¥r = !r(ml}J.L)= ~ I KJ.L'T]J.L = maJ.L'T]J.L= 0.1 Problem 12.39 KJ.LKJ.L= _(KO)2 + K. K. From Eq. 12.70, K. K = (1-~:/C2)' From Eq. 12.71: KO_~dE- 1 d ( me2 ) - me [-~ (-I/e2) 2u.a J -m (u.a)- e dr - evil - u2/e2 dt viI - u2/e2 - viI - u2/e2 2 (1 - u2/e2)3/2 - e (1 - u2/e2)2' m [ U2 (u .a) ] m( u .a)But (Eq.12.73): u.F=uFcosO= (u.a)+ 2( 2/2) = ( 2/2)3/2'SOviI - u2/e2 e 1 - u e 1 - u e KO = uFcosO evil - u2/C2; F2 u2 F2 COS20 [1 - (u2/e2) COS2OJK KJ.L= - = F2. qedJ.L (l-u2/e2) e2(I-u2/e2) (l-u2/e2) Problem 12.40 m [ u(u . a) J u(u . a) q / W= a+ 2 2 =q(E+uXB)=}a+(2 2)=-vl-u2/e2(E+uXB).viI - U2/ e2 e - u e - u m . u2(u .a) u .a q 2 2 Dotmu:(u.a)+ 2( 2/2)=( 2/2)=-vll-u/e[u.E+.u.(uvXB~;e l-u e l-u e m =0 u(u . a) q u(u . E) q 1 :. =-vll-u2/e2 . Soa=-vll-u2/e2 [E+uxB--u(u.E) ] . qed (e2 - U2) m e2 m e2 Problem 12.41 One way to see it is to look back at the general formula for E (Eq. 10.29). For a uniform infinite plane of charge, moving at constant velocity in the plane, j = 0 and p = 0, whilep (or rather, a) is independent of t (so retardation does nothing). Therefore the field is exactly the same as it would be for a plane at rest (except that a itself is altered by Lorentz contraction). A more elegant argument exploits the fact that E is a vector (whereas B is a pseudovector). This means that any given component changes sign if the configuration is reflected in a plane perpendicular to that direction. But in Fig. 12.35(b), if we reflect in the x y plane the configuration is unaltered, so the z component of E would 230 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY have to stay the same. Therefore it must in fact be zero. (By contrast, if you reflect in a plane perpendicular to the y direction the charges trade places, so it is perfectly appropriate that the y component of E should reverse its sign.) Problem 12.42 (a) Field is ao/Eo, and it points perpendicular to the positive plate, so: Eo = ao(cos45°x+sin45°y) = 1 ;'0 (-x+y). EO v 2 EO (b) From Eq. 12.108, Ex = Exo = -~; Ey = "IEyO= "I)i~o' So I E = ~(-x + "Iy).1 (c) From Frob. 12.10: tan 0 = "I,so 10 = tan-I "1.1 y (d) Let ii be a unit vector perpendicular to the plates in S-evidently ii = - sinOx + cosOy; lEI= ;'°'0 ';1 + "12. So the angle 4>between ii and E is: x E .ii 1. cosO 2"1_ lEI = cos4>= ~(smO+"IcosO) = ~(tanO+"I) = ~cosO1 + "12 V 1 + "12 1 + "12 But'" = tan 0 = sinlJ = v'I-cos21J= I 1 -1 => ...L20 = ",2 + 1 =>cosO = 1 . I coslJ coslJ Vcos21J cas I yI+'Y2 Evidently the field is ~ perpendicular to the plates in S. Problem 12.43 2 2 ~ ( ) E - ~ q(1 - v /c) R (E 1292)a - 4 (1 v2. 20 )3/2 R2 q. . =>7rEO - ~ sm So Icos 4> = C :"1"12 ). IE. da = q(1 - v2/c2) I R2 sin2O dO d4>47rEO R2(1 - ~ sin2 0)3/2 q(1 - V2/C2) 1 " sinO dO. . 