$B_0$ component of induction equation

The component of the induction equation in the direction of $\mathbf{B}_0$ is written as

$$\mathbf{B}_0 \cdot \mathbf{B}_1 =\mathbf{B}_0 \cdot \nabla \times (\ensuremath{\boldsymbol{\xi}} \times \mathbf{B}_0)$$ (88)

$$\Rightarrow Q_b =\mathbf{B}_0 \cdot \nabla \times (\ensuremath{\boldsymbol{\xi}} \times \mathbf{B}_0)$$ (89)

The term on right-hand side of the above equation is written as

$$\nabla \times (\ensuremath{\boldsymbol{\xi}} \times \mathbf{B}_0) = \nabla \times \left( \frac{\nabla \Psi \times \mathbf{B}_0}{\vert... ...\frac{\xi_s}{B^2} (\mathbf{B}_0 \times \nabla \Psi) \times \mathbf{B}_0 \right)$$

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

$$= \xi_{\psi} \nabla \times \frac{\nabla \Psi \times \mathbf{B}_0}{\... ...es \mathbf{B}_0}{\vert \nabla \Psi \vert^2} + \nabla \xi_s \times \nabla \Psi .$$ (90)

Using this, the right-hand side of Eq. (89) is written as

$$\mathbf{B}_0 \cdot \nabla \times (\ensuremath{\boldsymbol{\xi}} \times \mathbf{B}_0) = \xi_{\psi} \mathbf{B}_0 \cdot \nabla \times \frac{\nabla \Psi \ti... ... \cdot \nabla \xi_{\psi} + (\nabla \Psi \times \mathbf{B}_0) \cdot \nabla \xi_s$$

$$= \xi_{\psi} \mathbf{B}_0 \cdot \nabla \times \frac{\nabla \Psi \ti... ... \cdot \nabla \xi_{\psi} + (\nabla \Psi \times \mathbf{B}_0) \cdot \nabla \xi_s$$ (91)

$$= \xi_{\psi} \mathbf{B}_0 \cdot \left( -\mathbf{B}_0 \nabla \cdot \... ... \cdot \nabla \xi_{\psi} + (\nabla \Psi \times \mathbf{B}_0) \cdot \nabla \xi_s$$

$$= \xi_{\psi} \mathbf{B}_0 \cdot \left( \mathbf{B}_0 \cdot \nabla \f... ... \cdot \nabla \xi_{\psi} - (\mathbf{B}_0 \times \nabla \Psi) \cdot \nabla \xi_s$$ (92)

Before we try to simplify the above equation, we derive the expression for the divergence of $\ensuremath{\boldsymbol{\xi}}$, which is written as

$$\nabla \cdot \ensuremath{\boldsymbol{\xi}} = \nabla \cdot \left[ \frac{\nabla \Psi}{\vert \nabla \Psi \vert^2}... ..._0 \times \nabla \Psi)}{B^2_0} \xi_s + \frac{\mathbf{B}_0}{B^2_0} \xi_b \right]$$

$$= \frac{\nabla \Psi \cdot \nabla \xi_{\psi}}{\vert \nabla \Psi \ver... ...bla \Psi)}{B^2_0} +\mathbf{B}_0 \cdot \nabla \left( \frac{\xi_b}{B_0^2} \right)$$ (93)

It can be proved that the fourth term of the above equation can be written as (refer to (9.8) for the proof)

$$\nabla \cdot \frac{(\mathbf{B}_0 \times \nabla \Psi)}{B^2_0} = - 2 \frac{1}{B_0^2} \ensuremath{\boldsymbol{\kappa}} \cdot (\mathbf{B}_0 \times \nabla \Psi) .$$ (94)

Then Eq. (93) is written as

$$\nabla \cdot \ensuremath{\boldsymbol{\xi}}= \frac{\nabla \Psi \cd... ...imes \nabla \Psi) +\mathbf{B}_0 \cdot \nabla \left( \frac{\xi_b}{B_0^2} \right)$$ (95)

Eq. (95) agrees with Eq. (23) in Cheng's paper, but a $B_0^2$ factor is missed in the fourth term of Cheng's equation[3]. Using Eq. (95), Eq. (92) is written as

$$\mathbf{B}_0 \cdot \nabla \times (\ensuremath{\boldsymbol{\xi}} \... ...Psi) \xi_s + B_0^2 \mathbf{B}_0 \cdot \nabla \left( \frac{\xi_b}{B^2_0} \right)$$ (96)

It can be proved that

$$\mathbf{B}_0 \cdot \left( \mathbf{B}_0 \cdot \nabla \frac{\nabla ... ... \Psi \frac{B^2}{\vert \nabla \Psi \vert^2} +{\textmu}_0 \frac{d p_0}{d \Psi} .$$ (97)

(Refer to Sec. 9.10 for the proof.) Then Eq. (96) is written as

$$\mathbf{B}_0 \cdot \nabla \times (\ensuremath{\boldsymbol{\xi}} \... ...si) \xi_s + B_0^2 \mathbf{B}_0 \cdot \nabla \left( \frac{\xi_b}{B^2_0} \right),$$ (98)

and the component of the induction equation in the direction of $\mathbf{B}_0$ [Eq. (89)] is finally written as

$$Q_b = \xi_{\psi} \left( - \frac{2 B^2_0}{\vert \nabla \Psi \vert^... ...Psi) \xi_s + B_0^2 \mathbf{B}_0 \cdot \nabla \left( \frac{\xi_b}{B^2_0} \right)$$ (99)

Eq. (99) agrees with Eq. (22) in Cheng's paper[3].