Instructor Solution Manual Analytical Mechanics for Relativity and Quantum Mechanics, Second Edition

SOLUTION MANUAL

1 CHAPTER 1 EXERCISE SOLUTIONS d ! " r rO D rP rO C r rPO where the dot denotes differendt tiation with respect to time. Divide eqn (A.48), eqn (A.49), and eqn (A.50) by dt to obtain

Exercise 1.1 (a) By the product rule, v D

rPO D !O !P C "O sin ! "P P !O D !Or !P C "O cos ! "P # $ P "O D ! rO sin ! C !O cos ! "P

and hence

# $ v D rP D rP rO C r !O !P C "O sin ! "P

(1) (2) (3) (4)

(b) Again using the product rule # $ # $ P P a D vP D rR rO C rP rPO C rP !O !P C "O sin ! "P Cr !O !P C !O !R C "O sin ! "P C "O cos ! !P "P C "O sin ! "R (5) Substituting from eqn (1), eqn (2), eqn (3) and collecting terms then gives $ # $ # a D rO rR ! r !P 2 ! r sin2 ! "P 2 C !O r !R C 2rP !P ! r sin ! cos ! "P 2 # $ C "O r sin ! "R C 2r cos ! !P "P C 2rP sin ! "P (6)

Exercise 1.2 In Cartesian coordinates, r D x1 eO 1 Cx2 eO 2 Cx3 eO 3 and C D C1 eO 1 CC2 eO 2 CC3 eO 3 .

Hence (7) C " r D C1 x1 C C2 x2 C C3 x3 By assumption, the Ci components are not functions of r. Applying the gradient operator 3

rD to eqn (7) thus gives

X @ @ @ @ @ D eO i D eO 1 C eO 2 C eO 3 @r @xi @x1 @x2 @x3

(8)

r .C " r/ D eO 1 C1 C eO 2 C2 C eO 3 C3 D C

(9)

iD1

The demonstration that r v .C " v/ D C is the same, but with dots on the xi components and the assumption that C does not depend on v.

By the product rule, @ @ @ ! 2" @ v D .v " v/ D .v " v/ C .v " v/ (10) @v @v @v @v where in each term v denotes the factor to be treated as a constant in taking the derivative.

Applying eqn (A.78) with C replaced by v then gives @ ! 2" v D v C v D 2v (11) @v 1

2

CHAPTER 1 EXERCISE SOLUTIONS

Exercise 1.3 (a) The equation of a circular helix is r D a cosˇ eO 1 C a sinˇ eO 2 C cˇ eO 3

(12)

Differentiating and using the definitions from Section A.12 gives dr D !a sin ˇ eO 1 C a cos ˇ eO 2 C c eO 3 dˇ % % p % dr % ˚ 2 2 & ds % D a sin ˇ C a2 cos2 ˇ C c 2 1=2 D a2 C c 2 D% % % dˇ dˇ

tO D

˚ & d r=dˇ 1 dr D Dp !a sin ˇ eO 1 C a cos ˇ eO 2 C c eO 3 ds ds=dˇ a2 C c 2 ˚ & d tO 1 Dp !a cos ˇ eO 1 ! a sin ˇ eO 2 dˇ a2 C c 2

(13)

(14)

(15)

(16)

˚ & dt d t=dˇ 1 D D 2 !a cos ˇ eO 1 ! a sin ˇ eO 2 (17) 2 ds ds=dˇ a Cc % % %dt% a 1 dt % (18) #D% % ds % D a2 C c 2 hence nO D # ds D ! cos ˇ eO 1 ! sin ˇ eO 2 ˇ ˇ ˇ eO 1 eO 2 eO 3 ˇˇ ˇ ˚ & 1 ˇ !a sin ˇ a cos ˇ c ˇ D p 1 bO D tO # nO D p c sin ˇ eO 1 ! c cos ˇ eO 2 C aeO 3 ˇ ˇ a2 C c 2 ˇ ! cos ˇ ! sin ˇ 0 ˇ a2 C c 2 (19) ˚ & d b=dˇ c db !c D D 2 nO (20) cos ˇ eO 1 C sin ˇ eO 2 D 2 ds ds=dˇ a C c2 a C c2

Hence

$D

c a2 C c 2

(21)

(b) Since nO in eqn (18) has no eO 3 component, it is parallel to the x-y plane.

