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Chapter 05Vibrational Motion

P. J. Grandinetti

Chem. 4300

P. J. Grandinetti Chapter 05: Vibrational Motion

Simple Harmonic Oscillator

Simplest model for harmonic oscillator: mass attached to one end of spring while other end isheld fixed

+x-x 0

m

Mass at x = 0 corresponds to equilibrium positionx is displacement from equilibrium.Assume no friction and spring has no mass.


Simple Harmonic Oscillator

Pull mass and let go.

+x-x 0

m

Mass at x = 0 corresponds to equilibriumpositionx is displacement from equilibrium.Assume no friction and spring has nomass.

What happens? An oscillation.

Time for 1 complete cycle is T.How do we get and solve equation of motionfor this system?


Simple Harmonic OscillatorHooke’s law

For small displacements from equilibrium restoring force is F = −𝜅fxwhere 𝜅f is force constant for spring.

Use Newton’s 2nd law: F = ma = −𝜅fx

to obtain differential equation of motion: mx(t) + 𝜅fx(t) = 0

Propose solution

x(t) = A cos(𝜔t + 𝜙), x(t) = A𝜔 sin(𝜔t + 𝜙), x(t) = −A𝜔2 cos(𝜔t + 𝜙)

Substitute into differential equation:(𝜅f − m𝜔2)A cos(𝜔t + 𝜙) = 0

Satisfy solution for all t by making m𝜔2 = 𝜅f through definition: 𝜔 = 𝜔0 =√𝜅f∕m


Simple Harmonic OscillatorSolution is

x(t) = A cos(𝜔0t + 𝜙)

where 𝜔0 ≡ natural oscillation frequency given by 𝜔 = 𝜔0 =√𝜅f∕m

Velocity of mass isv(t) = x(t) = −𝜔0A sin(𝜔0t + 𝜙)

Make x(t) and x(t) equations satisfy initial conditions of

x(t = 0) = A and x(t = 0) = 0

by setting 𝜙 = 0 and A = x(0) to get final solution

x(t) = x(0) cos𝜔0t and v(t) = −𝜔0x(0) sin(𝜔0t)


Simple Harmonic Oscillator – Other trial solutions

Instead of using x(t) = A cos(𝜔t + 𝜙) as trial solution for mx(t) + 𝜅fx(t) = 0▶ Could have used

x(t) = A cos𝜔t + B sin𝜔t.

▶ Could also have used complex variables:★ replace x(t) with a complex variable, z(t),

mz(t) + 𝜅fz(t) = 0

★ Make an initial guess of

z(t) = Aei(𝜔t+𝜙) or z(t) = Aei𝜔t + Be−i𝜔t

★ Obtain real solution by taking real part of complex solution, z(t).


Complex Variables Review


Complex Variables ReviewComplex variables are a mathematical tool that simplifies equations describing oscillations.Consider the 2D motion of this vector.

How would you describe this mathematically?You probably would suggest:

x(t) = r cos𝜔t and y(t) = r sin𝜔tP. J. Grandinetti Chapter 05: Vibrational Motion

Complex Variables ReviewWith complex notation we combine two equations into one

Start with x(t) = r cos𝜔t and y(t) = r sin𝜔t

First we define the square root of −1 as

if i =√−1 then i2 = −1

Second we define complex variable z as

z = x + iy

x is real part and y is imaginary part of z.

Two circular motion equations become one circular motion equation

z(t) = r cos𝜔t + ir sin𝜔t


Complex Variables ReviewEuler’s formula

In 1748 Euler showed that ei𝜃 = cos 𝜃 + i sin 𝜃 Euler’s formula

With Euler’s formula z(t) = r cos𝜔t + ir sin𝜔t becomes z(t) = rei𝜔t

Calculate the product (cos𝜔at + i sin𝜔at)(cos𝜔bt + i sin𝜔bt).

(cos𝜔at+ i sin𝜔at)(cos𝜔bt+ i sin𝜔bt) = cos𝜔at cos𝜔bt+ i cos𝜔at sin𝜔bt+ i sin𝜔at cos𝜔bt−sin𝜔at sin𝜔bt

2 cos 𝜃 cos𝜙 = cos(𝜃 − 𝜙) + cos(𝜃 + 𝜙) 2 sin 𝜃 sin𝜙 = cos(𝜃 − 𝜙) − cos(𝜃 + 𝜙)2 sin 𝜃 cos𝜙 = sin(𝜃 + 𝜙) + sin(𝜃 − 𝜙) 2 cos 𝜃 sin𝜙 = sin(𝜃 + 𝜙) − sin(𝜃 − 𝜙)

