chapter 05 - vibrational motion - grandinetti
TRANSCRIPT
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.
P. J. Grandinetti Chapter 05: Vibrational Motion
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?
P. J. Grandinetti Chapter 05: Vibrational Motion
Simple Harmonic OscillatorHookeâs law
For small displacements from equilibrium restoring force is F = âð fxwhere ð f is force constant for spring.
Use Newtonâs 2nd law: F = ma = âð fx
to obtain differential equation of motion: mx(t) + ð fx(t) = 0
Propose solution
x(t) = A cos(ðt + ð), x(t) = Að sin(ðt + ð), x(t) = âAð2 cos(ðt + ð)
Substitute into differential equation:(ð f â mð2)A cos(ðt + ð) = 0
Satisfy solution for all t by making mð2 = ð f through definition: ð = ð0 =âð fâm
P. J. Grandinetti Chapter 05: Vibrational Motion
Simple Harmonic OscillatorSolution is
x(t) = A cos(ð0t + ð)
where ð0 â¡ natural oscillation frequency given by ð = ð0 =âð fâm
Velocity of mass isv(t) = x(t) = âð0A sin(ð0t + ð)
Make x(t) and x(t) equations satisfy initial conditions of
x(t = 0) = A and x(t = 0) = 0
by setting ð = 0 and A = x(0) to get final solution
x(t) = x(0) cosð0t and v(t) = âð0x(0) sin(ð0t)
P. J. Grandinetti Chapter 05: Vibrational Motion
Simple Harmonic Oscillator â Other trial solutions
Instead of using x(t) = A cos(ðt + ð) as trial solution for mx(t) + ð fx(t) = 0ⶠCould have used
x(t) = A cosðt + B sinðt.
ⶠCould also have used complex variables:â replace x(t) with a complex variable, z(t),
mz(t) + ð fz(t) = 0
â Make an initial guess of
z(t) = Aei(ðt+ð) or z(t) = Aeiðt + Beâiðt
â Obtain real solution by taking real part of complex solution, z(t).
P. J. Grandinetti Chapter 05: Vibrational Motion
Complex Variables Review
P. J. Grandinetti Chapter 05: Vibrational Motion
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 cosðt and y(t) = r sinðtP. J. Grandinetti Chapter 05: Vibrational Motion
Complex Variables ReviewWith complex notation we combine two equations into one
Start with x(t) = r cosðt and y(t) = r sinðt
First we define the square root of â1 as
if i =ââ1 then i2 = â1
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 cosðt + ir sinðt
P. J. Grandinetti Chapter 05: Vibrational Motion
Complex Variables ReviewEulerâs formula
In 1748 Euler showed that eið = cos ð + i sin ð Eulerâs formula
With Eulerâs formula z(t) = r cosðt + ir sinðt becomes z(t) = reiðt
Calculate the product (cosðat + i sinðat)(cosðbt + i sinðbt).
(cosðat+ i sinðat)(cosðbt+ i sinðbt) = cosðat cosðbt+ i cosðat sinðbt+ i sinðat cosðbtâsinðat sinðbt
2 cos ð cosð = cos(ð â ð) + cos(ð + ð) 2 sin ð sinð = cos(ð â ð) â cos(ð + ð)2 sin ð cosð = sin(ð + ð) + sin(ð â ð) 2 cos ð sinð = sin(ð + ð) â sin(ð â ð)
(cosðat + i sinðat)(cosðbt + i sinðbt) = cos(ða + ðb)t + i sin(ða + ðb)t
Using Eulerâs formula:(cosðat + i sinðat)(cosðbt + i sinðbt) = eiðateiðbt = ei(ða+ðb)t = cos(ða + ðb)t + i sin(ða + ðb)t
P. J. Grandinetti Chapter 05: Vibrational Motion
Complex Variables Review
Any complex number can be written in the form
z = x + iy = |z|eið
where |z| is the magnitude of the complex number
|z| = âx2 + y2
and ð is the argument of the complex number
tan ð = yâx
P. J. Grandinetti Chapter 05: Vibrational Motion
Complex Variables ReviewComplex Conjugate
Complex conjugate, zâ, of z is obtained by changing sign of imaginary part
if z = x + iy then zâ = x â iy
if z = 1 + 4i4 â 5i
then zâ = 1 â 4i4 + 5i
if z = 1 + ia4 â ib
then zâ = 1 â iaâ4 + ibâ
.
