2.1 Introducing the all-orders theory

In this section I will analyse the initial value problem for a toy model I will call two-oscillator drift kinetics. The motivation for studying this relatively simple model is that its initial value problem is structurally very similar to that of drift kinetics. In this first subsection I will give a derivation of the all-orders version of the model, which starts from the following system of two coupled one-dimensional oscillators,

(2.1) $$\begin{eqnarray}\displaystyle & \displaystyle \ddot{q}=-{\it\omega}_{1}^{2}q-2gq{\it\varphi} & \displaystyle\end{eqnarray}$$

(2.2) $$\begin{eqnarray}\displaystyle & \displaystyle \ddot{{\it\varphi}}=-{\it\omega}_{2}^{2}{\it\varphi}-gq^{2}. & \displaystyle\end{eqnarray}$$

$q,{\it\varphi}\in \mathbb{R}$

$q$

${\it\varphi}$

$g,{\it\omega}_{1},{\it\omega}_{2}$

(2.3) $$\begin{eqnarray}\displaystyle L(q,p,{\it\varphi},\dot{q},{\dot{p}},\dot{{\it\varphi}})=L_{p}(q,p,\dot{q},{\dot{p}})+L_{int}(q,{\it\varphi})+L_{f}({\it\varphi},\dot{{\it\varphi}}), & & \displaystyle\end{eqnarray}$$

where

(2.4) $$\begin{eqnarray}\displaystyle & \displaystyle L_{p}=p\,\dot{q}-\left({\textstyle \frac{1}{2}}p^{2}+{\textstyle \frac{1}{2}}{\it\omega}_{1}^{2}q^{2}\right) & \displaystyle\end{eqnarray}$$

(2.5) $$\begin{eqnarray}\displaystyle & \displaystyle L_{f}={\textstyle \frac{1}{2}}\dot{{\it\varphi}}^{2}-{\textstyle \frac{1}{2}}{\it\omega}_{2}^{2}{\it\varphi}^{2} & \displaystyle\end{eqnarray}$$

(2.6) $$\begin{eqnarray}\displaystyle & \displaystyle L_{int}=-gq^{2}{\it\varphi}, & \displaystyle\end{eqnarray}$$

$p$

$L_{p}$

$L_{p}$

$\tilde{L}_{p}=\dot{q}^{2}/2-{\it\omega}_{1}^{2}q^{2}/2$

$S=\int _{t_{1}}^{t_{2}}L\,\text{d}t$

$q,p,$

${\it\varphi}$

$t_{1}$

$t_{2}$

(2.7) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\partial L}{\partial q}-\frac{\text{d}}{\text{d}t}\frac{\partial L}{\partial \dot{q}}=0 & \displaystyle\end{eqnarray}$$

(2.8) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\partial L}{\partial p}-\frac{\text{d}}{\text{d}t}\frac{\partial L}{\partial {\dot{p}}}=0 & \displaystyle\end{eqnarray}$$

(2.9) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\partial L}{\partial {\it\varphi}}-\frac{\text{d}}{\text{d}t}\frac{\partial L}{\partial \dot{{\it\varphi}}}=0, & \displaystyle\end{eqnarray}$$

All-orders two-oscillator drift kinetics is obtained from this ‘particle-space’ model by passing from particle coordinates $(q,p)$ to ‘guiding centre coordinates’ $(Q,P)$ . Just as in drift kinetic theory, the guiding centre coordinates are related to the particle coordinates by a transformation of the particle phase space that depends parametrically on the first $N$ time derivatives of the field ${\it\varphi}$ , where $N$ is some positive integer. Let $j{\it\varphi}=({\it\varphi},\partial _{t}{\it\varphi},\ldots ,\partial _{t}^{N}{\it\varphi})$ and $T_{j{\it\varphi}}:\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ be the gyrocentre coordinate transformation. By definition,

(2.10) $$\begin{eqnarray}\displaystyle (Q,P)=T_{j{\it\varphi}}(q,p)\equiv T(q,p;j{\it\varphi}). & & \displaystyle\end{eqnarray}$$

Note that this transformation can be thought of as passing into a moving frame on the particle phase space; the frame moves as the field ${\it\varphi}$ evolves.

For the sake of concreteness and simplicity, assume $T_{j{\it\varphi}}$ is given by

(2.11) $$\begin{eqnarray}\displaystyle T_{j{\it\varphi}}(q,p)=(q+{\it\epsilon}\ddot{{\it\varphi}},p), & & \displaystyle\end{eqnarray}$$

where ${\it\epsilon}$ is a constant with dimensions of inverse frequency squared. The inverse of $T_{j{\it\varphi}}$ is therefore

(2.12) $$\begin{eqnarray}\displaystyle T_{j{\it\varphi}}^{-1}(Q,P)=(Q-{\it\epsilon}\ddot{{\it\varphi}},P). & & \displaystyle\end{eqnarray}$$

Using these explicit expressions for the guiding centre transformation, I will now determine the differential equations governing all-orders two-oscillator drift kinetics.

Suppose that $q,p,{\it\varphi}$ satisfy the dynamical equations of two-oscillator kinetics. I will find differential equations governing the quantities $Q=q+{\it\epsilon}\ddot{{\it\varphi}}$ , $P=p$ and ${\it\varphi}$ . Using the field equation for two-oscillator kinetics, it is straightforward to see that $Q$ is related to $q$ and ${\it\varphi}$ by

(2.13) $$\begin{eqnarray}\displaystyle Q=q-{\it\epsilon}gq^{2}-{\it\epsilon}{\it\omega}_{2}^{2}{\it\varphi}, & & \displaystyle\end{eqnarray}$$

which also gives a relationship between the time derivatives of $Q,q$ , and ${\it\varphi}$ according to

(2.14) $$\begin{eqnarray}\displaystyle \dot{Q}=(1-2{\it\epsilon}gq)\dot{q}-{\it\epsilon}{\it\omega}_{2}^{2}\dot{{\it\varphi}}. & & \displaystyle\end{eqnarray}$$

Therefore whenever $q<1/2{\it\epsilon}g$ ,

(2.15) $$\begin{eqnarray}\displaystyle \displaystyle q & = & \displaystyle \frac{1}{2{\it\epsilon}g}\left(1-\sqrt{1-4{\it\epsilon}gQ-4{\it\epsilon}^{2}{\it\omega}_{2}^{2}g{\it\varphi}}\right)\nonumber\\ \displaystyle \displaystyle & = & \displaystyle Q+{\it\epsilon}(gQ^{2}+{\it\omega}_{2}^{2}{\it\varphi})+O({\it\epsilon}^{2})\nonumber\\ \displaystyle \displaystyle & = & \displaystyle \hat{q}_{-}(Q,{\it\varphi}),\end{eqnarray}$$

and whenever $q>1/2{\it\epsilon}g$ ,

(2.16) $$\begin{eqnarray}\displaystyle \displaystyle q & = & \displaystyle \frac{1}{2{\it\epsilon}g}\left(1+\sqrt{1-4{\it\epsilon}gQ-4{\it\epsilon}^{2}{\it\omega}_{2}^{2}g{\it\varphi}}\right)\nonumber\\ \displaystyle \displaystyle & = & \displaystyle \frac{1}{{\it\epsilon}g}-Q-{\it\epsilon}(gQ^{2}+{\it\omega}_{2}^{2}{\it\varphi})+O({\it\epsilon}^{2})\nonumber\\ \displaystyle \displaystyle & = & \displaystyle \hat{q}_{+}(Q,{\it\varphi}).\end{eqnarray}$$

Now consider an interval of time when $q<1/2{\it\epsilon}g$ . Using the previous expressions, $P=p$ and $\dot{q}=p$ , it is straightforward to verify that $\dot{Q}$ and ${\dot{P}}$ are given by

(2.17) $$\begin{eqnarray}\displaystyle & \displaystyle {\dot{P}}=-({\it\omega}_{1}^{2}+2g{\it\varphi})\hat{q}_{-}(Q,{\it\varphi}) & \displaystyle\end{eqnarray}$$

(2.18) $$\begin{eqnarray}\displaystyle & \displaystyle \dot{Q}=(1-2{\it\epsilon}g\hat{q}_{-}(Q,{\it\varphi}))P-{\it\epsilon}{\it\omega}_{2}^{2}\dot{{\it\varphi}} & \displaystyle\end{eqnarray}$$

(2.19) $$\begin{eqnarray}\displaystyle & \displaystyle \ddot{{\it\varphi}}=-{\it\omega}_{2}^{2}{\it\varphi}-g\hat{q}_{-}^{2}(Q,{\it\varphi}). & \displaystyle\end{eqnarray}$$

$q<1/2{\it\epsilon}g$

$q>1/2{\it\epsilon}g$

(2.20) $$\begin{eqnarray}\displaystyle & \displaystyle {\dot{P}}=-({\it\omega}_{1}^{2}+2g{\it\varphi})\hat{q}_{+}(Q,{\it\varphi}) & \displaystyle\end{eqnarray}$$

(2.21) $$\begin{eqnarray}\displaystyle & \displaystyle \dot{Q}=(1-2{\it\epsilon}g\hat{q}_{+}(Q,{\it\varphi}))P-{\it\epsilon}{\it\omega}_{2}^{2}\dot{{\it\varphi}} & \displaystyle\end{eqnarray}$$

(2.22) $$\begin{eqnarray}\displaystyle & \displaystyle \ddot{{\it\varphi}}=-{\it\omega}_{2}^{2}{\it\varphi}-g\hat{q}_{+}^{2}(Q,{\it\varphi}). & \displaystyle\end{eqnarray}$$

$(P,Q,{\it\varphi},\dot{{\it\varphi}})$

$(P,Q,{\it\varphi},\dot{{\it\varphi}})$

$(p,q,{\it\varphi},\dot{{\it\varphi}})$

$(Q,{\it\varphi})$

(2.23) $$\begin{eqnarray}\displaystyle D=1-4{\it\epsilon}gQ-4{\it\epsilon}^{2}{\it\omega}_{2}^{2}g{\it\varphi}=0. & & \displaystyle\end{eqnarray}$$

On the ‘top’ sheet $\hat{q}_{+}$ is used in the equations of motion, while on the ‘bottom’ sheet $\hat{q}_{-}$ is used. When the system reaches the cut, one of two things happens. If $P=p=\dot{q}>0$ , then the subsequent evolution of the system occurs on the top sheet. If $P<0$ , then the subsequent evolution of the system occurs on the bottom sheet. With this knowledge of the sheet structure of $(P,Q,{\it\varphi},\dot{{\it\varphi}})$ -space together with our multi-valued dynamical vector field, the evolution equations for all-orders two-oscillator drift kinetics are completely specified.

I will conclude this subsection by presenting the Lagrangian formulation of the all-orders two-oscillator drift kinetic system. The Lagrangian is found by substituting the guiding centre transformation into the two-oscillator kinetics Lagrangian, which gives

(2.24) $$\begin{eqnarray}\displaystyle \bar{L}(Q,P,{\it\varphi},\dot{Q},{\dot{P}},\dot{{\it\varphi}},\ddot{{\it\varphi}},\dddot{{\it\varphi}}) & = & \displaystyle L(Q,P,{\it\varphi},\dot{Q},{\dot{P}},\dot{{\it\varphi}})+{\it\epsilon}(Q\ddot{{\it\varphi}}(2g{\it\varphi}+{\it\omega}_{1}^{2})-P\dddot{{\it\varphi}})\nonumber\\ \displaystyle & & \displaystyle -\,\frac{{\it\epsilon}^{2}}{2}\ddot{{\it\varphi}}^{2}(2g{\it\varphi}+{\it\omega}_{1}^{2}).\end{eqnarray}$$

Because this Lagrangian depends on the third time derivative of ${\it\varphi}$ , its associated Euler–Lagrange equations are given by

(2.25) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\partial \bar{L}}{\partial Q}-\frac{\text{d}}{\text{d}t}\frac{\partial \bar{L}}{\partial \dot{Q}}=0 & \displaystyle\end{eqnarray}$$

(2.26) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\partial \bar{L}}{\partial P}-\frac{\text{d}}{\text{d}t}\frac{\partial \bar{L}}{\partial {\dot{P}}}=0 & \displaystyle\end{eqnarray}$$

(2.27) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\partial \bar{L}}{\partial {\it\varphi}}-\frac{\text{d}}{\text{d}t}\frac{\partial \bar{L}}{\partial \dot{{\it\varphi}}}+\frac{\text{d}^{2}}{\text{d}t^{2}}\frac{\partial \bar{L}}{\partial \ddot{{\it\varphi}}}-\frac{\text{d}^{3}}{\text{d}t^{3}}\frac{\partial \bar{L}}{\partial \dddot{{\it\varphi}}}=0. & \displaystyle\end{eqnarray}$$

(2.28) $$\begin{eqnarray}\displaystyle & \displaystyle -{\dot{P}}-(Q-{\it\epsilon}\ddot{{\it\varphi}})(2g{\it\varphi}+{\it\omega}_{1}^{2})=0 & \displaystyle\end{eqnarray}$$

