proof: The derivation of Eq. (107)

Using Eq. (65), we obtain

$$\nabla \times \mathbf{B}_1 = \nabla \times \left( \frac{Q_{\psi}}{\vert \nabla \psi \vert^2} \... ...ert^2} (\mathbf{B}_0 \times \nabla \psi) + \frac{Q_b}{B^2} \mathbf{B}_0 \right)$$

$$= \nabla \frac{Q_{\psi}}{\vert \nabla \psi \vert^2} \times \nabla \... ...ac{Q_b}{B^2_0} \times \mathbf{B}_0 + \frac{Q_b}{B^2_0} {\textmu}_0 \mathbf{J}_0$$

$$= \nabla \frac{Q_{\psi}}{\vert \nabla \psi \vert^2} \times \nabla \... ...ac{Q_b}{B^2_0} \times \mathbf{B}_0 + \frac{Q_b}{B^2_0} {\textmu}_0 \mathbf{J}_0$$ (335)

Using this, the second term on the right-hand side of Eq. (106) is written

$$\nabla \psi \cdot {\textmu}_0^{- 1} (\nabla \times \mathbf{B}_1) \times \mathbf{B}_0 = {\textmu}_0^{- 1} \mathbf{B}_0 \cdot \nabla \psi \times (\nabla \times \mathbf{B}_1)$$

$$= {\textmu}_0^{- 1} \mathbf{B}_0 \cdot \nabla \psi \times \left( \n... ...2_0} \times \mathbf{B}_0 + \frac{Q_b}{B^2_0} {\textmu}_0 \mathbf{J}_0 \right) .$$ (336)

The first term of Eq. (336) is written

$${\textmu}_0^{- 1} \mathbf{B}_0 \cdot \nabla \psi \times \left( \nabla \frac{Q_{\psi}}{\vert \nabla \psi \vert^2} \times \nabla \psi \right) = {\textmu}_0^{- 1} \mathbf{B}_0 \cdot \left[ \vert \nabla \psi \ve... ...t \nabla \frac{Q_{\psi}}{\vert \nabla \psi \vert^2} \right) \nabla \psi \right]$$

$$= {\textmu}_0^{- 1} \vert \nabla \psi \vert^2 \mathbf{B}_0 \cdot \nabla \frac{Q_{\psi}}{\vert \nabla \psi \vert^2}$$ (337)

The last equality is due to $\mathbf{B}_0 \cdot \nabla \psi = 0$. The second term of Eq. (336) is written

$${\textmu}_0^{- 1} \mathbf{B}_0 \cdot \nabla \psi \times \frac{Q_s}{\vert \nabla \psi \vert^2} \nabla \times (\mathbf{B}_0 \times \nabla \psi) = {\textmu}_0^{- 1} \frac{Q_s}{\vert \nabla \psi \vert^2} \mathbf{B}_0 \cdot [\nabla \times (\nabla \psi \times \mathbf{B}_0)] \times \nabla \psi$$

$$= {\textmu}_0^{- 1} \frac{Q_s}{\vert \nabla \psi \vert^2} \mathbf{B... ...abla) \nabla \psi - (\nabla \psi \cdot \nabla) \mathbf{B}_0] \times \nabla \psi$$

$$= {\textmu}_0^{- 1} Q_s \frac{1}{\vert \nabla \psi \vert^2} \mathbf... ...abla) \nabla \psi - (\nabla \psi \cdot \nabla) \mathbf{B}_0] \times \nabla \psi$$

$$= {\textmu}_0^{- 1} Q_s \vert \nabla \psi \vert^2 S$$ (338)

The last equality is due to that the coefficients before $Q_s$, i.e.,

$$\frac{1}{\vert \nabla \psi \vert^2} \mathbf{B}_0 \cdot [(\mathbf{... ...la) \nabla \psi - (\nabla \psi \cdot \nabla) \mathbf{B}_0] \times \nabla \psi$$

is equal to $\vert \nabla \psi \vert^2 S$ (refer to Sec. 9.7 for the proof). The third term of Eq. (336) is written

$${\textmu}_0^{- 1} \mathbf{B}_0 \cdot \nabla \psi \times \left( \nabla \frac{Q_b}{B^2_0} \times \mathbf{B}_0 \right) = {\textmu}_0^{- 1} \mathbf{B}_0 \cdot \left[ 0 - \left( \nabla \psi \cdot \nabla \frac{Q_b}{B^2_0} \right) \mathbf{B}_0 \right]$$

$$= -{\textmu}_0^{- 1} \left( \nabla \frac{Q_b}{B^2_0} \cdot \nabla \psi \right) B^2_0 .$$ (339)

