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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 fourmomentum is Dpt" : (O,21mc). The same a,rgument gives the same value
for t he f inal t otal f ourmomentum, 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 fourvectors, 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 fourmomentum (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 zerocomponent 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 fourmomentum is Dpt" : (O,21mc). The same a,rgument gives the same value
for t he f inal t otal f ourmomentum, 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 fourvectors, 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 fourmomentum (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 zerocomponent 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 velocityadditionformula 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)1@)Qz (r)r(v)[u1 and the last gives t@')c: t V)[t(u)c  B g)1@)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 elocityaddition f ormula. S imilarly, t he s econdo f E qs. x
1
gives7 (u')u'r:1@)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 0e 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 01 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 0e 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 Bimu ")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: ( lI)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 t1  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 030k g. 1 .6 0x 1 0 1 3J (b) 1? :i i i f fi : 5.3 3xL 02 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 ,"rTn,: 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 ourvector 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 hreemomentum
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 ourvector i s f orwardt imelikei f a nd o nly i f l ql < q n. T he f ourmomentum
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 imelike.
(b) I f p a nd q a re f orward t imelike, 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 ourvector i s f orwardt imelikei f a nd o nly i f l ql < q n. T he f ourmomentum
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 imelike.
(b) I f p a nd q a re f orward t imelike, 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 ourmomentum p t"t i s f orward
timelike 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 threemomentums 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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This note was uploaded on 12/10/2011 for the course PHYS 4102 taught by Professor Fertig during the Spring '11 term at University of Georgia Athens.
 Spring '11
 FERTIG
 Special Relativity

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