(2.29) $$\begin{eqnarray}\displaystyle & \displaystyle -P+\dot{Q}-{\it\epsilon}\dddot{{\it\varphi}}=0 & \displaystyle\end{eqnarray}$$

(2.30) $$\begin{eqnarray}\displaystyle & \displaystyle -\ddot{{\it\varphi}}-{\it\omega}_{2}^{2}{\it\varphi}-gQ^{2}+{\it\epsilon}(\dddot{P}+4g\dot{Q}\dot{{\it\varphi}}+4gQ\ddot{{\it\varphi}}+\ddot{Q}(2g{\it\varphi}+{\it\omega}_{1}^{2})) & \displaystyle \nonumber\\ \displaystyle & \displaystyle \hspace{20.00003pt}-\,{\it\epsilon}^{2}(\ddddot{{\it\varphi}}(2g{\it\varphi}+{\it\omega}_{1}^{2})+3g\ddot{{\it\varphi}}^{2}+4g\dot{{\it\varphi}}\dddot{{\it\varphi}})=0. & \displaystyle\end{eqnarray}$$

$\dddot{P}$

$Q$

${\it\varphi}$

${\it\varphi}$

${\it\varphi}$

$\ddot{{\it\varphi}}$

$\ddot{{\it\varphi}}$

$Q$

${\it\varphi}$

$\ddot{{\it\varphi}}$

$P$

$Q$

2.2 Initial value problem for the all-orders model

Because I was able to write the differential equations governing all-orders two-oscillator drift kinetics explicitly, it is not difficult to see that the phase space for all-orders two-oscillator drift kinetics has the same dimension as the underlying particle-space theory, two-oscillator kinetics. In each case, four numbers must be specified at $t=0$ in order to uniquely determine a solution. Interestingly, in passing from particle-space to guiding centre phase space, the phase space for the field-particle system becomes folded, i.e. multi-sheeted.

While it is not simple to gain a complete understanding of the dynamical behaviour predicted by all-orders two-oscillator drift kinetics, it is straightforward to determine the characteristic frequencies of oscillations about the bottom-sheet equilibrium $(Q,P,{\it\varphi},\dot{{\it\varphi}})=0$ . Linearizing equations (2.17)–(2.19) about this equilibrium gives

(2.31) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}{\it\delta}P=-{\it\omega}_{1}^{2}({\it\delta}Q+{\it\epsilon}{\it\omega}_{2}^{2}{\it\delta}{\it\varphi}) & \displaystyle\end{eqnarray}$$

(2.32) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}{\it\delta}Q={\it\delta}P-{\it\epsilon}{\it\omega}_{2}^{2}{\it\delta}\dot{{\it\varphi}} & \displaystyle\end{eqnarray}$$

(2.33) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}{\it\delta}{\it\varphi}={\it\delta}\dot{{\it\varphi}} & \displaystyle\end{eqnarray}$$

(2.34) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}{\it\delta}\dot{{\it\varphi}}=-{\it\omega}_{2}^{2}{\it\delta}{\it\varphi}, & \displaystyle\end{eqnarray}$$

${\dot{z}}=Az$

$z=({\it\delta}P,{\it\delta}Q,{\it\delta}{\it\varphi},{\it\delta}\dot{{\it\varphi}})$

$A$

$4\times 4$

$A$

$\{\text{i}{\it\omega}_{1},-\text{i}{\it\omega}_{1},\text{i}{\it\omega}_{2},-\text{i}{\it\omega}_{2}\}$

2.3 Two oscillator drift kinetics and its initial value problem

Because I did not make any approximations in passing from the underlying particle-space model to the guiding centre model, all-orders two-oscillator drift kinetics is an exact theory. The same cannot be said of ordinary drift kinetic theory. In that theory, the coordinate transformation that relates guiding centre coordinates to particle coordinates is only known as an asymptotic series in ${\it\epsilon}={\it\rho}/L$ . As a result, the system Lagrangian for ordinary drift kinetics is not known exactly, but only asymptotically. In particular, the drift kinetic Lagrangian is always expanded in powers of ${\it\epsilon}$ and then truncated at some order.

For the sake of examining the consequences of this incomplete knowledge of the system Lagrangian in drift kinetic theory, let us now expand the two-oscillator drift kinetic Lagrangian in powers of ${\it\epsilon}$ and then drop terms that are $O({\it\epsilon}^{2})$ or higher, and then re-do the initial value analysis. I will refer to the dynamical system defined by the truncated Lagrangian simply as two-oscillator drift kinetics. The truncated Lagrangian is given by

(2.35) $$\begin{eqnarray}\displaystyle \bar{L}(Q,P,{\it\varphi},\dot{Q},{\dot{P}},\dot{{\it\varphi}},\ddot{{\it\varphi}})=L(Q,P,{\it\varphi},\dot{Q},{\dot{P}},\dot{{\it\varphi}})+{\it\epsilon}(Q\ddot{{\it\varphi}}(2g{\it\varphi}+{\it\omega}_{1}^{2})-P\dddot{{\it\varphi}}), & & \displaystyle\end{eqnarray}$$

which again depends on the third time derivative of ${\it\varphi}$ . The Euler–Lagrange equations are now

(2.36) $$\begin{eqnarray}\displaystyle & \displaystyle -{\dot{P}}-(Q-{\it\epsilon}\ddot{{\it\varphi}})(2g{\it\varphi}+{\it\omega}_{1}^{2})=0 & \displaystyle\end{eqnarray}$$

(2.37) $$\begin{eqnarray}\displaystyle & \displaystyle -P+\dot{Q}-{\it\epsilon}\dddot{{\it\varphi}}=0 & \displaystyle\end{eqnarray}$$

(2.38) $$\begin{eqnarray}\displaystyle & \displaystyle -\ddot{{\it\varphi}}-{\it\omega}_{2}^{2}{\it\varphi}-gQ^{2}+{\it\epsilon}(\dddot{P}+4g\dot{Q}\dot{{\it\varphi}}+4gQ\ddot{{\it\varphi}}+\ddot{Q}(2g{\it\varphi}+{\it\omega}_{1}^{2}))=0. & \displaystyle\end{eqnarray}$$

$\dddot{P}$

$Q$

${\it\varphi}$

${\it\varphi}$

${\it\varphi}$

(2.39) $$\begin{eqnarray}\displaystyle & \displaystyle {\dot{P}}=-(Q-{\it\epsilon}\ddot{{\it\varphi}})(2g{\it\varphi}+{\it\omega}_{1}^{2}) & \displaystyle\end{eqnarray}$$

(2.40) $$\begin{eqnarray}\displaystyle & \displaystyle \dot{Q}=P+{\it\epsilon}\dddot{{\it\varphi}} & \displaystyle\end{eqnarray}$$

(2.41) $$\begin{eqnarray}\displaystyle & \displaystyle \ddddot{{\it\varphi}}=\frac{1}{{\it\epsilon}^{2}(2g{\it\varphi}+{\it\omega}_{1}^{2})}(\ddot{{\it\varphi}}+gQ^{2}+{\it\omega}_{2}^{2}{\it\varphi}-2{\it\epsilon}gQ\ddot{{\it\varphi}}-2{\it\epsilon}^{2}g(\ddot{{\it\varphi}}^{2}+2\dot{{\it\varphi}}\dddot{{\it\varphi}})). & \displaystyle\end{eqnarray}$$

$(P,Q,{\it\varphi},\dot{{\it\varphi}},\ddot{{\it\varphi}},\dddot{{\it\varphi}})$

$(P,Q,{\it\varphi},\dot{{\it\varphi}},\ddot{{\it\varphi}},\dddot{{\it\varphi}})=0$

(2.42) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}{\it\delta}P=-{\it\omega}_{1}^{2}({\it\delta}Q-{\it\epsilon}{\it\delta}\ddot{{\it\varphi}}) & \displaystyle\end{eqnarray}$$

(2.43) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}{\it\delta}Q={\it\delta}P+{\it\epsilon}{\it\delta}\dddot{{\it\varphi}} & \displaystyle\end{eqnarray}$$

(2.44) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}{\it\delta}{\it\varphi}={\it\delta}\dot{{\it\varphi}} & \displaystyle\end{eqnarray}$$

(2.45) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}{\it\delta}\dot{{\it\varphi}}={\it\delta}\ddot{{\it\varphi}} & \displaystyle\end{eqnarray}$$

(2.46) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}{\it\delta}\ddot{{\it\varphi}}={\it\delta}\dddot{{\it\varphi}} & \displaystyle\end{eqnarray}$$

(2.47) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}{\it\delta}\dddot{{\it\varphi}}=\frac{1}{{\it\epsilon}^{2}{\it\omega}_{1}^{2}}({\it\delta}\ddot{{\it\varphi}}+{\it\omega}_{2}^{2}{\it\delta}{\it\varphi}), & \displaystyle\end{eqnarray}$$

${\dot{z}}=Az$

$z=({\it\delta}P,{\it\delta}Q,{\it\delta}{\it\varphi},{\it\delta}\dot{{\it\varphi}},{\it\delta}\ddot{{\it\varphi}},{\it\delta}\dddot{{\it\varphi}})$

$A$

$6\times 6$

$A$

$\{\text{i}{\it\omega}_{1},-\text{i}{\it\omega}_{1},i\tilde{{\it\omega}}_{2},-i\tilde{{\it\omega}}_{2},{\it\gamma}_{\times },-{\it\gamma}_{\times }\}$

(2.48) $$\begin{eqnarray}\displaystyle & \displaystyle \tilde{{\it\omega}}_{2}={\it\omega}_{2}\left(1-{\textstyle \frac{1}{2}}{\it\omega}_{1}^{2}{\it\omega}_{2}^{2}{\it\epsilon}^{2}\right)+O({\it\epsilon}^{3}) & \displaystyle\end{eqnarray}$$

(2.49) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\gamma}_{\times }=\frac{1}{{\it\epsilon}{\it\omega}_{1}}\left(1+{\textstyle \frac{1}{2}}{\it\omega}_{1}^{2}{\it\omega}_{2}^{2}{\it\epsilon}^{2}\right)+O({\it\epsilon}^{3}). & \displaystyle\end{eqnarray}$$

${\it\gamma}_{\times }$

As I will show in more detail in the next section, conventional Lagrangian drift kinetic theory suffers from problems similar to those that I have highlighted in the context of two-oscillator drift kinetics. In both systems, the root cause of the trouble is the combination of (1) employing a coordinate change on the single-particle phase space that depends on time derivatives of the electromagnetic fields and (2) truncating a ‘field $+$ particle’ system Lagrangian.

2.4 Rescuing phase space

This article is not meant to paint a depressing picture of Lagrangian drift kinetics. While the flaw with the conventional Lagrangian approach identified in the previous subsection is important to be aware of, it is equally important to know that the flaw is curable. In this subsection I will describe the cure in the context of two-oscillator drift kinetics.

In order to avoid the possibility of destroying the physical structure of phase space without resorting to the all-orders theory, I advocate truncating the phase-space Lagrangian for the entire two-oscillator drift kinetic system. See appendix A for a discussion of the precise meaning of the term ‘phase-space Lagrangian.’ I will explain why this is preferable to the Lagrangian approach described in the previous subsection by first showing that it works and then giving a brief abstract discussion.

