Expression of metric elements of magnetic coordinate system

To calculate the expression in Eq. (423), we need to calculate the metric elements of the coordinate system $(\psi , \theta , \phi )$. Note that, in this case, the coordinate system is $(\psi, \theta)$ while $R$ and $Z$ are functions of $\psi$ and $\theta$, i.e.,

$$R = R (\psi, \theta),$$ (305)

$$Z = R (\psi, \theta) .$$ (306)

Next, we express the elements of the metric matrix in terms of $R$, $Z$ and their partial derivatives with respect to $\psi$ and $\theta$. We note that

$$\nabla R = \hat{\mathbf{R}} = R_{\psi} \nabla \psi + R_{\theta} \nabla \theta$$ (307)

$$\nabla Z = \hat{\mathbf{Z}} = Z_{\psi} \nabla \psi + Z_{\theta} \nabla \theta$$ (308)

From Eqs. (307) and (308), we obtain

$$\nabla \psi = \frac{1}{R_{\psi} Z_{\theta} - Z_{\psi} R_{\theta}} (Z_{\theta} \hat{\mathbf{R}} - R_{\theta} \hat{\mathbf{Z}}),$$ (309)

$$\nabla \theta = \frac{1}{Z_{\psi} R_{\theta} - R_{\psi} Z_{\theta}} (Z_{\psi} \hat{\mathbf{R}} - R_{\psi} \hat{\mathbf{Z}}) .$$ (310)

We note that

$$\nabla R \times \nabla \phi \cdot \nabla Z = \frac{1}{R},$$ (311)

i.e.,

$$(R_{\psi} \nabla \psi + R_{\theta} \nabla \theta) \times \nabla \phi \cdot (Z_{\psi} \nabla \psi + Z_{\theta} \nabla \theta) = \frac{1}{R}$$ (312)

$$\Rightarrow (R_{\psi} \nabla \psi \times \nabla \phi + R_{\theta}... ...bla \phi) \cdot (Z_{\psi} \nabla \psi + Z_{\theta} \nabla \theta) = \frac{1}{R}$$ (313)

$$\Rightarrow R_{\theta} Z_{\psi} \mathcal{J}^{- 1} - R_{\psi} Z_{\theta} \mathcal{J}^{- 1} = \frac{1}{R}$$

$$\Longrightarrow \mathcal{J}= R (R_{\theta} Z_{\psi} - R_{\psi} Z_{\theta})$$ (314)

Using this, Eqs. (307) and (308) are written as

$$\nabla \psi = - \frac{R}{\mathcal{J}} (Z_{\theta} \hat{\mathbf{R}} - R_{\theta} \hat{\mathbf{Z}})$$ (315)

and

$$\nabla \theta = \frac{R}{\mathcal{J}} (Z_{\psi} \hat{\mathbf{R}} - R_{\psi} \hat{\mathbf{Z}}) .$$ (316)

Using Eqs. (315) and (316), the elements of the metric matrix are written as

$$\vert \nabla \psi \vert^2 = \frac{R^2}{\mathcal{J}^2} (Z_{\theta}^2 + R_{\theta}^2),$$ (317)

$$\vert \nabla \theta \vert^2 = \frac{R^2}{\mathcal{J}^2} (Z_{\psi}^2 + R_{\psi}^2),$$ (318)

and

$$\nabla \psi \cdot \nabla \theta = - \frac{R^2}{\mathcal{J}^2} (Z_{\theta} Z_{\psi} + R_{\theta} R_{\psi}) .$$ (319)

Eqs. (317), (318), and (319) can be used to express the elements of the metric matrix in terms of $R$, $R_{\psi}$, $R_{\theta}$, $Z_{\psi}$, and $Z_{\theta}$. [Combining the above results, we obtain

$$\frac{\nabla \psi \cdot \nabla \theta}{\vert \nabla \psi \vert^2}... ...\frac{Z_{\theta} Z_{\psi} + R_{\theta} R_{\psi}}{Z_{\theta}^2 + R_{\theta}^2} .$$ (320)

Equation (319) is used in GTAW code.]