2 2 = 27r 2 2 /' Let u = cos0, so du = - sm0 dO,sm 0 = 1 - u .47rEO 0 (1 - ~ sin 0)3 2 q(l-v2/c2) (I du q(l-v2/c2) ( C )3 {I du= 2Eo 1-1 [1- ~ + ~u2J3/2 = 2EO V J-I (C2 -1 + u2 )3/2'c c ~ 1 +1 2 ( V ) 3 2 The integral is: .ju 2 = ( c2 - 1) £. = -;; (1 - v2/ei) .(c2 - 1) c - 1+ U2 -1 ~ v~ ~ I q(1 - v2/C2) (C)3 (~)3 2 =. .(So E.da= 2EO v C (l-v2/c2) q 1 -.2..~/Loq2(I-v2/c2)2vsinO Rx(fi); (b) Using Eq. 12.111 and Eq. 12.92, S = /Lo(E X B) - /Lo47rEO47r R4(1 - ~sin20)3 (~ -(} 233 Now the inverse Lorentz transformations (Eq. 12.19) =* x = 'Y(x + vf) and t = 'Y(f + ; x), so kx-wt='Y[k(X+Vf)-w(f+ ;x)] ='Y[(k-~~)x-(w-kv)t] =kx-wf, - ( WV )where (recalling that k =wie): k =='Y k - ~ = 'Yk(1 - vie) = ak and w=='Yw(1 - vi c) =aw.. E(x, y, z, f) = Eocos(kx - wf)y, B(x, y, z, f) = Eo cos(kx - wf)z,c - - - - /1 - vlcwhereEo = aEo, k = ak, w= aw, and a = 1 I . 1 +v c Conclusion: (c) Iw = w 1 - vlc l+vlc' This is the I Doppler shift I for light. - 211" 211"- ~ The velocity of the\ - - - . A = k - ak a - w- w wave in S is v = -A = \" = @] 211" A same in any inertial system). I Yup, I this is exactly what I expected (the velocity of a light wave is the (d) Since intensity goes like E2, the ratio is i= ~~ =a2 = I ~ ~ ~~~. Dear AI, The amplitude, frequency, and intensity of the light wave will all I decrease to zero I as you run faster and faster. It'll get so faint you won't be able to see it, and so red-shifted even your night-vision goggles won't help. But it'll still be going 3 x 108mls relative to you. Sorry about that. Sincerely, David Problem 12.48 [02 = A~A;tA<7= A8A~tO2+ A~A~t12 = 'YtO2+ (-'Y{3)t12 = 'Y(tO2- {3t12). ~3 = A~A~tA<7 = A8A~tO3 + A~A~t13 = 'YtO3+ (-'Y{3)t13 = 'Y(tO3 - {3t13) = 'Y(tO3+ {3t31). ~f23 = A~A~tA<7 = A~A~t23 = t23. fH = A1A~tA<7 = A~AAt30 + A~Att31 = (-'Y{3)t30 + 'Yt31 = 'Y(t31 + {3tO3). f12 = AiA;tA<7 = AAA~tO2+ AtA~t12 = (-'Y{3)tO2+ 'Yt12= 'Y(t12- {3tO2). Problem 12.49 Suppose t"l-'= :HI-'" (+ for symmetric, - for antisymmetric). fl<A = A I<A Atl-'"I-' " fAI<= A;A~tl-'" = A;A~t"l-' [Because J.land v are both summed from 0 -+ 3, it doesn't matter which we call J.land and which call v.] [I used the symmetry of tl-"', and wrote the A's in the other order.]= A~A;(:f:tl-"') = :f:fI<A. qed 234 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY Problem 12.50 p"V P"V = pO°pOO- pOl pOl - pO2 pO2 - pO3 pO3 - plO plO - p20 p20 - p30 p30 + pH pH + pl2 p12 + p13 p13 + p21 p21 + p22 p22 + p23 p23 + p31 p31 + p32 p32 + p33 p33 = -(Ex/C)2 - (Ey/C)2- (Ez/C)2- (Ex/C)2- (Ey/C)2- (Ez/C)2+ B; + B; + B; + B; + B; + B; = 2B2 - 2E2 / C2 = 12 (B2 - ~:), which, apart from the constant factor -~, is the invariant we found in Prob. 12.46(b). I G"vG"" = 2(E2/c2 - B2) I (the same invariant). P"VG"v = -2 (pOlGOl + pO2GO2+ pO3GO3)+ 2 (P12G12 + p13G13 + p23G23) ( 1 1 1 )-2 -ExBx + -EyBy + -EzBz 2 [Bz(-Ez/c) + (-By)(Ey/c) + Bx( -Ex/c)]C C C 2 2 I 4 I-~(E. B) - ~(E. B) = -~(E. B), = = which, apart from the factor -4/ c, is the invariant of Prob. fundamental invariants you can construct from E and B.] Problem 12.51 2 } ( 0 COO ) E - L..2,\:x:- l!2.