Large aspect ratio expansion

Consider the case that the boundary flux surface is circular with radius $r = a$ and the center of the cirle at $(R = R_0, Z = 0)$. Consider the case $\varepsilon = a / R_0 \rightarrow 0$. Expanding $\Psi$ in the small parameter $\varepsilon$,

$$\Psi = \Psi_0 + \Psi_1$$ (383)

where $\Psi_0 \sim O (\varepsilon^0)$, $\Psi_1 \sim O (\varepsilon^1)$. Substituting Eq. (383) into Eq. (382), we obtain

$$\frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial \Psi_0}{... ...cos \theta)^2 P' (\Psi_0 + \Psi_1) - g' (\Psi_0 + \Psi_1) g (\Psi_0 + \Psi_1)$$

Multiplying the above equation by $R_0^2$, we obtain

$$R_0^2 \frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial \P... ...\theta)^2 P' (\Psi_0 + \Psi_1) - R_0^2 g' (\Psi_0 + \Psi_1) g (\Psi_0 + \Psi_1)$$ (384)

Further assume the following ordering

$$R_0 \frac{\partial \Psi}{\partial r} \sim O (\varepsilon^{- 1}) \Psi,$$ (385)

and

$$\frac{\partial \Psi}{\partial \theta} \sim O (\varepsilon^0) \Psi .$$ (386)

Using these orderings, the order of the terms in Eq. (384) can be estimated as

$$R_0^2 \frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial \Psi_0}{\partial r} \sim \frac{\Psi_0}{\varepsilon^2} \sim O (\varepsilon^{- 2})$$ (387)

$$R_0^2 \frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial \Psi_1}{\partial r} \sim \frac{\Psi_1}{\varepsilon^2} \sim O (\varepsilon^{- 1})$$ (388)

$$R_0^2 \frac{1}{r^2} \frac{\partial^2 \Psi_0}{\partial \theta^2} \sim \frac{1}{\varepsilon^2} \Psi_0 \sim O (\varepsilon^{- 2})$$ (389)

$$R_0^2 \frac{1}{r^2} \frac{\partial^2 \Psi_1}{\partial \theta^2} \sim \frac{\Psi_1}{\varepsilon^2} \sim O (\varepsilon^{- 1})$$ (390)

$$\frac{R_0}{R_0 + r \cos \theta} \approx 1 - \frac{r}{R_0} \cos \theta$$ (391)

$$R_0 \frac{\partial \Psi_0}{\partial r} \cos \theta \sim \frac{\Psi_0}{\varepsilon} \sim O (\varepsilon^{- 1})$$ (392)

$$R_0 \frac{\partial \Psi_0}{\partial \theta} \frac{1}{r} \sin \theta \sim \frac{\Psi_0}{\varepsilon} \sim O (\varepsilon^{- 1})$$ (393)

$$R_0 \frac{\partial \Psi_1}{\partial r} \cos \theta \sim \frac{\Psi_1}{\varepsilon} = O (\varepsilon^0)$$ (394)

$$R_0 \frac{\partial \Psi_1}{\partial \theta} \frac{1}{r} \sin \theta \sim \frac{\Psi_1}{\varepsilon} = O (\varepsilon^0)$$ (395)

The leading order ( $\varepsilon^{- 2}$ order) balance is given by the following equation:

$$R_0^2 \frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial \P... ...{\partial \theta^2} = - \mu_0 R_0^4 P' (\Psi_0) - R_0^2 g' (\Psi_0) g (\Psi_0),$$ (396)

It is reasonable to assume that $\Psi_0$ is independent of $\theta$ since $\Psi_0$ corresponds to the limit $a / R \rightarrow 0$. (The limit $a / R \rightarrow 0$ can have two cases, one is $r \rightarrow 0$, another is $R \rightarrow \infty$. In the former case, $\Psi$ must be independent of $\theta$ since $\Psi$ should be single-valued. The latter case corresponds to a cylinder, for which it is reasonable (really?) to assume that $\Psi_0$ is independent of $\theta$.) Then Eq. (396) is written

$$\frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial \Psi_0}{\partial r} = - \mu_0 R_0^2 P' (\Psi_0) - g' (\Psi_0) g (\Psi_0) .$$ (397)

(My remarks: The leading order equation (397) does not correponds strictly to a cylinder equilibrium because the magnetic field $\mathbf{B}= \nabla \Psi_0 \times \nabla \phi + g \nabla \phi$ depends on $\theta$.) The next order ( $\varepsilon^{- 1}$ order) equation is

$$R_0^2 \frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial \... ...0) - \mu_0 R_0^4 P'' (\Psi_0) \Psi_1 - R_0^2 [g' (\Psi_0) g (\Psi_0)]' \Psi_1$$

$$R_0^2 \frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial \P... ... r \cos \theta P' (\Psi_0) + R_0 \frac{\partial \Psi_0}{\partial r} \cos \theta$$ (398)

$$R_0^2 \frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial \P... ... r \cos \theta P' (\Psi_0) + R_0 \frac{\partial \Psi_0}{\partial r} \cos \theta$$ (399)

