Appendix A. Endless integrals

The second term in the $\unicode[STIX]{x1D705}\rightarrow 1$ limit of $\overline{\cos \unicode[STIX]{x1D703}}$ in (A 50) should be to the minus one power with a prefactor of $1/4$ rather than $1/2$.

The sign should be positive in the last form of (A 53) and both forms of (A 54).

Following (A 59), add the remark that ‘The $\unicode[STIX]{x1D717}$-independent contribution of order $\unicode[STIX]{x1D700}$ is neglected’.

Equation (A 61) should be

(A 61)$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70D} & = & \displaystyle \frac{30\sqrt{2}}{\unicode[STIX]{x03C0}}\left\{\int _{0}^{1}\frac{\text{d}\unicode[STIX]{x1D705}\unicode[STIX]{x1D705}[2\mathit{E}(\unicode[STIX]{x1D705})-\mathit{K}(\unicode[STIX]{x1D705})]^{2}}{(1-\unicode[STIX]{x1D700}+2\unicode[STIX]{x1D700}\unicode[STIX]{x1D705})^{7/2}\mathit{K}(\unicode[STIX]{x1D705})}+\int _{0}^{1}\frac{\text{d}\mathit{kk}[2\mathit{E}(\mathit{k})-(2-\mathit{k}^{2})\mathit{K}(\mathit{k})]^{2}}{[(1-\unicode[STIX]{x1D700})\mathit{k}^{2}+2\unicode[STIX]{x1D700}]^{7/2}\mathit{K}(\mathit{k})}\right\}\nonumber\\ \displaystyle & \simeq & \displaystyle \frac{30\sqrt{2}}{\unicode[STIX]{x03C0}}\left\{\int _{0}^{1}\frac{\text{d}\unicode[STIX]{x1D705}\unicode[STIX]{x1D705}}{\mathit{K}(\unicode[STIX]{x1D705})}[2\mathit{E}(\unicode[STIX]{x1D705})-\mathit{K}(\unicode[STIX]{x1D705})]^{2}+\int _{0}^{1}\frac{\text{d}\mathit{k}}{\mathit{k}^{6}\mathit{K}(\mathit{k})}[2\mathit{E}(\mathit{k})-(2-\mathit{k}^{2})\mathit{K}(\mathit{k})]^{2}\right\}\nonumber\\ \displaystyle & \simeq & \displaystyle 4.363.\end{eqnarray}$$

Equations (A 65) and (A 66) should read

(A 65)$$\begin{eqnarray}(\mathit{M}^{2}\text{}\underline{\mathit{B}}_{0}^{2}/2\mathit{nT}^{2}\mathit{B}_{0}^{2})\int \text{d}^{3}\mathit{vf}_{0}\mathit{v}_{\bot }^{2}(\mathit{v}_{\Vert }^{2}-\overline{\mathit{v}_{\Vert }^{2}})\simeq (4\unicode[STIX]{x1D700}-\unicode[STIX]{x1D70D}\unicode[STIX]{x1D700}^{3/2})\cos \unicode[STIX]{x1D703}+\cdots\end{eqnarray}$$

and

(A 66)$$\begin{eqnarray}(\mathit{M}^{2}\text{}\underline{\mathit{B}}_{0}^{2}/2\mathit{nT}^{2}\mathit{B}_{0}^{2})\int \text{d}^{3}\mathit{vf}_{0}\mathit{v}_{\bot }^{2}[\mathit{v}_{\Vert }\overline{\mathit{v}_{\Vert }}-(\mathit{v}_{\Vert }^{2}+\overline{\mathit{v}_{\Vert }^{2}})/2]\simeq -5\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D700}^{3/2}/2-(\unicode[STIX]{x1D70D}\unicode[STIX]{x1D700}^{3/2}/2)\cos \unicode[STIX]{x1D717}+\cdots \,.\end{eqnarray}$$

Equation (A 71) must be multiplied by 2, so that $\unicode[STIX]{x1D712}\simeq 0.2262$.

A more precise evaluation of the $\cos \unicode[STIX]{x1D717}$ coefficient in (A 80) replaces $\unicode[STIX]{x1D710}=5\unicode[STIX]{x1D6FE}\simeq 8.2$ by $\unicode[STIX]{x1D710}\equiv \unicode[STIX]{x03C0}^{-1}\unicode[STIX]{x1D700}^{-5/2}\oint \text{d}\unicode[STIX]{x1D703}\mathit{I}\cos \unicode[STIX]{x1D717}=7.96$.

