430 X A Character Study
whereDis a diagonal matrix with positive diagonal elements. By the linear transfor-
mationx=Tythe equations of motion are brought to the form
y ̈+Dy= 0.These ‘decoupled’ equations can be solved immediately: if
y=(η 1 ,...,ηn)t, D=diag [ω^21 ,...,ωn^2 ],withωk> 0 (k= 1 ,...,n),then
ηk=αkcosωkt+βksinωkt,whereαk,βk(k= 1 ,...,n)are arbitrary constants of integration. Hence there exist
vectorsak,bk∈Rnsuch that every solution of (13) is a linear combination of solu-
tions of the form
akcosωkt, bksinωkt (k= 1 ,...,n),the so-callednormal modesof oscillation. The eigenvalues of the matrixB−^1 Care the
squares of thenormal frequenciesω 1 ,...,ωn.
An important example is the system of particles formed by a molecule ofNatoms.
Since the displacement of each atom from its equilibrium position is specified by three
coordinates, the internal configuration of the molecule without regard to its position
and orientation in space may be specified byn= 3 N−6 internal coordinates. The de-
termination of the corresponding normal frequenciesω 1 ,...,ωnmay be a formidable
task even for moderate values ofN. However, the problem is considerably reduced by
taking advantage of the symmetry of the molecule.
Asymmetry operationis an isometry ofR^3 which sends the equilibrium position
of any atom into the equilibrium position of an atom of the same type. The set of all
symmetry operations is clearly a group under composition, thesymmetry groupof the
molecule.
For example, the methane moleculeCH 4 has four hydrogen atoms at the vertices
of a regular tetrahedron and a carbon atomat the centre, from which it follows that
the symmetry group ofCH 4 is isomorphic toS 4. Similarly, the ammonia molecule
NH 3 has three hydrogen atoms and a nitrogen atom at the four vertices of a regular
tetrahedron, and hence the symmetry group ofNH 3 is isomorphic toS 3.
We return now to the general case. IfGis the symmetry group of the molecule,
then to eachs∈Gthere corresponds a linear transformationA(s)of the configuration
spaceRn. Moreover the mapρ:s→A(s)is a representation ofG. Since the kinetic
and potential energies are unchanged by a symmetry operation, we have
A(s)tBA(s)=B, A(s)tCA(s)=C for everys∈G.It follows that
B−^1 CA(s)=A(s)B−^1 C for everys∈G.
Assume the notation chosen so that the distinctω’s areω 1 ,...,ωpandωkoccurs
mktimes in the sequenceω 1 ,...,ωn(k= 1 ,...,p). Thusn=m 1 +···+mp.IfVk
is the set of allv∈Rnsuch that
B−^1 Cv=ω^2 kv,