proof

In this subsection, we prove that

$$\Pi \equiv (\mathbf{B} \times \nabla \psi) \cdot \left[ Q_s \left... ...\mathbf{B}+ (\nabla \psi \cdot \nabla \mathbf{B}) \times \mathbf{B}] \right],$$

is zero.

$$\Pi = Q_s \left( \mathbf{B} \cdot \nabla \frac{1}{\vert \nabla \p... ...dot \nabla \mathbf{B}) \times \mathbf{B} \cdot (\mathbf{B} \times \nabla \psi)]$$

$$= Q_s \left( \mathbf{B} \cdot \nabla \frac{1}{\vert \nabla \psi \... ...mes \nabla \psi) \times \mathbf{B} \cdot (\nabla \psi \cdot \nabla \mathbf{B})]$$

$$= Q_s \left( \mathbf{B} \cdot \nabla \frac{1}{\vert \nabla \psi \... ...bla \nabla \psi) - B^2 \nabla \psi \cdot (\nabla \psi \cdot \nabla \mathbf{B})]$$ (352)

$$= Q_s \left( \mathbf{B} \cdot \nabla \frac{1}{\vert \nabla \psi \... ...nabla \nabla \psi) - (\nabla \psi) \cdot (\nabla \psi \cdot \nabla \mathbf{B})]$$

$$= Q_s \left( \mathbf{B} \cdot \nabla (\vert \nabla \psi \vert^2) ... ...nabla \nabla \psi) - (\nabla \psi) \cdot (\nabla \psi \cdot \nabla \mathbf{B})]$$

$$= 2 Q_s \nabla \psi \cdot \mathbf{B} \cdot \nabla \nabla \psi \le... ...nabla \nabla \psi) - (\nabla \psi) \cdot (\nabla \psi \cdot \nabla \mathbf{B})]$$

$$\left. = - Q_s \frac{B^2}{\vert \nabla \psi \vert^2} \nabla \psi ... ...\psi \vert^2} (\nabla \psi) \cdot (\nabla \psi \cdot \nabla \mathbf{B}) \right]$$

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

$$= - Q_s \frac{B^2}{\vert \nabla \psi \vert^2} \nabla \psi \cdot [... ...si \times \nabla \times \mathbf{B}+\mathbf{B} \times \nabla \times \nabla \psi]$$

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

$$= 0$$

where the last second equality is due to the fact that

$$\nabla (\mathbf{A} \cdot \mathbf{B}) = \mathbf{B} \cdot \nabla \mathbf{A}+\mathbf{A} \cdot \nabla \mathb... ...f{A} \times \nabla \times \mathbf{B}+\mathbf{B} \times \nabla \times \mathbf{A}$$ (353)