This change affects (5.15), (5.19)–(5.21), (5.31), (5.43), (5.45), (6.2), and (6.5)–(6.7). In (5.15), (5.19), (5.31), (5.43), and (6.2), let $5\unicode[STIX]{x1D6FE}\rightarrow \unicode[STIX]{x1D710}=7.96$. In (5.20) and (5.21), make the replacement $1-\unicode[STIX]{x1D700}^{1/2}/2\unicode[STIX]{x1D6FE}\mathit{q}^{2}\rightarrow [(\unicode[STIX]{x1D710}-3\unicode[STIX]{x1D6FE})/2\unicode[STIX]{x1D6FE}][1-(\unicode[STIX]{x1D710}-4\unicode[STIX]{x1D6FE})\unicode[STIX]{x1D700}^{1/2}/(\unicode[STIX]{x1D710}-3\unicode[STIX]{x1D6FE})\mathit{q}^{2}]$. In (6.5) and (6.6), the same replacement gives final forms with $1\rightarrow (\unicode[STIX]{x1D710}-3\unicode[STIX]{x1D6FE})/2\unicode[STIX]{x1D6FE}$. In (5.45), let $5\unicode[STIX]{x1D6FE}\tilde{\mathit{A}}_{\Vert }^{(0)}\rightarrow \unicode[STIX]{x1D710}\tilde{\mathit{A}}_{\Vert }^{(0)}$ and $5\unicode[STIX]{x1D6FE}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}\rightarrow [\unicode[STIX]{x1D710}(\unicode[STIX]{x1D710}-3\unicode[STIX]{x1D6FE})/2\unicode[STIX]{x1D6FE}]\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$, while in (6.2) let $5\unicode[STIX]{x1D6FE}\rightarrow \unicode[STIX]{x1D710}(\unicode[STIX]{x1D710}-3\unicode[STIX]{x1D6FE})/2\unicode[STIX]{x1D6FE}$.

3 Kinetic equation and solution

Two lines above (3.8), $\unicode[STIX]{x1D6E5}_{\mathit{s}}$ should be $\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime }$, so the result reads $\mathit{B}_{\mathit{p}}/\mathit{B}_{\mathit{t}}=(\unicode[STIX]{x1D700}/\mathit{q})[1-\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime }\cos \unicode[STIX]{x1D717}+\mathit{O}(\unicode[STIX]{x1D700}^{2})]$.

5 Quasineutrality and Ampere’s law

Remove the last sentence of the paragraph ending after (5.9).

The left side of (5.10) and the beginning of the paragraph including it should read as follows. Expanding for $\mathit{Q}=\mathit{k}_{\bot }\mathit{v}_{\Vert }/\unicode[STIX]{x1D6FA}_{\mathit{p}}\ll 1$, inserting $H$, and neglecting $\unicode[STIX]{x1D700}$ corrections to $\mathit{k}_{\bot }^{2}\unicode[STIX]{x1D70C}_{\mathit{i}}^{2}$ terms, but retaining all $\mathit{Q}^{2}$, $\unicode[STIX]{x1D6FD}$, and $\unicode[STIX]{x1D700}^{3/2}$ corrections to the $\tilde{\mathit{B}}_{\Vert }$ and $\tilde{\mathit{B}}_{\Vert }^{(0)}$ terms in (5.5), recalling $\tilde{\unicode[STIX]{x1D6F7}}=\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$, and using unperturbed quasineutrality, the perpendicular Ampere’s law becomes

(5.10)$$\begin{eqnarray}\displaystyle & & \displaystyle \tilde{\mathit{B}}_{\Vert }+\unicode[STIX]{x1D6F4}\frac{\unicode[STIX]{x03C0}\mathit{M}^{2}}{\mathit{TB}_{0}^{2}}\int \text{d}^{3}\mathit{vf}_{0}\mathit{v}_{\bot }^{4}\overline{\tilde{\mathit{B}}_{\Vert }}-\tilde{\mathit{B}}_{\Vert }^{(0)}\left\{\frac{\text{}\underline{\mathit{B}}_{0}}{\mathit{B}_{0}}-\unicode[STIX]{x1D700}\unicode[STIX]{x1D6F4}\frac{4\unicode[STIX]{x03C0}\mathit{M}}{\unicode[STIX]{x1D6FD}\mathit{B}_{0}^{2}}\int \text{d}^{3}\mathit{vf}_{0}\mathit{v}_{\bot }^{2}\overline{\cos \unicode[STIX]{x1D717}}\right.\nonumber\\ \displaystyle & & \displaystyle \quad +\left.\unicode[STIX]{x1D6F4}\frac{4\unicode[STIX]{x03C0}\mathit{ M}}{\unicode[STIX]{x1D6FD}\mathit{B}_{0}^{3}}\int \text{d}^{3}\mathit{vf}_{0}\mathit{v}_{\bot }^{2}\left[\mathit{Q}\overline{\mathit{QB}_{0}}-\frac{1}{2}\left(\mathit{Q}^{2}\overline{\mathit{B}_{0}}+\overline{\mathit{Q}^{2}\mathit{B}_{0}}\right)\right]\right\}\simeq \text{right side}.\qquad\end{eqnarray}$$

