2 Potentials, fields and currents

The standard electrostatic zonal flow residual calculation (Rosenbluth & Hinton Reference Rosenbluth and Hinton1998) assumes axisymmetry is preserved during the time evolution of the zonal flow. We seek to generalize this model to its fully electromagnetic counterpart in a tokamak, assuming that the magnetic field remains axisymmetric at all times. Unperturbed quantities are assumed to evolve slowly compared to the zonal flow relaxation.

The total magnetic field is

(2.1) $$\begin{eqnarray}\boldsymbol{B}=\boldsymbol{B}_{0}+\boldsymbol{B}_{1},\end{eqnarray}$$

where

(2.2) $$\begin{eqnarray}\boldsymbol{B}_{0}=I(\unicode[STIX]{x1D713})\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}+\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}\times \unicode[STIX]{x1D735}\unicode[STIX]{x1D713}=B_{0}\boldsymbol{b}\end{eqnarray}$$

is the background axisymmetric magnetic field and $\boldsymbol{B}_{1}$ is the perturbed magnetic field. Here $I=I(\unicode[STIX]{x1D713})$ must be a flux function to make the unperturbed radial current density vanish, $2\unicode[STIX]{x03C0}\unicode[STIX]{x1D713}$ is the unperturbed poloidal flux, $R\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}=\hat{\unicode[STIX]{x1D701}}$ is the toroidal unit vector, with $R$ the major radius, $B_{0}=|\boldsymbol{B}_{0}|$ is the magnitude of the unperturbed magnetic field, and $\boldsymbol{b}=\boldsymbol{B}_{0}/B_{0}$ is the unit vector in the direction of the unperturbed magnetic field.

We start by representing $\boldsymbol{B}_{1}$ in a form convenient to derive the gyroaveraged kinetic equation,

(2.3) $$\begin{eqnarray}\boldsymbol{B}_{1}=\unicode[STIX]{x1D735}\times \boldsymbol{A},\end{eqnarray}$$

where

(2.4) $$\begin{eqnarray}\boldsymbol{A}=A_{\Vert }\boldsymbol{b}+A_{\bot }\boldsymbol{e}_{\bot }+A_{\times }\boldsymbol{e}_{\times }.\end{eqnarray}$$

Here $\boldsymbol{e}_{\bot }=\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}/RB_{p}$ , $\boldsymbol{e}_{\times }=\boldsymbol{b}\times \boldsymbol{e}_{\bot }$ , $|\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}|=RB_{p}$ and $B_{p}$ is the unperturbed poloidal magnetic field. We assume that the characteristic length scale of $\boldsymbol{B}_{1}$ perpendicular to the background magnetic field $\boldsymbol{B}_{0}$ is small compared to the characteristic size of the device. Since we are considering zonal components, the perpendicular gradient is mainly in the radial direction. To describe this rapid radial variation of the perturbed fields, we use the eikonal form

(2.5) $$\begin{eqnarray}\{A_{\Vert },A_{\bot },A_{\times }\}=\{\tilde{A}_{\Vert },\tilde{A}_{\bot },\tilde{A}_{\times }\}\exp [\text{i}S(\unicode[STIX]{x1D713})],\end{eqnarray}$$

with $\unicode[STIX]{x1D735}S=S^{\prime }\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}=k_{\bot }\boldsymbol{e}_{\bot }$ . The coefficients with tilde are functions of time and are only allowed to be slow functions of $\unicode[STIX]{x1D713}$ and $\unicode[STIX]{x1D703}$ , for example, varying as $\cos \unicode[STIX]{x1D703}$ . Using (2.3), (2.4) and (2.5), we obtain

(2.6) $$\begin{eqnarray}\boldsymbol{B}_{1}\simeq B_{\Vert }\boldsymbol{b}-\text{i}k_{\bot }A_{\Vert }\boldsymbol{e}_{\times },\end{eqnarray}$$

where $B_{\Vert }=\text{i}k_{\bot }A_{\times }$ is the parallel component of $\boldsymbol{B}_{1}$ . The form for $\boldsymbol{B}_{1}$ in (2.6) is the most common way to express the perturbed magnetic field in gyrokinetic simulations. Note that the component $A_{\bot }$ of the vector potential never appears in the final expression for $\boldsymbol{B}_{1}$ , and can be safely ignored.

The form for $\boldsymbol{B}_{1}$ in (2.6) ensures that the magnetic field $\boldsymbol{B}_{1}$ is axisymmetric. We can make this more explicit by showing that (2.6) is equivalent to the axisymmetric form

(2.7) $$\begin{eqnarray}\boldsymbol{B}_{1}=\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}-\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}\times \unicode[STIX]{x1D735}A,\end{eqnarray}$$

where $\unicode[STIX]{x1D6FF}(\unicode[STIX]{x1D713},\unicode[STIX]{x1D703})$ , which need not be a flux function, is the perturbation to $I(\unicode[STIX]{x1D713})$ , and $-A(\unicode[STIX]{x1D713},\unicode[STIX]{x1D703})$ is the perturbation to $\unicode[STIX]{x1D713}$ . To match (2.6) and (2.7), we must realize that $A$ has an eikonal form similar to those in (2.5). Thus,

(2.8) $$\begin{eqnarray}\unicode[STIX]{x1D735}A\simeq \frac{\text{i}k_{\bot }A}{RB_{p}}\unicode[STIX]{x1D735}\unicode[STIX]{x1D713},\end{eqnarray}$$

and as a result, equation (2.7) gives

(2.9) $$\begin{eqnarray}\boldsymbol{B}_{1}\simeq \unicode[STIX]{x1D6FF}\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}-\frac{\text{i}k_{\bot }A}{RB_{p}}\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}\times \unicode[STIX]{x1D735}\unicode[STIX]{x1D713}.\end{eqnarray}$$

Equation (2.9) proves that $\boldsymbol{B}_{1}$ is, to lowest order in $(k_{\bot }a)^{-1}\ll 1$ , parallel to the flux surface. Then, it is easy to obtain the relations between $\unicode[STIX]{x1D6FF}$ , $A$ , $B_{\Vert }$ and $A_{\Vert }$ . From (2.6), we deduce that $B_{\Vert }=\boldsymbol{B}_{1}\boldsymbol{\cdot }\boldsymbol{b}$ and $A_{\Vert }=-\boldsymbol{B}_{1}\boldsymbol{\cdot }\boldsymbol{e}_{\times }/\text{i}k_{\bot }$ . Substituting into these two equations the form for $\boldsymbol{B}_{1}$ given in (2.9), and using

(2.10) $$\begin{eqnarray}\boldsymbol{e}_{\times }\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}=-\frac{B_{p}}{RB_{0}}\end{eqnarray}$$

and

(2.11) $$\begin{eqnarray}\boldsymbol{e}_{\times }\boldsymbol{\cdot }(\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}\times \unicode[STIX]{x1D735}\unicode[STIX]{x1D713})=\frac{IB_{p}}{RB_{0}},\end{eqnarray}$$

we obtain

(2.12) $$\begin{eqnarray}B_{\Vert }=\boldsymbol{B}_{1}\boldsymbol{\cdot }\boldsymbol{b}=\frac{I}{R^{2}B_{0}}\unicode[STIX]{x1D6FF}-\frac{\text{i}k_{\bot }B_{p}}{RB_{0}}A\end{eqnarray}$$

and

(2.13) $$\begin{eqnarray}A_{\Vert }=-\frac{1}{\text{i}k_{\bot }}\boldsymbol{B}_{1}\boldsymbol{\cdot }\boldsymbol{e}_{\times }=\frac{B_{p}}{\text{i}k_{\bot }RB_{0}}\unicode[STIX]{x1D6FF}+\frac{I}{R^{2}B_{0}}A.\end{eqnarray}$$

Similarly, from (2.9), we find $\unicode[STIX]{x1D6FF}=R^{2}\boldsymbol{B}_{1}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}$ and $A=-(R/\text{i}k_{\bot }B_{p})\boldsymbol{B}_{1}\boldsymbol{\cdot }(\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}\times \unicode[STIX]{x1D735}\unicode[STIX]{x1D713})$ . Substituting into these equations the form for $\boldsymbol{B}_{1}$ given in (2.6), and using (2.10) and (2.11), we obtain

(2.14) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}=R^{2}\boldsymbol{B}_{1}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}=\frac{I}{B_{0}}B_{\Vert }+\frac{\text{i}k_{\bot }RB_{p}}{B_{0}}A_{\Vert }\end{eqnarray}$$

and

(2.15) $$\begin{eqnarray}A=-\frac{R}{\text{i}k_{\bot }B_{p}}\boldsymbol{B}_{1}\boldsymbol{\cdot }(\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}\times \unicode[STIX]{x1D735}\unicode[STIX]{x1D713})=-\frac{RB_{p}}{\text{i}k_{\bot }B_{0}}B_{\Vert }+\frac{I}{B_{0}}A_{\Vert }.\end{eqnarray}$$

Expressions (2.12)–(2.15) allow us to change from the form most convenient for gyrokinetics, in (2.6), to the axisymmetric form in (2.9).

It is of interest to discuss the possible changes that the magnetic field can undergo. It is possible to change the direction of the magnetic field lines without changing the magnitude of $\boldsymbol{B}$ if $B_{\Vert }=0=I\unicode[STIX]{x1D6FF}-\text{i}k_{\bot }RB_{p}A$ . To avoid changing the direction of the field lines, the perturbation must satisfy $\boldsymbol{B}_{1}\times \boldsymbol{B}_{0}=0$ , or $A_{\Vert }=0=RB_{p}\unicode[STIX]{x1D6FF}+\text{i}k_{\bot }IA$ . Changing the local theta dependent direction of the magnetic field line does not necessarily imply a change in the safety factor $q(\unicode[STIX]{x1D713})=(2\unicode[STIX]{x03C0})^{-1}\int _{0}^{2\unicode[STIX]{x03C0}}(\boldsymbol{B}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}/\boldsymbol{B}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703})\,\text{d}\unicode[STIX]{x1D703}$ . To verify this we write the safety factor as

(2.16) $$\begin{eqnarray}q=q_{0}+q_{1},\end{eqnarray}$$

where

(2.17) $$\begin{eqnarray}q_{0}=\frac{1}{2\unicode[STIX]{x03C0}}\int _{0}^{2\unicode[STIX]{x03C0}}\frac{\boldsymbol{B}_{0}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}}{\boldsymbol{B}_{0}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}}\,\text{d}\unicode[STIX]{x1D703}\end{eqnarray}$$

is the unperturbed safety factor and

(2.18) $$\begin{eqnarray}q_{1}=\frac{1}{2\unicode[STIX]{x03C0}}\int _{0}^{2\unicode[STIX]{x03C0}}\left(\frac{\boldsymbol{B}_{1}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}}{\boldsymbol{B}_{0}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}}-\frac{\boldsymbol{B}_{1}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}}{\boldsymbol{B}_{0}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}}\right)\frac{\boldsymbol{B}_{0}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D701}}{\boldsymbol{B}_{0}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}}\,\text{d}\unicode[STIX]{x1D703}\end{eqnarray}$$

is the result of the perturbation. Note that our poloidal angle-like variable $\unicode[STIX]{x1D703}$ is not changed by the perturbations. Using (2.6) and (2.9), and recalling that $k_{\bot }/RB_{p}$ is independent of $\unicode[STIX]{x1D703}$ , the perturbation $q_{1}$ becomes

(2.19) $$\begin{eqnarray}q_{1}=\frac{1}{2\unicode[STIX]{x03C0}}\frac{\text{i}k_{\bot }}{RB_{p}}\int _{0}^{2\unicode[STIX]{x03C0}}\frac{A_{\Vert }}{\boldsymbol{b}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}}\,\text{d}\unicode[STIX]{x1D703}=\frac{1}{2\unicode[STIX]{x03C0}}\frac{\text{i}k_{\bot }}{RB_{p}}\int _{0}^{2\unicode[STIX]{x03C0}}\left(\frac{B_{p}\unicode[STIX]{x1D6FF}}{\text{i}k_{\bot }R}+\frac{IA}{R^{2}}\right)\frac{\text{d}\unicode[STIX]{x1D703}}{\boldsymbol{B}_{0}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}}.\end{eqnarray}$$

When $q_{1}$ changes, the lines in the flux surface change topology by switching between rational and irrational – the only form of reconnection allowed for axisymmetric perturbations.

3 Kinetic equation and solution

We need to solve the linearized gyrokinetic equation in which the unperturbed quantities are time independent or evolve slowly compared to the zonal flow relaxation. The unperturbed ion distribution function $f_{0}$ is assumed to be Maxwellian:

(3.1) $$\begin{eqnarray}f_{0}=n\left(\frac{M}{2\unicode[STIX]{x03C0}T}\right)^{3/2}\exp \left(-\frac{Mv^{2}}{2T}\right),\end{eqnarray}$$

with $n$ , $M$ and $T$ the ion density, mass and temperature, respectively. Then, as we shall consider a collisionless plasma, the linearized distribution function $f_{1}$ satisfies the Vlasov equation

(3.2) $$\begin{eqnarray}{\dot{f}}_{1}=\frac{\unicode[STIX]{x2202}f_{1}}{\unicode[STIX]{x2202}t}+\boldsymbol{v}\boldsymbol{\cdot }\unicode[STIX]{x1D735}f_{1}+\unicode[STIX]{x1D6FA}\boldsymbol{v}\times \boldsymbol{b}\boldsymbol{\cdot }\unicode[STIX]{x1D735}_{v}f_{1}=\frac{Ze}{T}\boldsymbol{E}_{1}\boldsymbol{\cdot }\boldsymbol{v}f_{0},\end{eqnarray}$$

where $\unicode[STIX]{x1D6FA}=ZeB_{0}/Mc$ with $Ze$ the ion charge and $c$ the speed of light. We have neglected the unperturbed electric field. For the perturbed electric field we use $\boldsymbol{E}_{1}=-\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F7}-c^{-1}\unicode[STIX]{x2202}\boldsymbol{A}/\unicode[STIX]{x2202}t$ , with $\unicode[STIX]{x1D6F7}$ the perturbed electrostatic potential. To remove the adiabatic piece we let $f_{1}=h-(Ze\unicode[STIX]{x1D6F7}/T)f_{0}$ to obtain

(3.3) $$\begin{eqnarray}{\dot{h}}=\frac{\unicode[STIX]{x2202}h}{\unicode[STIX]{x2202}t}+\boldsymbol{v}\boldsymbol{\cdot }\unicode[STIX]{x1D735}h+\unicode[STIX]{x1D6FA}\boldsymbol{v}\times \boldsymbol{b}\boldsymbol{\cdot }\unicode[STIX]{x1D735}_{v}h=\frac{Ze}{T}\left[\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D6F7}}{\unicode[STIX]{x2202}t}-\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\left(\frac{\boldsymbol{v}\boldsymbol{\cdot }\boldsymbol{A}}{c}\right)\right]f_{0}.\end{eqnarray}$$

This is the form that we will use to obtain the desired gyrokinetic equation.