O and bO form an orthogonal, right-handed triad of unit vectors, Exercise 1.4 Since tO, n, n o d tO ! # tO D $ tO C # bO # tO D # bO # tO D # nO D ds

n o d nO ! # nO D $ tO C # bO # nO D $ tO # nO C # bO # nO D $ bO ! # tO D ds n o d bO ! # bO D $ tO C # bO # bO D $ tO # bO D !$ nO D ds

where eqn (A.84), eqn (A.86), and eqn (A.85) have been used.

(22)

(23)

(24)

CHAPTER 1 EXERCISE SOLUTIONS

3

Exercise 1.5 Use a fixed coordinate system with origin at the location of the center of mass of the rocket at the instant of the explosion. (a) There are no external forces, so d P=dt D 0 and P=M D V is constant during the explosion and after. We are given V D .30 km/s/ eO 1 so at t D 10 s the center of mass of the system is located at (25) R .10 s/ D .300 km/ eO 1 The eqn (1.32) holds at all times. Solving it for r3 gives (assuming that the "location" of a piece is the location of its center of mass) 1 fM R .10 s/ ! m1 r1 .10 s/ ! m2 r2 .10 s/g m3 D 300 km eO 1 ! 100 km eO 2 ! 50 km eO 3

r3 .10 s/ D

(26)

(b) Each piece after the explosion is itself a collection with no external forces on it. Its velocity will be constant. To determine the constant velocity v3 write 1 .r3 .10 s/ ! D/ 10 s " 1 !

D 300 km eO 1 ! 100 km eO 2 ! 50 km eO 3 ! D 10 s

v3 D

(27)

where D is a vector of unknown direction but with magnitude less than 10 meters, the maximum dimension of the rocket. It is the location of the third piece at the instant of the explosion. Thus, at one hour the missing third piece will be at r3 .3600 s/ D r3 .10 s/ C .3590 s/ v3 ! " D 360 300 km eO 1 ! 100 km eO 2 ! 50 km eO 3 ! 359D

D 1:08 # 105 km eO 1 ! 3:6 # 104 km eO 2 ! 1:8 # 104 km eO 3 ! 359D

(28)

where 359D $ 3:59 km is the radius of the error sphere. (b) We have used only the law of momentum and the definition of R. The J, S, and L values have no effect. So the answers would not change at all.

Exercise 1.6 (a) From the axiom of Angular Momentum eqn (1.20) and the definitions in Section 1.5 we have 3 3 X X dJ D # .ext/ D # .ext/ D rn # f.ext/ (29) n n dt nD1 nD1 Using cylindrical polar coordinates with z D 0 assumed, r1 D 0 and, for n D 2; 3, Thus

r2 D b $O

r3 D 2b $O

# $ O O O D m g e D m g cos " $ ! sin " " f.ext/ n 1 n n

# $ # $ # .ext/ D b $O # m2 g cos " $O ! sin " "O C 2b $O # m3 g cos " $O ! sin " "O D !Oe3 .m2 C 2m3 / bg sin "

(30) (31)

(32)

4

CHAPTER 1 EXERCISE SOLUTIONS

JD

3 X

nD1

rn # mn vn D m2 b $O # b "P "O C 2m3 b $O # 2b "P "O

D b 2 "P .m2 C 4m3 / eO 3 where

v2 D b "P "O

v1 D 0

(33) v3 D b "P "O

(34)

were used. The first equality in eqn (29) then gives !g .m2 C 2m3 / sin " and "R D b .m2 C 4m3 / !

P2 1 d " Multiply both sides of the second of eqn (35) by "P and use "P "R D to write 2 dt b 2 "R .m2 C 4m3 / eO 3 D bg .m2 C 2m3 / sin " eO 3

d "P 2 dt

1 2

!

D

g .m2 C 2m3 / d cos " b .m2 C 4m3 / dt

2g .m2 C 2m3 / "P 2 D .cos " ! cos "max / b .m2 C 4m3 /

(35)

(36)

(37)

where the integration constant was chosen with the stick at rest at its maximum " value. (b) Differentiating rP 3 D v3 D 2b "P "O (38) gives P rR 3 D 2b "R "O C 2b "P "O

D 2b "R "O ! 2b "P 2 $O 4g .m2 C 2m3 / 2g .m2 C 2m3 / D! .cos " ! cos "max / $O ! sin " "O .m2 C 4m3 / .m2 C 4m3 /

(39)

C f.int/ and f3 D m3 rR 3 we have With f3 D f.ext/ 3 3 f.int/ D m3 rR 3 ! f.ext/ 3 3

# $ D m3 rR 3 ! m3 g cos " $O ! sin " "O

and hence f.int/ D 3

'

( !4m3 g .m2 C 2m3 / .cos " ! cos "max / ! m3 g cos " $O .m2 C 4m3 / ' ( !2m3 g .m2 C 2m3 / sin " C m3 g sin " "O C .m2 C 4m3 /

(40)

(41)

(c) We see from (b) that f.int/ is not central, as would be required by the strong form of the third 3 law. The $O component is central since it is along the line to the other masses. But the tangential

CHAPTER 1 EXERCISE SOLUTIONS

5

"O component is not central. Simplifying to the case with m2 D m3 D m, the "O component is !mg sin "=5 which is nonzero for " ¤ 0. So the strong form of the third law is violated here.

