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

The $\mathbf{B}_0 \times \nabla \Psi$ component of the linearized momentum equation is written

$$- \omega^2 \rho_0 \vert \nabla \Psi \vert^2 \xi_s = - (\mathbf{B}... ...Psi) \cdot {\textmu}_0^{- 1} (\nabla \times \mathbf{B}_0) \times \mathbf{B}_1 .$$ (111)

The last term of Eq. (111) is written

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

$$= -\mathbf{J}_0 \cdot (\mathbf{B}_0 \times \nabla \Psi) \times \mathbf{B}_1$$

$$= -\mathbf{J}_0 \cdot (\mathbf{B}_0 \times \nabla \Psi) \times \lef... ...\vert \nabla \Psi \vert^2} \nabla \Psi + \frac{Q_b}{B^2_0} \mathbf{B}_0 \right]$$

$$= -\mathbf{J}_0 \cdot \left[ \frac{Q_{\psi}}{\vert \nabla \Psi \ver... ... \Psi \vert^2 \mathbf{B}_0) + \frac{Q_b}{B^2_0} (B^2_0 \nabla \Psi - 0) \right]$$

$$= -\mathbf{J}_0 \cdot [- Q_{\psi} \mathbf{B}_0 + Q_b \nabla \Psi]$$

$$= \mathbf{B}_0 \cdot \mathbf{J}_0 Q_{\psi} .$$ (112)

Using Eq. (335), i.e.,

$$\nabla \times \mathbf{B}_1 = \nabla \frac{Q_{\psi}}{\vert \nabla ... ...{Q_b}{B^2_0} \times \mathbf{B}_0 + \frac{Q_b}{B^2_0} {\textmu}_0 \mathbf{J}_0 .$$ (113)

the second last term on the right-hand side of Eq. (111) is written

$$(\mathbf{B}_0 \times \nabla \Psi) \cdot {\textmu}_0^{- 1} (\nabla \times \mathbf{B}_1) \times \mathbf{B}_0 = {\textmu}_0^{- 1} (\mathbf{B}_0 \times \nabla \Psi) \cdot \left[ ... ...f{B}_0 + \frac{Q_b}{B^2_0} {\textmu}_0 \mathbf{J}_0 \right] \times \mathbf{B}_0$$

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

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

$$= {\textmu}_0^{- 1} (\mathbf{B}_0 \times \nabla \Psi) \cdot \left[ ... ...times \nabla \Psi) \times \mathbf{B}_0 - B^2_0 \nabla \frac{Q_b}{B^2_0} \right]$$ (114)

$$= {\textmu}_0^{- 1} \underbrace{B^2_0 (\mathbf{B}_0 \cdot \nabla Q_... ...times \nabla \Psi) \times \mathbf{B}_0 - B^2_0 \nabla \frac{Q_b}{B^2_0} \right]$$ (115)

The term $\nabla \times (\mathbf{B}_0 \times \nabla \Psi)$ in the above equation is written as

$$\nabla \times (\mathbf{B}_0 \times \nabla \Psi) = - \nabla \times (\nabla \Psi \times \mathbf{B}_0)$$

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

$$= \mathbf{B}_0 (\nabla^2 \Psi) - (\mathbf{B}_0 \cdot \nabla) \nabla \Psi + \nabla \Psi \cdot \nabla \mathbf{B}_0$$

$$\Rightarrow \frac{Q_s}{\vert \nabla \psi \vert^2} \nabla \times (\mathbf{B}_0 \times \nabla \Psi) \times \mathbf{B}_0 = \frac{Q_s}{\vert \nabla \Psi \vert^2} [- (\mathbf{B}_0 \cdot \nab... ...times \mathbf{B}_0 + \nabla \Psi \cdot \nabla \mathbf{B}_0 \times \mathbf{B}_0]$$ (116)

Gathering terms involving $Q_s$, excluding the first term, in expression (115) gives

$$(\mathbf{B}_0 \times \nabla \Psi) \cdot \left[ Q_s \left( \mathbf... ...{B}_0 + (\nabla \Psi \cdot \nabla \mathbf{B}_0) \times \mathbf{B}_0] \right],$$

which can be proved to be zero (refer to Sec. 9.6 for the proof). Gathering terms involving $Q_b$ in expression (115) gives

$$(\mathbf{B}_0 \times \nabla \Psi) \cdot \left[ - B^2_0 \nabla \frac{Q_b}{B^2_0} \right]$$

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

$$= \underbrace{- (\mathbf{B}_0 \times \nabla \Psi) \cdot \nabla Q_b} - B^2_0 Q_b (\mathbf{B}_0 \times \nabla \Psi) \cdot \nabla \frac{1}{B^2_0}$$

It can be proved that the second term of the above expression is equal to $2\ensuremath{\boldsymbol{\kappa}} \cdot (\mathbf{B}_0 \times \nabla \Psi) Q_b$ (refer to Sec. 9.9 for details). Thus, from Eq. (114), we obtain

$$(\mathbf{B}_0 \times \nabla \Psi) \cdot {\textmu}_0^{- 1} (\nabla... ...1} \ensuremath{\boldsymbol{\kappa}} \cdot (\mathbf{B}_0 \times \nabla \Psi) Q_b$$ (117)

Using Eqs. (112) and (117) in Eq. (111) yields

$$- \omega^2 \rho_0 \vert \nabla \Psi \vert^2 \xi_s = - (\mathbf{B}... ...mathbf{B}_0 \times \nabla \Psi) Q_b +\mathbf{B}_0 \cdot \mathbf{J}_0 Q_{\psi} .$$ (118)

Using $Q_b =\mathbf{B}_1 \cdot \mathbf{B}_0$, the above equation can be arranged as

$$- \omega^2 \rho_0 \vert \nabla \Psi \vert^2 \xi_s = - (\mathbf{B}... ...\mathbf{B}_0 \times \nabla \Psi) Q_b +\mathbf{B}_0 \cdot \mathbf{J}_0 Q_{\psi},$$ (119)

which agrees with Eq. (18) in Cheng's paper[3].