Further simplification of the perturbed Lagrangian $\mathcal{L}^{(1)}$

The expression $\mathcal{L}^{(1)}$ in Eq. (115) can be further simplified, by noticing that the term $m v_{\parallel} \ensuremath{\boldsymbol{b}}^{(1)} \cdot \dot{\ensuremath{\boldsymbol{X}}}^{(0)}$ is of the order $O (\delta^2)$ thus can be neglected, giving

$$\mathcal{L}^{(1)} = \frac{Z e}{c} \ensuremath{\boldsymbol{A}}^{(1... ...cdot \dot{\ensuremath{\boldsymbol{X}}}^{(0)} - \mu_0 B^{(1)} - Z e \phi^{(1)} .$$ (123)

Next, we provide the proof that the term $m v_{\parallel} \ensuremath{\boldsymbol{b}}^{(1)} \cdot \dot{\ensuremath{\boldsymbol{X}}}^{(0)}$ is on the order $O (\delta^2)$. The unperturbed velocity of guiding center is given by

$$\dot{\ensuremath{\boldsymbol{X}}}^{(0)} = v_{\parallel} \ensuremath{\boldsymbol{b}} + \frac{1}{m \Omega} \e... ...} \nabla B_0 + m v_{\parallel}^2 \ensuremath{\boldsymbol{\kappa}}^{(0)} \right)$$

$$= v_{\parallel} \ensuremath{\boldsymbol{b}} + \ensuremath{\boldsymbol{v}}_d$$ (124)

where $v_{\parallel} \sim O (\delta^0)$, $v_d \sim O (\delta)$. Next, we derive the expression of $\ensuremath{\boldsymbol{b}}^{(1)}$. Using

$$\frac{1}{B} = \frac{1}{\sqrt{B_0^2 + \left( B^{(1)} \right)^2 + 2 \ensuremath{\boldsymbol{B}}_0 \cdot \ensuremath{\boldsymbol{B}}^{(1)}}}$$

$$\approx \frac{1}{B_0} - \frac{1}{2 B_0^3} \left[ \left( B^{(1)} \right)^2 + 2 \ensuremath{\boldsymbol{B}}_0 \cdot \ensuremath{\boldsymbol{B}}^{(1)} \right]$$ (125)

we obtain

$$\ensuremath{\boldsymbol{b}} = \frac{\ensuremath{\boldsymbol{B}}_0 + \ensuremath{\boldsymbol{B}}^{(1)}}{B}$$

$$= \left( \ensuremath{\boldsymbol{B}}_0 + \ensuremath{\boldsymbol{B}... ...math{\boldsymbol{B}}_0 \cdot \ensuremath{\boldsymbol{B}}^{(1)} \right] \right\}$$

$$\approx \ensuremath{\boldsymbol{b}}^{(0)} - \frac{\ensuremath{\boldsymbol... ...h{\boldsymbol{B}}^{(1)} \right) + \frac{\ensuremath{\boldsymbol{B}}^{(1)}}{B_0}$$

$$= \ensuremath{\boldsymbol{b}}^{(0)} - \ensuremath{\boldsymbol{b}}^{... ...ymbol{B}}^{(1)}}{B_0} \right) + \frac{\ensuremath{\boldsymbol{B}}^{(1)}}{B_0} .$$

Therefore

$$\ensuremath{\boldsymbol{b}}^{(1)} = - \ensuremath{\boldsymbol{b}}^{(0)} \left( \ensuremath{\boldsymbo... ...dsymbol{B}}^{(1)}}{B_0} \right) + \frac{\ensuremath{\boldsymbol{B}}^{(1)}}{B_0}$$ (126)

Using this, the term $m v_{\parallel} \ensuremath{\boldsymbol{b}}^{(1)} \cdot \dot{\ensuremath{\boldsymbol{X}}}^{(0)}$ is written as

$$m v_{\parallel} \ensuremath{\boldsymbol{b}}^{(1)} \cdot \dot{\ensuremath{\boldsymbol{X}}}^{(0)} = - m v_{\parallel}^2 \ensuremath{\boldsymbol{b}}^{(0)} \cdot \frac... ...parallel} \ensuremath{\boldsymbol{b}}^{(1)} \cdot \ensuremath{\boldsymbol{v}}_d$$ (127)

$$= 0 + m v_{\parallel} \ensuremath{\boldsymbol{b}}^{(1)} \cdot \ensuremath{\boldsymbol{v}}_d,$$ (128)

where the first two terms on the right-hand side of Eq. (127), which are on the order $O (\delta)$, happen to cancel each other. Since $\ensuremath{\boldsymbol{b}}^{(1)} \sim O (\delta)$ and $\ensuremath{\boldsymbol{v}}_d \sim O (\delta)$, the product of these two terms are on the order $O (\delta^2)$. Therefore Eq. (134) indicates that the term $m v_{\parallel} \ensuremath{\boldsymbol{b}}^{(1)} \cdot \dot{\ensuremath{\boldsymbol{X}}}^{(0)}$ is on the order $O (\delta^2)$.

Next we show that, in the linear approximation, the perturbation in the strength of the magnetic field, $B^{(1)}$, is equal to $B_{\parallel}^{(1)}$, where $B_{\parallel}^{(1)} \equiv \ensuremath{\boldsymbol{b}}^{(0)} \cdot \ensuremath{\boldsymbol{B}}^{(1)}$. The total magnetic field can be written as

$$B = \sqrt{\ensuremath{\boldsymbol{B}} \cdot \ensuremath{\boldsymbol{B}}}$$

$$= \sqrt{\left( \ensuremath{\boldsymbol{B}}_0 + \ensuremath{\boldsym... ...eft( \ensuremath{\boldsymbol{B}}_0 + \ensuremath{\boldsymbol{B}}^{(1)} \right)}$$

$$= \sqrt{B_0^2 + \left( B^{(1)} \right)^2 + 2 \ensuremath{\boldsymbol{B}}_0 \cdot \ensuremath{\boldsymbol{B}}^{(1)}}$$ (129)

Expanding the right-hand side of the above equation at $B_0$, we obtain

$$B \approx \sqrt{B_0^2} + \frac{1}{2 \sqrt{B_0^2}} \left[ \left( B^{(1)} \ri... ...math{\boldsymbol{B}}_0 \cdot \ensuremath{\boldsymbol{B}}^{(1)} \right] + \ldots$$ (130)

Neglecting the second order term, the above equation is written as

$$B \approx B_0 + \frac{\ensuremath{\boldsymbol{B}}_0 \cdot \ensuremath{\boldsymbol{B}}^{(1)}}{B_0}$$

$$= B_0 + \ensuremath{\boldsymbol{b}}^{(0)} \cdot \ensuremath{\boldsymbol{B}}^{(1)}$$

$$= B_0 + B_{\parallel}^{(1)}$$ (131)

Thus we obtain

$$B^{(1)} = B_{\parallel}^{(1)} .$$ (132)

Eq. (132) seems strange at first glance (actually I think it is wrong at first glance, Dr. Fu let me know it is right and how to prove it (as given in the above)). Using Eq. (132), the Lagrangian in Eq. (123) is written as

$$\mathcal{L}^{(1)} = \frac{Z e}{c} \ensuremath{\boldsymbol{A}}^{(1... ...nsuremath{\boldsymbol{X}}}^{(0)} - \mu_0 B_{\parallel}^{(1)} - Z e \phi^{(1)} .$$ (133)