(cos𝜔at + i sin𝜔at)(cos𝜔bt + i sin𝜔bt) = cos(𝜔a + 𝜔b)t + i sin(𝜔a + 𝜔b)t

Using Euler’s formula:(cos𝜔at + i sin𝜔at)(cos𝜔bt + i sin𝜔bt) = ei𝜔atei𝜔bt = ei(𝜔a+𝜔b)t = cos(𝜔a + 𝜔b)t + i sin(𝜔a + 𝜔b)t


Complex Variables Review

Any complex number can be written in the form

z = x + iy = |z|ei𝜃

where |z| is the magnitude of the complex number

|z| = √x2 + y2

and 𝜃 is the argument of the complex number

tan 𝜃 = y∕x


Complex Variables ReviewComplex Conjugate

Complex conjugate, z∗, of z is obtained by changing sign of imaginary part

if z = x + iy then z∗ = x − iy

if z = 1 + 4i4 − 5i

then z∗ = 1 − 4i4 + 5i

if z = 1 + ia4 − ib

then z∗ = 1 − ia∗4 + ib∗

.

Related identities▶ |z| = √

zz∗, since

zz∗ = (x + iy)(x − iy) = x2 + iyx − ixy + y2 = x2 + y2 = |z|2▶ (z1z2z3 ⋯)∗ = z∗1z∗2z∗3 ⋯


Energy of simple harmonic oscillator


Energy of simple harmonic oscillatorTotal energy of simple harmonic oscillator is sum of kinetic and potential energy of mass andspring.

Kinetic energy is given by

K = 12

mv2, or K =p2

2m

Potential energy is energy stored in spring and equal to work done in extending andcompressing spring,

V(x) = −∫x

0F(x′)dx′ = ∫

x

0𝜅fx′dx′ = 1

2𝜅fx2

Expression above is work associated with extending spring.

For work in compressing spring just change integral limits to −x to 0 (get same result).


Energy of simple harmonic oscillatorPotential energy is given by

V(x) = 12𝜅fx2

V(x)

0x


Energy of simple harmonic oscillator

Although both K and V are time dependent during harmonic motion the total energy,E = K + V, remains constant.

E = 12

mv2(t) + 12𝜅fx2(t) or E =

p2(t)2m

+ 12𝜅fx2(t)

EnergyKinetic Energy

Potential EnergyTotal Energy


Energy of simple harmonic oscillatorSubstitute equation of motion into energy expression

E = 12

mv2(t) + 12𝜅fx2(t) = 1

2m𝜔2

0x2(0) sin2(𝜔0t) + 12𝜅fx2(0) cos2 𝜔0t

Recall that 𝜅f = m𝜔20 so

E = 12

m𝜔20x2(0)

[sin2(𝜔0t) + cos2 𝜔0t

]= 1

2m𝜔2

0x2(0)

Solve for initial amplitude, x(0), in terms of energy

x(0) = 1𝜔0

√2Em

and rewrite oscillation as

x(t) = x(0) cos𝜔0t =√

2E𝜔2

0mcos𝜔0t and v(t) = −𝜔0x(0) sin(𝜔0t) = −

√2Em

sin𝜔0t


Simple Harmonic Oscillator - Phase Space Trajectory

x(t) = x(0) cos𝜔0t =√

2E𝜔2

0mcos𝜔0t and v(t) = −𝜔0x(0) sin(𝜔0t) = −

√2Em

sin𝜔0t

State of oscillator specified by point in phase spaceE = constant means oscillator state always lies on ellipse

x2

2E∕𝜅f+ v2

2E∕m= 1

Trajectory moves in clockwise direction.Trajectories with different E can never cross.


Position probability distribution for harmonic oscillator


Position probability distribution for harmonic oscillatorScale x(t) by initial amplitude x(0), to obtain a function, y(t), that oscillates between y = −1and y = +1

y(t) = x(t)∕x(0) = cos𝜔0tCalculate normalized probability density, p(y), for finding the mass at any scaled positionbetween y = ±1.Probability of finding mass in interval dy at given y is proportional to time spent in dy interval,

p(y) dy = b dt = b dtdydy

= b dy dtdy

= by

dy

b ≡ proportionality constant, and y ≡ speed at a given yUse derivative of y(t)

y(t) = −𝜔0 sin𝜔0tto obtain

p(y) = by(t)

= b−𝜔0 sin𝜔0t

= b−𝜔0(1 − cos2 𝜔0t)1∕2

= b−𝜔0(1 − y2)1∕2


Position probability distribution for harmonic oscillator

p(y) = b−𝜔0(1 − y2)1∕2

Normalizing probability distribution gives

p(y) = 1𝜋(1 − y2)1∕2

+x-x 0

m-1.0 -0.5 0.0 0.5 1.0

2

0

4

6

8

10

Mass spends majority of time at maximumexcursions, that is, turning points where velocity isslowest and changes sign.