Related identitiesⶠ|z| = â
zzâ, since
zzâ = (x + iy)(x â iy) = x2 + iyx â ixy + y2 = x2 + y2 = |z|2ⶠ(z1z2z3 â¯)â = zâ1zâ2zâ3 â¯
P. J. Grandinetti Chapter 05: Vibrational Motion
Energy of simple harmonic oscillator
P. J. Grandinetti Chapter 05: Vibrational Motion
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) = ââ«x
0F(xâ²)dxâ² = â«
x
0ð fxâ²dxâ² = 1
2ð fx2
Expression above is work associated with extending spring.
For work in compressing spring just change integral limits to âx to 0 (get same result).
P. J. Grandinetti Chapter 05: Vibrational Motion
Energy of simple harmonic oscillatorPotential energy is given by
V(x) = 12ð fx2
V(x)
0x
P. J. Grandinetti Chapter 05: Vibrational Motion
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) + 12ð fx2(t) or E =
p2(t)2m
+ 12ð fx2(t)
EnergyKinetic Energy
Potential EnergyTotal Energy
P. J. Grandinetti Chapter 05: Vibrational Motion
Energy of simple harmonic oscillatorSubstitute equation of motion into energy expression
E = 12
mv2(t) + 12ð fx2(t) = 1
2mð2
0x2(0) sin2(ð0t) + 12ð fx2(0) cos2 ð0t
Recall that ð f = mð20 so
E = 12
mð20x2(0)
[sin2(ð0t) + cos2 ð0t
]= 1
2mð2
0x2(0)
Solve for initial amplitude, x(0), in terms of energy
x(0) = 1ð0
â2Em
and rewrite oscillation as
x(t) = x(0) cosð0t =â
2Eð2
0mcosð0t and v(t) = âð0x(0) sin(ð0t) = â
â2Em
sinð0t
P. J. Grandinetti Chapter 05: Vibrational Motion
Simple Harmonic Oscillator - Phase Space Trajectory
x(t) = x(0) cosð0t =â
2Eð2
0mcosð0t and v(t) = âð0x(0) sin(ð0t) = â
â2Em
sinð0t
State of oscillator specified by point in phase spaceE = constant means oscillator state always lies on ellipse
x2
2Eâð f+ v2
2Eâm= 1
Trajectory moves in clockwise direction.Trajectories with different E can never cross.
P. J. Grandinetti Chapter 05: Vibrational Motion
Position probability distribution for harmonic oscillator
P. J. Grandinetti Chapter 05: Vibrational Motion
Position probability distribution for harmonic oscillatorScale x(t) by initial amplitude x(0), to obtain a function, y(t), that oscillates between y = â1and y = +1
y(t) = x(t)âx(0) = cosð0tCalculate 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 â¡ proportionality constant, and y â¡ speed at a given yUse derivative of y(t)
y(t) = âð0 sinð0tto obtain
p(y) = by(t)
= bâð0 sinð0t
= bâð0(1 â cos2 ð0t)1â2
= bâð0(1 â y2)1â2
P. J. Grandinetti Chapter 05: Vibrational Motion
Position probability distribution for harmonic oscillator
p(y) = bâð0(1 â y2)1â2
Normalizing probability distribution gives
p(y) = 1ð(1 â y2)1â2
+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.
P. J. Grandinetti Chapter 05: Vibrational Motion
Diatomic molecule vibration as Harmonic oscillator
P. J. Grandinetti Chapter 05: Vibrational Motion
Diatomic molecule vibration as Harmonic oscillator
Dashed line is harmonic oscillator potential. Solid line is Morse potential.V(r) has minimum at reâwhere 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 â â there is no attractive force as V(r) has a slope of zero.
P. J. Grandinetti Chapter 05: Vibrational Motion
Diatomic molecule vibration as Harmonic oscillatorTaylor series expansion of V(r) about equilibrium bond length, r = re, gives
V(r) â V(re) +ᅵᅵᅵᅵ>
0dV(re)
dr(r â re) +
12!
d2V(re)dr2 (r â re)2 +
13!
d3V(re)dr3 (r â re)3 +â¯
V(re) is the potential energy at equilibrium bond length.1st-order term is zero since no restoring force, F = âdV(re)âdr, at r = re
Stop expansion at the 3rd-order term and define 2 constants
ð f =d2V(re)
dr2 and ðŸf =d3V(re)
dr3
and write potential expansion as
V(r) â V(re) â12ð f(r â re)2 +
16ðŸf(r â re)3 +â¯
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.
