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CHAPTER Contents Preface v Problems Solved in Student Solutions Manual vii Matrices, Vectors, and Vector Calculus Newtonian Mechanics—Single Particle 29 Oscillations Nonlinear Oscillations and Chaos Gravitation Some Methods in The Calculus of Variations Hamilton’s Principle—Lagrangian and Hamiltonian Dynamics Central-Force Motion Dynamics of a System of Particles 10 Motion in a Noninertial Reference Frame 11 Dynamics of Rigid Bodies 12 Coupled Oscillations 13 Continuous Systems; Waves 435 14 Special Theory of Relativity 461 79 127 149 165 181 233 277 333 353 397 iii iv CONTENTS CHAPTER Preface This Instructor’s Manual contains the solutions to all the end-of-chapter problems (but not the appendices) from Classical Dynamics of Particles and Systems, Fifth Edition, by Stephen T Thornton and Jerry B Marion It is intended for use only by instructors using Classical Dynamics as a textbook, and it is not available to students in any form A Student Solutions Manual containing solutions to about 25% of the end-of-chapter problems is available for sale to students The problem numbers of those solutions in the Student Solutions Manual are listed on the next page As a result of surveys received from users, I continue to add more worked out examples in the text and add additional problems There are now 509 problems, a significant number over the 4th edition The instructor will find a large array of problems ranging in difficulty from the simple “plug and chug” to the type worthy of the Ph.D qualifying examinations in classical mechanics A few of the problems are quite challenging Many of them require numerical methods Having this solutions manual should provide a greater appreciation of what the authors intended to accomplish by the statement of the problem in those cases where the problem statement is not completely clear Please inform me when either the problem statement or solutions can be improved Specific help is encouraged The instructor will also be able to pick and choose different levels of difficulty when assigning homework problems And since students may occasionally need hints to work some problems, this manual will allow the instructor to take a quick peek to see how the students can be helped It is absolutely forbidden for the students to have access to this manual Please not give students solutions from this manual Posting these solutions on the Internet will result in widespread distribution of the solutions and will ultimately result in the decrease of the usefulness of the text The author would like to acknowledge the assistance of Tran ngoc Khanh (5th edition), Warren Griffith (4th edition), and Brian Giambattista (3rd edition), who checked the solutions of previous versions, went over user comments, and worked out solutions for new problems Without their help, this manual would not be possible The author would appreciate receiving reports of suggested improvements and suspected errors Comments can be sent by email to stt@virginia.edu, the more detailed the better Stephen T Thornton