proof of $(B \times \nabla \psi ) \cdot \nabla \left ( \frac {1}{B^2} \right ) = - 2 \kappa \cdot (B \times \nabla \psi ) / B^2$

The left-hand side of the equation is written as

$$(\mathbf{B}_0 \times \nabla \psi) \cdot \nabla \frac{1}{B_0^2} = - \frac{2}{B_0^3} (\mathbf{B}_0 \times \nabla \psi) \cdot \nabla B_0$$

$$= - \frac{2}{B_0^3} (\mathbf{B}_0 \times \nabla \psi) \cdot \nabla (\mathbf{B}_0 \cdot \mathbf{b})$$ (361)

Using $\ensuremath{\boldsymbol{\kappa}}=\mathbf{b} \cdot \nabla \mathbf{b}= -\mathbf{b} \times \nabla \times \mathbf{b}$, the above equation is reduced to

$$(\mathbf{B}_0 \times \nabla \psi) \cdot \nabla \frac{1}{B_0^2} = - \frac{2}{B_0^3} (\mathbf{B}_0 \times \nabla \psi) \cdot [\mathb... ..._0 +\mathbf{b} \cdot \nabla \mathbf{B}_0 +\mathbf{B}_0 \cdot \nabla \mathbf{b}]$$

$$= - \frac{2}{B_0^3} (\mathbf{B}_0 \times \nabla \psi) \cdot [\mathbf{b} \times \nabla \times \mathbf{B}_0 +\mathbf{b} \cdot \nabla \mathbf{B}_0]$$

$$= - \frac{2}{B_0^3} (\mathbf{B}_0 \times \nabla \psi) \cdot [\mathbf{b} \times {\textmu}_0 \mathbf{J}_0 +\mathbf{b} \cdot \nabla \mathbf{B}_0]$$

$$= - \frac{2}{B_0^3} (\mathbf{B}_0 \times \nabla \psi) \cdot [\mathbf{b} \cdot \nabla \mathbf{B}_0]$$

$$= - \frac{2}{B_0^3} (\mathbf{B}_0 \times \nabla \psi) \cdot [\mathbf{b} \cdot \nabla B_0 \mathbf{b}]$$

$$= - \frac{2}{B_0^3} (\mathbf{B}_0 \times \nabla \psi) \cdot [(\mathbf{b} \cdot \nabla B_0) \mathbf{b}+ B_0 \mathbf{b} \cdot \nabla \mathbf{b}]$$

$$= - \frac{2}{B_0^3} (\mathbf{B}_0 \times \nabla \psi) \cdot [B_0 \mathbf{b} \cdot \nabla \mathbf{b}]$$

$$= - \frac{2}{B_0^2} (\mathbf{B}_0 \times \nabla \psi) \cdot \ensuremath{\boldsymbol{\kappa}}$$

Another way to prove that $(\mathbf{B} \times \nabla \psi) \cdot B^2 \nabla \frac{1}{B^2}$ is equal to $- 2\ensuremath{\boldsymbol{\kappa}} \cdot (\mathbf{B} \times \nabla \psi)$: The difference of these two terms is

$$- (\mathbf{B} \times \nabla \psi) \cdot B^2 \nabla \frac{1}{B^2} - 2\mathbf{K} \cdot (\mathbf{B} \times \nabla \psi)$$

$$= - (\mathbf{B} \times \nabla \psi) \cdot \left( B^2 \nabla \frac{1}{B^2} + 2\mathbf{K} \right)$$

$$= - (\mathbf{B} \times \nabla \psi) \cdot \left( B^2 \nabla \frac{1}{B^2} + 2 \frac{\mathbf{B}}{B} \cdot \nabla \frac{\mathbf{B}}{B} \right)$$

$$= - (\mathbf{B} \times \nabla \psi) \cdot \left( B^2 \nabla \frac... ...eft( 2 \frac{\mathbf{B}}{B} \cdot \nabla \right) \mathbf{B} \frac{1}{B} \right)$$

$$= - (\mathbf{B} \times \nabla \psi) \cdot \left( B^2 \nabla \frac... ...eft( 2 \frac{\mathbf{B}}{B} \cdot \nabla \right) \mathbf{B} \frac{1}{B} \right)$$

$$= - (\mathbf{B} \times \nabla \psi) \cdot \left( B^2 \nabla B^2 \... ...eft( 2 \frac{\mathbf{B}}{B} \cdot \nabla \right) \mathbf{B} \frac{1}{B} \right)$$

$$= - (\mathbf{B} \times \nabla \psi) \cdot \left( - \frac{1}{B^2} ... ...eft( 2 \frac{\mathbf{B}}{B} \cdot \nabla \right) \mathbf{B} \frac{1}{B} \right)$$

$$= - (\mathbf{B} \times \nabla \psi) \cdot \left( - \frac{1}{B^2} ... ...eft( 2 \frac{\mathbf{B}}{B} \cdot \nabla \right) \mathbf{B} \frac{1}{B} \right)$$

$$= - (\mathbf{B} \times \nabla \psi) \cdot \left[ - \frac{1}{B^2} ... ...f{B}) + \left( 2 \frac{\mathbf{B}}{B^2} \cdot \nabla \right) \mathbf{B} \right]$$

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

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

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

$$= (\mathbf{B} \times \nabla \psi) \cdot \left[ - \frac{1}{B^2} 2 P' \nabla \psi \right]$$

$$= 0$$

Thus we prove that

$$- Q_b (\mathbf{B} \times \nabla \psi) \cdot B^2 \nabla \frac{1}{B^2} = 2\ensuremath{\boldsymbol{\kappa}} \cdot (\mathbf{B} \times \nabla \psi) Q_b$$ (362)