proof

Try to prove that

$$\mathbf{B}_0 \cdot \left( \mathbf{B}_0 \cdot \nabla \frac{\nabla ... ... \Psi \frac{B^2}{\vert \nabla \Psi \vert^2} +{\textmu}_0 \frac{d p_0}{d \Psi} .$$ (363)

Proof: We have

$$\nabla p_0 = \frac{d p_0}{d \Psi} \nabla \Psi$$ (364)

Thus

$${\textmu}_0 \frac{d p_0}{d \Psi} = {\textmu}_0 \frac{\nabla \Psi \cdot \nabla p_0}{\vert \nabla \Psi \vert^2}$$

$$= {\textmu}_0 \frac{\nabla \Psi \cdot (\mathbf{J}_0 \times \mathbf{B}_0)}{\vert \nabla \Psi \vert^2}$$

$$= \frac{\nabla \Psi \cdot ((\nabla \times \mathbf{B}_0) \times \mathbf{B}_0))}{\vert \nabla \Psi \vert^2}$$

$$= - (\nabla \times \mathbf{B}_0) \cdot \left( \frac{\nabla \Psi}{\vert \nabla \Psi \vert^2} \times \mathbf{B}_0 \right)$$

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

$$= - \nabla \cdot \left( B_0^2 \frac{\nabla \Psi}{\vert \nabla \Psi ... ...left( \frac{\nabla \Psi}{\vert \nabla \Psi \vert^2} \times \mathbf{B}_0 \right)$$

$$= - \nabla \cdot \left( B_0^2 \frac{\nabla \Psi}{\vert \nabla \Psi ... ...\frac{\nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot \nabla \mathbf{B}_0 \right)$$

$$= - \nabla \cdot \left( B_0^2 \frac{\nabla \Psi}{\vert \nabla \Psi ... ...\frac{\nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot \nabla \mathbf{B}_0 \right)$$

$$= - \frac{\nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot \nabla B_0^... ...\frac{\nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot \nabla \mathbf{B}_0 \right)$$

$$= - \frac{\nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot \nabla \mat... ...\frac{\nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot \nabla \mathbf{B}_0 \right)$$

$$= - \frac{\nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot (2\mathbf{B... ...\frac{\nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot \nabla \mathbf{B}_0 \right)$$

$$= - \frac{\nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot (2\mathbf{B... ...si}{\vert \nabla \Psi \vert^2} \cdot (\mathbf{B}_0 \cdot \nabla B_0 \mathbf{b})$$ (365)

The last term on the right-hand side of Eq. (365) is written as

$$- \frac{2 \nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot (\mathbf{B}_0 \cdot \nabla B_0 \mathbf{b}) = - \frac{2 \nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot (B_0 \mathbf{B}_0 \cdot \nabla \mathbf{b}+ (\mathbf{B}_0 \cdot \nabla B)\mathbf{b})$$

$$= - \frac{2 \nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot (B_0 \mathbf{B}_0 \cdot \nabla \mathbf{b})$$

$$= - \frac{2 B^2 \nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot \ensuremath{\boldsymbol{\kappa}}.$$

Using this, Eq. (365) is written as

$${\textmu}_0 \frac{d p_0}{d \Psi} = \frac{2 \nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot ((\nabla \t... ... \nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot \ensuremath{\boldsymbol{\kappa}}$$

$$= 2{\textmu}_0 \frac{d p_0}{d \Psi} -\mathbf{B}_0 \cdot \left( \mat... ...\nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot \ensuremath{\boldsymbol{\kappa}},$$ (366)

i.e.,

$$\mathbf{B}_0 \cdot \left( \mathbf{B}_0 \cdot \nabla \frac{\nabla ... ...\frac{\nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot \nabla \mathbf{B}_0 \right) = {\textmu}_0 \frac{d p_0}{d \Psi} - \frac{2 B^2 \nabla \Psi}{\vert \nabla \Psi \vert^2} \cdot \ensuremath{\boldsymbol{\kappa}},$$ (367)

thus Eq. (363) is proved.