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 ${\displaystyle T_{1}}$ using a flip repetition time ${\displaystyle T_{R}}$, assuming transverse magnetization is eliminated between flips. The Ernst angle ${\displaystyle \theta }$ is calculated using the following relationship, derived by Richard R. Ernst, who won the 1991 Nobel Prize in Chemistry:^{[1] [2]}

$${\displaystyle \theta =\arccos(e^{-T_{R}/T_{1}})}$$

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 ${\displaystyle T_{R}}$ for complete ${\displaystyle T_{2}}$ decay (in NMR).

Derivation

Let ${\displaystyle M_{z}^{-}}$ denote the longitudinal magnetization just before an RF pulse. Assume the pulse rotates magnetization by angle ${\displaystyle \theta }$ about the transverse axis instantaneously. Longitudinal magnetization after the pulse will be:

$${\displaystyle M_{z}^{+}=M_{z}^{-}\cos(\theta )}$$

After relaxation for a time ${\displaystyle T_{R}}$ toward equilibrium ${\displaystyle M_{0}}$ with ${\displaystyle T_{1}}$ recovery the longitudinal component of magnetization will be:

$${\displaystyle M_{z}^{-(next)}=M_{0}-(M_{0}-M_{z}^{+})e^{-T_{R}/T_{1}}=M_{0}-(M_{0}-M_{z}^{-}\cos(\theta ))e^{-T_{R}/T_{1}}}$$

Assume that transverse magnetization now becomes zero via spoiling. At steady state ${\displaystyle M_{z}^{-(next)}=M_{z}^{-}}$. Substituting ${\displaystyle M_{z}^{-}}$ for ${\displaystyle M_{z}^{-(next)}}$ and solving for ${\displaystyle M_{z}^{-}}$ gives:

$${\displaystyle M_{z}^{-}={\frac {M_{0}(1-e^{-T_{R}/T_{1}})}{1-e^{-T_{R}/T_{1}}\cos(\theta )}}}$$

Transverse magnetization (signal intensity) right after the pulse during steady state is then:

$${\displaystyle M_{x}^{+}=M_{z}^{-}\sin(\theta )={\frac {M_{0}(1-e^{-T_{R}/T_{1}})\sin(\theta )}{1-e^{-T_{R}/T_{1}}\cos(\theta )}}}$$

To find the ${\displaystyle \theta }$ that maximizes ${\displaystyle M_{x}^{+}}$ solve for ${\displaystyle {\frac {dM_{x}^{+}}{d\theta }}=0}$:

$${\displaystyle {\begin{aligned}E&=e^{-T_{R}/T_{1}}\\A&=M_{0}(1-E)\\M_{x}^{+}&={\frac {A\sin(\theta )}{1-E\cos(\theta )}}\end{aligned}}}$$

Using the quotient rule with ${\displaystyle u=\sin(\theta )}$ and ${\displaystyle v=1-E\cos(\theta )}$ it can be shown that:

${\displaystyle {\frac {dM_{x}^{+}}{d\theta }}=A{\frac {\cos(\theta )-E}{(1-E\cos(\theta ))^{2}}}}$

${\displaystyle {\frac {dM_{x}^{+}}{d\theta }}=0}$ when ${\displaystyle \theta =\arccos(e^{-T_{R}/T_{1}})}$

References

1. "1991 Nobel Laureates in Chemistry".
2. Ernst, R. R. (1966). "Application of Fourier transform spectroscopy to magnetic resonance". Review of Scientific Instruments. 37 (1): 93–102. Bibcode:1966RScI...37...93E. doi:10.1063/1.1719961.