$\nabla \Psi$ component of induction equation

The induction equation is given by Eq. (37), i.e.,

$$\mathbf{B}_1 = \nabla \times (\ensuremath{\boldsymbol{\xi}} \times \mathbf{B}_0) .$$ (68)

Next, consider the $\nabla \Psi$ component of the above equation. Taking scalar product of the above equation with $\nabla \Psi$, we obtain

$$\mathbf{B}_1 \cdot \nabla \Psi = \nabla \times (\ensuremath{\boldsymbol{\xi}} \times \mathbf{B}_0) \cdot \nabla \Psi .$$ (69)

$$\Rightarrow Q_{\psi} = \nabla \times \left[ \frac{\xi_{\psi}}{\ve... ...(\mathbf{B}_0 \times \nabla \Psi) \times \mathbf{B}_0 \right] \cdot \nabla \Psi$$ (70)

$$\Rightarrow Q_{\psi} = \nabla \times \left[ \frac{\xi_{\psi}}{\ve... ...} \nabla \Psi \times \mathbf{B}_0 + \xi_s \nabla \Psi \right] \cdot \nabla \Psi$$ (71)

$$\Rightarrow Q_{\psi} = \left[ \nabla \times \left( \frac{\xi_{\ps... ...thbf{B}_0 \right) + \nabla \times (\xi_s \nabla \Psi) \right] \cdot \nabla \Psi$$ (72)

$$\Rightarrow Q_{\psi} = \left[ \nabla \times \left( \frac{\xi_{\ps... ...mathbf{B}_0 \right) + \nabla \xi_s \times \nabla \Psi \right] \cdot \nabla \Psi$$ (73)

$$\Rightarrow Q_{\psi} = \left[ \nabla \times \left( \frac{\xi_{\ps... ...\Psi \vert^2} \nabla \Psi \times \mathbf{B}_0 \right) \right] \cdot \nabla \Psi$$ (74)

$$\Rightarrow Q_{\psi} = \left[ (\mathbf{B}_0 \cdot \nabla) \left( ... ...a \Psi \vert^2} \nabla \Psi \cdot \nabla \mathbf{B}_0 \right] \cdot \nabla \Psi$$ (75)

$$Q_{\psi} = \left[ \frac{\xi_{\psi}}{\vert \nabla \Psi \vert^2} (\... ...a \Psi \vert^2} \nabla \Psi \cdot \nabla \mathbf{B}_0 \right] \cdot \nabla \Psi$$ (76)

$$Q_{\psi} = \left[ \frac{\xi_{\psi}}{\vert \nabla \Psi \vert^2} (\... ...a \Psi \vert^2} \nabla \Psi \cdot \nabla \mathbf{B}_0 \right] \cdot \nabla \Psi$$ (77)

$$Q_{\psi} = \left[ \frac{\xi_{\psi}}{\vert \nabla \Psi \vert^2} (\... ...\Psi \vert^2} (\nabla \Psi \cdot \nabla \mathbf{B}_0) \cdot \nabla \Psi \right]$$ (78)

Excluding $(\mathbf{B}_0 \cdot \nabla) \xi_{\psi}$ terms, the terms on the right hand side (r.h.s) of the above equation can be written

$$\frac{\xi_{\psi}}{\vert \nabla \Psi \vert^2} (\mathbf{B}_0 \cdot ... ... \nabla \Psi \vert^2} (\nabla \Psi \cdot \nabla \mathbf{B}_0) \cdot \nabla \Psi$$

$$= \frac{\xi_{\psi}}{\vert \nabla \Psi \vert^2} (\mathbf{B}_0 \cdo... ... \nabla \Psi \vert^2} (\nabla \Psi \cdot \nabla \mathbf{B}_0) \cdot \nabla \Psi$$

$$= \frac{\xi_{\psi}}{\vert \nabla \Psi \vert^2} (\mathbf{B}_0 \cdo... ... \nabla \Psi \vert^2} (\nabla \Psi \cdot \nabla \mathbf{B}_0) \cdot \nabla \Psi$$

$$= - \frac{\xi_{\psi}}{\vert \nabla \Psi \vert^2} \nabla \Psi \cdo... ... \nabla \Psi \vert^2} (\nabla \Psi \cdot \nabla \mathbf{B}_0) \cdot \nabla \Psi$$

$$= - \frac{\xi_{\psi}}{\vert \nabla \Psi \vert^2} \nabla \Psi \cdo... ...\mathbf{B}_0 \cdot \nabla) \nabla \Psi + \nabla \Psi \cdot \nabla \mathbf{B}_0]$$ (79)

Using $\nabla (\mathbf{A} \cdot \mathbf{B}) =\mathbf{B} \cdot \nabla \mathbf{A}+\math... ...f{A} \times \nabla \times \mathbf{B}+\mathbf{B} \times \nabla \times \mathbf{A}$, we obtain

$$\nabla (\nabla \Psi \cdot \mathbf{B}_0) = \mathbf{B}_0 \cdot \nabla \nabla \Psi + \nabla \Psi \cdot \nabla \mathbf{B}_0 + \nabla \Psi \times \nabla \times \mathbf{B}_0$$ (80)

$$= \nabla \Psi \cdot \nabla (\nabla \Psi \cdot \mathbf{B}_0) = \nabl... ...[\mathbf{B}_0 \cdot \nabla \nabla \Psi + \nabla \Psi \cdot \nabla \mathbf{B}_0]$$ 0 (81)

The r.h.s of the above equation is exactly the term appearing on the right-hand side of Eq. (79). Thus we obtain

$$Q_{\psi} =\mathbf{B}_0 \cdot \nabla \xi_{\psi},$$ (82)

which agrees with Eq. (20) in Cheng's paper[3].