First I will derive the phase-space Lagrangian for all-orders two-oscillator drift kinetics. See appendix A, which is a partial generalization of a discussion found in Marsden, Patrick & Shkoller (Reference Marsden, Patrick and Shkoller1998), for a rigorous justification of the procedure I will use to do this. In the derivation I will restrict attention to the bottom sheet (corresponding to $\hat{q}_{-}$ ); the top sheet can be treated in the same manner. Consider the free-endpoint variation of the all-orders two-oscillator drift kinetic Lagrangian. A simple calculation shows

(2.50) $$\begin{eqnarray}\displaystyle {\it\delta}\int _{t_{1}}^{t_{2}}\bar{L}\,\text{d}t & = & \displaystyle \int _{t_{1}}^{t_{2}}\left(\frac{\partial \bar{L}}{\partial Q}-\frac{\text{d}}{\text{d}t}\frac{\partial \bar{L}}{\partial \dot{Q}}\right){\it\delta}Q\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle +\,\int _{t_{1}}^{t_{2}}\left(\frac{\partial \bar{L}}{\partial P}-\frac{\text{d}}{\text{d}t}\frac{\partial \bar{L}}{\partial {\dot{P}}}\right){\it\delta}P\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle +\,\int _{t_{1}}^{t_{2}}\left(\frac{\partial \bar{L}}{\partial {\it\varphi}}-\frac{\text{d}}{\text{d}t}\frac{\partial \bar{L}}{\partial \dot{{\it\varphi}}}+\frac{\text{d}^{2}}{\text{d}t^{2}}\frac{\partial \bar{L}}{\partial \ddot{{\it\varphi}}}-\frac{\text{d}^{3}}{\text{d}t^{3}}\frac{\partial \bar{L}}{\partial \dddot{{\it\varphi}}}\right){\it\delta}{\it\varphi}\,\text{d}t\nonumber\\ \displaystyle & & \displaystyle +\,[{\it\vartheta}_{Q}{\it\delta}Q+{\it\vartheta}_{{\it\varphi}}{\it\delta}{\it\varphi}+{\it\vartheta}_{\dot{{\it\varphi}}}{\it\delta}\dot{{\it\varphi}}+{\it\vartheta}_{\ddot{{\it\varphi}}}{\it\delta}\ddot{{\it\varphi}}]_{t_{1}}^{t_{2}},\end{eqnarray}$$

where

(2.51) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\vartheta}_{Q}=\frac{\partial \bar{L}}{\partial \dot{Q}} & \displaystyle\end{eqnarray}$$

(2.52) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\vartheta}_{{\it\varphi}}=\frac{\partial \bar{L}}{\partial \dot{{\it\varphi}}}-\frac{\text{d}}{\text{d}t}\frac{\partial \bar{L}}{\partial \ddot{{\it\varphi}}}+\frac{\text{d}^{2}}{\text{d}t^{2}}\frac{\partial \bar{L}}{\partial \dddot{{\it\varphi}}} & \displaystyle\end{eqnarray}$$

(2.53) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\vartheta}_{\dot{{\it\varphi}}}=\frac{\partial \bar{L}}{\partial \ddot{{\it\varphi}}}-\frac{\text{d}}{\text{d}t}\frac{\partial \bar{L}}{\partial \dddot{{\it\varphi}}} & \displaystyle\end{eqnarray}$$

(2.54) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\vartheta}_{\ddot{{\it\varphi}}}=\frac{\partial \bar{L}}{\partial \dddot{{\it\varphi}}}. & \displaystyle\end{eqnarray}$$

${\it\delta}P$

$\bar{L}$

$P$

$1$

$(Q,P,\dot{Q},{\dot{P}},{\it\varphi},\dot{{\it\varphi}},\ddot{{\it\varphi}},\dddot{{\it\varphi}})$

(2.55) $$\begin{eqnarray}\displaystyle {\it\vartheta}^{\infty }={\it\vartheta}_{Q}\,\text{d}Q+{\it\vartheta}_{{\it\varphi}}\,\text{d}{\it\varphi}+{\it\vartheta}_{\dot{{\it\varphi}}}\,\text{d}\dot{{\it\varphi}}+{\it\vartheta}_{\ddot{{\it\varphi}}}\,\text{d}\ddot{{\it\varphi}} & & \displaystyle\end{eqnarray}$$

as the bare all-orders Lagrange $1$ -form. As described in Marsden et al. (Reference Marsden, Patrick and Shkoller1998), if the Lagrangian $\bar{L}$ for all-orders two-oscillator drift kinetics were non-degenerate, then ${\it\vartheta}^{\infty }$ would give the ‘symplectic part’ of the system phase space Lagrangian on $(Q,P,\dot{Q},{\dot{P}},{\it\varphi},\dot{{\it\varphi}},\ddot{{\it\varphi}},\dddot{{\it\varphi}})$ -space. However, because all higher time derivatives of ${\it\varphi}$ than $\ddot{{\it\varphi}}$ cancel out of the Euler–Lagrange equations, $\bar{L}$ is degenerate, and I cannot deduce the phase-space Lagrangian so easily. Moreover, it is a good thing that this technique does not work to determine the symplectic part of the system phase-space Lagrangian because $(Q,P,\dot{Q},{\dot{P}},{\it\varphi},\dot{{\it\varphi}},\ddot{{\it\varphi}},\dddot{{\it\varphi}})$ -space is not the correct phase space for all-orders two-oscillator drift kinetics! The correct symplectic part of the phase-space Lagrangian is given by the dressed Lagrange $1$ -form $\tilde{{\it\vartheta}}^{\infty }$ . Here ‘dressing’ means using the all-orders two-oscillator drift kinetic equations of motion, (2.17)–(2.19), to write the bare Lagrange $1$ -form in terms of only the true phase-space variables $(Q,P,{\it\varphi},\dot{{\it\varphi}})$ .

This calculation proceeds as follows. Formally, the dressed Lagrange 1-form is given by

(2.56) $$\begin{eqnarray}\displaystyle \tilde{{\it\vartheta}}^{\infty }=\tilde{{\it\vartheta}}_{Q}\,\tilde{\text{d}Q}+\tilde{{\it\vartheta}}_{{\it\varphi}}\,\tilde{\text{d}{\it\varphi}}+\tilde{{\it\vartheta}}_{\dot{{\it\varphi}}}\,\tilde{\text{d}\dot{{\it\varphi}}}+\tilde{{\it\vartheta}}_{\ddot{{\it\varphi}}}\,\tilde{\text{d}\ddot{{\it\varphi}}}, & & \displaystyle\end{eqnarray}$$

where the presence of a tilde indicates a dressed quantity. Because the differentials $\text{d}Q,\text{d}{\it\varphi},\text{d}\dot{{\it\varphi}}$ are already expressed as 1-forms on $(Q,P,{\it\varphi},\dot{{\it\varphi}})$ -space, dressing does nothing to them, i.e.

(2.57) $$\begin{eqnarray}\displaystyle & \displaystyle \tilde{\text{d}Q}=\text{d}Q & \displaystyle\end{eqnarray}$$

(2.58) $$\begin{eqnarray}\displaystyle & \displaystyle \tilde{\text{d}{\it\varphi}}=\text{d}{\it\varphi} & \displaystyle\end{eqnarray}$$

(2.59) $$\begin{eqnarray}\displaystyle & \displaystyle \tilde{\text{d}\dot{{\it\varphi}}}=\text{d}\dot{{\it\varphi}}. & \displaystyle\end{eqnarray}$$

$\text{d}\ddot{{\it\varphi}}$

$(Q,P,{\it\varphi},\dot{{\it\varphi}})$

$\tilde{\ddot{{\it\varphi}}}$

(2.60) $$\begin{eqnarray}\displaystyle \tilde{\text{d}\ddot{{\it\varphi}}} & = & \displaystyle -{\it\omega}_{2}^{2}\,\text{d}{\it\varphi}-2g\hat{q}_{-}\text{d}\hat{q}_{-}\nonumber\\ \displaystyle & = & \displaystyle -{\it\omega}_{2}^{2}\,\text{d}{\it\varphi}-\frac{2g\hat{q}_{-}}{\sqrt{D}}(\text{d}Q+{\it\epsilon}{\it\omega}_{2}^{2}\,\text{d}{\it\varphi})\nonumber\\ \displaystyle & = & \displaystyle -{\it\omega}_{2}^{2}\left(1+\frac{1-\sqrt{D}}{\sqrt{D}}\right)\,\text{d}{\it\varphi}-\frac{1-\sqrt{D}}{{\it\epsilon}\sqrt{D}}\,\text{d}Q\nonumber\\ \displaystyle & = & \displaystyle -\frac{{\it\omega}_{2}^{2}}{\sqrt{D}}\,\text{d}{\it\varphi}-\frac{1-\sqrt{D}}{{\it\epsilon}\sqrt{D}}\,\text{d}Q,\end{eqnarray}$$

which is a 1-form on $(Q,P,{\it\varphi},\dot{{\it\varphi}})$ space, as required. Here I have used the identity $2{\it\epsilon}g\hat{q}_{-}=1-\sqrt{D}$ . To dress the components of ${\it\vartheta}^{\infty }$ , the equations of motion must be used in an analogous fashion. Two of the components require no use of the equations of motion whatsoever, namely

(2.61) $$\begin{eqnarray}\displaystyle & \displaystyle \tilde{{\it\vartheta}}_{Q}=P & \displaystyle\end{eqnarray}$$

(2.62) $$\begin{eqnarray}\displaystyle & \displaystyle \tilde{{\it\vartheta}}_{\ddot{{\it\varphi}}}=-{\it\epsilon}P. & \displaystyle\end{eqnarray}$$

(2.63) $$\begin{eqnarray}\displaystyle Q-\hat{q}_{-}={\it\epsilon}\ddot{{\it\varphi}}, & & \displaystyle\end{eqnarray}$$

which follows from a second useful identity,

(2.64) $$\begin{eqnarray}\displaystyle 1-4{\it\epsilon}gQ-4{\it\epsilon}^{2}{\it\omega}_{2}^{2}g{\it\varphi}=D=1-4{\it\epsilon}g\hat{q}_{-}+4{\it\epsilon}^{2}g^{2}\hat{q}_{-}^{2}, & & \displaystyle\end{eqnarray}$$

and (2.19). Applying these identities to $\tilde{{\it\vartheta}}_{{\it\varphi}}$ gives

(2.65) $$\begin{eqnarray}\displaystyle \tilde{{\it\vartheta}}_{{\it\varphi}} & = & \displaystyle \frac{\partial \bar{L}}{\partial \dot{{\it\varphi}}}-\frac{\text{d}}{\text{d}t}\frac{\partial \bar{L}}{\partial \ddot{{\it\varphi}}}+\frac{\text{d}^{2}}{\text{d}t^{2}}\frac{\partial \bar{L}}{\partial \dddot{{\it\varphi}}}\nonumber\\ \displaystyle & = & \displaystyle \dot{{\it\varphi}}-\frac{\text{d}}{\text{d}t}({\it\epsilon}Q(2g{\it\varphi}+{\it\omega}_{1}^{2})-{\it\epsilon}^{2}\ddot{{\it\varphi}}(2g{\it\varphi}+{\it\omega}_{1}^{2}))+\frac{\text{d}}{\text{d}t}(-{\it\epsilon}{\dot{P}})\nonumber\\ \displaystyle & = & \displaystyle \dot{{\it\varphi}}-\frac{\text{d}}{\text{d}t}({\it\epsilon}Q(2g{\it\varphi}+{\it\omega}_{1}^{2})-{\it\epsilon}^{2}\ddot{{\it\varphi}}(2g{\it\varphi}+{\it\omega}_{1}^{2}))+\frac{\text{d}}{\text{d}t}({\it\epsilon}\hat{q}_{-}(2g{\it\varphi}+{\it\omega}_{1}^{2}))\nonumber\\ \displaystyle & = & \displaystyle \dot{{\it\varphi}}-\frac{\text{d}}{\text{d}t}((Q-\hat{q}_{-}-{\it\epsilon}\ddot{{\it\varphi}}){\it\epsilon}(2g{\it\varphi}+{\it\omega}_{1}^{2}))\nonumber\\ \displaystyle & = & \displaystyle \dot{{\it\varphi}}.\end{eqnarray}$$

Similarly, $\tilde{{\it\vartheta}}_{\dot{{\it\varphi}}}$ is given by

(2.66) $$\begin{eqnarray}\displaystyle \tilde{{\it\vartheta}}_{\dot{{\it\varphi}}} & = & \displaystyle \frac{\partial \bar{L}}{\partial \ddot{{\it\varphi}}}-\frac{\text{d}}{\text{d}t}\frac{\partial \bar{L}}{\partial \dddot{{\it\varphi}}}\nonumber\\ \displaystyle & = & \displaystyle ({\it\epsilon}Q(2g{\it\varphi}+{\it\omega}_{1}^{2})-{\it\epsilon}^{2}\ddot{{\it\varphi}}(2g{\it\varphi}+{\it\omega}_{1}^{2}))-(-{\it\epsilon}{\dot{P}})\nonumber\\ \displaystyle & = & \displaystyle (Q-\hat{q}_{-}-{\it\epsilon}\ddot{{\it\varphi}}){\it\epsilon}(2g{\it\varphi}+{\it\omega}_{1}^{2})\nonumber\\ \displaystyle & = & \displaystyle 0.\end{eqnarray}$$

It follows that the dressed Lagrange $1$ -form is given by

(2.67) $$\begin{eqnarray}\displaystyle \tilde{{\it\vartheta}}^{\infty } & = & \displaystyle P\,\text{d}Q+\dot{{\it\varphi}}\,\text{d}{\it\varphi}+0\,\text{d}\dot{{\it\varphi}}+{\it\epsilon}P\left(\frac{{\it\omega}_{2}^{2}}{\sqrt{D}}\,\text{d}{\it\varphi}+\frac{1-\sqrt{D}}{{\it\epsilon}\sqrt{D}}\,\text{d}Q\right)\nonumber\\ \displaystyle & = & \displaystyle P\left(1+\frac{1-\sqrt{D}}{\sqrt{D}}\right)\,\text{d}Q+\left(\dot{{\it\varphi}}+\frac{{\it\epsilon}P{\it\omega}_{2}^{2}}{\sqrt{D}}\right)\,\text{d}{\it\varphi}\nonumber\\ \displaystyle & = & \displaystyle \frac{P}{\sqrt{D(Q,{\it\varphi})}}\,\text{d}Q+\left(\dot{{\it\varphi}}+\frac{{\it\epsilon}{\it\omega}_{2}^{2}P}{\sqrt{D(Q,{\it\varphi})}}\right)\,\text{d}{\it\varphi}.\end{eqnarray}$$