Rather than perform a conventional gyrokinetic treatment (Catto Reference Catto1978) of (3.3), we use the canonical angular momentum $\unicode[STIX]{x1D713}_{\ast }$ for our radial variable when we change variables (Kagan & Catto Reference Kagan and Catto2008). Then using $\boldsymbol{v}=\boldsymbol{v}_{\bot }+v_{\Vert }\boldsymbol{b}$ gives

(3.4) $$\begin{eqnarray}\unicode[STIX]{x1D713}_{\ast }=\unicode[STIX]{x1D713}-\frac{Mc}{Ze}R\hat{\unicode[STIX]{x1D701}}\boldsymbol{\cdot }\boldsymbol{v}=\unicode[STIX]{x1D713}+\frac{1}{\unicode[STIX]{x1D6FA}}\boldsymbol{v}_{\bot }\times \boldsymbol{b}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D713}-\frac{Iv_{\Vert }}{\unicode[STIX]{x1D6FA}}.\end{eqnarray}$$

As for the vector potential in (2.5), we describe the rapid radial variation of the perturbed electrostatic potential $\unicode[STIX]{x1D6F7}$ by the eikonal expression

(3.5) $$\begin{eqnarray}\unicode[STIX]{x1D6F7}=\tilde{\unicode[STIX]{x1D6F7}}\exp [\text{i}S(\unicode[STIX]{x1D713})],\end{eqnarray}$$

where the coefficient $\tilde{\unicode[STIX]{x1D6F7}}$ is a function of time and is only allowed to be a slow function of $\unicode[STIX]{x1D703}$ and $\unicode[STIX]{x1D713}$ . Changing from $\unicode[STIX]{x1D713}$ , $\unicode[STIX]{x1D703}$ , $\unicode[STIX]{x1D701}$ and $\boldsymbol{v}$ variables to $\unicode[STIX]{x1D713}_{\ast }$ , $\unicode[STIX]{x1D703}$ , $\unicode[STIX]{x1D701}$ , $v$ , $\unicode[STIX]{x1D707}=v_{\bot }^{2}/2B_{0}$ , and gyrophase $\unicode[STIX]{x1D711}$ , with $\boldsymbol{v}_{\bot }=v_{\bot }(\boldsymbol{e}_{\bot }\cos \unicode[STIX]{x1D711}+\boldsymbol{e}_{\times }\sin \unicode[STIX]{x1D711})$ , and using $\dot{\unicode[STIX]{x1D713}}_{\ast }=0$ to remove the $\unicode[STIX]{x1D713}_{\ast }$ derivative (Kagan & Catto Reference Kagan and Catto2009) yields the lowest-order gyrokinetic equation

(3.6) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}h}{\unicode[STIX]{x2202}t}+v_{\Vert }\boldsymbol{b}\boldsymbol{\cdot }\unicode[STIX]{x1D735}h=\frac{Zef_{0}}{cT}\left\langle \exp [\text{i}S(\unicode[STIX]{x1D713})]\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}[c\tilde{\unicode[STIX]{x1D6F7}}-v_{\Vert }\tilde{A}_{\Vert }-\boldsymbol{v}_{\bot }\boldsymbol{\cdot }(\boldsymbol{e}_{\bot }\tilde{A}_{\bot }+\boldsymbol{e}_{\times }\tilde{A}_{\times })]\right\rangle _{\unicode[STIX]{x1D711}},\end{eqnarray}$$

where the $\unicode[STIX]{x1D703}$ dependence of $h$ is assumed slow, drift corrections to parallel streaming are neglected as small, and the gyroaverage $\langle \cdots \rangle _{\unicode[STIX]{x1D711}}=(2\unicode[STIX]{x03C0})^{-1}\oint \text{d}\unicode[STIX]{x1D711}(\cdots \,)$ is performed at fixed $\unicode[STIX]{x1D713}_{\ast }$ .

Next we write $h$ in the eikonal form

(3.7) $$\begin{eqnarray}h=h(\unicode[STIX]{x1D713}_{\ast },\unicode[STIX]{x1D703},v,\unicode[STIX]{x1D707},t)=\tilde{h}_{\ast }(\unicode[STIX]{x1D713}_{\ast },\unicode[STIX]{x1D703},v,\unicode[STIX]{x1D707},t)\exp [\text{i}S(\unicode[STIX]{x1D713}_{\ast })],\end{eqnarray}$$

where only a weak $\unicode[STIX]{x1D713}_{\ast }$ dependence of $\tilde{h}_{\ast }$ not captured by $S(\unicode[STIX]{x1D713}_{\ast })$ is allowed, $v=|\boldsymbol{v}|$ , and $\unicode[STIX]{x1D707}=v_{\bot }^{2}/2B_{0}$ . Then, we Taylor expand to obtain $S(\unicode[STIX]{x1D713})-S(\unicode[STIX]{x1D713}_{\ast })=Q-L+\cdots$ , where $L=(k_{\bot }v_{\bot }/\unicode[STIX]{x1D6FA})\sin \unicode[STIX]{x1D711}$ , $Q=S^{\prime }Iv_{\Vert }/\unicode[STIX]{x1D6FA}=k_{\bot }v_{\Vert }/\unicode[STIX]{x1D6FA}_{p}$ and $\unicode[STIX]{x1D6FA}_{p}=ZeB_{p}/Mc=\unicode[STIX]{x1D6FA}B_{p}/B_{0}$ . We retain the order $\unicode[STIX]{x1D716}$ Shafranov shift $\unicode[STIX]{x1D6E5}_{S}$ of the flux surfaces by writing $R=R_{0}(\unicode[STIX]{x1D713})+r(\unicode[STIX]{x1D713})\cos \unicode[STIX]{x1D703}$ with $R_{0}(\unicode[STIX]{x1D713}=0)=R_{0}(0)$ the location of the magnetic axis, $R_{0}(\unicode[STIX]{x1D713})=R_{0}(0)-\unicode[STIX]{x1D6E5}_{S}$ , and $r$ the minor radius for circular flux surfaces, then $\unicode[STIX]{x1D716}=r/R_{0}(\unicode[STIX]{x1D713})$ is a flux function and the ratio of the poloidal over the toroidal magnetic field is $B_{p}/B_{t}=(\unicode[STIX]{x1D716}/q)[1-\unicode[STIX]{x1D6E5}_{S}\cos \unicode[STIX]{x1D703}+O(\unicode[STIX]{x1D716}^{2})]$ with $I=RB_{t}$ . As a result, the lowest order gyrokinetic equation becomes

(3.8) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\tilde{h}_{\ast }}{\unicode[STIX]{x2202}t}+v_{\Vert }\boldsymbol{b}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\tilde{h}_{\ast }=\frac{Zef_{0}}{cT}\left\langle \exp [\text{i}Q-\text{i}L]\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}[c\tilde{\unicode[STIX]{x1D6F7}}-v_{\Vert }\tilde{A}_{\Vert }-\boldsymbol{v}_{\bot }\boldsymbol{\cdot }(\boldsymbol{e}_{\bot }\tilde{A}_{\bot }+\boldsymbol{e}_{\times }\tilde{A}_{\times })]\right\rangle _{\unicode[STIX]{x1D711}}.\end{eqnarray}$$

Performing the gyroaverages we obtain the desired form of the ion gyrokinetic equation

(3.9) $$\begin{eqnarray}\displaystyle \frac{\unicode[STIX]{x2202}\tilde{h}_{\ast }}{\unicode[STIX]{x2202}t}+v_{\Vert }\boldsymbol{b}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\tilde{h}_{\ast }=\frac{Zef_{0}}{cT}\exp (\text{i}Q)\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\left[\text{J}_{0}\left(\frac{k_{\bot }v_{\bot }}{\unicode[STIX]{x1D6FA}}\right)(c\tilde{\unicode[STIX]{x1D6F7}}-v_{\Vert }\tilde{A}_{\Vert })+\frac{2\text{J}_{1}(k_{\bot }v_{\bot }/\unicode[STIX]{x1D6FA})}{k_{\bot }v_{\bot }/\unicode[STIX]{x1D6FA}}\frac{v_{\bot }^{2}}{2\unicode[STIX]{x1D6FA}}\tilde{B}_{\Vert }\right],\hspace{-10.00002pt} & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

where $\tilde{B}_{\Vert }=\text{i}k_{\bot }\tilde{A}_{\times }$ , $\boldsymbol{b}\boldsymbol{\cdot }\unicode[STIX]{x1D735}=(qR)^{-1}\unicode[STIX]{x2202}/\unicode[STIX]{x2202}\unicode[STIX]{x1D703}$ , the coefficient of the $\tilde{A}_{\bot }$ term has gyroaveraged to zero and $\text{J}_{0}$ and $\text{J}_{1}$ denote Bessel functions of the first kind.

To lowest order the streaming term dominates for a weakly collisional plasma. The damping away of any initial $\unicode[STIX]{x1D703}$ dependence leads to the GAM behaviour observed during the early evolution to the final residual zonal flow steady state. We are not interested in the GAM behaviour so we annihilate the $\unicode[STIX]{x1D703}$ derivative in (3.9) to obtain the transit averaged gyrokinetic equation

(3.10) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\overline{\tilde{h}_{\ast }}}{\unicode[STIX]{x2202}t}=\frac{Zef_{0}}{cT}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\overline{\left[\text{J}_{0}\left(\frac{k_{\bot }v_{\bot }}{\unicode[STIX]{x1D6FA}}\right)(c\tilde{\unicode[STIX]{x1D6F7}}-v_{\Vert }\tilde{A}_{\Vert })+\frac{2\text{J}_{1}(k_{\bot }v_{\bot }/\unicode[STIX]{x1D6FA})}{k_{\bot }v_{\bot }/\unicode[STIX]{x1D6FA}}\frac{v_{\bot }^{2}}{2\unicode[STIX]{x1D6FA}}\tilde{B}_{\Vert }\right]\exp (\text{i}Q)},\end{eqnarray}$$

where $\overline{\tilde{h}_{\ast }}$ is the $\unicode[STIX]{x1D703}$ independent long time solution, while $\tilde{h}_{\ast }(t=0)$ is allowed to depend on $\unicode[STIX]{x1D703}$ so we can relate an initial $\unicode[STIX]{x1D703}$ dependent perturbation to the final steady state $\unicode[STIX]{x1D703}$ independent solution. The transit average of any quantity $X$ is defined as $\overline{X}=\oint \text{d}\unicode[STIX]{x1D70F}X/\!\oint \text{d}\unicode[STIX]{x1D70F}$ , with $\text{d}\unicode[STIX]{x1D70F}=\text{d}\unicode[STIX]{x1D703}/(v_{\Vert }\boldsymbol{b}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703})\approx qR\,\text{d}\unicode[STIX]{x1D703}/v_{\Vert }$ . The transit average is over a full bounce for trapped ions and over a complete poloidal circuit for the passing ones. The sign of $v_{\Vert }$ changes at turning points, while $d\unicode[STIX]{x1D70F}>0$ , giving $\overline{v_{\Vert }}=0$ for the trapped. For the passing $\overline{v_{\Vert }}\rightarrow \langle 1/v_{\Vert }\rangle ^{-1}$ in the large aspect ratio limit, where $\langle X\rangle =[\oint X\,\text{d}\unicode[STIX]{x1D703}/(\boldsymbol{B}_{0}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703})][\oint \text{d}\unicode[STIX]{x1D703}/(\boldsymbol{B}_{0}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703})]^{-1}$ denotes a flux surface average. It is easy to show that $\langle \int \text{d}^{3}v\,f_{0}X\rangle =\langle \int \text{d}^{3}v\,f_{0}\overline{X}\rangle$ . More details are given in appendix A.

Equation (3.10) generalizes the usual electrostatic result to include electromagnetic effects through $A_{\Vert }$ and $B_{\Vert }$ . It is important to realize that $\unicode[STIX]{x1D6F7}$ , $A_{\Vert }$ and $B_{\Vert }$ are allowed to be slow functions of $\unicode[STIX]{x1D703}$ and $\unicode[STIX]{x1D713}$ . Solving (3.10) by integrating from $t=0$ gives

(3.11) $$\begin{eqnarray}\displaystyle \overline{\tilde{h}_{\ast }} & = & \displaystyle \overline{\tilde{h}_{\ast }(t=0)}+\frac{Zef_{0}}{cT}\overline{\left[\text{J}_{0}(c\tilde{\unicode[STIX]{x1D6F7}}-v_{\Vert }\tilde{A}_{\Vert })+\frac{2\text{J}_{1}}{z}\frac{v_{\bot }^{2}}{2\unicode[STIX]{x1D6FA}}\tilde{B}_{\Vert }\right]\exp (\text{i}Q)}\nonumber\\ \displaystyle & & \displaystyle -\,\frac{Zef_{0}}{cT}\overline{\left\{\text{J}_{0}[c\tilde{\unicode[STIX]{x1D6F7}}^{(0)}-v_{\Vert }\tilde{A}_{\Vert }^{(0)}]+\frac{2\text{J}_{1}}{z}\frac{v_{\bot }^{2}}{2\unicode[STIX]{x1D6FA}}\tilde{B}_{\Vert }^{(0)}\right\}\exp (\text{i}Q)},\end{eqnarray}$$

where $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}=\tilde{\unicode[STIX]{x1D6F7}}(t=0)$ , $\tilde{A}_{\Vert }^{(0)}=\tilde{A}_{\Vert }(t=0)$ and $\tilde{B}_{\Vert }^{(0)}=\tilde{B}_{\Vert }(t=0)$ are allowed to be flux functions, $\text{J}_{0}\equiv \text{J}_{0}(z)$ and $\text{J}_{1}\equiv \text{J}_{1}(z)$ , and $z=k_{\bot }v_{\bot }/\unicode[STIX]{x1D6FA}$ . The strong poloidal variation in (3.11) is due to $v_{\Vert }$ and $Q\propto v_{\Vert }$ , while the poloidal variation of $B_{0}$ is weaker and sometimes unimportant. The electron response $\overline{\tilde{h}_{e}}$ is given by an equation similar to (3.11), but with $Z\rightarrow -1$ , $M\rightarrow m$ and $f_{0}\rightarrow f_{0e}$ . We will omit species subscripts to streamline notation except where they are needed to avoid confusion.

Next we form the perturbed quasineutrality equation

(3.12) $$\begin{eqnarray}0=\sum Ze\int \text{d}^{3}v\,f_{1},\end{eqnarray}$$

and the two components of Ampère’s law

(3.13) $$\begin{eqnarray}k_{\bot }^{2}A_{\Vert }\simeq (4\unicode[STIX]{x03C0}/c)\boldsymbol{J}_{1}\boldsymbol{\cdot }\boldsymbol{b}=(4\unicode[STIX]{x03C0}/c)\sum Ze\int \text{d}^{3}v\,f_{1}v_{\Vert }\end{eqnarray}$$

and

(3.14) $$\begin{eqnarray}k_{\bot }^{2}A_{\times }\simeq (4\unicode[STIX]{x03C0}/c)\boldsymbol{J}_{1}\boldsymbol{\cdot }\boldsymbol{e}_{\times }=(4\unicode[STIX]{x03C0}/c)\sum Ze\int \text{d}^{3}v\,f_{1}v_{\bot }\sin \unicode[STIX]{x1D711},\end{eqnarray}$$

where the integrals are taken at fixed $\unicode[STIX]{x1D713}$ , $\sum$ denotes a sum over ions and electrons. Notice that $\boldsymbol{J}_{1}\boldsymbol{\cdot }\boldsymbol{e}_{\bot }\simeq 0$ since $\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{J}_{1}=0=\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{A}$ . Using $f_{1}=\tilde{h}_{\ast }\exp [\text{i}S(\unicode[STIX]{x1D713}_{\ast })]-(Zef_{0}/T)\tilde{\unicode[STIX]{x1D6F7}}\exp [\text{i}S(\unicode[STIX]{x1D713})]$ and $\int \text{d}^{3}vh\{1,v_{\Vert },v_{\bot }\sin \unicode[STIX]{x1D711}\}=\int \text{d}^{3}v\tilde{h}_{\ast }\{1,v_{\Vert },v_{\bot }\sin \unicode[STIX]{x1D711}\}\exp [\text{i}(L-Q+S)]=\int \text{d}^{3}v\tilde{h}_{\ast }\{\text{J}_{0},v_{\Vert }\text{J}_{0},\text{i}v_{\bot }\text{J}_{1}\}\exp [\text{i}(S-Q)]$ , equations (3.12)–(3.14) give

(3.15) $$\begin{eqnarray}\displaystyle & \displaystyle \sum Ze\int \text{d}^{3}v\,\tilde{h}_{\ast }\text{J}_{0}\text{e}^{-\text{i}Q}-\sum \frac{Z^{2}\text{e}^{2}n}{T}\tilde{\unicode[STIX]{x1D6F7}}=0, & \displaystyle\end{eqnarray}$$

(3.16) $$\begin{eqnarray}\displaystyle & \displaystyle k_{\bot }^{2}\tilde{A}_{\Vert }=\frac{4\unicode[STIX]{x03C0}}{c}\sum Ze\int \text{d}^{3}v\,\tilde{h}_{\ast }v_{\Vert }\text{J}_{0}\text{e}^{-\text{i}Q} & \displaystyle\end{eqnarray}$$

and

(3.17) $$\begin{eqnarray}\frac{B_{0}\tilde{B}_{\Vert }}{4\unicode[STIX]{x03C0}}+\sum \int \text{d}^{3}v\,\tilde{h}_{\ast }\frac{Mv_{\bot }^{2}}{2}\frac{2\text{J}_{1}}{z}\text{e}^{-\text{i}Q}=0.\end{eqnarray}$$

Note that the perpendicular Ampère’s law (3.14) has become perpendicular pressure balance (3.17).

