A material's magnetic properties come from the magnetic moments arising from the angular momentum of its microscopic constituents, such as electrons and the nuclei. For some materials, these tiny magnetic moments are aligned along a particular direction to produce a net magnetic field, giving the material a non-zero magnetisation. For other materials, the individual magnetic moments are randomly oriented. However, when such a material is put in an external magnetic field, the magnetic moments rotate around the direction of the external magnetic field so on average the material has a non-zero magnetisation. If we turn off the external magnetic field, the magnetisation of the material gradually decreases to its original value. This occurs due to the interactions between the magnetic moments, or their interactions with other microscopic degrees of freedom such as lattice vibrations. This process is called relaxation, and is considered below using classical mechanics models.
We consider the problem of nuclear spin relaxation due to random fluctuations of other physical degrees of freedom, such as lattice vibrations or the electrons' magnetic dipole moments. In order to familiarise ourselves with randomness in the solutions of a classical mechanics system, let us start with a forced harmonic oscillator of mass $m$ and angular frequency $\omega_0$. The energy of the oscillator changes because of a time-dependent external force.
$$m\frac{{\rm d}^2q(t)}{{\rm d} t^2} + m\omega^2_0 q(t) = F(t)$$
where the external force is given by the following step-wise function.
$F(t) = \left\{ \begin{matrix} 0, & t<0 \\ +mf_0 ,& 0\le t<T_0/2 \\ - mf_0 ,& T_0/2\le t<T_0 \\ 0, & t\ge T_0 \end{matrix} \right.$
Here $\omega_0 = 2\pi/T_0$ is the angular frequency of the oscillator $q(t)$. Suppose that the initial condition is given as $q(0)=A\sin\delta,\, \dot q(0)=A\omega_0\cos\delta$.
Before turning on the external force, energy is conserved and its value is $E_0 = \frac{m}{2}\omega^2_0 A^2$. Without losing generality we assume $-\pi\le\delta < \pi$.
A2 1.20 Consider the total mechanical energy $E(t)= \frac{m(\dot q^2 + \omega^2_0 q^2)}{2}$. Calculate the difference of $E(t)$ between $t=T_0$ and $t=0$, due to the effect of the external force $F(t)$. In other words, calculate $\Delta E \equiv E(t\ge T_0) - E(t\le 0)$ and express it in terms of $A,\delta, f_0, \omega_0$.
A3 1.20 Suppose that $\delta$ is a random variable with a uniform distribution in the range $-\pi \le \delta < \pi$. In other words, we have a large number of identical forced harmonic oscillators which all follow the same Eq.1. Their initial conditions are given so that $A$ is the same, but $\delta$ is chosen randomly from $-\pi\le\delta < \pi$. Calculate the statistical average of the absorbed energy $\langle \Delta E \rangle$ , and also the 2nd moment $\langle (\Delta E)^2 \rangle$.
The energy of a magnetic dipole moment $\vec \mu$ under magnetic field $\vec B$ is given as
$E = - \vec \mu \cdot \vec B = - \gamma \vec S \cdot \vec B$
When we consider an infinitesimal rotation of the angular momentum $\vec S $ and equate the energy difference with the product of torque ($\vec\tau$) and angular displacement, we obtain the equation for $\vec S$:
$\vec\tau = \frac{{\rm d}\vec S}{{\rm d}t} = \gamma\vec S \times \vec B $
According to this equation, when $\vec B$ is constant the angular momentum $\vec S$ precesses around the direction of the magnetic field $\vec B$. This phenomenon is known as Larmor precession, and the frequency of the precession is given by $\gamma |\vec B|$. In particular, it is independent of the angle between $\vec S$ and $\vec B$.
Let us now consider turning on a magnetic field that contains an oscillating component in the xy-plane, in addition to a constant part along the z-direction. The magnetic energy is then
$$E = - \omega_0 S_z - \omega_1 \cos(\omega_2 t) S_x - \omega_1 \sin(\omega_2 t) S_y$$
where $\omega_0,\omega_1$ are determined by the relevant components of the magnetic field and $\gamma$, while $\omega_2$ is the frequency of the oscillating magnetic field. We assume $\omega_0,\omega_1,\omega_2$ are all positive. The equations for $\vec S$ are
$\dot S_x = +\omega_0 S_y - \omega_1 \sin(\omega_2 t) S_z\\ \dot S_y = -\omega_0 S_x + \omega_1 \cos(\omega_2 t) S_z\\ \dot S_z = - \omega_1 \cos (\omega_2 t) S_y +\omega_1 \sin (\omega_2 t) S_x$
It is convenient to write these equations in terms of $S_\pm \equiv S_x \pm i S_y$. We have
$\dot S_+ = - i\omega_0 S_+ + i \omega_1 e^{+i\omega_2 t }S_z \\ \dot S_- = + i\omega_0 S_- - i \omega_1 e^{-i\omega_2 t }S_z \\ \dot S_z = \frac{i\omega_1}{2} \left( e^{-i\omega_2t}S_+ - e^{+i\omega_2 t} S_- \right)$
For the next step, let us introduce spin in the rotating frame using $S_\pm \equiv e^{\pm i\omega_2 t} \Sigma_\pm, \, S_z \equiv \Sigma_z$. One can show that the equations for $\Sigma_x \equiv \tfrac{1}{2}(\Sigma_+ + \Sigma_-),\quad \Sigma_y \equiv \tfrac{i}{2}(\Sigma_- - \Sigma_+)$, and $\Sigma_z$can be written as
$$\frac{\rm d }{{\rm d}t} \vec \Sigma = \vec M \times \vec \Sigma$$
where $\vec \Sigma =(\Sigma_x, \Sigma_y, \Sigma_z)$ and $\vec M = (M_x, M_y,M_z)$.
The equations are simplified even further if we consider a static rotation in the xz-plane and define new variables $\Sigma_X,\Sigma_Y,\Sigma_Z$ as follows:
$\Sigma_X = \cos\Theta \Sigma_x - \sin\Theta\Sigma_z \\ \Sigma_Y = \Sigma_y \\ \Sigma_Z = \sin\Theta \Sigma_x + \cos\Theta \Sigma_z $
Then, in terms of the new variables in this doubly-rotating frame, the equations can be reduced to the following form,
$\dot\Sigma_X = +\Omega\Sigma_Y \\ \dot\Sigma_Y = -\Omega\Sigma_X \\ \dot\Sigma_Z = 0 $
if $\Omega$ and $\tan\Theta$ are chosen appropriately.
Let us now consider a large number of spins with a statistical distribution of initial configurations: at $t=0$, the average values satisfy $\langle S_x(0)\rangle = \langle S_y(0)\rangle = 0$ and $\langle S_z(0)\rangle>0$. The spins all satisfy the same equation derived from Eq.(2).