The phase-space action for all-orders two-oscillator drift kinetics is therefore

(2.68) $$\begin{eqnarray}\displaystyle S({\it\gamma})=\int _{{\it\gamma}}\tilde{{\it\vartheta}}^{\infty }-\int _{t_{1}}^{t_{2}}\tilde{H}^{\infty }({\it\gamma}(t))\,\text{d}t, & & \displaystyle\end{eqnarray}$$

where ${\it\gamma}$ is the curve in $(Q,P,{\it\varphi},\dot{{\it\varphi}})$ -space parameterized by the time interval $[t_{1},t_{2}]$ and $\tilde{H}^{\infty }$ is the dressed Hamiltonian. The dressed Hamiltonian is obtained from the usual ‘bare’ Hamiltonian,

(2.69) $$\begin{eqnarray}\displaystyle H(Q,P,\dot{Q},{\dot{P}},{\it\varphi},\dot{{\it\varphi}},\ddot{{\it\varphi}},\dddot{{\it\varphi}}) & = & \displaystyle \frac{\partial \bar{L}}{\partial \dot{Q}}\dot{Q}+\frac{\partial \bar{L}}{\partial {\dot{P}}}{\dot{P}}+\frac{\partial \bar{L}}{\partial \dddot{{\it\varphi}}}\dddot{{\it\varphi}}+\left(\frac{\partial \bar{L}}{\partial \ddot{{\it\varphi}}}-\frac{\text{d}}{\text{d}t}\frac{\partial \bar{L}}{\partial \dddot{{\it\varphi}}}\right)\ddot{{\it\varphi}}\nonumber\\ \displaystyle & & \displaystyle +\,\left(\frac{\partial \bar{L}}{\partial \dot{{\it\varphi}}}-\frac{\text{d}}{\text{d}t}\frac{\partial \bar{L}}{\partial \ddot{{\it\varphi}}}+\frac{\text{d}^{2}}{\text{d}t^{2}}\frac{\partial \bar{L}}{\partial \dddot{{\it\varphi}}}\right)\dot{{\it\varphi}}-\bar{L},\end{eqnarray}$$

by using the all-orders two-oscillator drift kinetic equations of motion (2.20)–(2.22) to express the bare Hamiltonian in terms of the true phase-space variables $(Q,P,{\it\varphi},\dot{{\it\varphi}})$ . A straightforward calculation, similar to the calculation of $\tilde{{\it\vartheta}}^{\infty }$ , shows that the dressed Hamiltonian is given by

(2.70) $$\begin{eqnarray}\displaystyle \tilde{H}^{\infty }(Q,P,{\it\varphi},\dot{{\it\varphi}})={\textstyle \frac{1}{2}}\dot{{\it\varphi}}^{2}+{\textstyle \frac{1}{2}}P^{2}+{\textstyle \frac{1}{2}}{\it\omega}_{2}^{2}{\it\varphi}^{2}+{\textstyle \frac{1}{2}}\hat{q}_{-}^{2}(Q,{\it\varphi})({\it\omega}_{1}^{2}+2g{\it\varphi}). & & \displaystyle\end{eqnarray}$$

By applying fixed-endpoint variations to the phase-space action, it is possible, yet tedious, to verify directly that all-orders two-oscillator drift kinetics admits this alternative variational principle. Indeed, for a general dressed Hamiltonian the Euler–Lagrange equations give

(2.71) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}Q=\sqrt{D}\frac{\partial \tilde{H}^{\infty }}{\partial P}-{\it\epsilon}{\it\omega}_{2}^{2}\frac{\partial \tilde{H}^{\infty }}{\partial \dot{{\it\varphi}}} & \displaystyle\end{eqnarray}$$

(2.72) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}P=-\sqrt{D}\frac{\partial \tilde{H}^{\infty }}{\partial Q} & \displaystyle\end{eqnarray}$$

(2.73) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}{\it\varphi}=\frac{\partial \tilde{H}^{\infty }}{\partial \dot{{\it\varphi}}} & \displaystyle\end{eqnarray}$$

(2.74) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}\dot{{\it\varphi}}=-\frac{\partial \tilde{H}^{\infty }}{\partial {\it\varphi}}+{\it\epsilon}{\it\omega}_{2}^{2}\frac{\partial \tilde{H}^{\infty }}{\partial Q}. & \displaystyle\end{eqnarray}$$

With this phase-space action principle for all-orders two-oscillator drift kinetics in hand, I will now use it to construct an alternative truncated theory that I will refer to as renormalized two-oscillator drift kinetics (I will explain the use of the term ‘renormalized’ shortly). Renormalized two-oscillator drift kinetics is defined simply by replacing $\tilde{{\it\vartheta}}^{\infty }$ and $\tilde{H}^{\infty }$ in (2.68) with the corresponding first-order Taylor expansions in ${\it\epsilon}$ , $\tilde{{\it\vartheta}}$ and $\tilde{H}$ , where

(2.75) $$\begin{eqnarray}\displaystyle & \displaystyle \tilde{{\it\vartheta}}=P(1+{\it\epsilon}2gQ)\,\text{d}Q+(\dot{{\it\varphi}}+{\it\epsilon}P{\it\omega}_{2}^{2})\,\text{d}{\it\varphi} & \displaystyle\end{eqnarray}$$

(2.76) $$\begin{eqnarray}\displaystyle & \displaystyle \tilde{H}={\textstyle \frac{1}{2}}\dot{{\it\varphi}}^{2}+{\textstyle \frac{1}{2}}P^{2}+{\textstyle \frac{1}{2}}{\it\omega}_{2}^{2}{\it\varphi}^{2}+{\textstyle \frac{1}{2}}({\it\omega}_{1}^{2}+2g{\it\varphi})Q^{2} & \displaystyle \nonumber\\ \displaystyle & \displaystyle \hspace{50.00008pt}+\,{\it\epsilon}({\it\omega}_{1}^{2}{\it\omega}_{2}^{2}Q{\it\varphi}+2g{\it\omega}_{2}^{2}Q{\it\varphi}^{2}+[2g^{2}{\it\varphi}+g{\it\omega}_{1}^{2}]Q^{3}). & \displaystyle\end{eqnarray}$$

(2.77) $$\begin{eqnarray}\displaystyle S({\it\gamma})=\int _{{\it\gamma}}\tilde{{\it\vartheta}}-\int _{t_{1}}^{t_{2}}\tilde{H}({\it\gamma}(t))\,\text{d}t, & & \displaystyle\end{eqnarray}$$

and the associated equations of motion are

(2.78) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}Q=\frac{1}{1+2{\it\epsilon}gQ}\left(\frac{\partial \tilde{H}}{\partial P}-{\it\epsilon}{\it\omega}_{2}^{2}\frac{\partial \tilde{H}}{\partial \dot{{\it\varphi}}}\right) & \displaystyle\end{eqnarray}$$

(2.79) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}P=-\frac{1}{1+2{\it\epsilon}gQ}\frac{\partial \tilde{H}}{\partial Q} & \displaystyle\end{eqnarray}$$

(2.80) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}{\it\varphi}=\frac{\partial \tilde{H}}{\partial \dot{{\it\varphi}}} & \displaystyle\end{eqnarray}$$

(2.81) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}\dot{{\it\varphi}}=-\frac{\partial \tilde{H}}{\partial {\it\varphi}}+\frac{{\it\epsilon}{\it\omega}_{2}^{2}}{1+2{\it\epsilon}gQ}\frac{\partial \tilde{H}}{\partial Q}. & \displaystyle\end{eqnarray}$$

$O({\it\epsilon})$

$(Q,P,{\it\varphi},\dot{{\it\varphi}})=0$

(2.82) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}{\it\delta}Q={\it\delta}P-{\it\epsilon}{\it\omega}_{2}^{2}{\it\delta}\dot{{\it\varphi}} & \displaystyle\end{eqnarray}$$

(2.83) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}{\it\delta}P=-{\it\omega}_{1}^{2}({\it\delta}Q+{\it\epsilon}{\it\omega}_{2}^{2}{\it\delta}{\it\varphi}) & \displaystyle\end{eqnarray}$$

(2.84) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}{\it\delta}{\it\varphi}={\it\delta}\dot{{\it\varphi}} & \displaystyle\end{eqnarray}$$

(2.85) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}}{\text{d}t}{\it\delta}\dot{{\it\varphi}}=-{\it\omega}_{2}^{2}(1-{\it\epsilon}^{2}{\it\omega}_{1}^{2}{\it\omega}_{2}^{2}){\it\delta}{\it\varphi}, & \displaystyle\end{eqnarray}$$

${\dot{z}}=Az$

$z=({\it\delta}Q,{\it\delta}P,{\it\delta}{\it\varphi},{\it\delta}\dot{{\it\varphi}})$

$A$

$4\times 4$

$A$

$\{\text{i}{\it\omega}_{1},-\text{i}{\it\omega}_{1},\text{i}{\it\omega}_{2}\sqrt{1-{\it\epsilon}^{2}{\it\omega}_{1}^{2}{\it\omega}_{2}^{2}},-\text{i}{\it\omega}_{2}\sqrt{1-{\it\epsilon}^{2}{\it\omega}_{1}^{2}{\it\omega}_{2}^{2}}\}$

How did this modified Lagrangian procedure succeed in preserving the correct phase space? The answer is that by implementing approximations at the level of the system phase-space Lagrangian it is impossible for the phase space of all-orders two-oscillator drift kinetics to be modified (modulo non-perturbative modifications of the phase-space sheet structure). As long as the $1$ -form appearing in the phase-space action has a non-degenerate exterior derivative, the Euler–Lagrange equations that follow from the phase-space action principle are guaranteed to define a first-order ordinary differential equation (ODE) on the space where the $1$ -form is defined. This non-degeneracy condition is morally guaranteed by the fact that the transformation between the particle coordinates and the guiding centre coordinates is near-identity.

In what sense is renormalized two-oscillator drift kinetics a ‘renormalized’ theory? While I do not believe the renormalization group is playing any role here, the procedure described in this subsection bears a strong resemblance to the procedure used to extract sensible results from perturbative quantum field theories. I began with a system with a nonsensical phase space, namely two-oscillator drift kinetics. In a loose sense, this system contained an unphysical infinity. To eliminate the absurd features of this model, I regularized it by passing to all-orders two-oscillator drift kinetics. This step can be thought of as inserting a ‘cutoff’; I inserted higher-order ${\it\epsilon}$ -dependent terms into the two-oscillator drift kinetic Lagrangian that guaranteed the dimension of phase space was sensible. Finally, I devised a method to vary the cutoff (i.e. change which higher-order terms were in the theory) that would not alter the value of the dimension of phase space. This step is analogous to tailoring the values of the coupling constants in a perturbative quantum field theory (Wentzel Reference Wentzel1949) in order to render the propagator cutoff-independent.

4.1 The all-orders phase-space Lagrangian

The all-orders version of MSST drift kinetics is derived as follows. The starting point is the magnetized Vlasov–Poisson system expressed in a special coordinate system on single-particle phase space,

(4.1) $$\begin{eqnarray}\displaystyle & \displaystyle \partial _{t}f_{s}+\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }([\boldsymbol{u}_{E}+\boldsymbol{w}]f_{s})+{\rm\nabla}_{\boldsymbol{w}}\boldsymbol{\cdot }(\dot{\boldsymbol{w}}_{s}f_{s})=0 & \displaystyle\end{eqnarray}$$

(4.2) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\epsilon}_{o}{\rm\Delta}{\it\varphi}=-{\it\rho}(f) & \displaystyle\end{eqnarray}$$

(4.3) $$\begin{eqnarray}\displaystyle & \displaystyle \dot{\boldsymbol{w}}_{s}=-\partial _{t}\boldsymbol{u}_{E}-\boldsymbol{u}_{E}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}\boldsymbol{u}_{E}-\boldsymbol{w}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}\boldsymbol{u}_{E}+\frac{q_{s}}{m_{s}}\boldsymbol{w}\times \boldsymbol{B}, & \displaystyle\end{eqnarray}$$

${\it\rho}(f)=\sum _{s}q_{s}\int f_{s}\,\text{d}^{3}\boldsymbol{w}$

$\boldsymbol{B}$

$\boldsymbol{w}$

$\boldsymbol{v}$

$\boldsymbol{v}=\boldsymbol{u}_{E}+\boldsymbol{w}$

$f$

$(\boldsymbol{x},\boldsymbol{w})$

$\mathbb{R}^{3}$

${\it\varphi}$

${\it\varphi}$

${\it\Delta}$

$G={\it\Delta}^{-1}$

$G$

$f$

(4.4) $$\begin{eqnarray}\displaystyle & \displaystyle \partial _{t}f_{s} & \displaystyle =-\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }(\dot{\boldsymbol{x}}(f)f_{s})-{\rm\nabla}_{\boldsymbol{w}}\boldsymbol{\cdot }(\dot{\boldsymbol{w}}_{s}(f)f_{s})\end{eqnarray}$$