Equation (3.11) represents the formal solution of an initial value problem for a time scale long compared to the periodic gyromotion about guiding centres. It depends on the initial value $\tilde{h}_{\ast }(0)$ , which we are free to choose arbitrarily as long as it satisfies Maxwell’s equations. Our particular choice for $\tilde{h}_{\ast }(0)$ will be motivated in the next section.

4 Choice of initial condition

We must pick $\tilde{h}_{\ast }$ to satisfy the Maxwell equations at $t=0$ , while also obtaining a convenient and sensible form for $\overline{\tilde{h}_{\ast }}$ to evaluate the long time relaxation behaviour of the zonal flow response. The non-transit averaged initial condition for $\tilde{h}_{\ast }(t=0)$ must depend on $\unicode[STIX]{x1D703}$ as well as $\unicode[STIX]{x1D713}_{\ast }$ , $v$ , $\unicode[STIX]{x1D707}$ to generate a transiently evolving zonal flow (the GAM) or else it will be a homogeneous solution to the non-transit averaged Vlasov equation. However, we do not allow $\tilde{h}_{\ast }(t=0)$ to depend on gyrophase since we assume any initial transient response associated with gyromotion has already damped to its classical polarization level.

We require Maxwell’s equations to be satisfied at $t=0$ , which we take to mean after many gyrations, but much less than the time for a poloidal bounce or transit to be completed. Consequently, at $t=0$ we must satisfy (3.15)–(3.17).

A GAM develops on a time scale of the order of a transit or bounce time and much longer than a gyration period. It oscillates as it damps away to the residual zonal flow level. The initial conditions are sometimes viewed as approximating the turbulent sources (Rosenbluth & Hinton Reference Rosenbluth and Hinton1998; Sugama & Watanabe Reference Sugama and Watanabe2005) of charge and current densities on a time much less than a transit or bounce time, but after many gyrations. When the long time behaviour of the system is studied electrostatically we may use the transit average result

(4.1) $$\begin{eqnarray}f_{1}=\overline{\tilde{h}_{\ast }(t)}\text{e}^{\text{i}S(\unicode[STIX]{x1D713}_{\ast })}-\frac{Zef_{0}}{T}\tilde{\unicode[STIX]{x1D6F7}}\text{e}^{\text{i}S(\unicode[STIX]{x1D713})},\end{eqnarray}$$

where $\overline{\tilde{h}_{\ast }(t)}$ is given by the electrostatic limit of (3.11).

Next we explain how we choose $\tilde{h}_{\ast }(0)$ to include magnetic perturbations. For our results to have the required generality we do not flux surface average quasineutrality and the two components of Ampère’s law. Moreover, we desire forms at $t=0$ in which $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ , $\tilde{A}_{\Vert }^{(0)}$ and $\tilde{B}_{\Vert }^{(0)}$ terms only contribute to quasineutrality, and the parallel and perpendicular components of Ampère’s law, respectively. To avoid the need for complicated velocity space structure, we do not consider arbitrary wavenumbers (Xiao & Catto Reference Xiao and Catto2006b ; Xiao et al. Reference Xiao, Catto and Dorland2007a ). We do manipulations consistent with an expansion in $k_{\bot }v_{\bot }/\unicode[STIX]{x1D6FA}\ll 1$ , but to avoid lengthy expressions, we prefer to keep the finite Larmor radius terms in the form of Bessel functions for a while and make the expansions explicit later. To simplify our treatment and properly recover the $k_{\bot }v_{\bot }/\unicode[STIX]{x1D6FA}\ll Q\ll 1$ limit electrostatically we can use the simple form

(4.2) $$\begin{eqnarray}\displaystyle \tilde{h}_{\ast }(0)=\frac{Zef_{0}\text{e}^{\text{i}Q}}{T\text{J}_{0}}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}, & & \displaystyle\end{eqnarray}$$

where since we are only interested in $k_{\bot }v_{\bot }/\unicode[STIX]{x1D6FA}\ll 1$ the zeroes of the Bessel function are of no concern. A nice discussion of the difference between treating electrostatic zonal flows as an initial value problem rather than as a turbulent source in quasineutrality is presented in § 3 of Monreal et al. (Reference Monreal, Calvo, Sánchez, Parra, Bustos, Könies, Kleiber and Görler2016). Moreover, their paper and the references therein should be consulted to understand the differences between the residual zonal flow behaviour in stellarators and tokamaks.

The preceding expression motivates us to assume

(4.3) $$\begin{eqnarray}\displaystyle \tilde{h}_{\ast }(0)=\frac{Zef_{0}\text{e}^{\text{i}Q}}{T\text{J}_{0}}\left[H\tilde{\unicode[STIX]{x1D6F7}}^{(0)}+\frac{k_{\bot }^{2}c}{\unicode[STIX]{x1D714}_{p}^{2}}v_{\Vert }\tilde{A}_{\Vert }^{(0)}-\frac{2Tu}{ZeB_{0}\unicode[STIX]{x1D6FD}}\tilde{B}_{\Vert }^{(0)}\right], & & \displaystyle\end{eqnarray}$$

where we have introduced the species dependent quantities

(4.4) $$\begin{eqnarray}H=1-\frac{3\unicode[STIX]{x1D6FC}}{2}\left(\frac{Mv^{2}}{3T}-1\right)\end{eqnarray}$$

and

(4.5) $$\begin{eqnarray}\unicode[STIX]{x1D6FC}=\frac{\displaystyle 2\int \text{d}^{3}vf_{0}v_{\bot }^{2}\left[\frac{2\text{J}_{1}}{z\text{J}_{0}}-1\right]}{\displaystyle \int \text{d}^{3}vf_{0}v_{\bot }^{2}\frac{2\text{J}_{1}}{z\text{J}_{0}}\left(\frac{Mv^{2}}{T}-3\right)},\end{eqnarray}$$

with $z=k_{\bot }v_{\bot }/\unicode[STIX]{x1D6FA}$ , $2\text{J}_{1}/z\text{J}_{0}\approx 1+z^{2}/8$ giving $\unicode[STIX]{x1D6FC}\approx k_{\bot }^{2}T/2M\unicode[STIX]{x1D6FA}^{2}\ll 1$ for $z\ll 1$ , and the species independent quantity

(4.6) $$\begin{eqnarray}u=\frac{\sum nT}{\displaystyle \sum \int \text{d}^{3}vf_{0}Mv_{\bot }^{2}\frac{\text{J}_{1}}{z\text{J}_{0}}}.\end{eqnarray}$$

In the preceding $\unicode[STIX]{x1D6FD}=8\unicode[STIX]{x03C0}\sum nT/B_{0}^{2}=8\unicode[STIX]{x03C0}(n_{i}T_{i}+n_{e}T_{e})/B_{0}^{2}=\unicode[STIX]{x1D6FD}_{i}+\unicode[STIX]{x1D6FD}_{e}$ and $\unicode[STIX]{x1D714}_{p}^{2}=4\unicode[STIX]{x03C0}\sum Z^{2}\text{e}^{2}n/M=4\unicode[STIX]{x03C0}\text{e}^{2}[(Z^{2}n_{i}/M)+(n_{e}/m)]=\unicode[STIX]{x1D714}_{pi}^{2}+\unicode[STIX]{x1D714}_{pe}^{2}$ . We will also make use of the definitions $v_{i}=(2T_{i}/M)^{1/2}$ , $\unicode[STIX]{x1D70C}_{i}=v_{i}/\unicode[STIX]{x1D6FA}_{i}$ , $\unicode[STIX]{x1D6FA}_{i}=ZeB_{0}/(Mc)$ and $\unicode[STIX]{x1D70C}_{pi}=q\unicode[STIX]{x1D70C}_{i}/\unicode[STIX]{x1D716}$ .

The functional form of $H$ in (4.3) is chosen so at $t=0$ it does not alter quasineutrality (3.15) since

(4.7) $$\begin{eqnarray}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}\sum \frac{Z^{2}\text{e}^{2}}{T}n=\tilde{\unicode[STIX]{x1D6F7}}^{(0)}\sum \frac{Z^{2}\text{e}^{2}}{T}\int \text{d}^{3}vf_{0}H=\tilde{\unicode[STIX]{x1D6F7}}^{(0)}\sum \frac{Z^{2}\text{e}^{2}}{T}\int \text{d}^{3}vf_{0}.\end{eqnarray}$$

Moreover, the $\unicode[STIX]{x1D6FC}$ in $H$ is chosen so $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ will not enter perpendicular Ampère’s law by taking

(4.8) $$\begin{eqnarray}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}\sum \frac{ZeM}{2T}\int \text{d}^{3}vf_{0}v_{\bot }^{2}\left[H\frac{2\text{J}_{1}}{z\text{J}_{0}}-1\right]=0.\end{eqnarray}$$

The $\tilde{A}_{\Vert }^{(0)}$ term does not enter quasineutrality because its integral is odd in $v_{\Vert }$ , and $\tilde{B}_{\Vert }^{(0)}$ does not enter because $u$ and $\unicode[STIX]{x1D6FD}$ are species independent as we can use unperturbed quasineutrality. We can also see that parallel Ampère’s law, equation (3.16), is satisfied at $t=0$ since the $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ and $\tilde{B}_{\Vert }^{(0)}$ integrals are odd in $v_{\Vert }$ leaving

(4.9) $$\begin{eqnarray}\tilde{A}_{\Vert }^{(0)}=\frac{4\unicode[STIX]{x03C0}}{\unicode[STIX]{x1D714}_{p}^{2}}\tilde{A}_{\Vert }^{(0)}\sum \frac{Z^{2}\text{e}^{2}}{T}\int \text{d}^{3}vf_{0}v_{\Vert }^{2}=\frac{\tilde{A}_{\Vert }^{(0)}}{\unicode[STIX]{x1D714}_{p}^{2}}\sum \frac{4\unicode[STIX]{x03C0}Z^{2}\text{e}^{2}n}{M}.\end{eqnarray}$$

To satisfy the perpendicular Ampère’s law, equation (3.17), at $t=0$ we require (4.4)–(4.6) to be satisfied as can be seen from

(4.10) $$\begin{eqnarray}B_{0}\tilde{B}_{\Vert }^{(0)}=-4\unicode[STIX]{x03C0}\sum \int \text{d}^{3}vf_{0}Mv_{\bot }^{2}(\text{J}_{1}/z\text{J}_{0})\left[\frac{Ze}{T}H\tilde{\unicode[STIX]{x1D6F7}}^{(0)}-\frac{2u}{B_{0}\unicode[STIX]{x1D6FD}}\tilde{B}_{\Vert }^{(0)}\right]=\tilde{B}_{\Vert }^{(0)}\frac{u}{\unicode[STIX]{x1D6FD}}\frac{8\unicode[STIX]{x03C0}\sum nT}{B_{0}u},\end{eqnarray}$$

where we use unperturbed quasineutrality as well as $\int \text{d}^{3}vf_{0}v_{\bot }^{2}[(2\text{J}_{1}H/z\text{J}_{0})-1]=0$ .

Apart from the need to satisfy Maxwell’s equations, $\tilde{h}_{\ast }(t=0)$ is arbitrary. Other choices for $\tilde{h}_{\ast }(t=0)$ may be used, but (4.3) is sufficient for our purpose. It has the transit average

(4.11) $$\begin{eqnarray}\displaystyle \overline{\tilde{h}_{\ast }(0)}=\frac{Zef_{0}}{T}\left[H\unicode[STIX]{x1D6F7}^{(0)}\overline{\text{e}^{\text{i}Q}/\text{J}_{0}}-\frac{2Tu}{ZeB_{0}^{2}\unicode[STIX]{x1D6FD}}\tilde{B}_{\Vert }^{(0)}\overline{B_{0}\text{e}^{\text{i}Q}/\text{J}_{0}}+\frac{k_{\bot }^{2}c}{\unicode[STIX]{x1D714}_{p}^{2}}\tilde{A}_{\Vert }^{(0)}\overline{v_{\Vert }\text{e}^{\text{i}Q}/\text{J}_{0}}\right], & & \displaystyle\end{eqnarray}$$

where the weak $\unicode[STIX]{x1D703}$ dependences of $u$ and $\unicode[STIX]{x1D6FC}$ are neglected.

Before closing this section we prove that $\tilde{\unicode[STIX]{x1D6F7}}$ is a lowest order flux function for fully electromagnetic initial conditions. For this demonstration we will only be interested in the $L\sim k_{\bot }v_{\bot }/\unicode[STIX]{x1D6FA}\ll Q\sim k_{\bot }v_{\Vert }/\unicode[STIX]{x1D6FA}_{p}\ll 1$ limit. Consequently, to lowest order we let $k_{\bot }\rightarrow 0$ to make $\text{e}^{\text{i}Q}\rightarrow 1$ , $\text{J}_{0}\rightarrow 1$ , $\text{J}_{1}\rightarrow k_{\bot }v_{\bot }/2\unicode[STIX]{x1D6FA}$ , $\unicode[STIX]{x1D6FC}\rightarrow 0$ , $H\rightarrow 1$ and $u\rightarrow 1$ in (3.15)–(3.17), and then use $\tilde{h}_{\ast }\approx \overline{\tilde{h}_{\ast }}$ to obtain quasineutrality in the form

(4.12) $$\begin{eqnarray}\tilde{\unicode[STIX]{x1D6F7}}\sum \frac{Z^{2}\text{e}^{2}}{T}n=\sum \frac{Ze}{T}\int \text{d}^{3}v\overline{\tilde{h}_{\ast }},\end{eqnarray}$$

with

(4.13) $$\begin{eqnarray}\overline{\tilde{h}_{\ast }(t)}=\overline{\tilde{h}_{\ast }(0)}+\frac{Ze}{cT}\,f_{0}[(c\overline{\tilde{\unicode[STIX]{x1D6F7}}}-\overline{v_{\Vert }\tilde{A}_{\Vert }}+v_{\bot }^{2}\overline{\tilde{B}_{\Vert }}/2\unicode[STIX]{x1D6FA})-(c\tilde{\unicode[STIX]{x1D6F7}}^{(0)}-\overline{v_{\Vert }}\tilde{A}_{\Vert }^{(0)}+\tilde{B}_{\Vert }^{(0)}v_{\bot }^{2}/2\unicode[STIX]{x1D6FA})],\end{eqnarray}$$

and

(4.14) $$\begin{eqnarray}\overline{\tilde{h}_{\ast }(0)}=\frac{Ze}{T}f_{0}\left[\tilde{\unicode[STIX]{x1D6F7}}^{(0)}+\frac{k_{\bot }^{2}c}{\unicode[STIX]{x1D714}_{p}^{2}}\overline{v_{\Vert }}\tilde{A}_{\Vert }^{(0)}-\frac{2T\overline{B_{0}}}{ZeB_{0}^{2}\unicode[STIX]{x1D6FD}}\tilde{B}_{\Vert }^{(0)}\right],\end{eqnarray}$$

where we have set $\unicode[STIX]{x1D6FC}=0$ and $u=1$ since we let $k_{\bot }\rightarrow 0$ and used $\overline{v_{\bot }^{2}/2B_{0}}=\unicode[STIX]{x1D707}=v_{\bot }^{2}/2B_{0}$ . The integrals odd in $v_{\Vert }$ do not contribute to quasineutrality, of course. In addition, unperturbed quasineutrality prevents the $\tilde{B}_{\Vert }^{(0)}$ and $\tilde{B}_{\Vert }$ from contributing to perturbed quasineutrality as well since they result in the integrals $\sum Ze\int \text{d}^{3}vf_{0}\overline{B_{0}}=(B_{0}/2\text{}\underline{B}_{0})(\int _{0}^{\text{}\underline{B}_{0}/B_{0}}\text{d}\unicode[STIX]{x1D706}\overline{B_{0}}/\unicode[STIX]{x1D709})\sum Zen=0$ and

