Classical Mechanics. Small Oscillations Examples

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PHYS 705: Classical Mechanics
Small Oscillations: Example
A Linear Triatomic Molecule
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A Linear Triatomic Molecule
k
m
k
M
b
x1
x2
m
b
x3
x
Experimentally, one might be interested in the radiation resulted from the
intrinsic oscillation modes from these triatomic molecule.
The potential energy for the two springs is,
V
k
k
2
2
x

x

b

x

x

b
 2 1 
 3 2 
2
2
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A Linear Triatomic Molecule
k
k
m
x1
M
b
m
b
x2
x3
Now, we will introduce generalized coordinates relative to their
equilibrium positions, x01 , x02 , x03 :
 j  x j  x0 j
Note:
j  1, 2, 3
x02  x01  x03  x02  b
x
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A Linear Triatomic Molecule
Expanding the potential energy about its equilibrium position, we have:
x2  x1  b  x2  x1   x02  x01 
k
k
2
2
V   2  1   3   2 
2
2
 x2  x02  x1  x01
  2  1
Multiplying the squares out, we have:
k 2
k
1  21 2   22    22  2 23  32 

2
2
k
V  12  1 2  21  222  23  32  32 
2
V
In matrix form, this quadratic
form has this form:
 k

V   k
 0

k
2k
k
0 

k 
k 
1
V  V jk  j k
2
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A Linear Triatomic Molecule
  2V 
Direct method to evaluate V jk using V jk  
 q q 
 j k 0
k
k
2
2
Recall, we have V   x2  x1  b    x3  x2  b 
2
2
x j  x0 j,
By taking the partial derivatives directly and evaluating at
V11 :
V k
  2  x2  x1  b  
q1 2
V12 :
V k
k
  2  x2  x1  b     2  x3  x2  b  
q2 2
2
V
 k
q1q2
2
and
and

 2V
 k
q1q1
 k

V   k
 0

k
2k
k
0 

k 
k 
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A Linear Triatomic Molecule
The kinetic energy for the three mass is given by:
T
m 2
M 2
2


x

x

x2

1
3 
2
2
Substituting our generalized coordinates
 j  x j  x0 j or  j  x j,
m
M
T  12  32   22
2
2
In matrix form, this quadratic
form has this form:
m 0

T0 M
0 0

1
T  T jk  j k
2
0

0
m 
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A Linear Triatomic Molecule
Combining these two quadratic forms into the characteristic equation,
k   2m
k
V   T  V   2 T  k
2k   2 M
0
k
0
k
0
k   2m
Explicitly evaluating this determinant, we have the following equation,
 k   m   2k   M   2k  k   m   0
 k   m   k   m  2k   M   2k   0
2
2
 k   m   2k
2
2
2
2
2
2
2
2
2
 2k 2  k 2 M  2k 2 m   4 mM   0

 2  k   2 m    k  M  2m    2 Mm   0
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A Linear Triatomic Molecule
And, this has three distinct solutions (eigenfrequencies):
1  0
2 
k
m
3 
k  2m 
1 

m
M 
Note:
The
1  0 solution means that the corresponding normal coordinate  j will
have the following trivial ODE:
1   0   1  0
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)
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A Linear Triatomic Molecule
Now, we find the eigenvectors for each solved eigenvalues using:
 V   T a
2
r
 k  r2 m
k

2

k
2
k


M
r

 0
k

r
0
  a1r 
 
 k   a2 r   0
k  r2 m   a3r 
0
Normal Mode #1: r  1
1  0
a11k  a21k  0


a11k  a21 2k  a31k  0

a21k  a31k  0

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A Linear Triatomic Molecule
Normal Mode #1: r  1
a11k  a21k  0


 a11k  a21 2k  a31k  0

 a21k  a31k  0

Solving 1st and 3rd equations, we have a21  a31 and a11  a21
Putting them together, we have a11  a21  a31
(Note that the solution a11  a21  a31 also satisfies the 2nd equation.)
So, the eigenvector for
1  0 is:
 1
 
a1  a11 1
 1
 
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A Linear Triatomic Molecule
Normal Mode #1: r  1
The eigenvector needs to be normalized with respect to T:
a 1Ta1  1
This gives:
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a
m 0

a112 1 1 1  0 M
0 0

 M  2m   1
Finally, we have the normalized
eignvector for
1  0 :
a11 
a1 
0 1 
 
0 1   1

m 
 1
1
M  2m
 1
1
 
1

M  2m  
 1
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A Linear Triatomic Molecule
Normal Mode #2: r  2
k
2 
m
k

