2 Sect. 6.3: Free Vibration Frequencies & Normal Coordinates We just saw, for a system displaced slightly from equilib:Eqtns of motion: Tijηj + Vijηj = 0 (i = 1,2,3, … n) (1)are satisfied by oscillatory solutions: ηj Caje-iωtIn general: n eigenfrequencies ωk (n allowed oscillation frequencies). (1) is a linear differential equation. The complete solution of (1) is a linear combination (superposition) of oscillations at all allowed frequencies. A system displaced slightly from equilib performs small oscillations about equilib with frequencies:ω1,ω2, ... ωn. Frequencies of “free” vibration Resonant or natural frequencies of the system
3 Tijηj + Vijηj = 0 (i = 1,2,3, … n) (1a) Equations of motion:Tijηj + Vijηj = 0 (i = 1,2,3, … n) (1a)Or: (Vijaj - ω2Tijaj ) = 0 (i = 1,2,3, … n) (1b)General solution (summation convention):ηj(t) = Ckajkexp(-iωkt) (2)ajk eigenvector elements (make up columns & rows of “eigenvector matrix” A). Ck a scale factor (complex).(1b) & (2) For each eigenvalue λk = (ωk)2 Seems as if there may be 2 resonant frequencies: ωk. But recall, the actual motion is the real part of (2). Text (p. 250) goes into great detail about how this means that there is only one physical resonant frequency.
4 General solution (summation convention): ηj(t) = Ckajkexp(-iωkt) (2)ajk eigenvector elements (make up columns & rows of “eigenvector matrix” A). Ck a scale factor (complex).Ck: Determined by the initial conditions. (As the solution to 2nd order differential equations, need 2 initial conditions to specify the solution). Text (p ) derives relations between the complex factors Ck & the initial generalized displacements ηj(0) & velocities ηj(0):ηj(0) = ∑kRe(Ckajk) ηj(0) = ∑kIm(Ckajkωk)andRe(Ci) = ∑(j,k)ajiTjkηk(0) Im(Ci) = (ωi)-1∑(j,k)ajiTjkηk(0)
5 General solution (summation convention): ηj(t) = Ckajkexp(-iωkt) (2)= sum of simple harmonic oscillations at all frequencies ωk satisfying the eqtn of motion (secular eqtn). Eqtns of motion for the ηj = coupled simple harmonic oscillator eqtnsIt is possible, & quite useful, to transform from the generalized coords ηj(t) to a new set of generalized coords ζj(t) called the Normal Coordinates, in which each ζj oscillates separately at only one of the resonant frequencies ωj. Equations of motion for the ζj = uncoupled simple harmonic oscillator eqtns. That is, each ζj satisfies a simple harmonic oscillator equation of motion at a different ωj & there is no coupling between them!
6 T = (½)ζiζi V = (½)(ωi)2ζiζi Normal Coordinates, ζj(t): Text (p. 251) shows that these coords are obtained from the ηj by:η = Aζ or ζ = A-1η = Ãηwhere A = matrix of the eigenvectors & η,ζ = column vectors of the generalized displacements.Text (p ) shows: Both the KE T & the PE V are diagonal in the Normal Coordinate representation. That is, a similarity transformation with A diagonalizes both T & V (summation convention):T = (½)ζiζi V = (½)(ωi)2ζiζiNote: with this definition, factors of the masses are included in thedefinition of ζi, so the mass doesn’t appear explicitly in either T or V!
7 In terms of Normal Coordinates: T = (½)ζkζk V = (½)(ωk)2ζkζk The Lagrangian is:L = T - V = (½)[ζkζk - (ωk)2 ζkζk] Lagrange’s eqtns of motion:(d/dt)[(L/ζk)] - (L/ζk) = (k = 1,2,3, … n)Give: ζk + (ωk)2ζk = (k = 1,2,3, … n)With solutions: ζk = Ckexp(-iωkt)n uncoupled SHO equations! Each Normal Coordinate is a sinusoidal function of only one of the resonant frequencies!