Diatomic molecule vibration as Harmonic oscillator


Diatomic molecule vibration as Harmonic oscillator

Dashed line is harmonic oscillator potential. Solid line is Morse potential.V(r) has minimum at re—where restoring force is zero.V(r) causes repulsive force at r < re and attractive force at r > re.V(r) increases steeply at r < re but levels out to constant at r > re.At r → ∞ there is no attractive force as V(r) has a slope of zero.


Diatomic molecule vibration as Harmonic oscillatorTaylor series expansion of V(r) about equilibrium bond length, r = re, gives

V(r) ≈ V(re) +��>

0dV(re)

dr(r − re) +

12!

d2V(re)dr2 (r − re)2 +

13!

d3V(re)dr3 (r − re)3 +⋯

V(re) is the potential energy at equilibrium bond length.1st-order term is zero since no restoring force, F = −dV(re)∕dr, at r = re

Stop expansion at the 3rd-order term and define 2 constants

𝜅f =d2V(re)

dr2 and 𝛾f =d3V(re)

dr3

and write potential expansion as

V(r) − V(re) ≈12𝜅f(r − re)2 +

16𝛾f(r − re)3 +⋯

For small displacements drop 3rd-order term and potential looks like simple harmonic oscillator.For slightly larger displacements keep 3rd-order term to account for vibration anharmonicity.


Diatomic molecule vibration equations of motion

Make harmonic oscillator approximation taking force on m1 and m2 as

F1 = −𝜅f(r1 − r2 + re) and F2 = −𝜅f(r2 − r1 − re)

Equations of motion are 2 coupled differential equations:

m1d2r1

dt2= 𝜅f(r2 − r1 − re) and m2

d2r2

dt2= −𝜅f(r2 − r1 − re)

Transform to center of mass frame: M = m1 + m2 and R = 1M (m1r1 + m1r2)

Obtain 2 uncoupled differential equations,

M d2Rdt2

= 0, and 𝜇d2Δrdt2

= −𝜅fΔr

Δr = r2 − r1 − re and 𝜇 is the reduced mass given by 1𝜇

= 1m1

+ 1m2


Diatomic molecule vibration equations of motionDifferential equation of motion describing the vibration

𝜇dΔr2(t)

dt2 + 𝜅fΔr(t) = 0

Same differential equation of motion as simple harmonic oscillator.

Solutions takes the same form,

Δr(t) = Δr(0) cos𝜔0t where 𝜔0 =√𝜅f∕𝜇

In spectroscopy vibrational frequencies are given in terms of the spectroscopic wavenumber,

�� =𝜔0

2𝜋c0=

√𝜅f∕𝜇

2𝜋c0, rearranges to 𝜅f = 𝜇

(2𝜋c0��

)2


Force constants for selected diatomic molecules

𝜅f = 𝜇(2𝜋c0��

)2

Bond 𝜅f/(N/m) 𝜇∕10−28 kg ��/cm−1 Bond length/pmH–H 570 8.367664 4401 74.1D–D 527 16.72247 2990 74.1

H–35Cl 478 16.26652 2886 127.5H–79Br 408 16.52430 2630 141.4H–127I 291 16.60347 2230 160.9

35Cl–35Cl 319 290.3357 554 198.879Br–79Br 240 655.2349 323 228.4127I–127I 170 1053.649 213 266.7

16O=16O 1142 132.8009 1556 120.714N≡14N 2243 116.2633 2331 109.412C≡16O 1857 113.8500 2143 112.814N=16O 1550 123.9830 1876 115.123Na–35Cl 117 230.3282 378 236.139K–35Cl 84 306.0237 278 266.7

23Na–23Na 17 190.8770 158 307.8


Coupled Harmonic Oscillators and Normal Modes


Two coupled oscillators constrained to move in one dimension

m m

+x1-x1 0 +x2-x2 0

2 equal masses connected with 3 springs with force constants 𝜅1 and 𝜅2.3 springs are at equilibrium lengths when x1 = x2 = 0.Middle spring is also at equilibrium length whenever x2 = x1.Restoring force acting on left mass: F1 = −𝜅1x1 + 𝜅2(x2 − x1)Restoring force acting on right mass: F2 = −𝜅1x2 − 𝜅2(x2 − x1)Coupled differential equations of motion for each mass:

mx1 + (𝜅1 + 𝜅2)x1 − 𝜅2x2 = 0 and mx2 − 𝜅2x1 + (𝜅1 + 𝜅2)x2 = 0


Two coupled oscillators: Normal ModesCoupled differential equations: mx1 + (𝜅1 + 𝜅2)x1 − 𝜅2x2 = 0 and mx2 − 𝜅2x1 + (𝜅1 + 𝜅2)x2 = 0