P. J. Grandinetti Chapter 05: Vibrational Motion
Diatomic molecule vibration equations of motion
Make harmonic oscillator approximation taking force on m1 and m2 as
F1 = âð f(r1 â r2 + re) and F2 = âð f(r2 â r1 â re)
Equations of motion are 2 coupled differential equations:
m1d2r1
dt2= ð f(r2 â r1 â re) and m2
d2r2
dt2= âð f(r2 â r1 â re)
Transform to center of mass frame: M = m1 + m2 and R = 1M (m1r1 + m1r2)
Obtain 2 uncoupled differential equations,
M d2Rdt2
= 0, and ðd2Îrdt2
= âð fÎr
Îr = r2 â r1 â re and ð is the reduced mass given by 1ð
= 1m1
+ 1m2
P. J. Grandinetti Chapter 05: Vibrational Motion
Diatomic molecule vibration equations of motionDifferential equation of motion describing the vibration
ðdÎr2(t)
dt2 + ð fÎr(t) = 0
Same differential equation of motion as simple harmonic oscillator.
Solutions takes the same form,
Îr(t) = Îr(0) cosð0t where ð0 =âð fâð
In spectroscopy vibrational frequencies are given in terms of the spectroscopic wavenumber,
ᅵᅵ =ð0
2ðc0=
âð fâð
2ðc0, rearranges to ð f = ð
(2ðc0ᅵᅵ
)2
P. J. Grandinetti Chapter 05: Vibrational Motion
Force constants for selected diatomic molecules
ð f = ð(2ðc0ᅵᅵ
)2
Bond ð f/(N/m) ðâ10â28 kg ᅵᅵ/cmâ1 Bond length/pmHâH 570 8.367664 4401 74.1DâD 527 16.72247 2990 74.1
Hâ35Cl 478 16.26652 2886 127.5Hâ79Br 408 16.52430 2630 141.4Hâ127I 291 16.60347 2230 160.9
35Clâ35Cl 319 290.3357 554 198.879Brâ79Br 240 655.2349 323 228.4127Iâ127I 170 1053.649 213 266.7
16O=16O 1142 132.8009 1556 120.714Nâ¡14N 2243 116.2633 2331 109.412Câ¡16O 1857 113.8500 2143 112.814N=16O 1550 123.9830 1876 115.123Naâ35Cl 117 230.3282 378 236.139Kâ35Cl 84 306.0237 278 266.7
23Naâ23Na 17 190.8770 158 307.8
P. J. Grandinetti Chapter 05: Vibrational Motion
Coupled Harmonic Oscillators and Normal Modes
P. J. Grandinetti Chapter 05: Vibrational Motion
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 ð 1 and ð 2.3 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 = âð 1x1 + ð 2(x2 â x1)Restoring force acting on right mass: F2 = âð 1x2 â ð 2(x2 â x1)Coupled differential equations of motion for each mass:
mx1 + (ð 1 + ð 2)x1 â ð 2x2 = 0 and mx2 â ð 2x1 + (ð 1 + ð 2)x2 = 0
P. J. Grandinetti Chapter 05: Vibrational Motion
Two coupled oscillators: Normal ModesCoupled differential equations: mx1 + (ð 1 + ð 2)x1 â ð 2x2 = 0 and mx2 â ð 2x1 + (ð 1 + ð 2)x2 = 0
Uncoupled by coordinate transformation: ð1 = x1 + x2 and ð2 = x1 â x2
ð1 and ð2 are called normal mode coordinates
Gives 2 uncoupled equations: mð1 + (ð 1 + 2ð 2)ð1 = 0 and mð2 + ð 1ð2 = 0
Substitute 2 proposed solutions: ð1(t) = A1 cos(ð1t + ð1) and ð2(t) = A2 cos(ð2t + ð2)
Obtain:((ð 1 + 2ð 2) â mð2
1)
A1 cos(ð1t + ð1) = 0 and(ð 1 â mð2
2)
A2 cos(ð2t + ð2) = 0
Equations true for all values of t if: (ð 1 + 2ð 2) = mð21 and ð 1 = mð2
2
yields 2 normal mode frequencies: ð1 =â(ð 1 + 2ð 2)âm and ð2 =
âð 1âm Note: ð1 > ð2
Normal mode is collective oscillation where all masses move with same oscillation frequency.