(4.15) $$\begin{eqnarray}\displaystyle \sum \frac{ZeM}{T}\int \text{d}^{3}vf_{0}\unicode[STIX]{x1D707}\overline{\tilde{B}_{\Vert }} & = & \displaystyle \frac{3B_{0}}{2\overline{B_{0}}^{2}}\sum Zen\left(\mathop{\int }_{0}^{\text{}\underline{B}_{0}/B_{0}}\text{d}\unicode[STIX]{x1D706}\unicode[STIX]{x1D706}\frac{\overline{\tilde{B}_{\Vert }}}{\unicode[STIX]{x1D709}}\right)\nonumber\\ \displaystyle & = & \displaystyle \frac{3B_{0}}{2\overline{B_{0}}^{2}}\left(\int _{0}^{\text{}\underline{B}_{0}/B_{0}}\text{d}\unicode[STIX]{x1D706}\unicode[STIX]{x1D706}\frac{\overline{\tilde{B}_{\Vert }}}{\unicode[STIX]{x1D709}}\right)\sum Zen=0,\end{eqnarray}$$

where $B_{0}/\text{}\underline{B}_{0}\simeq R_{0}(\unicode[STIX]{x1D713})/R=1-\unicode[STIX]{x1D716}(\unicode[STIX]{x1D713})\cos \unicode[STIX]{x1D703}+O(\unicode[STIX]{x1D716}^{2})$ , $\text{}\underline{B}_{0}=\langle B_{0}\rangle$ , $\unicode[STIX]{x1D716}=r(\unicode[STIX]{x1D713})/R_{0}(\unicode[STIX]{x1D713})$ , $\unicode[STIX]{x1D706}=2\unicode[STIX]{x1D707}\text{}\underline{B}_{0}/v^{2}$ , $\unicode[STIX]{x1D709}=\sqrt{1-\unicode[STIX]{x1D706}B_{0}/\text{}\underline{B}_{0}}$ and $\text{d}^{3}v\rightarrow \sum _{sgn\,v_{\Vert }}\unicode[STIX]{x03C0}B_{0}v^{3}\,\text{d}v\,\text{d}\unicode[STIX]{x1D706}/\text{}\underline{B}_{0}|v_{\Vert }|$ (where $\sum _{sgn\,v_{\Vert }}$ can be replaced by a factor 2 when the integrand is even in $\text{sgn}\,v_{\Vert }$ ). Consequently, assuming $k_{\bot }v_{\bot }/\unicode[STIX]{x1D6FA}\ll Q\ll 1$ the lowest order perturbed quasineutrality equation becomes

(4.16) $$\begin{eqnarray}\tilde{\unicode[STIX]{x1D6F7}}\sum \frac{Z^{2}\text{e}^{2}}{T}n=\sum \frac{Ze}{T}\int \text{d}^{3}v\overline{\tilde{h}_{\ast }}=\sum \frac{Z^{2}\text{e}^{2}}{T}\int \text{d}^{3}vf_{0}\overline{\tilde{\unicode[STIX]{x1D6F7}}},\end{eqnarray}$$

which requires

(4.17) $$\begin{eqnarray}\tilde{\unicode[STIX]{x1D6F7}}=\frac{\displaystyle \int \text{d}^{3}v\,\overline{\tilde{\unicode[STIX]{x1D6F7}}}f_{0}}{\displaystyle \int \text{d}^{3}vf_{0}},\end{eqnarray}$$

and is satisfied if $\tilde{\unicode[STIX]{x1D6F7}}=\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$ . Therefore, we may safely assume $\tilde{\unicode[STIX]{x1D6F7}}$ is a flux function to lowest order. To see this more rigorously, we multiply (4.16) by $\tilde{\unicode[STIX]{x1D6F7}}$ to form $\tilde{\unicode[STIX]{x1D6F7}}^{2}\int \text{d}^{3}vf_{0}=\int \text{d}^{3}vf_{0}\tilde{\unicode[STIX]{x1D6F7}}^{2}=\tilde{\unicode[STIX]{x1D6F7}}\int \text{d}^{3}vf_{0}\overline{\tilde{\unicode[STIX]{x1D6F7}}}=\int \text{d}^{3}vf_{0}\overline{\tilde{\unicode[STIX]{x1D6F7}}}\tilde{\unicode[STIX]{x1D6F7}}$ , then flux surface average to obtain

(4.18) $$\begin{eqnarray}\left\langle \int \text{d}^{3}vf_{0}\tilde{\unicode[STIX]{x1D6F7}}^{2}\right\rangle =\left\langle \int \text{d}^{3}vf_{0}\overline{\tilde{\unicode[STIX]{x1D6F7}}}\tilde{\unicode[STIX]{x1D6F7}}\right\rangle .\end{eqnarray}$$

As a result, using $\langle \int \text{d}^{3}vf_{0}\tilde{\unicode[STIX]{x1D6F7}}^{2}\rangle =\langle \int \text{d}^{3}vf_{0}\overline{\tilde{\unicode[STIX]{x1D6F7}}^{2}}\rangle$ and $\langle \int \text{d}^{3}vf_{0}\overline{\tilde{\unicode[STIX]{x1D6F7}}}\tilde{\unicode[STIX]{x1D6F7}}\rangle =\langle \int \text{d}^{3}vf_{0}\overline{\tilde{\unicode[STIX]{x1D6F7}}}^{2}\rangle$ we find

(4.19) $$\begin{eqnarray}0=\left\langle \int \text{d}^{3}vf_{0}(\overline{\tilde{\unicode[STIX]{x1D6F7}}^{2}}-\overline{\tilde{\unicode[STIX]{x1D6F7}}}^{2})\right\rangle =\left\langle \int \text{d}^{3}vf_{0}\overline{(\overline{\tilde{\unicode[STIX]{x1D6F7}}}-\tilde{\unicode[STIX]{x1D6F7}})^{2}}\right\rangle .\end{eqnarray}$$

Therefore, we need

(4.20) $$\begin{eqnarray}\tilde{\unicode[STIX]{x1D6F7}}=\overline{\tilde{\unicode[STIX]{x1D6F7}}},\end{eqnarray}$$

but $\tilde{\unicode[STIX]{x1D6F7}}=\tilde{\unicode[STIX]{x1D6F7}}(\unicode[STIX]{x1D713},\unicode[STIX]{x1D703})$ , while $\overline{\tilde{\unicode[STIX]{x1D6F7}}}=\overline{\tilde{\unicode[STIX]{x1D6F7}}}(\unicode[STIX]{x1D713},\unicode[STIX]{x1D706})$ , so it must be that

(4.21) $$\begin{eqnarray}\tilde{\unicode[STIX]{x1D6F7}}=\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle\end{eqnarray}$$

to lowest order.

To generalize our results for quasineutrality and to treat both components of Ampère’s law, we extend our initial condition (4.14) to include finite orbit effects by employing (4.11) in (3.11).

In the calculation of the electrostatic zonal flow residual (Rosenbluth & Hinton Reference Rosenbluth and Hinton1998) the non-adiabatic electron response could be neglected as small in $(m/M)^{1/2}$ , where $m$ is the electron mass, however here it sometimes needs to be retained, since the electromagnetic terms in the non-adiabatic response are proportional to the thermal speed of the species. In the next section we will use the preceding results to form quasineutrality and Ampère’s law.

5 Quasineutrality and Ampère’s law

In this section we form and consider the non-flux surface averaged components of Ampère’s law to demonstrate that poloidal variation of the parallel vector potential and the parallel magnetic field must be retained. Indeed, we will find that there are cases for which these field responses have strong poloidal variation. In addition, we will perform a more complete evaluation of quasineutrality once we have examined the two components of Ampère’s law.

To perform the derivation of the two components of Ampère’s law we must realize that $\tilde{A}_{\Vert }$ and $\tilde{B}_{\Vert }$ are not normally flux functions. Indeed, even when $\tilde{B}_{\Vert }^{(0)}=0=\tilde{A}_{\Vert }^{(0)}$ we will find they both have poloidal variation. Retaining $\text{e}^{\text{i}Q}$ modifications, but ignoring $L\sim k_{\bot }v_{\bot }/\unicode[STIX]{x1D6FA}$ corrections as unimportant except in the $\tilde{\unicode[STIX]{x1D6F7}}$ and $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ terms, we use

(5.1) $$\begin{eqnarray}\displaystyle \overline{\tilde{h}_{\ast }(t)} & = & \displaystyle \overline{\tilde{h}_{\ast }(0)}+\frac{Ze}{cT}\,f_{0}\left[c\overline{\tilde{\unicode[STIX]{x1D6F7}}\text{J}_{0}\text{e}^{\text{i}Q}}+\frac{v_{\bot }^{2}}{2\unicode[STIX]{x1D6FA}}\overline{\tilde{B}_{\Vert }\text{e}^{\text{i}Q}}-\overline{\tilde{A}_{\Vert }v_{\Vert }\text{e}^{\text{i}Q}}-c\tilde{\unicode[STIX]{x1D6F7}}^{(0)}\overline{\text{J}_{0}\text{e}^{\text{i}Q}}\right.\nonumber\\ \displaystyle & & \displaystyle \left.-\,\frac{v_{\bot }^{2}}{2\unicode[STIX]{x1D6FA}}\tilde{B}_{\Vert }^{(0)}\overline{\text{e}^{\text{i}Q}}+\tilde{A}_{\Vert }^{(0)}\overline{v_{\Vert }\text{e}^{\text{i}Q}}\right],\end{eqnarray}$$

with

(5.2) $$\begin{eqnarray}\overline{\tilde{h}_{\ast }(0)}=\frac{Ze}{T}f_{0}\left[\tilde{\unicode[STIX]{x1D6F7}}^{(0)}\overline{(\text{e}^{\text{i}Q}H/\text{J}_{0})}+\frac{k_{\bot }^{2}c}{\unicode[STIX]{x1D714}_{p}^{2}}\overline{v_{\Vert }\text{e}^{\text{i}Q}}\tilde{A}_{\Vert }^{(0)}-\frac{2T}{ZeB_{0}^{2}\unicode[STIX]{x1D6FD}}\tilde{B}_{\Vert }^{(0)}\overline{B_{0}\text{e}^{\text{i}Q}}\right].\end{eqnarray}$$

5.1 Ampère’s law

Inserting the preceding into

(5.3) $$\begin{eqnarray}\tilde{B}_{\Vert }=-\sum \frac{2\unicode[STIX]{x03C0}M}{B_{0}}\int \text{d}^{3}v\overline{\tilde{h}_{\ast }}v_{\bot }^{2}\text{e}^{-\text{i}Q}\end{eqnarray}$$

and

(5.4) $$\begin{eqnarray}k_{\bot }^{2}\tilde{A}_{\Vert }=\frac{4\unicode[STIX]{x03C0}}{c}\sum Ze\int \text{d}^{3}v\overline{\tilde{h}_{\ast }}v_{\Vert }\text{e}^{-\text{i}Q},\end{eqnarray}$$

we obtain

(5.5) $$\begin{eqnarray}\displaystyle \tilde{B}_{\Vert } & = & \displaystyle -\sum \frac{4\unicode[STIX]{x03C0}Ze}{B_{0}}\int \text{d}^{3}vf_{0}\frac{Mv_{\bot }^{2}}{2T}\frac{2\text{J}_{1}}{z}\text{e}^{-\text{i}Q}[\overline{\tilde{\unicode[STIX]{x1D6F7}}\text{J}_{0}\text{e}^{\text{i}Q}}+\tilde{\unicode[STIX]{x1D6F7}}^{(0)}(\overline{\text{e}^{\text{i}Q}H/\text{J}_{0}}-\overline{\text{J}_{0}\text{e}^{\text{i}Q}})]\nonumber\\ \displaystyle & & \displaystyle +\,\sum \frac{4\unicode[STIX]{x03C0}Ze}{cB_{0}}\int \text{d}^{3}vf_{0}\frac{Mv_{\bot }^{2}}{2T}\text{e}^{-\text{i}Q}\left[\overline{v_{\Vert }\tilde{A}_{\Vert }\text{e}^{\text{i}Q}}-\tilde{A}_{\Vert }^{(0)}\left(1+\frac{k_{\bot }^{2}c^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}\right)\overline{v_{\Vert }\text{e}^{\text{i}Q}}\right]\nonumber\\ \displaystyle & & \displaystyle -\,\sum \frac{4\unicode[STIX]{x03C0}T}{\mathop{B}_{0}^{2}}\int \text{d}^{3}vf_{0}\frac{Mv_{\bot }^{2}}{2T}\text{e}^{-\text{i}Q}\left[\frac{Mv_{\bot }^{2}}{2T}(\overline{\tilde{B}_{\Vert }\text{e}^{\text{i}Q}}-\tilde{B}_{\Vert }^{(0)}\overline{\text{e}^{\text{i}Q}})-\frac{2\tilde{B}_{\Vert }^{(0)}}{B_{0}\unicode[STIX]{x1D6FD}}\overline{B_{0}\text{e}^{\text{i}Q}}\right]\end{eqnarray}$$

and

(5.6) $$\begin{eqnarray}\displaystyle \tilde{A}_{\Vert } & = & \displaystyle -\sum \frac{4\unicode[STIX]{x03C0}Z^{2}\text{e}^{2}}{\mathop{k}_{\bot }^{2}c^{2}T}\int \text{d}^{3}vf_{0}v_{\Vert }\text{e}^{-\text{i}Q}\left[\overline{v_{\Vert }\tilde{A}_{\Vert }\text{e}^{\text{i}Q}}-\tilde{A}_{\Vert }^{(0)}\left(1+\frac{k_{\bot }^{2}c^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}\right)\overline{v_{\Vert }\text{e}^{\text{i}Q}}\right]\nonumber\\ \displaystyle & & \displaystyle +\,\sum \frac{4\unicode[STIX]{x03C0}Z^{2}\text{e}^{2}}{\mathop{k}_{\bot }^{2}cT}\int \text{d}^{3}vf_{0}v_{\Vert }\text{e}^{-\text{i}Q}\text{J}_{0}[\overline{\tilde{\unicode[STIX]{x1D6F7}}\text{J}_{0}\text{e}^{\text{i}Q}}+\tilde{\unicode[STIX]{x1D6F7}}^{(0)}(\overline{\text{e}^{\text{i}Q}H/\text{J}_{0}}-\overline{\text{J}_{0}\text{e}^{\text{i}Q}})]\nonumber\\ \displaystyle & & \displaystyle +\,\sum \frac{4\unicode[STIX]{x03C0}Ze}{\mathop{k}_{\bot }^{2}cB_{0}}\int \text{d}^{3}vf_{0}v_{\Vert }\text{e}^{-\text{i}Q}\left[\frac{Mv_{\bot }^{2}}{2T}(\overline{\tilde{B}_{\Vert }\text{e}^{\text{i}Q}}-\tilde{B}_{\Vert }^{(0)}\overline{\text{e}^{\text{i}Q}})-\frac{2\tilde{B}_{\Vert }^{(0)}}{B_{0}\unicode[STIX]{x1D6FD}}\overline{B_{0}\text{e}^{\text{i}Q}}\right].\end{eqnarray}$$

Recalling that $\tilde{A}_{\Vert }$ and $\tilde{B}_{\Vert }$ need not be flux functions and that the unperturbed magnetic field gives rise to $\cos \unicode[STIX]{x1D703}$ dependence, we Fourier decompose and retain only the leading poloidal dependence by writing

(5.7) $$\begin{eqnarray}\tilde{A}_{\Vert }=\langle \tilde{A}\rangle +a\cos \unicode[STIX]{x1D703}+\cdots \simeq \left[\langle \tilde{A}_{\Vert }\rangle -\frac{a}{2}(\unicode[STIX]{x1D716}-\unicode[STIX]{x1D6E5}_{S}^{\prime })\right]+a\cos \unicode[STIX]{x1D703}\end{eqnarray}$$

and

(5.8) $$\begin{eqnarray}\tilde{B}_{\Vert }=\langle \tilde{B}\rangle +b\cos \unicode[STIX]{x1D703}+\cdots \simeq \left[\langle \tilde{B}_{\Vert }\rangle -\frac{b}{2}(\unicode[STIX]{x1D716}-\unicode[STIX]{x1D6E5}_{S}^{\prime })\right]+b\cos \unicode[STIX]{x1D703},\end{eqnarray}$$

with the coefficients $\langle \tilde{A}_{\Vert }\rangle$ , $\langle \tilde{B}_{\Vert }\rangle$ , $\langle \tilde{A}\rangle$ , $\langle \tilde{B}\rangle$ , $a$ , and $b$ flux functions and $\boldsymbol{B}_{0}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}\propto 1-(\unicode[STIX]{x1D716}-\unicode[STIX]{x1D6E5}_{S}^{\prime })\cos \unicode[STIX]{x1D703}$ in the flux surface averages for our Shafranov shifted circular flux surface model, where $\unicode[STIX]{x1D6E5}_{S}^{\prime }=\text{d}\unicode[STIX]{x1D6E5}_{S}/\text{d}r\sim \unicode[STIX]{x1D716}$ . We do not assume an ordering of $a$ and $b$ relative to $\langle \tilde{A}_{\Vert }\rangle$ and $\langle \tilde{B}_{\Vert }\rangle$ as we will calculate all four of these coefficients. In addition, we recall the electrostatic result of Rosenbluth & Hinton (Reference Rosenbluth and Hinton1998),

(5.9) $$\begin{eqnarray}\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle =\frac{\tilde{\unicode[STIX]{x1D6F7}}^{(0)}}{1+\unicode[STIX]{x1D6FE}q^{2}/\unicode[STIX]{x1D716}^{1/2}},\end{eqnarray}$$

with $\unicode[STIX]{x1D6FE}\approx 1.64$ . The preceding allows us to assume $q^{2}\unicode[STIX]{x1D716}^{-1/2}\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle \sim \tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ for $\unicode[STIX]{x1D6FD}\ll 1$ and $\unicode[STIX]{x1D716}\ll 1$ .