Charlottesville, Virginia v vi PREFACE CHAPTER Matrices, Vectors, and Vector Calculus 1-1 x2 = x2′ x1′ 45˚ x1 45˚ x3 x3′ Axes x′1 and x′3 lie in the x1 x3 plane The transformation equations are: x1′ = x1 cos 45° − x3 cos 45° x2′ = x2 x3′ = x3 cos 45° + x1 cos 45° x1′ = 1 x1 − x3 2 x2′ = x2 x3′ = 1 x1 − x3 2 So the transformation matrix is:        2 − 1  2      CHAPTER 1-2 a) x3 D E γ O β x2 B α A C x1 From this diagram, we have OE cos α = OA OE cos β = OB (1) OE cos γ = OD Taking the square of each equation in (1) and adding, we find 2 OE cos α + cos β + cos γ  = OA + OB + OD (2) But 2 2 2 OA + OB = OC (3) and OC + OD = OE (4) Therefore, 2 OA + OB + OD = OE (5) Thus, cos α + cos β + cos γ = (6) b) x3 D E D′ O A A′ x1 E′ θ B′ B C x2 C′ First, we have the following trigonometric relation: OE + OE′ − 2OE OE′ cos θ = EE′ 2 (7) MATRICES, VECTORS, AND VECTOR CALCULUS But, 2       EE′ = OB′ − OB  + OA′ − OA  + OD′ − OD        2     = OE′ cos β ′ − OE cos β  + OE′ cos α ′ − OE cos α        + OE′ cos γ ′ − OE cos γ    2 (8) or, 2 EE′ = OE′ cos α ′ + cos β ′ + cos γ ′  + OE  cos α + cos β + cos γ  − 2OE′ OE cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′  = OE′ + OE2 − 2OE OE′ cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′  (9) Comparing (9) with (7), we find cos θ = cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′ (10) 1-3 x3 e3 O e1 e2 e2′ e3 A x2 e2 e1′ e3′ e1 x1 Denote the original axes by x1 , x2 , x3 , and the corresponding unit vectors by e1 , e2 , e3 Denote the new axes by x′1 , x′2 , x′3 and the corresponding unit vectors by e1′ , e2′ , e3′ The effect of the rotation is e1 → e3′ , e2 → e1′ , e3 → e2′ Therefore, the transformation matrix is written as:  cos ( e1′ , e1 ) cos ( e1′ , e2 ) cos ( e1′ , e3 )  0    λ = cos ( e′2 , e1 ) cos ( e′2 , e2 ) cos ( e′2 , e3 )  = 0    1 0    cos ( e′3 , e1 ) cos ( e′3 , e2 ) cos ( e′3 , e3 )   1-4 a) Let C = AB where A, B, and C are matrices Then, Cij = ∑ Aik Bkj (1) k (C ) t ij = C ji = ∑ Ajk Bki = ∑ Bki Ajk k k CHAPTER ( ) Identifying Bki = Bt ik ( ) and Ajk = At , kj (C ) = ∑ (B ) ( A ) t t ij t ik k (2) kj or, C t = ( AB) = Bt At (3) ( AB) B−1 A−1 = I = ( B−1 A−1 ) AB (4) ( AB) B−1 A−1 = AIA−1 = AA−1 = I (5) (B (6) t b) To show that ( AB) = B−1 A−1 , −1 That is, 1-5 −1 ) A−1 ( AB) = B−1 IB = B−1B = I Take λ to be a two-dimensional matrix: λ = λ11 λ12 = λ11λ 22 − λ12 λ 21 λ 21 λ 22 (1) Then, 2 2 2 λ = λ112 λ 22 − 2λ11λ 22 λ12 λ 21 + λ12 λ 21 + ( λ11 λ 21 + λ122 λ 22 ) − ( λ112 λ 212 + λ122 λ 222 ) ( ) ( ) ( 2 2 = λ 22 λ112 + λ122 + λ 21 λ112 + λ122 − λ112 λ 21 + 2λ11λ 22λ12 λ 21 + λ12 λ 22 ( )( ) 2 2 = λ11 + λ12 λ 22 + λ 21 − ( λ11λ 21 + λ12 λ 22 ) But since λ is an orthogonal transformation matrix, ∑λ λ ij ) (2) kj = δ ik j Thus, 2 λ112 + λ122 = λ 21 + λ 22 =1 (3) λ11λ 21 + λ12 λ 22 = Therefore, (2) becomes λ =1 1-6 (4) The lengths of line segments in the x j and x ′j systems are L= ∑x j j ; L′ = ∑ x′ i i (1) MATRICES, VECTORS, AND VECTOR CALCULUS If L = L ′ , then ∑ x = ∑ x′ j i (2) xi′ = ∑ λ ij x j (3) j i The transformation is j Then,  ∑ x = ∑  ∑ λ j j i ik k   xk   ∑ λ iA xA    