L1 · Chapter 1

How proton magnetization precesses, tips over and recovers

Sandbox Physics · 2026-10-29

The water in the body is full of protons, and each proton is a tiny bar magnet. In a strong field their combined magnetization turns about the field like a tilted spinning top. This chapter describes that motion with the equations Bloch wrote down in 1946: how a radio-frequency pulse tips the magnetization over, how the tipped magnetization induces a voltage in a coil, and how it returns to rest with two time constants, $T_1$ and $T_2$ . Every imaging method in the later chapters rests on these few equations.

Magnetization precesses about the field

Point the static field along $+z$ and call it $\mathbf B_0=B_0\hat z$ . The sum of all proton magnetic moments in a unit volume is the magnetization $\mathbf M$ , in A/m. A proton’s magnetic moment points along its spin angular momentum, and the ratio between them is the gyromagnetic ratio $\gamma$ .

The field exerts a torque $\mathbf M\times\mathbf B$ on the magnetization, the torque changes the angular momentum, and the result is Eq. 11 of Bloch’s paper. The cross product is always perpendicular to $\mathbf M$ , so the magnetization keeps its length and only swings round $\mathbf B_0$ . That swinging is precession, and its angular frequency is the Larmor frequency.

For the proton $\gamma/2\pi=42.576\,\mathrm{MHz/T}$ ; the MIT handout rounds this to 42.58 MHz/T. The frequency is proportional to the field, and Chapter 3 turns that proportionality into an image.

\frac{d\mathbf M}{dt}=\gamma\,\mathbf M\times\mathbf B

Bloch’s Eq. 11 in SI units, with the applied field only and no relaxation.

\omega_0=\gamma B_0,\qquad f_0=\frac{\gamma}{2\pi}B_0

For the proton

\gamma>0

, and seen from

+z

the magnetization turns clockwise.

The magnetization precesses about the static field at the Larmor frequency

f_0=\bar\gamma B_0

, 127.7 MHz at 3 T. Its rotating flux induces the voltage that the receiver coil records. Original schematic.

Source: F. Bloch · J. Hrabe · MIT 6.3000 Signal Processing

A radio-frequency pulse tips the magnetization over

At equilibrium the magnetization lies along $+z$ , parallel to the field. It does not precess, and a coil cannot detect it. To make it precess we first push it off the $z$ axis with a weak perpendicular field $B_1$ that rotates at $\omega_0$ : a radio-frequency pulse.

Seen from a frame that turns about $z$ at $\omega_0$ , the rotation cancels the static field and $B_1$ stands still. The magnetization then turns slowly about $B_1$ alone, at $\omega_1=\gamma B_1$ . Bloch makes exactly this change of variables in his Eqs. 40–42, which turns the equations into a system with nearly constant coefficients.

After a pulse of length $\tau$ the magnetization has turned through the flip angle. After 90° it lies entirely in the transverse plane and, in the laboratory frame, precesses at $\omega_0$ . Its changing flux induces a voltage in the receive coil. The MIT handout calls this voltage the signal from which an MR image is derived.

\alpha=\gamma B_1\tau

As long as the pulse frequency equals

\omega_0

, the flip angle depends only on the pulse area

B_1\tau

Source: F. Bloch · MIT 6.3000 Signal Processing

Two relaxations: T1 and T2

Tipped magnetization does not precess forever. Bloch describes its return with the two simplest exponential laws. The longitudinal part $M_z$ returns to its equilibrium value $M_0$ with time constant $T_1$ (Eq. 32); for that the spins must hand energy to the surrounding molecules. The transverse parts $M_x$ and $M_y$ decay with $T_2$ (Eq. 37); for that the protons only need to fall out of step with one another, and no energy changes hands.

Bloch lumps the causes of this loss of step into an effective field irregularity $H'$ and gets $T_2\approx1/(\gamma H')$ (Eq. 33). For neighbouring nuclei held still he estimates $H'$ of about 1 gauss and $T_2$ of about $10^{-4}\,\mathrm s$ . He notes that $T_2$ can be many orders of magnitude shorter than $T_1$ and that assuming the two equal can be a serious error.

Adding both relaxations to the torque gives Bloch’s Eqs. 38a–c, now called the Bloch equations. Below are their solutions in the static field alone.

\frac{dM_z}{dt}=\frac{M_0-M_z}{T_1},\qquad \frac{dM_{x,y}}{dt}=-\frac{M_{x,y}}{T_2}

Bloch’s Eqs. 32 and 37. The full equations also contain the precession term of the first section.

M_z(t)=M_0+\bigl[M_z(0)-M_0\bigr]e^{-t/T_1},\qquad |M_{xy}(t)|=|M_{xy}(0)|\,e^{-t/T_2}

The longitudinal part recovers towards equilibrium and the transverse part decays to zero; precession only turns the transverse part.

Source: F. Bloch · MIT 6.3000 Signal Processing

Checking the equations by numerical integration

The closed-form solutions can be checked against direct integration. The benchmark program behind the frequency-encoding experiment integrates the Bloch equations with a fourth-order Runge–Kutta method. After a resonant rectangular pulse the computed flip angle differs from $\gamma B_1\tau$ by less than $10^{-10}$ ; without a pulse the magnetization turns in the sense given above; and the transverse part decays as the exponential solution says.

The program also splits each water tube into 400 small pieces, integrates each one and adds their contributions in the coil. The voltage obtained this way agrees with the Fourier formula of Chapter 3 to better than one part in a thousand, which is why the later chapters may use the formula instead of integrating point by point.

