Ernst angle

In nuclear magnetic resonance spectroscopy and magnetic resonance imaging, the Ernst angle is the flip angle that maximizes the steady-state signal for a spin with a spin–lattice relaxation time T_1 using a flip repetition time T_R , assuming transverse magnetization is eliminated between flips. The Ernst angle \theta is calculated using the following relationship, derived by Richard R. Ernst, who won the 1991 Nobel Prize in Chemistry: \theta=\arccos(e^{-T_R/T_1}) The derivation of the Ernst angle equation explicitly assumes that all transverse magnetization is completely eliminated between repetition times. This elimination is achieved in practice through spoiler gradients (in MRI) or by using a long enough T_R for complete T_2 decay (in NMR).

In nuclear magnetic resonance spectroscopy and magnetic resonance imaging, the Ernst angle is the flip angle that maximizes the steady-state signal for a spin with a spin–lattice relaxation time T_1 using a flip repetition time T_R , assuming transverse magnetization is eliminated between flips. The Ernst angle \theta is calculated using the following relationship, derived by Richard R. Ernst, who won the 1991 Nobel Prize in Chemistry: \theta=\arccos(e^{-T_R/T_1}) The derivation of the Ernst angle equation explicitly assumes that all transverse magnetization is completely eliminated between repetition times. This elimination is achieved in practice through spoiler gradients (in MRI) or by using a long enough T_R for complete T_2 decay (in NMR).

Derivation

Let M_z^- denote the longitudinal magnetization just before an RF pulse. Assume the pulse rotates magnetization by angle \theta about the transverse axis instantaneously. Longitudinal magnetization after the pulse will be: M_z^+=M_z^-\cos(\theta) After relaxation for a time T_R toward equilibrium M_0 with T_1 recovery the longitudinal component of magnetization will be: M_z^{-(next)}=M_0-(M_0-M_z^+)e^{-T_R/T_1}=M_0-(M_0-M_z^-\cos(\theta))e^{-T_R/T_1} Assume that transverse magnetization now becomes zero via spoiling. At steady state M_z^{-(next)}=M_z^- . Substituting M_z^- for M_z^{-(next)} and solving for M_z^- gives: M_z^-=\frac{M_0(1-e^{-T_R/T_1})}{1-e^{-T_R/T_1}\cos(\theta)} Transverse magnetization (signal intensity) right after the pulse during steady state is then: M_x^+=M_z^-\sin(\theta)=\frac{M_0(1-e^{-T_R/T_1})\sin(\theta)}{1-e^{-T_R/T_1}\cos(\theta)} To find the \theta that maximizes M_x^+ solve for \frac{dM_x^+}{d\theta}=0 : \begin{align} E&=e^{-T_R/T_1}\\ A&=M_0(1-E)\\ M_x^+&=\frac{A\sin(\theta)}{1-E\cos(\theta)} \end{align} Using the quotient rule with u=\sin(\theta) and v=1-E\cos(\theta) it can be shown that: \frac{dM_x^+}{d\theta}=A\frac{\cos(\theta)-E}{(1-E\cos(\theta))^2} \frac{dM_x^+}{d\theta}=0 when \theta=\arccos(e^{-T_R/T_1})

References

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