A (4)   = ∑ xk xA  ∑ λik λ iA   i  k ,A But this can be true only if ∑λ ik λ iA = δ k A (5) i which is the desired result 1-7 x3 (0,0,1) (1,0,1) (0,0,0) x1 (0,1,1) (1,1,1) (1,0,0) (0,1,0) x2 (1,1,0) There are diagonals: D1 , from (0,0,0) to (1,1,1), so (1,1,1) – (0,0,0) = (1,1,1) = D1 ; D , from (1,0,0) to (0,1,1), so (0,1,1) – (1,0,0) = (–1,1,1) = D ; D , from (0,0,1) to (1,1,0), so (1,1,0) – (0,0,1) = (1,1,–1) = D ; and D , from (0,1,0) to (1,0,1), so (1,0,1) – (0,1,0) = (1,–1,1) = D The magnitudes of the diagonal vectors are D1 = D = D = D = The angle between any two of these diagonal vectors is, for example, D1 ⋅ D (1,1,1) ⋅ ( −1,1,1) = cos θ = = D1 D 3 CHAPTER so that θ = cos−1   = 70.5°  3 Similarly, D1 ⋅ D D ⋅D D ⋅D D ⋅D D ⋅D = = = = =± D1 D D1 D D2 D3 D2 D D3 D 1-8 Let θ be the angle between A and r Then, A ⋅ r = A2 can be written as Ar cos θ = A2 or, r cos θ = A (1) This implies QPO = π Therefore, the end point of r must be on a plane perpendicular to A and passing through P 1-9 a) A = i + 2j − k B = −2i + 3j + k A − B = 3i − j − 2k 2 A − B = ( 3) + ( −1) + (−2)2  12 A − B = 14 b) B θ A component of B along A The length of the component of B along A is B cos θ A ⋅ B = AB cos θ B cos θ = A ⋅ B −2 + − = = or A 6 The direction is, of course, along A A unit vector in the A direction is ( i + 2j − k ) (2) 478 CHAPTER 14 Thus v = i v cos ω t − j v sin ω t Then r=i The path is a circle of radius v ω sin ω t + j v ω cos ω t v ω r= v q B0 γ m = γ mv q B0 = p q B0 From problem 14-22  T2  p =  2Tm +  c   12 So  T2  Tm +  c  r= q B0 14-26 below 12 Suppose a photon traveling in the x-direction is converted into an e − and e + as shown e+ θ θ e– before after Cons of energy gives pp c = 2Ee where pp = momentum of the photon Ee = energy of e + = energy of e − Cons of px gives pp = pe cos θ Dividing gives (p e = momentum of e + , e − ) 479 THE SPECIAL THEORY OF RELATIVITY pp c pp =c= Ee pe cos θ or pe2 c cos θ = Ee2 (1) But Ee2 > pe2 c , so (1) cannot be satisfied for cos θ ≤ An isolated photon cannot be converted into an electron-positron pair This result can also be seen by transforming to a frame where px = after the collision But, before the collision, px = pp c ≠ in any frame moving along the x-axis So, without another object nearby, momentum cannot be conserved; thus, the process cannot take place 14-27 The minimum energy required occurs when the p and p are at rest after the collision By conservation of energy 2Ee = ( 938 MeV ) Ee = 938 MeV = T + E0 Since Ee = 0.5 MeV , Te+ = Te− = 937.5 MeV 14-28 Tclassical = mv 2 Trel = (γ − 1) mc ≥ Tclassical We desire Trel − Tclassical ≤ 0.01 Trel mv 2 1− ≤ 0.01 (γ − 1) mc2 v ≥ 0.99 (γ − 1) c β2 γ −1 ( Putting γ = − β ) −1 and solving gives ≥ 1.98 480 CHAPTER 14 v ≤ 0.115 c The classical kinetic energy will be within 1% of the correct value for ≤ v ≤ 3.5 × 107 m/sec, independent of mass E = γ E0 14-29 For E = 30 × 109 eV E0  0.51 × 106 eV, γ  5.88 × 10 γ= 1 − β2 β  1− 2γ ( or β = − γ −2 ) 12 = − 1.4 × 10 −10 ( ) v = − 1.4 × 10 −10 c = 0.99999999986 c 14-30 A neutron at rest has an energy of 939.6 MeV