Metric elements

Let $\psi = r$ and define $h^{\psi R} = \nabla \psi \cdot \hat{\mathbf{R}}$, $h^{\alpha R} = \nabla \alpha \cdot \hat{\mathbf{R}}$, etc. Explicit expressions for these elements can be written as

$$h^{\psi R} = \nabla \psi \cdot \hat{\mathbf{R}}$$

$$= \nabla r \cdot \hat{\mathbf{R}}$$

$$= - \frac{R}{\mathcal{J}} (Z_{\theta} \hat{\mathbf{R}} - R_{\theta} \hat{\mathbf{Z}}) \cdot \hat{\mathbf{R}}$$

$$= - \frac{R}{\mathcal{J}} Z_{\theta}$$

$$= \cos \theta$$ (361)

$$h^{\psi Z} = \nabla \psi \cdot \hat{\mathbf{Z}}$$

$$= \nabla r \cdot \hat{\mathbf{Z}}$$

$$= - \frac{R}{\mathcal{J}} (Z_{\theta} \hat{\mathbf{R}} - R_{\theta} \hat{\mathbf{Z}}) \cdot \hat{\mathbf{Z}}$$

$$= \frac{R}{\mathcal{J}} R_{\theta}$$

$$= \sin \theta$$

$$h^{\psi \phi} = \nabla \psi \cdot \hat{\ensuremath{\boldsymbol{\phi}}}$$

$$=$$ 0

$$h^{\theta R} = \nabla \theta \cdot \hat{\mathbf{R}}$$

$$= \frac{R}{\mathcal{J}} (Z_{\psi} \hat{\mathbf{R}} - R_{\psi} \hat{\mathbf{Z}}) \cdot \hat{\mathbf{R}}$$

$$= \frac{R}{\mathcal{J}} Z_{\psi}$$

$$= - \frac{1}{r} \sin \theta$$

$$h^{\theta Z} = \nabla \theta \cdot \hat{\mathbf{Z}}$$

$$= \frac{R}{\mathcal{J}} (Z_{\psi} \hat{\mathbf{R}} - R_{\psi} \hat{\mathbf{Z}}) \cdot \hat{\mathbf{Z}}$$

$$= - \frac{R}{\mathcal{J}} R_{\psi}$$

$$= \frac{1}{r} \cos \theta$$

$$h^{\theta \phi} = \nabla \theta \cdot \hat{\ensuremath{\boldsymbol{\phi}}}$$

$$=$$ 0

$$h^{\alpha R} = \nabla \alpha \cdot \hat{\mathbf{R}}$$

$$= \left[ \frac{\hat{\ensuremath{\boldsymbol{\phi}}}}{R} + \left( \f... ...\mathcal{J}} R_{\theta} \right) \hat{\mathbf{Z}} \right] \cdot \hat{\mathbf{R}}$$

$$= \frac{\partial \overline{\delta}}{\partial \psi} \frac{R}{\mathca... ...rac{\partial \overline{\delta}}{\partial \theta} \frac{R}{\mathcal{J}} Z_{\psi}$$

$$= \frac{\partial \overline{\delta}}{\partial \psi} \frac{R}{\mathca... ...{\partial \overline{\delta}}{\partial \theta} \frac{R}{\mathcal{J}} \sin \theta$$

$$= - \frac{\partial \overline{\delta}}{\partial \psi} \cos \theta + \frac{\partial \overline{\delta}}{\partial \theta} \frac{1}{r} \sin \theta$$

$$= - \frac{\partial \overline{\delta}}{\partial r} \cos \theta + \hat{q} \frac{1}{r} \sin \theta$$

$$h^{\alpha Z} = \nabla \alpha \cdot \hat{\mathbf{Z}}$$

$$= \left[ \frac{\hat{\ensuremath{\boldsymbol{\phi}}}}{R} + \left( \f... ...\mathcal{J}} R_{\theta} \right) \hat{\mathbf{Z}} \right] \cdot \hat{\mathbf{Z}}$$

$$= \frac{\partial \overline{\delta}}{\partial \theta} \frac{R}{\math... ...rac{\partial \overline{\delta}}{\partial \psi} \frac{R}{\mathcal{J}} R_{\theta}$$

$$= - \hat{q} \frac{1}{r} \cos \theta - \frac{\partial \overline{\delta}}{\partial r} \sin \theta$$

$$h^{\alpha \phi} = \frac{1}{R} .$$ (362)

Using expression (359), $d \overline{\delta} / d r$ can be evaluated analytically, yielding

$$\frac{d \overline{\delta}}{d r} = 2 \frac{d q}{d r} \arctan \left( \frac{(R_0 - r)}{\sqrt{R_0^2 - r... ...a}{2} \right) \frac{d}{d r} \left[ \frac{(R_0 - r)}{\sqrt{R_0^2 - r^2}} \right]$$

$$= 2 \frac{d q}{d r} \arctan \left( \frac{(R_0 - r)}{\sqrt{R_0^2 - r... ...\tan \left( \frac{\theta}{2} \right) \frac{- R_0}{(R_0 + r) \sqrt{R_0^2 - r^2}}$$

where use has been made of

$$\frac{d}{d x} \arctan (x) = \frac{1}{1 + x^2}$$

(I did not remeber this formula and I use SymPy to get this.) These expressions are used to benchmark the numerical code that assume general flux surface shapes. The results show that the code gives correct result when concentric circular flux surfaces are used.