Next, consider the gradient of the generalized toroidal angle $\alpha$, which is defined by Eq. (282), i.e., $\alpha = \phi - \overline {\delta }$, where $\overline{\delta} = \int_0^{\theta} \mathbf{B} \cdot \nabla \phi / \left( \mathbf{B} \cdot \nabla \theta \right) d \theta$. The gradient of $\alpha$ is written as

$$\nabla \alpha = \nabla \phi - \nabla \overline{\delta}$$

$$= \frac{\hat{\ensuremath{\boldsymbol{\phi}}}}{R} - \frac{\partial \... ... \nabla \psi - \frac{\partial \overline{\delta}}{\partial \theta} \nabla \theta$$

$$= \frac{\hat{\ensuremath{\boldsymbol{\phi}}}}{R} + \frac{\partial \... ...} \frac{R}{\mathcal{J}} (Z_{\psi} \hat{\mathbf{R}} - R_{\psi} \hat{\mathbf{Z}})$$

$$= \frac{\hat{\ensuremath{\boldsymbol{\phi}}}}{R} + \left( \frac{\pa... ...ta}}{\partial \psi} \frac{R}{\mathcal{J}} R_{\theta} \right) \hat{\mathbf{Z}} .$$ (321)

Then

$$\nabla \psi \cdot \nabla \alpha = \left( - \frac{R}{\mathcal{J}} Z_{\theta} \hat{\mathbf{R}} + \fra... ...partial \psi} \frac{R}{\mathcal{J}} R_{\theta} \right) \hat{\mathbf{Z}} \right]$$

$$= - \frac{R}{\mathcal{J}} Z_{\theta} \left( \frac{\partial \overlin... ...al \overline{\delta}}{\partial \psi} \frac{R}{\mathcal{J}} R_{\theta} \right) .$$ (322)

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

$$= \left( \frac{\partial \overline{\delta}}{\partial \psi} \frac{R}{... ...\psi} \frac{R}{\mathcal{J}} R_{\theta} \right) \frac{R}{\mathcal{J}} R_{\psi} .$$ (323)

Using the above results, $h^{\alpha \beta}$ are written as

$$h^{\psi \psi} = \frac{\mathcal{J}}{R^2} \vert \nabla \psi \vert^2 = \frac{1}{\mathcal{J}} (Z_{\theta}^2 + R_{\theta}^2)$$ (324)

$$h^{\theta \theta} = \frac{\mathcal{J}}{R^2} \vert \nabla \theta \vert^2 = \frac{1}{\mathcal{J}} (Z_{\psi}^2 + R_{\psi}^2),$$ (325)

$$h^{\psi \theta} = \frac{\mathcal{J}}{R^2} \nabla \psi \cdot \nabla \theta = - \frac{1}{\mathcal{J}} (Z_{\theta} Z_{\psi} + R_{\theta} R_{\psi})$$ (326)

[As a side product of the above results, we can calculate the arc length in the poloidal plane along a constant $\psi$ surface, $d \ell$, which is expressed as

$$d \ell = \sqrt{(d R)^2 + (d Z)^2}$$

$$= \sqrt{(R_{\psi} d \psi + R_{\theta} d \theta)^2 + (Z_{\psi} d \psi + Z_{\theta} d \theta)^2} .$$

Note that $d \psi = 0$ since we are considering the arc length along a constant $\psi$ surface in $(R, Z)$ plane. Then the above equation is reduced to

$$d \ell = \sqrt{(R_{\theta} d \theta)^2 + (Z_{\theta} d \theta)^2}$$

$$= \sqrt{R_{\theta}^2 + Z_{\theta}^2} d \theta$$

$$= \frac{\vert\mathcal{J} \nabla \psi \vert}{R} d \theta,$$ (327)

which agrees with Eq. (188).]