$\nabla \Psi$ component of momentum equation

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

$$- \omega^2 \rho_0 \xi_{\psi} = - \nabla \Psi \cdot \nabla p_1 + \... ...\Psi \cdot {\textmu}_0^{- 1} (\nabla \times \mathbf{B}_0) \times \mathbf{B}_1 .$$ (106)

After some algebra (the details are given in Sec. (9.5)), Eq. (106) is written

$$- \omega^2 \rho_0 \xi_{\psi} = - \nabla \Psi \cdot \nabla P_1 +{\textmu}_0^{- 1} \vert \nabla \P... ...hbf{B}_0 \cdot \nabla \left( \frac{Q_{\psi}}{\vert \nabla \Psi \vert^2} \right)$$

$$+ ({\textmu}_0^{- 1} \vert \nabla \Psi \vert^2 S -\mathbf{B}_0 \c... ..._s + 2{\textmu}_0^{- 1} \ensuremath{\boldsymbol{\kappa}} \cdot \nabla \Psi Q_b,$$ (107)

where

$$P_1 \equiv p_1 + \frac{\mathbf{B}_1 \cdot \mathbf{B}_0}{{\textmu}_0},$$ (108)

and $\ensuremath{\boldsymbol{\kappa}} \equiv \mathbf{b} \cdot \nabla \mathbf{b}$ is the magnetic field curvature with $\mathbf{b}=\mathbf{B}_0 / B_0$ the unit vector along equilibrium magnetic field. Equation (107) agrees with Eq. (17) in Cheng's paper[3]. In passing, let us examine the physical meaning of $P_1$ defined by (108). In linear approximation, we have

$$B^2 = \vert\mathbf{B}_0 +\mathbf{B}_1 \vert^2$$

$$= B_0^2 + B_1^2 + 2\mathbf{B}_0 \cdot \mathbf{B}_1$$

$$\approx B_0^2 + 2\mathbf{B}_0 \cdot \mathbf{B}_1$$ (109)

This indicates the perturbation in the square of the magnetic strength is $2\mathbf{B}_0 \cdot \mathbf{B}_1$. Therefore, the perturbation in magnetic pressure is written

$$\Delta P_{\ensuremath{\operatorname{mag}}} \equiv \frac{\Delta (B... ...thbf{B}_1}{2{\textmu}_0} = \frac{\mathbf{B}_0 \cdot \mathbf{B}_1}{{\textmu}_0},$$ (110)

which indicate $P_1$ defined by Eq. (108) is the total perturbation in the thermal and magnetic pressure.