Uncoupled by coordinate transformation: 𝜉1 = x1 + x2 and 𝜉2 = x1 − x2

𝜉1 and 𝜉2 are called normal mode coordinates

Gives 2 uncoupled equations: m𝜉1 + (𝜅1 + 2𝜅2)𝜉1 = 0 and m𝜉2 + 𝜅1𝜉2 = 0

Substitute 2 proposed solutions: 𝜉1(t) = A1 cos(𝜔1t + 𝜙1) and 𝜉2(t) = A2 cos(𝜔2t + 𝜙2)

Obtain:((𝜅1 + 2𝜅2) − m𝜔2

1)

A1 cos(𝜔1t + 𝜙1) = 0 and(𝜅1 − m𝜔2

2)

A2 cos(𝜔2t + 𝜙2) = 0

Equations true for all values of t if: (𝜅1 + 2𝜅2) = m𝜔21 and 𝜅1 = m𝜔2

2

yields 2 normal mode frequencies: 𝜔1 =√(𝜅1 + 2𝜅2)∕m and 𝜔2 =

√𝜅1∕m Note: 𝜔1 > 𝜔2

Normal mode is collective oscillation where all masses move with same oscillation frequency.


Two coupled oscillators: Normal Modes

Imposing initial conditions:

𝜉1(0) = x1(0) + x2(0) and 𝜉2(0) = x1(0) − x2(0), and ��1(0) = ��2(0) = 0

on solutions: 𝜉1(t) = A1 cos(𝜔1t + 𝜙1) and 𝜉2(t) = A2 cos(𝜔2t + 𝜙2)gives:

𝜙1 = 𝜙2 = 0 and A1 = 𝜉1(0) and A2 = 𝜉2(0)

𝜉1(t) = 𝜉1(0) cos(𝜔1t) and 𝜉2(t) = 𝜉2(0) cos(𝜔2t)

Converting normal mode coordinates back to original coordinates:

x1(t) =12

[𝜉2(0) cos(𝜔2t) + 𝜉1(0) cos(𝜔1t)

]and x2(t) =

12

[𝜉2(0) cos(𝜔2t) − 𝜉1(0) cos(𝜔1t)

]Oscillation of each mass is linear combination of two normal mode oscillations.


Two coupled oscillators: Normal ModesConsider two cases where only one normal mode is active:

𝜉2(0) = 0 or 𝜉1(0) = 0

x1(t) = 12𝜉1(0) cos(𝜔1t)

x2(t) = − 12𝜉1(0) cos(𝜔1t)

⎫⎪⎬⎪⎭ out-of-phase,x1(t) = 1

2𝜉2(0) cos(𝜔2t)

x2(t) = 12𝜉2(0) cos(𝜔2t)

⎫⎪⎬⎪⎭ in-phase.

m m

m m

out of phase normal mode

m m

in phase normal mode

m m

m

m m

m

All motion in this system can be decomposed into linear combination of these two normal modes.P. J. Grandinetti Chapter 05: Vibrational Motion

Normal mode analysisConsider a harder problem

m1 m2

+x1-x1 0 +x2-x2 0

Both masses and all 3 spring constants are different.Simple coordinate transformation 𝜉1 = x1 + x2 and 𝜉2 = x1 − x2 will not transform probleminto uncoupled differential equations.Need systematic approach for determining coordinate transformation (normal modes) thattransform coupled linear differential equations into uncoupled differential equations.


Normal mode analysis

m1 m2

+x1-x1 0 +x2-x2 0

Define mass-weighted coordinates: q1 = m1∕21 x1 and q2 = m1∕2

2 x2Coupled differential equations are

q1 +(𝜅1 + 𝜅2)

m1q1 −

𝜅2

(m2m1)1∕2q2 = 0 and q2 −

𝜅2

(m2m1)1∕2q1 −

(𝜅3 − 𝜅2)m2

q2 = 0

In matrix form:

⎡⎢⎢⎣q1

q2

⎤⎥⎥⎦⏟⏟⏟

q

+

⎡⎢⎢⎢⎢⎢⎣

(𝜅1 + 𝜅2)m1

−𝜅2

(m2m1)1∕2

−𝜅2

(m2m1)1∕2

(𝜅3 + 𝜅2)m2

⎤⎥⎥⎥⎥⎥⎦⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

K

⎡⎢⎢⎣q1

q2

⎤⎥⎥⎦⏟⏟⏟

q

= 0


Normal mode analysisAny coupled harmonic oscillator problem can be written in form: q + Kq = 0

K is a square n × n real symmetric (i.e., K12 = K21) matrix, q and q are n × 1 matrices.Similarity transformation of real n × n symmetric matrix into diagonal matrix, 𝚲.