P. J. Grandinetti Chapter 05: Vibrational Motion
Two coupled oscillators: Normal Modes
Imposing initial conditions:
ð1(0) = x1(0) + x2(0) and ð2(0) = x1(0) â x2(0), and ᅵᅵ1(0) = ᅵᅵ2(0) = 0
on solutions: ð1(t) = A1 cos(ð1t + ð1) and ð2(t) = A2 cos(ð2t + ð2)gives:
ð1 = ð2 = 0 and A1 = ð1(0) and A2 = ð2(0)
ð1(t) = ð1(0) cos(ð1t) and ð2(t) = ð2(0) cos(ð2t)
Converting normal mode coordinates back to original coordinates:
x1(t) =12
[ð2(0) cos(ð2t) + ð1(0) cos(ð1t)
]and x2(t) =
12
[ð2(0) cos(ð2t) â ð1(0) cos(ð1t)
]Oscillation of each mass is linear combination of two normal mode oscillations.
P. J. Grandinetti Chapter 05: Vibrational Motion
Two coupled oscillators: Normal ModesConsider two cases where only one normal mode is active:
ð2(0) = 0 or ð1(0) = 0
x1(t) = 12ð1(0) cos(ð1t)
x2(t) = â 12ð1(0) cos(ð1t)
â«âªâ¬âªâ out-of-phase,x1(t) = 1
2ð2(0) cos(ð2t)
x2(t) = 12ð2(0) cos(ð2t)
â«âªâ¬âªâ 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 ð1 = x1 + x2 and ð2 = x1 â 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.
P. J. Grandinetti Chapter 05: Vibrational Motion
Normal mode analysis
m1 m2
+x1-x1 0 +x2-x2 0
Define mass-weighted coordinates: q1 = m1â21 x1 and q2 = m1â2
2 x2Coupled differential equations are
q1 +(ð 1 + ð 2)
m1q1 â
ð 2
(m2m1)1â2q2 = 0 and q2 â
ð 2
(m2m1)1â2q1 â
(ð 3 â ð 2)m2
q2 = 0
In matrix form:
â¡â¢â¢â£q1
q2
â€â¥â¥âŠâââ
q
+
â¡â¢â¢â¢â¢â¢â£
(ð 1 + ð 2)m1
âð 2
(m2m1)1â2
âð 2
(m2m1)1â2
(ð 3 + ð 2)m2
â€â¥â¥â¥â¥â¥âŠâââââââââââââââââââââââââââââââââââââââââââââââ
K
â¡â¢â¢â£q1
q2
â€â¥â¥âŠâââ
q
= 0
P. J. Grandinetti Chapter 05: Vibrational Motion
Normal mode analysisAny coupled harmonic oscillator problem can be written in form: q + Kq = 0
K is a square n à n real symmetric (i.e., K12 = K21) matrix, q and q are n à 1 matrices.Similarity transformation of real n à n symmetric matrix into diagonal matrix, ð².