Expanding for $Q=k_{\bot }v_{\Vert }/\unicode[STIX]{x1D6FA}_{p}\ll 1$ , inserting $H$ , neglecting $\unicode[STIX]{x1D716}$ corrections to $k_{\bot }^{2}\unicode[STIX]{x1D70C}_{i}^{2}$ terms, ignoring all $Q^{2}$ , $\unicode[STIX]{x1D6FD}$ and $\unicode[STIX]{x1D716}^{3/2}$ corrections to the $\tilde{B}_{\Vert }$ and $\tilde{B}_{\Vert }^{(0)}$ terms in (5.5), recalling $\tilde{\unicode[STIX]{x1D6F7}}=\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$ , and using unperturbed quasineutrality, perpendicular Ampère’s law becomes

(5.10) $$\begin{eqnarray}\displaystyle \tilde{B}_{\Vert }-\tilde{B}_{\Vert }^{(0)}\frac{\text{}\underline{B}_{0}}{B_{0}} & = & \displaystyle -\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle \sum \frac{2\unicode[STIX]{x03C0}ZeM}{TB_{0}}\!\int \text{d}^{3}vf_{0}v_{\bot }^{2}\!\left[\overline{Q}Q-\frac{(Q^{2}+\overline{Q^{2}})}{2}\right]+(\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle -\tilde{\unicode[STIX]{x1D6F7}}^{(0)})\frac{3c\unicode[STIX]{x1D6FD}_{i}k_{\bot }^{2}}{4\unicode[STIX]{x1D6FA}_{i}}\nonumber\\ \displaystyle & & \displaystyle +\,\text{i}k_{\bot }\sum \frac{2\unicode[STIX]{x03C0}qM^{2}}{\unicode[STIX]{x1D716}TB_{0}}\int \text{d}^{3}vf_{0}v_{\bot }^{2}(\overline{B_{0}^{-1}v_{\Vert }^{2}\tilde{A}_{\Vert }}-B_{0}^{-1}v_{\Vert }\overline{v_{\Vert }\tilde{A}_{\Vert }})\nonumber\\ \displaystyle & & \displaystyle -\,\text{i}k_{\bot }\tilde{A}_{\Vert }^{(0)}\left(1+\frac{k_{\bot }^{2}c^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}\right)\sum \frac{2\unicode[STIX]{x03C0}qM^{2}}{\unicode[STIX]{x1D716}TB_{0}}\int \text{d}^{3}vf_{0}v_{\bot }^{2}(\overline{B_{0}^{-1}v_{\Vert }^{2}}-B_{0}^{-1}v_{\Vert }\overline{v_{\Vert }}).\end{eqnarray}$$

Here and hereafter, we use $Q\propto v_{\Vert }/B_{0}$ and then treat $k_{\bot }$ as a flux function, except in $k_{\bot }^{2}\tilde{A}_{\Vert }^{(0)}$ terms from (5.2) and $k_{\bot }^{2}\tilde{A}_{\Vert }$ from (5.4).

In (5.6) we neglect $Q$ corrections for $\tilde{A}_{\Vert }$ and $\tilde{A}_{\Vert }^{(0)}$ terms, insert $H$ , and continue to use $\tilde{\unicode[STIX]{x1D6F7}}=\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$ , to find that parallel Ampère’s law is

(5.11) $$\begin{eqnarray}\displaystyle & & \displaystyle \tilde{A}_{\Vert }+\sum \frac{4\unicode[STIX]{x03C0}Z^{2}\text{e}^{2}}{\mathop{k}_{\bot }^{2}c^{2}T}\int \text{d}^{3}vf_{0}v_{\Vert }\overline{v_{\Vert }\tilde{A}_{\Vert }}-\tilde{A}_{\Vert }^{(0)}\left(1+\frac{k_{\bot }^{2}c^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}\right)\sum \frac{4\unicode[STIX]{x03C0}Z^{2}\text{e}^{2}}{\mathop{k}_{\bot }^{2}c^{2}T}\int \text{d}^{3}vf_{0}v_{\Vert }\overline{v_{\Vert }}\nonumber\\ \displaystyle & & \displaystyle \quad =\text{i}\sum \frac{4\unicode[STIX]{x03C0}qM}{k_{\bot }\unicode[STIX]{x1D716}B_{0}}\int \text{d}^{3}vf_{0}\frac{Mv_{\bot }^{2}}{2T}v_{\Vert }[(\overline{B_{0}^{-1}v_{\Vert }\tilde{B}_{\Vert }}-B_{0}^{-1}v_{\Vert }\overline{\tilde{B}_{\Vert }})-\tilde{B}_{\Vert }^{(0)}(\overline{B_{0}^{-1}v_{\Vert }}-B_{0}^{-1}v_{\Vert })]\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\text{i}\frac{2}{\unicode[STIX]{x1D6FD}}\tilde{B}_{\Vert }^{(0)}\sum \frac{4\unicode[STIX]{x03C0}qM}{k_{\bot }\unicode[STIX]{x1D716}\mathop{B}_{0}^{2}}\int \text{d}^{3}vf_{0}v_{\Vert }(\overline{v_{\Vert }}-v_{\Vert })\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\text{i}\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle \sum \frac{2\unicode[STIX]{x03C0}Z^{2}\text{e}^{2}}{\mathop{k}_{\bot }^{2}cT}\int \text{d}^{3}vf_{0}v_{\Vert }\left[Q\overline{Q^{2}}-Q^{2}\overline{Q}+\frac{Q^{3}-\overline{Q^{3}}}{3}\right]\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\text{i}\sum \frac{2\unicode[STIX]{x03C0}cM^{2}}{T\mathop{B}_{0}^{2}}\left[(\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle -\tilde{\unicode[STIX]{x1D6F7}}^{(0)})\int \text{d}^{3}vf_{0}v_{\Vert }(Q-\overline{Q})v_{\bot }^{2}\right.\nonumber\\ \displaystyle & & \displaystyle \left.\qquad +\,\tilde{\unicode[STIX]{x1D6F7}}^{(0)}\frac{T}{M}\int \text{d}^{3}vf_{0}v_{\Vert }(Q-\overline{Q})\right],\end{eqnarray}$$

where we used (A 22), (A 30), and (A 37) and similar integrals for $\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$ and $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ terms to see that due to unperturbed quasineutrality $\sum (Z^{2}\text{e}^{2}/T)\int \text{d}^{3}vf_{0}v_{\Vert }(Q-\overline{Q})=0$ . The zonal flow responses of the perturbed parallel magnetic field and vector potential in (5.10) and (5.11) are generated by the polarizations terms associated with poloidally varying departures from flux surfaces.

Next, we flux surface average using $\langle \int \text{d}^{3}vf_{0}\overline{X}Y\rangle =\langle \int \text{d}^{3}vf_{0}X\overline{Y}\rangle$ , to obtain the flux surface averaged perpendicular Ampère’s law

(5.12) $$\begin{eqnarray}\displaystyle & & \displaystyle \langle \tilde{B}_{\Vert }\rangle -\tilde{B}_{\Vert }^{(0)}\simeq \frac{c\unicode[STIX]{x1D6FD}_{i}k_{\bot }^{2}\unicode[STIX]{x1D70C}_{i}}{4v_{i}}[(5\unicode[STIX]{x1D6FE}q^{2}\unicode[STIX]{x1D716}^{-1/2}+3)\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle -3\tilde{\unicode[STIX]{x1D6F7}}^{(0)}]\nonumber\\ \displaystyle & & \displaystyle +\,\frac{\text{i}q\unicode[STIX]{x1D6FD}k_{\bot }}{2}\left\{\frac{5\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}}{2}\left[\langle \tilde{A}_{\Vert }\rangle -\tilde{A}_{\Vert }^{(0)}\left(1+\frac{k_{\bot }^{2}c^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}\right)\right]+a\frac{3\unicode[STIX]{x1D716}}{2}\right\},\end{eqnarray}$$

where we used (A 44), (A 63), (A 66), (A 67), and (A 72). The $B_{0}$ factors in (5.10) must be treated with care, even though $\overline{v_{\Vert }\cos \unicode[STIX]{x1D703}}=0$ for the trapped and $\overline{v_{\Vert }\cos \unicode[STIX]{x1D703}}\sim \unicode[STIX]{x1D716}v$ for the passing, since they alter the coefficient. For now we retain the $\unicode[STIX]{x1D716}^{1/2}$ smaller $\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$ and $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ terms compared to the $\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{-1/2}q^{2}\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$ term.

In the parallel Ampère’s law for now we keep $O(\unicode[STIX]{x1D716}^{3/2})$ corrections to the typically large $\unicode[STIX]{x1D714}_{p}^{2}/k_{\bot }^{2}c^{2}$ terms obtained by using (A 30). By keeping the $\unicode[STIX]{x1D716}^{3/2}$ correction to $\tilde{A}_{\Vert }$ terms we are making the easily satisfied assumption that $\unicode[STIX]{x1D716}^{3/2}\unicode[STIX]{x1D714}_{p}^{2}/k_{\bot }^{2}c^{2}\gg q^{2}\unicode[STIX]{x1D6FD}$ , or roughly $M/m\gg \unicode[STIX]{x1D716}^{1/2}k_{\bot }^{2}\unicode[STIX]{x1D70C}_{pi}^{2}$ , since we already neglected the more complicated $Q^{2}$ corrections to the same terms. The flux surface averaged parallel Ampère’s law then becomes

(5.13) $$\begin{eqnarray}\displaystyle & & \displaystyle \langle \tilde{A}_{\Vert }\rangle \left[1+\frac{\unicode[STIX]{x1D714}_{p}^{2}}{k_{\bot }^{2}c^{2}}(1-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{3/2})\right]-\tilde{A}_{\Vert }^{(0)}\left[\left(1+\frac{\unicode[STIX]{x1D714}_{p}^{2}}{k_{\bot }^{2}c^{2}}\right)(1-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{3/2})\right]\nonumber\\ \displaystyle & & \displaystyle \qquad -\,a\left[\frac{\unicode[STIX]{x1D716}\unicode[STIX]{x1D714}_{p}^{2}}{2k_{\bot }^{2}c^{2}}(1-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{3/2})+\unicode[STIX]{x1D6E5}_{S}^{\prime }\right]\nonumber\\ \displaystyle & & \displaystyle \quad \simeq \,\frac{\text{i}q\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}\tilde{B}_{\Vert }^{(0)}}{k_{\bot }}+\text{i}\frac{c\unicode[STIX]{x1D6FD}_{i}qk_{\bot }\unicode[STIX]{x1D70C}_{i}}{4v_{i}}[(\unicode[STIX]{x1D70E}\unicode[STIX]{x1D716}^{-1/2}q^{2}+5\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}+4\unicode[STIX]{x1D716})\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle -4(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}+\unicode[STIX]{x1D716})\tilde{\unicode[STIX]{x1D6F7}}^{(0)}],\hspace{15.60004pt}\end{eqnarray}$$

where the constant $\unicode[STIX]{x1D70E}\approx 5.3$ is evaluated in appendix A, and we make use of (A 30), (A 35), (A 37), (A 44), (A 73) and (A 80). We use (5.7) and keep the Shafranov shift, $\unicode[STIX]{x1D6E5}_{S}$ , terms from $k_{\bot }\propto RB_{p}\propto 1-\unicode[STIX]{x1D6E5}_{S}^{\prime }\cos \unicode[STIX]{x1D703}$ in the $\langle \tilde{A}_{\Vert }\rangle$ and $\tilde{A}_{\Vert }^{(0)}$ terms that arise from (5.2) and the left hand side of (5.4). To simplify the flux surface averaged parallel Ampère’s law we ignored $\unicode[STIX]{x1D6FD}\tilde{B}_{\Vert }$ and $\unicode[STIX]{x1D6FD}\tilde{B}_{\Vert }^{(0)}$ terms. Also, we have neglected all $b$ terms because they are multiplied by $\unicode[STIX]{x1D6FD}$ . Based on (5.9) we must retain the $\unicode[STIX]{x1D70E}q^{2}\unicode[STIX]{x1D716}^{-1/2}\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$ term in (5.13), but will ignore the $\unicode[STIX]{x1D716}$ and $\unicode[STIX]{x1D716}^{3/2}$ smaller term $(5\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}+4\unicode[STIX]{x1D716})\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$ , as well as the $(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}+\unicode[STIX]{x1D716})\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ term, all associated with the Bessel function corrections.

The terms $\unicode[STIX]{x1D6FE}$ and $\unicode[STIX]{x1D70E}$ in (5.12) and (5.13) arise from the trapped particle responses to any initial perturbation, while the $a\unicode[STIX]{x1D716}$ and $a\unicode[STIX]{x1D6E5}_{S}^{\prime }$ terms are passing responses. We estimate the different behaviour of the passing and trapped (and barely passing) particles by using $\overline{v_{\Vert }}=0$ and estimating $v_{\Vert }\sim \unicode[STIX]{x1D716}^{1/2}v$ and $\int \text{d}^{3}v\propto \unicode[STIX]{x1D716}^{1/2}$ for the trapped particles, while using $\overline{v_{\Vert }}\simeq v_{\Vert }[1+O(\unicode[STIX]{x1D716})]$ for the passing ones.

Next, we subtract the flux surface averaged equations from the full equations. For the perpendicular Ampère’s law we use (A 63), (A 66), (A 67), and (A 72) to find

(5.14) $$\begin{eqnarray}\displaystyle & & \displaystyle b\simeq \unicode[STIX]{x1D716}\tilde{B}_{\Vert }^{(0)}-\frac{\text{i}q\unicode[STIX]{x1D6FD}k_{\bot }}{2}(3-\unicode[STIX]{x1D70D}\unicode[STIX]{x1D716}^{1/2})\left[\langle \tilde{A}_{\Vert }\rangle -\tilde{A}_{\Vert }^{(0)}\left(1+\frac{k_{\bot }^{2}c^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}\right)\right]\nonumber\\ \displaystyle & & \displaystyle \quad -\,\frac{\text{i}\unicode[STIX]{x1D712}\unicode[STIX]{x1D716}^{3/2}q\unicode[STIX]{x1D6FD}k_{\bot }a}{4}+\frac{c\unicode[STIX]{x1D70D}\unicode[STIX]{x1D6FD}_{i}q^{2}k_{\bot }^{2}\unicode[STIX]{x1D70C}_{i}}{2\unicode[STIX]{x1D716}^{1/2}v_{i}}\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle ,\end{eqnarray}$$

where Bessel function terms are ignored since any $\cos \unicode[STIX]{x1D703}$ dependence is smaller by $\unicode[STIX]{x1D716}$ and the constants $\unicode[STIX]{x1D712}\approx 0.11$ and $\unicode[STIX]{x1D70D}\approx 5.3$ are evaluated in appendix A. The $\langle \tilde{A}_{\Vert }\rangle$ , $\tilde{A}_{\Vert }^{(0)}$ , and $\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$ terms are from passing contributions, while the $\unicode[STIX]{x1D716}^{3/2}a$ term is a trapped contribution.

