Flux Surface Functions

Next, examine the meaning of the following volume integral

$$D (\psi) \equiv \int_V \mathbf{B} \cdot \nabla \theta d \tau,$$ (218)

where the volume $V = V (\psi)$, which is the volume within the magnetic surface labeled by $\psi$. Using $\nabla \cdot \mathbf{B}= 0$, the quantity $D$ can be further written as

$$D = \int_V \nabla \cdot (\theta \mathbf{B}) d \tau .$$ (219)

Note that $\theta$ is not a single-value function of the spacial points. In order to evaluate the integration in Eq. (219), we need to select one branch of $\theta$, which can be chosen to be $0 \leqslant \theta < 2 \pi$. Note that function $\theta = \theta (R, Z)$ is not continuous in the vicinity of the contour of $\theta = 0$. Next, we want to use the Gauss's theorem to convert the above volume integration to surface integration. Noting the discontinuity of the integrand $\theta \mathbf{B}$ in the vicinity of the contour of $\theta = 0$, the volume should be cut along the contour, thus, generating two surfaces. Denote these two surfaces by $S_1$ and $S_2$, then equation (219) is written as

$$D = \int_{S_1} \theta \mathbf{B} \cdot d\mathbf{S}+ \int_{S_2} \theta \mathbf{B} \cdot d\mathbf{S}+ \int_{S_3} \theta \mathbf{B} \cdot d\mathbf{S},$$

where the direction of surface $S_1$ is in the negative direction of $\theta$, the direction of $S_2$ is in the positive direction of $\theta$, and the surface $S_3$ is the toroidal magnetic surface $\psi = \psi_0$. The surface integration through $S_3$ is obviously zero since $\mathbf {B}$ lies in this surface. Therefore, we have

$$D = \int_{S_1} \theta \mathbf{B} \cdot d\mathbf{S}+ \int_{S_2} \theta \mathbf{B} \cdot d\mathbf{S}+ 0$$

$$= \int_{S_1} 0\mathbf{B} \cdot d\mathbf{S}+ \int_{S_2} 2 \pi \mathbf{B} \cdot d\mathbf{S}$$

$$= 2 \pi \int_{S_2} \mathbf{B} \cdot d\mathbf{S}.$$ (220)

Eq. (220) indicates that $D$ is $2 \pi$ times the magnetic flux through the $S_2$ surface. Thus, the poloidal flux through $S_2$ is written as

$$\Psi_p = \frac{1}{2 \pi} D = \frac{1}{2 \pi} \int_V \mathbf{B} \cdot \nabla \theta d \tau .$$ (221)

Using the expression of the volume element $d \tau = \vert\mathcal{J}\vert d \theta d \phi d \psi$, $\Psi _p$ can be further written in terms of flux surface averaged quantities.

$$\Psi_p = \frac{1}{2 \pi} \int_V \mathbf{B} \cdot \nabla \theta \vert\mathcal{J}\vert d \theta d \phi d \psi$$

$$= \int_0^{\psi} d \psi \int_0^{2 \pi} \mathbf{B} \cdot \nabla \theta \vert\mathcal{J}\vert d \theta$$

$$= \int_0^{\psi} d \psi \int_0^{2 \pi} \Psi' \nabla \psi \times \nabla \phi \cdot \nabla \theta \vert\mathcal{J}\vert d \theta$$

$$= - \ensuremath{\operatorname{sign}} (\mathcal{J}) \int_0^{\psi} d \psi \int_0^{2 \pi} \Psi' (\psi) d \theta$$

$$= - 2 \pi \ensuremath{\operatorname{sign}} (\mathcal{J}) \int_0^{\psi} \Psi' (\psi) d \psi$$

$$= - 2 \pi \ensuremath{\operatorname{sign}} (\mathcal{J}) [\Psi (\psi) - \Psi (0)] .$$ (222)

Note that the sign of the Jacobian appears in Eq. (222), which is due to the positive direction of surface $S_2$ is determined by the positive direction of $\theta$, which in turn is determined by the sign of the Jacobian (In my code, however, the positive direction of $\theta$ is chosen by me and the sign of the Jacobian is determined by the positive direction of $\theta$). We can verify the sign of Eq. (222) is exactly consistent with that in Eq. (19).