LT KL =
â¡â¢â¢â¢â£l11 l21 ⊠ln1l12 l22 ⊠ln2â® â® â± â®l1n l2n ⊠lnn
â€â¥â¥â¥âŠâ¡â¢â¢â¢â£
K11 K12 ⊠K1nK21 K22 ⊠K2nâ® â® â± â®
Kn1 Kn2 ⊠Knn
â€â¥â¥â¥âŠâ¡â¢â¢â¢â£
l11 l12 ⊠l1nl21 l22 ⊠l2nâ® â® â± â®ln1 ln2 ⊠lnn
â€â¥â¥â¥âŠ =â¡â¢â¢â¢â£
ð1 0 ⊠00 ð2 ⊠0â® â® â± â®0 0 ⊠ðn
â€â¥â¥â¥âŠ = ð²
ⶠEigenvalues, ði, of K along diagonal of ð².ⶠEigenvectors, li, of K along columns of L and along rows of LT .ⶠL is orthogonal matrix: product with its transpose, LT , is identity matrix, 1,
LLT =
â¡â¢â¢â¢â£l11 l12 ⊠l1nl21 l22 ⊠l2nâ® â® â± â®ln1 ln2 ⊠lnn
â€â¥â¥â¥âŠâââââââââââââââââââââââââââââââ
L
â¡â¢â¢â¢â£l11 l21 ⊠ln1l12 l22 ⊠ln2â® â® â± â®l1n l2n ⊠lnn
â€â¥â¥â¥âŠâââââââââââââââââââââââââââââââ
LT
=
â¡â¢â¢â¢â£1 0 ⊠00 1 ⊠0â® â® â± â®0 0 ⊠1
â€â¥â¥â¥âŠâââââââââââââââââââââââââ
1
= 1
P. J. Grandinetti Chapter 05: Vibrational Motion
Normal mode analysisInsert identity matrix as 1 = LLT before and after K in matrix form of equations of motion:
q + Kq = q + (LLT )âââ
1
K (LLT )âââ
1
q = q + L (LTKL)âââ
ð²
(LTq) = 0
Multiplying both sides by LT , regroup, and define new matrix variables:
LT
âââââq + L (LTKL)âââ
ð²
LTqâââââ = LT q
âââQ
+ (LTKL)âââ
ð²
(LT q)âââ
Q
= Q + ð² Q = 0
Q = LT q is linear transformation of q into Q, i.e., normal mode coordinatesDifferential equations in Q and ð² are uncoupled for each normal mode coordinate: Qi + ðiQi = 0Substitute in proposed solutions: Qi(t) = Ai cos(ðit+ði) gives (ði âð2
i ) cos(ðit+ði) = 0
Identify the ith normal mode frequency as ði =âði
With initial conditions, solutions become Qi(t) = Qi(0) cos(ðit
)P. J. Grandinetti Chapter 05: Vibrational Motion
How do we determine eigenvalues, ði, and the eigenvectors, li, from K?ith row in L is li eigenvector for ði eigenvalue.
Kli =â¡â¢â¢â¢â£
K11 K12 ⊠K1nK21 K22 ⊠K2nâ® â® â± â®
Kn1 Kn2 ⊠Knn
â€â¥â¥â¥âŠâ¡â¢â¢â¢â£
li1li2â®lin
â€â¥â¥â¥âŠ = ðili = ði
â¡â¢â¢â¢â£li1li2â®lin
â€â¥â¥â¥âŠ = ði
â¡â¢â¢â¢â£1 0 ⊠00 1 ⊠0â® â® â± â®0 0 ⊠1
â€â¥â¥â¥âŠâââââââââââââââââââââââ
1
â¡â¢â¢â¢â£li1li2â®lin
â€â¥â¥â¥âŠâââââââââ
â¡â¢â¢â¢â£K11 K12 ⊠K1nK21 K22 ⊠K2nâ® â® â± â®
Kn1 Kn2 ⊠Knn
â€â¥â¥â¥âŠâââââââââââââââââââââââââââââââââââââ
K
âði
â¡â¢â¢â¢â£1 0 ⊠00 1 ⊠0â® â® â± â®0 0 ⊠1
â€â¥â¥â¥âŠâââââââââââââââââââââââââ
1
âââââââââ
â¡â¢â¢â¢â£li1li2â®lin
â€â¥â¥â¥âŠâââ
li
= 0
Compactly written as(K â ði1) li = 0
Encountered same problem when diagonalizing moment of inertia tensor to determine PAS componentsand orientation.
P. J. Grandinetti Chapter 05: Vibrational Motion
How do we determine eigenvalues, ði, and the eigenvectors, li, from K?
Obtain eigenvalues by expanding determinant
|K â ði1| = 0
to get nth order polynomial equation in ð whose roots are eigenvalues.
Substitute each eigenvalue back into (K â ði1) 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.
P. J. Grandinetti Chapter 05: Vibrational Motion
P. J. Grandinetti Chapter 05: Vibrational Motion
Normal mode analysisBack to our original problem
â¡â¢â¢â£q1
q2
â€â¥â¥âŠâââ
q
+
â¡â¢â¢â¢â¢â¢â£
(ð 1 + ð 2)m1
âð 2
(m2m1)1â2
âð 2
(m2m1)1â2
(ð 3 + ð 2)m2
â€â¥â¥â¥â¥â¥âŠâââââââââââââââââââââââââââââââââââââââââââââ
K
â¡â¢â¢â£q1
q2
â€â¥â¥âŠâââ
q
= 0
To solve this problem we willdetermine eigenvalues, ði and eigenvectors, li, of Kobtain normal mode frequencies from ði =
âði
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 ð 3 = ð 1.