LT KL =

⎡⎢⎢⎢⎣l11 l21 … ln1l12 l22 … ln2⋮ ⋮ ⋱ ⋮l1n l2n … lnn

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣

K11 K12 … K1nK21 K22 … K2n⋮ ⋮ ⋱ ⋮

Kn1 Kn2 … Knn

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣

l11 l12 … l1nl21 l22 … l2n⋮ ⋮ ⋱ ⋮ln1 ln2 … lnn

⎤⎥⎥⎥⎦ =⎡⎢⎢⎢⎣

𝜆1 0 … 00 𝜆2 … 0⋮ ⋮ ⋱ ⋮0 0 … 𝜆n

⎤⎥⎥⎥⎦ = 𝚲

▶ Eigenvalues, 𝜆i, of K along diagonal of 𝚲.▶ Eigenvectors, li, of K along columns of L and along rows of LT .▶ L is orthogonal matrix: product with its transpose, LT , is identity matrix, 1,

LLT =

⎡⎢⎢⎢⎣l11 l12 … l1nl21 l22 … l2n⋮ ⋮ ⋱ ⋮ln1 ln2 … lnn

⎤⎥⎥⎥⎦⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

L

⎡⎢⎢⎢⎣l11 l21 … ln1l12 l22 … ln2⋮ ⋮ ⋱ ⋮l1n l2n … lnn

⎤⎥⎥⎥⎦⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

LT

=

⎡⎢⎢⎢⎣1 0 … 00 1 … 0⋮ ⋮ ⋱ ⋮0 0 … 1

⎤⎥⎥⎥⎦⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

1

= 1


Normal mode analysisInsert identity matrix as 1 = LLT before and after K in matrix form of equations of motion:

q + Kq = q + (LLT )⏟⏟⏟

1

K (LLT )⏟⏟⏟

1

q = q + L (LTKL)⏟⏟⏟

𝚲

(LTq) = 0

Multiplying both sides by LT , regroup, and define new matrix variables:

LT

⎛⎜⎜⎜⎝q + L (LTKL)⏟⏟⏟

𝚲

LTq⎞⎟⎟⎟⎠ = LT q

⏟⏟⏟Q

+ (LTKL)⏟⏟⏟

𝚲

(LT q)⏟⏟⏟

Q

= Q + 𝚲 Q = 0

Q = LT q is linear transformation of q into Q, i.e., normal mode coordinatesDifferential equations in Q and 𝚲 are uncoupled for each normal mode coordinate: Qi + 𝜆iQi = 0Substitute in proposed solutions: Qi(t) = Ai cos(𝜔it+𝜙i) gives (𝜆i −𝜔2

i ) cos(𝜔it+𝜙i) = 0

Identify the ith normal mode frequency as 𝜔i =√𝜆i

With initial conditions, solutions become Qi(t) = Qi(0) cos(𝜔it

)P. J. Grandinetti Chapter 05: Vibrational Motion

How do we determine eigenvalues, 𝜆i, and the eigenvectors, li, from K?ith row in L is li eigenvector for 𝜆i eigenvalue.

Kli =⎡⎢⎢⎢⎣

K11 K12 … K1nK21 K22 … K2n⋮ ⋮ ⋱ ⋮

Kn1 Kn2 … Knn

⎤⎥⎥⎥⎦⎡⎢⎢⎢⎣

li1li2⋮lin

⎤⎥⎥⎥⎦ = 𝜆ili = 𝜆i

⎡⎢⎢⎢⎣li1li2⋮lin

⎤⎥⎥⎥⎦ = 𝜆i

⎡⎢⎢⎢⎣1 0 … 00 1 … 0⋮ ⋮ ⋱ ⋮0 0 … 1

⎤⎥⎥⎥⎦⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

1


⎤⎥⎥⎥⎦⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎡⎢⎢⎢⎣K11 K12 … K1nK21 K22 … K2n⋮ ⋮ ⋱ ⋮

Kn1 Kn2 … Knn

⎤⎥⎥⎥⎦⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

K

−𝜆i

⎡⎢⎢⎢⎣1 0 … 00 1 … 0⋮ ⋮ ⋱ ⋮0 0 … 1

⎤⎥⎥⎥⎦⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

1

⎞⎟⎟⎟⎟⎟⎟⎟⎠


⎤⎥⎥⎥⎦⏟⏟⏟

li

= 0

Compactly written as(K − 𝜆i1) li = 0

Encountered same problem when diagonalizing moment of inertia tensor to determine PAS componentsand orientation.


How do we determine eigenvalues, 𝜆i, and the eigenvectors, li, from K?

Obtain eigenvalues by expanding determinant

|K − 𝜆i1| = 0

to get nth order polynomial equation in 𝜆 whose roots are eigenvalues.

Substitute each eigenvalue back into (K − 𝜆i1) li = 0 to get corresponding eigenvector.

Determining eigenvalues and eigenvectors of a matrix is usually done numerically. Softwarepackages utilizing various algorithms for matrix diagonalization are available in various problemsolving environments such as MatLab, Mathematica, or Python notebooks.


