Analytical Mechanics ( AM )

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Ἑστία Αργυριάδης
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1 Analytical Mechanics ( AM ) lecture notes part 10, Summary Olaf Scholten KVI, kamer v3.008 tel. nr scholten@kvi.nl Web page: Book Classical Dynamics of Particles and Systems, Stephen T. Thornton & Jerry B. Marion 5th Edition ISBN-10: ISBN-13: 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,...

2 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) pendulum Tue Mar :19: Tune up Tune down x 3 oscil Frequency Initial conditions at slowly tune up or down frequency

3 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)/ amplitude Phase x Poincaré time [0.01 sec] NL-pendulum Wed Apr :00:0

4 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 )

5 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

6 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

7 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

8 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

9 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

10 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

11 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

12 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

13 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

14 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

15 Traveling Light.

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