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= 0.1Problem 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 = uFcosOevil - u2/C2;

F2 u2 F2 COS20

[1 - (u2/e2) COS2O

JK KJ.L= - = F2. qed

J.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 2Dotmu:(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 12.41One 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 (exceptthat 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 thatany 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 perpendicularto the y direction the charges trade places, so it is perfectly appropriate that the y component of E shouldreverse 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 (~

-(}

231

q2 (l-v2je2)2vsinOAS = 1- 2 2 2 (J.

1611"£0 R4(1- ~ sin 0)3

Problem 12.44

(a) Fields of A at B: E = 4;fO ~Yj B = O. So force on qB is IF = 4:£0 q~;B Y.

*-x~-L y (b) (i) From Eq. 12.68: IF = ~Ty.1 (Note: here the particle is at rest in S.)

W 2 2d(..

) Fr E 12 92. h 0 - 90°. E- - 1 qA(1 - v je ) 1 A - 'Y qA A

- 11 am q. . , WIt - . - 4-

( 2 j 2)3/2 .ny - _4 .n yX 11"£01 - v e u- 11"£0u-

V qA (this also follows from Eq. 12.108).

B i- 0, but since VB = 0 in S, there is no magnetic force anyway,and IF = ~TY I (as before).Problem 12.45

Here 0 = 90°, ~ = y, lb = Z,Iz-= r, so (using e2 = IjJ-lo£o):

E=-~ 'Y A

411"£0r2 y,B = -~ v 'Y A

411"£0 C2 r2 z,

1

where 'Y= VI - v2je2.

Note that (E2 - B2e2) = (~)2'Y2(1-~) = (~)2 is invariant, because it doesn't depend on v. We canuse this as a check.

System A: VA = v, so E = - q 1 A

411"£0 r2 y,

1

q . ~ 1 Z, where 1 = Vl- v2jc2B --- 22- 411"£0c r

2- 2 2- 2F = q[E + (-vx) X B] = !Ll [y - ~(x X z)] = !Ll (1+ ~)y.~~~ ~ ~~~ ~

System B:

~

VB= v+v - 2vl+v2je2 - (1+V2jc2)

'YB = - 14v2/c2 = (1 + v2jc2) = (1 + v2jc2) = -2 V2.vI (t+V2/C2)2 V1-2~+V4 (l-v2je2) 'Y (1+

C2),Vb'YB=2v12.c C4

:.E=-~~12 (I+v2 ) A.B- q 2v12A411"£0r2 e2 y, - -- 4

--z.11"£0 e2 r2

[2 2 2 4 4 2 2 I 2

]Check: E2 - B2C2 = (-L- ) ,:y4(1 + v + v - v ) = (-L- ) 14 - = (-L- ) . ./41rfor I ~ C4 ~ 41rfor ¥ 41TfOr

q2 12 v2F = qE = --4 "2(1 + "2)Y. (+q at rest => no magnetic force). [Check: Eq. 12.68 => FA = J-FB. ./]11"£0r c -y

SystemC: ve=O. E=-~~Yj B=O; F=qE=_L~y.411"£0r2 411"£0r2

[The relative velocity of Band C is 2vj(1 + v2jc2), and the corresponding 'Yis 12(1 + v2je2).

=> Fe = -y2(t+~2/c2)FB'./]

So Eq. 12.68

232 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

Summary:

} ~ ~ ";'-~'I"

Problem 12.46

(a) From Eq. 12.108:

-- -- -- -- 2 V vE. B = ExBx + EyBy + EzBz = ExBx + "/ (Ey - vBz)(By + zEJ + ,,/(Ez+ vBy)(Bz - ZEy)c c

v ~ v2 V ~ V2= ExBx + ,,/2{EyBy + ci/'yEz - vP(Bz - C2EzBz + EzBz - ci/'yEz + vP(Bz - c2EyBy}

= ExBx + ,,/2 [EyBy (1 - ~:) + EzBz (1 - ~:)] = ExBx + EyBy + EzBz = E .B. qed

(b) P;2 - C2jp = [E; + ,,/2(Ey - VBz)2 + ,,/2(Ez + VBy)2] - c2 [B; + ,,/2(By + ; Ez)2 + ,,/2(Bz - ; Ey)2]

