Theoretical Dynamics September 16, Homework 2. Taking the point of support as the origin and the axes as shown, the coordinates are

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1 Teoretical Dynaics Septeber 16, 2010 Instructor: Dr. Toas Coen Hoework 2 Subitte by: Vivek Saxena 1 Golstein 1.22 Taking te point of support as te origin an te axes as sown, te coorinates are x 1, y 1 = l 1 sin θ 1, l 1 cos θ 1 1 x 2, y 2 = l 1 sin θ 1 l 2 sin θ 2, l 1 cos θ 1 l 2 cos θ 2 2 Te Lagrangian is L = T V 3 were T = 1 2 1ẋ 2 ẏ ẋ ẏ 2 2 = 1 2 1l 2 1 θ l 2 1 θ 2 l 2 2 θ 2 2 2l 1 l 2 θ1 θ2 cosθ θ 2 4 an V = 1 gl 1 cos θ 1 2 gl 1 cos θ l 2 cos θ 2 5 So, L = l 2 1 θ l 2 2 θ l 1 l 2 θ1 θ2 cosθ θ gl 1 cos θ 2 gl 1 cos θ l 2 cos θ

2 Te erivatives are θ 1 = 2 l 2 1 θ 1 2 l 1 l 2 θ2 cosθ θ 2, θ 2 = 2 l 2 2 θ 2 2 l 1 l 2 θ1 cosθ θ 2 7 θ 1 = 2 l 1 l 2 θ1 θ2 sinθ θ 2 1 gl 1 sin θ 1 2 gl 1 sin θ 1, θ 1 θ 2 Te Euler-Lagrange equations are tat is, θ 2 = 2 l 1 l 2 θ1 θ2 sinθ θ 2 2 gl 2 sin θ 2 = 2 l 2 1 θ 1 2 l 1 l 2 θ2 cosθ θ l 1 l 2 θ2 θ θ 2 sinθ θ 2 9 = 2 l 2 2 θ 2 2 l 1 l 2 θ1 cosθ θ l 1 l 2 θ1 θ θ 2 sinθ θ 2 10 θ 1 θ 2 θ 1 = 0 θ 2 = 0 2 l 2 1 θ 1 2 l 1 l 2 θ2 cosθ θ l 1 l 2 θ2 2 sinθ θ gl 1 sin θ 1 = Golstein l 2 1 θ 2 2 l 1 l 2 θ1 cosθ θ l 1 l 2 θ2 1 sinθ θ gl 2 sin θ 2 = Kinetic Energy T = 1 2 Mẋ2 1 2 ẋ2 2 + ẏ Potential Energy V = gy

3 Constraint: Lagrangian: Constraine Lagrangian: Gx 1, x 2, y 2 = y 2 x 2 x 1 tan α = 0 15 L = T V = 1 2 Mẋ2 1 2 ẋ2 2 + ẏ 2 2 gy 2 16 L c = T V λg = 1 2 Mẋ2 1 2 ẋ2 2 + ẏ 2 2 gy 2 λ[y 2 x 2 x 1 tan α] 17 Te Euler-Lagrange equation, give Aing 21 an 22 we get ẋ 1 ẋ 2 ẏ 2 x 1 = 0 18 x 2 = 0 19 y 2 = 0 20 Mẍ λ tan α = 0 21 ẍ 2 λ tan α = 0 22 ÿ 2 + g + λ = 0 23 Mẍ ẍ 2 = 0 24 wic upon one integration wrt tie, yiels te expecte result tat te linear oentu of te block + wege syste in te X-irection is constant. Multiplying 23 trougout by tan α, using 15 to write ÿ 2 = ẍ 2 ẍ 1 tan α an substituing λ tan α = Mẍ 1 fro 21 we get ẍ 2 ẍ 1 tan α + g + λ = 0 = M + ẍ 1 tan 2 α + g tan α Mẍ 1 = 0 So, ẍ 1 = g tan α M M tan 2 25 g tan α ẍ 2 = M tan 2 26 ÿ 2 = g tan 2 α M M tan 2 27 g λ = M tan 2 28 Te signs are consistent: as te particle escens te slope of te wege, it oves to te left in te lab frae, as te wege oves to te rigt, conserving linear oentu in te orizontal irection. Also, as /M 0, we recover te solution for a particle oving own a stationary wege: ẍ 1 = 0, ẍ 2 = g sin α cos α, ÿ 2 = g sin 2 α so tat te acceleration of te particle along te incline is ẍ2 2 + ÿ2 2 = g sin α. 2-3