(4.5) $$\begin{eqnarray}\displaystyle & \displaystyle & \displaystyle \equiv {\dot{f}}_{s}(f),\end{eqnarray}$$

(4.6) $$\begin{eqnarray}\displaystyle & \displaystyle \dot{\boldsymbol{x}}(f)=\boldsymbol{u}_{E}(f)+\boldsymbol{w} & \displaystyle\end{eqnarray}$$

(4.7) $$\begin{eqnarray}\displaystyle & \displaystyle \dot{\boldsymbol{w}}_{s}(f)=-\dot{\boldsymbol{u}}_{E}(f)-\boldsymbol{u}_{E}(f)\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}\boldsymbol{u}_{E}(f)-\boldsymbol{w}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}\boldsymbol{u}_{E}(f)+\frac{q_{s}}{m_{s}}\boldsymbol{w}\times \boldsymbol{B}, & \displaystyle\end{eqnarray}$$

$\boldsymbol{u}_{E}(f)=\boldsymbol{{\mathcal{E}}}(f)\times b/B$

$\dot{\boldsymbol{u}}_{E}(f)=\boldsymbol{{\mathcal{E}}}^{(1)}(f)\times b/B$

(4.8) $$\begin{eqnarray}\displaystyle & \displaystyle \boldsymbol{{\mathcal{E}}}(f)=\frac{1}{{\it\epsilon}_{o}}\boldsymbol{{\rm\nabla}}G[{\it\rho}(f)] & \displaystyle\end{eqnarray}$$

(4.9) $$\begin{eqnarray}\displaystyle & \displaystyle \boldsymbol{{\mathcal{E}}}^{(1)}(f)=-\frac{1}{{\it\epsilon}_{o}}\boldsymbol{{\rm\nabla}}G[\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }(\boldsymbol{u}_{E}(f)\,{\it\rho}(f)+\boldsymbol{J}(f))], & \displaystyle\end{eqnarray}$$

$\boldsymbol{J}(f)=\sum _{s}q_{s}\int \boldsymbol{w}f_{s}\,\text{d}^{3}\boldsymbol{w}$

$\partial _{t}{\it\rho}(f)=-\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }(\boldsymbol{u}_{E}(f){\it\rho}(f)+\boldsymbol{J}(f))$

Next, a near-identity field-dependent phase-space coordinate transformation, ${\it\Phi}_{s}(j\boldsymbol{E}):P\rightarrow P$ , is identified that decouples the rapid gyromotion from the slower drift motion. Here $j\boldsymbol{E}=(\boldsymbol{E},\partial _{t}\boldsymbol{E},\partial _{t}^{2}\boldsymbol{E},\ldots )$ is an infinite sequence of vector fields, and $P$ is the six-dimensional position-velocity phase space. There are many possible ways to choose this coordinate transformation, and MSST drift kinetics is defined by the particular choice described in Miyato et al. (Reference Miyato, Scott, Strintzi and Tokuda2009).

Using this coordinate transformation, the guiding centre distribution function $F$ is defined as

(4.10) $$\begin{eqnarray}\displaystyle F_{s}={\it\Phi}_{s}(j\boldsymbol{E})_{\ast }f_{s}, & & \displaystyle\end{eqnarray}$$

where $f_{s}$ is the particle distribution function, and the subscripted $\ast$ denotes the pushforward. Assuming that $\boldsymbol{E}$ and $f_{s}$ satisfy the magnetized Vlasov–Poisson equations, the right-hand-side of (4.10) can be expressed entirely in terms of $f$ because $j\boldsymbol{E}=(j\boldsymbol{{\mathcal{E}}})(f)=(\boldsymbol{{\mathcal{E}}}(f),\boldsymbol{{\mathcal{E}}}^{(1)}(f),\boldsymbol{{\mathcal{E}}}^{(2)}(f),\ldots )$ , where

(4.11) $$\begin{eqnarray}\displaystyle & \displaystyle \boldsymbol{{\mathcal{E}}}(f)=\frac{1}{{\it\epsilon}_{o}}\boldsymbol{{\rm\nabla}}G[{\it\rho}(f)] & \displaystyle\end{eqnarray}$$

(4.12) $$\begin{eqnarray}\displaystyle & \displaystyle \boldsymbol{{\mathcal{E}}}^{(1)}(f)=-\frac{1}{{\it\epsilon}_{o}}\boldsymbol{{\rm\nabla}}G[\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }(\boldsymbol{u}_{E}(f)\,{\it\rho}(f)+\boldsymbol{J}(f))] & \displaystyle\end{eqnarray}$$

(4.13) $$\begin{eqnarray}\displaystyle & \displaystyle \boldsymbol{{\mathcal{E}}}^{(2)}(f)=\ldots , & \displaystyle\end{eqnarray}$$

${\mathcal{E}}^{(n)}$

$n>1$

$n$

$\boldsymbol{E}$

$f$

(4.14) $$\begin{eqnarray}\displaystyle F_{s} & = & \displaystyle {\it\Phi}_{s}(j\boldsymbol{{\mathcal{E}}}(f))_{\ast }f_{s}\end{eqnarray}$$

(4.15) $$\begin{eqnarray}\displaystyle & \equiv & \displaystyle {\mathcal{F}}_{s}^{-1}(f).\end{eqnarray}$$

${\mathcal{F}}^{-1}$

$P$

$P$

${\mathcal{F}}^{-1}:\oplus _{s}C^{\infty }(P)\rightarrow \oplus _{s}C^{\infty }(P)$

$\oplus _{s}$

Because ${\it\Phi}$ is near-identity, ${\mathcal{F}}_{s}^{-1}(f)\approx f_{s}$ . This implies that ${\mathcal{F}}^{-1}$ has a unique perturbatively defined inverse ${\mathcal{F}}$ that satisfies

(4.16) $$\begin{eqnarray}\displaystyle F_{s}={\it\Phi}_{s}(j\boldsymbol{{\mathcal{E}}}({\mathcal{F}}(F)))_{\ast }{\mathcal{F}}_{s}(F). & & \displaystyle\end{eqnarray}$$

I will explain the procedure for obtaining an asymptotic expansion of ${\mathcal{F}}$ in the next subsection. Therefore,

(4.17) $$\begin{eqnarray}\displaystyle & \displaystyle \partial _{t}{\mathcal{F}}(F)=\partial _{t}f & \displaystyle\end{eqnarray}$$

(4.18) $$\begin{eqnarray}\displaystyle & \displaystyle \Rightarrow D{\mathcal{F}}[\partial _{t}F]={\dot{f}}(f) & \displaystyle\end{eqnarray}$$

(4.19) $$\begin{eqnarray}\displaystyle & \displaystyle \Rightarrow \partial _{t}F=D{\mathcal{F}}^{-1}[{\dot{f}}({\mathcal{F}}(F))]. & \displaystyle\end{eqnarray}$$

$D{\mathcal{F}}$

${\mathcal{F}}$

${\it\Phi}$

$F$

$F$

A variational principle for all-orders MSST drift kinetics can be derived by transforming the following variational principle for the magnetized Vlasov–Poisson system into guiding centre coordinates. The Vlasov–Poisson Lagrangian is given by

(4.20) $$\begin{eqnarray}\mathfrak{L}(f,U,{\it\varphi},\boldsymbol{E})=\mathop{\sum }_{s}\int l_{s}(z,U_{s}(z),{\it\varphi},\boldsymbol{E})\,f_{s}\,\text{d}{\it\omega}+\frac{{\it\epsilon}_{o}}{2}\int |\boldsymbol{{\rm\nabla}}{\it\varphi}|^{2}\,\text{d}^{3}\boldsymbol{x},\end{eqnarray}$$

where the single-particle phase-space Lagrangian is

(4.21) $$\begin{eqnarray}\displaystyle l_{s}(z,{\dot{z}},{\it\varphi},\boldsymbol{E})=(m_{s}\boldsymbol{w}+m_{s}\boldsymbol{u}_{E}+q_{s}\boldsymbol{A})\boldsymbol{\cdot }\dot{\boldsymbol{x}}-H_{s}(z,{\it\varphi},\boldsymbol{E}), & & \displaystyle\end{eqnarray}$$

the single-particle Hamiltonian is

(4.22) $$\begin{eqnarray}\displaystyle H_{s}(z,{\it\varphi},\boldsymbol{E})={\textstyle \frac{1}{2}}m_{s}(\boldsymbol{w}+\boldsymbol{u}_{E})^{2}+q_{s}{\it\varphi}, & & \displaystyle\end{eqnarray}$$

the volume element $\text{d}{\it\omega}=\text{d}^{3}\boldsymbol{x}\,\text{d}^{3}\boldsymbol{w}$ , and $U$ is the Eulerian phase-space fluid velocity. The equations of motion corresponding to this system Lagrangian may be obtained by varying ${\it\varphi}$ , $f$ and $U$ in the action integral $S=\int _{t_{1}}^{t_{2}}L\,\text{d}t$ subject to the constraints

(4.23) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\delta}U_{s}=\dot{{\it\xi}}_{s}+L_{U_{s}}{\it\xi}_{s} & \displaystyle\end{eqnarray}$$

(4.24) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\delta}f_{s}=-\text{div}({\it\xi}_{s}f_{s}), & \displaystyle\end{eqnarray}$$

${\it\xi}$

$L_{U_{s}}$

$U_{s}$

Using the MSST prescription for choosing ${\it\Phi}$ , the transformed system Lagrangian is given by

(4.25) $$\begin{eqnarray}\displaystyle L^{\infty }(F,V,{\it\varphi})=\mathop{\sum }_{s}\int \ell _{s}^{\infty }(z,V_{s}(z),{\it\varphi},j\boldsymbol{E})\,F_{s}\,\text{d}{\it\Omega}_{s}+\frac{{\it\epsilon}_{o}}{2}\int |\boldsymbol{{\rm\nabla}}{\it\varphi}|^{2}\,\text{d}^{3}\boldsymbol{x}, & & \displaystyle\end{eqnarray}$$

where the all-orders guiding centre phase-space Lagrangian is given by

(4.26) $$\begin{eqnarray}\displaystyle \ell _{s}^{\infty }(z,{\dot{z}},{\it\varphi},j\boldsymbol{E})=\left(q_{s}\boldsymbol{A}+m_{s}v_{\Vert }\hat{b}-\frac{m_{s}}{q_{s}}{\it\mu}\boldsymbol{W}\right)\boldsymbol{\cdot }\dot{\boldsymbol{X}}+\frac{q_{s}}{m_{s}}{\it\mu}\,\dot{{\it\zeta}}-h_{s}^{\infty }(z,{\it\varphi},j\boldsymbol{E}), & & \displaystyle\end{eqnarray}$$

and the all-orders guiding centre Hamiltonian is of the form

(4.27) $$\begin{eqnarray}\displaystyle h_{s}^{\infty }(z,{\it\varphi},j\boldsymbol{E})=K_{s}^{\infty }(j\boldsymbol{E})+q_{s}{\it\varphi}, & & \displaystyle\end{eqnarray}$$

where $K_{s}^{\infty }=K_{os}+K_{1s}+K_{2s}+\cdots \,$ and $K_{os},K_{1s}$ are given in (3.4) and (3.5). Note that the MSST method for finding the guiding centre transformation ${\it\Phi}$ produces a guiding centre phase-space Lagrangian with a simple time-independent ‘symplectic part’. The symplectic part of $\ell _{s}^{\infty }$ , $\ell _{s}^{symp}$ , is

(4.28) $$\begin{eqnarray}\displaystyle \ell _{s}^{symp}(z,{\dot{z}}) & = & \displaystyle \ell _{s}^{\infty }+h_{s}^{\infty }\nonumber\\ \displaystyle & = & \displaystyle \left(q_{s}\boldsymbol{A}+m_{s}v_{\Vert }\hat{b}-\frac{m_{s}}{q_{s}}{\it\mu}\boldsymbol{W}\right)\boldsymbol{\cdot }\dot{\boldsymbol{X}}+\frac{q_{s}}{m_{s}}{\it\mu}\dot{{\it\zeta}},\end{eqnarray}$$

and achieving this time-independence was one of the principal goals of Miyato et al. (Reference Miyato, Scott, Strintzi and Tokuda2009). The variational principle associated with $L^{\infty }$ is very similar to the one introduced earlier for first-order MSST drift kinetics. The action $S^{\infty }=\int _{t_{1}}^{t_{2}}L^{\infty }\,\text{d}t$ is varied subject to the constraints given by

(4.29) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\delta}V_{s}=\dot{{\it\xi}}_{s}+L_{V_{s}}{\it\xi}_{s} & \displaystyle\end{eqnarray}$$

(4.30) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\delta}F_{s}=-\frac{1}{{\mathcal{J}}_{s}}\text{div}({\mathcal{J}}_{s}{\it\xi}_{s}F_{s}). & \displaystyle\end{eqnarray}$$

Using this variational principle for all-orders MSST drift kinetics, the all-orders system phase-space Lagrangian can finally be derived using the method described in § 2. The dressed Lagrange $1$ -form and the dressed Hamiltonian must each be found.