= E2 + ",2 (E2 - 2E £B + v2 B2 + E2 + 2E £B + v2 B2 - C2B2 - c22 vk Ex / Y "')fv, z z z 7v, y y y ;li Uy z

V2 ~ V2-c2-E2-C2B2+C22 E -c2-E2 ) -C2B2

c4Z z zy c4Y x

2 2 2 2

[

2 (V2

) 2 (V2

) 2 2 (V2

) 2 2 (v2

)]=E -cB +"/ E 1-- +E 1-- -cB 1-- -cB I--

x x Y c2 z C2 Y c2 z C2

= (E2 + E2 + E2 ) - c2(B2 + B2 + B2 ) = E2 - B2C2 q edx Y z x Y z .

(c) I No.1 For if B = 0 in one system, then (E2 - C2B2) is positive. Since it is invariant, it must be positive inany system. Therefore E =I- 0 in all systems.Problem 12.47

(a) Making the appropriate modifications in Eq. 9.48 (and picking 8 = 0 for convenience),

E(x, y, z, t) = Eo cos(kx - wt) Y, B(x, y, z, t) = Eo cos(kx - wt) z, where k ==~.c c

(b) Using Eq. 12.108 to transform the fields:

Ex = Ez = 0, Ey = ,,/(Ey- vBz) = ,,/Eo[cos(kx- wt) - ~ cos(kx- Wi)]= o:Eo cos(kx - wi),

Bx = By = 0,- v

[

1 v

]

EoBz = ,,/(Bz - ZEy) = ,,/Eo - cos(kx - wt) - z cos(kx - wt) = 0:- cos(kx - wi),c c c c

where0:=="/(1-~)=

1 - vlcl+vlc.

(-),y (-),y2(1 + )y (- 47r;Or2)Y

( - 47rfQor2) "/ Z ( )2V 2A 0- 47rfOr 2"/ Z

(- 47r:r2),(1 + )Y () 2(1 v2) A (- 47r;or2)Y- 47rfOr "/ + 2 Y

233

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 = w1 - vlcl+vlc' This is the I Doppler shift I for light. - 211" 211"- ~ The velocity of the\ - - - .

A = k - ak a

- w- wwave 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 yourun faster and faster. It'll get so faint you won't be able to see it, and so red-shifted even yournight-vision goggles won't help. But it'll still be going 3 x 108mls relative to you. Sorry aboutthat.

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 12.528vP"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. qedProblem 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. Nowthisequation stands for 64 separate equations (JL= 0 ~ 3, v = 0 ~ 3, A = 0 ~ 3, and 4 x 4 x 4 = 64). But manyof 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 andP,,'\ = - P'\", so this is trivial: 0 = O. To get anything significant, then, JL,v, A must all be different. Theycould 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 883Pl2 + 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 minusit).

~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 ycomponent; 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 to8vGp.v = O. qedProblem 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 &m8t 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. qedProblem 12.56

According to Prob. 12.53, 8ff;"~= 0 is equivalent to Eq. 12.129. Using Eq. 12.132, we find (in the notationof 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

2:~6 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

Problem 12.57y

Step 1: rotate from xy to XY, using Eq. 1.29:

x = cos ifJx + sin ifJy

Y = - sin ifJx + cosifJy<{

~y

Step 2: Lorentz-transform from XY to XY, usingEq. 12.18:

X = 1'(X - vt) = 1'[cosifJx+ sinifJy - .Bet]

Y = Y = - sin ifJ x + cos ifJYZ=Z=z

et = 1'(et - .BX) = 1'[et- .B(cosifJx+ sin ifJy)]

Step 3: Rotate from XY to xii, using Eq. 1.29 with negative ifJ:

x

(p x

x = cosifJX - sin ifJY = 1'cosifJ[cosifJx+ sinifJy - .Bet]- sinifJ[- sin4>x + cos4>Y]

= C'YCOS2ifJ+ sin2 ifJ)x + C'Y..:... 1) sin ifJcos ifJy - ,.B cos ifJ (et)

ii = sin ifJX + cos ifJY = l' sin ifJ(cos ifJx + sin ifJy - .Bet) + cos ifJ(- sin ifJe + cos ifJ y)

= C'Y- 1) sinifJcosifJx + C'Ysin2ifJ+ COS24»y - ,.Bsin 4>(ct)

(

C

) (

l' -1'.B cos ifJ -1'.B sin ifJ

In matrix form: ~ = -1'.B c~s ifJ C'YCOS2ifJ+ sin2 ifJ) C'Y- 1) sin 4>cos 4>

~ -1'.BsmifJ C'Y-1)sin4>cos4> C'Ysin24>+cos2ifJ)zOO 0 ~) m.

