Nonexponential Relaxation of Rotating Three-Spin Systems in Molecules of a Liquid

The longitudinal and transverse nuclear magnetic relaxation of a system of three identical spin‐$\frac{1}{2}$ nuclei at the corners of an equilateral triangle which undergoes hindered rotation with respect to a molecule which undergoes rotational Brownian motion has been calculated on the assumption that the relaxation is due to dipole–dipole interactions between the three nuclei, including the effects of cross correlations. The longitudinal relaxation is in general the sum of three decaying exponentials. If the rotational Brownian motion is rapid compared to the Larmor frequency, the longitudinal and transverse relaxations are both proportional to $A_1\exp {\mathrm{(s}}_1{\mathrm{t)\hspace{0.17em}+\hspace{0.17em}A}}_2\exp {\mathrm{(s}}_2\mathrm{t)}$ . For a molecule of arbitrary shape, the expressions for the coefficients $A_1$ and $A_2$ and the negative exponents $s_1$ and $s_2$ are quite complicated. The expressions simplify somewhat for the case of a symmetric top molecule, and even more for the case of a spherical molecule. It is shown for a molecule of arbitrary shape that, if the hindered rotation is rapid compared to the rotational Brownian motion, which in turn is rapid compared to the Larmor frequency, then the predicted relaxation is distinctly nonexponential.