Linearized drift kinetic equation

Next, we derive the linearized version of Eq. (77). The perturbation in electromagnetic field causes perturbation in both distribution function and particle orbits $\dot{\ensuremath{\boldsymbol{X}}}$, $\dot{v_{\parallel}}$, and $\dot{y}$. Thus we write

$$f = f_0 + f_1$$ (81)

$$\dot{\ensuremath{\boldsymbol{X}}} = \dot{\ensuremath{\boldsymbol{X}}}^{(0)} + \dot{\ensuremath{\boldsymbol{X}}}^{(1)},$$ (82)

$$\dot{v_{\parallel}} = \dot{v_{\parallel}}^{(0)} + \dot{v_{\parallel}}^{(1)},$$ (83)

$$\dot{y} = \dot{y}^{(0)} + \dot{y}^{(1)} .$$ (84)

and substitute this into Eq. (77), we obtain

$$\frac{\partial f_0}{\partial t} + \frac{\partial f_1}{\partial t}... ...eft( \dot{y}^{(0)} + \dot{y}^{(1)} \right) \frac{\partial f_1}{\partial y} = 0.$$ (85)

The zero order equation is

$$\frac{\partial f_0}{\partial t} + \dot{\ensuremath{\boldsymbol{X}... ...0}{\partial v_{\parallel}} + \dot{y}^{(0)} \frac{\partial f_0}{\partial y} = 0.$$ (86)

The first order equation is

$$\frac{\partial f_1}{\partial t} + \dot{\ensuremath{\boldsymbol{X}... ...\partial v_{\parallel}} + \dot{y}^{(1)} \frac{\partial f_0}{\partial y} \right]$$ (87)

Define

$$\frac{D}{D t} = \frac{\partial}{\partial t} + \dot{\ensuremath{\b... ...{\partial}{\partial v_{\parallel}} + \dot{y}^{(0)} \frac{\partial}{\partial y},$$ (88)

which is the unperturbed orbit propagator, then Eq. (87) is written as

$$\frac{D f_1}{D t} = - \left[ \dot{\ensuremath{\boldsymbol{X}}}^{(... ...partial v_{\parallel}} + \dot{y}^{(1)} \frac{\partial f_0}{\partial y} \right],$$ (89)

which agrees with Eq. (21) in Porcelli's paper[1].

At this point, I would like to discuss the equilibrium distribution function. We know that functions of constants of the motion are solutions to the kinetic equation. Noting that $P_{\varphi 0}$, $\varepsilon_0$, and $\mu_0$ are constants of the motion in equilibrium field. Then $f_0 = F (P_{\varphi 0}, \varepsilon_0, \mu_0)$ is a solution to the kinetic equation Eq. (77). Noting that the right-hand side of Eq. (89) contains partial derivative with respect to variables $( \ensuremath{\boldsymbol{X}}, v_{\parallel}, y)$, we would like to transform the partial derivatives with respect to $( \ensuremath{\boldsymbol{X}}, v_{\parallel}, y)$ to one with respect to $(P_{\varphi 0}, \varepsilon_0, \mu_0)$. The right-hand side of Eq. (89) can be written term by term as

$$\dot{\ensuremath{\boldsymbol{X}}}^{(1)} \cdot \nabla f_0 = \dot{\ensuremath{\boldsymbol{X}}}^{(1)} \cdot \left( \frac{\parti... ...} \nabla \varepsilon_0 + \frac{\partial F}{\partial \mu_0} \nabla \mu_0 \right)$$

$$= \dot{\ensuremath{\boldsymbol{X}}}^{(1)} \cdot \left( \frac{\parti... ...a \phi_0 - \frac{\partial F}{\partial \mu_0} \frac{y}{B^2_0} \nabla B_0 \right)$$ (90)

$$\dot{v_{\parallel}}^{(1)} \frac{\partial f_0}{\partial v_{\parallel}} = \dot{v_{\parallel}}^{(1)} \left( \frac{\partial F}{\partial P_{\v... ...artial F}{\partial \mu_0} \frac{\partial \mu_0}{\partial v_{\parallel}} \right)$$

$$= \dot{v_{\parallel}}^{(1)} \left( \frac{\partial F}{\partial P_{\v... ...\parallel}} + \frac{\partial F}{\partial \varepsilon_0} m v_{\parallel} \right)$$ (91)

$$\dot{y}^{(1)} \frac{\partial f_0}{\partial y} = \dot{y}^{(1)} \left( \frac{\partial F}{\partial P_{\varphi 0}} \f... ...} + \frac{\partial F}{\partial \mu_0} \frac{\partial \mu_0}{\partial y} \right)$$

$$= \dot{y}^{(1)} \left( \frac{\partial F}{\partial \varepsilon_0} + \frac{\partial F}{\partial \mu_0} \frac{1}{B_0} \right)$$ (92)