The first step toward finding the dressed Lagrange $1$ -form $\tilde{{\it\vartheta}}^{\infty }$ is to identify the bare Lagrange $1$ -form ${\it\vartheta}^{\infty }$ . This is done by examining the endpoint contributions to free-endpoint variations of $S^{\infty }$ . A straightforward calculation shows that the first free-endpoint variation of $S^{\infty }$ about a solution of all-orders MSST drift kinetics is given by

(4.31) $$\begin{eqnarray}\displaystyle {\it\delta}S^{\infty }={\it\vartheta}^{\infty }[{\it\xi},j{\it\delta}\boldsymbol{E}]|_{t_{1}}^{t_{2}}, & & \displaystyle\end{eqnarray}$$

where the bare Lagrange $1$ -form ${\it\vartheta}^{\infty }$ is given by

(4.32) $$\begin{eqnarray}\displaystyle {\it\vartheta}^{\infty }[{\it\xi},{\it\delta}j\boldsymbol{E}] & = & \displaystyle v\mathop{\sum }_{s}\int \ell _{s}^{symp}(z,{\it\xi}_{s}(z))\,F_{s}\,\text{d}{\it\Omega}_{s}\nonumber\\ \displaystyle & & \displaystyle -\,\mathop{\sum }_{k=1}^{\infty }\mathop{\sum }_{j=1}^{k}\int ({\it\delta}\partial _{t}^{k-j}\boldsymbol{E})\,\boldsymbol{\cdot }\left((-\partial _{t})^{j-1}\frac{{\it\delta}{\mathcal{K}}^{\infty }}{{\it\delta}\partial _{t}^{k}\boldsymbol{E}}\right)\,\text{d}^{3}\boldsymbol{X},\end{eqnarray}$$

with ${\mathcal{K}}^{\infty }(F,j\boldsymbol{E})=\sum _{s}\int K_{s}^{\infty }(j\boldsymbol{E})\,F_{s}\,\text{d}{\it\Omega}_{s}$ . In deriving this expression, the all-orders Euler–Lagrange equations,

(4.33) $$\begin{eqnarray}\displaystyle & \displaystyle V_{s}^{\infty i}=\{z^{i},h_{s}^{\infty }\}_{s}^{gc} & \displaystyle\end{eqnarray}$$

(4.34) $$\begin{eqnarray}\displaystyle & \displaystyle \boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }\boldsymbol{D}={\it\rho}(F) & \displaystyle\end{eqnarray}$$

(4.35) $$\begin{eqnarray}\displaystyle & \displaystyle \boldsymbol{D}={\it\epsilon}_{o}\boldsymbol{E}-\frac{{\it\delta}{\mathcal{K}}^{\infty }}{{\it\delta}\boldsymbol{E}}-\mathop{\sum }_{k=1}^{\infty }(-\partial _{t})^{k}\frac{{\it\delta}{\mathcal{K}}^{\infty }}{{\it\delta}\partial _{t}^{k}\boldsymbol{E}}, & \displaystyle\end{eqnarray}$$

${\it\delta}S^{\infty }$

$1$

$\tilde{{\it\vartheta}}$

${\it\vartheta}$

$F$

${\it\xi}$

$\boldsymbol{E}$

$F$

(4.36) $$\begin{eqnarray}\displaystyle j\boldsymbol{E}=j\boldsymbol{{\mathcal{E}}}({\mathcal{F}}(F))\equiv j\tilde{\boldsymbol{{\mathcal{E}}}}(F). & & \displaystyle\end{eqnarray}$$

Similarly, the variations of the time derivatives of $\boldsymbol{E}$ must be expressed in terms of ${\it\xi}$ using

(4.37) $$\begin{eqnarray}\displaystyle \partial _{t}^{k-j}\boldsymbol{E}=\boldsymbol{{\mathcal{E}}}^{(k-j)}({\mathcal{F}}(F))\equiv \tilde{\boldsymbol{{\mathcal{E}}}}^{(k-j)}(F), & & \displaystyle\end{eqnarray}$$

which gives

(4.38) $$\begin{eqnarray}\displaystyle {\it\delta}\partial _{t}^{k-j}\boldsymbol{E} & = & \displaystyle -D\tilde{\boldsymbol{{\mathcal{E}}}}^{(k-j)}[{\mathcal{J}}^{-1}\text{div}({\mathcal{J}}F{\it\xi})]\nonumber\\ \displaystyle & = & \displaystyle -\mathop{\sum }_{s}D_{s}\tilde{\boldsymbol{{\mathcal{E}}}}^{(k-j)}[{\mathcal{J}}_{s}^{-1}\text{div}({\mathcal{J}}_{s}F_{s}{\it\xi}_{s})].\end{eqnarray}$$

Here $D_{s}$ denotes a partial Fréchet derivative with respect to the species- $s$ distribution function. When this dressing procedure is complete, the all-orders dressed Lagrange $1$ -form can be expressed as

(4.39) $$\begin{eqnarray}\displaystyle \tilde{{\it\vartheta}}^{\infty }[{\it\xi}]=\mathop{\sum }_{s}\int \tilde{\ell }_{s}^{symp,\infty }(z,{\it\xi}_{s}(z);F)\,F_{s}\,\text{d}{\it\Omega}_{s}, & & \displaystyle\end{eqnarray}$$

where

(4.40) $$\begin{eqnarray}\displaystyle \tilde{\ell }_{s}^{symp,\infty }(z,{\dot{z}};F)=\ell _{s}^{symp}(z,{\dot{z}})+{\dot{z}}^{i}\partial _{z^{i}}{\it\alpha}_{s}^{\infty }(F), & & \displaystyle\end{eqnarray}$$

and the function ${\it\alpha}_{s}^{\infty }(F):P\rightarrow \mathbb{R}$ is given by

(4.41) $$\begin{eqnarray}\displaystyle {\it\alpha}_{s}^{\infty }(F)=-\mathop{\sum }_{k=1}^{\infty }\mathop{\sum }_{j=1}^{k}\left(D_{s}\tilde{\boldsymbol{{\mathcal{E}}}}^{(k-j)\dagger }(-\partial _{t})^{j-1}\frac{{\it\delta}{\mathcal{K}}^{\infty }}{{\it\delta}\partial _{t}^{k}\boldsymbol{E}}\right). & & \displaystyle\end{eqnarray}$$

Here, $(-\partial _{t})^{j-1}({\it\delta}{\mathcal{K}}/{\it\delta}\partial _{t}^{k}\boldsymbol{E})$ must be expressed in terms of $F$ by evaluating the functional derivative, applying the time derivative, and then using (4.36); and the adjoint operation, which is indicated by $\cdot ^{\dagger }$ , is defined by the following identity,

(4.42) $$\begin{eqnarray}\displaystyle \int {\it\delta}\boldsymbol{E}\boldsymbol{\cdot }D_{s}\tilde{\boldsymbol{{\mathcal{E}}}}^{(k-j)}[{\it\delta}F_{s}]\,\text{d}^{3}\boldsymbol{X}=\int D_{s}\tilde{\boldsymbol{{\mathcal{E}}}}^{(k-j)\dagger }[{\it\delta}\boldsymbol{E}]\,{\it\delta}F_{s}\,\text{d}{\it\Omega}_{s}. & & \displaystyle\end{eqnarray}$$

In terms of the dressed Lagrange $1$ -form, the dressed Hamiltonian is

(4.43) $$\begin{eqnarray}\displaystyle \tilde{{\mathcal{H}}}^{\infty }=\tilde{{\it\vartheta}}^{\infty }[\tilde{V}^{\infty }]-\tilde{L}^{\infty }, & & \displaystyle\end{eqnarray}$$

where $\tilde{V}^{\infty }$ and $\tilde{L}^{\infty }$ are given by dressing equations (4.33) and (4.25) using (4.36). More explicitly, $\tilde{{\mathcal{H}}}^{\infty }$ is given by

(4.44) $$\begin{eqnarray}\displaystyle \tilde{{\mathcal{H}}}(F) & = & \displaystyle \mathop{\sum }_{s}\int (K_{s}^{\infty }(j\tilde{\boldsymbol{{\mathcal{E}}}}(F))+\tilde{V}_{s}^{\infty i}\partial _{z^{i}}{\it\alpha}_{s}^{\infty }(F))\,F_{s}\,\text{d}{\it\Omega}_{s}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{1}{2}\int {\it\epsilon}_{o}\tilde{\boldsymbol{{\mathcal{E}}}}\boldsymbol{\cdot }\tilde{\boldsymbol{{\mathcal{E}}}}\,\text{d}^{3}\boldsymbol{X}-\int \left(\mathop{\sum }_{k=0}^{\infty }(-\partial _{t})^{k}\frac{{\it\delta}{\mathcal{K}}^{\infty }}{{\it\delta}\partial _{t}^{k}\boldsymbol{E}}\right)\boldsymbol{\cdot }\tilde{\boldsymbol{{\mathcal{E}}}}\,\text{d}^{3}\boldsymbol{X}.\end{eqnarray}$$

The all-orders system phase-space Lagrangian is therefore

(4.45) $$\begin{eqnarray}\displaystyle {\mathcal{L}}^{\infty }(F,V)=\tilde{{\it\vartheta}}^{\infty }[V]-\tilde{{\mathcal{H}}}^{\infty }(F). & & \displaystyle\end{eqnarray}$$

When varying this Lagrangian, the now-familiar constraints,

(4.46) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\delta}V_{s}=\dot{{\it\xi}}_{s}+L_{V_{s}}{\it\xi}_{s} & \displaystyle\end{eqnarray}$$

(4.47) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\delta}F_{s}=-\frac{1}{{\mathcal{J}}_{s}}\text{div}({\mathcal{J}}_{s}{\it\xi}_{s}F_{s}), & \displaystyle\end{eqnarray}$$

4.2 The truncated system phase-space Lagrangian

As it has been presented so far, the all-orders system phase-space Lagrangian is a purely formal result. However, I will now show that it can be exploited to build an attractive alternative theory to conventional first-order MSST drift kinetics. This alternative theory, which I will refer to as renormalized first-order MSST drift kinetics, will have the following desirable features: (1) a system phase space with the correct size, (2) first-order (in magnetic non-uniformity) accuracy, and (3) a Lagrangian formulation.

The procedure I will use to derive renormalized MSST drift kinetics mirrors the one I used earlier to derive renormalized two-oscillator drift kinetics. First the all-orders system phase-space Lagrangian must be expanded in powers of ${\it\epsilon}={\it\rho}/L$ , which is formally equivalent to expanding in powers of the inverse elementary charge, $|e|$ (note that $q_{s}=Z_{s}|e|$ , where $Z_{s}$ is some integer). For the sake of achieving first-order accuracy, both the dressed Lagrange $1$ -form and the dressed Hamiltonian will be expanded up to first order if $|e|^{-1}$ .