8 In terms of Normal Coordinates: L = T - V = (½)[ζkζk - (ωk)2 ζkζk] Eqtns of motion: ζk + (ωk)2ζk = (k = 1,2,3, … n)With solutions: ζk = Ckexp(-iωkt)Each Normal Coordinate corresponds to a vibration of the entire system with only one frequency.It’s common to speak of Normal Modes of Vibration.In one of these normal modes, each particle in the entire system is vibrating with the same frequency.Further, they are either vibrating in phase (if the correspondingeigenvectors have the same sign) or π out of phase (if the eigenvectorshave opposite signs).Relative amplitudes of particle displacements are given by appropriate ajk.NORMAL MODES:Collective properties of the system, not of any one particle!
9 Sect. 6.4: Free Vibrations of a Linear Triatomic Molecule Detailed example of this formalism: Normal modes of vibration for a linear, triatomic molecule.The figure shows the equilibrium configuration of a molecule:Symmetric! Atomic masses: m1 = m3 = m, m2 = M. Generalized coords x1, x2, x3. Equilib distances: x02 - x01 = b = x03 - x02. A very simple model of the interatomic potential between atoms: Assume ideal springs of spring constant k. Consider only longitudinal vibrations (displacements in the x direction).The PE is: V = (½)k(x2 - x1 -b)2 + (½)k(x3 - x2 -b)2
10 Or: V = (½)k[(η1)2 + 2(η2)2 +(η3)2 - 2η1η2 - 2η2η3] V = (½)k(x2 - x1 -b)2 + (½)k(x3 - x2 -b)2Displacement coordinates:ηj xj - x0j Note again: x02 - x01 = b = x03 - x02 V = (½)k(η2 - η1)2 + (½)k(η3 - η2)2Or: V = (½)k[(η1)2 + 2(η2)2 +(η3)2 - 2η1η2 - 2η2η3]k -k 0Tensor or matrix V = -k 2k -k0 -k kThe KE: T = (½) m[(x1)2 + (x3)2] + (½)M(x2)2Or: T = (½)m[(η1)2 + (η3)2] + (½)M(η2)2mTensor or matrix T = M Diagonal!m
11 (Vijaj - ω2Tijaj ) = 0 (i = 1,2,3) (1) We needn’t derive the eqtns of motion from the Lagrangian. We could instead go directly to the secular matrix eqtn:(Vijaj - ω2Tijaj ) = (i = 1,2,3) (1)Its instructive, however, to briefly derive the differential equations of motion from the Lagrangian: L = T - V =(½)m[(η1)2 + (η3)2] + (½)M(η2)2- (½)k[(η1)2 + 2(η2)2 + (η3)2 - 2η1η2 - 2η2η3]Lagrange’s Eqtns: (d/dt)[(L/ηi)] - (L/ηi) = 0 mη1 + k(η1 - η2) = 0Mη2 + k(2η2 - η1 - η3) = 0mη3 + k(η3 - η2) = 0
12 (Vijaj - ω2Tijaj ) = 0 (i = 1,2,3) (1) However, the power of our matrix (tensor) formalism is that we needn’t derive these AND we needn’t try to solve them directly. We have the general solution in terms of the matrices & the secular or eigenvalue equation: So, go directly to the secular matrix equation:(Vijaj - ω2Tijaj ) = (i = 1,2,3) (1)Using k -k mV = k 2k -k T = M 00 -k k m(1) |V - ω2T| = 0 or:k - ω2m k-k k - ω2M k = 0k k - ω2m
13 Evaluating the determinant (algebra): ω2(k - ω2m)[k(M + 2m) - ω2Mm] = 0A cubic equation for the resonant or natural frequencies (3 solutions of course!):ω1 = 0, ω2 = (k/m)½, ω3 = [(k/m)(1 + 2m/M)]½PHYSICS: ω1 = 0. At first, this may seem strange. No oscillation! The corresponding equation of motion for the normal coordinate is: ζ1 + (ω1)2ζ1 = 0 = ζ1 The normal coordinate ζ1 is actually a uniform translation of the molecule along the x axis! We can translate the molecule in this way without changing the PE. That is, this is a neutral equilibrium! We’ve assumed the molecule has 3 vibrational degrees of freedom. In reality, it has one translational degree of freedom & 2 vibrational degrees of freedom. One translational normal mode & 2 vibrational normal modes.