Using these results, Eq. (89) is written as

$$\frac{D f_1}{D t} = - \left[ \dot{\ensuremath{\boldsymbol{X}}}^{(... ...epsilon_0} + \frac{\partial F}{\partial \mu_0} \frac{1}{B_0} \right) \right],$$

which can be arranged in the form

$$\frac{D f_1}{D t} = - \left[ \left( \dot{\ensuremath{\boldsymbol{... ...{X}}}^{(1)} \cdot \nabla B_0 \right) \frac{\partial F}{\partial \mu_0} \right],$$ (93)

which agrees with Eq. (22) in Porcelli's paper[1]. Next, we need to express $\dot{\ensuremath{\boldsymbol{X}}}^{(1)}$, $\dot{v}_{\parallel}^{(1)}$, and $\dot{y}^{(1)}$ in terms of the perturbed electromagnetic field. Let us first consider the coefficient befor the $\partial F / \partial \varepsilon_0$ term in Eq. (93). We note that Eq. (35) takes the following form:

$$m v_{\parallel} \dot{v_{\parallel}} + \dot{y} + Z e \dot{\phi} = ... ...rac{y}{B} \frac{\partial B}{\partial t} + Z e \frac{\partial \phi}{\partial t},$$ (94)

whose linearized version is (noting that $\ensuremath{\boldsymbol{A}}^{(0)}$, $\ensuremath{\boldsymbol{b}}^{(0)}$, and $B^{(0)}$ is time independent, thus $\partial \ensuremath{\boldsymbol{A}}^{(0)} / \partial t$, $\partial \ensuremath{\boldsymbol{b}}^{(0)} / \partial t$, and $\partial B^{(0)} / \partial t$ are all zeros)

$$m v_{\parallel} \dot{v_{\parallel}}^{(1)} + \dot{y}^{(1)} + Z e \... ...rac{\partial B^{(1)}}{\partial t} + Z e \frac{\partial \phi^{(1)}}{\partial t},$$ (95)

where $\mu_0 = y / B_0$. Using

$$\dot{\phi}^{(1)} \equiv \left( \frac{d \phi}{d t} \right)^{(1)}$$

$$= \left( \frac{\partial \phi}{\partial t} + \dot{\ensuremath{\boldsymbol{X}}} \cdot \nabla \phi + 0 + 0 + 0 \right)^{(1)}$$

$$= \frac{\partial \phi^{(1)}}{\partial t} + \dot{\ensuremath{\boldsy... ...la \phi^{(1)} + \dot{\ensuremath{\boldsymbol{X}}}^{(1)} \cdot \nabla \phi^{(0)}$$

$$= \frac{D \phi^{(1)}}{D t} + \dot{\ensuremath{\boldsymbol{X}}}^{(1)} \cdot \nabla \phi^{(0)}$$ (96)

(Note that $\frac{D}{D t}$ here denotes total time derivative along the unperturbed orbit, instead of the perturbed orbit ) in Eq. (95) gives

$$m v_{\parallel} \dot{v_{\parallel}}^{(1)} + \dot{y}^{(1)} + Z e \... ...}}}^{(1)} \cdot \nabla \phi^{(0)} = - L_t^{(1)} - Z e \frac{D \phi^{(1)}}{D t},$$ (97)

where, for notation ease, we have defined

$$- \mathcal{L}_t^{(1)} = - \left( \frac{Z e}{c} \frac{\partial \en... ...ac{\partial B^{(1)}}{\partial t} + Z e \frac{\partial \phi^{(1)}}{\partial t} .$$ (98)

Equation (97) agrees with Eq. (30) in Porcelli's paper[1]. The right-hand side of Eq. (97) provides the desired expression for the coefficent before the $\partial F / \partial \varepsilon_0$ term of Eq. (93).

