Diffusion encoding: how a gradient pair measures how far water moves

A gradient pair: still water returns, moving water keeps a phase

Hrabe and co-workers explain the principle with a diagram. A water molecule at $x$ gains the phase $\gamma Gx\delta$ during the first gradient pulse of duration $\delta$. After a wait, an equal and opposite gradient is applied. If the molecule did not move in between, all phase differences are reversed and the magnetization is fully restored.

In a spin-echo sequence the second pulse has the same sign as the first, but the 180° pulse between them flips every phase, which has the same effect as reversing the gradient. With short pulses, a molecule that moves from $x_1$ to $x_2$ ends with a phase that depends only on its displacement. Define $q=\gamma G\delta$, which is $2\pi$ times the spatial frequency $k$ of Chapter 4: the gradient pair walks out to $q$ in k-space and back to the origin.

The coil measures the average over all molecules. A uniform shift of everything changes no magnitude; only the spread of displacements makes the phases cancel. So the diffusion-weighted echo is sensitive only to random motion.

Source: J. Hrabe · F. Szczepankiewicz and J. Hutter

A random walk and the echo attenuation

Hrabe and co-workers start from the simplest random walk: every $\Delta t$ a molecule jumps $\Delta x$ to the left or right with equal probability. After $n$ steps the variance of its position is $2Dn\Delta t$, with $D=\Delta x^2/(2\Delta t)$. Letting $\Delta t\to0$ with $D$ fixed gives continuous diffusion with $\sigma^2=2Dt$.

Each jump changes the phase by $\pm q\Delta x$. The $n$ jumps are independent, so the average phase factor over all molecules is $\cos^n(q\Delta x)$. Substituting $\Delta x=\sqrt{2D\Delta t}$ and letting $n\to\infty$ gives the exponential decay $e^{-bD}$, where $b=q^2\tau$ for narrow pulses.

The course benchmark checks both steps. The discrete walk, with 40,000 molecules and 20, 50 and 200 steps, agrees with $\cos^n$ to within 1.04 standard errors. In the continuous limit with $bD=1.07$, $n=10\,000$ steps give 0.342995 against the limit 0.343009.

Source: J. Hrabe · F. Szczepankiewicz and J. Hutter

The b-value from the gradient waveform

When the pulses are not short, molecules also move during them, and $b$ has to be computed from the whole waveform. The ISMRM 2019 problem set gives the general definition: $q(t)$ is the gradient area accumulated up to time $t$ (times $\gamma$), and $b$ is the time integral of $q^2$. For two rectangular pulses the integral splits into three parts: each pulse contributes $q^2\delta/3$ and the gap between them $q^2(\Delta-\delta)$, which adds up to the Stejskal–Tanner formula.

The answer key also notes that the diffusion time has no simple definition, because it depends on the tissue architecture; experiments on time-dependent diffusion should report the whole waveform, such as $\Delta$ and $\delta$. When all three gradient axes play the same waveform, the gradient grows by $\sqrt3$ and $b$ by 3, to about $1.6\,\mathrm{ms/\mu m^2}$.

The diffusion-encoding experiment accumulates each molecule’s phase step by step and does not use the $b$ formula; the benchmark integrates the experiment’s timing numerically and agrees with the closed form above to $7.7\times10^{-13}$ or better.

Source: F. Szczepankiewicz and J. Hutter · J. Hrabe

Worked example: 40 mT/m, 10 ms, 50 ms

Problem: the settings of the ISMRM problem and the default of the diffusion-encoding experiment. A 40 mT/m gradient, 10 ms pulses 50 ms apart, and free water with $D=2.0\,\mathrm{\mu m^2/ms}$ (the value the experiment assumes). Find $q$, $b$, the echo ratio, the standard deviation of the phase and the typical displacement between the pulses.

$\gamma G=2.675\times10^8\times0.04=1.07\times10^7\,\mathrm{rad\,s^{-1}m^{-1}}$, or $0.0107\,\mathrm{rad\,ms^{-1}\mu m^{-1}}$, the number in the problem set. Then $q=0.0107\times10=0.107\,\mathrm{rad/\mu m}$: a molecule must move 59 µm for its phase to turn once. $b=0.107^2\times(50-10/3)=0.534\,\mathrm{ms/\mu m^2}=534\,\mathrm{s/mm^2}$; the answer key gives 0.53.

The echo ratio is $e^{-bD}=e^{-1.0686}=0.3435$. The phase is close to Gaussian with variance $2bD$, so its standard deviation is $\sqrt{2.137}=1.46\,\mathrm{rad}$. The typical displacement along the gradient is $\sqrt{2D\Delta}=\sqrt{200}=14.1\,\mathrm{\mu m}$, about one cell, yet enough to cut the echo to a third.

The experiment simulates 6000 molecules and reads an echo ratio of 0.356 with a standard error of 0.012, about one standard error from 0.3435; fitting nine gradient strengths gives an apparent diffusion coefficient of $1.94\,\mathrm{\mu m^2/ms}$. The benchmark checks 20,000 molecules at six gradient strengths, with a largest deviation of 0.64 standard errors.

Source: F. Szczepankiewicz and J. Hutter · J. Hrabe

Water between membranes: the echo decays more slowly

Hrabe and co-workers recall that Tanner and Stejskal first proposed measuring restricted diffusion by varying the delay between the pulses: the signal from water held in by cell walls decays less than that from free water, and they used this to estimate the diameter of yeast cells. Hrabe and co-workers also note that with a very short diffusion time most molecules hit no wall and the process is close to free diffusion; a longer time lets molecules explore the structure. Results from different diffusion times cannot be mixed, and the diffusion time should always be reported.

The diffusion-encoding experiment models this with parallel membranes a distance $L$ apart that reflect any molecule hitting them. With the default gradient and $L=20\,\mu\mathrm m$, the echo ratio rises from 0.3435 for free water to 0.739 and the apparent diffusion coefficient falls to 0.588. Lengthening the separation to 90 ms would take free water down to 0.137, but water between membranes only to 0.713, and the apparent diffusion coefficient to 0.367. With membranes 5 µm apart the echo ratio is 0.995, almost no decay.

Under restriction a single exponential no longer fits. The experiment holds out the largest usable $b$ and predicts it from an exponential fitted to the other points: the miss is 2.5% for free water, −5.1% between membranes and −12.6% after the longer wait. When the wait is very long, a molecule’s positions at the two pulses are independent and each is uniform between the membranes, so the echo ratio tends to $\operatorname{sinc}^2(qL/2)$. This derivation is the course’s own. The benchmark checks it with $L=10\,\mu\mathrm m$, 0.2 ms pulses and a 100 ms separation: the largest difference is 0.0045, and the root-mean-square displacement is 4.081 µm against the limit $L/\sqrt6=4.082\,\mu\mathrm m$.

Source: J. Hrabe · F. Szczepankiewicz and J. Hutter