Component of the momentum equation parallel to $B_0$

The three components of the linearized momentum equation can be obtained by taking scalar product of Eq. (35) with $\nabla \Psi$, $\mathbf{B}_0 \times \nabla \psi$, and $\mathbf{B}_0$, respectively. We first consider the $\mathbf{B}_0$ component of the momentum equation, which is written as

$$- \omega^2 \rho_0 \ensuremath{\boldsymbol{\xi}} \cdot \mathbf{B}_... ...}_0 \cdot [{\textmu}_0^{- 1} (\nabla \times \mathbf{B}_0) \times \mathbf{B}_1],$$ (100)

i.e.,

$$\omega^2 \rho_0 \xi_b =\mathbf{B}_0 \cdot \nabla p_1 +\mathbf{J}_0 \cdot (\mathbf{B}_0 \times \mathbf{B}_1) .$$ (101)

The last term on the right-hand side of Eq. (101) can be written as

$$\mathbf{J}_0 \cdot (\mathbf{B}_0 \times \mathbf{B}_1) = \mathbf{J}_0 \cdot \mathbf{B}_0 \times \left[ \frac{Q_{\psi}}{\ve... ...\frac{Q_s}{\vert \nabla \Psi \vert^2} (\mathbf{B}_0 \times \nabla \Psi) \right]$$

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

$$= \mathbf{J}_0 \cdot \left[ \frac{Q_{\psi}}{\vert \nabla \Psi \vert... ...s \nabla \Psi - \frac{Q_s}{\vert \nabla \Psi \vert^2} B^2_0 \nabla \Psi \right]$$ (102)

$$= \mathbf{J}_0 \cdot \left[ \frac{Q_{\psi}}{\vert \nabla \Psi \vert^2} \mathbf{B}_0 \times \nabla \Psi \right] - 0$$

$$= \nabla \Psi \cdot \left[ \frac{Q_{\psi}}{\vert \nabla \Psi \vert^2} \mathbf{J}_0 \times \mathbf{B}_0 \right]$$

$$= \nabla \Psi \cdot \left[ \frac{Q_{\psi}}{\vert \nabla \Psi \vert^2} \nabla p_0 \right]$$

$$= \nabla \Psi \cdot \left[ \frac{Q_{\psi}}{\vert \nabla \Psi \vert^2} p'_0 \nabla \Psi \right]$$

$$= Q_{\psi} p'_0 .$$

where $p_0' \equiv d p_0 / d \Psi$. Using Eq. (82), i.e., $Q_{\psi} =\mathbf{B}_0 \cdot \nabla \xi_{\psi}$, the above equation is written as

$$\mathbf{J}_0 \cdot \mathbf{B}_0 \times \mathbf{B}_1 = (\mathbf{B}_0 \cdot \nabla \xi_{\psi}) p'_0$$ (103)

Substituting Eq. (103) into Eq. (101) gives

$$\omega^2 \rho_0 \xi_b =\mathbf{B}_0 \cdot \nabla p_1 + (\mathbf{B}_0 \cdot \nabla \xi_{\psi}) p'_0$$ (104)

Using $\mathbf{B}_0 \cdot \nabla p'_0 = 0$ in the above equation gives

$$\omega^2 \rho_0 \xi_b =\mathbf{B}_0 \cdot \nabla (p_1 + p'_0 \xi_{\psi}),$$ (105)

which agrees with Eq. (19) in Cheng's paper[3].