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Final Exam Review Question Solutions

# Final Exam Review Question Solutions - CHAPTER 15 SPECIAL...

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Unformatted text preview: CHAPTER 15. SPECIAL RELATIVITY 247 15.5L * * ( a) T he t wo i ncoming p articlqs h ave t he s ame m &ss, T Tts: 1 116:r n. S ince t heir velocities are equal and opposite, vo - -v6 : v, say, they have the same value of 7, that is, T o: ^ ru:'y, s ay. I t f ollows t 'hat p o: ( 'ymv,.yrnc)a nd p 6 - ( -"1*u,1'rnc).Therefore t he total initial four-momentum is Dpt" : (O,21mc). The same a,rgument gives the same value for t he f inal t otal f our-momentum, a nd w e've s hown t hat ! pi" : D pn., i n f rame,S. (b) Since the two sides of the last equation are four-vectors, the truth of the equation in one inertial frame automatically assuresits truth in all such frames. L5.52 ** (a) Let Q be the change in the total four-momentum (between any two timm of interest), Q : P nn- P ir,. W e a re t old t hat t he s pace p art Q i zzero i n a ll f rames, a nd w e have to prove that the same is true of Q4 To do this, consider an arbitrary frame 5 and a s econd f rame 5 ', o btained f rom 5 b y a s tandard b oost. T hen Q 'r: . y(\$ - p Qa), a nd, since Q " : Q r : 0, i t f ollows t hat Q a : 0. S ince t he f rame . S w as a rbitrary, t his p rovest hat Qq:0 i n a ll f rames. (b) Using the same notation as in part (a), we a^re old that one component of Q is zero t in all frames. By the zero-component theorem, this guarantees that all components a.rezero in all frames, and we're home. 15.53 ** The quantity po.pois invariant, so can be evaluatedin any convenientlychosen : frame. I n t he r est f rame o f a , p o : ( 0, m ,c) a md o : ( po, F a61"),o p o'.pa - rnoEb. S till p s in t he s amef rame, t he s peedo f b i s u r"t, s o i ts e nergyi s E o : 7 (a*1)m6c2, nd p o.pa : a -mom5c21(r*r). Finally, working in the sameway, but in the rest frame of b, we find that Po'Pb: - rnoEo' 15.54 r.r-* The velocity of frame 5' relative to 5 is V : € along the r anis. Thus, applying CHAPTER 15. SPECIAL RELATIVITY 247 15.5L * * ( a) T he t wo i ncoming p articlqs h ave t he s ame m &ss, T Tts: 1 116:r n. S ince t heir velocities are equal and opposite, vo - -v6 : v, say, they have the same value of 7, that is, T o: ^ ru:'y, s ay. I t f ollows t 'hat p o: ( 'ymv,.yrnc)a nd p 6 - ( -"1*u,1'rnc).Therefore t he total initial four-momentum is Dpt" : (O,21mc). The same a,rgument gives the same value for t he f inal t otal f our-momentum, a nd w e've s hown t hat ! pi" : D pn., i n f rame,S. (b) Since the two sides of the last equation are four-vectors, the truth of the equation in one inertial frame automatically assuresits truth in all such frames. L5.52 ** (a) Let Q be the change in the total four-momentum (between any two timm of interest), Q : P nn- P ir,. W e a re t old t hat t he s pace p art Q i zzero i n a ll f rames, a nd w e have to prove that the same is true of Q4 To do this, consider an arbitrary frame 5 and a s econd f rame 5 ', o btained f rom 5 b y a s tandard b oost. T hen Q 'r: . y(\$ - p Qa), a nd, since Q " : Q r : 0, i t f ollows t hat Q a : 0. S ince t he f rame . S w as a rbitrary, t his p rovest hat Qq:0 i n a ll f rames. (b) Using the same notation as in part (a), we a^re old that one component of Q is zero t in all frames. By the zero-component theorem, this guarantees that all components a.rezero in all frames, and we're home. 15.53 ** The quantity po.pois invariant, so can be