P. J. Grandinetti Chapter 05: Vibrational Motion
Normal mode analysis - Find the Eigenvalues
Obtain eigenvalues by solving the determinant,|||||||||ð 1 + ð 2
mâ ð â
ð 2m
âð 2m
ð 1 + ð 2m
â ð
|||||||||=(ð 1 + ð 2
mâ ð
)(ð 1 + ð 2
mâ ð
)â
ð 22
m2 = 0
obtainingð2 = ð =
ð 1 + ð 2 ± ð 2m
ð1 = (ð 1 + 2ð 2)âm and ð2 = ð 1âm
P. J. Grandinetti Chapter 05: Vibrational Motion
Normal mode analysis - Find the EigenvectorsObtain first eigenvector, associated with ð1 = (ð 1 + 2ð 2)âm, by expandingâ¡â¢â¢â£
(ð 1 + ð 2)âm â ð1 âð 2âm
âð 2âm (ð 1 + ð 2)âm â ð1
â€â¥â¥âŠâ¡â¢â¢â£
l11
l12
â€â¥â¥âŠ = 0,
to get 2 identical equations:((ð 1 + ð 2)âm â ð1
)l11 â (ð 2âm)l12 = 0, and â (ð 2âm)l11 +
((ð 1 + ð 2)âm â ð1
)l12 = 0,
which simplify to l11 = âl12, i.e. two elements have equal magnitude and opposite sign.
l1 =[
l11l12
]= b
[1â1
]where b is some unknown constant.
Insisting that l1 be normalized, i.e.,â
l211 + l212 = 1, gives b = 1ââ
2 and our 1st eigenvector is:
l1 =[
l11l12
]=
[1â
â2
â1ââ
2
]P. J. Grandinetti Chapter 05: Vibrational Motion
Normal mode analysis - Find the EigenvectorsSimilarly, second eigenvector, associated with ð2 = ð 1âm, is given by
l2 =â¡â¢â¢â£
l21
l22
â€â¥â¥âŠ =â¡â¢â¢â£
1ââ
2
1ââ
2
â€â¥â¥âŠTaking these two results together, we construct the eigenvector matrix
L =[
l11 l21l12 l22
]=â¡â¢â¢â£
1ââ
2 1ââ
2
â1ââ
2 1ââ
2
â€â¥â¥âŠFinally, we obtain solutions, in terms of the mass-weighted coordinates, as
q = LQ =â¡â¢â¢â£
q1
q2
â€â¥â¥âŠ =â¡â¢â¢â£
1ââ
2 1ââ
2
â1ââ
2 1ââ
2
â€â¥â¥âŠâ¡â¢â¢â£
Q1(0) cosð1t
Q2(0) cosð2t
â€â¥â¥âŠP. J. Grandinetti Chapter 05: Vibrational Motion
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P. J. Grandinetti Chapter 05: Vibrational Motion
Polyatomic Molecule Vibrations
P. J. Grandinetti Chapter 05: Vibrational Motion
Polyatomic Molecule VibrationsNormal mode analysis of molecule with N atoms
Define each atomâs displacement coordinates in PAS of moleculeâs moment of inertia tensor withcenter of mass at origin.For example, for water molecule define
105°
a
b
ð1 = aO â aO,e, ð2 = bO â bO,e, ð3 = cO â cO,e,ð4 = aH1
â aH1,e, ð5 = bH1â bH1,e, ð6 = cH1
â cH1,e,ð7 = aH2
â aH2,e, ð8 = bH2â bH2,e, ð9 = cH2
â cH2,e,
subscript e represents equilibrium coordinate.ⶠFurther define mass-weighted displacement coordinates, q1,⊠, q3N , e.g.,
qi = m1â2i ði where
m1 = m2 = m3 = mO, m4 = m5 = m6 = mH , and m7 = m8 = m9 = mH .Derive equations of motion from kinetic and potential energy
ddt
(ðK(q1,⊠, q3N)
ðq
)+
ðV(q1,⊠, q3N)ðqi
= 0
P. J. Grandinetti Chapter 05: Vibrational Motion
Polyatomic Molecule VibrationsKinetic and Potential Energy in mass-weighted coordinates
Kinetic energy becomes
K = 12