Using Eqs. (337)-(339) in Eq. (336) yields

$$\nabla \psi \cdot {\textmu}_0^{- 1} (\nabla \times \mathbf{B}_1) ... ... B^2_0 + \frac{Q_b}{B^2_0} \mathbf{B}_0 \cdot (\nabla \psi \times \mathbf{J}_0)$$ (340)

Now we calculate the $\nabla \psi \cdot {\textmu}_0^{- 1} (\nabla \times \mathbf{B}_0) \times \mathbf{B}_1$ term appearing in Eq. (106), which can be written as

$$\nabla \psi \cdot {\textmu}_0^{- 1} (\nabla \times \mathbf{B}_0) \times \mathbf{B}_1 = -\mathbf{J}_0 \cdot \nabla \psi \times \mathbf{B}_1$$

$$= -\mathbf{J}_0 \cdot \nabla \psi \times \left( \frac{Q_s}{\vert \n... ...t^2} (\mathbf{B}_0 \times \nabla \psi) + \frac{Q_b}{B^2_0} \mathbf{B}_0 \right)$$

$$= -\mathbf{J}_0 \cdot \left( \frac{Q_s}{\vert \nabla \psi \vert^2} ... ...\times \nabla \psi) + \frac{Q_b}{B^2_0} \nabla \psi \times \mathbf{B}_0 \right)$$

$$= -\mathbf{J}_0 \cdot \left( \frac{Q_s}{\vert \nabla \psi \vert^2} ... ...vert^2 \mathbf{B}_0 + \frac{Q_b}{B^2_0} \nabla \psi \times \mathbf{B}_0 \right)$$

$$= -\mathbf{J}_0 \cdot \mathbf{B}_0 Q_s -\mathbf{J}_0 \cdot \left( \frac{Q_b}{B^2_0} \nabla \psi \times \mathbf{B}_0 \right) .$$ (341)

Gathering the terms involving $Q_b$ in Eqs. (340) and (341), we obtain

$$-{\textmu}_0^{- 1} \left( \nabla \frac{Q_b}{B^2_0} \cdot \nabla \... ...{J}_0) + \frac{Q_b}{B^2_0} \mathbf{B}_0 \cdot (\nabla \psi \times \mathbf{J}_0)$$

$$= - \left[ \left( \frac{1}{B^2} \nabla Q_b + Q_b \nabla \frac{1}{... ... B^2_0 + 2 \frac{Q_b}{B^2} \mathbf{B}_0 \cdot [\nabla \psi \times \mathbf{J}_0]$$

$$= -{\textmu}_0^{- 1} \nabla Q_b \cdot \nabla \psi -{\textmu}_0^{-... ...{B}_0 \cdot [\nabla \psi \times {\textmu}_0^{- 1} (\nabla \times \mathbf{B}_0)]$$

$$={\textmu}_0^{- 1} \left\{ - \nabla Q_b \cdot \nabla \psi - Q_b B... ...0 \cdot \nabla) \nabla \psi - (\nabla \psi \cdot \nabla) \mathbf{B}_0] \right\}$$ (342)

The second term of Eq. (342) is written

$$- Q_b B^2_0 (\nabla \psi \cdot \nabla) \frac{1}{B^2_0} = - \left( 0 - Q_b \frac{1}{B^2_0} (\nabla \psi \cdot \nabla) B^2_0 \right)$$

$$= Q_b \frac{1}{B^2_0} (\nabla \psi \cdot \nabla) B^2_0$$ (343)

The last term of Eq. (342) is written

$$- 2 \frac{Q_b}{B^2_0} \mathbf{B}_0 \cdot (\nabla \psi \cdot \nabla) \mathbf{B}_0 = - \frac{Q_b}{B^2_0} (\nabla \psi \cdot \nabla) (\mathbf{B}_0 \cdot \mathbf{B}_0)$$

$$= - \frac{Q_b}{B^2_0} (\nabla \psi \cdot \nabla) B^2_0$$ (344)

The terms in Eqs. (343) and (344) exactly cancel each other. Thus the expression in (342) now reduces to

$${\textmu}_0^{- 1} \left\{ - \nabla Q_b \cdot \nabla \psi + 2 \fra... ...{B^2_0} \mathbf{B}_0 \cdot [- (\mathbf{B}_0 \cdot \nabla) \nabla \psi] \right\}$$ (345)