~:X: - 4'/1"£0X - 2'/1"x p"v = JLoA -c 0 0 -v . B - 1& 2Av ~ - l!2.'\v ~ 21TX 0 0 0 0 - 4'/1"--xy - 2'/1"-xy 0 V 0 0 12.46(a). [These are, incidentally, the only Problem 12.52 8vP"v = JLoJ". Differentiate: 8,,8vP"" = JLo8"J". But 8"8,, = 8v8" (the combination is symmetric) while pv" = -P"v (antisymmetric). .'. 8"8,,P"" = O. [Why? Well, these indices are both summed from 0 ~ 3, so it doesn't matter which we call JL,which v: 8"8,,P"v = 8v8"pv" = 8,,8,,(- P"") = -8,,8vP"v. But if a quantity is equal to minus itself, it must be zero.] Conclusion: 8"J" = O. qed Problem 12.53 We know that 8"G"" = 0 is equivalent to the two homogeneousMaxwellequations, V.B = 0 and VxE = - ~~. All we have to show, then, is that 8,\P"v + 8"Pv'\ + 8"P,\" = 0 is also equivalent to them. Nowthis equation stands for 64 separate equations (JL= 0 ~ 3, v = 0 ~ 3, A = 0 ~ 3, and 4 x 4 x 4 = 64). But many of them are redundant, or trivial. Suppose two indices are the same (say, JL= v). Then 8,\P"" + 8"P,,'\ + 8"P,\" = O. But P"" = 0 and P,,'\ = - P'\", so this is trivial: 0 = O. To get anything significant, then, JL,v, A must all be different. They could be all spatial (JL,v, A = 1,2,3 = x, y, z - or some permutation thereof), or one temporal and two spatial (JL= 0, v, A = 1,2 or 2,3, or 1,3 - or some permutation). Let's examine these two cases separately. All spatial: say, JL= 1,v = 2, A = 3 (other permutations yield the same equation, or minus it). 8 8 8 83Pl2 + 81P23+ 82P31=0 => 8z (Bz) + 8x (B:zJ+ 8y (By) = 0 =>V .B = O. 235 One temporal: say, J1.= 0,v = 1,A= 2 (other permutations of these indices yield the same result, or minus it). ~FOl + 8oFl2 + 8lF20 = 0 ~ :y (- ~:r:)+ 8(~) (B%)+ :x (~y)= 0, or -!!..fft+ (§j: - 8ff:r:.)= 0, which is the z component of -~~ = VxE. (If J1.= 0,v = 1,A= 2,we get the y component; for v = 2,A= 3 we get the x component.) Conclusion: 8>.Fp.v+ 8p.Fv>.+ 8vF>.p.= 0 is equivalent to V.B = 0 and ~~ = - V X E, and hence to 8vGp.v = O. qed Problem 12.54 Ko = q'f/vFov= q('f/lFo1+ 'f/2Fo2+ 'f/3FO3)= q(TJ.E)je = I ~1'u, E.! Now from Eq. 12.71 we know that Ko = ~ dd'; ,. where W is the energy of the particle. Since dr = ~dt,we have: 1 dW q IdW I -1'- = -1'(u. E) ~ - =q(u. E).edt e dt This says the power delivered to the particle is force (qE) times velocity (u) - which is as it should be. Problem 12.55 8°1jJ = ~1jJ = _! 8_1jJ= _!(81jJ8~+ 81jJ8~+ 81jJ8~+ 81jJ8:).