在笛状近似（Flute reduction）的情况下，$k_\parallel \sim \epsilon k_\perp$。其中$\epsilon$是小量，此时可忽略平行方向的磁场扰动。
此时除了平衡量，扰动量之外，扰动量也分为一阶和二阶量。
分析时一阶量不可忽略，二阶量可忽略。
比如$\bm{u} \simeq \bm{u}_\perp = \bm{b} \cp \qty(\bm{u} \cp \bm{b})$
只关注垂直磁场方向,为一阶扰动量；
而$\bm{b} \cdot \bm{u}$为二阶量，可忽略。
涡量方程
将$\bm{b} \cdot \curl$作用在动量方程上，则有：
$$
\bm{b} \cdot \curl \qty(\rho \dv{\bm{u}}{t} + \div \bm\Pi) =
– \bm{b} \cdot \curl \qty(\grad P)
+ \bm{b} \cdot \curl \qty(\bm{J} \cp \bm{B})
$$
令
$$
\bm{f} = \rho \dv{\bm{u}}{t} + \div \bm \Pi
$$
且右侧第一项为零，则有
\begin{equation}
\begin{aligned}
\bm{b} \cdot \curl \bm{f} &=
\bm{b} \cdot \curl \rho \dv{\bm{u}}{t}
= \bm{b} \cdot \curl \qty(\bm{J} \cp \bm{B}) \\
&= \frac{1}{B} \bm{B} \cdot \curl \qty(\bm{J} \cp \bm{B}) \\
&= \frac{1}{B} \div \qty[\qty(\bm{J} \cp \bm{B}) \cp \bm{B}]
+ \qty(\bm{J} \cp \bm{B}) \cdot \curl \bm{B} \\
&= \frac{1}{B} \div \qty[\qty(\bm{J} \cp \bm{B}) \cp \bm{B}]
+ \qty(\bm{J} \cp \bm{B}) \cdot \mu_0 \bm{J} \\
&= \frac{1}{B} \div \qty[B^2\qty(\bm{J} \cp \bm{b}) \cp \bm{b}] \\
&= \frac{1}{B} \div \qty{B^2\qty[\qty(\bm{J} \cdot \bm{b})\bm{b} – \bm{J}]} \\
&= \frac{1}{B} \div \qty[B^2\qty(\frac{J_\parallel}{B}\bm{B} – \bm{J})] \\
\end{aligned}
\end{equation}
$\div$分别对每一项的作用：
$$
\begin{aligned}
\bm{b} \cdot \curl \bm{f}
=& \frac{J_\parallel}{B} \bm{b} \cdot \grad B^2
+ B^2 \bm{b} \cdot \grad \frac{J_\parallel}{B}
+ J_\parallel \div \bm{B} \\
& – \frac{\bm{J}}{B} \cdot \grad B^2
– B \div \bm{J}
\end{aligned}
$$
磁场无源，所以右侧第三项为零。
同时$\mu_0\bm{J} = \curl \bm{B}$，可知右侧第五项也为零。剩下的合并同类项有
\begin{equation}
\label{eq:vorticity_mid}
\begin{aligned}
\bm{b} \cdot \curl \bm{f}
&=
B^2 \bm{b} \cdot \grad \frac{J_\parallel}{B}
+ \frac{J_\parallel}{B} \bm{b} \cdot \grad B^2
– \frac{\bm{J}}{B} \cdot \grad B^2 \\
&=
B^2 \bm{b} \cdot \grad \frac{J_\parallel}{B}
– \frac{\bm{J}_\perp}{B} \cdot \grad B^2 \\
&=
B^2 \bm{b} \cdot \grad \frac{J_\parallel}{B}
– \frac{\bm{J}_\perp}{B} \cdot \grad_\perp B^2 \\
&=
B^2 \bm{b} \cdot \grad \frac{J_\parallel}{B}
– \frac{1}{B^2} \bm{b} \cp \qty(\bm{J} \cp \bm{B}) \cdot \grad_\perp B^2 \\
&=
B^2 \bm{b} \cdot \grad \frac{J_\parallel}{B}
– \frac{2}{B} \bm{b} \cp \qty(\bm{J} \cp \bm{B}) \cdot \grad_\perp B
\end{aligned}
\end{equation}
此时需要把$\grad P$重新引入回来，观察$\bm{J} \cp \bm{B}$项有下面两个式子：
\begin{equation*}
\begin{aligned}
\bm{J} \cp \bm{B}
&= \frac{1}{\mu_0} \qty(\curl \bm{B}) \cp \bm{B} \\
&= \frac{B^2}{\mu_0} \qty(\curl \bm{b}) \cp \bm{b}
+ \frac{B}{\mu_0}\qty(\grad B \cp \bm{b}) \cp \bm {b} \\
&= \frac{B^2}{\mu_0} \bm{b} \cdot \grad \bm{b}
+ \frac{B}{\mu_0}\qty(\grad B \cp \bm{b}) \cp \bm {b} \\
&= \frac{B^2}{\mu_0} \bm{\kappa}
– \frac{B}{\mu_0} \grad_\perp B
\end{aligned}
\end{equation*}
同时有：
\begin{equation*}
