# Appendix. Various Details and Explanations

## Kinematic Equations in Mechanics

Kinematic equations are constrained equations found in dynamics of body. In the mechanics, the constraint that is incurred on is to have constant acceleration on a body with mass m.

 $v= v 0 +at$ A1
 $x= x 0 + v 0 t+ 1 2 a t 2$ A2
 $v 2 = v 0 2 +2a( x- x 0 )$ A3
 $v ¯ = v+ v 0 2$ A4

## Determinant

Determinant is used in eigenvalue equations and other important. The simplest case is 2 x 2 determinant,
 $| a 1 a 2 a 3 a 4 |= a 1 a 4 − a 2 a 3$ A5
In the case of 3 x 3 determinant,
 $| a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 9 |= a 1 ( a 5 a 9 − a 6 a 8 )− a 2 ( a 4 a 9 − a 6 a 7 )+ a 3 ( a 4 a 8 − a 5 a 7 )$ A6
Larger determinants can be solved by similar techique used in going from 2 x 2 to 3 x 3 determinant.

## Lagrangian Dynamics

Setting up equations of motion in terms of force can be rather nuisance. It is particularly true when there are many two-body interactions. The general procedure is to set up Lagrangian equations of motion to make them into matrix form.

The Lagrangian function is defined by

 $L( q i , q ˙ i ,t )=T( q i , q ˙ i ,t )−V( q i )$ A7
where T is kinetic energy and V is potential energy. The variables are in terms of ith coordinate qi, and ${\stackrel{˙}{q}}_{i}$ is the velocity associated with qi (q with a dot and a double-dot on top indicates first derivative in time and second derivative, respecitively), and t is time.
 $T( q i , q ˙ i ,t )= 1 2 m q ˙ i 2$ A8 $V( q i )= 1 2 k ( q i − q 0 ) 2$ A9
The above equation for potential energy is for harmonic motion of masses in linear Hooke's law forces between them.

Then, the dynamical equation, called Lagrange's equation, is obtained by the following.

 $d dt ( ∂L ∂ q ˙ i )− ∂L ∂ q i =0$ A10
Let us now use an example. Let us write dynamical equation for molecular vibration in one dimension. Easiest one perhaps is the CO2, as shown below.

Figure A1. CO2 molecule and its coordinates

The Lagrangian is written as

 $L( q i , q ˙ i ,t )= 1 2 m O x ˙ 1 2 + 1 2 m C x ˙ 2 2 + 1 2 m O x ˙ 3 2 − 1 2 k ( x 2 − x 1 ) 2 − 1 2 k ( x 3 − x 2 ) 2$ A11

We now need to subject Eqn. A11 to Eqn. A10 for each of ith component. The first term on the left-hand side of A10 is

 $d dt ( ∂L ∂ q ˙ i )= ∑ i m i x ¨ i$ A12
where mi is the mass associated with xi coordinate. You can easily see it that by taking the derivative of the first three terms of the right-hand side of Eqn. A11, and subsequently taking the time-derivative.

The second term on the left-hand side of A10 is

 $∂L ∂ q i =−k( x 2 − x 1 )+k( x 2 − x 1 )−k( x 3 − x 2 )$ A13
We can rewrite Eqns. A12 and A13 into three simultaneous linear equations containing given i component,
 $m O x ¨ 1 −k( x 2 − x 1 )=0 m C x ¨ 2 +k( x 2 − x 1 )−k( x 3 − x 2 )=0 m O x ¨ 3 −k( x 3 − x 2 )=0$ A14
This gives the equations of motion to be solved.