Abstract
In the limit of molecular rotation sufficiently rapid to give a "white spectrum" to the dipole interaction, it is shown that the longitudinal and transverse components of nuclear magnetization relax in an identical manner for a system of an arbitrary number of identical nuclei with arbitrary spin, provided Boltzmann-like initial conditions are assumed. This gives generalization to the specific result obtained by Hubbard for four equivalently located spin ½ nuclei as well as to the familiar T1=T2 for two identical spins. The Hamiltonian studied consists of Zeeman and dipole-dipole terms. If chemical shifts or scalar spin-spin interactions are included, the results remain valid for equivalent spins but cannot be applied to nonequivalent spins. As an example, Hubbard's three-spin calculation is repeated to include the transverse component, and it is illustrated that if other than Boltzmann-like initial conditions are used, the components need not relax identically.