Subtracting the rest energies of the proton (938.3 MeV) and the electron (0.5 MeV) leaves 0.8 MeV Other than rest energies 0.8 MeV is available 14-31 θ 0.98c θ Conservation of energy gives Eπ = 2Ep where Ep = energy of each photon (Cons of py implies that the photons have the same energy) 481 THE SPECIAL THEORY OF RELATIVITY Thus γ E0 = 2Ep Ep = γ E0 = 135 MeV − 0.98 = 339 MeV The energy of each photon is 339 MeV Conservation of px gives γ mv = pp cos θ where pp = momentum of each photon (135 Mev/c ) ( 0.98 c) cos θ = − 0.98 ( 339 MeV/c ) = 0.98 θ = cos −1 0.98 = 11.3° 14-32 From Eq (14.67) we have E2 − E02 = p c With E = E0 + T , this reduces to 2E0 T + T = p c Using the quadratic formula (taking the + root since T ≥ 0) gives T = E02 + p c − E0 Substituting pc = 1000 MeV E0 ( electron ) = 0.5 MeV E0 ( proton ) = 938 MeV gives Telectron = 999.5 MeV Tproton = 433 MeV 482 CHAPTER 14 14-33 e 120˚ p n 120˚ ν before after Conservation of py gives pe sin 60° = pν sin 60° or pe = pν Conservation of px gives pp = pe cos 60° + pν cos 60° = pe So pe = pp = pν ≡ p Conservation of energy gives E0 n = Ee + Ep + Eν E0 n = E02e + p c + E02p + p c + pc Substituting E0 n = 939.6 MeV E0 p = 938.3 MeV E0 e = 0.5 MeV and solving for pc gives p = 0.554 MeV/c pp = pe = pν = 0.554 MeV/c Substituting into T = E − E0 = E02 + p c − E0 gives ( E0ν = ) Tν = 0.554 MeV Tp = × 10 −4 MeV, or 200 eV Te = 0.25 MeV (1) 483 THE SPECIAL THEORY OF RELATIVITY ∆s′ = − c 2t ′ + x1′ + x2′ + x3′ 14-34 Using the Lorentz transformation this becomes ∆s′ = v x12 + x1vt x12 + v 2t − 2x1vt c2 + + x22 + x32 − v2 c2 − v2 c2 −c 2t −  v x12   v 2   x1 − c  − c t − c t    + x2 + x2  = 2 1− v c = − c 2t + x12 + x22 + x32 So ∆s′ = ∆s2 14-35 Let the frame of Saturn be the unprimed frame, and let the frame of the first spacecraft be the primed frame From Eq (14.17a) (switch primed and unprimed variables and change the sign of v) u1 = u1′ + v u ′v + 12 c Substituting v = 0.9 c u1′ = 0.2 c gives u1 = 0.93 c 14-36 Since Fµ = d dτ  dX µ   m dτ  and X µ = ( x1 , x2 , x3 , ict ) we have F1 = d dτ F2 = m F4 = d dτ d x1  dx1  = m m  dτ  dτ d x2 dτ F3 = m d x3 dτ d 2t  d ( ict )  = m icm  dτ  dτ  484 CHAPTER 14 Thus F1′ = m d x1 d2 γ ( x1 − vt )  m − dτ dτ  =γm F2′ = m d x1′ d 2t γ mv = = γ ( F1 + iβ F4 ) dτ dτ d x2′ d x2 m = = F2 ; F3′ = F3 dτ dτ F4′ = icm d dτ = γ icm   vx1   γ  t − c     d 2t d x1 γ β − i m dτ dτ = γ ( F4 − iβ F1 ) Thus the required transformation equations are shown 14-37 From the Lagrangian ( ) L = mc − − β − kx (1) we compute ∂L = − kx ∂x (2) ∂ L ∂β ∂ L β = = mc ∂v ∂ v ∂β − β2 (3) Then, from (2) and (3), the Lagrange equation of motion is d dt  mcβ    + kx =  − β  (4) from which mcβ (1 − β ) 32 + kx = (5) Using the relation cβ = we can rewrite (4) as dv dv dx dv = =v dt dx dt dx (6) 485 THE SPECIAL THEORY OF RELATIVITY mc β (1 − β ) 32 dβ + kx = dx (7) kx = E (8) This is easily integrated to give mc 1− β + where E is the constant of integration The value of E is evaluated for some particular point in phase space, the easiest being x = a; β = 0: ka (9) kx = mc + ka 2 (10) E = mc + From (8) and (9), mc 1− β + Eliminating β from (10), we have β2 = − m2 c  2  mc + k a − x   ( ( = k a2 − x ) )  k 2   mc + a − x   k 2  mc a x + −   ( ) ( (11) ) and, therefore, ( k a2 − x dx β= = c dt ) ( ( mc + k a − x ) mc + k a − x 2 ) (12) The period will then be four times the integral of dt = dt(x) from x = to x = a: m τ=4 k a ∫ k  2  1 + 2mc a − x  dx k 2 2 a − x 1+ a −x mc ( ) ( ) (13) Since x varies between and a, the variable x a takes on values in the interval to 1, and therefore, we can define sin φ = from which x a (14) 486 CHAPTER 14 a2 − x a (15) dx = a − x dφ (16) cos φ = and We also define the dimensionless parameter, a κ≡ k mc (17) Using (14) – (17), (13) transforms into τ= 2a κc π ∫ (1 + 2κ cos φ + κ cos φ ) dφ (18) Since ka mc  for the weakly relativistic case, we can expand the integrand of (18) in a series of powers of κ : (1 + 2κ cos φ ) ≅ + 2κ ( (1 + κ cos φ ) 2 2 12  κ2  cos φ  − cos φ    ) 1  ≅ +  −  κ cos φ 2  = + κ cos φ (19) Substitution of (19) into (18) yields 2a τ≅ κc = π  ∫ 1 + κ aπ 3κ a + κ c 2c  cos φ  dφ  π   φ + sin 2φ  (20) Evaluating (20) and substituting the expression for κ from (17), we obtain τ = 2π m 3π a + k 8c k m (21) or,  τ = τ 1 +  ka  16 mc  (22) 487 THE SPECIAL THEORY OF RELATIVITY 14-38 F= dp d = (γ mu) dt dt =m d (γ u) dt =m  d  u   dt  − u c    (for m = constant)  2  1− u c = m   ( ( = m − u2 c ) ) 12  u − u  −  − u2 c  c  − u2 c2 ( ) ( −3 ) −1   du   dt  du dt Thus F=m 14-39 ( du − u2 c dt ) −3 The kinetic energy is T = p c + m02 c − m0 c (1) Tproton = 10 + ( 931) − 931 ≅ 936 − 931 = MeV (2) Telectron = 10 + ( 0.51) − 0.51 ≅ 100 − 0.5 = 99.5 MeV (3) For a momentum of 100 MeV/c, 2 In order to obtain γ and β, we use the relation E = mc = γ m0 c = m0 c − β2 (4) so that γ= E m0 c (5) and β = 1− γ electron = γ2 100 ≅ 200 0.51 (6) (7) 488 CHAPTER 14  βelectron = −   ≅ 0.999988 200  (8) This is a relativistic velocity γ proton = 936 ≅ 1.0054 931 (9)  β proton = −   ≅ 0.1 1.0053  (10) This is a nonrelativistic velocity 14-40 If we write the velocity components of the center-of-mass system as v j , the transformation of pα , j into the center-of-mass system becomes v j Eα   pα′ , j = γ  pα , j −  c   where γ = 1− v 2j Since in the center-of-mass system, ∑ pα′ α (1) ,j = must be satisfied, we have c2 ∑ pα′ α ,j v j Eα   = ∑ γ  pα , j −  = c  α  (2) or, ∑ pα c = α ∑ Eα (3) T1 E1 − m0 c = T0 E0 − m0 c (1) vj c ,j α 14-41 We want to compute where T and E represent the kinetic and total energy in the laboratory system, respectively, the subscripts and indicate the initial and final states, and m0 is the rest mass of the incident particle The expression for E0 in terms of γ is E0 − m0 c 2γ (2) E1 can be related to E′1 (total energy of particle in the center of momentum reference frame after the collision) through the Lorentz transformation [cf Eq (14.92)] (remembering that for the inverse transformation we switch the primed and unprimed variables and change the sign of v): 489 THE SPECIAL THEORY OF RELATIVITY E1 = γ 1′ ( E1′ + cβ1′p1′ cos θ ) (3) where p1′ = m0 cβ1γ 1′ and E1′ = m0 c 2γ 1′ : ( ) (4) T1 γ 1′ + γ 