Although the rotational and vibrational motions of the nuclei in a molecule cannot be fully separated, the Eckart conditions minimize the coupling close to a reference (usually equilibrium) configuration.
These equilibrium coordinates of the nuclei—with masses MA—are expressed with respect to a fixed orthonormal principal axes frame and hence satisfy the relations Here λi0 is a principal inertia moment of the equilibrium molecule.
The triplets RA0 = (RA10, RA20, RA30) satisfying these conditions, enter the theory as a given set of real constants.
Since we take the origin of the Eckart frame in the instantaneous center of mass, the following relation holds.
These rotational conditions follow from the specific construction of the Eckart frame, see Biedenharn and Louck, loc.
Finally, for a better understanding of the Eckart frame it may be useful to remark that it becomes a principal axes frame in the case that the molecule is a rigid rotor, that is, when all N displacement vectors are zero.
The Eckart conditions give an orthogonal direct sum decomposition of this space The elements of the 3N-6 dimensional subspace Rint are referred to as internal coordinates, because they are invariant under overall translation and rotation of the molecule and, thus, depend only on the internal (vibrational) motions.
The elements of the 6-dimensional subspace Rext are referred to as external coordinates, because they are associated with the overall translation and rotation of the molecule.
To that end we introduce the following 6 vectors (i=1,2,3): An orthogonal, unnormalized, basis for Rext is, A mass-weighted displacement vector can be written as For i=1,2,3, where the zero follows because of the translational Eckart conditions.
belongs to the orthogonal complement of Rext, so that it is an internal vector.
could be Wilson's s-vectors or could be obtained in the harmonic approximation by diagonalizing the Hessian of V. We next introduce internal (vibrational) modes, The physical meaning of qr depends on the vectors
We already saw that the corresponding external modes are zero because of the Eckart conditions, The vibrational (internal) modes are invariant under translation and infinitesimal rotation of the equilibrium (reference) molecule if and only if the Eckart conditions apply.
Under rotation we have, Rotational invariance follows if and only if The external modes, on the other hand, are not invariant and it is not difficult to show that they change under translation as follows: where M is the total mass of the molecule.
They change under infinitesimal rotation as follows where I0 is the inertia tensor of the equilibrium molecule.
The vibrational energy is written in terms of the displacement coordinates, which are linearly dependent because they are contaminated by the 6 external modes, which are zero, i.e., the dA's satisfy 6 linear relations.
It is possible to write the vibrational energy solely in terms of the internal modes qr (r =1, ..., 3N-6) as we will now show.
This form of the kinetic energy of vibration enters Wilson's GF method.
It is of some interest to point out that the potential energy in the harmonic approximation can be written as follows where H is the Hessian of the potential in the minimum and F, defined by this equation, is the F matrix of the GF method.
In the harmonic approximation to the nuclear vibrational problem, expressed in displacement coordinates, one must solve the generalized eigenvalue problem where H is a 3N × 3N symmetric matrix of second derivatives of the potential
It can be shown that the invariance of V under simultaneous translation over t of all nuclei implies that vectors T = (t, ..., t) are in the kernel of H. From the invariance of V under an infinitesimal rotation of all nuclei around s, it can be shown that also the vectors S = (s x R10, ..., s x RN0) are in the kernel of H : Thus, six columns of C corresponding to eigenvalue zero are determined algebraically.
Thus, T and S correspond to the overall (external) motions: translation and rotation, respectively.
They are zero-energy modes because space is homogeneous (force-free) and isotropic (torque-free).
applied to the "internal" (non-zero eigenvalue) and "external" (zero-eigenvalue) columns of C are equivalent to the Eckart conditions.