Laplacian operator in general coordinates $(\psi , \theta , \zeta )$

Using the transformation between the two kinds of basis vectors given in Eq. (152), the gradient of $f$ can be written as

$$\nabla f = \frac{\partial f}{\partial \psi} \nabla \psi + \frac{\partial f}{\partial \theta} \nabla \theta + \frac{\partial f}{\partial \zeta} \nabla \zeta$$

$$= \frac{\partial f}{\partial \psi} (g_{11} \nabla \theta \times \na... ...s \nabla \psi \mathcal{J}+ g_{13} \nabla \psi \times \nabla \theta \mathcal{J})$$

$$+ \frac{\partial f}{\partial \theta} (g_{21} \nabla \theta \times \... ...s \nabla \psi \mathcal{J}+ g_{23} \nabla \psi \times \nabla \theta \mathcal{J})$$

$$+ \frac{\partial f}{\partial \zeta} (g_{31} \nabla \theta \times \n... ...s \nabla \psi \mathcal{J}+ g_{33} \nabla \psi \times \nabla \theta \mathcal{J})$$

$$= \left( \frac{\partial f}{\partial \psi} g_{11} + \frac{\partial f... ...f}{\partial \zeta} g_{31} \right) \nabla \theta \times \nabla \zeta \mathcal{J}$$

$$+ \left( \frac{\partial f}{\partial \psi} g_{12} + \frac{\partial f... ...l f}{\partial \zeta} g_{32} \right) \nabla \zeta \times \nabla \psi \mathcal{J}$$

$$+ \left( \frac{\partial f}{\partial \psi} g_{13} + \frac{\partial f... ... f}{\partial \zeta} g_{33} \right) \nabla \psi \times \nabla \theta \mathcal{J}$$

Using the above expression of $\nabla f$ and the formula for the divergence operator, the Laplacian operator is written as

$$\nabla^2 f = \nabla \cdot (\nabla f)$$

$$= \frac{1}{\mathcal{J}} \frac{\partial \left( \frac{\partial f}{\pa... ...+ \frac{\partial f}{\partial \zeta} g_{33} \right) \mathcal{J}}{\partial \zeta}$$

$$= \frac{1}{\mathcal{J}} \frac{\partial \mathcal{J}}{\partial \psi} ... ...heta} g_{21} + \frac{\partial f}{\partial \zeta} g_{31} \right)}{\partial \psi}$$

$$+ \frac{1}{\mathcal{J}} \frac{\partial \mathcal{J}}{\partial \theta... ...ta} g_{22} + \frac{\partial f}{\partial \zeta} g_{32} \right)}{\partial \theta}$$

$$+ \frac{1}{\mathcal{J}} \frac{\partial \mathcal{J}}{\partial \zeta}... ...eta} g_{23} + \frac{\partial f}{\partial \zeta} g_{33} \right)}{\partial \zeta}$$

(wrong! The derivative of the equilibrium quantities are of order $\delta^1$ and thus every term involves these derivatives are of higher order and can be dropped. Then the above expression reduces to

$$\nabla^2 f \approx g_{11} \frac{\partial^2 f}{\partial^2 \psi} + g_{21} \frac{\parti... ...theta \partial \psi} + g_{31} \frac{\partial^2 f}{\partial \zeta \partial \psi}$$

$$+ g_{12} \frac{\partial^2 f}{\partial \psi \partial \theta} + g_{22... ...\partial^2 \theta} + g_{32} \frac{\partial^2 f}{\partial \zeta \partial \theta}$$

$$+ g_{13} \frac{\partial^2 f}{\partial \psi \partial \zeta} + g_{23}... ...{\partial \theta \partial \zeta} + g_{33} \frac{\partial^2 f}{\partial \zeta^2}$$ (137)

wrong!)

In field aligned coordinates, and $k_{\parallel} \ll k_{\perp}$,

$$\nabla^2 f \approx \frac{1}{\mathcal{J}} \frac{\partial \mathcal{J}}{\partial \psi} ... ..._{31} + \frac{\partial f}{\partial \zeta} \frac{\partial g_{31}}{\partial \psi}$$ (138)

$$+ \frac{1}{\mathcal{J}} \frac{\partial \mathcal{J}}{\partial \theta... ...ial f}{\partial \psi} g_{12} + \frac{\partial f}{\partial \zeta} g_{32} \right)$$

$$+ \frac{\partial^2 f}{\partial \psi \partial \zeta} g_{13} + \frac{\partial^2 f}{\partial \zeta^2} g_{33}$$

$$=$$