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Sandbox Physics

Longitudinal recovery in nuclear induction

NMR Inversion Recovery Relaxometer

How do you measure how fast nuclear magnetization recovers after it is flipped? Flip it, wait, then tip it sideways with a right-angle pulse and measure the signal, starting from equilibrium for each wait. The signal starts negative, crosses zero and climbs back to equilibrium; the rate of that climb gives the longitudinal relaxation time. Keep the sign: taking the magnitude folds the curve. The zero sits at 0.69 times that time only for a perfect flip; fit the whole curve.

Turn the equilibrium magnetization upside down, wait, then read it out with a right-angle pulse. Repeat for a series of delays, each prepared from scratch, and infer the longitudinal relaxation without assuming a perfect inversion or a known receiver amplitude.

Interactive modelNMR Inversion Recovery Relaxometer

Vertical static field; horizontal transmit/receive coil. A switch isolates the receiver during the pulse, then routes the induced voltage to a phase-referenced quadrature receiver. Cable motion is schematic. B0=0.35 T,fref≃14.902 MHzB_0=0.35\,\mathrm{T},\quad f_{\rm ref}\simeq14.902\,\mathrm{MHz}

Pulse sequence and averaged receiver voltages

Flip the magnetization, wait, and read its sign

Equilibrium resetαx\alpha_xIdeal crushertdt_d(π/2)x(\pi/2)_xReceiver blankingSigned quadratures

Each of 41 delays uses a fresh equilibrium ensemble. After inversion and ideal transverse spoiling, a calibrated right-angle pulse maps the recovered longitudinal component into induction. Every signed sample is taken 0.4 milliseconds after readout. Repetitions are fully reset; playback is accelerated.

Ready. Predict, then acquire.

Raw voltages, with the source expectation shown separately

Two quadrature channels keep the sign

td/s,I,Q  (μV)t_d/\mathrm{s},\quad I,Q\;(\mathrm{\mu V})

Teal: in-phase voltage. Gold: quadrature voltage. Dots: retained observations. Thin lines: source expectations. Purple dashes: fit using records only. Gray: saved observations.

Signed recovery, inferred from the record alone

A zero crossing carries phase information.

td/s,I/μVt_d/\mathrm{s},\quad I/\mathrm{\mu V}

The deficit and equilibrium voltage are inferred together with recovery time. The quadrature noise channel is retained above; it does not directly measure longitudinal magnetization.

Inference without the source parameters

Inferred longitudinal time

—\text{—}

Inferred preparation deficit ratio

—\text{—}

Inferred equilibrium voltage

—\text{—}

Retained complex samples

00

Voltage residual RMS

—\text{—}

The fit is a global search within bounds, with the amplitude profiled out. The weak-information and boundary checks are rough diagnostics, not confidence intervals on the parameters. Recovering the inputs from synthetic data does not independently prove the physical law.

Latest raw observations
Indextd/st_d/\mathrm{s}I/μVI/\mathrm{\mu V}Q/μVQ/\mathrm{\mu V}

Saved comparison and model limits

Single noninteracting resonance; no molecular chemical shifts, coupling multiplets, diffusion, radiation damping, coil loading or realistic pulse bandwidth. Gain and phase are calibrated; no drift or digitizer quantization. Instantaneous hard-pulse approximation; the Cauchy static-offset model is not a computed field map.

Physics tutorial

NMR Inversion Recovery Relaxometer

BackgroundTurn the equilibrium magnetization upside down, wait, then read it out with a right-angle pulse. Repeat for a series of delays, each prepared from scratch, and infer the longitudinal relaxation without assuming a perfect inversion or a known receiver amplitude.

Why it mattersThe constant is inferred from a calibrated instrument response and the observations you keep.

Start with the essentials

Focus question
How do you measure how fast nuclear magnetization recovers after it is flipped?
One-sentence intuition
Keep what the simulation assumes about the source apart from what you estimate from the recorded data.

Core mathematical model

Recovery of the longitudinal magnetization

My(td)M0=1−(1−cos⁡α)e−td/T1\frac{M_y(t_d)}{M_0}=1-(1-\cos\alpha)e^{-t_d/T_1}

The vertical static field sets the longitudinal direction. After the preparation, an ideal crusher wipes out any leftover transverse coherence. Every delay point starts again from a fresh sample at equilibrium.

Reading the sign through induction

⟨I(td)⟩=a−be−td/T1b/a=1−cos⁡α\begin{aligned}\langle I(t_d)\rangle&=a-be^{-t_d/T_1}\\b/a&=1-\cos\alpha\end{aligned}

A calibrated right-angle readout pulse tips the longitudinal magnetization into the plane the receiver senses. The fixed receiver dead time and gain go into the unknown overall scale. The receiver cannot pick up a static longitudinal component directly.

Zero crossing after a perfect inversion

tzero=T1ln⁡2(α=π)t_{\rm zero}=T_1\ln 2\quad(\alpha=\pi)

Fit the equilibrium scale, the deficit and the recovery time together. An imperfect inversion moves the zero crossing, so in general the zero crossing alone does not give the recovery time.

Common difficulties

A simulation already contains its law

Typical misconceptionGetting a source parameter back out of the data independently confirms the law.

Better mental modelThe synthetic data test a measurement and inference procedure. They are not historical observations, they give no confidence interval for the parameter, and they are not external scientific certification.

Run the experiment

  1. 01

    Keep the sign of the signal

    Scan all the delay points and look at the in-phase voltage, which starts out negative.

    What to observe: The phase flips when the recovering longitudinal component passes through zero. Plotting only the magnitude would fold the curve over at that point.
  2. 02

    Change the preparation only

    Save the run, then choose Imperfect inversion and acquire again.

    What to observe: The estimated deficit changes, but the recovery time still agrees with the earlier value. The fit is never told the source preparation angle.
  3. 03

    A flat record has no time scale

    Choose No preparation and repeat the same scan.

    What to observe: Fitting an exponential to the data is not enough to show a relaxation time. The fit is compared with a constant response, and an estimate that barely beats it is not reported.

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