Vibrations & Oscillations

Chapter 3 Vibrations & Oscillations The topic of vibrations and oscillations is typically discussed in some detail in a course on waves (at MIT this is 8.03). Our goal for this chapter is to revisit aspects of oscillation phenomena using generalized coordinates. Many equations of motion we have encountered have been nonlinear. Here, we will expand about a minimum of the potential V (q1 , . . . , qn ), yielding linear equations. Let us take qi = q0i + ηi , where ~q0 minimizes V (q), and expand in the ηi . Henceforth and until further notice, repeated indices will implicitly be summed over. Then V (q1 , . . . , qn ) = V (q01 , . . . , q0n ) + ∂V ∂qi ηi + 0 1 ∂ 2V 2 ∂qi ∂qj ηi ηj + . . . , (3.1) 0 where |0 means “evaluate the quantity at ~q0 ”. We already know that ∂V ∂qi = 0 as by 0 definition ~q0 minimizes V (q). As a matter of convention, we choose V (q0 ) = 0, since this just corresponds to picking the convention for the zero of the Energy. Finally, we define the 2V constants Vij ≡ ∂q∂i ∂q . There is no time dependence in the definition of our generalized j 0 coordinates, so the kinetic energy is 1 1 T = mij (q1 , . . . , qn )q̇i q̇j = mij (q01 , . . . , q0n )η̇i η̇j + O(ηη̇ 2 ), 2 2 (3.2) where mij (q01 , . . . , q0n ) ≡ Tij are constants, and terms of O(ηη̇ 2 ) and beyond are neglected. Thus, the Lagrangian to quadratic order in the ηi s is 1 (Tij η̇i η̇j − Vij ηi ηj ) . 2 From this, the equations of motion are L= Tij η̈j + Vij ηj = 0 to the same order. These are coupled linear equations of motion. (3.3) (3.4) 3.1 Simultaneous Diagonalization of Tˆ and Vˆ To solve Eq. (3.4) lets try ηi = ai e−iωt (3.5) where ai ∈ C for all i ∈ {1, . . . , n}, and the frequency ω is the same along all directions in the generalized coordinate space. Notationally, i and j will denote coordinate indices, while √ i = + −1 is the imaginary unit. This gives Vij aj = ω 2 Tij aj (3.6) which can be rewritten in matrix form as V̂ · ~a = λ Tˆ · ~a (3.7) with λ = ω 2 . This looks like an eigenvalue equation except that when we act with the linear operator Vˆ on ~a we get back Tˆ ·~a instead of just the eigenvector ~a. This can be rewritten as (Vˆ − λTˆ) · ~a = 0 (3.8) where Vˆ and Tˆ are real and symmetric n×n matrices. In order to have a non-trivial solution of this equation we need det(Vˆ − λTˆ) = 0 (3.9) which is an nth order polynomial equation with n solutions eigenvalues λα with α ∈ {1, . . . , n}. The solutions of (Vˆ − λα Tˆ) · ~a(α) = 0 are the eigenvectors ~a(α) . This means V̂ · ~a(α) = λα Tˆ · ~a(α) , (3.10) and the solutions are much like a standard eigenvalue problem. Here and henceforth, there will be no implicit sum over repeated eigenvalue indices α (so any sums that are needed will be made explicit), but we will retain implicit sums over repeated coordinate indices i & j. There are two cases we will consider. 