Similarly, the toroidal flux within a flux surface is written as

$$\Psi_t = \frac{1}{2 \pi} \int_V \mathbf{B} \cdot \nabla \phi d \tau,$$ (223)

the poloidal current within a flux surface is written as

$$K (\psi) = \frac{1}{2 \pi} \int_V \mathbf{J} \cdot \nabla \theta d \tau,$$ (224)

and toroidal current within a flux surface is written as

$$I (\psi) = \frac{1}{2 \pi} \int_V \mathbf{J} \cdot \nabla \phi d \tau .$$ (225)

(**check**)The toroidal magnetic flux is written as

$$\Psi_t = \frac{1}{2 \pi} \int \mathbf{B} \cdot \nabla \phi \vert\mathcal{J}\vert d \theta d \phi d \psi$$

$$= \int_0^{\psi} d \psi \int_0^{2 \pi} g \frac{1}{R^2} \vert\mathcal{J}\vert d \theta$$

$$= \int_0^{\psi} \left[ g \frac{V'}{2 \pi} \left\langle \frac{1}{R^2} \right\rangle \right] d \psi .$$ (226)

$$\Rightarrow \Psi_t' = g \frac{V'}{2 \pi} \left\langle \frac{1}{R^2} \right\rangle$$

$$\Rightarrow \frac{d \Psi_t}{d V} = g \frac{1}{2 \pi} \left\langle \frac{1}{R^2} \right\rangle$$

$$\Rightarrow \frac{d \Psi}{d V} = \frac{g}{2 \pi q} \frac{1}{2 \pi} \left\langle \frac{1}{R^2} \right\rangle .$$ (227)

Next, calculate the derivative of the toroidal flux with respect to the poloidal flux.

$$\frac{d \Psi_t}{d \Psi_p} = \frac{\Psi_t'}{\Psi_p'}$$

$$= - \frac{g V'}{(2 \pi)^2 \Psi'} \left\langle \frac{1}{R^2} \right\rangle,$$ (228)

Comparing this result with Eq. (471) indicates that it is equal to the safety factor, i.e.,

$$\frac{d \Psi_t}{d \Psi_p} = q (\psi) .$$ (229)

$$- \ensuremath{\operatorname{sgn}} (\mathcal{J}) \frac{V'}{(2 \pi)^2} \frac{g}{\Psi'} \langle R^{- 2} \rangle$$

By using the contravariant representation of current density (331), the poloidal current within a magnetic surface is written as

$$K (\psi) = \frac{1}{2 \pi} \int \mathbf{J} \cdot \nabla \theta \mathcal{J}d \theta d \phi d \psi$$

$$= \frac{1}{\mu_0} \int (- g') \nabla \phi \times \nabla \psi \cdot \nabla \theta \mathcal{J}d \theta d \psi$$

$$= - \frac{1}{\mu_0} \int g' d \theta d \psi$$

$$= - \frac{2 \pi}{\mu_0} \int_0^{\psi} g' d \psi$$

$$= - \frac{2 \pi}{\mu_0} [g (\psi) - g (0)] .$$ (230)

Note that the poloidal current is proportional to $g$, which explains why $g$ is sometimes called poloidal current function in tokamak literature.

$$- \left[ \left( \Psi' \frac{\mathcal{J}}{R^2} \vert \nabla \psi \... ...\right] \nabla \psi \times \nabla \theta - g' \nabla \phi \times \nabla \psi,$$

