Analytical Mechanics ( AM ) lecture notes part 10, Summary Olaf Scholten KVI, kamer v3.008 tel. nr. 363-355 email: scholten@kvi.nl Web page: http://www.kvi.nl/~scholten Book Classical Dynamics of Particles and Systems, Stephen T. Thornton & Jerry B. Marion 5th Edition ISBN-10: 0534408966 ISBN-13: 9780534408961 Intro Introduce an abstract reference-frame independent formulation of Mechanics via the Hamiltonian and Lagrangian formalisms. This lends itself to generalizations to problems in Statistical Mechanics, Quantum Mechanics, Relativistic Mechanics, Field Theory,...
January 17, 014 Advanced Analytical Mechanics- - Summary.1 - Damped & Driven Oscillators Driving force F (t) = m A 0 cos(ωt), eom is real part of ẍ + βẋ + ω rx = A 0 e iωt trial solution: x(t) = Ae iωt gives special solution: A = A 0 (ω r ω ) + iβω = A e iδ Full sol. = Special sol. + sol. homogeneous eq. Non-linear effects Frequency doubling (tripling) & Hysteresis & Chaos Oscillator with memory effect Max Ampl Phase ẍ + βẋ + ω rx + ω rc x 3 = A 0 cos(ω d t) 3 1 0-1 - 4 3 1 0 pendulum Tue Mar 31 009 3:19:3 0 1 3 4 5 6...... Tune up Tune down....... x 3 oscil 0 1 3 4 5 6 Frequency Initial conditions at slowly tune up or down frequency
January 17, 014 Advanced Analytical Mechanics-3 - Summary. - Chaos Computer simulations-6 Driven and damped non-linear Oscillator inverted linear part ẍ + βẋ + ωrx + ωrc x 3 = A 0 cos(ω d t) parameters ω r =, β = 0.1, A 0 = 4., ω d = 4.3, C = 1.1 (dx/dt)/ 0. 0........... - - amplitude Phase -4-3 - -1 0 1 3 x Poincaré -4-3 - -1 0 1 3 0 1000 000 3000 4000 3. 1 0-1 - 3 1. 0-1 - -3 4000 5000 6000 7000 8000 time [0.01 sec] NL-pendulum Wed Apr 01 009 3:00:0
January 17, 014 Advanced Analytical Mechanics-4 - Summary.3 - EOM of the type: ẍ + βẋ + ω rx }{{} L[x] F (t) = n= Fourier transform = F (t)/m A n e i nω 0t with ω 0 = π/τ, A n = A n and A n = 1 τ τ/ τ/ F (t) e i nω 0t dt gives the special solution x(t) = 1 A n m (ωr n ω0 ) + i nβω 0 n= Greens function e i nω 0t The Greens function is the response to F (t) = δ(t t ) G(t; t ) = { 1 ω 1 m e β(t t ) sin ω 1 (t t ) for t > t 0 for t < t If F (t) = dt δ(t t )F (t ) then x(t) = dt G(t; t )F (t )
January 17, 014 Advanced Analytical Mechanics-5 - Summary.4 - Variational Calculus Brachystochrone & Soapfilms & Geodesic A = da where da = πxds 1 ds = dx + dy = dx 1 + Y = dy 1 + X Lagrange multiplier λ(x) with G(Y 1,, Y N ; x) = 0 F d Y i dx Alternative form d dx F Y i [ F Y F Y ] + λ(x) G Y i = 0 = F x i = 1,, N
January 17, 014 Advanced Analytical Mechanics-6 - Summary.5 - Hamilton s principle of least action Fully equivalent to Newtonian Dynamics S 1, = Lagrangian t t 1 L(x, ẋ; t) dt L(x, ẋ; t) = T (ẋ) V (x) Constraint g(x i, t) = 0 gives L q i d dt L + λ(t) g = 0 with g(q i ; t) = 0 q i q i where force exerted by the constraint is Q i = λ(t) g q i Hamiltonian eq. of motion, momentum p i = L q i H(q, p, t) def = p i q i L(q, q, t) q i = H p i ; Poisson Brackets ṗ i = H q i ; {F, G} = F q n G p n F p n G q n L t = H t = dh dt
January 17, 014 Advanced Analytical Mechanics-7 - Summary.6 - Conservation Laws If L t = 0 then de dt = 0 with E = q i p i L = T + V total energy is conserved L invariant under translation r r + δ r then d dt N p a = 0 a=1 total linear momentum is constant of motion L invariant under rotation δ r = δ φ r d dt a M N a=1 = 0 ; M = r p total angular momentum is Constant Of Motion