4 Work one by te constraint forces Te tree constraint forces are F x1 = λ G g tan α = λ tan α = x 1 M tan 2 F x2 = λ G g tan α = λ tan α = x 2 M tan 2 F y2 = λ G g = λ = y 2 M tan 2 Te accelerations foun above are constant, so te velocity varies linearly wit tie. Assuing tat at t = 0, te wege an particle bot ave zero velocity, te work one by te constraint force on te wege is W 1 = F x1 x 1 = 1 2 F x 1 ẍ 1 t 2 = 1 2 g tan α M tan 2 2 M g tan α M tan 2 t 2 32 = 1 M g2 tan 2 α [ 2 M tan 2 ] 2 t2 33 Siilarly, te work one by te constraint force on te particle is W 2 = F x2 x 2 + F y2 y 2 = 1 2 F x 2 ẍ 2 t F y 2 ÿ 2 t 2 = 1 g tan α g tan α 2 M tan 2 M tan 2 t g 2 M tan 2 g tan 2 α M M tan 2 = 1 g 2 tan 2 α [ 2 tan 2 ] 2 t = 1 2 M 2 [ M M g 2 tan 2 α [ M tan 2 ] 2 t2 M g2 tan 2 α tan 2 ] 2 t2 34 We note tat W W 2 = 0, confiring te fact tat te total work one on te syste by te constraint forces in tie t is zero. Tis is consistent wit te fact tat te constraint forces are internal to te syste, an te constraint G = 0 is inepenent of tie. t 2 2-4

5 3 Golstein 13.4 Te given Lagrangian ensity is L = 2 8π 2 ψ ψ + V ψ ψ + 4πi ψ ψ ψ ψ 35 Te Euler-Lagrange equation for ψ is µ µ ψ ψ = 0 36 tat is, Te erivatives are Substituting into 37, we get or ψ 4πi ψ + + ψ ψ = 0 37 Π = ψ = 4πi ψ = 4πi ψ ψ ψ ψ = = 2 8π 2 ψ 2 8π 2 2 ψ ψ = V ψ 4πi ψ 2 8π 2 2 ψ V ψ + 4πi ψ = 0 38 i ψ 2π = 2 8π 2 2 ψ + V ψ 39 wic is Scroinger s equation. Te oentu canonically conjugate to ψ is So, te Hailtonian ensity is Π = ψ = 4πi ψ 40 H = Π ψ + Π ψ L 41 = 4πi ψ ψ 4πi ψ ψ 2 8π 2 ψ ψ V ψ ψ 4πi ψ ψ + 4πi ψ ψ = 2 8π 2 ψ ψ V ψ ψ

6 4 Proble 1 Te equations of otion are ẍ + ω 2 x = 0 43 ÿ + αω 2 y = 0 44 Part a Te energy is So, E Hence te energy is conserve. Part b Hence is conserve. Part c E = 1 2 ẋ2 + ẏ ω2 x 2 + αy 2 45 = ẋẍ + ẏÿ + ω 2 xẋ + αyẏ = ẋ ω 2 x + ẏ αω 2 y + ω 2 xẋ + αyẏ using 43 an 44 = 0 46 = ẋẍ ẏÿ + ω 2 xẋ αyẏ = ẋ ω 2 x ẏ αω 2 y + ω 2 xẋ αyẏ using 43 an 44 = 0 47 It can be sown tat for a olonoic ecanical syste, te kinetic energy is always a bilinear for of te generalize coorinates, aking ters of te for / q necessarily linear in te generalize velocities, wenever te potential is inepenent of te generalize velocity. In particular, for te given Lagrangian, ẋ ẏ = ẋ 48 = ẏ 49 Since Q 1 x, y; ɛ an Q 2 x, y; ɛ are point transforations, tey are inepenent of velocities. Terefore te quantity Γ = Q 1 ẋ ɛ + Q 2 ɛ=0 ẏ ɛ 50 ɛ=0 = ẋ Q 1 ɛ + ẏ Q 2 ɛ=0 ɛ 51 ɛ=0 necessarily linear in te velocities, ẋ an ẏ. 2-6