14 Some more points about ω1 = 0 & ζ1 uniform translation being one of the solutions. 1. Corresponds to the case where all displacements ηi are equal:η1 = η2 = η3. Corresponds to the case where V = 0.2. Since ω1 = 0 does not affect the vibrational properties, we could have done the problem from the beginning to exclude this root. This could be done by imposing the constraint that the CM of the molecule be stationary at the origin. Or:m(x1 + x3) + Mx2 = 0 Constraint eqtnCould use constraint eqtn to eliminate one of the coords xi (or one of the displacements ηi) from T & V Reducing the system to 2 degrees of freedom.
15 A general point about zero frequencies: ω = 0. One more point about ω1 = 0 & ζ1 uniform translation being one of the solutions.3. Motion restriction along molecular axis results in 1 type of uniform translational motion. For real molecules in 3d, vibrations in all 3 directions & must be considered. Will find 6 vanishing frequencies corresponding to 3 translational degrees of freedom & 3 rotational degrees of freedom :A general point about zero frequencies: ω = 0.Could also happen if both (V/qi)0 = 0 and Vij = (2V/qiqj)0 = 0 at equilibrium. In this case, it may not be possible to make the small oscillation approximation. Could do so if the 4th derivatives vanish, but then motion will not be simple harmonic!
16 ω2(k - ω2m)[k(M + 2m) - ω2Mm] = 0 Solutions: Back to resonant frequencies for the linear triatomic molecule. Solutions to: ω2(k - ω2m)[k(M + 2m) - ω2Mm] = 0Solutions:ω1 = 0, ω2 = (k/m)½, ω3 = [(k/m)(1 + 2m/M)]½ Consider ω2: From elementary physics, this is the frequency of oscillation for a single mass m suspended from a single spring of constant k. Based on this, we speculate that only end the atoms, m, are involved in the oscillations associated with this mode.Expect the normal mode displacement ζ2 for eigenvalue ω2 has end atoms moving & center atom M stationary. So, look at the eigenvectors for 3 modes.
17 ω1 = 0, ω2 = (k/m)½, ω3 = [(k/m)(1 + 2m/M)]½ To get the normal mode displacements (eigenvectors) put these solutions back into secular equation. In this case, we get 3 eqtns for each of the 3 eigenvalues ωj:[k - (ωj)2m]a1j ka2j = 0- ka1j [2k - (ωj)2M]a2j ka3j = 0ka2j [k - (ωj)2m]a3j = 0In addition, have normalization: m[(a1j)2 + (a3j)2] + M(a1j)2 = 1
18 Algebra gives the eigenvector components: For ω1 = 0: (The mode corresponding to uniform translation). a11 = a21 = a31 = (2m +M)-½. All 3 equal is what we expect for uniform translational motion. As in the figure:For ω2 = (k/m)½ : (The mode expected to correspond to the end atoms m vibrating & the center atom M not moving). Finda12 = (2m)-½ = - a32. a22 = 0. Consistent with the expectation!a22 = 0 the center atom is at rest. a12 = - a32 The 2 end atoms vibrate exactly π out of phase. As in figure:
19 Algebra gives the eigenvector components: For ω3 = [(k/m)(1 + 2m/M)]½ :a13 = a33 = [2m(1 + 2m/M)]-½ a23 = -2[2M(2 + M/m)]-½Interpretation: The outer 2 atoms vibrate with the same amplitude & phase. The inner atom vibrates out of phase with them & with a different amplitude. As in the figure:
20 Given the eigenvector components, we can compute the normal mode displacements (algebra). Use ζ = A-1η = Ãη. Get:For ω1 = 0: (Uniform translation).ζ1 = (2m +M)-½[(m)½η1 + (M)½η2 + (m)½η3].Clearly corresponds to:For ω2 = (k/m)½ : (The end atoms vibrating, the center atom not moving). ζ2 = (2)-½[η1 - η3]. Clearly corresponds to:
21 Use ζ = A-1η = Ãη. Get:For ω3 = [(k/m)(1 + 2m/M)]½ :ζ3 = (2m +M)-½[(M/2)½(η1 + η3) - (2m)½η2].Clearly corresponding to:Any general longitudinal vibration (excluding uniform translation) of the molecule involves some linear combination of the normal mode ζ2 at ω2 = (k/m)½ :ζ2 = (2)-½[η1 - η3]exp(-iω2t) & the normal modeζ3 at ω3 = [(k/m)(1 + 2m/M)]½ :ζ3 = (2m +M)-½[(M/2)½(η1 + η3) - (2m)½η2]exp(-iω3t)For example, D = C2 ζ2 + C3 ζ3 , C2, C3 complex. Relative sizes & phases determined by initial conditions.