D f31 C f32 , then at least one of the f3n must have a component not If it were true that f.int/ 3 along the line between the masses, thus violating the assumption used to prove the strong form of the third law. dJ D # .ext/ to solve the problem was correct. The law of angular momentum But our use of dt is more general than the model with internal forces coming from other masses upon which the strong form argument is based.

Exercise 1.7 The line of contact of the cylinder with the plane is instantaneously at rest at all times. So the friction and normal forces do no work on the cylinder. We may use energy conservation, ignoring these forces. (a) The gravitational potential of the cylinder with center of mass at R D X eO 1 CY eO 2 CZ eO 3 is U D MgY . Initially the kinetic energy is zero, so E.0/ D 0 C MgY.0/. After the cylinder has fallen a distance h, E D T C Mg .Y .0/ ! h/. Equating E and E.0/, MgY.0/ D E.0/ D E D T C Mg .Y .0/ ! h/

and so

T D Mgh

(42)

1 1P From Section 1.14, T D To C TI where To D M V 2 and TI D mn #Pn2 . To get #Pn 2 2 n we note that an observer standing on the center of mass is moving with velocity V whereas a mass at the point of contact is at rest. Hence he sees that mass as moving with velocity !V. That center of mass observer also sees all masses on the circumference of the rigid cylinder as having that same speed relative to him (even though not the same relative velocity directions).

So #Pn D V for all n and hence !

1 X 1 mn V 2 D M V 2 and T D MV 2 (43) TI D 2 n 2 Equating eqn (42) and eqn (43) and cancelling the common M factor then gives p V D gh

(44)

Let dO D cos ˛ eO 1 ! sin ˛ eO 2 be a fixed unit vector pointing down the plane. Then p p ! " P D M V D M gh dO D M gh cos ˛ eO 1 ! sin ˛ eO 2 (45) P P n . The vector $n points radially From eqn (1.44), the spin vector is S D n mn $n # $ outward from the center of the cylinder to a mass mn at its circumference. At that same point, the vector $P n is tangential with magnitude V . Hence, for all masses $n # $P n D !#n #Pn eO 3 D !aV eO 3 . Therefore !

X p mn aV eO 3 D !Ma gh eO 3 (46) SD! n

(b) Use the law of momentum in eqn (1.14) to get F(ext) D

where

1 dP 1 dh p D M g dO p D Mg sin ˛ dO dt 2 2 h dt

p dh D V sin ˛ D gh sin ˛ was used. dt

(47)

6

CHAPTER 1 EXERCISE SOLUTIONS

Similarly, eqn (1.52) is used to find the external torque (ext)

#s

D

1 1 dh dS p D !Ma g eO 3 p D ! Mag sin ˛ eO 3 dt dt 2 2 h

(48)

Exercise 1.8 Expanding eqn (1.54) using the linerity of dot products gives N X

N N 1X 1X T D Tn D mn vn " vn D mn .V C $P n / " .V C $P n / 2 nD1 2 nD1 nD1

n o 1X D mn V2 C $P n " V C V " $P n C k$P n k2 2 nD1 ! ! !) ( N N N X X X 1 2 2 D mn V C 2 mn $P n " V C mn k$P n k 2 nD1 nD1 nD1 N

(49)

Note that the quantities not depending on n have been factored out of the sums. Due to Lemma 1.9.1 the middle term vanishes. Thus T D To C T I where To D

1 MV 2 2

and

TI D

(50) N X

nD1

mn k$P n k2

(51)

Exercise 1.9 If f D f .x; y; z; t / and x, y, and z are functions of t only, then the chain rule of Section D.14 gives @f .x; t; z; t / @f .x; t; z; t / @f .x; t; z; t / @f .x; t; z; t / df .x; t; z; t / D xP C yP C zP C dt @x @y @z @t

(52)