1) Let us start by considering the case when Tˆ is diagonal. In particular, let us consider the even easier case proportional to the unit matrix, where Tij = mδij . This means mη̈i + Vij ηj = 0 . (3.11) Here we have the standard eigenvalue equation V̂ · ~a(α) = mλα~a(α) . (3.12) The eigenvalues λα are real and nonnegative as λα = ωα2 ; the quantities ωα are the normal mode frequencies. The eigenvectors ~a(α) are orthogonal, and we can choose their normalization so that m ~a(β)† · ~a(α) = δβα (or ~a(β)† · ~a(α) = δβα ) . (3.13) This implies that λα = ~a(α)† · Vˆ · ~a(α) (or mλα = ~a(α)† · Vˆ · ~a(α) ) . (3.14) The time-dependent eigenvectors are then (α) ~η (α) = ~a(α) e−iωα t , or ηi (α) = ai e−iωα t . (3.15) These are the normal mode solutions for the n coordinates labeled by i, and there are n such solutions labeled by α. The general solution of a linear equation is a superposition of the independent normal mode solutions: X ~η = Cα ~η (α) (3.16) α where Cα ∈ C are fixed by initial conditions. To find real coordinate solutions, we take the real parts of these equations. Lets prove the statements made above. Again, there will be no implicit sum over the eigenvalue index α. Dotting in ~a(β)† into Eq. (3.12) gives ~a(β)† · V̂ · ~a(α) = mλα~a(β)† · ~a(α) , (3.17) taking the Hermitian conjugate of both sides, noting that Vˆ † = Vˆ , and then swapping α ↔ β gives ~a(β)† · Vˆ · ~a(α) = mλ?β~a(β)† · ~a(α) . Taking the difference of these results gives (λα − λ?β )~a(β)† · ~a(α) = 0, (3.18) and if α = β then (λα − λα? )~a(α)† · ~a(α) = 0 implies the eigenvalues are real λα ∈ R. For λα 6= λβ , Eq. (3.18) then implies ~a(β)† · ~a(α) = 0 so the eigenvectors are orthogonal. If by chance λα = λβ for some α = 6 β then we can always simply choose the corresponding eigenvectors to be orthogonal. By convention, we then normalize the eigenvectors so that they satisfy Eq. (3.13). Finally, if α = β then Eq. (3.17) now gives λα = ~a(α)† · Vˆ · ~a(α) . The statement that we are at a local minimum of the multivariable potential and not a saddle point or a maximum implies then that λα ≥ 0 (we have positive second derivatives in each eigenvector direction). 2) Let us now consider when Tˆ is not diagonal and summarize which parts of the result are the same and where there are differences. Here we have (Vˆ − λTˆ) · ~a = 0. Again, the eigenvalues λα are real and nonnegative, with λα = ωα2 . Now, however, ~a(β)† · T̂ · ~a(α) = 0 , (3.19) for α 6= β, and we can replace the old normalization condition by a new one stating that ~a(β)† · T̂ · ~a(α) = δαβ , (3.20) which up to an overall prefactor reduces to the old orthonormality condition when Tˆ = m1̂. Here again, λα = ~a(α)† · V̂ · ~a(α) , (3.21) and the αth normal mode solution is ~η (α) = ~a(α) e−iωα t . The general solution is again the superposition X ~η = Cα ~η (α) , (3.22) (3.23) α with the complex coefficients Cα fixed by the initial conditions (and a real part taken to get real coordinates). Lets repeat the steps of our proof for this case. Dotting ~a(β)† into Eq. (3.10) gives ~a(β)† · V̂ · ~a(α) = λα~a(β)† · T̂ · ~a(α) . (3.24) Taking the Hermitian conjugate of both sides yields ~a(β)† · V̂ · ~a(α) = λ?β~a(β)† · Tˆ · ~a(α) . Subtracting the two results this gives (λα − λ?