Example: Normal modes of a linear triatomic

 

A - B - A à x

 

  1. Consider only motion along x for this example.
  2. Assume that the force constant for the A - B bond is k.

 

The coupled equations have the form:

In mass-weighted Cartesian coordinates:

Substituting in the values:

We obtain the equations:

These equations can be written in matrix form as:

Note that we have assumed that a13 = 0 for simplicity.

 

 

The determinant can be expanded as:

 

One solution is:

The second solution is obtained by expansion of the terms inside the brackets {}:

 

This can be simplified as follows:

 

The eigenvectors are the matrix terms that transform from mass-weighted Cartesian space to normal coordinate space. The eigenvectors can be found by substituting in each of the eigenvalues into the equation:

 

For example, for l = a11 we have:

 

Thus,

Finally, we use normalization to determine the magnitude of the coefficients.

The eigenvector corresponding to this normal mode is:

The normal mode corresponds to a symmetric stretch.

The normal coordinate is:

The central atom B does not move in this normal mode. This illustrates the point that in the normal mode description the center of mass does not move.

We can repeat the calculation using the roots l+ and l-. The eigenvectors in these cases are:

Where M = 2mA + mB.

In l2 the negative sign on the central atom B indicates that the displacement is opposite to the end atoms A. Thus, this mode is an asymmetric stretch.

The third eigenvector represents translation. Translation has not been separated out since we are working in Cartesian coordinates.