Using eqn (A.75) and eqn (A.73) * ) ! " @f @f @f v " r f D xP eO 1 C yP eO 2 C zP eO 3 " eO 1 C eO 2 C eO 3 @x @y @z @f @f @f df @f D xP C yP C zP D ! @x @y @z dt @t

(53)

Thus

@f @f @f df C D v " rf C D @r @t @t dt Exercise 1.10 For rigid bodies (only) we have from eqn (1.69) v"

N X dT D f.ext/ " vn n dt nD1

(54)

(55)

If there is a potential U .ext/ .r1 ; : : : ; rN ; t / from which the external forces can be derived as D! f.ext/ n

@U .ext/ .r1 ; : : : ; rN ; t / @rn

(56)

CHAPTER 1 EXERCISE SOLUTIONS

7

then N X @U .ext/ .r1 ; : : : ; rN ; t / @U .ext/ .r1 ; : : : ; rN ; t / d U .ext/ .r1 ; : : : ; rN ; t / dT D! C vn " D! dt @rn dt @t nD1 (57) and hence, with E D T C U .ext/ , " ! d T C U .ext/ @U .ext/ .r1 ; : : : ; rN ; t / dE D D (58) dt dt @t

Thus E is conserved whenever the external potential is not explicitly time dependent. (ext)

Exercise 1.11 (a) Given fn

D mn g, try the potential

U .r1 ; r2 ; : : : ; rN ; t / D !

N X

nD1

mn rn " g

(59)

Using eqn (A.76) with C replaced by g and rk in place of r, and noting that the partial derivative is zero when k ¤ n, gives !

@U .r1 ; r2 ; : : : ; rN ; t / @ (ext) D mk .rk " g/ D mk g D fk @rk @rk

(60)

as was to be shown. From the definition of center of mass in eqn (1.32) the potential may also be written as !

N X U .r1 ; r2 ; : : : ; rN ; t / D ! mn rn " g D !M R " g (61) nD1

(b) From eqn (1.13),

(ext)

F

D

From eqn (1.49),

and # .ext/ D s

N X

nD1

(ext)

fn

D

N X

nD1

!

mn g D M g

(62)

D R # F.ext/ D R # M g # .ext/ o

(63)

!

(64)

N X

nD1

$n # f.ext/ D n

N X

nD1

mn $n # g D 0

where eqn (1.38) was used to prove the bracketed factor equal to zero. (c) Since U is not an explicit function of time, E D T C U is conserved. The law of dL dP D F(ext) D M g shows that P is not conserved. Similarly, eqn (1.50) D momentum dt dt dS # .ext/ D # .ext/ D R # M g shows that L is not conserved. Also D 0 shows that S is o s dt d To conserved. From eqn (1.58), D V " F.ext/ D V " M g and so To is not conserved. Since dt PN d TI (ext) Pn D there are no internal forces, fn D fn here and eqn (1.59) gives D nD1 fn " $ dt #P $ N P n " g D 0 where eqn (1.38) was used. Thus, TI is conserved. nD1 mn $

8

CHAPTER 1 EXERCISE SOLUTIONS

.a/ Exercise 1.12 Let the nth mass of the ath sub-collection m.a/ n have location rn . Then the total mass of the ath sub-collection is .a/ N X .a/ m.a/ (65) M D n nD1

its center of mass is .a/

R

.a/

N 1 X .a/ .a/ D .a/ mn rn M nD1

(66)

and its total momentum is P

.a/

D

.a/ N X

.a/ m.a/ n vn

(67)

nD1

d r.a/ n . dt (a) The center of mass of the whole system is thus 0 .a/ 1 .a/ C N C N C X X X X 1 1 X .a/ .a/ 1 .a/ .a/ .a/ A @ RD D m.a/ r D m r M R M aD1 nD1 n n M aD1 nD1 n n M aD1 where v.a/ n D

where

.a/

M D

C N X X

aD1 nD1

m.a/ n D

C X

aD1

The total momentum of the whole system is .a/

PD

C N X X

aD1 nD1

.a/ m.a/ n vn D

C X

aD1

0 @ 0 @

.a/ N X

nD1

.a/ N X

nD1

where differentiating eqn (66) and denoting V.a/ D M

V

.a/ .a/

D

.a/ N X

nD1

1

AD m.a/ n

C X

M .a/

(68)

(69)

aD1

1

.a/ A m.a/ D n vn

C X

P.a/

(70)

aD1

d R.a/ gives dt

.a/ .a/ m.a/ n vn D P

(71)

(b) Following the pattern of Lemma 1.9.1, .a/ N X

nD1

.a/ m.a/ n $n D

.a/ N X

nD1

D

# $ .a/ r.a/ m.a/ n n !R

.a/ N X

nD1

0

.a/ @ m.a/ n rn !

