Guiding-center motion

The phase-space Lagrangian for guiding-center motion was first given in Littlejohn's paper[1], which takes the following form

$$\mathcal{L} ( \ensuremath{\boldsymbol{X}}, v_{\parallel}, y, \alp... ...\frac{1}{\Omega} y \dot{\alpha} - \frac{1}{2} m v_{\parallel}^2 - y - Z e \phi,$$ (1)

where $\ensuremath{\boldsymbol{X}}$ is the location of the guiding-center, $v_{\parallel}$ is the parallel (to magnetic field) velocity of the particle (will be proved later that $v_{\parallel}$ is also the parallel velocity of the guiding center) $y \equiv m v_{\perp}^2 / 2$ with $v_{\perp}$ the perpendicular (to magnetic field) velocity of the particle, $\alpha$ is the gyrophase, $\Omega = Z e B / (m c)$ with $Z$ being the charge number and $e$ being the elementary charge, $\ensuremath{\boldsymbol{b}} = \ensuremath{\boldsymbol{B}} / B$. Note that here the (phase-space) Lagrangian of guiding center is considered to be a function of variables $\ensuremath{\boldsymbol{X}}, v_{\parallel}, y, \alpha, \dot{\ensuremath{\boldsymbol{X}}}, \dot{v}_{\parallel}, \dot{y}, \dot{\alpha}$, and $t$. Also note that three variables, $\alpha$, $\dot{v}_{\parallel}$ and $\dot{y}$ happens not to appear in Eq. (1). Further note that the explicit dependence of $\mathcal{L}$ on $\ensuremath{\boldsymbol{X}}$ and $t$ is through the electromagnetic field $\ensuremath{\boldsymbol{A}}$, $\phi$, and the frequency $\Omega$. The Euler-Lagrange equation corresponding to variable $y$ is written as

$$\frac{d}{d t} \left( \frac{\partial \mathcal{L}}{\partial \dot{y}} \right) = \frac{\partial \mathcal{L}}{\partial y},$$ (2)

which, after evaluating the partial derivatives, is reduced to

$$\dot{\alpha} = \Omega,$$ (3)

which indicates as expected that the time change rate of the gyrophase $\alpha$ equals the cylcotron frequency $\Omega$. The Euler-Lagrange equation corresponding to the variable $\alpha$ is written as

$$\frac{d}{d t} \left( \frac{\partial \mathcal{L}}{\partial \dot{\alpha}} \right) = \frac{\partial \mathcal{L}}{\partial \alpha},$$ (4)

which can be written as

$$\frac{d}{d t} \left( \frac{y}{\Omega} \right) = 0,$$ (5)

which indicats that the magnetic moment $\mu \equiv y / B$ is a constant of the motion. The Euler-Lagrange equation for the variable $v_{\parallel}$ is

$$\frac{d}{d t} \left( \frac{\partial \mathcal{L}}{\partial \dot{v}_{\parallel}} \right) = \frac{\partial \mathcal{L}}{\partial v_{\parallel}},$$ (6)

which can be simplified to

$$v_{\parallel} = \ensuremath{\boldsymbol{b}} \cdot \dot{\ensuremath{\boldsymbol{X}}},$$ (7)

which indicates that $v_{\parallel}$ is also the parallel velocity of the guiding center. Next, consider the Euler-Lagrange equation corresponding to the coordinate $\ensuremath{\boldsymbol{X}}$, which is given by

$$\frac{d}{d t} \left( \frac{\partial \mathcal{L}}{\partial \dot{\e... ...}} \right) = \frac{\partial \mathcal{L}}{\partial \ensuremath{\boldsymbol{X}}},$$ (8)

which should be understood as a shorthand of the three Euler-Lagrange equations coorresponding to three coordinates. Is the above equation still valid in arbitrary coordinates system if we consider $\partial / \partial \dot{\ensuremath{\boldsymbol{X}}}$ and $\partial / \partial \ensuremath{\boldsymbol{X}}$ as gradient operators? The answer is yes. However it is not trivial for me to find the proof (the proof is provided in Sec. (4)). Using Eq. (1) and vector identies, we obtain