In the $\tilde{\mathit{B}}_{\Vert }^{(0)}/\unicode[STIX]{x1D6FD}$ term of (5.11), the factor $\overline{\mathit{v}_{\Vert }}-\mathit{v}_{\Vert }$ should be replaced by $\overline{\mathit{v}_{\Vert }}-\mathit{B}_{0}^{-1}\mathit{v}_{\Vert }\overline{\mathit{B}_{0}}$ and in the last term we use $\int \text{d}^{3}\mathit{vf}_{0}\mathit{v}_{\Vert }(\mathit{Q}-\overline{\mathit{Q}})(\mathit{Mv}^{2}/2\mathit{T}-3/2)=\int \text{d}^{3}\mathit{vf}_{0}\mathit{v}_{\Vert }(\mathit{Q}-\overline{\mathit{Q}})$.

Correct the ‘$a$’ term on the right side of (5.12) by the replacement $3\unicode[STIX]{x1D700}\mathit{a}/2\rightarrow 3\mathit{a}/2$ and then fix the left side of (5.12) by keeping $\unicode[STIX]{x1D6FD}_{\mathit{i}}$ and $\unicode[STIX]{x1D6FD}$ corrections to read

(5.12)$$\begin{eqnarray}(1+\unicode[STIX]{x1D6FD})(\langle \tilde{\mathit{B}}_{\Vert }\rangle -\tilde{\mathit{B}}_{\Vert }^{(0)})+\frac{5\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD}_{\mathit{i}}}{4\unicode[STIX]{x1D6FD}}\unicode[STIX]{x1D700}^{3/2}\mathit{k}_{\bot }^{2}\unicode[STIX]{x1D70C}_{\mathit{pi}}^{2}\tilde{\mathit{B}}_{\Vert }^{(0)}\simeq \text{corrected right side}.\end{eqnarray}$$

Remove the last sentence of the paragraph that ends after (5.12).

In (5.13), remove the $\mathit{a}\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime }$ term on the left side and on the right remove the term proportional to $\unicode[STIX]{x1D6FD}_{\mathit{i}}\unicode[STIX]{x1D700}(\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle -\tilde{\unicode[STIX]{x1D6F7}}^{(0)})$ and other $\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$ and $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ terms that are negligible in what follows, to obtain

(5.13)$$\begin{eqnarray}\displaystyle & & \displaystyle \langle \tilde{\mathit{A}}_{\Vert }\rangle \left[1+\frac{\unicode[STIX]{x1D714}_{\mathit{p}}^{2}}{\mathit{k}_{\bot }^{2}\mathit{c}^{2}}(1-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D700}^{3/2})\right]-\tilde{\mathit{A}}_{\Vert }^{(0)}\left[\left(1+\frac{\unicode[STIX]{x1D714}_{\mathit{p}}^{2}}{\mathit{k}_{\bot }^{2}\mathit{c}^{2}}\right)(1-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D700}^{3/2})\right]\nonumber\\ \displaystyle & & \displaystyle \quad \simeq \mathit{a}\frac{\unicode[STIX]{x1D700}\unicode[STIX]{x1D714}_{\mathit{p}}^{2}}{2\mathit{k}_{\bot }^{2}\mathit{c}^{2}}(1-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D700}^{3/2})+\frac{\text{i}\unicode[STIX]{x1D6FE}\mathit{q}\unicode[STIX]{x1D700}^{1/2}}{\mathit{k}_{\bot }}\tilde{\mathit{B}}_{\Vert }^{(0)}+\text{i}\frac{\mathit{c}\unicode[STIX]{x1D70E}}{4\mathit{v }_{\mathit{i}}}\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{q}^{3}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{i}}\unicode[STIX]{x1D700}^{-1/2}\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle .\end{eqnarray}$$