.a/ N X

nD1

1

A R.a/ D M .a/ R.a/ ! M .a/ R.a/ D 0 (72) m.a/ n

CHAPTER 1 EXERCISE SOLUTIONS

9

where eqn (66) was used. Differentiating this result with respect to time then gives .a/ N X

nD1

P .a/ m.a/ n D0 n $

(73)

Similarly, C X

M

.a/

%

.a/

aD1

D

C X

aD1

# $ M .a/ R.a/ ! R D

C X

M

.a/

R

.a/

aD1

C X

!

M

.a/

aD1

!

R D M R ! M R D 0 (74)

and C X

aD1

M .a/ %P .a/ D 0

(75)

where eqn (68) was used.

P .a/ , write To show that J D L C K C C aD1 S .a/

JD

C N X X

aD1 nD1

.a/ .a/ m.a/ n rn # vn .a/

C N $ # $ X X# .a/ .a/ .a/ P P R C % .a/ C $.a/ # m V C % (76) D C $ n n n aD1 nD1

Expanding and factoring the terms, and performing the sums over masses when appropriate,

J D R # MV C CR# CR#

C X

aD1

C X

aD1

0 @

C X

M

.a/

%

.a/

aD1

!

M .a/ %P .a/ C

.a/ N X

nD1

1

!

#VC

C X

aD1

AC P .a/ m.a/ n n $

C X

aD1

0 @

.a/ N X

nD1

% .a/ # M .a/ %P .a/ C

C X

aD1

0

% .a/ # @

.a/ N X

nD1

1

.a/ A m.a/ #V n $n C X

aD1

0 @

1

.a/ N X

nD1

AC P .a/ m.a/ n n $

1

.a/ A m.a/ # $P .a/ n n $n .a/

C N X X

aD1 nD1

.a/ .a/ Pn $.a/ n # mn $

(77)

The eqn (72) through eqn (75) can be used to show all bracketed terms to be zero. Identifying the remaining terms in eqn (1.82) gives the desired result. (c) Denote by a D 1 the sub-collection with four masses, and by a D 2 the sub-collection with six masses. Then M .1/ D 4m and M .2/ D 6m. The centers of masses of the two subcollections are at their centers, which are at the ends of the stick. The center of mass of the

10

CHAPTER 1 EXERCISE SOLUTIONS

whole system is then given by eqn (68) as RD

$ $ $ 1 # .1/ 1 # .1/ .1/ 3 # .2/ M R C M .2/ R.2/ D 4R C 6R.2/ D R.1/ C R ! R.1/ M 10 5

(78)

3 where M D M .1/ C M .2/ D 10m. The center of mass is thus on the line of the stick and is b 5 from the end with the center of the square attached to it.

The plane square and the hexagon need not be assumed coplanar. So long as their centers are attached to the ends of the stick, their orientation has no effect on the center of mass location.

ADDENDUM: From Axiom 1.5.1 we have .a/

dJ D # (ext) dt

where

#

(ext)

D

C N X X

aD1 nD1

.a/(ext) r.a/ n # fn

(79)

and the "(ext)" superscript here means forces coming from sources external to the whole system.

Then eqn (1.79) gives .a/

#

(ext)

C N $ X X# (ext) R C % .a/ C $.a/ # f.a/ D n n aD1 nD1

(ext)

DR#F

C

C X

%

.a/

aD1

.a/(ext)

#F

C

.a/

C N X X

aD1 nD1

.a/(ext) $.a/ n # fn

(80)

PN .a/ .a/(ext) is the total force on the ath sub-collection coming from forces nD1 fn PC PN .a/ .a/(ext) (ext) outside the whole system and F D is the total force on the whole nD1 fn aD1 system coming from sources outside of it.

P .a/ , we can write the first of eqn (79) as Using J D L C K C C aD1 S where F.a/(ext) D

C d K X d S.a/ dL C C dt dt dt aD1

(ext)

DR#F

C

C X

.a/

%

.a/

aD1

.a/(ext)

#F

C

C N X X

aD1 nD1

.a/(ext) $.a/ n # fn

(81)

and use eqn (1.50) to cancel the first term on each side of the equal sign. The result is d dt

KC

C X

aD1

S

.a/

!