β ) ~a(β)† · T̂ · ~a(α) = 0, (3.25) and if α = β then (λα − λα? )~a(α)† · Tˆ · ~a(α) = 0 implies λα ∈ R since Tˆ = Tˆ? = Tˆ> and physically we know that the kinetic energy T = ~η˙ · Tˆ · ~η˙ > 0 for any ~η˙ 6= 0. For λα 6= λβ , then the condition instead implies ~a(β)† · T̂ · ~a(α) = 0 so the eigenvectors are orthogonal; if by chance λα = λβ for some α 6= β then we can choose the corresponding eigenvectors to be orthogonal. By convention, we normalize the eigenvectors so that they will be orthonormal as in Eq. (3.20). Finally, if α = β then λα = ~a(α)† · Vˆ · ~a(α) , which is positive, so λα > 0 also. The statement that we are at a local minimum of the potential and not a saddle point or a maximum implies then that λα ≥ 0. 3.2 Vibrations and Oscillations with Normal Coordinates Given these results, it is natural to ask whether a different set of generalized coordinates might be better for studying motion about the minimum? We form the matrix A by placing the eigenvectors in columns   A = ~a(1) ~a(2) . . . ~a(n) (3.26) ˆ = diag(λ1 , λ2 , . . . , λn ). The matrix A can be and construct a diagonal eigenvalue matrix λ (α) ensured to be real because each ~a only has at most an overall phase1 , and these can be removed by putting them into the coefficients Cα . The matrix A simultaneously diagonalizes Tˆ and Vˆ since ˆ A> TˆA = 1 and A> Vˆ A = λ. (3.27) We choose new normal coordinates ξ~ by letting ˙ ~η = A ξ~ and ~η˙ = A ξ~ (3.28) so that the Lagrangian 1 1 L = ~η˙ · T̂ · ~η˙ − ~η · Vˆ · ~η 2 2 1 ~˙ ˙ 1 = ξ · (A> T̂ A) · ξ~ − ξ~ · (A> V̂ A) · ξ~ 2 2  1 X  ˙2 = ξα − ωα2 ξα2 . 2 α (3.29) This gives the simple equations of motion for each α: ξ¨α + ωα2 ξα = 0 . (3.30) Thus, each normal coordinate describes the oscillations of the system with normal mode frequency ωα . Example: Let us consider the triatomic molecule CO2 shown in Figure 3.1. We can picture it as a carbon atom of mass M in the middle of two oxygen atoms each of mass m. For the three particles there are 9 coordinates given by r1 , r2 , and r3 . Six of these coordinates correspond to translations and rotations of the mass system treated as a rigid body. This leaves 3 coordinates that correspond to internal motions of the system. To model the potential we connect each oxygen atom to the carbon atom with a spring of constant k and relaxed length b. This does not add any cost to relative motion of the atoms with fixed spring length, which we will address below by adding another potential term in order to favor the linear configuration. It is straightforward to guess what the normal modes could be: Why is it just an overall phase? The equation Vˆ · ~a(α) = λα Tˆ · ~a(α) alone does not fix the normalization (α) (α) (α) of ~a . Let us say we pick ai ∈ R for some i. Then Vkj aj = λα Kkj aj is a set of equations with all real 1 (α) (α) (α) coefficients and one real term in the sums. Hence the solutions aj /ai that at most there is an overall phase in ~a(α) . ∈ R for all j ∈ {1, . . . , n}, implying Figure 3.1: The CO2 molecule. • The oxygen atoms moving in the same direction along the line and the carbon atom moving in the opposite direction. This is a longitudinal oscillation. • The oxygen atoms opposing each other along the line while the carbon atom remains at rest. This is a longitudinal oscillation. • The