In (5.14), make the replacements $\mathit{a}\unicode[STIX]{x1D712}\unicode[STIX]{x1D700}^{3/2}\rightarrow \mathit{a}\unicode[STIX]{x1D712}\unicode[STIX]{x1D700}^{1/2}$ and $\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle \rightarrow \langle \tilde{\unicode[STIX]{x1D6F7}}\rangle /2$ and fix the $\tilde{\mathit{B}}_{\Vert }^{(0)}$ and ‘$b$’ terms to read

(5.14)$$\begin{eqnarray}\displaystyle & & \displaystyle \left(1+\frac{\unicode[STIX]{x1D70D}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D700}^{1/2}}{4}\right)\mathit{b}\simeq \unicode[STIX]{x1D700}\tilde{\mathit{B}}_{\Vert }^{(0)}\left(1-\frac{\unicode[STIX]{x1D70D}}{5}\unicode[STIX]{x1D700}^{1/2}-\frac{\unicode[STIX]{x1D70D}\unicode[STIX]{x1D6FD}_{\mathit{i}}}{4\unicode[STIX]{x1D6FD}}\unicode[STIX]{x1D700}^{1/2}\mathit{k}_{\bot }^{2}\unicode[STIX]{x1D70C}_{\mathit{pi}}^{2}\right)+\frac{\mathit{c}\unicode[STIX]{x1D70D}\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{q}^{2}\mathit{k}_{\bot }^{2}\unicode[STIX]{x1D70C}_{\mathit{i}}}{4\unicode[STIX]{x1D700}^{1/2}\mathit{v}_{\mathit{i}}}\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle \nonumber\\ \displaystyle & & \displaystyle \quad -\frac{\text{i}\mathit{q}\unicode[STIX]{x1D6FD}\mathit{k}_{\bot }}{2}(3-\unicode[STIX]{x1D70D}\unicode[STIX]{x1D700}^{1/2})\left[\langle \tilde{\mathit{A}}_{\Vert }\rangle -\tilde{\mathit{A}}_{\Vert }^{(0)}\left(1+\frac{\mathit{k}_{\bot }^{2}\mathit{c}^{2}}{\unicode[STIX]{x1D714}_{\mathit{p}}^{2}}\right)\right]-\frac{\text{i}\unicode[STIX]{x1D712}\unicode[STIX]{x1D700}^{1/2}\mathit{q}\unicode[STIX]{x1D6FD}\mathit{k}_{\bot }\mathit{a}}{4}.\end{eqnarray}$$

Following (5.14), correct the constants to read $\unicode[STIX]{x1D712}\simeq 0.226$ and $\unicode[STIX]{x1D70D}\simeq 4.36$.

Correcting (5.15), because $\mathit{k}_{\bot }^{2}$ in (4.11) must be a flux function, whereas on the left side of (5.4) $\mathit{k}_{\bot }^{2}\rightarrow \mathit{k}_{\bot }^{2}(1+2\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime }\cos \unicode[STIX]{x1D717})$, and multiplying the $\tilde{\mathit{B}}_{\Vert }^{(0)}$ term by 2 gives

(5.15)$$\begin{eqnarray}\displaystyle & & \displaystyle \mathit{a}\left[1+\frac{\unicode[STIX]{x1D700}(\unicode[STIX]{x1D700}+2\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime })\unicode[STIX]{x1D714}_{\mathit{p}}^{2}}{2\mathit{k}_{\bot }^{2}\mathit{c}^{2}}\right]\simeq (\unicode[STIX]{x1D700}+2\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime })\left[\frac{\unicode[STIX]{x1D714}_{\mathit{p}}^{2}}{\mathit{k}_{\bot }^{2}\mathit{c}^{2}}\langle \tilde{\mathit{A}}_{\Vert }\rangle -\left(\frac{\unicode[STIX]{x1D714}_{\mathit{p}}^{2}}{\mathit{k}_{\bot }^{2}\mathit{c}^{2}}+1\right)\tilde{\mathit{A}}_{\Vert }^{(0)}\right]\nonumber\\ \displaystyle & & \displaystyle \quad +\,\frac{\text{i}2\mathit{q}}{\mathit{k}_{\bot }}\tilde{\mathit{B}}_{\Vert }^{(0)}+\frac{\text{i}\mathit{c}}{4\mathit{v }_{\mathit{i}}}\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{qk}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{i}}[(5\unicode[STIX]{x1D6FE}\mathit{q}^{2}\unicode[STIX]{x1D700}^{-1/2}+4)\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle -3\tilde{\unicode[STIX]{x1D6F7}}^{(0)}].\end{eqnarray}$$

Equations (5.16) and (5.17) are the same as (5.15) and (5.13), but with the $\tilde{\mathit{B}}_{\Vert }^{(0)}$ terms removed.