3Nâi=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) +3Nâi=1 ᅵᅵᅵᅵᅵᅵ*
0(ðV(0)ðqi
)qi + 1
2
3Nâi=1
3Nâk=1
(ð2V(0)ðqiðqk
)qiqk +â¯
Rewriting as
V(q) â V(0) + 12
qT q where qT =[q1,⊠, q3N
], and q is its transpose
where is 3N Ã 3N matrix with elements given by
i,k =(ð2V(0)ðqiðqk
)P. J. Grandinetti Chapter 05: Vibrational Motion
Polyatomic Molecule VibrationsNormal mode analysis of molecule with N atoms
Derive equations of motion from kinetic and potential energy
ddt
(ðK(q1,⊠, q3N)
ðq
)+
ðV(q1,⊠, q3N)ðqi
= 0
and obtain a set of 3N coupled differential equations (in matrix notation),
q + q = 0As before, heart of the problem is determining eigenvalues, ði, and eigenvectors, li of .Orthogonal eigenvector matrix, L, transforms problem into 3N uncoupled differential equations,
q + q = q + (LLT )(LLT )q = 0 ⶠLT q + (LTL) (LTq) = Q + ð²Q = 0
ⶠsolutions: Qi(t) = Qi(0) cos(ðit)ⶠnormal mode frequencies: ði =
âði
ⶠnormal mode coordinates: Q = LTqNormal modes with identical frequencies are called degenerate modes.
P. J. Grandinetti Chapter 05: Vibrational Motion
Polyatomic Molecule VibrationsNormal mode analysis of molecule with N atoms
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 â Ve â12
QTð²Q = 12
3Nâ6âi
ðiQ2i , and K = 1
2QTQ = 1
2
3Nâ6âi
Q2i .
Keep in mind that approximations were made along the way.ⶠIgnored 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 ânormalmodes.â
ⶠCoordinate 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
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.
ð1 = aA â aA,e, ð2 = bA â bA,e = 0, ð3 = cA â cA,e = 0,ð4 = aB â aB,e, ð5 = bB â bB,e = 0, ð6 = cB â cB,e = 0,
From 6 displacement coordinates, define mass-weighted coordinates,qi = m1â2
i ði, where m1 = m2 = m3 = mA, and m4 = m5 = m6 = mB.
Since Îr = r â re = ð4 â ð1 = q1âm1â21 â q4âm1â2
4 we obtain potential energy function
V â Ve â12ð f(r â re)2 = 1
2ð f
(q1
m1â21
âq4
m1â24
)2
= 12ð f
(q2
1m1
â2q1q4
(m1m4)1â2+
q24
m4
)Identify non-zero matrix elements of as
11 = ð fâm1, and 44 = ð fâm4, and 14 = âð fâ(m1m4
)1â2
P. J. Grandinetti Chapter 05: Vibrational Motion
Normal mode analysis of diatomic moleculeEquations of motion in matrix form is
â¡â¢â¢â¢â¢â¢â¢â¢â¢â¢â¢â¢â¢â¢â£
q1
q2
q3
q4
q5
q6
â€â¥â¥â¥â¥â¥â¥â¥â¥â¥â¥â¥â¥â¥âŠâââ
q
+
â¡â¢â¢â¢â¢â¢â¢â¢â¢â¢â¢â¢â¢â¢â¢â¢â¢â£
ð f
m10 0 â
ð f(m1m4
)1â20 0
0 0 0 0 0 0
0 0 0 0 0 0
âð f(
m1m4)1â2
0 0ð f
m40 0
0 0 0 0 0 0
0 0 0 0 0 0
â€â¥â¥â¥â¥â¥â¥â¥â¥â¥â¥â¥â¥â¥â¥â¥â¥âŠâââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââââ
â¡â¢â¢â¢â¢â¢â¢â¢â¢â¢â¢â¢â¢â¢â£
q1
q2
q3
q4
q5
q6
â€â¥â¥â¥â¥â¥â¥â¥â¥â¥â¥â¥â¥â¥âŠâââ
q
= 0
Note: Four equations are already uncoupled and have eigenvalues of ði = 0.