oxygen atoms move in the same direction perpendicular to the line and the carbon atom moving in the opposite direction. This is a transverse oscillation. These three normal modes are shown in Figure 3.2. Figure 3.2: The three Normal Modes of the CO2 molecule We pick the body frame axes as follows: • the three particles are in the xy-plane fixing 3 coordinates zi = 0 for i ∈ {1, 2, 3}, • the origin is the CM so m(x1 + x3 ) + M x2 = m(y1 + y3 ) + M y2 = 0, which fixes two more coordinates, • the axes are oriented so that y1 = y3 , which fixes one coordinate. m Defining the mass ratio as ρ ≡ M , then x2 = −ρ(x1 + x3 ) and y2 = −2ρy1 can be eliminated. Altogether this fixes 6 coordinates, leaving the coordinates (x1 , x3 , y1 ). This setup is shown in Figure 3.3. For the potential we take V = k λb2 2 k (s1 − b)2 + (s2 − b)2 + (α1 + α22 ) . 2 2 2 (3.31) Figure 3.3: The orientation of the CO2 molecule on xy plane The first two terms are the springs discussed previously, and the last two provide a quadratic energy cost to the springs rotating away from the linear configuration, with strength given by λ. The spring lengths are q p 2 2 (3.32) s1 = (x1 − x2 ) + (y1 − y2 ) = [x1 + ρ(x1 + x3 )]2 + (1 + 2ρ)2 y12 q p s2 = (x2 − x3 )2 + (y2 − y3 )2 = [x3 + ρ(x1 + x3 )]2 + (1 + 2ρ)2 y12 , and the two angles are   y −y  (1 + 2ρ)y1 3 2 −1 , α1 = tan = tan (1 + ρ)x1 + x3 x3 − x 2     (1 + 2ρ)y1 −1 y2 − y1 −1 α2 = tan = tan . x2 − x 1 (1 + ρ)x3 + x1 −1 These results give V = V (x1 , x3 , y1 ). [to be continued] (3.33) For the kinetic energy we have
M
M m 2 m 2 ẋ1 + ẋ23 + ẋ22 + ẏ1 + ẏ32 + ẏ22 , (3.34) 2 2 2 2 which after eliminating coordinates becomes
m T = (1 + ρ) ẋ21 + ẋ32 + mρẋ1 ẋ3 + m (1 + 2ρ) ẏ12 . (3.35) 2 Equilibrium comes from taking y1 = 0, x3 = −x1 = b, which implies α1 = α2 = 0, s1 = s2 = b, and V = 0. We define coordinates for expanding about this potential minimum ∂V =0 as η1 = x1 + b, η3 = x3 − b, and η2 = y1 . Then as V (−b, b, 0) = 0 in equilibrium and ∂η i T = then V = 12 Vij ηi ηj + . . . where   k(1 + 2ρ + 2ρ2 ) 0 2kρ(1 + ρ) 2 ∂ V  0 2λ(1 + 2ρ)2 0 = Vij = ∂ηi ∂ηj 0 2 2kρ(1 + ρ) 0 k(1 + 2ρ + 2ρ ) 0 (3.36) for this system. Additionally,   m(1 + ρ) 0 mρ  0 2m(1 + 2ρ) 0 Tij =  mρ 0 m(1 + ρ) (3.37) for this system. Since there are no off-diagonal terms in the 2nd row or 2nd column in either Vˆ or Tˆ, the transverse and the longitudinal modes decouple. For the transverse mode, we are left with 2λ(1 + 2ρ)2 y¨1 + y1 = 0 , (3.38) 2m(1 + 2ρ) which is a simple harmonic oscillator. For the longitudinal modes, we have ~η = (η1 , η3 ). The frequencies come from   k(1 + 2ρ + 2ρ2 ) − λm(1 + ρ) 2kρ(1 + ρ) − λmρ det =0 (3.39) 2kρ(1 + ρ) − λmρ k(1 + 2ρ + 2ρ2 ) − λm(1 + ρ) The solutions give the normal mode frequencies λ1 = ω12 = k , m with associated eigenvectors   1 1 (1) ~a = √ , 2m −1 λ2 = ω22 = (2) ~a k (1 + 2ρ) , m =p 1 2m(1 + 2ρ)  (3.40) 1 1  , (3.41) which were chosen to satisfy ~a(α) · T̂ · ~a(β) = δαβ . Thus, the normal coordinates for the longitudinal modes are ξ1 ∝ x1 − x3 and ξ2 ∝ x1 + x3 . Oscillations in these coordinates correspond to the normal mode motions in Fig. 3.2(b) and Fig. 3.2(a) respectively.