Noting that

$$\mathbf{B}_0 \cdot [- (\mathbf{B}_0 \cdot \nabla) \nabla \psi] = - (\mathbf{B}_0 \cdot \nabla) [\mathbf{B}_0 \cdot \nabla \psi] + [(\mathbf{B}_0 \cdot \nabla) \mathbf{B}_0] \cdot \nabla \psi$$

$$= [(\mathbf{B}_0 \cdot \nabla) \mathbf{B}_0] \cdot \nabla \psi,$$ (346)

the expression (345) is further written as

$${\textmu}_0^{- 1} \left\{ - \nabla Q_b \cdot \nabla \psi + 2 \frac{Q_b}{B^2} [\mathbf{B}_0 \cdot \nabla \mathbf{B}_0] \cdot \nabla \psi \right\}$$

$$={\textmu}_0^{- 1} \left\{ - \nabla Q_b \cdot \nabla \psi + 2 Q_b... ... \left( \frac{\mathbf{B}_0}{B_0} B_0 \right) \right] \cdot \nabla \psi \right\}$$

$$={\textmu}_0^{- 1} \left\{ - \nabla Q_b \cdot \nabla \psi + 2 Q_b... ...ac{1}{B_0} [\mathbf{b} \cdot \nabla (\mathbf{b}B_0)] \cdot \nabla \psi \right\}$$

$$={\textmu}_0^{- 1} \{ - \nabla Q_b \cdot \nabla \psi + 2 Q_b [\mathbf{b} \cdot \nabla (\mathbf{b})] \cdot \nabla \psi + 0 \}$$

$$={\textmu}_0^{- 1} \{ - \nabla Q_b \cdot \nabla \psi + 2 Q_b \ensuremath{\boldsymbol{\kappa}} \cdot \nabla \psi \},$$

where $\mathbf{b}=\mathbf{B}_0 / B$ is the unit vector along the direction of equilibrium magnetic field, and $\ensuremath{\boldsymbol{\kappa}}=\mathbf{b} \cdot \nabla \mathbf{b}$ is the magnetic field curvature. Using these results, we obtain

$$\nabla \psi \cdot {\textmu}_0^{- 1} (\nabla \times \mathbf{B}_1) ... ...a \psi \cdot {\textmu}_0^{- 1} (\nabla \times \mathbf{B}_0) \times \mathbf{B}_1$$

$$={\textmu}_0^{- 1} \vert \nabla \psi \vert^2 \mathbf{B}_0 \cdot \... ...\cdot \nabla \psi + 2 Q_b \ensuremath{\boldsymbol{\kappa}} \cdot \nabla \psi) .$$ (347)

Using the above results, the radial component equation

$$- \omega^2 \rho_{m 0} \xi_{\psi} = - \nabla \psi \cdot \nabla p_1... ... \psi \cdot {\textmu}_0^{- 1} (\nabla \times \mathbf{B}_0) \times \mathbf{B}_1,$$ (348)

is written as

$$- \omega^2 \rho_{m 0} \xi_{\psi} = - \nabla \psi \cdot \nabla p_1... ...b \cdot \nabla \psi + 2 Q_b \ensuremath{\boldsymbol{\kappa}} \cdot \nabla \psi)$$ (349)

which can be arranged in the form

$$- \omega^2 \rho_{m 0} \xi_{\psi} = - \nabla \psi \cdot \nabla (p_1 +{\textmu}_0^{- 1} \mathbf{B}_1 \... ...hbf{B}_0 \cdot \nabla \left( \frac{Q_{\psi}}{\vert \nabla \psi \vert^2} \right)$$

$$+ ({\textmu}_0^{- 1} \vert \nabla \psi \vert^2 S -\mathbf{B}_0 \c... ...s + 2{\textmu}_0^{- 1} \ensuremath{\boldsymbol{\kappa}} \cdot \nabla \psi Q_b .$$ (350)

Define $P_1 = p_1 +{\textmu}_0^{- 1} \mathbf{B}_1 \cdot \mathbf{B}_0$, the above equation is written as

$$- \omega^2 \rho_{m 0} \xi_{\psi} = - \nabla \psi \cdot \nabla P_1 +{\textmu}_0^{- 1} \vert \nabla \p... ...hbf{B}_0 \cdot \nabla \left( \frac{Q_{\psi}}{\vert \nabla \psi \vert^2} \right)$$

$$+ ({\textmu}_0^{- 1} \vert \nabla \psi \vert^2 S -\mathbf{B}_0 \c... ..._s + 2{\textmu}_0^{- 1} \ensuremath{\boldsymbol{\kappa}} \cdot \nabla \psi Q_b,$$ (351)

which is identical with Eq. (107).