~ em emm &m ~m &m 8t 8x 8y 8z From Eq. 12.19, we have: 8t = 1', 8t = 1'V, 8t = 8t = O. - 1 81jJ 81jJ. - 81jJ v81jJ So 8°1jJ = --1'(_8 + v-8 ) or (smce et = XO = -xo): 8°1jJ=1'(_8 - --8 l) =l' [(8oljJ) - ~(811jJ)].e t x Xo e x 8lIjJ = ~1> = 81jJ8t + 81>8x + 81jJ8y + 81>8z = 1'3!..8cp + 81>=1'( 81jJ - ~ 8cp) = [(8lljJ) - ~(8°1jJ)].8x 8t 8Xl 8x 8x 8y 8x 8z 8x e2 8t l' 8x 8Xl e 8xo l' ~1jJ=~=~m+~&+~~+~&=~=~~ 8y 8t 8y 8x 8y 8y 8y 8z 8y 8y ~1jJ=~=~m+~&+~~+~&=~=~~ 8z 8t 8z 8x 8z 8y 8z 8z 8z 8z 1-(:onclusion: 8p.1>transforms in the same way as ap. (Eq. 12.27)-and hence is a contravariant 4-vector. qed Problem 12.56 According to Prob. 12.53, 8ff;"~= 0 is equivalent to Eq. 12.129. Using Eq. 12.132, we find (in the notation of Prob. 12.55): 8Fp.v 8Fv>. 8F>.p. 8 8 8- 8 \ + -8 + _8 = >.Fp.v+ p.Fv>.+ vF>.p.x" xp. XV = 8>.(8p.Av - 8vAp.) + 8p.(8vA>. - 8>.Av) + 8v(8>.Ap. - 8p.A>.) = (8>.8p.Av - 8p.8>.Av) + (8p.8vA>. - 8v8p.A>.) + (8v8>.Ap. - 8>.8vAp.) = O. qed [Note that 8>.8p.Av= 88~v = 88~v >. = 8p.8>.Av,by equality of cross-derivatives.]x xp. x x 238 CHAPTER 12,ELECTRODYNAMICS AND RELATIVITY w cos () = V(1 + cot2 ~+ wcot2 ~)(1 + tan2 ~ + wtan2~) (where w ==,.,? - 1) , () - r2 - 4 (1 + ) - 4 2 t () - -2r..sm - sin"" r - i? w - ~"( , SO an - (-y2-1)sin"" ( ) . 2 I 2c2Or, since ("(2- 1) = "(2 1 -? = "(2~, tan () = "(V2 sin </>. w sin 1!.cos 1!.2 2 - V(CSc2~ + wcot2 ~)(sec2 ~ + wtan2~) - V(1+ WCOS2 ~)(1+ wsin2~) - ~wsin</> - sin</> - V[1 + ~w(1 + cos</»][1 + ~w(l- cos</»]- V[(~ + 1) + cos</>][(~ + 1) - cos</>] - sin </> - sin </> - 1 h 2 - 4 4- - - , were r - - + -, V(~+1)2-cos2</> V~+t+sin2</> VI + (rj sin </»2 W2 w /$." . \ <0'. 0/ ,x/ fL 1 w r j sin </> Problem 12.60 d dt dt - 1 '= mu ,* =K (a constant) =>'!lfdT= K, But dr - Vl-u2/c2' P Vl-u2/c2 'J!.. ( u ) = Kyl-u2jc2, Multiply by ~; =~:' 'dt Vl-u2/c2 m dt ~ ( u )= !£( u )= K -/1- u2jc2 Let w = ~dx dt VI - U2j C2 dx VI - U2j c2 m u' VI - U2j C2. dw = K ~; wdw = ~!£W2 = !:.-; d(W2) = 2K => d(w2) = 2K (dx),dx mw dx 2dx m dx m m :. w2 = 2Kx+ constant. But at t = 0, x = 0 and u = 0 (so w = 0), and hence the constant is O.m 2K U2 W2 = -x = 1 2j 2 jm -u c 2 2Kxjm - c2 , U = 1 + 2Kx - 1 + (mc2 ) , mc2 2Kx 2 2Kx 2Kx 2 u =---u ;m mc2 dx c ; dt = VI + (~;~) u2(1 + 2Kx ) = 2Kx,mc2 m J mc2 ct = 1+ (2Kx) dx. Let mc2 = a2,2K - , ct= J~ d..;x x . Let x ==y2j dx = 2ydyj ..;x = y. o~ ct = J -/y2y+ a2 2ydy = 2JVy2 + a2 dy = [YVy2 + a2 + a2ln(y + Vy2 + a2)] + constant, At t = 0, x = 0 => y = 0, so 0 = a2Ina+ constant =>constant = -a2 Ina. :. ct = YVy2 + a2 + a2ln(yja + v(yja)2 + 1) = a2 [ (~) V (~) 2 + 1 + In (~+/ (~)2 + 1) ], Let: z ==yja = ..;xv -?!b= V~'!j, Then I ¥!d = z~ + In(z+ ~),I 239 Problem 12.61 ......--........ (a) x(t) = f; [VI + (at)2 -1], where a = ::c' The force of +q on -q will be the mirror image of the force of -q on +q (in the x axis), so the net force is in the x direction (the net magnetic force is zero). All we need is the x component of E. The field at +q due to -q is: (Eq. 10.65) q ,z E=--( )3 [u(c2_v2)+u(~'a)-a(~'u)].47f/Oo~. U U = Cot- v ~ Ux = ci - v = t(cl- v,z)j~. U= 0'1-- ~. v = (0'1--Iv); ~. a = la. So: Ex = -~ ,z [ ~(cl- m)(c2 - V2) + ~(cl - )4)la -" a(O'1-- jAJ)]47f/Oo(0'1-- vl)3 ,z "z , ~ tca(12 - ~2) = -carP /~ =-- 4 q ( 1 1)3[(cl-m)(c2-v2)-cacP.]7f/Oo 0'1-- V x ............-..-........ -d/2""".""'.:'q The force on +q is qEx, and there is an equal force on -q, so the net force on the dipole is: 2q2 1 F = - 47f/Oo(c,z -lv)3 [(cl - V,z)(C2 - V2) - cad2]x. It remains to determine ,z, 1, v, and a, and plug these in. dx c 1 1 2 cat catr / v(t) = -d = -- 2 V 2a t = V ; v = v(tr) = -T ' where T ==vI + (atr)2.t a 1 + (at)2 1 - (at)2 dv ca ( 1 ) 2a2tr ca [ 2 2] ca a(tr) = dtr= T + catr -2 ~ = T3 1+ (atr) - (atr) = f3' Now calculate tr: c2(t - tr)2 =,z2 = 12+ rP;1= x(t) - x(tr) = f; [VI + (at)2 - VI + (atr)2] , so y - 2ttr +Yr = ~ [1 + (~2 + 1 + (a}{)2- 2Vl + (at)2vl + (atr)2] + (d/C)2 (*) VI + (at)2vl + (atr)2 = 1+ a2ttr + ~ (':,d)2. Square both sides: .4.h.4.h 1 ( ad ) 4 ( ad ) 2 ( ad ) 2 ,X + (at)2+ (atr)2+ a/ t; =,X+ a/ t; +:1 ~ + 2a2ttr+ ~ + a2ttr ~ 2 2 ( ad ) 2 ( d ) 2 a2 ( d ) 4 t + tr - 2ttr - ttr ~ - ~ - 4 ~ = O. At this point we could solve for tr in terms of t, but since v and a are already expressed in terms of tr it is simpler to solve for t (in terms of tr), and express everything in terms of tr: t2-ttr[2+(:d)2]+[t;_(~)2 _~2(~)4] =O~ t = ~{tr [2+ (:dr] ~ t~r +4(acdr + (:df] - ~ +4(~r + a2(~f} = tr [1+ ~ (:dr] ~ [1+ (atr)2](~r [1+ (~:r] 240 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY Which sign? For small a we want t ~ tr + die, so we need the + sign: [ 1 ( ad ) 2 ] d I ( ad ) 2 t = tr 1+"2 7 + ~TD, where D ==VI + 2e . So1-= e(t - tr) =?1-= ~(Q:)2 + dTD. Now go back to Eq. (*) and solvefor 0+(at)2: VI + (at)2 = ~ { 1 + ~ (aedf + a2tr [tr (1 + ~ (:d f) + ~TD] } = ~{ II+ ~atr)2J[1 + ~ (aedf] + a2~rdTD } = [1 + ~ (aedf] T + a2~rdD. T2 e [ ] e { rv 1 ( ad ) 2 ] a2trd } ( d. )l = a VI + (at2)- VI + (atr)2.= a r +"2 7 T + --;;-D -;t =ad 2eT + trD . Putting all this in, the numerator in square brackets in F becomes: [ ]= {cad(:eT+trD) - e~r [e~r(:df +dTD]}(e2 - e2;:t~) -e~~d2 [ d d(atr)2 ] e2 2 2 e2a~ = cad 2eT + vb - 2eT - vb T2 [1 + (¥'r) - (¥'r) ] - 'f3 e2a~ [ 1 1 ] e2ad2 [ ] c2a~ = 'f3 "2T2- "2(atr)2- 1 = 2T3 1+ (atr)2- (atr)2 - 2 =- 2T3 . q2 c2ad2 ~ . F = 3x. . . 