Orientation distribution function INDEX

The molecular origin of these relaxations has been established for dipolar molecular liquids by Debye [122] who has shown that the applied electric field perturbs the orientational distribution function for the dipolar molecules, leading to a static relative permittivity So greater than n, where n is the optical refractive index, and a dispersion for c (/) accompanied by a peak in c" f). 

This expression can be combined with equation (4.12) to obtain the following expression for the refractive index tensor as a function of the orientation distribution of the rods making up the bulk sample ... 

In another type of application a low molecular fluorescent probe is added to a system containing macromolecules. As would be expected, the rotation of a small species is insensitive to the molecular weight of high polymers, but depends on the "microscopic viscosity" which is a function of free volume. For instance, Nishijima has shown that the microscopic viscosity of liquid paraffin hydrocarbons levels off for molecular weights above 1000 and that the microscopic viscosity of polystyrene containing 10 volume"/ benzene is only 200 times as high as that of benzene (15). Nishijima also showed that the emission anisotropy is a useful index of molecular orientation. Since both the excitation and the emission are anisotropic, the method yields the fourth moment of the distribution function of orientations, while other optical properties (dichroism, birefringence) depend on the second moment (15). 

Suppose we wish to know the dipole moment of, say, the HCl molecule, the quantity that tells us important information about the charge distribution. We look up the output and we do not find anything about dipole moment. The reason is that all molecules have the same dipole moment in any of their stationary state y, and this dipole moment equals to zero, see, e.g., Piela (2007) p. 630. Indeed, the dipole moment is calculated as the mean value of the dipole moment operator i.e., ft = (T l/i l ) = ( F (2, q/r,) T), index i runs over all electrons and nuclei. This integral can be calculated very easily the integrand is antisymmetric with respect to inversion and therefore ft = 0. Let us stress that our conclusion pertains to the total wave function, which has to reflect the space isotropy leading to the zero dipole moment, because all orientations in space are equally probable. If one applied the transformation r -r only to some particles in the molecule (e.g., electrons), and not to the other ones (e.g., the nuclei), then the wave function will show no parity (it would be neither symmetric nor antisymmetric). We do this in the adiabatic or Born-Oppenheimer approximation, where the electronic wave function depends on the electronic coordinates only. This explains why the integral ft = ( F F) (the integration is over electronic coordinates only) does not equal zero for some molecules (which we call polar). Thus, to calculate the dipole moment we have to use the adiabatic or the Born-Oppenheimer approximation. 

---