To form the difference equation for the parallel Ampère’s law we ignore $\unicode[STIX]{x1D6FD}\langle \tilde{B}_{\Vert }\rangle$ , $\unicode[STIX]{x1D6FD}\tilde{B}_{\Vert }^{(0)}$ and $\unicode[STIX]{x1D6FD}b$ terms. As a result, the poloidally varying parallel Ampère’s law reduces to

(5.15) $$\begin{eqnarray}\displaystyle a\left(1+\frac{\unicode[STIX]{x1D716}^{2}\unicode[STIX]{x1D714}_{p}^{2}}{2k_{\bot }^{2}c^{2}}\right) & \simeq & \displaystyle \left(\frac{\unicode[STIX]{x1D716}\unicode[STIX]{x1D714}_{p}^{2}}{k_{\bot }^{2}c^{2}}+2\unicode[STIX]{x1D6E5}_{S}^{\prime }\right)(\langle \tilde{A}_{\Vert }\rangle -\tilde{A}_{\Vert }^{(0)})-\unicode[STIX]{x1D716}\tilde{A}_{\Vert }^{(0)}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{\text{i}q\tilde{B}_{\Vert }^{(0)}}{k_{\bot }}+\text{i}\frac{c\unicode[STIX]{x1D6FD}_{i}qk_{\bot }\unicode[STIX]{x1D70C}_{i}}{4v_{i}}[(5\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{-1/2}q^{2}+4)\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle -3\tilde{\unicode[STIX]{x1D6F7}}^{(0)}],\end{eqnarray}$$

where we make use of use (A 30), (A 35), (A 37), (A 44), (A 73), and (A 80), and account for $k_{\bot }\propto 1-\unicode[STIX]{x1D6E5}_{S}^{\prime }\cos \unicode[STIX]{x1D703}$ . Only passing contributions enter for $\langle \tilde{A}_{\Vert }\rangle$ , $\tilde{A}_{\Vert }^{(0)}$ , $a$ , and $\tilde{B}_{\Vert }^{(0)}$ terms, and only Bessel terms enter for $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ , while for $\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$ both trapped and Bessel contributions enter.

5.2 Parallel Ampère’s law limit

Simulations are sometimes run ignoring the perpendicular Ampère’s law. If we do the same by dropping all $b$ and $\tilde{B}_{\Vert }^{(0)}$ terms in parallel Ampère’s law, then (5.15) becomes

(5.16) $$\begin{eqnarray}\displaystyle a\left(1+\frac{\unicode[STIX]{x1D716}^{2}\unicode[STIX]{x1D714}_{p}^{2}}{2k_{\bot }^{2}c^{2}}\right) & \simeq & \displaystyle \left(\frac{\unicode[STIX]{x1D716}\unicode[STIX]{x1D714}_{p}^{2}}{k_{\bot }^{2}c^{2}}+2\unicode[STIX]{x1D6E5}_{S}^{\prime }\right)(\langle \tilde{A}_{\Vert }\rangle -\tilde{A}_{\Vert }^{(0)})-\unicode[STIX]{x1D716}\tilde{A}_{\Vert }^{(0)}\nonumber\\ \displaystyle & & \displaystyle +\,\text{i}\frac{c\unicode[STIX]{x1D6FD}_{i}qk_{\bot }\unicode[STIX]{x1D70C}_{i}}{2v_{i}}[1-(\unicode[STIX]{x1D716}^{1/2}/2\unicode[STIX]{x1D6FE}q^{2})]\tilde{\unicode[STIX]{x1D6F7}}^{(0)},\end{eqnarray}$$

and (5.13) reduces to

(5.17) $$\begin{eqnarray}\displaystyle & & \displaystyle \langle \tilde{A}_{\Vert }\rangle \left[1+\frac{\unicode[STIX]{x1D714}_{p}^{2}}{k_{\bot }^{2}c^{2}}(1-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{3/2})\right]-\tilde{A}_{\Vert }^{(0)}\left[\left(1+\frac{\unicode[STIX]{x1D714}_{p}^{2}}{k_{\bot }^{2}c^{2}}\right)(1-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{3/2})\right]\nonumber\\ \displaystyle & & \displaystyle \qquad -\,a\left[\frac{\unicode[STIX]{x1D716}\unicode[STIX]{x1D714}_{p}^{2}}{2k_{\bot }^{2}c^{2}}(1-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{3/2})+\unicode[STIX]{x1D6E5}_{S}^{\prime }\right]\nonumber\\ \displaystyle & & \displaystyle \quad \simeq \,\text{i}\frac{c\unicode[STIX]{x1D70E}\unicode[STIX]{x1D6FD}_{i}qk_{\bot }\unicode[STIX]{x1D70C}_{i}}{4\unicode[STIX]{x1D6FE}v_{i}}[1-(\unicode[STIX]{x1D716}^{1/2}/\unicode[STIX]{x1D6FE}q^{2})-(4\unicode[STIX]{x1D6FE}^{2}\unicode[STIX]{x1D716}^{1/2}/\unicode[STIX]{x1D70E})]\tilde{\unicode[STIX]{x1D6F7}}^{(0)},\end{eqnarray}$$

where we have used $\unicode[STIX]{x1D6FD}\ll 1$ so we can use (5.9) to eliminate $\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$ .

If $\unicode[STIX]{x1D6FD}\rightarrow 0$ at finite $k_{\bot }$ so that $\unicode[STIX]{x1D714}_{p}^{2}/k_{\bot }^{2}c^{2}\rightarrow 0$ then (5.16) and (5.17) reduce to the large skin depth results

(5.18) $$\begin{eqnarray}\frac{v_{i}}{c}\langle \tilde{A}_{\Vert }\rangle \rightarrow \frac{v_{i}}{c}\tilde{A}_{\Vert }^{(0)}(1-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{3/2})+\text{i}\frac{\unicode[STIX]{x1D70E}\unicode[STIX]{x1D6FD}_{i}qk_{\bot }\unicode[STIX]{x1D70C}_{i}}{4\unicode[STIX]{x1D6FE}}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}\end{eqnarray}$$

and

(5.19) $$\begin{eqnarray}\frac{v_{i}}{c}a\rightarrow -\frac{v_{i}}{c}\unicode[STIX]{x1D716}\tilde{A}_{\Vert }^{(0)}+\text{i}\frac{\unicode[STIX]{x1D6FD}_{i}qk_{\bot }\unicode[STIX]{x1D70C}_{i}}{2}\tilde{\unicode[STIX]{x1D6F7}}^{(0)},\end{eqnarray}$$

where we neglect $\unicode[STIX]{x1D716}^{1/2}$ corrections to $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ terms. Interestingly, for $\tilde{A}_{\Vert }^{(0)}=0$ we see that $\langle \tilde{A}_{\Vert }\rangle /a\simeq \unicode[STIX]{x1D70E}/2\unicode[STIX]{x1D6FE}\simeq O(1)$ , meaning that substantial poloidal variation occurs when $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}\neq 0$ . When $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}=0$ , $\langle \tilde{A}_{\Vert }\rangle \simeq \tilde{A}_{\Vert }^{(0)}(1-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{3/2})$ and $a\simeq -\unicode[STIX]{x1D716}\langle \tilde{A}_{\Vert }\rangle$ , and poloidal variation is somewhat weak.

More interestingly, we consider finite $\unicode[STIX]{x1D6FD}\ll 1$ by allowing $\unicode[STIX]{x1D716}^{3/2}\gg k_{\bot }^{2}c^{2}/\unicode[STIX]{x1D714}_{p}^{2}$ or $q^{2}\unicode[STIX]{x1D6FD}_{i}\unicode[STIX]{x1D716}^{1/2}\gg Zmk_{\bot }^{2}\unicode[STIX]{x1D70C}_{pi}^{2}/M$ . Continuing to neglect $\unicode[STIX]{x1D716}^{1/2}$ corrections in this small skin depth limit we find

(5.20) $$\begin{eqnarray}\frac{v_{i}}{c}\langle \tilde{A}_{\Vert }\rangle \simeq \frac{v_{i}}{c}\tilde{A}_{\Vert }^{(0)}\left(1-\frac{\unicode[STIX]{x1D716}^{2}}{2}\right)+\frac{\text{i}\unicode[STIX]{x1D716}\unicode[STIX]{x1D6FD}_{i}k_{\bot }\unicode[STIX]{x1D70C}_{pi}}{4}\tilde{\unicode[STIX]{x1D6F7}}^{(0)},\end{eqnarray}$$

and

(5.21) $$\begin{eqnarray}\frac{v_{i}}{c}a\simeq -\frac{v_{i}}{c}\unicode[STIX]{x1D716}\tilde{A}_{\Vert }^{(0)}+\frac{\text{i}\unicode[STIX]{x1D716}\unicode[STIX]{x1D6FD}_{i}k_{\bot }\unicode[STIX]{x1D70C}_{pi}}{2}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}.\end{eqnarray}$$

When $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}=0$ , $\langle \tilde{A}_{\Vert }\rangle \simeq \tilde{A}_{\Vert }^{(0)}\simeq -a/\unicode[STIX]{x1D716}$ so only weak poloidal variation occurs. However, for $\tilde{A}_{\Vert }^{(0)}\simeq 0$ , we find that $a/\langle \tilde{A}_{\Vert }\rangle \simeq 2$ , which results in strong poloidal variation.

In both the small and large skin depth limits, when $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}=0$ , only small changes in $\tilde{A}_{\Vert }$ occur since $\tilde{A}_{\Vert }\simeq \langle \tilde{A}_{\Vert }\rangle \simeq \tilde{A}_{\Vert }^{(0)}\gg a$ . However, for $\tilde{A}_{\Vert }^{(0)}\simeq 0$ , we see that poloidal variation in $\tilde{A}_{\Vert }$ arises due to $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ with $a/\langle \tilde{A}_{\Vert }\rangle \sim 1$ in both limits.

5.3 Full Ampère’s law

The general case retains the perpendicular Ampère’s law. Using (5.13) and (5.15) with (5.9) inserted and ignoring $\unicode[STIX]{x1D716}^{1/2}$ corrections we find

(5.22) $$\begin{eqnarray}a\left(1+\frac{\unicode[STIX]{x1D716}^{2}\unicode[STIX]{x1D714}_{p}^{2}}{2k_{\bot }^{2}c^{2}}\right)\simeq \left(\frac{\unicode[STIX]{x1D716}\unicode[STIX]{x1D714}_{p}^{2}}{k_{\bot }^{2}c^{2}}+2\unicode[STIX]{x1D6E5}_{S}^{\prime }\right)(\langle \tilde{A}_{\Vert }\rangle -\tilde{A}_{\Vert }^{(0)})-\unicode[STIX]{x1D716}\tilde{A}_{\Vert }^{(0)}+\text{i}\frac{c\unicode[STIX]{x1D6FD}_{i}qk_{\bot }\unicode[STIX]{x1D70C}_{i}}{2v_{i}}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}+\text{i}qk_{\bot }^{-1}\tilde{B}_{\Vert }^{(0)}\end{eqnarray}$$

and

(5.23) $$\begin{eqnarray}\displaystyle & & \displaystyle (\langle \tilde{A}_{\Vert }\rangle -\tilde{A}_{\Vert }^{(0)})\left(1+\frac{\unicode[STIX]{x1D714}_{p}^{2}}{k_{\bot }^{2}c^{2}}\right)-a\left[\frac{\unicode[STIX]{x1D716}\unicode[STIX]{x1D714}_{p}^{2}}{2k_{\bot }^{2}c^{2}}(1-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{3/2})+\unicode[STIX]{x1D6E5}_{S}^{\prime }\right]\nonumber\\ \displaystyle & & \displaystyle \quad \simeq \,\text{i}\frac{c\unicode[STIX]{x1D70E}\unicode[STIX]{x1D6FD}_{i}qk_{\bot }\unicode[STIX]{x1D70C}_{i}}{4\unicode[STIX]{x1D6FE}v_{i}}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}+\text{i}q\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}k_{\bot }^{-1}\tilde{B}_{\Vert }^{(0)}.\end{eqnarray}$$

In addition, equations (5.12) and (5.14) reduce to

(5.24) $$\begin{eqnarray}\displaystyle \langle \tilde{B}_{\Vert }\rangle -\tilde{B}_{\Vert }^{(0)} & \simeq & \displaystyle \frac{c\unicode[STIX]{x1D6FD}_{i}k_{\bot }^{2}\unicode[STIX]{x1D70C}_{i}}{2v_{i}}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{\text{i}q\unicode[STIX]{x1D6FD}k_{\bot }}{2}\left\{\!\frac{5\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}}{2}\left[\langle \tilde{A}_{\Vert }\rangle -\tilde{A}_{\Vert }^{(0)}\left(1+\frac{k_{\bot }^{2}c^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}\right)\right]+a\frac{3\unicode[STIX]{x1D716}}{2}\!\right\}\end{eqnarray}$$

and

(5.25) $$\begin{eqnarray}\displaystyle b & \simeq & \displaystyle \unicode[STIX]{x1D716}\tilde{B}_{\Vert }^{(0)}-\frac{\text{i}q\unicode[STIX]{x1D6FD}k_{\bot }}{2}(3-\unicode[STIX]{x1D70D}\unicode[STIX]{x1D716}^{1/2})\left[\langle \tilde{A}_{\Vert }\rangle -\tilde{A}_{\Vert }^{(0)}\left(1+\frac{k_{\bot }^{2}c^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}\right)\right]\nonumber\\ \displaystyle & & \displaystyle -\,\frac{\text{i}\unicode[STIX]{x1D712}\unicode[STIX]{x1D716}^{3/2}q\unicode[STIX]{x1D6FD}k_{\bot }a}{4}+\frac{c\unicode[STIX]{x1D70D}\unicode[STIX]{x1D6FD}_{i}k_{\bot }^{2}\unicode[STIX]{x1D70C}_{i}}{2\unicode[STIX]{x1D6FE}v_{i}}\tilde{\unicode[STIX]{x1D6F7}}^{(0)},\end{eqnarray}$$

where we used the lowest order version of (5.9),

(5.26) $$\begin{eqnarray}\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle \simeq \frac{\unicode[STIX]{x1D716}^{1/2}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}}{\unicode[STIX]{x1D6FE}q^{2}}.\end{eqnarray}$$

When $\tilde{B}_{\Vert }^{(0)}=0$ we may employ (5.18)–(5.21) to find $\langle \tilde{B}_{\Vert }\rangle$ and $b$ from (5.24) and (5.25). In the $\unicode[STIX]{x1D714}_{p}^{2}/k_{\bot }^{2}c^{2}\rightarrow 0$ limit (5.18) and (5.19) are used along with (5.24) and (5.25) to find

(5.27) $$\begin{eqnarray}\frac{v_{i}\langle \tilde{B}_{\Vert }\rangle }{k_{\bot }c}\rightarrow \frac{\unicode[STIX]{x1D6FD}_{i}k_{\bot }\unicode[STIX]{x1D70C}_{i}}{2}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}-\text{i}\frac{5\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}q\unicode[STIX]{x1D6FD}k_{\bot }^{2}c^{2}}{4\unicode[STIX]{x1D714}_{p}^{2}}\frac{v_{i}}{c}\tilde{A}_{\Vert }^{(0)}\end{eqnarray}$$

and

(5.28) $$\begin{eqnarray}\frac{v_{i}b}{k_{\bot }c}\rightarrow \text{i}q\unicode[STIX]{x1D6FD}\frac{3k_{\bot }^{2}c^{2}}{2\unicode[STIX]{x1D714}_{p}^{2}}\frac{v_{i}}{c}\tilde{A}_{\Vert }^{(0)}+\frac{\unicode[STIX]{x1D70D}\unicode[STIX]{x1D6FD}_{i}k_{\bot }\unicode[STIX]{x1D70C}_{i}}{2\unicode[STIX]{x1D6FE}}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}.\end{eqnarray}$$

Consequently, when $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}=0$ we see that $\langle \tilde{B}_{\Vert }\rangle /b\simeq -5\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}/6$ so strong poloidal variation occurs in $\tilde{B}_{\Vert }$ . For $\tilde{A}_{\Vert }^{(0)}=0$ the poloidal variation of $\tilde{B}_{\Vert }$ is strong since $b/\langle \tilde{B}_{\Vert }\rangle \simeq \unicode[STIX]{x1D70D}/\unicode[STIX]{x1D6FE}\approx 3.2$ .