$$\frac{\partial \mathcal{L}}{\partial \dot{\ensuremath{\boldsymbol... ...}{c} \ensuremath{\boldsymbol{A}} + m v_{\parallel} \ensuremath{\boldsymbol{b}},$$ (9)

and

$$\frac{\partial \mathcal{L}}{\partial \ensuremath{\boldsymbol{X}}}... ...}} \right) - \frac{1}{\Omega^2} y \dot{\alpha} \nabla \Omega - Z e \nabla \phi,$$ (10)

where $\nabla \equiv \partial / \partial \ensuremath{\boldsymbol{X}}$. Using Eqs. (9) and (10), Eq. (8) is written as

$$\frac{Z e}{c} \left( \frac{\partial \ensuremath{\boldsymbol{A}}}{... ...b}} \right) - \frac{1}{\Omega^2} y \dot{\alpha} \nabla \Omega - Z e \nabla \phi$$ (11)

Using $\Omega = B Z e / (m c)$ and $\dot{\alpha} = \Omega$, the second last term can be reduced to $- \mu \nabla B$. Then Eq. (11) is written as

$$\frac{Z e}{c} \frac{\partial \ensuremath{\boldsymbol{A}}}{\partia... ...\parallel} \ensuremath{\boldsymbol{b}} \right) - \mu \nabla B - Z e \nabla \phi$$ (12)

Noting that $m v_{\parallel} \nabla \ensuremath{\boldsymbol{b}} = \nabla (m v_{\parallel} \ensuremath{\boldsymbol{b}})$ (this is because $\nabla \equiv (\partial / \partial \ensuremath{\boldsymbol{X}})_{v_{\parallel}}$, i.e., holding $v_{\parallel}$ constant), so that the second term on the right-hand side of the above equation is canceled by terms on the right-hand, yielding

$$\frac{Z e}{c} \frac{\partial \ensuremath{\boldsymbol{A}}}{\partia... ...a \times \ensuremath{\boldsymbol{b}} \right) - \mu \nabla B - Z e \nabla \phi .$$ (13)

Equation (13) can be further written in compact form by defining new magnetic-like and electric-like quantities. Define

$$\ensuremath{\boldsymbol{A}}^{\star} = \ensuremath{\boldsymbol{A}} + \frac{m c}{Z e} v_{\parallel} \ensuremath{\boldsymbol{b}},$$ (14)

and

$$\ensuremath{\boldsymbol{B}}^{\star} = \nabla \times \ensuremath{\boldsymbol{A}}^{\star},$$ (15)

then

$$\frac{\partial \ensuremath{\boldsymbol{A}}^{\star}}{\partial t} =... ...m c}{Z e} v_{\parallel} \frac{\partial \ensuremath{\boldsymbol{b}}}{\partial t}$$ (16)

$$\ensuremath{\boldsymbol{B}}^{\star} = \ensuremath{\boldsymbol{B}} + \frac{m c}{Z e} v_{\parallel} \nabla \times \ensuremath{\boldsymbol{b}},$$ (17)

(Note that the time partial differential does not operate on $v_{\parallel}$ because it is an independent variables.) Using these, Eq. (13) is written as

$$\frac{Z e}{c} \frac{\partial \ensuremath{\boldsymbol{A}}^{\star}}... ...}}} \times \ensuremath{\boldsymbol{B}}^{\star} - \mu \nabla B - Z e \nabla \phi$$ (18)

$$\Longrightarrow \frac{Z e}{c} \dot{\ensuremath{\boldsymbol{X}}} \... ...{\partial t} = \mu \nabla B + m \dot{v}_{\parallel} \ensuremath{\boldsymbol{b}}$$ (19)

Define

$$\ensuremath{\boldsymbol{E}}^{\star} = - \nabla \phi - \frac{1}{c} \frac{\partial \ensuremath{\boldsymbol{A}}^{\star}}{\partial t}$$ (20)

then Eq. (19) is written as

$$Z e \left( \ensuremath{\boldsymbol{E}}^{\star} + \frac{1}{c} \dot... ...ar} \right) = \mu \nabla B + m \dot{v}_{\parallel} \ensuremath{\boldsymbol{b}},$$ (21)

which agrees with Eq. (23) of Porcelli's paper[1].