Magnetic Resonance — Question 6

Section 40-5 · Physics pages 6662–6663

Question 6 Longitudinal recovery reveals the relaxation time from a null crossing After a perfect inversion pulse, a sample has Mz(0) = −M0. It obeys dMz/dt = (M0 −Mz)/T1. The signal crosses zero at t0 = .420 s. t Mz Schematic; labels, not drawing lengths, specify the data. (a) Solve the recovery equation. (b) Infer T1 and find when the signal first reaches 0.90M0.
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Question 6 - Solution Longitudinal recovery reveals the relaxation time from a null crossing t Mz (a) Method and calculation Mz −M0 = [Mz(0) −M0]e−t/T1. Mz(t) = M0(1 −2e−t/T1) . (b) Interpretation and further calculation 0 = 1 −2e−t0/T1 ⇒T1 = t0/ ln 2 = .6059 s . .90 = 1 −2e−t/T1 ⇒t = T1 ln 20 = 1.815 s . Physical check The zero crossing is not one relaxation time. The specified inversion initial condition accounts for the factor two in the recovery curve.

Question 6

Longitudinal recovery reveals the relaxation time from a null crossing

After a perfect inversion pulse, a sample has Mz(0)=−M0M_z(0)=-M_0. It obeys dMz/dt=(M0−Mz)/T1dM_z/dt=(M_0-M_z)/T_1. The signal crosses zero at t0=.420st_0=.420\,\mathrm s.

Diagram shown in the original SVG worksheet.

  1. Solve the recovery equation.

  2. Infer T1T_1 and find when the signal first reaches 0.90M00.90M_0.

Question 6 - Solution

Longitudinal recovery reveals the relaxation time from a null crossing

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Mz−M0=[Mz(0)−M0]e−t/T1.M_z-M_0=[M_z(0)-M_0]e^{-t/T_1}.Mz(t)=M0(1−2e−t/T1).\boxed{M_z(t)=M_0(1-2e^{-t/T_1})}.

(b) Interpretation and further calculation

0=1−2e−t0/T1⇒T1=t0/ln⁡2=.6059s.0=1-2e^{-t_0/T_1}\Rightarrow\boxed{T_1=t_0/\ln2=.6059\,\mathrm s}..90=1−2e−t/T1⇒t=T1ln⁡20=1.815s..90=1-2e^{-t/T_1}\Rightarrow\boxed{t=T_1\ln20=1.815\,\mathrm s}.

Physical check

The zero crossing is not one relaxation time. The specified inversion initial condition accounts for the factor two in the recovery curve.