$B_0 \times \nabla \Psi$ component of induction equation

The $\mathbf{B}_0 \times \nabla \Psi$ component of the induction equation is given by

$$\mathbf{B}_1 \cdot \mathbf{B}_0 \times \nabla \Psi = \nabla \time... ...\boldsymbol{\xi}} \times \mathbf{B}_0) \cdot (\mathbf{B}_0 \times \nabla \Psi),$$ (83)

which can be further written as

$$B^2_0 Q_s = \nabla \times \left( \xi_{\psi} \frac{\nabla \Psi \times \mathbf{... ...si \vert^2} + \xi_s \nabla \Psi \right) \cdot (\mathbf{B}_0 \times \nabla \Psi)$$

$$= \left[ \nabla \times \left( \xi_{\psi} \frac{\nabla \Psi \times \... ...\nabla \xi_s \times \nabla \Psi \right] \cdot (\mathbf{B}_0 \times \nabla \Psi)$$

$$= \left[ \xi_{\psi} \nabla \times \frac{\nabla \Psi \times \mathbf{... ...\nabla \xi_s \times \nabla \Psi \right] \cdot (\mathbf{B}_0 \times \nabla \Psi)$$

$$= \left[ \xi_{\psi} \nabla \times \frac{\nabla \Psi \times \mathbf{... ...\nabla \xi_s \times \nabla \Psi \right] \cdot (\mathbf{B}_0 \times \nabla \Psi)$$

$$= - \vert \nabla \psi \vert^2 S \xi_{\psi} + (\nabla \xi_s \times \nabla \Psi) \cdot (\mathbf{B}_0 \times \nabla \Psi),$$ (84)

where

$$S = \left( \nabla \times \frac{\mathbf{B}_0 \times \nabla \Psi}{\... ...ght) \cdot \frac{(\mathbf{B}_0 \times \nabla \Psi)}{\vert \nabla \Psi \vert^2},$$ (85)

is the negative local magnetic shear. Using $(\mathbf{A} \times \mathbf{B}) \cdot (\mathbf{C} \times \mathbf{D}) = (\mathbf... ...\cdot \mathbf{D}) - (\mathbf{A} \cdot \mathbf{D}) (\mathbf{B} \cdot \mathbf{C})$, equation (84) is written as

$$B^2_0 Q_s = - \vert \nabla \Psi \vert^2 S \xi_{\psi} + \vert \nabla \Psi \vert^2 \mathbf{B}_0 \cdot \nabla \xi_s,$$ (86)

$$\Rightarrow Q_s = \left( \frac{\vert \nabla \Psi \vert}{B_0} \right)^2 (\mathbf{B}_0 \cdot \nabla \xi_s - S \xi_{\psi}) .$$ (87)

Eq. (87) agrees with Eq. (21) in Cheng's paper[3].