47r€o[(e1--lv)T] It remains to compute the denominator: { [ ctr ( ad ) 2 ] ( d . ) eatr }(c1--lv)T= e""2 7 +dTD -ad 2eT+trD T T =[~¥d2 +cdTD- ~¥d2 - Cd(~r)2D]T = edD[~ ] = dcD. 1+(%)2 _(%)2 q2 e2d2a ~ I q2 a X :. F = 47r€o c3d3D3 X = 47"€ocd[I + (adI2e)2]3/2 (a = :J. Energy must come from the "reservoir" of energy stored in the electromagnetic fields. 1 q2 a [ ( ad ) 2 ]3/2 q2 Jloq2(b) F = mea = - - =? 1+ - = = -. 2 47r€ocd[I + (adI2e)2]3/2 2e 81f€omc2d 87rmdt (force on one end only) . - 2e ( Jloq2 )2/3 - - 2mc2 ( JlOq2 )2/3 -. . a - d 87rmd 1, so F - d 87rmd 1. 243 x Just after: Field lines outside sphere of radius ct emanate from position particle would have reached, had it kept going on its original "flight plan". Inside the sphere E = O. On the sur- face the lines connect up (since they cannot simply terminate in empty space), as suggested in the figure. This produces a dense cluster of tangentially-directed field lines, which expand with the spherical shell. This is a pic- torial way of understanding the generation of electromagnetic radiation. Problem 12.66 Equation 12.68 assumes the particle is (instantaneously) at rest in S. Here the particle is at rest in 5. So 1- - -- Fl..=-Fl.., Fli=Fli. UsingF=qE,then,I Fx = Fx = qE:z:, 1 - 1- Fy = -Fy = -qEy, I I 1 - 1- Fz = -Fz = -qEz. I I Invoking Eq. 12.108: Fx = qEx, 1 Fy = -ql(Ey - vBz) = q(Ey - vBz), I 1 Fz = -ql(Ez + vBy) = q(Ez + vBy). I But v x B = -vBz x + vByz, so F = q(E+ v x B). Qed Problem 12.67 Z.I. .z RewriteEq. 12.108withx -+ y, y -+ Z, Z-+ x: y fi Ey = Ey By = By Ez = I(Ez - vEx) . Bz=I(Bz+ ;Ex) Ex = I(Ex + vBz) Bx = I (Bx - ; Ez) 1E ~- x v x This gives the fields in system 5 moving in the y direction at speed v. Now E = (0,0, Eo); B = (Bo, 0,0), so Ey = 0, Ez = I(Eo - vBo), Ex = O. If we want E = 0, we must pick v so that Eo - vBo = 0; i.e. Iv = Eo/Bo.1 (The condition Eo/ Bo < c guarantees that there is no problem getting to such a system.) Withthis,By = 0,Bz = 0,Bx = leBo- .z.Eo)= IBo(I-~) = IBo~ = ~Bo;113= ~Box.1 z The trajectory in 5: Since the particle started out at rest at the origin in S, it started out with velocity -vy in 5. According to Eq. 12.72 it will move in a circle of radius R, given by ( 1 ) ~ p=qBR, orlmv=q -:rBo R=}~ fi x The actual trajectory is given by I x = 0; fi = -Rsinwl; z = R(1 - coswt); I where Iw = ~ .1 244 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY The trajectory in S: The Lorentz transformations Eqs. 12.18 and 12.19, for the case of relative motion in the y direction, read: x=x x=x y = ,(y - vt) Y= ,('f)+ vi) z=z z=z t=,(t-~y) t=,(t+~y) So the trajectory in S is given by: x=o; y=,(-RSinwt+vi)=,{-Rsin[w,(t- ;y)] +v,(t- ;y)}, or ( V2 ) [ ( V )]Y 1+,2- = "'?vt-,Rsinw, t--y~ c2 } (y-vth=-Rsin[w,(t-~y)]; 'Y2y(l-~+~)='Y2y z = R( 1 - COS2 wi) = R [1 - cosw, (t - ; y)] . So: Ix=o; y=vt-~sin[w,(t-~y)]; z=R-Rcos[w,(t- ;y)]. We can get rid of the trigonometric terms