- Spin-Lattice Relaxation Time

Spin-Lattice Relaxation Time

Definition: (Lattice) The material in which the nuclear magnets are embedded is generally referred to as the lattice, whether it be solid, liquid or gas.

Let us now consider an assembly of identical spin-1/2 nuclei in the presence of a steady magnetic field $H_{_{0}}$. Assume that there is only weak coupling between the nuclear spins. This assumption allows us to consider that the spin energy levels of the assembly are the same as that for a single spin. Although we assume weak coupling between the spins the coupling must not be zero. That is, the weak coupling between the spins must at least be sufficient enough to bring about thermal equilibrium between the spins. Furthermore, suppose that the interaction between the nuclear spins and the lattice is even smaller than the interaction between spins. This means that, apart from the feeble paramagnetism produced by the nuclei, the material concerned must be diamagnetic (water is diamagnetic).

Each nucleus has two possible energy levels separated by a gap of $2\mu H_{_{0}}$. Suppose we apply radiation of the resonant frequency polarized in a direction perpendicular to $H_{_{0}}$, transitions between energy levels then take place. From the theory of the Einstein coefficients the probability of an upwards transition (by absorption) equals the probability of a downward transition (by stimulated emission). In comparison, the probability of a downward transition (by spontaneous emission) is found to be negligible (Purcell 1946).

If the numbers of nuclei in each energy level were equal, the average rate of transitions up and down would be equal and no net effect would be observed on the system. However, since the nuclear spins are in equilibrium at temperature $T_{_{S}}$ the population of the lower level exceeds that of the upper level by the Boltzmann Factor $e^{\frac {2\mu H_{_{0}}}{kT_{_{S}}}}$, where $k$ is Boltzmann's constant.

At room temperature, for protons in a field of 5000 Gauss, this factor has the value of $e^{\frac {2\mu H_{_{0}}}{kT_{_{S}}}}\approx 1+{\frac {2\mu H_{_{0}}}{kT_{_{S}}}}=1+4\times 10^{-6}$. Because of this typically small, but finite, excess population in the lower energy state, there is a net absorption of energy from the transverse RF field.