proof 2

We want to prove that

$$\mathbf{B}_0 \cdot [(\mathbf{B}_0 \cdot \nabla) \nabla \psi - (\n... ...es \frac{\nabla \psi}{\vert \nabla \psi \vert^2} = \vert \nabla \psi \vert^2 S,$$ (354)

where $S = \left( \mathbf{B}_0 \times \frac{\nabla \psi}{\vert \nabla \psi \vert^2} \... ...left( \mathbf{B}_0 \times \frac{\nabla \psi}{\vert \nabla \psi \vert^2} \right)$. The right-hand side of Eq. (354) is written as

$$\vert \nabla \psi \vert^2 \left( \mathbf{B}_0 \times \frac{\nabla... ...left( \mathbf{B}_0 \times \frac{\nabla \psi}{\vert \nabla \psi \vert^2} \right) = (\mathbf{B}_0 \times \nabla \psi) \cdot \left[ - (\mathbf{B}_0 \c... ...abla \psi}{\vert \nabla \psi \vert^2} \cdot \nabla \right) \mathbf{B}_0 \right]$$

$$= - (\mathbf{B}_0 \times \nabla \psi) \cdot (\mathbf{B}_0 \cdot \na... ...vert \nabla \psi \vert^2} \right) \cdot (\nabla \psi \cdot \nabla) \mathbf{B}_0$$

$$= \left[ (\mathbf{B}_0 \cdot \nabla) \frac{\nabla \psi}{\vert \nabl... ...\times \frac{\nabla \psi}{\vert \nabla \psi \vert^2} \right] \cdot \mathbf{B}_0$$ (355)

$$= \left[ \frac{1}{\vert \nabla \psi \vert^2} (\mathbf{B}_0 \cdot \n... ...\times \frac{\nabla \psi}{\vert \nabla \psi \vert^2} \right] \cdot \mathbf{B}_0$$

$$= \mathbf{B}_0 \cdot [(\mathbf{B}_0 \cdot \nabla) \nabla \psi - (\n... ...cdot \nabla)\mathbf{B}_0] \times \frac{\nabla \psi}{\vert \nabla \psi \vert^2},$$ (356)

which is exactly the left-hand side of Eq. (354). Thus Eq. (354) is proved.