by the usual trick: ,(y - vt) = -Rsin [w,(t - ~y)] } ::} 1,2(y - vt)2 + (2'- R)2 = R2.1z - R = -Rcos [w,(t - ~y)] Absent the ,2, this would be the cycloid we found back in Ch. 5 (Eq. 5.9). The,2 makes it, as it were, an elliptical cycloid - same picture as p. 206, but with the horizontal axis stretched out. Problem 12.68 (a) D = foE + P suggests E -+ l..D } .. .. . . H 1 B M t B fO H but 1t'S a httle cleaner 1fwe d1v1deby Po while we're at 1t, so that= JloO - sugges s -+ Po E -+ ~D = c2D B -+ H. Then:Jloofo ' Then (following the derivation on p. 539): d 8 8 1 8 1 8DJloV ~Dov = cV.D = CPt= Jfo; ~D1v = -!5""(-cD.,) + (VxH)x = (Jf)x j so ~ = JfJlo,uxV uxV c ut uxV where 1 Jj = (cPt,J f ).1 Meanwhile, the homogeneous Maxwell equations (V.B = 0, E = - 0::)are unchanged, and hence I ~ = 0.1 (b) IHJloV = { =~: -Hz H., Hy 0 -cDz cDz 0 -cDy cD., Hz } cDy -cOD., D"" = {-D' cD., cDy CD,} 0 Hz -Hy -cDy -Hz 0 H., -cDz Hy -Hz 0 245 (c) If the material is at rest, TJv= (-c, 0,0,0), and the sum over v collapsesto a single term: E D/JoTJO= C2€p/JOTJO~ D/Jo = C2€p/JO~ -cD = -C2€- ~ D = €E (Eq. 4.32) .(c ' 1 1 1 1 H/JoTJO= -G/JoTJo~ H/Jo= -G/Jo ~ -H = --B ~ H = -B (Eq. 6.31). .( JL JL JL JL (d) In general, TJv= ')'(-c, u), so, for JL= 0: DOvTJv = DOlTJl + DO2TJ2 + DO3TJ3 = cDx(')'ux) + cDy(')'uy) + cDz(')'uz) = ')'c(D . u), povTJv = pOlTJl+ pO21J2+ pO3TJ3= Ex (')'ux) + Ey (')'Uy) + Ez (')'uz) = 1:(E. u), soc c c c DovTJv= ~€povTJv~ ')'c(D .u) =C2€ (~) (E. u) ~ Do u =€(E .u). [1] y HOvTJv= HOlTJl+ HO2TJ2+ HO3TJ3= Hx(')'ux) + Hy(')'uy) + Hz (')'uz) = ')'(H on), GovTJv= GOlTJl+ GO2TJ2+ GO3TJ3= Bx(')'ux) + By (')'Uy)+ Bz(')'uz) = ')'(B. u), so 1 11 HovTJv= -GovTJv~ ')'(H .u) = -(')')(B 0u) ~ H. u = -(B. u). JL JL JL Similarly,for JL=1: Dlv1Jv = DlO1JO+ Dl2TJ2+ Dl31J3= (-cDx)( -')'c) + Hz(')'uy)+ (-Hy)(')'uz) = ')'(c2 Dx + uyHz - uzHy) = ')' [c2D + (u x H)Jz, -E plvTJv = plOTJO+ pl2TJ2+ pl3TJ3= =.(-')'c) + Bz(')'uy) + (-By)(')'uz) = ')'(Ex + uyBz - uzBy)c = ')' [E + (u x B)]x' so DlvTJv= C2€plV1Jv~ [2] 1 ')' [c2D + (u x H)Jz = C2€(')') [E + (u x B)]x ~ D + C2(u x H) = € [E + (u x B)]. [3] HlvTJv = HlOTJO+ Hl2TJ2+ Hl3TJ3= (-Hx)( -')'c) + (-cDzH')'uy) + (cDy)(')'uz) = ')'c(Hx- uyDz + uzDy) = ')'C[H- (u x D)]x, GlvTJv = GlOTJo+ Gl21J2 + Gl3TJ3 = (-BxH -')'c) + (- ~z ) (')'Uy) + (~y) (')'uz) = 1:(C2Bx - uyEz + uzEy) = 1: [c2B - (u x E)] , so HlvTJv = ..!..GlV1Jv~ c c x JL ')'C[H - (u x D)]x = ..!..1:[c2B- (u x E)] ~ H - (u x D) = ..!.. [ B - 12(u x E) ] 0 JLc x JL c Use Eq. [4] as an expression for H, plug this into Eq. [3], and solve for D: D+ :2UX {(UXD)+~ [B- c~(UXE)]} =€[E+(uxB)]; 1 1 1 D + 2" [(u. D)u - u2D] = € [E + (u x B)] - ~(u x B) + -du x (u x E)].c ~ ~ [4]
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