1′ β1′ cos θ − = T0 γ1 −1 (5) E1 = m0 c 2γ 1′ + β1′ cos θ Then, from (1), (2), and (4), For the case of collision between two particles of equal mass, we have, from Eq (14.127), γ 1′ = 1+ γ1 (6) and, consequently, γ 1′ β1′ = γ 1′ − = γ1 −1 (7) Thus, with the help of (6) and (7), (5) becomes T1 γ − + (γ − 1) cos θ = T0 (γ − 1) = + cos θ (8) We must now relate the scattering angle θ in the center of momentum system to the angle ψ in the lab system Squaring Eq (14.128), which is valid only for m1 = m2 , we obtain an equation quadratic in cos θ Solving for cos θ in terms of tan ψ , we obtain cos θ = − γ1 +1 1+ tan ψ ± γ1 +1 tan ψ (9) One of the roots given in (9) corresponds to θ = π, i.e., the incident particle reverses its path and is projected back along the incident direction Substitution of the other root into (8) gives cos ψ T1 = = T0 + γ + tan ψ cos ψ + (γ + 1) sin ψ An elementary manipulation with the denominator of (10), namely, (10) 490 CHAPTER 14 ( ) cos ψ + (γ + 1) sin ψ = cos ψ + γ 1 − cos ψ + sin ψ = γ + sin ψ + cos ψ − γ cos ψ + cos ψ = γ + − γ cos ψ + cos ψ = (γ + 1) − (γ − 1) cos ψ (11) provides us with the desired result: cos ψ T1 = T0 (γ + 1) − (γ − 1) cos ψ (12) Notice that the shape of the curve changes when T1 > m0 c , i.e., when γ > T1 T0 1.0 T1 = 0.1 GeV 0.8 T1 = GeV 0.6 T1 = 10 GeV 0.4 0.2 0˚ 30˚ 60˚ ψ 90˚ 14-42 y γmec2 hν x φ θ hν′ From conservation of energy, we have hν + me c = γ me c + hν ′ (1) Momentum conservation along the x axis gives hν hν ′ cos θ + γ me v cos φ = c c (2) Momentum conservation along the y axis gives γ me v sin φ = hν ′ sin θ c (3) 491 THE SPECIAL THEORY OF RELATIVITY In order to eliminate φ, we use (2) and (3) to obtain  hν hν ′  cos θ   −  c γ me v  c   hν ′  sin φ = sin θ  γ me v cos φ = (4) Then, cos φ + sin φ = = 2 γ me v Since γ = 1− v c2 and v = c γ   hν   hν ′    hν   hν ′  cos θ − +   +     c   c   c   c   (5) γ − we have γ v = c ( γ − 1) (6) Substituting γ from (1) into (6), we have γ 2v2 = 2h h2 ( ν − ν ′ ) + 2 (ν − ν ′ ) me me c (7) From (5) and (7), we can find the equation for ν′: h2  hν   hν ′   hν   hν ′  hm cos θ ν ν ν − ν ′) + − = − + ′ ( ) e (  c   c   c   c  c (8)  2me c  2me c 2 cos ν θ ν ν + − = ′ ( )  h h   (9)     ν ν′ =   + hν (1 − cos θ )    me c (10) 2 or, Then, or,   E − cos θ )  E′ = E 1 + (  me c  The kinetic energy of the electron is −1 (11) 492 CHAPTER 14   T = γ me c − me c = hν − hν ′ = E 1 −  E − cos θ )   1+ ( me c   T= − cos θ E2    me c + E − cos θ  ( )  me c   (12) ... contains the solutions to all the end-of-chapter problems (but not the appendices) from Classical Dynamics of Particles and Systems, Fifth Edition, by Stephen T Thornton and Jerry B Marion It is... intended for use only by instructors using Classical Dynamics as a textbook, and it is not available to students in any form A Student Solutions Manual containing solutions to about 25% of the end-of-chapter... to this manual Please not give students solutions from this manual Posting these solutions on the Internet will result in widespread distribution of the solutions and will ultimately result in