Subject: normal vibrations

From: [email-masked] (Dr. James Stewart)
Date: Thu, 31 Aug 95 10:15:53 PDT

Ferenc Molnar writes:
> The eigenvalue number k correspond to (w_k)^2, where the w_ks
> are the vibrational frequencies, this means:
> 
>  (w_k)^2=K_k/M_k 
> 
> K_k: force constant of the kth normal mode
> M_k: reduced mass of the k_th normal mode
> 
> Now my question is, if (w_k)^2 determines only the ratio
> of K_k and M_k, then how are the reduced masses, reported
> in the "vibrational analysis" section of the MOPAC output 
> file, calculated? Is there a convention, which "fraction" of
> (w_k)^2 to use for K_k and which for M_k?

The reduced mass is like atomic charges, in that it is not an observable,
however, for specific systems - mainly homonuclear diatomics - the reduced 
mass does have meaning.   The reduced mass definition used in MOPAC can be
understood as follows:

Each vibration can be modelled by a mass, M_k, at the end of a spring of
force constant K_k, attached to an infinite mass.

  Inf   |       K_k        M_k
  Mass  |
        |^^^^^^^^^^^^^^^^^^O
        |
        |

The contribution to the mass is proportional to the amount each atom
contributes to the normal mode, and is proportional to the fraction of
the atomic mass contributed by each atom.

Put in more formal terms, the contribution of each atom to the effective mass
of a vibration is proportional to the product of the intensity on that
atom times the mass-weighted intensity.

 rho = sum_A <c_A|c_A><c_A|M_A|c_A> = sum_A (c_A_x**2+c_A_y**2+c_A_z**2)**2*M_A

where c_A are the coefficients of the normal modes.  

Consider H2: c_1 = 0.7071*H_1+0.7071*H_2

     rho_1 = 0.7071**4*1 + 0.7071**4*1 =0.5

Consider M-H, M being an atom of very large mass, say 1000:

             c_1 = 0.0316*M+0.9995*H

     rho_1 = 0.0316**4*1000 + 0.9995**4*1 = 0.9990

[0.0316 ~ sqrt(1/1000); 0.9995 ~sqrt(1-1/1000)]

Consider N2: c_1 = 0.7071*N_1+0.7071*N_2

     rho_1 = 0.7071**4*14+0.7071**4*14 = 7.0

Jimmy Stewart