Chapter 4 Small oscillations and normal modes 4.1 Linear oscillations Discuss a generalization of the harmonic oscillator problem: oscillations of a system of several degrees of freedom near the position of equilibrium remember for s = 1 L = 1 Mq q Vq, T > 0 4.1 } {{} T q 0 minimum of the potential energy, x = q q 0 displacement expand Vq and Mq Vq = Vq 0 + dv }{{} dq q q 0 + 1 d V q=q0 dq q q 0 +... q=q0 const }{{} =0 }{{} k>0 Vq = const+ 1 kx +Ox 3 if k = 0 non-linear oscillation, higher derivatives important Mq = Mq 0 +Ox = m+ox restrict to orders Ox,Oẋ in general also Oxẋ in L for linear oscillations L = 1 mẋ 1 kx procedure is called: linearization of the original Lagrangian 4.1 results: linear differential equation with constant coefficients mẍ+kx = 0 6
ansatz for the solution x = Acosωt+ϕ algebraic equation mω +k = 0 ω = k m m > 0,k > 0: ω > 0 A amplitude, ω frequency, ωt+ϕ phase, ϕ initial phase now consider s degrees of freedom L = T Vq 1,...q s, T = 1 M ij q q i q j since q i q j is symmetric relative to i j, the coefficients M ij can be chosen symmetric M ij q = M ji q denote by q i0,i = 1,...,s the point of the minimum for the potential energy V expand V relative to x i = q i q i0 Vq = const+ 1 k ij x i x j +Ox 3 k, k ij = V q i q j = k ji qi0 i,j for x i 0 the potential energy increases with respect to its minimum at x i = 0 quadratic form k ij x i x j 0 expand M ij q i,j i,j M ij q = m ij +Ox k, m ij = M ij q k0 = m ji since T > 0 we have another quadratic form m ij ẋ i ẋ j 0 again restrict ourselves to Ox i x j,oẋ i ẋ j i,j L = 1 m ij ẋ i ẋ j k ij x i x j i,j 63
Introduce a matrix notation column vector x = x 1. x s and the transposed vector T denotes transposition as a raw vector mass matrix ˆm and matrix of elasticity ˆk m 11... m 1s ˆm =... m s1... m ss introduce the scalar product x T = x 1,..., x s ˆk = k 11... k 1s... k s1... k ss x,y x T y = s x i y i i=1 form of the Lagrangian with ẋ, ˆmẋ 0 and x, ˆkx 0 ˆk and ˆm are symmetric matrices: equation of motion L = 1 ẋ, ˆmẋ 1 x, ˆkx ˆm T = ˆm, ˆkT = ˆk ansatz for the solution d L dt ẋ = L x ˆmẍ+ˆkx = 0 x = Acosωt+ϕ non-trivial solutions for ω ˆm+ˆkA = 0 eigenvalue equation det ω ˆm+ˆk = 0 characteristic equation 64
assume ω 1,...,ω s eigenvalues solutions of the characteristic equation put the eigenvalues ω α into the eigenvalue equation, find the eigenvectors A α if A α is a solution, then aa α is also a solution introduce the normal mode vector x α t ω α ˆm+ˆkA α = 0 4. x α t = A α Q α t, Q α t = a α cosω α t+ϕ α Q α normal coordinate, ω α normal frequency a α arbitrary amplitude, ϕ α arbitrary phase the complete solution in components xt = s x α t = α=1 x i t = s α=1 s A α Q α t α=1 A α i Q α t the solution contains s arbitrary amplitudes a α and phases ϕ α which have to be found from the initial conditions x0 and ẋ0 In the language of vector algebra the transformation from vector x to Q is a linear transformation x = Û Q, x i = α U iα Q α and U iα A α i with x = x 1. x s, Q = Q 1. Q s, Û = A 1 1... A s 1. A 1 s..... A s s for that transformation both quadratic forms for T and V become diagonal: s L = L α, L α = 1 M Q α α 1 K αq α 4.3 α=1 M α = A α, ˆmA α, K α = A α, ˆkA α 65
the Lagrangian has the form of s non-interacting oscillations we get the one-dimensional equations of motion for s independent normal coordinates Q α Note that normal frequencies are real: M α Qα +K α Q α = 0, α = 1,...,s ω α = K α M α = Aα, ˆmA α A α, ˆkA α 0 So, each normal coordinate Q α corresponds to an oscillation with one frequency ω α : Q α +ω αq α = 0 the oscillation is called normal mode Thesolutionintheoriginalvariablesx i isalinearsuperpositionofoscillationswithdifferent frequencies 66