D

C X

aD1

.a/

%

.a/

.a/(ext)

#F

C

C N X X

aD1 nD1

.a/(ext) $.a/ n # fn

(82)

But we can’t subdivide further and say, for example, that the first term on the left in eqn (82) is equal to the first term on the right. Forces internal to the whole collection (for example, tidal forces between stars in a globular cluster) can change K and the S.a/ . Thus F.a/(ext) D 0 does dK not imply that D 0 separately. dt

CHAPTER 1 EXERCISE SOLUTIONS

11

Exercise 1.13 (a) Axiom 1.4.1 implies F(int) D 0. Also F(int) D f12 C f21 , hence f12 C f21 D 0. (b) Axiom 1.5.1 impies that # (int) D 0. Also # (int) D r1 # f12 C r2 # f21 D r # f21

(83)

Hence r # f21 D 0 which in turn implies that f21 must be parallel or anti-parallel to r and thus central. (c) With the given U.r1 ; r2 /, p @U du.r/ @ r " r du.r/ 1 @ .r " r/ f12 D !

D !U0 D !U0 p @r1 dr @r1 dr 2 r " r @r1 du.r/ r du.r/ rO D U0 D U0 p dr dr r"r

(84)

Similarly, f21 D !

du.r/ @U rO D !f12 D !U0 @r2 dr

(85)

Thus the forces are equal and opposite, and along the line rO between the two masses, as required. (d) The center of mass is 1 .m1 r1 C m2 r2 / (86) RD M where M D m1 C m2 is the constant total mass, and hence ) * 1 d 2 r1 d 2 r2 d 2R 1 m1 2 C m2 2 D D (87) .f21 C f12 / D 0 dt 2 M dt dt M dR is constant.

It follows that V D dt The relative position is r D r2 ! r1 . Thus d 2r d 2 r2 d 2 r1 1 1 D !

D f21 ! f12 D dt 2 dt 2 dt 2 m2 m1 and so

d 2r % 2 D f21 dt

where

%D

)

)

1 1 C m2 m1

1 1 C m2 m1

*!1

D

*

f21 D

1 f21 %

m1 m2 m1 C m2

(88)

(89)

(e) We have M R D m2 r2 C m1 r1 r D r2 ! r1

(90)

Multiplying the second of eqn (90) by m1 and adding gives M R C m1 r D .m1 C m2 / r2 and hence eqn (89) r2 D R C

m1 r M

(91)

(92)

12

CHAPTER 1 EXERCISE SOLUTIONS

Similarly r1 D R !

Then

$2 D r 2 ! R D

m1 r M

m2 r M

(93)

$1 D r 1 ! R D !

and

m2 r M

(94)

(f) From eqn (1.40), P D M V. From eqn (1.44), L D R # P D R # M V and * ) m1 m22 m21 m2 S D $1 # .m1 $P 1 / C $2 # .m2 $P 2 / D r # rP D r # %Pr C M2 M2 From eqn (1.56) To D 12 M V 2 and

$ 1 1# TI D m1 k$P 1 k2 C m2 k$P 2 k2 D 2 2

)

m1 m22 m21 m2 C M2 M2

*

rP 2 D

1 2 %Pr 2

(95)

(96)

P dS .ext/ D # .ext/ , where eqn (1.49) gives # .ext/ D 2nD1 $n # f.ext/ D0 From eqn (1.52), s s n . But fn dt here, so S is conserved. We can also see this directly by differentiating the last expression in eqn (95) and using eqn (89) dS d 2r D rP # %Pr C r # % 2 D 0 C r # f21 dt dt Since f21 is parallel or anti-parallel to r, this last expression is zero.

Exercise 1.14 (a) We are given f21 D !f12 D f .r/Or, where r D r2 ! r1 and Z r1 Z r2 U.r1 ; r2 / D ! f12 " d r1 ! f21 " d r2 r10

r20

In terms of the center of mass R and the relative position vector r, m2 r1 D R ! r m1 C m2 m1 r2 D R C r m1 C m2

(97)

(98)

(99)

Changing to the integration variables R and r, eqn (98) becomes * Z R Z r) m2 m1 U.r1 ; r2 / D ! .f12 C f21 / " d R C f12 ! f21 " d r m1 C m2 m1 C m2 R r0 Z r0 Z r D! f21 " d r D ! f .r/Or " d r (100) r0

But

r0

! " 2rdr D d r 2 D d .r " r/ D 2r " d r

and hence rO " d r D dr and, writing u.r/ for the integral of f .r/, Z r U.r1 ; r2 / D ! f .r/dr D ! u.r/jrr0 D !u.r/ C u.r0 / r0

which depends only on the scalar separation variables r and r0 . Thus U.r1 ; r2 / is independent of the integration path in eqn (100) and is single valued.