Problem 12.58

In center-of-momentum system, threshold occurs when incident ener-gy is just sufficient to cover the rest energy of the resulting particles,with none "wasted" as kinetic energy. Thus, in lab system, we wantthe outgoing K and E to have the same velocity, at threshold:

1f P0-- --0

00KE

before (CM)

after (CM)

0-- 01f P

Before

(X)---+-KE

After

Initial momentum: P7T;initial energy of 1f: E2 - p2e2 = m2e4 :::} E; = m~c4+ p~c2.Total initial energy: mpe2 + vm;c4 + p;e2. These are also the final energy and momentum: E2 - p2e2 =

(mK + m~:Ye4.

(mpe2 + vm;c4 + p;c2) 2 - p;C2= (mK + m!Yc4

2m e2

m;!, + -;f-vm;e2 + p; e + m;!' + rJ!c2- rJ!c2 = (mK + mE)2!,

2:p vm;e2 + p; = (mK + mE)2- m; -m;

237

(2 2 2

)4m; ( )4 2(

2 2)( )

2 4 4 2 2 2m".c +p". ~ = mK +mI; - mp+m". mK +mI; +mp+m". + mpm".c

4m2-TP; = (mK + mI;)4 - 2(m; + m;)(mK + mI;)2 + (m; - m;)2c

P". = 2~ v(mK + mI;)4 - 2(m~ + m~)(mK + mI;)2 + (m~ - m~)2p

= (2m~c2)cv(mKc2 + mI;c2)4 - 2[(mpc2)2+ (m".c2)2](mKc2 + mI;c2)2 + [(mpc2)2 - (m".c2)2]2

= 2C(~OO)V(1700)4 - 2[(900)2 + (150)2](1700)2+ [(900)2 - (150)2]2

= ~V(8.35 X 1012)- (4.81 X 1012)+ (0.62 x 1012)= ~(2.04 X 106)= 11133MeVIc.!Problem 12.59

In CM:P P

0-- --0

Before

rP

u;;;~u

sp)After

Lx (p=magnitude of 3-momentumin CM, 4> = CM scattering angle)

Outgoing4-momenta: rP = (~,pcos4>,psin4>,O)jsP = (~,-pcos4>,-psin4>,O)o

In Lab: 0---+- 0

TP

~-" S

Problem: calculate 0, in terms of p, 4>.

Before

Lorentz transformation: Tx = -y(rx - {3rO)j Ty = rvj Bx= -y(sx - {3s0)jBy= By.

Now E = -ymc2jP = --ymv (v here is to the left); E2 - p2c2 = m2c4, so {3= -1jfo

:. Tx= -y(pcos 4> + 1jf~) = -yp(1+ COg4»j Tv =psin 4>jBx= -yp(1- COg 4»; By = -psin 4>.

r .S -y2p2(1 - COS2 4» - p2 sin2 4>cosO = - =TB Vh2p2(1 + COg4»2 + p2 sin2 4>][-y2p2(1 - cos 4»2 + p2 sin2 4>]

- (-y2 - I) sin2 4>

- Vh2(1 + cos4»2 + sin24>]h2(1- cos4»2 + sin2 4>]

- (-y2- I) - (-y2- I)

- [-y2e;i~o:<p)2+ I] [-y2e~i~o:<p)2+ I] - V (-y2cot2~ + I) (-y2tan2~ + I)

238 CHAPTER 12,ELECTRODYNAMICS AND RELATIVITY

wcos () =

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/

fL1

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 2u =---u ;m mc2

dx c ;

dt = VI + (~;~)

u2(1 + 2Kx ) = 2Kx,mc2 m

Jmc2

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