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The Bloch equations describe an ensemble magnetization $\mathbf{M}$ with Larmor precession plus phenomenological recovery and decay:Here $T_1,T_2>0$ are model parameters, not a microscopic explanation of relaxation. The precession term follows from the torque on a magnetic moment; the two dissipative terms prescribe exponential decay. Bloch explicitly introduced this macroscopic relaxation description as a simplifying approximation.^[ref]

For positive $\gamma$ and no transverse drive, $M_x+i{M}_y$ rotates as $e^{-i\gamma B_{0}t}$: its signed rotation is negative about $z$, although the resonance angular frequency ${\omega }_0=\gamma B_0$ is positive. This distinction prevents a hidden sign change when choosing a rotating frame.

Free induction decay is a receiving signal from freely precessing transverse magnetization after excitation ends. Faraday's law of induction gives coil voltage proportional to the time derivative of linked magnetic flux, with sign set by the winding. At ${\omega }_0{T}_2\gg 1$ its envelope follows the transverse magnetization; a fixed receiver phase can absorb the derivative's near-quadrature shift. We define the calibrated complex signal below to represent magnetization, not raw coil voltage.

For an original teaching model, take positive $\gamma$, a circular drive $B_1(\cos \omega t\hat{\mathbf{x}}-\sin \omega t\hat{\mathbf{y}})$, and axes rotating at $-\omega$ about $z$. Define $u+iv=e^{i\omega t}(M_x+i{M}_y)$, $w=M_z$, $\Delta ={\omega }_0-\omega$ and ${\omega }_1=\gamma B_1$. The Bloch equations becomeA linearly oscillating field of peak amplitude $2B_1$ has a co-rotating component of amplitude $B_1$; neglecting its counter-rotating component requires ${\omega }_1\ll {\omega }_0$ near resonance.

At resonance, a short pulse starting from $(0,0,M_0)$ gives $(u,v,w)=(0,M_0\sin \theta ,M_0\cos \theta )$ with $\theta =\gamma B_1\tau$, neglecting relaxation during duration $\tau$. A receiver phase rotation can make a $\pi /2$ pulse's initial transverse signal real. After the pulse, direct integration givesThese are the two components extracted by quadrature demodulation. Differentiating verifies $\dot{S}=-(T_2^{-1}+i\Delta )S$; taking its magnitude separates decay from detuning.

Choose rounded illustrative $\gamma /(2\pi )=42.58MHz/T$, $B_0=0.200\mathrm{T}$, $B_1=10\mu \mathrm{T}$, $T_1=0.800\mathrm{s}$ and $T_2=0.120\mathrm{s}$. Then $f_0=8.516\mathrm{MHz}$ and the $\pi /2$ duration is $\tau =1/(4(\gamma /(2\pi ))B_1)=0.587\mathrm{ms}$, short relative to both relaxation times. For receiver detuning $\Delta /(2\pi )=5\mathrm{Hz}$ and $S_0=1$, $S(0.050\mathrm{s})=-0.65924i$ and $S(0.100\mathrm{s})=-0.43460$. These are calculated examples, not 1946 measurements.

With no transverse drive, the Bloch equations giveLongitudinal nuclear spin relaxation restores the population imbalance and thus the magnetization parallel to the static magnetic field. It corresponds to population relaxation for the two-level model. Starting from $M_z(0)=0$, waiting $T_1$ recovers $1-e^{-1}$ of $M_0$, whereas waiting $5T_1$ recovers about $99.3\%$. This is why repeated measurements require a recovery-time decision.

Drive reduces the longitudinal population imbalance when excitation competes with recovery. The dimensionless $s$ in the steady Bloch equations measures that competition.