CHAPTER 1 EXERCISE SOLUTIONS

13

(b) Referring to eqn (84) and eqn (85) with U0 set equal to minus one here, we have du.r/ @U rO D f .r/Or D f21 D @r2 dr @U du.r/ rO D !f .r/Or D f12 !

D! @r1 dr !

Exercise 1.15 (a) Before collision, at t D 0!, P D m3 v0 D m3 v0 eO 3

(101)

So the velocity of the center of mass is VD

m3 v0 1 P D eO 3 D v0 eO 3 M m1 C m2 C m3 3

(102)

Since there are no external forces, this velocity is constant for all time. At t D 0, the center of mass is at " 1 ! a R.0/ D !m1 a eO 1 C m2 a eO 1 C m3 a eO 1 D eO 1 (103) M 3 Thus for t > 0 1 a (104) R D R.0/ C Vt D eO 1 C v0 t eO 3 3 3 Before collision, the m2 , m3 system has p23 .0!/ D m3 v0 eO 3 . Just after collision, it has p23 .0C/ D .m2 C m3 / v23 .0C/. But p23 .0C/ D p23 .0!/ Thus v23 .0C/ D

1 v0 eO 3 2

(105)

where the subscript "23" denotes the combined m2 , m3 blob.

For all t > 0 after the collision is complete, %Rr D !kr is the reduced mass and

where

%D

m1 m23 2 D m m1 C m23 3

(106)

r D r2 ! r1 D x eO 1 C y eO 2 C z eO 3

(107)

Thus xR D !! 2 x 2

yR D !! y

H)

x D A1 cos !t C B1 sin !t

H)

z D A3 cos !t C B3 sin !t

y D A2 cos !t C B2 sin !t

H)

2

zR D !! z

where !D

s

k D %

r

3k 4m

The initial conditions at t D 0C are r.0C/ D 2a eO 1

and

(108)

v.0C/ D v23 .0C/ ! v1 .0C/ D

(109)

v0 eO 3 2

(110)

14

CHAPTER 1 EXERCISE SOLUTIONS

v0 , and B1 D B2 D 0. Thus 2! v0 r D 2a cos !t eO 1 C sin !t eO 3 2!

Hence A1 D 2a, A2 D A3 D 0, B3 D

(111)

(b) To get the position of m23 multiply eqn (107) by m1 and add it to M R D m1 r1 C m23 r2 from the definition of the center of mass. The result is m1 r C M R D .m1 C m23 / r2

and so

1 r2 D R C r 3

(112)

Similarly, multiplying eqn (107) by m2 and subtracting it from the definition of the center of mass gives 2 (113) r1 D R ! r 3 Thus * ) $ 1# a 1 v0 r2 D eO 1 C v0 t eO 3 C 2a cos !t eO 1 C sin !t eO 3 3 3 3 2! ) * a 1 v0 D .1 C 2 cos !t / eO 1 C tC sin !t eO 3 (114) 3 3 2! and * $ 2# a 1 v0 eO 1 C v0 t eO 3 !

2a cos !t eO 1 C sin !t eO 3 3 3 3 2! ) * a 1 v0 D .1 ! 4 cos !t / eO 1 C t ! sin !t eO 3 3 3 !

r1 D

)

(115)

(c) Taking the derivatives of eqn (114) and eqn (115)with respect to time gives 2a! v0 1 sin !t eO 1 C .1 C cos !t /Oe3 3 3 2 4a! v0 sin !t eO 1 C .1 ! cos !t / eO 3 v1 D 3 3 v2 D !

(116)

At !tn D 2& n, where n is a positive integer, the velocity v1 vanishes. But the z components of v1 and v2 are non-negative for all t > 0, so neither m1 nor m23 will ever return to the eO 1 axis where z D 0.

Exercise 1.16 From eqn (A.3) !

"

V D Vk C V ?

(117)

O So, where Vk D nO nO " V is parallel to unit vector nO and V? D V ! Vk is perpendicular to n. since nO and Vk are parallel, we have ! " V # nO D Vk C V? # nO D 0 C V? # nO (118)

Then

! " ! " " ! " ! nO # V # nO D nO # V? # nO D nO " nO V? ! nO " V? nO D V?

since nO " nO D 1 and nO " V? D 0.