P. J. Grandinetti Chapter 05: Vibrational Motion
Normal mode analysis of diatomic moleculeRemaining equations of motion are[
q1q4
]+â¡â¢â¢â£
ð fâm1 âð fâ(m1m4
)1â2
âð fâ(m1m4
)1â2 ð fâm4
â€â¥â¥âŠ[
q1q4
]= 0
Get eigenvalues from solving|||||||ð fâm1 â ð âð fâ
(m1m4
)1â2
âð fâ(m1m4
)1â2 ð fâm4 â ð
||||||| = (ð fâm1 â ð)(ð fâm4 â ð) â ð 2f â
(m1m4
)= 0
gives ð = 0 andð fð
where 1ð
= 1m1
+ 1m4
=m1 + m4
m1m4= M
m1m4, or ð = (m1m4)âM,
ð is the reduced mass, and M = m1 + m4 is the total mass.5 out of 6 eigenvalues are zeroâassociated with 3 translational and 2 rotational degrees offreedom.1 out of 6 degrees of freedom is associated with vibration with ð1 =
âð fâð, and normal mode
solution Q1(t) = Q1(0) cos(ð1t
)P. J. Grandinetti Chapter 05: Vibrational Motion
Normal mode analysis of diatomic molecule
Turning to normal mode eigenvectors. If we substitute the eigenvalue ð1 = ð fâð into
( â ð11)l1 =
â¡â¢â¢â¢â£ð fâm1 â ð1 âð fâ
(m1m4
)1â2
âð fâ(m1m4
)1â2 ð fâm4 â ð1
â€â¥â¥â¥âŠâ¡â¢â¢â£
l11
l14
â€â¥â¥âŠ = 0,
we obtain two equivalent equations, which simplify to l11 = â(m4âm1
)1â2 l14Write eigenvector as
l1 =â¡â¢â¢â£
l11
l14
â€â¥â¥âŠ = bâ¡â¢â¢â£â(m4âm1
)1â2
1
â€â¥â¥âŠ ⶠl1 =
â¡â¢â¢â¢â£(ðâm1
)1â2
â(ðâm4
)1â2
â€â¥â¥â¥âŠwhere b =
(m4âM
)1â2 after normalization.
Same approach gives
l2 =
â¡â¢â¢â¢â£(ðâm4
)1â2
(ðâm1
)1â2
â€â¥â¥â¥âŠP. J. Grandinetti Chapter 05: Vibrational Motion
Normal mode analysis of diatomic moleculeBring eigenvectors together to create transformation between mass-weighted and normalcoordinates:
Q = LTq =â¡â¢â¢â£
Q1
Q4
â€â¥â¥âŠ =â¡â¢â¢â¢â¢â¢â£
(ð
m1
)1â2â(
ðm4
)1â2
(ð
m4
)1â2 (ð
m1
)1â2
â€â¥â¥â¥â¥â¥âŠâ¡â¢â¢â£
q1
q4
â€â¥â¥âŠThe Q1 coordinate expands to
Q1 = (ð)1â2 ( q1âm1â21
âââð1
â q4âm1â24
âââð4
)= (ð)1â2 ( ð1 â ð4
âââÎr
)= (ð)1â2 Îr
Rearranging gives vibrational motion solution:
Îr(t) = (ð)â1â2 Q1(t) = (ð)â1â2 Q1(0) cos(ð1t
)P. J. Grandinetti Chapter 05: Vibrational Motion
Normal mode analysis of diatomic moleculeQ4 normal mode (with ð4 = 0) can be expanded as
Q4 = (ð)1â2(
q1âm1â24 + q4âm1â2
1
)= 1
M1â2
(m1ð1 + m4ð4
)= 1
M1â2
(mA(aA â aA,e) + mB(aB â aB,e)
)Recalling center of mass along the a axis,
aCOM = (mAaA + mBaB)âM and aCOM,e = (mAaA,e + mBaB,e)âM = 0
where center of mass when atoms are at equilibrium coordinates is zero, gives
Q4 = M1â2
(mAaA + mBaB
)M
= M1â2aCOM
Since ð4 = 0, equation of motion is Q4 = 0
initial condition of aCOM = Q4 = 0 means aCOM cannot change in time.
ði = 0 for 5 normal modes associated with translation and rotation, so no change in center ofmass or PAS orientation as molecule vibrates.
P. J. Grandinetti Chapter 05: Vibrational Motion
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P. J. Grandinetti Chapter 05: Vibrational Motion