evaluatedin any convenientlychosen : frame. I n t he r est f rame o f a , p o : ( 0, m ,c) a md o : ( po, F a61"),o p o'.pa - rnoEb. S till p s in t he s amef rame, t he s peedo f b i s u r"t, s o i ts e nergyi s E o : 7 (a*1)m6c2, nd p o.pa : a -mom5c21(r*r). Finally, working in the sameway, but in the rest frame of b, we find that Po'Pb: - rnoEo' 15.54 r.r-* The velocity of frame 5' relative to 5 is V : € along the r anis. Thus, applying the relativistic velocity-additionformula to the initial velocity of ball a, we find ,,:f f i --#: o and r L- - f f i: #m (where0 : €lc) and similarly with the other three velocities.The initial and final valuesof D*r' are shownin last column of the table below (whereI haveomitted the z components, h a all of which are zero). While the r components re equal, the y components ave opposite is not conserved. signsand ate not equal. Thus D*4 First particle (v!) Before: After: or -' 7l( 1 - p ') p Second article (vi) ml, + mv't -.r u ,s -ts "YG 1t+ B z1', + 92) ( -2e n \ (0,1ftu,t) \(r+BT'Fol) G + p 2)'. ,t(t- p 4) ( - z*e - z*n|'\ + \(1 PT'Gm) B (v)[email protected])Qz (r)r(v)[u1 and the last gives [email protected]')c: t V)[t(u)c - B g)[email protected])ur]: t @)t(V)c[r- u lVf c 2l. Dividing E q.(xi) b y E q.(xii), w e f ind ( x:: ,, _ V ,-r : ,,. L - u 1V/c2 which i s t he f irst c omponent o f t he v elocity-addition f ormula. S imilarly, t he s econdo f E qs. x 1 gives7 (u')u'r:[email protected])ur & nd, d ividing t his b y E q.(xii), w e f ind u'r: . ==!2-' - u lVf c 2) jV)Q whichi s t he s econd omponent.T he t hird w orksi n t he s amew ay,a nd w e're h ome. c 15. 5 6* ( a ) S inc eM i c2 * Ti :M r c 2 + 4, wes eet h at L,M : - S e Ylc2 : L , Mc2: - L T - -beV . T h u s - 5.4 x 1 0-e u . (b) S ince t he i nitial m ass o f t wo H 2 a nd o ne 0 2 m oleculesi s 3 6 u , t he f ractional c hang. in m assi s A ,M /M : - ( 5. 4 x 1 0- e)/ 36 : - 1. 5 x 1 0-1 0. (c) W hatever t he i nitial m ass, t he f ractional c hangew ill b e t he s ame, s o, w ith 1 0 g ranr-. initially, t he c hange w ill b e A ,M : - 1.5 x 1 0-e g ram. P retty s mall! 15,57 * W ith t he i nitial a tom a t r est, c onservationo f e nergy i mplies t hat M ic2 : M rc2l Tr. so 4 - ( M r- M t ) "': ( m e.r _ l r r Bi-mu ")c 2: ( 0. 0 087 )c2 : 8. 1 M eV : 1 . 3 x 1 0 -1 2J . t 15.58* : T+mc2 : 2 mc2 nd7 : 2 . T hereforc : ( a) I f T : T nc2,then.E a B \ /L - L l 4:0.866. (b) I f E : n mc2, hen. y: n a ndp : t | - t lrz - - | - I lnz 15.59* T he c orrect elativistic E i s 7 : E - mc2: ( l-I)mc2. T he q uestions: C ould r K i L o1mu2? t hesew eret he s ame,t hen i t w ould h avet o b e t hat : this b e t he s amea s | muuru2 If h - D "' : | 'yu' o t-1 - t / Q : i u' / "') w hich i s c ertainlyf alse. p : p clE : ( 4 M eV)/(5 M eV) : 0 .8, T herefore : 0 .8c. u 15. 62 * E :T*m c2:13 p M e V . T he re f o re c : r/ E = mP : 5 M e V o rp :5M eY f c, p and : p cfE : 5 l13: 0 .38, ou : 0 .38c. s 1 5.6 3 * ( ") 1 MeV MeV 1.6 0x 1 0 -13 J c2 (3.00x 1 08m /s)2 : 1 .78x 1 0-30k g. 1 .6 0x 1 0- 1 3J (b) 1? :i i i f fi : 5.3 3xL 0-2 2 kg. m/s .( T h e fi n a l 3 l r e r e co n t a in sa rounding e rror a nd s hould r eally b e a 4 .) L5.64 * T he q uantity u .p i s i nvariant, s o h as t he s ame v alue i n a ll i nertial f rames. I n t he fr a me S ' ,u': ( 0,c) a n d d : ( p ',E '1" ). T h e ref ore u ' p: 't ! , ' ' p ' : - E ' a n d E t : - u ' p, 8 's claimed. 