(119)

CHAPTER 1 EXERCISE SOLUTIONS

15

Exercise 1.17 From eqn (A.68),

and hence

A # .B # C/ D B .A " C/ ! C .A " B/

(120)

.B # C/ # A D !A # .B # C/ D C .A " B/ ! B .A " C/

(121)

.A # B/ # C D B .A " C/ ! A .B " C/

(122)

Making the substitutions ABC !! CAB, and using the property that the dot product of two vectors in independent of their order, gives

Note that eqn (119) with the parentheses around .B # C/ is not equal to eqn (122) with the parentheses around .A # B/. The triple cross product is not associative.

Although the "BAC-CAB" mnemonic is often used for eqn (120), it is not correct for eqn !!!!! !!!!! !!!!

(122). A mnemonic that is correct for both cases labels the vectors as outside, middle, and other for the vector outside the parenthesis, the vector in the middle, and the other vector inside the parentheses. Then the right sides of both eqn (120) and eqn (122) may be correctly written as !!!!! #!!!!! !!!!$ !!!! #!!!!! !!!!!$ middle outside " other ! other outside " middle (123) Exercise 1.18 (a) The left side of eqn (A.54) may be expanded as 3 X

nD1

ak ck n D ak ck1 C ak ck2 C ak ck3 D ak .ck2 C ck2 C ck3 /

(124)

which is the expansion of the right side. (b) The first equality in eqn (1.84) follows from substituting aij D !aj i . The second equality follows from changing dummy variables using ij !! j 0 i 0 and then dropping the primes on the new indices. The third equality follows from changing the order of summation over unrelated indices. (c) The first and last expressions in eqn (1.84) are equal. So either is equal to one-half their sum.

Thus 1 0 3 3 3 3 X 3 3 X X X 1 @X X ri aij D ri aij ! rj aij A 2 iD1 j D1

iD1 j D1

iD1 j D1

D

1 2

3 3 X X

iD1 j D1

!

3 3 " " 1 XX! ri aij ! rj aij D ri ! rj aij 2

(125)

iD1 j D1

where eqn (A.53) was used to get the second equality. (d) The ck does not depend on indices i or j and hence may be factored from the sum. Then aik does not depend on index j and can also be factored out of the j sum. Thus eqn (A.59) may be written as 3 3 X X

iD1 j D1

aik bij ck D ck

3 3 X X

iD1 j D1

aik bij D ck

3 X iD1

aik

3 X

j D1

bij

(126)

16

CHAPTER 1 EXERCISE SOLUTIONS

Exercise 1.19 Changing the dummy indices ij to j 0 i 0 and then dropping the primes on the dummy indices, the left side of eqn (1.86) becomes 3 3 X X

iD1 j D1

aij xi xj D

3 3 X X

j D1 iD1

aj i xj xi D

3 3 X X

j D1 iD1

3 3 X X

aj i xi xj D

(127)

aj i xi xj

iD1 j D1

where the last equality is obtained by exchaning the order of finite sums with unrelated indices.

Then write the left side of eqn (127) as one-half the sum of two equal quantities, as 3 3 X X

iD1 j D1

aij xi xj D

0

1@ 2 D

3 3 X X

iD1 j D1

aij xi xj C

3 3 X X

iD1 j D1

1

aj i xi xj A

3 3 3 3 " " 1 XX! 1 XX! aij xi xj C aj i xi xj D aij C aj i xi xj 2 2 iD1 j D1

(128)

iD1 j D1

(b) The eqn (A.62) is F .x/ D

3 3 X X

iD1 j D1

aij xi xj D

3 3 X X ! iD1 j D1

3 3 3 X 3 X X X " bij C cij xi xj D bij xi xj C cij xi xj iD1 j D1

iD1 j D1

(129) But cij C cj i D 0 and hence eqn (128) implies that the second sum on the right in eqn (129) is zero, thus verifying the first of eqn (A.63).

Taking partial derivatives and using @xi =@xk D ıik gives 3

3

3

3

3

3

XX XX ! " XX @F .x/ D bij ıik xj C xi ıj k D bij ıik xj C bij xi ıj k @xk iD1 j D1

D

3 X

j D1

iD1 j D1

bkj xj C

3 X iD1

iD1 j D1

bik xi D

3 X

j D1

bkj xj C

3 X

bki xi

(130)

iD1

where bik D bki was used to get the last equality. But the two terms of the final expression differ only in the choice of dummy index i or j . Thus they are equal and their sum is twice either one, giving 3 X @F .x/ D2 bkj xj (131) @xk j D1

which verifies the second of eqn (A.63). " 1! aij C aj i D 0. Thus the quadratic form F .x/ and its deriva(c) If aj i D !aij then bij D 2 tive are zero.