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Introduction to Electrodynamics - ch12

# Introduction to Electrodynamics - ch12 - Chapter 12...

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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. Assumingmass is conserved, (mA + mE) = (mc + mD), it followsthat mAUA+ mBuB = mcuc + mDuD, so momentum is conservedin 5. (b) 1 2 1 2 - 1 2 1 2 2mAUA + 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 d 0 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. 12.3says: /3 = f~%~;2' or: /32= /3f + 2/31/32+ /3~ = 1 + 2/31/32+ /3f/3~ - (1 + /3f/3~- /3f - /3~) = 1 - (1 - /3f)(l - /3~) = 1 - A (1 + 2/31/32+ /3f/3~) (1 + 2/31/32+ /3f/3~) (1 + 2/31/32+ /3f/3~) (1 + /31/32)2 ' 219

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~ 220 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY where ~ ==(1 - ,8f)(1 - ,8~)/(1+ ,81,82)2 is 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. Velocityofgetawaycar: ~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 Velocityof 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 clockis 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 0 dO - (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, eventhough v < 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. Thus 1-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 1 d v = 1- e2; v d + e2 = 1; v = (rld)2 + (1/e)2' Te - (2 x 10-6)(3 x 108) - ~-~.
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