For $\tilde{B}_{\Vert }^{(0)}=0$ , $\unicode[STIX]{x1D6FD}\ll 1$ , and $\unicode[STIX]{x1D716}^{3/2}\gg k_{\bot }^{2}c^{2}/\unicode[STIX]{x1D714}_{p}^{2}$ we use (5.20) and (5.21) to find the small skin depth forms

(5.29) $$\begin{eqnarray}\frac{v_{i}\langle \tilde{B}_{\Vert }\rangle }{k_{\bot }c}\simeq \frac{\unicode[STIX]{x1D716}\unicode[STIX]{x1D6FD}_{i}k_{\bot }\unicode[STIX]{x1D70C}_{pi}}{2q}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}-\frac{\text{i}3\unicode[STIX]{x1D716}^{2}\unicode[STIX]{x1D6FD}qv_{i}}{4c}\tilde{A}_{\Vert }^{(0)}\end{eqnarray}$$

and

(5.30) $$\begin{eqnarray}\frac{v_{i}b}{k_{\bot }c}\simeq \frac{\text{i}3\unicode[STIX]{x1D6FD}qv_{i}}{4c}\left(\unicode[STIX]{x1D716}^{2}+\frac{2k_{\bot }^{2}c^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}\right)\tilde{A}_{\Vert }^{(0)}+\frac{\unicode[STIX]{x1D70D}\unicode[STIX]{x1D716}\unicode[STIX]{x1D6FD}_{i}k_{\bot }\unicode[STIX]{x1D70C}_{pi}}{2\unicode[STIX]{x1D6FE}q}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}.\end{eqnarray}$$

When $\tilde{A}_{\Vert }^{(0)}=0$ the poloidal variation of $\tilde{B}_{\Vert }$ is strong since $b/\langle \tilde{B}_{\Vert }\rangle \simeq \unicode[STIX]{x1D70D}/\unicode[STIX]{x1D6FE}\approx 3.2$ . For $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}=0$ it is also strong with $b/\langle \tilde{B}_{\Vert }\rangle \simeq -[1+(2k_{\bot }^{2}c^{2}/\unicode[STIX]{x1D716}^{2}\unicode[STIX]{x1D714}_{p}^{2})]$ . Consequently, in the small skin depth limit, for $\tilde{B}_{\Vert }^{(0)}=0$ the poloidal variation of $\tilde{B}_{\Vert }$ is strong when $\tilde{A}_{\Vert }^{(0)}=0$ and weak when $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}=0$ .

Next we consider the case $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}=0$ and $\tilde{A}_{\Vert }^{(0)}=0$ when $\tilde{B}_{\Vert }^{(0)}\neq 0$ by using (5.22)–(5.25). The $\unicode[STIX]{x1D714}_{p}^{2}/k_{\bot }^{2}c^{2}\rightarrow 0$ limit gives

(5.31) $$\begin{eqnarray}\displaystyle & \displaystyle k_{\bot }a\rightarrow \text{i}q\tilde{B}_{\Vert }^{(0)}, & \displaystyle\end{eqnarray}$$

(5.32) $$\begin{eqnarray}\displaystyle & \displaystyle k_{\bot }\langle \tilde{A}_{\Vert }\rangle \rightarrow \text{i}q\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}\tilde{B}_{\Vert }^{(0)} & \displaystyle\end{eqnarray}$$

(5.33) $$\begin{eqnarray}\displaystyle & \displaystyle \langle \tilde{B}_{\Vert }\rangle \rightarrow [1-(3+5\unicode[STIX]{x1D6FE}^{2})(\unicode[STIX]{x1D716}q^{2}\unicode[STIX]{x1D6FD}/4)]\tilde{B}_{\Vert }^{(0)}, & \displaystyle\end{eqnarray}$$

and

(5.34) $$\begin{eqnarray}b\rightarrow (\unicode[STIX]{x1D716}+3\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}q^{2}\unicode[STIX]{x1D6FD}/2)\tilde{B}_{\Vert }^{(0)}.\end{eqnarray}$$

Consequently, the poloidal variation of $\tilde{B}_{\Vert }$ is weak (with $b/\langle \tilde{B}_{\Vert }\rangle \simeq \unicode[STIX]{x1D716}$ ), while the poloidal variation of $\tilde{A}_{\Vert }$ is strong (with $\langle \tilde{A}_{\Vert }\rangle /a\simeq \unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}$ ) in this large skin depth limit.

In the small skin depth limit $\unicode[STIX]{x1D716}^{3/2}\gg k_{\bot }^{2}c^{2}/\unicode[STIX]{x1D714}_{p}^{2}$ for finite $\unicode[STIX]{x1D6FD}\ll 1$ and $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}=0=\tilde{A}_{\Vert }^{(0)}$ we use (5.22) and (5.23) to find $\unicode[STIX]{x1D716}a\simeq 2(1+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{3/2})\langle \tilde{A}_{\Vert }\rangle$ . Then (5.22) to (5.25) give

(5.35) $$\begin{eqnarray}\displaystyle & \displaystyle k_{\bot }a\simeq \frac{\text{i}q\tilde{B}_{\Vert }^{(0)}}{1+(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{7/2}\unicode[STIX]{x1D714}_{p}^{2}/2k_{\bot }^{2}c^{2})}, & \displaystyle\end{eqnarray}$$

(5.36) $$\begin{eqnarray}\displaystyle & \displaystyle k_{\bot }\langle \tilde{A}_{\Vert }\rangle \simeq \frac{\text{i}\unicode[STIX]{x1D716}q\tilde{B}_{\Vert }^{(0)}}{2+(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{7/2}\unicode[STIX]{x1D714}_{p}^{2}/k_{\bot }^{2}c^{2})} & \displaystyle\end{eqnarray}$$

(5.37) $$\begin{eqnarray}\displaystyle & \displaystyle \langle \tilde{B}_{\Vert }\rangle \simeq \tilde{B}_{\Vert }^{(0)}\left\{1-\frac{3\unicode[STIX]{x1D716}\unicode[STIX]{x1D6FD}q^{2}}{2[2+(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{7/2}\unicode[STIX]{x1D714}_{p}^{2}/k_{\bot }^{2}c^{2})]}\right\}, & \displaystyle\end{eqnarray}$$

and

(5.38) $$\begin{eqnarray}b\simeq \tilde{B}_{\Vert }^{(0)}\left\{\unicode[STIX]{x1D716}+\frac{3\unicode[STIX]{x1D716}\unicode[STIX]{x1D6FD}q^{2}}{2[2+(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{7/2}\unicode[STIX]{x1D714}_{p}^{2}/k_{\bot }^{2}c^{2})]}\right\}.\end{eqnarray}$$

As a result, for this $\tilde{B}_{\Vert }^{(0)}\neq 0$ case with $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}=0$ and $\tilde{A}_{\Vert }^{(0)}=0$ , the poloidal variation of $\tilde{A}_{\Vert }$ is strong with $\langle \tilde{A}_{\Vert }\rangle /a\simeq \unicode[STIX]{x1D716}/2$ , while the poloidal variation of $\tilde{B}_{\Vert }$ varies from weak to strong with $b/\langle \tilde{B}_{\Vert }\rangle \simeq \unicode[STIX]{x1D716}$ to $1$ .

5.4 Quasineutrality

To complete our treatment of the Maxwell equations we need to form quasineutrality with its finite $\unicode[STIX]{x1D6FD}$ effects retained using

(5.39) $$\begin{eqnarray}\tilde{\unicode[STIX]{x1D6F7}}\sum \frac{Z^{2}\text{e}^{2}n}{T}=\sum Ze\int \text{d}^{3}v\overline{\tilde{h}_{\ast }(t)}\text{e}^{\text{i}(L-Q)}=\sum Ze\int \text{d}^{3}v\overline{\tilde{h}_{\ast }(t)}\text{J}_{0}\text{e}^{-\text{i}Q}.\end{eqnarray}$$

After inserting (3.11) and (4.11), we only retain Bessel function modifications to $\tilde{\unicode[STIX]{x1D6F7}}\simeq \langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$ and $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ terms. We then find

(5.40) $$\begin{eqnarray}\displaystyle & & \displaystyle \tilde{\unicode[STIX]{x1D6F7}}\sum \frac{Z^{2}\text{e}^{2}n}{T}=(\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle -\tilde{\unicode[STIX]{x1D6F7}}^{(0)})\sum \frac{Z^{2}\text{e}^{2}}{T}\int \text{d}^{3}vf_{0}\text{J}_{0}\text{e}^{-\text{i}Q}\overline{\text{J}_{0}\text{e}^{\text{i}Q}}\nonumber\\ \displaystyle & & \displaystyle \quad +\,\tilde{\unicode[STIX]{x1D6F7}}^{(0)}\sum \frac{Z^{2}\text{e}^{2}}{T}\int \text{d}^{3}vf_{0}H\text{J}_{0}\text{e}^{-\text{i}Q}\overline{(\text{e}^{\text{i}Q}/\text{J}_{0})}\nonumber\\ \displaystyle & & \displaystyle \quad -\,\sum \frac{Z^{2}\text{e}^{2}}{cT}\int \text{d}^{3}vf_{0}\text{e}^{-\text{i}Q}[\overline{\tilde{A}_{\Vert }v_{\Vert }\text{e}^{\text{i}Q}}-\tilde{A}_{\Vert }^{(0)}(1+k_{\bot }^{2}c^{2}/\unicode[STIX]{x1D714}_{p}^{2})\overline{v_{\Vert }\text{e}^{\text{i}Q}}]\nonumber\\ \displaystyle & & \displaystyle \quad -\,\tilde{B}_{\Vert }^{(0)}\frac{2\text{}\underline{B}_{0}}{\unicode[STIX]{x1D6FD}B_{0}^{2}}\sum Ze\int \text{d}^{3}vf_{0}\text{e}^{-\text{i}Q}\overline{\text{e}^{\text{i}Q}}+\sum \frac{ZeM}{2TB_{0}}\int \text{d}^{3}vf_{0}v_{\bot }^{2}\text{e}^{-\text{i}Q}(\overline{\tilde{B}_{\Vert }\text{e}^{\text{i}Q}}-\tilde{B}_{\Vert }^{(0)}\overline{\text{e}^{\text{i}Q}}).\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Expanding the Bessel functions and the exponentials in $Q$ , inserting $H$ with $\unicode[STIX]{x1D6FC}=k_{\bot }^{2}T/2M\unicode[STIX]{x1D6FA}^{2}$ , and using unperturbed quasineutrality leaves

(5.41) $$\begin{eqnarray}\displaystyle 0 & = & \displaystyle \langle \tilde{\unicode[STIX]{x1D6F7}}\rangle \sum \frac{Z^{2}\text{e}^{2}}{T}\int \text{d}^{3}vf_{0}\left[Q\bar{Q}-\frac{Q^{2}+\overline{Q^{2}}}{2}\right]-(\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle -\tilde{\unicode[STIX]{x1D6F7}}^{(0)})\sum \frac{Z^{2}\text{e}^{2}n\mathop{k}_{\bot }^{2}}{M\unicode[STIX]{x1D6FA}^{2}}\nonumber\\ \displaystyle & & \displaystyle -\,\text{i}\sum \frac{Z^{2}\text{e}^{2}}{2cT}\int \text{d}^{3}vf_{0}\left[\overline{\tilde{A}_{\Vert }v_{\Vert }Q}Q^{2}-\overline{\tilde{A}_{\Vert }v_{\Vert }Q^{2}}Q-\frac{\overline{\tilde{A}_{\Vert }v_{\Vert }}Q^{3}-\overline{\tilde{A}_{\Vert }v_{\Vert }Q^{3}}}{3}\right]\nonumber\\ \displaystyle & & \displaystyle +\,\tilde{A}_{\Vert }^{(0)}\left(1+\frac{k_{\bot }^{2}c^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}\right)\text{i}\sum \frac{Z^{2}\text{e}^{2}}{2cT}\int \text{d}^{3}vf_{0}\left[\overline{v_{\Vert }Q}Q^{2}-\overline{v_{\Vert }Q^{2}}Q-\frac{\overline{v_{\Vert }}Q^{3}-\overline{v_{\Vert }Q^{3}}}{3}\right]\nonumber\\ \displaystyle & & \displaystyle -\,\tilde{B}_{\Vert }^{(0)}\frac{2\text{}\underline{B}_{0}}{\unicode[STIX]{x1D6FD}B_{0}^{2}}\sum Ze\int \text{d}^{3}vf_{0}[Q\bar{Q}-(Q^{2}+\overline{Q^{2}})/2]\nonumber\\ \displaystyle & & \displaystyle +\,\sum \frac{ZeM}{2TB_{0}}\int \text{d}^{3}vf_{0}v_{\bot }^{2}\left\{\left[Q\overline{\tilde{B}_{\Vert }Q}-\frac{Q^{2}\overline{\tilde{B}_{\Vert }}+\overline{Q^{2}\tilde{B}_{\Vert }}}{2}\right]-\tilde{B}_{\Vert }^{(0)}\left[Q\bar{Q}-\frac{Q^{2}+\overline{Q^{2}}}{2}\right]\right\}.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

We have already shown that the poloidal variation of $\tilde{\unicode[STIX]{x1D6F7}}$ is negligible so we only require the flux surface average of (5.41), which is

(5.42) $$\begin{eqnarray}\displaystyle & & \displaystyle \langle \tilde{\unicode[STIX]{x1D6F7}}\rangle \left[\sum \frac{Z^{2}\text{e}^{2}}{T}\left\langle \int \text{d}^{3}vf_{0}Q(Q-\bar{Q})\right\rangle +\sum \frac{Z^{2}\text{e}^{2}n\mathop{k}_{\bot }^{2}}{M\unicode[STIX]{x1D6FA}^{2}}\right]=\tilde{\unicode[STIX]{x1D6F7}}^{(0)}\sum \frac{Z^{2}\text{e}^{2}n\mathop{k}_{\bot }^{2}}{M\unicode[STIX]{x1D6FA}^{2}}\nonumber\\ \displaystyle & & \displaystyle -\text{i}\sum \frac{Z^{2}\text{e}^{2}}{2cT}\left\langle \left[\tilde{A}_{\Vert }-\tilde{A}_{\Vert }^{(0)}\left(1+\frac{k_{\bot }^{2}c^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}\right)\right]\int \text{d}^{3}vf_{0}v_{\Vert }\left[\overline{Q^{2}}Q-\overline{Q}Q^{2}+\frac{Q^{3}-\overline{Q^{3}}}{3}\right]\right\rangle \nonumber\\ \displaystyle & & \displaystyle +\,\tilde{B}_{\Vert }^{(0)}\frac{2\text{}\underline{B}_{0}}{\unicode[STIX]{x1D6FD}B_{0}^{2}}\unicode[STIX]{x1D6F4}Ze\left\langle \int \text{d}^{3}vf_{0}Q(Q-\bar{Q})\right\rangle ,\end{eqnarray}$$

where we neglect $\unicode[STIX]{x1D6FD}\langle \tilde{B}_{\Vert }\rangle$ , $\unicode[STIX]{x1D6FD}\tilde{B}_{\Vert }^{(0)}$ and $\unicode[STIX]{x1D6FD}b$ corrections.