Source: F. Bloch · J. Hrabe

Worked example: one excitation in a 3 T scanner

Problem: in a $B_0=3\,\mathrm T$ scanner a 1 ms rectangular radio-frequency pulse tips the magnetization by 90°. Find the precession frequency, the $B_1$ required, the decay of the signal and the recovery of the magnetization. Take $T_2^*=30\,\mathrm{ms}$ (the default of the frequency-encoding experiment) and $T_1=1\,\mathrm s$ ; both are assumed only for the calculation.

The precession frequency is $f_0=42.576\times3=127.7\,\mathrm{MHz}$ : one turn takes 7.83 ns, and in 1 ms the magnetization turns 127,728 times. A flip of $\pi/2$ needs $B_1=\pi/(2\gamma\tau)=5.87\,\mu\mathrm T$ , about two millionths of $B_0$ . So weak a field works because it turns in step with the precession and its effect builds up turn after turn.

After the pulse the envelope of the coil voltage decays as $e^{-t/T_2^*}$ and is down to $e^{-1}=0.368$ after 30 ms. Meanwhile the longitudinal magnetization recovers from zero. If the excitation is repeated every 2 s, $M_z=M_0(1-e^{-2})=0.865M_0$ just before the next pulse, so in the steady state each excitation gives about 86.5 % of the fully recovered signal.

Source: J. Hrabe · MIT 6.3000 Signal Processing

Try it in the Lab

Read T1 from the zero crossing

Choose the “Full inversion” preset. Predict first: after a perfect inversion $M_z=M_0(1-2e^{-t/T_1})$ , so the read-out signal changes sign at $t=T_1\ln2$ .
Find the delay at which the read-out signal turns from negative to positive and estimate T1 from it.
Switch to the “Slower recovery” preset, see which way the sign change moves and explain why.

One thin tube, one tone

Choose the “One thin tube, one tone” preset and look at the real and imaginary parts of the coil record: two oscillations at the same frequency, a quarter period apart.
Switch to the “Fast decay blurs the edges” preset and compare how fast the envelope of the record dies away.

Exercises with explained answers

What is the proton precession frequency in a 1.5 T scanner?
Show the answer
$f_0=42.576\,\mathrm{MHz/T}\times1.5\,\mathrm T=63.86\,\mathrm{MHz}$ , exactly half the 3 T value.
A resonant pulse has $B_1=5.87\,\mu\mathrm T$ . How long must it last to flip the magnetization by 180°?
Show the answer
The flip angle is proportional to $B_1\tau$ . The same $B_1$ needs 1 ms for 90° and therefore 2 ms for 180°, when $\gamma B_1\tau=\pi$ .
After a 90° pulse $M_z=0$ . With $T_1=1\,\mathrm s$ , how long until $M_z$ is back to $M_0/2$ ?
Show the answer
From $1-e^{-t/T_1}=1/2$ we get $t=T_1\ln2=0.693\,\mathrm s$ .
With $T_2=80\,\mathrm{ms}$ , how much transverse magnetization is left 50 ms after excitation, and how long until it is down to 1 %?
Show the answer
$e^{-50/80}=0.535$ is left after 50 ms. It falls to 1 % after $t=T_2\ln100=368\,\mathrm{ms}$ , far sooner than the longitudinal recovery takes with $T_1=1\,\mathrm s$ .

Sources and what we recomputed

MIT 6.3000 Signal Processing, Magnetic Resonance Imaging lecture handout (6 May 2025), data and examples from E. Adalsteinsson
Read in full. Page 9: the proton gyromagnetic ratio rounded to 42.58 MHz/T. Pages 11–13: gradient fields, the phase integral and the definition of k as the time integral of the gradient. Pages 15–18: k-space data, the complex image and its magnitude. Page 19: 256 lines at 2 s each, about 8.5 minutes. Pages 20–21: keeping only even columns of k-space. Pages 24–36: two-coil unfolding, including the coil pair whose sensitivities cannot separate the folded pixels. The handout uses the sign convention of the Fourier kernel that this course follows.
F. Bloch, Nuclear induction, Physical Review 70, 460 (1946), doi:10.1103/PhysRev.70.460
Read in full from a scanned copy. Eq. 11 (precession of the polarization), Eq. 32 (longitudinal relaxation time), Eqs. 33–35 (transverse relaxation time from an effective field irregularity, with the estimate of about 1e-4 s), Eq. 37 and Eqs. 38a–c (the equations with both relaxation times), Eqs. 40–42 (the rotating-frame variables). Bloch writes in Gaussian units with H for the field; the course restates the equations in SI units with B.
J. Hrabe, G. Kaur and D. N. Guilfoyle, Principles and limitations of NMR diffusion measurements, Journal of Medical Physics 32(1), 34–42 (2007), doi:10.4103/0971-6203.31148
Read in full (PubMed Central). Introduction: Hahn’s 1950 observation, the Stejskal–Tanner pulsed gradients and Eqs. 1–3. Eqs. 4–10 (the binomial random walk, D from Eq. 7, Einstein’s relation in Eqs. 8 and 10). Eq. 11 (Larmor frequency, with 2π × 42.576 MHz/T for the proton). Eqs. 12–14 (signal of the lattice walk, the b-value of Eq. 13 and the discrete Stejskal–Tanner form) and Eqs. 15–17 (the continuous limit). The pitfalls section: imaging gradients, local gradients, dependence on the diffusion time and tissue compartments. The 1950 and 1965 papers it cites were not opened.