4. Orthogonality of normal modes Assume two eigenvectors A α,a β are given with ω α ω β the corresponding normal mode vectors are x α = A α Q α and x β = A β Q β there is no reason that the scalar products of two eigenvectors or normal mode vectors have to vanish: A β, A α 0, x β, x α 0 Consider the eigenvector equations [see 4.] ω α,β ˆmA α,β = ˆkA α,β multiply from the left with A β,α ωα ωβ A β, ˆmA α = A α, ˆmA β = A β, ˆkA α A α, ˆkA β 4.4 and subtract A β, ˆmA α ωβ ω α A α, ˆmA β = A β, ˆkA α A α, ˆkA β it is known that for each real matrix ˆn: A,ˆnB = ˆn T A,B in our case ˆk and ˆm are real and symmetric ˆk T = ˆk, ˆm T = ˆm using A,B = B,A the difference vanishes ωα A β, ˆmA α ωβ ˆmA α, A β = A β, ˆkA α ˆkA α, A β 0 we get ωα ωβ A β, ˆmA α = 0 Therefore: A β, ˆmA α = 0 and [see 4.4] A β, ˆkA α = 0 67
Conclusions: The normal mode vectors x α and x β with ω α ω β are orthogonal to each other, if their scalar product is defined using the so called metric tensors ˆm or ˆk x α and x β are orthogonal in the metric of mass or elasticity x β, ˆmx α = 0, x β, ˆkx α = 0 In case of degenerate frequencies: e.g. the normal mode vectors x 1 and x with ω 1 = ω the linear combination ax 1 +bx is also a solution with the same frequency the space of solution is the plane given by the vectors x 1 and x choose a pair of independent vectors which satisfy the orthogonality condition in the mass metric togetherwithallothernormalmodevectorswithnon-degeneratefrequenciestheybuildthe basisforthenormalcoordinatesleadingtothediagonallagrangianofdecoupledoscillators 4.3 68
4.3 Thedouble pendulumin the fieldof constant gravity an example s =, use as generalized coordinates θ 1 and θ for the notations, see the figure Cartesian coordinates and the generalized coordinates are related via x 1 = l 1 sinθ 1, x = l 1 sinθ 1 +l sinθ z 1 = l 1 cosθ 1, z = l 1 cosθ 1 +l cosθ the Lagrangian can be found as follows see Landau/Lifshitz, 5, problem 1 L = 1 m 1 +m l1 θ 1 + 1 m l θ +m l 1 l θ1 θ cosθ 1 θ +m 1 +m gl 1 cosθ 1 +m gl cosθ Consider the special case m 1 = m = m, l 1 = l, l = l L = 4ml θ 1 + 1 ml θ +ml θ1 θ cosθ 1 θ +4mgl cosθ 1 +mgl cosθ linearize the Lagrangian θ i 1 L = 1 ml 8 θ 1 +4 θ 1 θ + θ 1 mgl4θ 1 +θ identify the matrices of mass and elasticity 8 4 0 ˆm = ml, ˆk = mgl 1 0 1 the equations of motion 8 θ 1 + θ +4ω 0θ 1 = 0, θ 1 + θ +ω 0θ = 0, ω 0 = 69 g l
Search the solution in the form θ1 x = = A cosωt+ϕ = θ A1 A cosωt+ϕ the eigenvalue equation ω ˆm+ˆkA = 0 [ 8 ml ω +mgl 1 4 0 0 1 ] A1 A = 0 divide the matrix equation by ml and introduce ω0 = g l [ ] 8 4 0 ω +ω0 A1 1 0 1 A = 0 8ω +4ω 0 ω ω ω +ω 0 A1 A = 0 solve the characteristic equation 8ω det +4ω0 ω ω ω +ω0 = 0 ω 1, = 3 5 ω 0 determine the eigenvectors A 1 and A from 8ω i +4ω 0 ω i ω i ω i +ω 0 A i 1 A i = 0 the A i are defined up to a normalization for A 1 we get the two equations 8ω 1 +4ω 0A 1 1 ω 1A 1 = 0, ω 1A 1 1 + ω 1 +ω 0A 1 = 0 from the second equation we obtain same result from the first equation choose A 1 1 = 1 A 1 = the normal mode vectors x 1, = A 1, Q 1, = A 1 = ω 1 A 1 ω0 ω1 1 = 5 1A 1 1 1 1, analogously A = 5 1 5+1 ±1 ±1 a 5 1 1, cosω 1, t+ϕ 1, = Q 5 1 1, 70
complete solution x = x 1 +x Q = 1 Q 5 1Q 1 + 5+1Q with the angles θ 1 t = a 1 cosω 1 t+ϕ 1 a cosω t+ϕ θ1 θ t = 5 1a 1 cosω 1 t+ϕ 1 + 5+1a cosω t+ϕ the constants a 1,a,ϕ 1,ϕ are found from the initial conditions Let us also check that the Lagrangian becomes diagonal using the normal coordinates Q α 1 L = M Q α α 1 K αq α and ωα = K α M α we find identify α=1 mgl4θ1 +θ = [ 5 5 1Q 1 + ] 5+1Q ml 8 θ 1 +4 θ 1 θ + θ = [ 5 5+1 Q 1 + 5 1 Q ] θ mgl ml M 1, = 5 5±1ml, K 1, = 5 5 1mgl 5 1 ω1, g = 5±1 l = 5 1 g 4 l = 3 5 ω0 M 1, and K 1, can be found also from M α = A α, ˆmA α, K α = A α, ˆkA α In a plane with orthogonal axes θ 1,θ the eigenvectors A 1 and A give the directions of the new axes of normal coordinates Q 1 and Q the axes Q 1 and Q are not orthogonal to each other A 1,A = 1, 1 5 1 = 3 0 5+1 but the vectors A 1 and A are orthogonal in the metric of mass or elasticity A 1, ˆk mgl A = 1, 4 0 5 1 0 1 1 = 1, 4 5 1 = 0 5+1 5+1 71