22 This simple treatment: Considered only longitudinal (x directed) displacements. A real molecule: also has normal vibrational modes molecular axis. In general, 9 degrees of freedom 9 normal modes. Complicated, but doable algebra. The basic physics is similar to what we just went through. Instead of detailed math, we now do a qualitative physics discussion of expected results.In general, for a molecule with n atoms there are 3n degrees of freedom. The vibrational frequency eigenvalue problem will have 3n frequencies as solutions. In general, there will be 6 frequencies for which ω = 0. 3 of these correspond to the 3 translational degrees of freedom, in which the normal modes are 3 uniform translations along 3 mutually axes. Three (usually) correspond to the 3 rotational degrees of freedom in which the normal modes are 3 rigid rotations about 3 mutually axes.
23 4 vibrational frequencies & 4 (non-zero) For a linear molecule, things are a bit more complicated:There are 3 translational degrees of freedom, corresponding to uniform translations along 3 mutually axes. However, there are only 2 rotational degrees of freedom, in which the normal modes are 2 rigid rotations about the 2 mutually axes molecule itself. Rotation with the rotation axis being the molecular (x) axis itself makes no sense, since the molecule (in this approximation!) has no extent in directions x. (This is an artifact of our simple model!). So, there will only be 2 normal modes corresponding to rigid rotations. # vibrational degrees of freedom = (trans) (rot) = 4 Expect 4 true vibrational degrees of freedom. vibrational frequencies & 4 (non-zero)solutions to the eigenvalue problem.
24 The linear triatomic molecule: 4 true vibrational degrees of freedom The linear triatomic molecule: 4 true vibrational degrees of freedom. 4 vibrational frequencies & 4 (non-zero) solutions to the eigenvalue problem.2 of these 4 modes correspond to longitudinal vibrations. These are the ones we’ve just computed. In fact, if coupling between vibrations along the molecular (x) axis & vibrations that direction is neglected, the solutions for the frequencies and for the eigenvectors are EXACTLY the ones we’ve already computed!2 of these 4 modes correspond to transverse vibrations (vibrations x direction). Molecule is axially symmetric: No reason to expect a preference for one direction (say y) over another (say z). The 2 frequencies corresponding to these 2 transverse vibrations must be equal. The 2 transverse modes are degenerate.
25 The 2 transverse modes are degenerate. They have the same frequency The 2 transverse modes are degenerate. They have the same frequency. So, to determine the eigenvectors, we cannot directly follow the procedure we used up to now, as degeneracy was excluded. In computing the eigenvectors, any 2 mutually orthogonal axes x axis can be chosen. The motion of the molecule x axis depends on the amplitudes & relative phases of these 2 degenerate modes.If both modes are excited, & if they vibrate exactly in phase, the atoms will move in a straight line passing through the equilibrium configuration. But, if they are out of phase, the total transverse motion is an elliptical Lissajous figure (like the 2d isotropic SHO). In this case, the motion in these modes is actually a rotation rather than a vibration!
26 From molecular symmetry: In all modes, the amplitudes of the displacements of the end atoms must be equal. Detailed calculation of the degenerate, transverse vibrational (rotational) modes shows that, in this mode, the end atoms travel in the same direction along a Lissajous figure. In order to conserve angular momentum, the center atom must move (revolve) in the opposite direction. See figure (which is for the 2 degenerate modes which are 90° out of phase).See the book’s discussion of use of group theory & modern computational techniques to find the vibrational normal modes of real, multi-atom molecules!