\begin{aligned}
\bm{J} \cp \bm{B}
&= \grad P + \bm{f}
\end{aligned}
\end{equation*}
将$\bm{J} \cp \bm{B}$的表达式代入\eqref{eq:vorticity_mid}有：
\begin{equation*}
\begin{aligned}
\bm{b} \cdot \curl \bm{f}
&=
B^2 \bm{b} \cdot \grad \frac{J_\parallel}{B}
– \frac{2}{B} \bm{b} \cp
\qty(\frac{B^2}{\mu_0} \bm{\kappa} – \frac{B}{\mu_0} \grad_\perp B)
\cdot \grad_\perp B \\
&=
B^2 \bm{b} \cdot \grad \frac{J_\parallel}{B}
– \frac{2}{B} \bm{b} \cp
\qty(\frac{B^2}{\mu_0} \bm{\kappa})
\cdot \frac{\mu_0}{B}
\qty[\frac{B^2}{\mu_0} \bm{\kappa}
– \qty(\grad P + \bm{f})] \\
&=
B^2 \bm{b} \cdot \grad \frac{J_\parallel}{B}
+ \frac{2 \mu_0}{B^2} \bm{b} \cp
\qty(\frac{B^2}{\mu_0} \bm{\kappa})
\cdot \qty(\grad P + \bm{f}) \\
&=
B^2 \bm{b} \cdot \grad \frac{J_\parallel}{B}
+ 2 \bm{b} \cp \bm{\kappa} \cdot \grad P
+ 2 \bm{b} \cdot \bm{\kappa} \cp \bm{f} \\
\end{aligned}
\end{equation*}
最后得到
\begin{equation*}
\begin{aligned}
\bm{b} \cdot \qty(\curl \bm{f} – 2 \bm{\kappa} \cp \bm{f})
&=
B^2 \bm{b} \cdot \grad \frac{J_\parallel}{B}
+ 2 \bm{b} \cp \bm{\kappa} \cdot \grad P \\
\end{aligned}
\end{equation*}
其中左侧主要是第一项为主，定义涡量
$$
U \equiv \bm{b} \cdot \curl \bm{u}
$$
可以得到涡量方程
\begin{equation}
\label{eq:vorticity_1}
\begin{aligned}
\rho \dv{U}{t}
&=
B^2 \bm{b} \cdot \grad \frac{J_\parallel}{B}
+ 2 \bm{b} \cp \bm{\kappa} \cdot \grad P \\
\end{aligned}
\end{equation}
势能方程
为了推导出标量方程，需要使用磁矢势$\bm{A}$和电标量势$\phi$来替换原电磁方程。
\begin{equation}
\label{eq:potential_B}
\begin{aligned}
\bm{B} &= \curl \bm{A} \\
\end{aligned}
\end{equation}
\begin{equation}
\label{eq:potential_E}
\begin{aligned}
\bm{E} &= – \grad \phi – \pdv{\bm{A}}{t}
\end{aligned}
\end{equation}
对式\eqref{eq:potential_B}，将磁矢势沿磁场方向分解
$$
\bm{B} = \curl \qty(A_\parallel \bm{b_0}) + \curl \bm{A}_\perp
$$
右边第一项垂直与磁场方向，第二项平行于磁场方向。
由笛状近似，平行磁场方向为高阶小项，$\bm{B_1} \perp \bm{B_0}$，同时用$\psi = A_\parallel / B_0$来代替$A_\parallel$：
\begin{equation}
\label{eq:B_1}
\begin{aligned}
\bm{B_1} &= \curl \psi \bm{B_0} \\
&= \psi \curl \bm{B_0} + \grad \psi \cp \bm{B_0} \\
& \simeq \grad \psi \cp \bm{B_0}
\end{aligned}
\end{equation}
进一步由式\eqref{eq:potential_E}，笛状近似下
$$
E_\parallel = – \nabla_\parallel \phi – \pdv{A_\parallel}{t} = – \nabla_\parallel \phi – B_0\pdv{\psi}{t} = 0
$$
所以有：
\begin{equation}
\label{eq:vorticity_2}
\begin{aligned}
\pdv{\psi}{t} = – \frac{1}{B_0}\nabla_\parallel \phi
\end{aligned}
\end{equation}
其中
\begin{equation}
\label{eq:nabla_parallel}
\begin{aligned}
\nabla_\parallel \equiv \bm{b} \cdot \grad
&= \frac{\bm{B_0} + \bm{B_1}}{B_0 + B_1} \cdot \grad \\
&\simeq \frac{\bm{B_0} + \bm{B_1}}{B_0} \cdot \grad \\
&= \bm{b_0} \cdot \grad + \grad \psi \cp \bm{b_0} \cdot \grad\\
&= \bm{b_0} \cdot \grad – \bm{b_0} \cp \grad \psi \cdot \grad\\