k

m
k

m

kM
 k
2k 

m

 0
k

 a22 k  0


M



a
k

a
k
2

 12
22 
  a32 k  0
m



 a22 k  0


  a12 
 
 k   a22   0

  a32 
k
k  m 
m 
0
 a22  0

a12   a32
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A Linear Triatomic Molecule
Normal Mode #2: r  2
 a22  0

a12   a32
1
 
a 2  a12  0 
 1 
 
Again, we need to normalized with respect to T:
a 2 Ta 2  1
This gives:
a122  2m   1
m 0

a122 1 0 1  0 M
0 0

0  1 
 
0  0  1
m   1
1
1  
a2 
0

2m  
 1 
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A Linear Triatomic Molecule
Normal Mode #3: r  3
k  2m 
3 
1 

m
M 
 k  32 m
k

2

k
2
k


3M

 0
k

  a13 
 
 k   a23   0
k  32 m   a33 
0
First, let try to simply the matrix elements first:
 k  2m  
k   m  k   1 
 m
M 
m
2m 

 k k1 

M 

2km

M
2
3
 k  2m  
2k   M  2k    1 
M
M 
m
 kM

 2k  
 2k 
 m

kM

m
2
3
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A Linear Triatomic Molecule
Normal Mode #3: r  3
Putting these values back into the matrix, we have
  2km M

 k

0

k
0
  a13 
 
 kM m
k
  a23   0
k
 2km M   a33 
2km

  a13 M  a23 k  0

kM

 a33 k  0
 a13 k  a23
m

2km


a
k

a
0
23
33

M

a23 
2m
a13
M
2 m
M
 a13 
a13
 a33  0
M
m
a13  a33
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A Linear Triatomic Molecule
Normal Mode #3: r  3
2m

a

a13
 23
M

 a33  a13
 1 


a3  a13  2m M 
 1 


Again, we need to normalized with respect to T:
a 3Ta3  1
m 0
2m  

a132 1 
1  0 M
M


0 0
 1 
0 

2
m

 1
0 
 M 

m

1


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A Linear Triatomic Molecule
Normal Mode #3: r  3
m 0
2m  

a132 1 
1  0 M
M


0 0
 1 
0 

  2m 
0 
1
 M 

m

 1 
 m 
2m  
4m 2 

2 
2 
a13 1 
1  2m   a13  2m 
 1
M
M 




 m 
This gives:
a13 
1
4m 2
2m 
M
1


1


a3 

2
m
M

2m 1  2m M  

1


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A Linear Triatomic Molecule
Then, our general solution is given by:
 j  a jr r
and
a1
 r  Cr e
ir t
a2
*  ir t
r
C e
(the complex coefficient Cr
will be determined by IC)
a3
1 
1
1
1
1 
2 
3
M  2m
2m
2m 1  2m M 
2 
1
1 
M  2m
3 
1
1
1
1 
2 
3
M  2m
2m
2m 1  2m M 
0

2
2M  2  M m 
3
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Longitudinal Normal Modes
So, if a single normal mode
coordinates
 r is active, the motion of the three generalized
  will looks like the following,
j
1 ,2 ,3
 1 : 1  0 

k 
 2 :  1 

m


k  2m  
 3 :  1 
1 
 

m
M



1
 1 ,  1 ,  1
M  2m
m
 2 , 0,   2 
2
mM
 3 ,  2  3 ,  3 
2  2m  M 
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Longitudinal Normal Modes
Animations for the two non-rigid-translational modes:
 2 and  3
http://cis.poly.edu/~mleung/CS4744/f03/ch04/L
inMole/LinMole.html
(from Polytechnics Institute of NY Univ: K. Ming Leung)
https://www.youtube.com/watch?v=W5gi
mZlFY6I
(from YouTube kyoroskichannnel)
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Summary
1. Pick generalized coordinates and find T ( q j ) and V ( q j )
2. Expand T and V about equilibrium q0 j
 This gives two real symmetric quadratic forms: T jk ( j ) V jk ( j )
with
 j  q j  q0 j
3. Calculate eigenfrequencies
r  r2 from characteristic equation
det V jk  r2T jk   0
4. Calculate eigenvectors for each eigenfrequencies using
5. Normalize eigenvectors with respect to T: T jk a jr akr  1
V
jk
 r2T jk  a jr  0
6. General solutions are in terms of the normal modes
 r  t   Cr ei t  Cr*e i t
Cr determined by IC
7. Original coordinates are related back through  j  a jr r
r
r