It is obvious that the simple poloidal dependence of $\cos \theta$ will satisfy the above equation. Therefore, we consider $\Psi _1$ of the form

$$\Psi_1 = \Delta (r) \frac{d \Psi_0 (r)}{d r} \cos \theta,$$ (400)

where $\Delta (r)$ is a new function to be determined. Substitute this into the Eq. (), we obtain an equation for $\Delta (r)$,

$$R_0^2 \frac{1}{r} \frac{d}{d r} \left( r \frac{d \Psi_0}{d r} \fr... ...\Psi_0 (r)}{d r} = - \mu_0 R_0^2 2 R_0 r P' (\Psi_0) + R_0 \frac{d \Psi_0}{d r}$$ (401)

$$R_0^2 \frac{1}{r} \frac{d}{d r} \left( r \frac{d \Psi_0}{d r} \fr... ...Psi_0) \} \Delta = - \mu_0 R_0^2 2 R_0 r P' (\Psi_0) + R_0 \frac{d \Psi_0}{d r}$$ (402)

$$R_0^2 \frac{1}{r} \frac{d}{d r} \left( r \frac{d \Psi_0}{d r} \fr... ...Psi_0) \} \Delta = - \mu_0 R_0^2 2 R_0 r P' (\Psi_0) + R_0 \frac{d \Psi_0}{d r}$$ (403)

$$R_0^2 \frac{1}{r} \frac{d}{d r} \left( r \frac{d \Psi_0}{d r} \fr... ...Psi_0) \} \Delta = - \mu_0 R_0^2 2 R_0 r P' (\Psi_0) + R_0 \frac{d \Psi_0}{d r}$$ (404)

Using the identity

$$\frac{d}{d r} \left[ \frac{1}{r} \frac{d}{d r} \left( r \frac{d \... ... r \frac{d^2 \Psi_0}{d r^2} \right) - \frac{1}{r^2} \frac{d \Psi_0 (r)}{d r},$$

equation () is written as

$$R_0^2 \frac{1}{r} \frac{d}{d r} \left( r \frac{d \Psi_0}{d r} \fr... ...Psi_0) \} \Delta = - \mu_0 R_0^2 2 R_0 r P' (\Psi_0) + R_0 \frac{d \Psi_0}{d r}$$ (405)

Using the leading order equation (), we know that the second and fourth term on the l.h.s of the above equation cancel each other, giving

$$R_0^2 \frac{1}{r} \frac{d}{d r} \left( r \frac{d \Psi_0}{d r} \fr... ...c{d \Delta}{d r} = - \mu_0 R_0^2 2 R_0 r P' (\Psi_0) + R_0 \frac{d \Psi_0}{d r}$$ (406)

$$\Rightarrow \frac{1}{r} \frac{d}{d r} \left( r \frac{d \Psi_0}{d ... ...\Delta}{d r} = - \mu_0 2 R_0 r P' (\Psi_0) + \frac{1}{R_0} \frac{d \Psi_0}{d r}$$ (407)

Using the identity

$$\frac{1}{r} \frac{d r}{d \Psi_0} \frac{d}{d r} \left[ r \left( \f... ...} \frac{d \Delta}{d r} \right) + \frac{d^2 \Psi_0}{d r^2} \frac{d \Delta}{d r},$$ (408)

equation (407) is written

$$\frac{1}{r} \frac{d r}{d \Psi_0} \frac{d}{d r} \left[ r \left( \f... ... r} \right] = - \mu_0 2 R_0 r P' (\Psi_0) + \frac{1}{R_0} \frac{d \Psi_0}{d r},$$ (409)

$$\Rightarrow \frac{1}{r} \frac{d}{d r} \left[ r \left( \frac{d \Ps... ... 2 R_0 r \frac{d P}{d r} + \frac{1}{R_0} \left( \frac{d \Psi_0}{d r} \right)^2,$$ (410)

Using

$$B_{\theta 0} = \frac{1}{R_0} \frac{d \Psi_0}{d r}$$ (411)

equation (410) is written

$$\frac{1}{r} \frac{d}{d r} \left[ r B_{\theta 0}^2 \frac{d \Delta}... ...ght] = - \mu_0 2 \frac{1}{R_0} r \frac{d P}{d r} + \frac{B_{\theta 0}^2}{R_0} .$$ (412)

$$\Rightarrow \frac{d}{d r} \left[ r B_{\theta 0}^2 \frac{d \Delta}... ...c{r}{R_0} \left( - 2 \mu_0 r \frac{d P (\Psi_0)}{d r} + B_{\theta 0}^2 \right),$$ (413)

which agrees with equation (3.6.7) in Wessson's book[12].