&= \bm{b_0} \cdot \grad – \bm{b_0} \cdot \grad \psi \cp \grad\\
\end{aligned}
\end{equation}
涡量电势方程
假设零阶速度（本底速度）$\bm{u_0} = 0$，一阶速度为电漂移速度$\bm{u_1} = \bm{u_E}$，
由欧姆定律有：
\begin{equation*}
\begin{aligned}
\bm{E_1}
&= – \qty(\bm{u_1} \cp \bm{B_0}) \\
\bm{E_1} \cp \bm{B_0}
&= \bm{B_0} \cp \qty(\bm{u_1} \cp \bm{B_0}) \\
&= B_0^2 \bm{u_1} – \qty(\bm{u_1} \cdot \bm{B_0})\bm{B_0}
\end{aligned}
\end{equation*}
根据笛状近似，$\bm{u_1} \cdot \bm{B_0} \simeq 0$，另外用静电势$\phi$表示$E_1$有：
\begin{equation*}
\begin{aligned}
\bm{u} \simeq \bm{u_1} = \bm{u_E} = \frac{\bm{b}}{B} \cp \grad \phi
\end{aligned}
\end{equation*}
带入涡量方程表达式有
\begin{equation}
\label{eq:vorticity_mid2}
\begin{aligned}
U &= \bm{b} \cdot \curl \bm{u} \\
&= \bm{b} \cdot \curl \qty(\frac{\bm{b}}{B} \cp \grad \phi) \\
&= \bm{b} \cdot \curl \qty(\frac{\bm{b}}{B} \cp \grad \phi) \\
&= \bm{b} \cdot
\qty[
\frac{\bm{b}}{B}\div\grad \phi
– \grad \phi \qty(\div \frac{\bm{b}}{B})
+ \qty(\grad \phi \cdot \grad) \frac{\bm{b}}{B}
– \qty(\frac{\bm{b}}{B} \cdot \grad) \grad \phi
]
\end{aligned}
\end{equation}
在笛状近似下$k_\parallel \ll k_\perp$，平行方向$\bm{b} \cdot \grad$被忽略，
同时$\bm{b}$作为平衡量求导后$\grad \bm{b}$相比于第一项$\nabla^2_\perp$可以忽略不计：
\begin{equation}
\label{eq:vorticity_3}
\begin{aligned}
U = \frac{1}{B} \nabla_{\perp}^2 \phi
\end{aligned}
\end{equation}
电流方程
由式\eqref{eq:B_1}，可推导出电流表达式为：
\begin{equation*}
\begin{aligned}
\bm{J} &= \bm{J_0} + \frac{1}{\mu_0}\curl \bm{B_1} \\
&= \bm{J_0} + \frac{1}{\mu_0}\curl \qty(\grad \psi \cp \bm{B_0}) \\
&= \bm{J_0} + \frac{1}{\mu_0} \qty(\grad \cdot \bm{B_0})\grad \psi
– \frac{\bm{B_0}}{\mu_0} \nabla^2 \psi
+ \frac{1}{\mu_0} \qty(\bm{B_0} \cdot \grad) \grad \psi
– \frac{1}{\mu_0} \grad \psi \cdot \grad\bm{B_0}
\end{aligned}
\end{equation*}
类似涡量电势方程的推导\eqref{eq:vorticity_mid2}，忽略平行方向的微分（右边第二、四、五项），并且取平行方向的分量有：
\begin{equation}
\label{eq:vorticity_4}
\begin{aligned}
J_\parallel = J_{\parallel 0} – \frac{1}{\mu_0} B_0 \nabla^2_\perp \psi
\end{aligned}
\end{equation}
压强方程
用守恒方程与质量方程可推导出
$$
\pdv{t} P + \bm{u} \cdot \grad P
+ \gamma P
\div \bm{u} = 0
$$
忽略第三项，将电漂移速度带入有：
\begin{equation}
\label{eq:vorticity_5}
\begin{aligned}
\pdv{t} P = – \frac{\bm{b_0}}{B_0} \cp \grad \phi \cdot \grad P
\end{aligned}
\end{equation}
总结
根据上面的推导，我们得到了三场方程如下：
\begin{equation}
\label{eq:3field}
\begin{aligned}
\rho \dv{U}{t}
&=
B^2 \bm{b} \cdot \grad \frac{J_\parallel}{B}
+ 2 \bm{b} \cp \bm{\kappa} \cdot \grad P \\
\pdv{\psi}{t} &= – \frac{1}{B_0}\nabla_\parallel \phi \\
\pdv{t} P &= – \frac{\bm{b_0}}{B_0} \cp \grad \phi \cdot \grad P \\
U &= \frac{1}{B} \nabla_{\perp}^2 \phi \\
J_\parallel &= J_{\parallel 0} – \frac{1}{\mu_0} B_0 \nabla^2_\perp \psi \\
\end{aligned}
\end{equation}
