Guiding center drift

Next, we derive the guiding center drift. Using equation (13), we obtain

$$- Z e \ensuremath{\boldsymbol{E}} + \mu \nabla B + m v_{\parallel... ...dsymbol{B}} + m v_{\parallel} \nabla \times \ensuremath{\boldsymbol{b}} \right)$$ (57)

Taking cross product of the above equation with $\ensuremath{\boldsymbol{b}} / (m \Omega)$, we obtain

$$\frac{1}{m \Omega} \ensuremath{\boldsymbol{b}} \times \left( - Z ... ...B}} + m v_{\parallel} \nabla \times \ensuremath{\boldsymbol{b}} \right) \right]$$ (58)

The right-hand side of the above equation is simplified as

$$- \frac{1}{m \Omega} \left( \ensuremath{\boldsymbol{b}} \cdot \do... ...s \ensuremath{\boldsymbol{b}} \right) \right] \dot{\ensuremath{\boldsymbol{X}}}$$

$$= - v_{\parallel} \left( \ensuremath{\boldsymbol{b}} + \frac{v_{\... ...la \times \ensuremath{\boldsymbol{b}} \right] \dot{\ensuremath{\boldsymbol{X}}}$$

$$= - v_{\parallel} \ensuremath{\boldsymbol{b}} - \frac{1}{\Omega} ... ...la \times \ensuremath{\boldsymbol{b}} \right) \dot{\ensuremath{\boldsymbol{X}}}$$ (59)

Using this, Eq. (58) is written as

$$\frac{1}{m \Omega} \ensuremath{\boldsymbol{b}} \times \left( - Z ... ... \times \ensuremath{\boldsymbol{b}} \right) \dot{\ensuremath{\boldsymbol{X}}}$$

$$\Rightarrow \dot{\ensuremath{\boldsymbol{X}}} = v_{\parallel} \en... ...la \times \ensuremath{\boldsymbol{b}} \right) \dot{\ensuremath{\boldsymbol{X}}}$$ (60)

$$\Rightarrow \dot{\ensuremath{\boldsymbol{X}}} = v_{\parallel} \en... ...s \ensuremath{\boldsymbol{b}} \right) \dot{\ensuremath{\boldsymbol{X}}} \right]$$ (61)

Equation (61) contains $\ensuremath{\boldsymbol{E}} \times \ensuremath{\boldsymbol{B}}$ drift and $\nabla B$ drift. Next we examine the terms in the square bracket of Eq. (61), which can be further written as

$$\frac{1}{m \Omega} \left[ \ensuremath{\boldsymbol{b}} \times m v_... ...s \ensuremath{\boldsymbol{b}} \right) \dot{\ensuremath{\boldsymbol{X}}} \right]$$

$$= \frac{1}{m \Omega} \left\{ \ensuremath{\boldsymbol{b}} \times m... ...math{\boldsymbol{b}} \right) \dot{\ensuremath{\boldsymbol{X}}} \right] \right\}$$

$$= \frac{1}{m \Omega} \left\{ \ensuremath{\boldsymbol{b}} \times m... ...oldsymbol{b}} \right) \times \dot{\ensuremath{\boldsymbol{X}}} \right] \right\}$$

$$= \frac{1}{m \Omega} \ensuremath{\boldsymbol{b}} \times \left\{ m... ...emath{\boldsymbol{b}} \right) \times \dot{\ensuremath{\boldsymbol{X}}} \right\}$$

$$= \frac{1}{m \Omega} \ensuremath{\boldsymbol{b}} \times \left\{ m... ...th{\boldsymbol{b}} + \dot{\ensuremath{\boldsymbol{X}}}_{\perp} \right) \right\}$$

$$= \frac{1}{m \Omega} \ensuremath{\boldsymbol{b}} \times \left\{ m... ...l{b}} \right) \times \dot{\ensuremath{\boldsymbol{X}}}_{\perp} \right] \right\}$$

Using this, Eq. (61) is written as

$$\Rightarrow \dot{\ensuremath{\boldsymbol{X}}} = v_{\parallel} \en... ...l{b}} \right) \times \dot{\ensuremath{\boldsymbol{X}}}_{\perp} \right] \right\}$$ (62)

Besides $\ensuremath{\boldsymbol{E}} \times \ensuremath{\boldsymbol{B}}$ drift, $\nabla B$ drift, and magnetic curvature drift, Eq. (62) also contains additional terms (the terms in the square bracket). The first term in the square bracket is usually small compared with other terms, so this term is usually ignored (this term appears in equation (181) of Boozer's paper[2], and is said to be ignorably small in his paper). We now exxamine the last term, i.e.,

$$\frac{1}{m \Omega} \ensuremath{\boldsymbol{b}} \times m v_{\paral... ...emath{\boldsymbol{b}} \right) \times \dot{\ensuremath{\boldsymbol{X}}}_{\perp},$$ (63)

which can be written as

$$- \frac{v_{\parallel}}{\Omega} \left( \ensuremath{\boldsymbol{b}}... ...\ensuremath{\boldsymbol{b}} \right) \dot{\ensuremath{\boldsymbol{X}}}_{\perp} .$$ (64)

In Eq. (66) of Ref. [3], it is pointed out that

$$\nabla \times \ensuremath{\boldsymbol{b}} \approx - \ensuremath{\... ...bol{b}}) = \ensuremath{\boldsymbol{b}} \times \ensuremath{\boldsymbol{\kappa}},$$ (65)

which is correct in the order considered here (I do not check this). This indicates that $\ensuremath{\boldsymbol{b}} \cdot \nabla \times \ensuremath{\boldsymbol{b}} \approx 0$. Using this, we know the expression in Eq. (64) is approximately zero. Thus the terms in the square bracket of Eq. (62) can be dropped. Then Eq. (62) reduces to

$$\dot{\ensuremath{\boldsymbol{X}}} = v_{\parallel} \ensuremath{\bo... ...E}} + \mu \nabla B + m v_{\parallel}^2 \ensuremath{\boldsymbol{\kappa}} \right)$$ (66)