7 Part As justifie above, any invariant quantity resulting fro te syetry of te Lagrangian uner a point transforation is necessarily linear in te velocities. Since is a quaratic for in te velocities, we cannot fin a point transforation wic leaves te Lagrangian invariant an correspons to a Noeterian conserve current tat is equal to. Tis proves tat wile every invariance of a Lagrangian uner a continuous point transforation yiels an associate conserve quantity, te converse is not necessarily true. For olonoic ecanical systes, te stronger stateent is: For every invariance of a Lagrangian uner a continuous point transforation, tere is an associate conserve quantity linear in te generalize oenta, an vice versa. Part e For α = 1, te syste becoes an isotropic aronic oscillator in 2D wit te Lagrangian, Due to rotational syetry, te angular oentu L = 1 2 ẋ2 + ẏ ω2 x 2 + y 2 52 J z = xẏ ẋy 53 is an invariant, wic is of te for Γ. As J z is linear in te velocities, it cannot be written as a linear cobination of E an wic ave no linear ters in ẋ an ẏ at all. 5 Proble 2 Part a Te action is S = L = x L 54 Te Lagragian ensity is not explicitly epenent on te fiel, but only on its erivatives. So, te variation in te action is δs = δ x L 55 = x δl 56 = x t φ δ tφ + x φ δ xφ 57 = x µ φ δ µφ for µ = 0, 1 58 [ ] = x µ µ φ δφ µ δφ 59 µ φ = x µ δφ 60 µ φ 2-7

8 since te first ter in 60 can be converte to a surface integral over te bounary of te 1-spacetie region, were δφ = 0 over te bounary. So, Hailton s principle δs = 0 yiels te Euler-Lagrange equation µ = 0 61 µ φ or Part b + t t φ x x φ = 0 62 = 2 t c 2 2 xφx, t = 0 63 Fro te inverse Lorentz transforations, t = γ t + βc x 64 x = γ x + βct 65 we ave t = t t t + x t x = γ t + γβc x x = t x t + x x x = γβ c t + γ x So, te Lagrangian ensity in te transfore frae is [ L = 1 2 ] 2 2 t c 2 x [ = γ 2 + γ 2 β 2 c 2 + 2γ 2 βc 2 t x t x c 2 γ 2 β 2 2 ] 2 c 2 + γ 2 + 2γ2 β t x c t x [ = 1 2 ] 2 γ 2 1 β 2 γ 2 c 2 1 β 2 2 t x [ = 1 2 ] 2 c 2 as γ = 1 2 t x 1 β 2 = L Hence te Lagrangian ensity is invariant uner te Lorentz transforation. 2-8

9 Part c x = = x x t x γ γβ c x t t x 71 t γβc γ x = γ 2 1 β 2 x = x 72 So te volue eleent in 1-spacetie is Lorentz invariant. Since te Lagrangian ensity is also Lorentz invariant, terefore te action S = x L is also a Lorentz invariant quantity. Part Te Euler-Lagrange equation is obtaine by extreizing te action, i.e. via δs = 0. As L, te 1- spacetie volue eleent as well as te Lagrangian ensity L are all Lorentz invariant quantities, δs = δ x L = 0 δs = δ x L = 0 = δs 73 Repeating te steps carrie out in part a wit all quantities replace by teir prie counterparts, we arrive at te Euler Lagrange equation, t + t φ x = 0 74 x φ in te transfore frae. Tis proves tat te Euler-Lagrange equations are for invariant, i.e. covariant. In particular, using 67 an 68 we ave so tat 2 φ t 2 = γ 2 t 2 φ + 2γ 2 βc xtφ 2 + γ 2 β 2 c 2 x φ x 2 = γ2 β 2 c 2 2 t φ + 2γ2 β 2 c xtφ + γ 2 β 2 c x φ t 2 c2 2 φ x 2 = γ 2 1 β 2 2 t φ γ 2 1 β 2 2 x 77 So, we conclue tat te Euler-Lagrange equation is also invariant. = 2 φ t 2 c2 2 φ x

10 6 Proble 3 Te oifie action is S = = = L 79 [ 1 2 ẋ2 q φ Λ ] + A + Λ ẋ 80 L + q Λ q Λ ẋ 81 = S + q = S + q = S + q = S Λ q Λ q t= Λ q Λ ẋ 82 x Λ 83 t= Λ 84 So, uner a tie epenent gauge transforation, te action is left invariant, inepenent of te pat