Normal mode analysisBack to our original problem

⎡⎢⎢⎣q1

q2

⎤⎥⎥⎦⏟⏟⏟

q

+

⎡⎢⎢⎢⎢⎢⎣

(𝜅1 + 𝜅2)m1

−𝜅2

(m2m1)1∕2

−𝜅2

(m2m1)1∕2

(𝜅3 + 𝜅2)m2

⎤⎥⎥⎥⎥⎥⎦⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

K

⎡⎢⎢⎣q1

q2

⎤⎥⎥⎦⏟⏟⏟

q

= 0

To solve this problem we willdetermine eigenvalues, 𝜆i and eigenvectors, li, of Kobtain normal mode frequencies from 𝜔i =

√𝜆i

obtain relationship between mass-weighted and normal-mode coordinates from Q = LTq

To make solution more manageable, only illustrate the case where m1 = m2 = m and 𝜅3 = 𝜅1.


Normal mode analysis - Find the Eigenvalues

Obtain eigenvalues by solving the determinant,|||||||||𝜅1 + 𝜅2

m− 𝜆 −

𝜅2m

−𝜅2m

𝜅1 + 𝜅2m

− 𝜆

|||||||||=(𝜅1 + 𝜅2

m− 𝜆

)(𝜅1 + 𝜅2

m− 𝜆

)−

𝜅22

m2 = 0

obtaining𝜔2 = 𝜆 =

𝜅1 + 𝜅2 ± 𝜅2m

𝜆1 = (𝜅1 + 2𝜅2)∕m and 𝜆2 = 𝜅1∕m


Normal mode analysis - Find the EigenvectorsObtain first eigenvector, associated with 𝜆1 = (𝜅1 + 2𝜅2)∕m, by expanding⎡⎢⎢⎣

(𝜅1 + 𝜅2)∕m − 𝜆1 −𝜅2∕m

−𝜅2∕m (𝜅1 + 𝜅2)∕m − 𝜆1

⎤⎥⎥⎦⎡⎢⎢⎣

l11

l12

⎤⎥⎥⎦ = 0,

to get 2 identical equations:((𝜅1 + 𝜅2)∕m − 𝜆1

)l11 − (𝜅2∕m)l12 = 0, and − (𝜅2∕m)l11 +

((𝜅1 + 𝜅2)∕m − 𝜆1

)l12 = 0,

which simplify to l11 = −l12, i.e. two elements have equal magnitude and opposite sign.

l1 =[

l11l12

]= b

[1−1

]where b is some unknown constant.

Insisting that l1 be normalized, i.e.,√

l211 + l212 = 1, gives b = 1∕√

2 and our 1st eigenvector is:

l1 =[

l11l12

]=

[1∕

√2

−1∕√

2

]P. J. Grandinetti Chapter 05: Vibrational Motion

Normal mode analysis - Find the EigenvectorsSimilarly, second eigenvector, associated with 𝜆2 = 𝜅1∕m, is given by

l2 =⎡⎢⎢⎣

l21

l22

⎤⎥⎥⎦ =⎡⎢⎢⎣

1∕√

2

1∕√

2

⎤⎥⎥⎦Taking these two results together, we construct the eigenvector matrix

L =[

l11 l21l12 l22

]=⎡⎢⎢⎣

1∕√

2 1∕√

2

−1∕√

2 1∕√

2

⎤⎥⎥⎦Finally, we obtain solutions, in terms of the mass-weighted coordinates, as

q = LQ =⎡⎢⎢⎣

q1

q2

⎤⎥⎥⎦ =⎡⎢⎢⎣

1∕√

2 1∕√

2

−1∕√

2 1∕√

2

⎤⎥⎥⎦⎡⎢⎢⎣

Q1(0) cos𝜔1t

Q2(0) cos𝜔2t

⎤⎥⎥⎦P. J. Grandinetti Chapter 05: Vibrational Motion

Web Apps by Paul Falstad

Coupled Oscillators

Web App Link: Coupled Oscillators


http://www.falstad.com/coupled

Polyatomic Molecule Vibrations


Polyatomic Molecule VibrationsNormal mode analysis of molecule with N atoms

Define each atom’s displacement coordinates in PAS of molecule’s moment of inertia tensor withcenter of mass at origin.For example, for water molecule define

105°

a

b

𝜂1 = aO − aO,e, 𝜂2 = bO − bO,e, 𝜂3 = cO − cO,e,𝜂4 = aH1

− aH1,e, 𝜂5 = bH1− bH1,e, 𝜂6 = cH1

− cH1,e,𝜂7 = aH2

− aH2,e, 𝜂8 = bH2− bH2,e, 𝜂9 = cH2

− cH2,e,

subscript e represents equilibrium coordinate.▶ Further define mass-weighted displacement coordinates, q1,… , q3N , e.g.,

qi = m1∕2i 𝜂i where

m1 = m2 = m3 = mO, m4 = m5 = m6 = mH , and m7 = m8 = m9 = mH .Derive equations of motion from kinetic and potential energy

ddt

(𝜕K(q1,… , q3N)