1 5 .6 5* * T : m c 2( 1 - 1 ) : r n c 2 f (l- p 2 )-r/ 2 - r l : r n c2 t + | g ' + & p n * Bu + ' .) - t ] + l( : r nc2iB, + & po * 6u + . .. ] + l is (a) T he f irst t erm r s | mc2p2: | mu2: ? n.. T he d ifference T ,"r-Tn,: m c'[l7n + " ] ' 4 "'= 4 ' = u # : l ong f ractionald iff...n"" i . (b) T he * P2,and t his i s l ess t han T Toas 4t 4"r + P2 a s0 < 0 .1 2c. - 0 Elr) 15.66 * * T hat p i s a f our-vector r equires,a mong o ther t hings, t hat p \:'Y(h If w e a dd a n a rbitrary c onstant t o E , t his a dds a c onstant t o p \, w hich i s c ertainly n ot permissible. ( For i nstance, i f t he b ody i s s tationary i n t he f rame 3 ', P \ m ust c ertainly b e zero. ) 25O cHAprER tb. spEcrAL RELATrvrry 15.67 * W ith u : 0 .8c,' ,t : 5 13. S incet he i nitial v elocitiesa re e qual a nd o pposite,t be sameis true of the momenta. Thereforethe total momentumis zero. and the final bodv is : at r est. T hus M : E nn/cz E in/c2: 2.ym: 3 .33rn. 15.68 * I n t he C M f rame p l" : - pl, s o t he t wo i nitial m omenta a re e qual i n m agnitudq p^ : p f : p i', s ay. ( I'll u se t he i talic p t o d enote t he m agnitude o f t he t hree-momentum here.) By conservation of momentum, the final total momentum is also zero, so the same argument applies to the final momenta and d" : d" : pfin, say. Now the initial total energr is Ei " -rM + (pi"")' I ( m6c2)2 with a similar expression or .Efr". Notice that .Ei" is a monotonically increasingfunction f o of pin (and likewiseEn"). Thus conservation f energy, Bfr" - Ein, requiresthat pfi" -- f t and h encet hat p !' : * pf. H ere t he p lus s ign c orrespondso t he i nitial s tate, b efore the collision. After the collision the minus sign must apply: In the CM frame the three i momentumof particle a (and likewiseb) simply reversestself. q 15.69 * ( ") A f our-vector i s f orwardt ime-likei f a nd o nly i f l ql < q n. T he f our-momentum p 6 of a m assive articlei s d efined p : T uLL:lm(v,c), a nd, s incer n > 0 , l tl ( c a nd h ence p lpl < p n. T herefore i s f orward t ime-like. (b) I f p a nd q a re f orward t ime-like, l pl < p a a nd l ql < q n. I t f ollowst hat l p+ql S p + q i s f orwardt imelike. 1 p1-,+q r: ( p+ + q 1 5.69 * ( ") A f our-vector i s f orwardt ime-likei f a nd o nly i f l ql < q n. T he f our-momentum p 6 of a m assive articlei s d efined p : T uLL:lm(v,c), a nd, s incer n > 0 , l tl ( c a nd h ence p lpl < p n. T herefore i s f orward t ime-like. (b) I f p a nd q a re f orward t ime-like, l pl < p a a nd l ql < q n. I t f ollowst hat l p+ql S lpl + l ql 1 p1-,+q r: ( p+ i la.Therefore,p + q i s f orwardt imelike. 15.70 * ( a) F lom P roblem 1 5.69 i t f ollows t hat t he t otal f our-momentum p t"t i s f orward time-like a nd s o l pl"'l < p \"t. B y r otating o ur a xes i f n ecessaryw e c an p ut p to'along t he positive r a xis, s o t hat p tot: ( p|"t,0,0,p1ot) w ith p !"t < p |t. N ow c onsider a s tandard b oost to a frame S' in which p'r.'o':ilp?t - Z pPt). p If w e c hoose - p ftlp\"t ( which i s l esst han 1 b ecause 'ft < pl"t), t hen i n t he f rame5 'the B i total three-momentums zero. (b) It is alreadyclear from the abovethat in the original frame 5 the velocity of the CM frame has to be given by I : p'"' lpft : Dpcl D E. t 15.71.* (a) The total energyneeded o produceany givenfinal set of particlesmust satisfy : E > Dlrlfinc2. The unique featureof the CM frame is that with p1o1 0, the final particles can all be at rest and the inequality can actually be an equality. Thus the thresholdenergy I in the CM frame is ! rn6,,c2. f the particle d is much heavierthan all the others, this gives for E"^ N n'Ldc2 the threshold. ...
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