Energy conservation

Doting Eq. (21) with $\dot{\ensuremath{\boldsymbol{X}}}$, we obtain

$$Z e \ensuremath{\boldsymbol{E}}^{\star} \cdot \dot{\ensuremath{\b... ...parallel} \ensuremath{\boldsymbol{b}} \cdot \dot{\ensuremath{\boldsymbol{X}}} .$$ (28)

Using

$$\ensuremath{\boldsymbol{E}}^{\star} = - \nabla \phi - \frac{1}{c}... ...{m v_{\parallel}}{Z e} \frac{\partial \ensuremath{\boldsymbol{b}}}{\partial t},$$ (29)

Eq. (28) is written as

$$m \dot{v}_{\parallel} \ensuremath{\boldsymbol{b}} \cdot \dot{\ens... ...b}}}{\partial t} + \mu \nabla B \right) \cdot \dot{\ensuremath{\boldsymbol{X}}}$$ (30)

Using Eq. (7), i.e., $\ensuremath{\boldsymbol{b}} \cdot \dot{\ensuremath{\boldsymbol{X}}} = v_{\parallel}$, the above equation is written as

$$m v_{\parallel} \dot{v}_{\parallel} = - \left( Z e \nabla \phi + ... ...}}}{\partial t} + \mu \nabla B \right) \cdot \dot{\ensuremath{\boldsymbol{X}}},$$ (31)

which gives the time change rate of the parallel velocity $v_{\parallel}$. By using $\dot{\mu} = 0$ and $y = \mu B$, the time change rate of the perpendicular velocity is written as

$$\dot{y} = \mu \frac{d B}{d t}$$

$$= \mu \left( \frac{\partial B}{\partial t} + \dot{\ensuremath{\boldsymbol{X}}} \cdot \nabla B \right)$$ (32)

Next, calculate the total time derivative of the particle energy. The particle energy $\varepsilon$ is the sum of the kinetic and potential energy, i.e.,

$$\varepsilon = \frac{1}{2} m v_{\parallel}^2 + y + Z e \phi,$$ (33)

from which we obtain

$$\dot{\varepsilon} = m v_{\parallel} \dot{v_{\parallel}} + \dot{y} + Z e \dot{\phi} .$$ (34)

Using Eqs. (31) and (32), the right-hand side of Eq. (34) is written as

$$m v_{\parallel} \dot{v_{\parallel}} + \dot{y} + Z e \dot{\phi} = - \left( \frac{Z e}{c} \frac{\partial \ensuremath{\boldsymbol{A}}... ...} + \dot{\ensuremath{\boldsymbol{X}}} \cdot \nabla B \right) + Z e \dot{\phi} .$$

$$= - \left( \frac{Z e}{c} \frac{\partial \ensuremath{\boldsymbol{A}}... ...}} + \mu \frac{\partial B}{\partial t} + Z e \frac{\partial \phi}{\partial t} .$$ (35)

For the equilibrium case, electromagnetic field is independent of time, so the result of the above expression is zero, indicating that energy is a constant of the motion.