𝜕q

)+

𝜕V(q1,… , q3N)𝜕qi

= 0


Polyatomic Molecule VibrationsKinetic and Potential Energy in mass-weighted coordinates

Kinetic energy becomes

K = 12

3N∑i=1

(dqidt

)2= 1

2qT q where qT =

[dq1dt

,… ,dq3N

dt

]and q is its transpose

Potential energy expanded as Taylor-series expansion

V(q) = V(0) +3N∑i=1 ��*

0(𝜕V(0)𝜕qi

)qi + 1

2

3N∑i=1

3N∑k=1

(𝜕2V(0)𝜕qi𝜕qk

)qiqk +⋯

Rewriting as

V(q) ≈ V(0) + 12

qT q where qT =[q1,… , q3N

], and q is its transpose

where is 3N × 3N matrix with elements given by

i,k =(𝜕2V(0)𝜕qi𝜕qk



Derive equations of motion from kinetic and potential energy

ddt

(𝜕K(q1,… , q3N)

𝜕q

)+

𝜕V(q1,… , q3N)𝜕qi

= 0

and obtain a set of 3N coupled differential equations (in matrix notation),

q + q = 0As before, heart of the problem is determining eigenvalues, 𝜆i, and eigenvectors, li of .Orthogonal eigenvector matrix, L, transforms problem into 3N uncoupled differential equations,

q + q = q + (LLT )(LLT )q = 0 ⟶ LT q + (LTL) (LTq) = Q + 𝚲Q = 0

▶ solutions: Qi(t) = Qi(0) cos(𝜔it)▶ normal mode frequencies: 𝜔i =

√𝜆i

▶ normal mode coordinates: Q = LTqNormal modes with identical frequencies are called degenerate modes.



Since translational and rotational coordinates are included, 5 or 6 normal modefrequencies, depending on whether molecule is linear or not, respectively, will be zero.Potential and kinetic energy in terms of the vibrational normal modes coordinates are

V − Ve ≈12

QT𝚲Q = 12

3N−6∑i

𝜆iQ2i , and K = 1

2QTQ = 1

2

3N−6∑i

Q2i .

Keep in mind that approximations were made along the way.▶ Ignored anharmonic (3rd and higher order) terms ignored in Taylor-series expansion of V(q).

Can still use matrix, but anharmonicities cause small couplings between the “normalmodes.”

▶ Coordinate system fixed on molecule, i.e., PAS of moment of inertia tensor, is a rotatingframe. We neglected fictitious forces of such a frame. Coriolis forces will cause smallcouplings between the normal modes.

Primary task: obtain matrix and find its eigenvalues and eigenvectors.P. J. Grandinetti Chapter 05: Vibrational Motion

Normal mode analysis of diatomic molecule



a

b Define displacements in PAS of its moment of inertia tensor.

𝜂1 = aA − aA,e, 𝜂2 = bA − bA,e = 0, 𝜂3 = cA − cA,e = 0,𝜂4 = aB − aB,e, 𝜂5 = bB − bB,e = 0, 𝜂6 = cB − cB,e = 0,

From 6 displacement coordinates, define mass-weighted coordinates,qi = m1∕2

i 𝜂i, where m1 = m2 = m3 = mA, and m4 = m5 = m6 = mB.

Since Δr = r − re = 𝜂4 − 𝜂1 = q1∕m1∕21 − q4∕m1∕2

4 we obtain potential energy function

V − Ve ≈12𝜅f(r − re)2 = 1

2𝜅f

(q1

m1∕21

−q4

m1∕24

)2

= 12𝜅f

(q2

1m1

−2q1q4

(m1m4)1∕2+

q24

m4

)Identify non-zero matrix elements of as

11 = 𝜅f∕m1, and 44 = 𝜅f∕m4, and 14 = −𝜅f∕(m1m4

)1∕2


Normal mode analysis of diatomic moleculeEquations of motion in matrix form is

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

q1

q2

q3

q4

q5

q6

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⏟⏟⏟

q

+

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

𝜅f

m10 0 −

𝜅f(m1m4

)1∕20 0

0 0 0 0 0 0

0 0 0 0 0 0

−𝜅f(

m1m4)1∕2

0 0𝜅f

m40 0

0 0 0 0 0 0

0 0 0 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

q1

q2

q3

q4

q5

q6

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⏟⏟⏟

q

= 0

Note: Four equations are already uncoupled and have eigenvalues of 𝜆i = 0.