In order to calculate $\tilde{{\it\vartheta}}$ and $\tilde{{\mathcal{H}}}$ with the desired level of accuracy, it is necessary to have explicit expression for some parts of the operator $j\tilde{\boldsymbol{{\mathcal{E}}}}(F)$ . To be precise, if $\tilde{\boldsymbol{{\mathcal{E}}}}$ and $\tilde{\boldsymbol{{\mathcal{E}}}}^{(1)}$ are expanded in powers of ${\it\epsilon}$ as

(4.48) $$\begin{eqnarray}\displaystyle & \displaystyle \tilde{\boldsymbol{{\mathcal{E}}}}=\tilde{\boldsymbol{{\mathcal{E}}}}_{o}+\tilde{\boldsymbol{{\mathcal{E}}}}_{1}+\cdots & \displaystyle\end{eqnarray}$$

(4.49) $$\begin{eqnarray}\displaystyle & \displaystyle \tilde{\boldsymbol{{\mathcal{E}}}}^{(1)}=\tilde{\boldsymbol{{\mathcal{E}}}}_{o}^{(1)}+\tilde{\boldsymbol{{\mathcal{E}}}}_{1}^{(1)}+\cdots \,, & \displaystyle\end{eqnarray}$$

${\it\epsilon}$

$\tilde{\boldsymbol{{\mathcal{E}}}}_{o}$

$\tilde{\boldsymbol{{\mathcal{E}}}}_{1}$

$\tilde{\boldsymbol{{\mathcal{E}}}}_{o}^{(1)}$

${\mathcal{F}}$

$j\boldsymbol{{\mathcal{E}}}\circ {\mathcal{F}}$

(4.50) $$\begin{eqnarray}\displaystyle \boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }({\it\epsilon}_{o}\tilde{\boldsymbol{{\it\epsilon}}}\boldsymbol{\cdot }\tilde{\boldsymbol{{\mathcal{E}}}}_{o})={\it\rho}(F), & & \displaystyle\end{eqnarray}$$

where the dyad $\tilde{\boldsymbol{{\it\epsilon}}}$ is given by

(4.51) $$\begin{eqnarray}\displaystyle \tilde{\boldsymbol{{\it\epsilon}}}=1+\left(\mathop{\sum }_{s}\frac{c^{2}{\it\mu}_{o}m_{s}N_{s}}{B^{2}}\right)(1-\hat{b}\hat{b}). & & \displaystyle\end{eqnarray}$$

Because $\boldsymbol{{\rm\nabla}}\times \tilde{\boldsymbol{{\mathcal{E}}}}_{o}=0$ , this equation uniquely determines $\tilde{\boldsymbol{{\mathcal{E}}}}_{o}$ . Indeed, if

(4.52) $$\begin{eqnarray}\displaystyle G_{\bot }=({\it\Delta}+{\it\Delta}_{\bot })^{-1}, & & \displaystyle\end{eqnarray}$$

where the negative-definite elliptic operator ${\it\Delta}+{\it\Delta}_{\bot }$ is given by $({\it\Delta}+{\it\Delta}_{\bot }){\it\varphi}=\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }(\tilde{\boldsymbol{{\it\epsilon}}}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\varphi})$ , (4.50) prescribes $\tilde{\boldsymbol{{\mathcal{E}}}}_{o}$ as

(4.53) $$\begin{eqnarray}\displaystyle \tilde{\boldsymbol{{\mathcal{E}}}}_{o}(F)=\frac{1}{{\it\epsilon}_{o}}\boldsymbol{{\rm\nabla}}G_{\bot }[{\it\rho}(F)]. & & \displaystyle\end{eqnarray}$$

Note that because the operator $G_{\bot }$ depends on $F$ through the $N_{s}$ in ${\it\Delta}_{\bot }$ , $\tilde{\boldsymbol{{\mathcal{E}}}}_{o}$ is a nonlinear functional of $F$ . Similarly, a dominant balance of the time derivative of (4.50) gives

(4.54) $$\begin{eqnarray}\displaystyle \tilde{\boldsymbol{{\mathcal{E}}}}_{o}^{(1)}(F)=-\frac{1}{{\it\epsilon}_{o}}\boldsymbol{{\rm\nabla}}G_{\bot }[\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }\text{Im}(F)], & & \displaystyle\end{eqnarray}$$

where

(4.55) $$\begin{eqnarray}\displaystyle \text{Im}(F) & = & \displaystyle \mathop{\sum }_{s}\left(\vphantom{\frac{m_{s}}{B^{2}}}q_{s}\langle \dot{\boldsymbol{X}}_{os}+\dot{\boldsymbol{X}}_{1s}\rangle _{s}N_{s}\right.\nonumber\\ \displaystyle & & \displaystyle \left.-\,\frac{m_{s}}{B^{2}}\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }(\langle \dot{\boldsymbol{X}}_{os}\rangle _{s}N_{s})(1-\hat{b}\hat{b})\boldsymbol{\cdot }\tilde{\boldsymbol{{\mathcal{E}}}}_{o}\right);\end{eqnarray}$$

the angle brackets denote velocity-space averages, $\langle \boldsymbol{Q}\rangle _{s}=N_{s}^{-1}\int \boldsymbol{Q}F_{s}{\mathcal{J}}_{s}\,\text{d}v_{\Vert }\,\text{d}{\it\mu}\,\text{d}{\it\zeta}$ ; and

(4.56) $$\begin{eqnarray}\displaystyle & \displaystyle \dot{\boldsymbol{X}}_{os}=v_{\Vert }\hat{b}+\boldsymbol{u}_{Eo} & \displaystyle\end{eqnarray}$$

(4.57) $$\begin{eqnarray}\displaystyle & \displaystyle \dot{\boldsymbol{X}}_{1s}=\left(\frac{u_{Eo}^{2}}{{\it\omega}_{cs}}{\it\tau}+\frac{2v_{\Vert }}{{\it\omega}_{cs}}\hat{b}\times \boldsymbol{u}_{Eo}\boldsymbol{\cdot }\boldsymbol{{\it\kappa}}-\frac{\boldsymbol{u}_{Eo}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}\times \boldsymbol{u}_{Eo}}{2\,{\it\omega}_{cs}}\right)\hat{b} & \displaystyle \nonumber\\ \displaystyle & \displaystyle +\,\frac{{\it\mu}\,\hat{b}\times \boldsymbol{{\rm\nabla}}B}{q_{s}B}+\frac{v_{\Vert }^{2}\hat{b}\times \boldsymbol{{\it\kappa}}}{{\it\omega}_{cs}}-\frac{\hat{b}\times \boldsymbol{{\rm\nabla}}u_{Eo}^{2}}{2\,{\it\omega}_{cs}}, & \displaystyle\end{eqnarray}$$

$\boldsymbol{u}_{Eo}=\tilde{\boldsymbol{{\mathcal{E}}}}_{o}(F)\times \hat{b}B^{-1}$

$O(1)$

$O({\it\epsilon})$

$\tilde{\boldsymbol{{\mathcal{E}}}}_{1}$

(4.58) $$\begin{eqnarray}\displaystyle \boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }({\it\epsilon}_{o}\tilde{\boldsymbol{{\it\epsilon}}}\boldsymbol{\cdot }\tilde{\boldsymbol{{\mathcal{E}}}}_{1})=-\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }\unicode[STIX]{x1D6D1}_{1}(F), & & \displaystyle\end{eqnarray}$$

where

(4.59) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6D1}_{1}(F) & = & \displaystyle -\frac{{\it\delta}{\mathcal{K}}_{1}}{{\it\delta}\boldsymbol{E}}+\partial _{t}\frac{{\it\delta}{\mathcal{K}}_{1}}{{\it\delta}\dot{\boldsymbol{E}}}\nonumber\\ \displaystyle & = & \displaystyle -\mathop{\sum }_{s}\left(\frac{q_{s}N_{s}}{2{\it\omega}_{cs}^{2}}\dot{\boldsymbol{u}}_{Eo}-\frac{q_{s}}{2{\it\omega}_{cs}^{2}}\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }(\langle \boldsymbol{X}_{os}\rangle _{s}N_{s})\boldsymbol{u}_{Eo}\right)-\frac{{\it\delta}{\mathcal{K}}_{1}}{{\it\delta}\boldsymbol{E}}(\tilde{\boldsymbol{{\mathcal{E}}}}_{o},\tilde{\boldsymbol{{\mathcal{E}}}}_{o}^{(1)}),\end{eqnarray}$$

${\mathcal{K}}_{1}=\sum _{s}\int K_{1s}F_{s}\,\text{d}{\it\Omega}_{s}$

$\dot{\boldsymbol{u}}_{Eo}=\tilde{\boldsymbol{{\mathcal{E}}}}_{o}^{(1)}(F)\times \hat{b}B^{-1}$

(4.60) $$\begin{eqnarray}\displaystyle \tilde{\boldsymbol{{\mathcal{E}}}}_{1}(F)=-\frac{1}{{\it\epsilon}_{o}}\boldsymbol{{\rm\nabla}}G_{\bot }[\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }\unicode[STIX]{x1D6D1}_{1}(F)]. & & \displaystyle\end{eqnarray}$$

Because doing so will lead to considerable simplifications of the renormalized drift kinetic theory later on, it is also useful to introduce the ‘second-order electric field’ $\tilde{\boldsymbol{{\mathcal{E}}}}_{2}(F)$ , which is specified by the equation

(4.61) $$\begin{eqnarray}\displaystyle \boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }({\it\epsilon}_{o}\tilde{\boldsymbol{{\it\epsilon}}}\boldsymbol{\cdot }\tilde{\boldsymbol{{\mathcal{E}}}}_{2}+\unicode[STIX]{x1D6D1}(F)-\unicode[STIX]{x1D6D1}_{1}(F))=0, & & \displaystyle\end{eqnarray}$$

where

(4.62) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6D1}(F)=-\frac{{\it\delta}{\mathcal{K}}_{1}}{{\it\delta}\boldsymbol{E}}(\underline{\boldsymbol{{\mathcal{E}}}},\underline{\boldsymbol{{\mathcal{E}}}}^{(1)})+\frac{{\it\delta}^{2}{\mathcal{K}}_{1}}{{\it\delta}\boldsymbol{E}{\it\delta}\dot{\boldsymbol{E}}}[\underline{\boldsymbol{{\mathcal{E}}}}^{(1)}]+\frac{{\it\delta}^{2}{\mathcal{K}}_{1}}{{\it\delta}F{\it\delta}\dot{\boldsymbol{E}}}(\underline{\boldsymbol{{\mathcal{E}}}})[{\dot{F}}^{\ast }] & & \displaystyle\end{eqnarray}$$

and

(4.63) $$\begin{eqnarray}\displaystyle & \displaystyle {\dot{F}}_{s}^{\ast }=-{\mathcal{J}}_{s}^{-1}\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }({\mathcal{J}}_{s}(\dot{\boldsymbol{X}}_{os}+\dot{\boldsymbol{X}}_{1s})F_{s}) & \displaystyle\end{eqnarray}$$

(4.64) $$\begin{eqnarray}\displaystyle & \displaystyle \underline{\boldsymbol{{\mathcal{E}}}}=\tilde{\boldsymbol{{\mathcal{E}}}}_{o}+\tilde{\boldsymbol{{\mathcal{E}}}}_{1} & \displaystyle\end{eqnarray}$$

(4.65) $$\begin{eqnarray}\displaystyle & \displaystyle \underline{\boldsymbol{{\mathcal{E}}}}^{(1)}=D\underline{\boldsymbol{{\mathcal{E}}}}[{\dot{F}}^{\ast }]. & \displaystyle\end{eqnarray}$$

$\unicode[STIX]{x1D6D1}-\unicode[STIX]{x1D6D1}_{1}=O({\it\epsilon}^{2})$

$\tilde{\boldsymbol{{\mathcal{E}}}}_{2}=O({\it\epsilon}^{2})$

With these expressions for $\tilde{\boldsymbol{{\mathcal{E}}}}_{o}$ , $\tilde{\boldsymbol{{\mathcal{E}}}}_{1}$ and $\tilde{\boldsymbol{{\mathcal{E}}}}_{o}^{(1)}$ in hand, the truncated system phase-space Lagrangian can be found. The truncated dressed Lagrange $1$ -form is given by

(4.66) $$\begin{eqnarray}\displaystyle \tilde{{\it\vartheta}}[{\it\xi}]=\mathop{\sum }_{s}\int \tilde{\ell }_{s}^{symp}(z,{\it\xi}_{s}(z);F)F_{s}\,\text{d}{\it\Omega}_{s}, & & \displaystyle\end{eqnarray}$$

where

(4.67) $$\begin{eqnarray}\displaystyle \tilde{\ell }_{s}^{symp}(z,{\dot{z}};F)=\ell _{s}^{symp}(z,{\dot{z}})+\dot{\boldsymbol{X}}\boldsymbol{\cdot }\boldsymbol{{\rm\nabla}}{\it\alpha}_{s}, & & \displaystyle\end{eqnarray}$$

the function ${\it\alpha}_{s}$ is given by

(4.68) $$\begin{eqnarray}\displaystyle {\it\alpha}_{s} & = & \displaystyle -D_{s}\tilde{\boldsymbol{{\mathcal{E}}}}_{o}^{\dagger }\frac{{\it\delta}{\mathcal{K}}}{{\it\delta}\dot{\boldsymbol{E}}}\nonumber\\ \displaystyle & = & \displaystyle D_{s}\tilde{\boldsymbol{{\mathcal{E}}}}_{o}^{\dagger }\left[\mathop{\sum }_{\bar{s}}\frac{m_{s}N_{s}}{2B^{2}{\it\omega}_{cs}}\tilde{\boldsymbol{{\mathcal{E}}}}_{o}\times \hat{b}\right],\end{eqnarray}$$

and ${\mathcal{K}}={\mathcal{K}}_{o}+{\mathcal{K}}_{1}=\sum _{s}\int (K_{os}+K_{1s})F_{s}\,\text{d}{\it\Omega}_{s}$ . The truncated dressed Hamiltonian is given by

(4.69) $$\begin{eqnarray}\displaystyle \tilde{{\mathcal{H}}}(F)=\mathop{\sum }_{s}\int (\tilde{K}_{s}(F)F_{s}+{\it\alpha}_{s}^{\ast }{\dot{F}}_{s}^{\ast })\,\text{d}{\it\Omega}_{s}+\frac{1}{2}\int {\it\epsilon}_{o}\tilde{\boldsymbol{{\mathcal{E}}}}\boldsymbol{\cdot }\tilde{\boldsymbol{{\mathcal{E}}}}\,\text{d}^{3}\boldsymbol{X}+\int \tilde{\boldsymbol{P}}\boldsymbol{\cdot }\tilde{\boldsymbol{{\mathcal{E}}}}\,\text{d}^{3}\boldsymbol{X}, & & \displaystyle\end{eqnarray}$$

where $\tilde{\boldsymbol{{\mathcal{E}}}}=\tilde{\boldsymbol{{\mathcal{E}}}}_{o}+\tilde{\boldsymbol{{\mathcal{E}}}}_{1}+\tilde{\boldsymbol{{\mathcal{E}}}}_{2}$ ;

