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PHYS 705: Classical MechanicsSmall Oscillations: ExampleA Linear Triatomic Molecule

1

A Linear Triatomic Molecule

Experimentally, one might be interested in the radiation resulted from the

intrinsic oscillation modes from these triatomic molecule.

1x

Mm m

kk

2x 3x xbb

2 22 1 3 22 2

k kV x x b x x b

The potential energy for the two springs is,

2


1, 2,3j 0j j jx x

1x

Mm m

kk

2x 3x x

Now, we will introduce generalized coordinates relative to their

equilibrium positions, :01 02 03, ,x x x

b b

Note: 02 01 03 02x x x x b

3


Expanding the potential energy about its equilibrium position, we have:

2 22 1 3 22 2

k kV

In matrix form, this quadratic

form has this form:

2 2 2 21 1 2 2 2 2 3 32 2

2 2k kV

02

0

k kk k k

k k

V

Multiplying the squares out, we have:

12 jk j kV V 2 2 2

1 1 2 2 1 2 2 3 3 2 322kV

2 1

2 1 2 1 02 01x bx x x x x

2 02 1 01x x x x

4

Recall, we have


Direct method to evaluate using

02

0

kkk k

k kk

V

By taking the partial derivatives directly and evaluating at ,

jkV

2 22 1 3 22 2

k kV x x b x x b

2

0

jkj k

VVq q

0j jx x

2 11

22

V k x x bq

and 11 :V1

2

1

V kq q

2 1 3 22

2 22 2

V k kx x b x x bq

and

12 :V

1

2

2

V kq q

5


The kinetic energy for the three mass is given by:

2 2 21 3 22 2

m MT x x x

In matrix form, this quadratic

form has this form:

0 00 00 0

mM

m

T

Substituting our generalized coordinates or ,0j j jx x

2 2 21 3 22 2

m MT 12 jk j kT T

j jx

6


Combining these two quadratic forms into the characteristic equation,

Explicitly evaluating this determinant, we have the following equation,

2

2 2

2

02 0

0

k m kk k M k

k k m

V T V T

2 2 22 22 2 0k Mk m k mk

22 2 02k M m Mk m m

2 2 22 2 2 0k m k M kk m

2 22k m k 22k 2 2 4 02k M k m mM

7


And, this has three distinct solutions (eigenfrequencies):

The solution means that the corresponding normal coordinate will

have the following trivial ODE: 1 0

1 0 2

km

321k m

m M

Note:

1 10 0

j

or 1 0 (uniform

translational motion)

The entire molecule will simply move uniformly to the right or left; no

oscillations (not quite interesting motion by itself)

8


Now, we find the eigenvectors for each solved eigenvalues using:

1 0 11 21

11 21 31

21 31

02 0

0

a k a ka k a k a k

a k a k

Normal Mode #1:

2 0r r V T a

21

22

23

02 0

0

r r

r r

r r

k m k ak k M k a

k k m a

1r

9


Solving 1st and 3rd equations, we have and

11 21

11 21 31

21 31

02 0

0

a k a ka k a k a k

a k a k

21 31a a 11 21a a

Putting them together, we have 11 21 31a a a

(Note that the solution also satisfies the 2nd equation.)11 21 31a a a

So, the eigenvector for is: 1 11

111

a

a1 0

Normal Mode #1: 1r

10

Finally, we have the normalized

eignvector for :


1 1 a Ta 1

211 2 1a M m This gives:

1

11 1

2 1M m

a

The eigenvector needs to be normalized with respect to T:

1 0

211

0 0 11 1 1 0 0 1 1

0 0 1

ma M

m

111

2a

M m

Normal Mode #1: 1r

11


22

12 22 32

22

0

2 0

0

a kMa k a k a km

a k

Normal Mode #2:

12

22

32

0 0

2 0

0 0

kk m km a

kMk k k am

akk k mm

2r

2km

22

12 32

0aa a

12

A Linear Triatomic MoleculeNormal Mode #2: 2r

22

12 32

0aa a

2 12

101

a

a

2 2 a Ta 1

Again, we need to normalized with respect to T:

212

0 0 11 0 1 0 0 0 1

0 0 1

ma M

m

212 2 1a m This gives: 2

11 02 1m

a

13

A Linear Triatomic MoleculeNormal Mode #3:

23 13

23 23

23 33

02 0

0

k m k ak k M k a

k k m a

3r

321k m

m M

First, let try to simply the matrix elements first:

23

kk m km

21 m mM

k 1k 2

2

mM

kmM

23

22 2 1k mk M k Mm M

2k 2kM km

kMm

14


13 23

13 23 33

23 33

2 0

0

2 0

kma a kM

kMa k a a km

kma k aM

Normal Mode #3:

13

23

33

2 00

0 2

km M k ak kM m k a

k km M a

3r

Putting these values back into the matrix, we have

23 132ma aM

132 ma

M 13

Mam 33 0a

13 33a a

15


3 13

12

1a m M

a

3 3 a Ta 1

Again, we need to normalized with respect to T:

213

10 02 21 1 0 0 1

0 0 1

mm ma M

M Mm

23 13

33 13

2ma aM

a a

16


213

10 02 21 1 0 0 1

0 0 1

mm ma M

M Mm

13 2

142

ammM

This gives:

3

11 2

2 1 2 1m M

m m M

a

22 213 13

2 41 1 2 2 1m

m ma m a mM M

m

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A Linear Triatomic MoleculeThen, our general solution is given by:

j jr ra *r ri t i tr r rC e C e

(the complex coefficient

will be determined by IC)and

rC

1 1 2 31

2 21

2 1 21m m mm MM

2 1 302

12

22M M mM m

3 1 2 31

21 1

2 1 22m mM mm M

1a 2a 3a

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Longitudinal Normal ModesSo, if a single normal mode is active, the motion of the three generalized

coordinates will looks like the following,

1 1 11 , ,

2m M m

M m

j

1 1: 0

19

r

1 2 3, ,

2 2,0,2m

2 1: km

3 12: 1k m

m M

3 3 3, 2 ,2 2

nMm M

Longitudinal Normal Modes

Animations for the two non-rigid-translational modes: 2 3 and

http://cis.poly.edu/~mleung/CS4744/f03/ch04/LinMole/LinMole.html

(from Polytechnics Institute of NY Univ: K. Ming Leung)

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Summary

1. Pick generalized coordinates and find and

2. Expand T and V about equilibrium

This gives two real symmetric quadratic forms:

( )jT q ( )jV q

0 jq( )jk jT ( )jk jV

0j j jq q with

3. Calculate eigenfrequencies from characteristic equation

4. Calculate eigenvectors for each eigenfrequencies using

5. Normalize eigenvectors with respect to T:

6. General solutions are in terms of the normal modes

7. Original coordinates are related back through

2r r

2det 0jk r jkV T

2 0jk r jk jrV T a 1jk jr krT a a

*r ri t i tr r rt C e C e

j jr ra rC determined by IC

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PHYS 705: Classical Mechanics - George Mason Universitycomplex.gmu.edu/ Small Oscillations... · PHYS 705: Classical Mechanics Small Oscillations: Example A Linear Triatomic Molecule