Normal mode analysis of diatomic moleculeRemaining equations of motion are[

q1q4

]+⎡⎢⎢⎣

𝜅f∕m1 −𝜅f∕(m1m4

)1∕2

−𝜅f∕(m1m4

)1∕2 𝜅f∕m4

⎤⎥⎥⎦[

q1q4

]= 0

Get eigenvalues from solving|||||||𝜅f∕m1 − 𝜆 −𝜅f∕

(m1m4

)1∕2

−𝜅f∕(m1m4

)1∕2 𝜅f∕m4 − 𝜆

||||||| = (𝜅f∕m1 − 𝜆)(𝜅f∕m4 − 𝜆) − 𝜅2f ∕

(m1m4

)= 0

gives 𝜆 = 0 and𝜅f𝜇

where 1𝜇

= 1m1

+ 1m4

=m1 + m4

m1m4= M

m1m4, or 𝜇 = (m1m4)∕M,

𝜇 is the reduced mass, and M = m1 + m4 is the total mass.5 out of 6 eigenvalues are zero—associated with 3 translational and 2 rotational degrees offreedom.1 out of 6 degrees of freedom is associated with vibration with 𝜔1 =

√𝜅f∕𝜇, and normal mode

solution Q1(t) = Q1(0) cos(𝜔1t



Turning to normal mode eigenvectors. If we substitute the eigenvalue 𝜆1 = 𝜅f∕𝜇 into

( − 𝜆11)l1 =

⎡⎢⎢⎢⎣𝜅f∕m1 − 𝜆1 −𝜅f∕

(m1m4

)1∕2

−𝜅f∕(m1m4

)1∕2 𝜅f∕m4 − 𝜆1

⎤⎥⎥⎥⎦⎡⎢⎢⎣

l11

l14

⎤⎥⎥⎦ = 0,

we obtain two equivalent equations, which simplify to l11 = −(m4∕m1

)1∕2 l14Write eigenvector as

l1 =⎡⎢⎢⎣

l11

l14

⎤⎥⎥⎦ = b⎡⎢⎢⎣−(m4∕m1

)1∕2

1

⎤⎥⎥⎦ ⟶ l1 =

⎡⎢⎢⎢⎣(𝜇∕m1

)1∕2

−(𝜇∕m4

)1∕2

⎤⎥⎥⎥⎦where b =

(m4∕M

)1∕2 after normalization.

Same approach gives

l2 =

⎡⎢⎢⎢⎣(𝜇∕m4

)1∕2

(𝜇∕m1

)1∕2

⎤⎥⎥⎥⎦P. J. Grandinetti Chapter 05: Vibrational Motion

Normal mode analysis of diatomic moleculeBring eigenvectors together to create transformation between mass-weighted and normalcoordinates:

Q = LTq =⎡⎢⎢⎣

Q1

Q4

⎤⎥⎥⎦ =⎡⎢⎢⎢⎢⎢⎣

(𝜇

m1

)1∕2−(

𝜇m4

)1∕2

(𝜇

m4

)1∕2 (𝜇

m1

)1∕2

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎣

q1

q4

⎤⎥⎥⎦The Q1 coordinate expands to

Q1 = (𝜇)1∕2 ( q1∕m1∕21

⏟⏟⏟𝜂1

− q4∕m1∕24

⏟⏟⏟𝜂4

)= (𝜇)1∕2 ( 𝜂1 − 𝜂4

⏟⏟⏟Δr

)= (𝜇)1∕2 Δr

Rearranging gives vibrational motion solution:

Δr(t) = (𝜇)−1∕2 Q1(t) = (𝜇)−1∕2 Q1(0) cos(𝜔1t


Normal mode analysis of diatomic moleculeQ4 normal mode (with 𝜆4 = 0) can be expanded as

Q4 = (𝜇)1∕2(

q1∕m1∕24 + q4∕m1∕2

1

)= 1

M1∕2

(m1𝜂1 + m4𝜂4

)= 1

M1∕2

(mA(aA − aA,e) + mB(aB − aB,e)

)Recalling center of mass along the a axis,

aCOM = (mAaA + mBaB)∕M and aCOM,e = (mAaA,e + mBaB,e)∕M = 0

where center of mass when atoms are at equilibrium coordinates is zero, gives

Q4 = M1∕2

(mAaA + mBaB

)M

= M1∕2aCOM

Since 𝜆4 = 0, equation of motion is Q4 = 0

initial condition of aCOM = Q4 = 0 means aCOM cannot change in time.

𝜆i = 0 for 5 normal modes associated with translation and rotation, so no change in center ofmass or PAS orientation as molecule vibrates.


Web Videos

Web Video Link: Vibration of Polyatomic MoleculesWeb Links: Crash Course Physics: Simple Harmonic Motion


https://www.youtube.com/watch?v=3RqEIr8NtMI

https://www.youtube.com/watch?v=jxstE6A_CYQ