(4.70) $$\begin{eqnarray}\displaystyle & \displaystyle \tilde{K}_{s}(F)=K_{os}(\tilde{\boldsymbol{{\mathcal{E}}}})+K_{1s}(\underline{\boldsymbol{{\mathcal{E}}}},\underline{\boldsymbol{{\mathcal{E}}}}^{(1)}); & \displaystyle\end{eqnarray}$$

(4.71) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\alpha}_{s}^{\ast }=-D_{s}\underline{\boldsymbol{{\mathcal{E}}}}^{\dagger }\frac{{\it\delta}{\mathcal{K}}_{1}}{{\it\delta}\dot{\boldsymbol{E}}}(\underline{\boldsymbol{{\mathcal{E}}}}); & \displaystyle\end{eqnarray}$$

$\tilde{\boldsymbol{P}}$

(4.72) $$\begin{eqnarray}\displaystyle \tilde{\boldsymbol{P}}=-\frac{{\it\delta}{\mathcal{K}}_{o}}{{\it\delta}\boldsymbol{E}}(\tilde{\boldsymbol{{\mathcal{E}}}})+\unicode[STIX]{x1D6D1}(F). & & \displaystyle\end{eqnarray}$$

Note that $\boldsymbol{{\rm\nabla}}\boldsymbol{\cdot }({\it\epsilon}_{o}\tilde{\boldsymbol{{\mathcal{E}}}}+\tilde{\boldsymbol{P}})={\it\rho}(F)$ , which can be carefully applied to show that the functional derivative of $\tilde{{\mathcal{H}}}$ is given by

(4.73) $$\begin{eqnarray}\displaystyle \frac{{\it\delta}\tilde{{\mathcal{H}}}}{{\it\delta}F_{s}}=H_{s}+\mathop{\sum }_{\bar{s}}{\it\eta}_{s\bar{s}}^{\ast }[{\dot{F}}_{\bar{s}}^{\ast }]+D_{s}\tilde{\boldsymbol{{\mathcal{E}}}}_{2}^{\dagger }[\unicode[STIX]{x1D6D1}], & & \displaystyle\end{eqnarray}$$

where $H_{s}=q_{s}\tilde{{\it\varphi}}+\tilde{K}_{s}$ , $-\boldsymbol{{\rm\nabla}}\tilde{{\it\varphi}}=\tilde{\boldsymbol{{\mathcal{E}}}}$ , and ${\it\eta}_{s\bar{s}}^{\ast }=D_{s}{\it\alpha}_{\bar{s}}^{\ast \dagger }-D_{\bar{s}}{\it\alpha}_{s}^{\ast }$ . The truncated system phase-space Lagrangian is therefore

(4.74) $$\begin{eqnarray}\displaystyle {\mathcal{L}}(F,V)=\tilde{{\it\vartheta}}[V]-\tilde{{\mathcal{H}}}. & & \displaystyle\end{eqnarray}$$

Upon applying the constrained variations

(4.75) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\delta}V_{s}=\dot{{\it\xi}}_{s}+L_{V_{s}}{\it\xi}_{s} & \displaystyle\end{eqnarray}$$

(4.76) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\delta}F_{s}=-\frac{1}{{\mathcal{J}}_{s}}\text{div}({\mathcal{J}}_{s}{\it\xi}_{s}F_{s}) & \displaystyle\end{eqnarray}$$

(4.77) $$\begin{eqnarray}\displaystyle V_{s}^{i}=\left\{z^{i},{\it\varepsilon}_{s}+\frac{{\it\delta}\tilde{{\mathcal{H}}}}{{\it\delta}F_{s}}\right\}_{s}^{gc}, & & \displaystyle\end{eqnarray}$$

where

(4.78) $$\begin{eqnarray}\displaystyle {\it\varepsilon}_{s} & = & \displaystyle \mathop{\sum }_{\bar{s}}(D_{s}{\it\alpha}_{\bar{s}}^{\dagger }-D_{\bar{s}}{\it\alpha}_{s})[{\mathcal{J}}_{\bar{s}}^{-1}\text{div}({\mathcal{J}}_{\bar{s}}V_{\bar{s}}F_{\bar{s}})]\nonumber\\ \displaystyle & \equiv & \displaystyle \mathop{\sum }_{\bar{s}}{\it\eta}_{s\bar{s}}[{\mathcal{J}}_{\bar{s}}^{-1}\text{div}({\mathcal{J}}_{\bar{s}}V_{\bar{s}}F_{\bar{s}})].\end{eqnarray}$$

Note that because ${\it\varepsilon}_{s}$ depends on $V$ , (4.77) is an implicit equation for $V$ . In order to solve for $V$ , first note that

(4.79) $$\begin{eqnarray}\displaystyle {\it\varepsilon}_{s} & = & \displaystyle \mathop{\sum }_{\bar{s}}{\it\eta}_{s\bar{s}}[\{F_{\bar{s}},{\it\varepsilon}_{\bar{s}}\}_{\bar{s}}^{gc}+\{F_{\bar{s}},{\it\delta}\tilde{{\mathcal{H}}}/{\it\delta}F_{\bar{s}}\}_{\bar{s}}^{gc}]\nonumber\\ \displaystyle & \equiv & \displaystyle \mathop{\sum }_{\bar{s}}{\it\chi}_{s\bar{s}}[{\it\varepsilon}_{\bar{s}}+{\it\delta}\tilde{{\mathcal{H}}}/{\it\delta}F_{\bar{s}}].\end{eqnarray}$$

Therefore,

(4.80) $$\begin{eqnarray}\displaystyle {\mathcal{D}}[{\it\varepsilon}]={\it\chi}[{\it\delta}\tilde{{\mathcal{H}}}/{\it\delta}F], & & \displaystyle\end{eqnarray}$$

where the matrix of operators ${\mathcal{D}}$ is given by

(4.81) $$\begin{eqnarray}\displaystyle ({\mathcal{D}}[{\it\varepsilon}])_{s}=\mathop{\sum }_{\bar{s}}{\mathcal{D}}_{s\bar{s}}[{\it\varepsilon}_{\bar{s}}]=\mathop{\sum }_{\bar{s}}({\it\delta}_{s\bar{s}}-{\it\chi}_{s\bar{s}})[{\it\varepsilon}_{\bar{s}}]. & & \displaystyle\end{eqnarray}$$

In appendix B, I argue that ${\mathcal{D}}$ should be invertible as a mapping from multi-species phase-space functions into itself when ${\it\epsilon}$ is small. Thus,

(4.82) $$\begin{eqnarray}\displaystyle {\it\varepsilon}={\mathcal{D}}^{-1}{\it\chi}[{\it\delta}\tilde{{\mathcal{H}}}/{\it\delta}F], & & \displaystyle\end{eqnarray}$$

which, in conjunction with (4.77), determines the Eulerian phase-space fluid velocity $V$ explicitly,

(4.83) $$\begin{eqnarray}\displaystyle V^{i} & = & \displaystyle \left\{z^{i},{\mathcal{D}}^{-1}{\it\chi}\left[\frac{{\it\delta}\tilde{{\mathcal{H}}}}{{\it\delta}F}\right]+\frac{{\it\delta}\tilde{{\mathcal{H}}}}{{\it\delta}F}\right\}^{gc}\nonumber\\ \displaystyle & = & \displaystyle \left\{z^{i},{\mathcal{D}}^{-1}\frac{{\it\delta}\tilde{{\mathcal{H}}}}{{\it\delta}F}\right\}^{gc}.\end{eqnarray}$$

The time evolution equation for $F$ in renormalized MSST drift kinetics is therefore

(4.84) $$\begin{eqnarray}\displaystyle \partial _{t}F=-\left\{F,{\mathcal{D}}^{-1}\frac{{\it\delta}\tilde{{\mathcal{H}}}}{{\it\delta}F}\right\}^{gc}, & & \displaystyle\end{eqnarray}$$

which is manifestly a first-order ODE on $F$ -space. Recall that the functional derivative ${\it\delta}\tilde{{\mathcal{H}}}/{\it\delta}F$ is given in (4.73).

I will conclude this section by analysing the linearization of renormalized MSST drift kinetics about the same type of slab equilibria used in § 3. Again I will assume the background magnetic field is uniform and $z$ -directed. It is straightforward to show that when $F_{s}$ is given by (3.15) with $N_{os}=n(1+x/L)$ , $\tilde{\boldsymbol{{\mathcal{E}}}}_{o}=\tilde{\boldsymbol{{\mathcal{E}}}}_{1}=\tilde{\boldsymbol{{\mathcal{E}}}}_{o}^{(1)}=0$ , the functional derivatives ${\it\delta}\tilde{{\mathcal{H}}}/{\it\delta}F_{s}=(1/2)m_{s}v_{\Vert }^{2}+{\it\mu}B$ , and ${\it\chi}_{s\bar{s}}[{\it\delta}\tilde{{\mathcal{H}}}/{\it\delta}F_{\bar{s}}]=0$ . Therefore the equilibria from § 3 are indeed equilibria of renormalized MSST drift kinetics. When renormalized MSST drift kinetics is linearized about this type of equilibrium, the dynamical equations governing $z$ -independent fluctuations reduce to the following pair of linear integro-differential equations,

(4.85) $$\begin{eqnarray}\displaystyle & \displaystyle -{\it\epsilon}_{o}({\it\Delta}+{\it\Delta}_{\bot }){\it\delta}\dot{{\it\varphi}}_{1}=\frac{{\it\alpha}}{2}{\rm\nabla}_{\bot }^{2}\partial _{y}{\it\delta}{\it\varphi}_{A}+{\it\gamma}{\rm\nabla}_{\bot }^{4}\partial _{y}{\it\delta}{\it\varphi}_{B} & \displaystyle\end{eqnarray}$$

(4.86) $$\begin{eqnarray}\displaystyle & \displaystyle -{\it\epsilon}_{o}({\it\Delta}+{\it\Delta}_{\bot }){\it\delta}\dot{{\it\varphi}}_{o}=\frac{{\it\alpha}}{2}{\rm\nabla}_{\bot }^{2}\partial _{y}{\it\delta}{\it\varphi}_{B}, & \displaystyle\end{eqnarray}$$

${\it\varphi}_{o}$

${\it\varphi}_{1}$

$0th$

$1st$

(4.87) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\delta}{\it\varphi}_{A}={\it\delta}{\it\varphi}_{o}+{\it\delta}{\it\varphi}_{1}+\frac{{\it\alpha}{\it\beta}}{2}\frac{G_{\bot }}{{\it\epsilon}_{o}}\partial _{y}^{2}{\rm\nabla}_{\bot }^{2}\frac{G_{\bot }}{{\it\epsilon}_{o}}({\it\delta}{\it\varphi}_{1}+{\it\delta}{\it\varphi}_{2}) & \displaystyle \nonumber\\ \displaystyle & \displaystyle \hspace{20.00003pt}-\,\frac{{\it\alpha}{\it\beta}}{2}\left(\frac{G_{\bot }}{{\it\epsilon}_{o}}\right)^{2}{\rm\nabla}_{\bot }^{2}\partial _{y}^{2}({\it\delta}{\it\varphi}_{o}+{\it\delta}{\it\varphi}_{1}-{\it\delta}{\it\varphi}_{2}) & \displaystyle\end{eqnarray}$$

(4.88) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\delta}{\it\varphi}_{B}={\it\delta}{\it\varphi}_{o}+{\it\delta}{\it\varphi}_{1}-{\it\delta}{\it\varphi}_{2} & \displaystyle\end{eqnarray}$$

(4.89) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\delta}{\it\varphi}_{2}=-\frac{{\it\alpha}{\it\beta}}{2}\left(\frac{G_{\bot }}{{\it\epsilon}_{o}}\right)^{2}{\rm\nabla}_{\bot }^{2}\partial _{y}^{2}{\it\delta}{\it\varphi}_{o}. & \displaystyle\end{eqnarray}$$

$k_{x}=0$

(4.90) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\omega}_{d}=-\frac{\displaystyle v_{\ast }k_{y}}{\displaystyle 1+\frac{x}{L}+\frac{v_{Ao}^{2}}{c^{2}}}+O({\it\epsilon}^{2}) & \displaystyle\end{eqnarray}$$

(4.91) $$\begin{eqnarray}\displaystyle & \displaystyle {\it\omega}_{s}=O({\it\epsilon}^{2}). & \displaystyle\end{eqnarray}$$

${\it\omega}_{d}$

${\it\omega}_{\times }$

${\it\omega}_{s}$