Performing the integrals using (A 30), (A 76), (A 79), (A 82), and (A 83), and noting that only the trapped ion and Bessel contributions matter, we obtain

(5.43) $$\begin{eqnarray}\displaystyle & & \displaystyle (\unicode[STIX]{x1D6FE}q^{2}\unicode[STIX]{x1D716}^{-1/2}+1)\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle =\tilde{\unicode[STIX]{x1D6F7}}^{(0)}+\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}q\unicode[STIX]{x1D70C}_{pi}\frac{v_{i}}{c\unicode[STIX]{x1D6FD}}\tilde{B}_{\Vert }^{(0)}\nonumber\\ \displaystyle & & \displaystyle \quad -\,\text{i}\frac{v_{i}}{4c}\unicode[STIX]{x1D716}^{1/2}q^{2}k_{\bot }\unicode[STIX]{x1D70C}_{pi}\left\{\unicode[STIX]{x1D70E}\left[\langle \tilde{A}_{\Vert }\rangle -\tilde{A}_{\Vert }^{(0)}\left(1+\frac{k_{\bot }^{2}c^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}\right)\right]+\frac{5\unicode[STIX]{x1D6FE}a}{2}\right\}.\end{eqnarray}$$

In the limit in which we take $\unicode[STIX]{x1D6FD}\rightarrow 0$ at finite $k_{\bot }$ (such that $\unicode[STIX]{x1D714}_{p}^{2}/k_{\bot }^{2}c^{2}\rightarrow 0$ ), equation (5.43) with (5.19) and (5.20) inserted gives the large skin depth expression

(5.44) $$\begin{eqnarray}\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle \rightarrow \left(\frac{\unicode[STIX]{x1D6FE}q^{2}}{\unicode[STIX]{x1D716}^{1/2}}+1\right)^{-1}\left[\tilde{\unicode[STIX]{x1D6F7}}^{(0)}+\text{i}\frac{v_{i}\tilde{A}_{\Vert }^{(0)}}{4c}\frac{k_{\bot }^{2}c^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}(\unicode[STIX]{x1D70E}\unicode[STIX]{x1D716}^{1/2}q^{2}k_{\bot }\unicode[STIX]{x1D70C}_{pi})+\frac{\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}}{\unicode[STIX]{x1D6FD}}\frac{q\unicode[STIX]{x1D70C}_{pi}v_{i}\tilde{B}_{\Vert }^{(0)}}{c}\right].\end{eqnarray}$$

In the more interesting finite limit, for which $\unicode[STIX]{x1D716}^{3/2}\gg k_{\bot }^{2}c^{2}/\unicode[STIX]{x1D714}_{p}^{2}$ , using (5.20), (5.21), (5.35), and (5.36) we find the small skin depth result

(5.45) $$\begin{eqnarray}\displaystyle \langle \tilde{\unicode[STIX]{x1D6F7}}\rangle & \simeq & \displaystyle \frac{\tilde{\unicode[STIX]{x1D6F7}}^{(0)}}{(\unicode[STIX]{x1D6FE}q^{2}\unicode[STIX]{x1D716}^{-1/2}+1)}\left[1+\frac{\unicode[STIX]{x1D70E}+5\unicode[STIX]{x1D6FE}}{16}\unicode[STIX]{x1D6FD}_{i}\unicode[STIX]{x1D716}^{3/2}q^{2}k_{\bot }^{2}\unicode[STIX]{x1D70C}_{pi}^{2}\right]\nonumber\\ \displaystyle & & \displaystyle +\,\frac{\text{i}5\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{3/2}q^{2}k_{\bot }\unicode[STIX]{x1D70C}_{pi}v_{i}\tilde{A}_{\Vert }^{(0)}}{8(\unicode[STIX]{x1D6FE}q^{2}\unicode[STIX]{x1D716}^{-1/2}+1)c}+\frac{\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}qk_{\bot }\unicode[STIX]{x1D70C}_{pi}v_{i}\tilde{B}_{\Vert }^{(0)}}{(\unicode[STIX]{x1D6FE}q^{2}\unicode[STIX]{x1D716}^{-1/2}+1)\unicode[STIX]{x1D6FD}k_{\bot }c}.\end{eqnarray}$$

Clearly, the modification of the electrostatic limit is small when $\tilde{B}_{\Vert }^{(0)}=0=\tilde{A}_{\Vert }^{(0)}$ . From (5.45), we see that initial magnetic perturbations as large as $\unicode[STIX]{x1D716}^{1/2}q\unicode[STIX]{x1D70C}_{pi}v_{i}\tilde{B}_{\Vert }^{(0)}/c\sim \unicode[STIX]{x1D6FD}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ and/or $\unicode[STIX]{x1D716}^{3/2}q^{2}k_{\bot }\unicode[STIX]{x1D70C}_{pi}v_{i}\tilde{A}_{\Vert }^{(0)}/c\sim \tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ are required to obtain order unity corrections to the electrostatic response. For the same size initial perturbations in the expression (5.44) the response to $\tilde{B}_{\Vert }^{(0)}$ remains the same, but that due to $\tilde{A}_{\Vert }^{(0)}$ is very small.

6 Zonal flow responses in terms of initial field values: 15 test cases

In this section we present a summary of our results in the small skin depth limit. Before doing so we note that

(6.1) $$\begin{eqnarray}\frac{k_{\bot }^{2}c^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}=\frac{Z(1+\unicode[STIX]{x1D70F})\unicode[STIX]{x1D716}^{2}m}{q^{2}\unicode[STIX]{x1D6FD}M}k_{\bot }^{2}\unicode[STIX]{x1D70C}_{pi}^{2}=\frac{Z(1+\unicode[STIX]{x1D70F})m}{\unicode[STIX]{x1D6FD}M}k_{\bot }^{2}\unicode[STIX]{x1D70C}_{i}^{2}=\frac{(1+\unicode[STIX]{x1D70F})}{\unicode[STIX]{x1D70F}\unicode[STIX]{x1D6FD}}k_{\bot }^{2}\unicode[STIX]{x1D70C}_{e}^{2},\end{eqnarray}$$

with $\unicode[STIX]{x1D6FD}=\unicode[STIX]{x1D6FD}_{i}(1+\unicode[STIX]{x1D70F})$ and $\unicode[STIX]{x1D70F}=ZT_{e}/T_{i}$ . We, of course, neglect mass ratio corrections in the plasma frequency.

The large skin depth limit of $\unicode[STIX]{x1D6FD}\rightarrow 0$ requires $q^{2}\unicode[STIX]{x1D6FD}\ll Zmk_{\bot }^{2}\unicode[STIX]{x1D70C}_{pi}^{2}/M\ll Zm/M$ since we must keep $k_{\bot }^{2}\unicode[STIX]{x1D70C}_{pi}^{2}\ll 1$ for our analysis to hold. Consequently, the more interesting limit is that of small skin depth $\unicode[STIX]{x1D716}^{2}\sim \unicode[STIX]{x1D716}\unicode[STIX]{x1D6E5}_{S}^{\prime }\gg k_{\bot }^{2}c^{2}/\unicode[STIX]{x1D714}_{p}^{2}$ and small $\unicode[STIX]{x1D6FD}_{i}$ ( $1\gg \unicode[STIX]{x1D6FD}_{i}\gtrsim Zmk_{\bot }^{2}\unicode[STIX]{x1D70C}_{pi}^{2}/q^{2}M$ ), which will be our focus. In the following we summarize our results for this limit. We will find that electromagnetic perturbations give an electrostatic potential response comparable to the electrostatic limit of $\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle \simeq \unicode[STIX]{x1D716}^{1/2}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}/\unicode[STIX]{x1D6FE}q^{2}$ when $v_{i}\tilde{A}_{\Vert }^{(0)}\unicode[STIX]{x1D716}^{3/2}q^{2}k_{\bot }\unicode[STIX]{x1D70C}_{pi}/c\sim \tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ or $\unicode[STIX]{x1D716}^{1/2}q\unicode[STIX]{x1D70C}_{pi}v_{i}\tilde{B}_{\Vert }^{(0)}/c\sim \unicode[STIX]{x1D6FD}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ .

6.1 Electromagnetic response to an electrostatic initial perturbation ( $\tilde{B}_{\Vert }^{(0)}=0=\tilde{A}_{\Vert }^{(0)}$ )

In this case $\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle /\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ has a very small, linear in $\unicode[STIX]{x1D6FD}$ , correction to the electrostatic Rosenbluth and Hinton form as seen from (5.45),

(6.2) $$\begin{eqnarray}\left(\frac{\unicode[STIX]{x1D6FE}q^{2}}{\unicode[STIX]{x1D716}^{1/2}}+1\right)\frac{\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle }{\tilde{\unicode[STIX]{x1D6F7}}^{(0)}}\simeq 1+\frac{\unicode[STIX]{x1D70E}+5\unicode[STIX]{x1D6FE}}{16}\unicode[STIX]{x1D6FD}_{i}\unicode[STIX]{x1D716}^{3/2}q^{2}k_{\bot }^{2}\unicode[STIX]{x1D70C}_{pi}^{2},\end{eqnarray}$$

where $\unicode[STIX]{x1D70E}\approx 5.3$ . Moreover, $\langle \tilde{B}_{\Vert }\rangle \unicode[STIX]{x1D70C}_{i}v_{i}/c\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ and $b\unicode[STIX]{x1D70C}_{i}v_{i}/c\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ are linear in $\unicode[STIX]{x1D6FD}$ and are seen from (5.29) and (5.30) to be given by

(6.3) $$\begin{eqnarray}\frac{v_{i}\langle \tilde{B}_{\Vert }\rangle }{k_{\bot }c\tilde{\unicode[STIX]{x1D6F7}}^{(0)}}\simeq \frac{\unicode[STIX]{x1D716}\unicode[STIX]{x1D6FD}_{i}k_{\bot }\unicode[STIX]{x1D70C}_{pi}}{2q}\end{eqnarray}$$

and

(6.4) $$\begin{eqnarray}\frac{v_{i}b}{k_{\bot }c\tilde{\unicode[STIX]{x1D6F7}}^{(0)}}\simeq \frac{\unicode[STIX]{x1D70D}\unicode[STIX]{x1D716}\unicode[STIX]{x1D6FD}_{i}k_{\bot }\unicode[STIX]{x1D70C}_{pi}}{2\unicode[STIX]{x1D6FE}q}.\end{eqnarray}$$

From (5.20) and (5.21) we find

(6.5) $$\begin{eqnarray}\frac{v_{i}\langle \tilde{A}_{\Vert }\rangle }{\text{i}c\tilde{\unicode[STIX]{x1D6F7}}^{(0)}}\simeq \frac{\unicode[STIX]{x1D716}\unicode[STIX]{x1D6FD}_{i}k_{\bot }\unicode[STIX]{x1D70C}_{pi}}{4},\end{eqnarray}$$

and

(6.6) $$\begin{eqnarray}\frac{v_{i}a}{\text{i}c\tilde{\unicode[STIX]{x1D6F7}}^{(0)}}\simeq \frac{\unicode[STIX]{x1D716}\unicode[STIX]{x1D6FD}_{i}k_{\bot }\unicode[STIX]{x1D70C}_{pi}}{2},\end{eqnarray}$$

where all electromagnetic responses are proportional to $\unicode[STIX]{x1D6FD}$ . The correction to the electrostatic result is very small, but all electromagnetic responses are larger and of the same order so strong poloidal variation occurs.

6.2 Electromagnetic response to a plucked field line initial condition ( $\tilde{B}_{\Vert }^{(0)}=0=\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ )

Using (5.45) gives the electrostatic potential response for this $\tilde{A}_{\Vert }^{(0)}\neq 0$ case to be

(6.7) $$\begin{eqnarray}\frac{c\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle }{\text{i}v_{i}\tilde{A}_{\Vert }^{(0)}}\simeq \frac{5\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{3/2}q^{2}k_{\bot }\unicode[STIX]{x1D70C}_{pi}}{8(\unicode[STIX]{x1D6FE}q^{2}\unicode[STIX]{x1D716}^{-1/2}+1)}\simeq \frac{5\unicode[STIX]{x1D716}^{2}k_{\bot }\unicode[STIX]{x1D70C}_{pi}}{8}.\end{eqnarray}$$

For this case, the parallel vector potential responses from (5.20) and (5.21) are given by

(6.8) $$\begin{eqnarray}\frac{\langle \tilde{A}_{\Vert }\rangle }{\tilde{A}_{\Vert }^{(0)}}\simeq 1-\frac{\unicode[STIX]{x1D716}^{2}}{2}\end{eqnarray}$$

and

(6.9) $$\begin{eqnarray}\frac{a}{\tilde{A}_{\Vert }^{(0)}}\simeq -\unicode[STIX]{x1D716},\end{eqnarray}$$

showing that $\tilde{A}_{\Vert }$ will only be very slightly perturbed from $\tilde{A}_{\Vert }^{(0)}$ with weak poloidal variation due to field line plucking satisfying $R\tilde{A}_{\Vert }\simeq R_{0}(\unicode[STIX]{x1D713})\tilde{A}_{\Vert }^{(0)}$ . The $\unicode[STIX]{x1D6FD}$ proportional $\tilde{B}_{\Vert }$ responses from (5.29) and (5.30) give $b$ and $\langle \tilde{B}_{\Vert }\rangle$ as comparable, as seen from

(6.10) $$\begin{eqnarray}\frac{\text{i}\langle \tilde{B}_{\Vert }\rangle }{k_{\bot }\tilde{A}_{\Vert }^{(0)}}\simeq \frac{3\unicode[STIX]{x1D716}^{2}\unicode[STIX]{x1D6FD}q}{4}\end{eqnarray}$$

and

(6.11) $$\begin{eqnarray}\frac{b}{\text{i}k_{\bot }\tilde{A}_{\Vert }^{(0)}}\simeq \frac{3\unicode[STIX]{x1D6FD}q}{4}\left(\unicode[STIX]{x1D716}^{2}+\frac{2k_{\bot }^{2}c^{2}}{\unicode[STIX]{x1D714}_{p}^{2}}\right).\end{eqnarray}$$

6.3 Electromagnetic response to a compressed field line initial condition ( $\tilde{A}_{\Vert }^{(0)}=0=\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ )

The response $ck_{\bot }\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle /v_{i}\tilde{B}_{\Vert }^{(0)}$ is proportional to $1/\unicode[STIX]{x1D6FD}$ as seen from (5.45) so in this $\tilde{B}_{\Vert }^{(0)}\neq 0$ case we multiply through by $\unicode[STIX]{x1D6FD}$ to form

(6.12) $$\begin{eqnarray}\frac{c\unicode[STIX]{x1D6FD}k_{\bot }\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle }{v_{i}\tilde{B}_{\Vert }^{(0)}}\simeq \frac{\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{1/2}qk_{\bot }\unicode[STIX]{x1D70C}_{pi}}{\unicode[STIX]{x1D6FE}q^{2}\unicode[STIX]{x1D716}^{-1/2}+1}\simeq \frac{\unicode[STIX]{x1D716}k_{\bot }\unicode[STIX]{x1D70C}_{pi}}{q}.\end{eqnarray}$$

The responses $\langle \tilde{B}_{\Vert }\rangle /\tilde{B}_{\Vert }^{(0)}$ and $b/\tilde{B}_{\Vert }^{(0)}$ are given by (5.37) and (5.38) for this field line stretching case:

(6.13) $$\begin{eqnarray}\frac{\langle \tilde{B}_{\Vert }\rangle }{\tilde{B}_{\Vert }^{(0)}}\simeq 1-\frac{3\unicode[STIX]{x1D716}q^{2}\unicode[STIX]{x1D6FD}}{2[2+(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{7/2}\unicode[STIX]{x1D714}_{p}^{2}/k_{\bot }^{2}c^{2})]}\simeq 1\end{eqnarray}$$

and

(6.14) $$\begin{eqnarray}\frac{b}{\tilde{B}_{\Vert }^{(0)}}\simeq \unicode[STIX]{x1D716}+\frac{3\unicode[STIX]{x1D716}q^{2}\unicode[STIX]{x1D6FD}}{2[2+(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{7/2}\unicode[STIX]{x1D714}_{p}^{2}/k_{\bot }^{2}c^{2})]}\simeq \unicode[STIX]{x1D716}.\end{eqnarray}$$

The responses $k_{\bot }\langle \tilde{A}_{\Vert }\rangle /\text{i}\tilde{B}_{\Vert }^{(0)}$ and $k_{\bot }a/\text{i}\tilde{B}_{\Vert }^{(0)}$ are obtained from (5.35) and (5.36)

(6.15) $$\begin{eqnarray}\frac{k_{\bot }\langle \tilde{A}_{\Vert }\rangle }{\text{i}\tilde{B}_{\Vert }^{(0)}}\simeq \frac{\unicode[STIX]{x1D716}q}{2+(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{7/2}\unicode[STIX]{x1D714}_{p}^{2}/k_{\bot }^{2}c^{2})}\end{eqnarray}$$

and

(6.16) $$\begin{eqnarray}\frac{k_{\bot }a}{\text{i}\tilde{B}_{\Vert }^{(0)}}\simeq \frac{q}{1+(\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D716}^{7/2}\unicode[STIX]{x1D714}_{p}^{2}/2k_{\bot }^{2}c^{2})}.\end{eqnarray}$$

Stretching or compressing field lines causes very strong poloidal variation in $\tilde{A}_{\Vert }$ but weak poloidal variation in $\tilde{B}_{\Vert }$ satisfying $R_{0}(\unicode[STIX]{x1D713})\tilde{B}_{\Vert }\simeq R\tilde{B}_{\Vert }^{(0)}$ .