 Solution

For an exponentially correlated local field, a single-frequency illustrative relaxation rate is $T_1^{-1}=C{\tau }_c/(1+{\omega }_0^2{\tau }_c^2)$. Its derivative with respect to ${\tau }_c$ changes sign at ${\omega }_0{\tau }_c=1$. Slow motion gives $T_1\propto {\tau }_c$, and fast motion gives $T_1\propto 1/{\tau }_c$; an Arrhenius plot consequently has a minimum of $T_1$ between opposite-slope limits.

For identical spin-one-half nuclei with homonuclear dipolar relaxation, include both transition frequencies:Writing $x={\omega }_0{\tau }_c$, the maximum of the rate satisfiesLet $\ell$ be jump length, ${\tau }_j$ the total waiting time, ${\tau }_c=c{\tau }_j$, and $f_D$ the diffusion jump correlation factor. In three dimensions $D=f_D{\ell }^2/(6{\tau }_j)=f_{D}c{\ell }^2/(6{\tau }_c)$. Combining this with $D=D_0\exp [-E_D/(k_{B}T)]$ givesThe numerical constant is one for the single-frequency illustration and 0.616 for the homonuclear dipolar model. Assumptions are a single activated process, exponential field correlations, a temperature-independent prefactor and coupling amplitude, and an established relation between local correlation and translational motion. Defect creation can enter the measured activation energy; it cannot automatically be identified with a bare migration barrier. The fluctuation-spectrum approach originates in the relaxation theory of Nicolaas Bloembergen, Edward Mills Purcell and Robert Pound.^[ref]