The first order equation of Eq. (52) is written as (Noting that we are considering toroidal symmetrical equilibrium, thus terms such as $\partial A^{(0)}_R / \partial \varphi$, $\partial b^{(0)}_R / \partial \varphi$, and $\partial B^{(0)} / \partial \varphi$ are all zeros.)

$$\left( \frac{\partial \mathcal{L}}{\partial \varphi} \right)^{(1)... ...B^{(1)}}{\partial \varphi} - Z e \frac{\partial \phi^{(1)}}{\partial \varphi} .$$ (99)

Using this in the Euler equation (43), we obtain

$$\left( \dot{P}_{\varphi} \right)^{(1)} = \left( \frac{\partial \mathcal{L}}{\partial \varphi} \right)^{(1)}$$ (100)

Note that

$$P_{\varphi} = \frac{Z e}{c} A_{\varphi} R + m R v_{\parallel} \frac{B_{\varphi}}{B},$$ (101)

then

$$\dot{P}_{\varphi} = \frac{\partial P_{\varphi}}{\partial t} + \dot{\ensuremath{\bolds... ...rphi} + \dot{v}_{\parallel} \frac{\partial P_{\varphi}}{\partial v_{\parallel}}$$

Using this in Eq. (100), we obtain

$$\left( \frac{\partial P_{\varphi}}{\partial t} + \dot{\ensuremath... ...ht)^{(1)} = \left( \frac{\partial \mathcal{L}}{\partial \varphi} \right)^{(1)},$$ (102)

which can be further written as

$$\dot{\ensuremath{\boldsymbol{X}}}^{(1)} \cdot \nabla P_{\varphi}^... ...rallel}} = \left( \frac{\partial \mathcal{L}}{\partial \varphi} \right)^{(1)} .$$ (103)

$$\Longrightarrow \dot{\ensuremath{\boldsymbol{X}}}^{(1)} \cdot \na... ... \mathcal{L}}{\partial \varphi} \right)^{(1)} - \frac{D P_{\varphi}^{(1)}}{D t}$$ (104)

The right-hand side of Eq. (104) gives desired expression for the coefficient before the term $\partial F / \partial P_{\varphi 0}$ of Eq. (93). The linearized version of Eq. (32)

$$\dot{y} = \mu \left( \frac{\partial B}{\partial t} + \dot{\ensuremath{\boldsymbol{X}}} \cdot \nabla B \right),$$ (105)

is written as

$$\dot{y}^{(1)} = \mu_0 \left( \frac{\partial B_1}{\partial t} + \d... ... \right) + \mu^{(1)} \dot{\ensuremath{\boldsymbol{X}}}^{(0)} \cdot \nabla B_0 .$$ (106)

Noting that $\mu_0 = y / B_0$, $\mu^{(1)} = - (y / B_0^2) B_1$, the above equation is written as

$$\frac{\dot{y}^{(1)}}{B_0} = \frac{y}{B_0} \left( \frac{\partial B... ...0^2} B_1 \dot{\ensuremath{\boldsymbol{X}}}^{(0)} \cdot \frac{1}{B_0} \nabla B_0$$ (107)

$$\frac{\dot{y}^{(1)}}{B_0} - y \frac{1}{B_0^2} \dot{\ensuremath{\b... ...\ensuremath{\boldsymbol{X}}}^{(0)} \cdot \frac{1}{B_0^2} \nabla B^{(0)} \right)$$ (108)

$$\frac{\dot{y}^{(1)}}{B_0} - y \frac{1}{B_0^2} \dot{\ensuremath{\b... ...} \nabla B^{(1)} - B^{(1)} \cdot \frac{1}{B_0^2} \nabla B^{(0)} \right) \right]$$ (109)

$$\frac{\dot{y}^{(1)}}{B_0} - y \frac{1}{B_0^2} \dot{\ensuremath{\b... ...\ensuremath{\boldsymbol{X}}}^{(0)} \cdot \nabla \frac{B^{(1)}}{B^{(0)}} \right)$$ (110)

$$\frac{\dot{y}^{(1)}}{B_0} - y \frac{1}{B_0^2} \dot{\ensuremath{\b... ...}^{(1)} \cdot \nabla B_0 = \mu_0 \frac{D}{D t} \left( \frac{B_1}{B_0} \right) .$$ (111)

Eq. (111) agrees with Eq. (31) in Porceli's paper. The right-hand side of Eq. (111) provide the desired expression for the coefficient before $\partial F / \partial \mu_0$ term in Eq. (93). Using Eqs. (97), (104), and (111), Eq. (93) is finally written as

$$\frac{D f_1}{D t} = - \left\{ \left[ \left( \frac{\partial \mathc... ...D t} \left( \frac{B_1}{B_0} \right) \frac{\partial F}{\partial \mu_0} \right\},$$ (112)