## Thursday, July 29, 2021

### More Advanced Lagrangian Applications: Modeling A Linear Tri-atomic Molecule

We have already seen how Lagrangian dynamics applies to simple mechanical systems, e.g.

• Lagrangian Dynamics Revisited (1)
• We now examine a more advanced application, namely in terms of modeling a linear triatomic molecule.  We sketch the basic model system below:
We see the linear molecule is composed of three components: m1, m2, and M,  with the smaller masses m1 = m2 = m.  We also apply the condition they can only move along the linear dimension with atoms specified at positions: x1, x2 and x3.  We designate the equilibrium position as b.

We recall here the Lagrangian L = T - V.

And obtain the kinetic energy. T and potential energy V:

Kinetic energy:

T   =   ½  ( m x1’ 2   -   M x2 2    + m x3’ 2 )

Potential energy:

V =   ½  k[ (x2 – x1) – b] 2   +  ½  k[ (x3 – x2) – b] 2

T   =

½  [(m….0…..0)

[(0….M…..0)

[(0….0… ..m)

And make use here of the refined position parameter in terms of displacements h i:

h i        =  x i  – x o i

So:   h 1           =  x 1  – x o 1

x 02  – x o 1   =  x 03  – x o2 =   b

Then the potential energy is:

V =   ½  k[ (h 2 – h 1) 2   +  ½  k[ (h 3 – h 2)  2 ]

And the kinetic energy:

T   =   ½  ( m h 1’ 2   -   M h 2 2    + m h 3’ 2 )

Rewriting V:

V   =   ½  k[ (2 h 2 2    + h 1 2   +  h 3 2  -  2  h 1 h 2  - 2  h 3 h 2 )]

In tensor form:

V   =

½  [(k… .-k…..0)

[(-k….2k…. -k)

[(0….-k… .. k)

T  =  h  ×  T ×   h

By Lagrange’s eqns.(Tensor form)

T ×   h   +     V ×   h   =  0

Assume a solution of form:  exp (iw t)

w 2    T ×   h   +     V ×   h   =  0

And the determinant of coefficients must = 0 for non-trivial soln.

è   | w 2 T  +   V  |  =  0

è

Secular determinant:

(k - w 2  …-k…   . .0)

(-k …..2 k - w 2 m  ..  -k)

(0…. ….-k….. k - w 2 m)

From here and working through some messy algebra (solving the secular equations from the determinant), we can arrive at three solutions for the angular frequency:

w1  =  0,     w2  = Ö(k/ m),   w3  = Ö{ k/ m (1 + 2m/ M)}

For amplitudes, three solutions are obtained:

h 1  =   A1 exp (iw t)

h 2  =   A2 exp (iw t)

h 3  =   A3 exp (iw t)

Suggested Problem:

For each of the frequency solutions applied to the linear tri-atomic molecule, show the relation between the normalized amplitudes: A1, A2 and A3