The toroidal current is written as

$$I_{\phi} (\psi) = \frac{1}{2 \pi} \int \mathbf{J} \cdot \nabla \phi \mathcal{J}d \theta d \phi d \psi$$

$$= - \frac{1}{2 \pi \mu_0} \int \left[ \left( \Psi' \frac{\mathcal{J... ...a \psi \times \nabla \theta \cdot \nabla \phi \mathcal{J}d \theta d \phi d \psi$$

$$= - \frac{1}{\mu_0} \int \left[ \left( \Psi' \frac{\mathcal{J}}{R^2... ...}{R^2} \nabla \psi \cdot \nabla \theta \right)_{\theta} \right] d \theta d \psi$$

$$= - \frac{1}{\mu_0} \int \left[ \left( \Psi' \frac{\mathcal{J}}{R^2} \vert \nabla \psi \vert^2 \right)_{\psi} \right] d \theta d \psi$$

$$= - \frac{1}{\mu_0} \int_0^{2 \pi} d \theta \int_0^{\psi} d \psi \left( \Psi' \frac{\mathcal{J}}{R^2} \vert \nabla \psi \vert^2 \right)_{\psi}$$

$$= - \frac{1}{\mu_0} \int_0^{2 \pi} d \theta \left( \Psi' \frac{\mathcal{J}}{R^2} \vert \nabla \psi \vert^2 - 0 \right) .$$ (231)

The last equality is due to $\nabla \psi = 0$ at $\psi = 0$. By using the flux surface average operator, Eq. (231) is written

$$I_{\phi} (\psi) = - \frac{V' \Psi'}{2 \pi \mu_0} \left\langle \frac{\vert \nabla \psi \vert^2}{R^2} \right\rangle .$$ (232)

Next, calculate another useful surface-averaged quantity,

$$\frac{\langle \mathbf{J} \cdot \mathbf{B} \rangle}{\langle \mathbf{B} \cdot \nabla \phi \rangle} = \frac{\left\langle \frac{g^2}{\mathcal{J}} \left[ \left( \frac{1}... ...J}}{R^2} \right)_{\theta} \right] \right\rangle}{\mu_0 \langle g / R^2 \rangle}$$

$$= \frac{\frac{2 \pi}{V'} \int_0^{2 \pi} d \theta g^2 \left[ \left( ... ...ac{\mathcal{J}}{R^2} \right)_{\theta} \right]}{\mu_0 g \langle R^{- 2} \rangle}$$

$$= \frac{\frac{2 \pi}{V'} g^2 \int_0^{2 \pi} d \theta \left[ \left( ... ...ac{\mathcal{J}}{R^2} \right)_{\theta} \right]}{\mu_0 g \langle R^{- 2} \rangle}$$

$$= \frac{\frac{2 \pi}{V'} g \int_0^{2 \pi} d \theta \left[ \left( \f... ...vert \nabla \psi \vert^2 \right)_{\psi} \right]}{\mu_0 \langle R^{- 2} \rangle}$$

$$= \frac{\frac{2 \pi}{V'} g \int_0^{2 \pi} d \theta \left[ \left( \f... ...vert \nabla \psi \vert^2 \right)_{\psi} \right]}{\mu_0 \langle R^{- 2} \rangle}$$ (233)

The differential with respect to $\psi$ and the integration with respect to $\theta$ can be interchanged, yielding

$$\frac{\langle \mathbf{J} \cdot \mathbf{B} \rangle}{\langle \mathbf{B} \cdot \nabla \phi \rangle} = \frac{\frac{2 \pi}{V'} g \left[ \frac{1}{g} \Psi' \left( \int_0^{... ...vert \nabla \psi \vert^2 \right) \right]_{\psi}}{\mu_0 \langle R^{- 2} \rangle}$$

$$= \frac{\frac{1}{V'} g \left[ \frac{1}{g} \Psi' V' \left\langle \fr... ...\psi \vert^2}{R^2} \right\rangle \right]_{\psi}}{\mu_0 \langle R^{- 2} \rangle}$$

$$= \frac{g}{\mu_0 V' \langle R^{- 2} \rangle} \left[ \frac{\Psi' V'}... ...\left\langle \frac{\vert \nabla \psi \vert^2}{R^2} \right\rangle \right]_{\psi}$$ (234)