January 17, 014 Advanced Analytical Mechanics-8 - Summary.7 - Central potential L = 1 µ(ṙ + r θ ) V (r) L cyclic in θ, conjugate momentum is constant of motion p θ = L/ θ = µr θ = Jz A) Conservation of energy: E = 1 µṙ + 1 l µr +V (r) t t 0 = r r 0 µ (E V (r )) l dr µ r B) find θ(r): dθ = θ θ(r) = ± ṙ dr gives r l/µr dr µ (E V (r )) l µ r C) Euler-Lagrange: µ( r r θ ) = V r d 1 dθ r + 1 r = µr l F (r) = F (r) Small amplitude expansion Stability
January 17, 014 Advanced Analytical Mechanics-9 - Summary.8 - Kepler s problem; Planetary motion F = k r ˆr or V = k/r with k > 0 α r = 1 + ɛ cos θ with α = l /µk and eccentricity ɛ = 1 + Eα/k Kepler s first law: ɛ = 0 E = k α circle ɛ < 1 E < 0 ellipse ɛ = 1 E = 0 parabola ɛ > 1 E > 0 hyperbola Interplanetary travel & stability circular orbit Effective potential Absorb parts of kinetic energy in an Effective Potential
January 17, 014 Advanced Analytical Mechanics-10 - Summary.9 - Non inertial frame Inertial frame ê i, rotating frame, ê i, same origine Vectors r = (r 1, r, r 3 ) and r = (r 1, r, r 3) point to point p = (r 1ê 1, r ê, r 3ê 3) = (r 1 ê 1, r ê, r 3 ê 3 ) v = v + ω r a = a + ω v + ω r + ω ( ω r ) Thus F = F inert m ω v m ω r m ω ( ω r ) Deviation of falling mass from plumb line F = F m g = m ω v Foucault pendulum: Solution x = A x (t)e iω 0t, y = A y (t)e iω 0t with ω 0 = g l A x = A sin ω r t & A y = A cos ω r t with ω r = ω E sin λ
January 17, 014 Advanced Analytical Mechanics-11 - Summary.10 - T = 1M V + T r with T r = 1 I ij = [ m α δ ij ( α k Inertial Tensor i,j I ijω i ω j and ] rα,k) r α,i r α,j Integral from: I ij = ) ρ( r) (δ ij r r i r j Steiner parallel axis theorem: I O ij = Icm ij d 3 r ] + M [δ ij a a i a j L i = j I ij ω j Coordinate Transformation: r i = j λ ijr j and r i = j λ ij r j with λ 1 = ê ê 1 = r r and λ ij = ê i ê j = λ ji thus r r = { λ} = {λ} 1 = {λ} T = [ r r ] T L = {λ} L and {I } = {λ}{i}{ λ} Euler angles r = λ ψ λ θ λ φ r = λ r
January 17, 014 Advanced Analytical Mechanics-1 - Summary.11 - Equation of motion L = N Rotating Body in Body-fixed (rotating) system: {I} ω + ω L = N Use principal axes I 1 ω 1 + (I 3 I )ω ω 3 = N 1 I ω + (I 1 I 3 )ω 3 ω 1 = N I 3 ω 3 + (I I 1 )ω 1 ω = N 3 Euler s equations for Rigid Body Symmetric top, N=0 ω 1 (t) = A cos(ωt + φ), ω (t) = A sin(ωt + φ) with Ω = ω 3 (I 3 I 1 )/I 1 Symmetric top, tip fixed -I L = 1 I 1( φ sin θ + θ ) + 1 I 3( φ cos θ + ψ) Mgh cos θ Euler-Lagrange eq in φ: p φ = L = constant φ Euler-Lagrange eq in ψ: p ψ = L = constant ψ
January 17, 014 Advanced Analytical Mechanics-13 - Summary.1 - The Euler-Lagrange eqn L ψ d dt Continuous string L ψ d dx String: Lagrangian density L ψ = 0 L(ψ, ψ, ψ ) = ρ ψ τ ψ resulting finally in the wave equation ρ d dt ψ τ d dx ψ = 0 with solutions (k/ω) = ρ/τ ψ(x, t) = dk f(k) e i(kx ωt) Proca Lagrangian: L(ψ, ψ, ψ ) = ρ ψ τ ψ µ ψ with ψ(x, t) = e i(kx ωt) we get dispersion µ + ρω τk = 0
January 17, 014 Advanced Analytical Mechanics-14 - EDy.1 - Particle in Electromagnetic Field choice coordinates: particle at position a(t) EM field given by A µ (x) or F µν (x) S = Ldt = Ld 4 x with L = c Ld 3 x L is L-invariant L = mc δ3 ( x a) γ 1 c Aµ (x)j µ (x) 1 16πc F µν F µν Euler-Lagrange equation for fields gives Maxwell µ F µν = 4π c jν Lagrangian for particle eom: L( a, a; t) = mc 1 a /c e [ φ(a) A(a) a/c ] d ( e γm v + A dt c ) + e φ a e c A v a = 0
Traveling Light.