The large skin depth limit of (5.16) and (5.17) are

(5.18)$$\begin{eqnarray}\frac{\mathit{v}_{\mathit{i}}}{\mathit{c}}\langle \tilde{\mathit{A}}_{\Vert }\rangle \rightarrow \frac{\mathit{v}_{\mathit{i}}}{\mathit{c}}\tilde{\mathit{A}}_{\Vert }^{(0)}(1-\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D700}^{3/2})+\text{i}\frac{\unicode[STIX]{x1D70E}\unicode[STIX]{x1D6FD}_{\mathit{i}}}{4\unicode[STIX]{x1D700}^{1/2}}\mathit{q}^{3}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{i}}\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle\end{eqnarray}$$

and

(5.19)$$\begin{eqnarray}\frac{\mathit{v}_{\mathit{i}}}{\mathit{c}}\mathit{a}\rightarrow -\frac{\mathit{v}_{\mathit{i}}}{\mathit{c}}(\unicode[STIX]{x1D700}+2\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime })\tilde{\mathit{A}}_{\Vert }^{(0)}+\text{i}\frac{\unicode[STIX]{x1D6FD}_{\mathit{i}}}{4}\mathit{qk}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{i}}[(5\unicode[STIX]{x1D6FE}\mathit{q}^{2}\unicode[STIX]{x1D700}^{-1/2}+4)\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle -3\tilde{\unicode[STIX]{x1D6F7}}^{(0)}].\end{eqnarray}$$

The small skin depth limits ($\mathit{k}_{\bot }^{2}\mathit{c}^{2}/\unicode[STIX]{x1D714}_{\mathit{p}}^{2}\ll 1$) of (5.16) and (5.17), (5.20) and (5.21), should read

(5.20)$$\begin{eqnarray}\frac{\mathit{v}_{\mathit{i}}}{\mathit{c}}\langle \tilde{\mathit{A}}_{\Vert }\rangle \simeq \frac{\mathit{v}_{\mathit{i}}}{\mathit{c}}\left[1-\frac{1}{2}\unicode[STIX]{x1D700}(\unicode[STIX]{x1D700}+2\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime })\right]\tilde{\mathit{A}}_{\Vert }^{(0)}+\text{i}\frac{1}{4}\unicode[STIX]{x1D6FD}_{\mathit{i}}\unicode[STIX]{x1D700}^{2}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}\left(1-\frac{\unicode[STIX]{x1D700}^{1/2}}{2\unicode[STIX]{x1D6FE}\mathit{q}^{2}}\right)\tilde{\unicode[STIX]{x1D6F7}}^{(0)}\end{eqnarray}$$

and

(5.21)$$\begin{eqnarray}\frac{\mathit{v}_{\mathit{i}}}{\mathit{c}}\mathit{a}\simeq -\frac{\mathit{v}_{\mathit{i}}}{\mathit{c}}(\unicode[STIX]{x1D700}+2\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime })\tilde{\mathit{A}}_{\Vert }^{(0)}+\text{i}\frac{1}{2}\unicode[STIX]{x1D6FD}_{\mathit{i}}\unicode[STIX]{x1D700}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}\left(1-\frac{\unicode[STIX]{x1D700}^{1/2}}{2\unicode[STIX]{x1D6FE}\mathit{q}^{2}}\right)\tilde{\unicode[STIX]{x1D6F7}}^{(0)},\end{eqnarray}$$

upon using (5.9) to calculate the response to $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$, because the corrections to $\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$ due to $\tilde{\mathit{A}}_{\Vert }^{(0)}$ are negligible (as will be found from quasineutrality shortly). Notice that at small skin depth the relation $\langle \tilde{\mathit{A}}_{\Vert }\rangle \simeq \tilde{\mathit{A}}_{\Vert }^{(0)}+\unicode[STIX]{x1D700}\mathit{a}/2$ holds. Keeping $\unicode[STIX]{x1D700}^{1/2}$ corrections, we note that for $\tilde{\mathit{A}}_{\Vert }^{(0)}=0$ stronger poloidal variation occurs, as $\langle \tilde{\mathit{A}}_{\Vert }\rangle /\mathit{a}\simeq \unicode[STIX]{x1D700}/2$.

Correcting the full parallel Ampere’s law without inserting (5.9), but otherwise making the changes already noted, (5.22) and (5.23) are the same as the corrected (5.15) and (5.13).

Evaluating (5.12) and (5.14) without inserting (5.9), (5.24) and (5.25) should read as follows:

(5.24)$$\begin{eqnarray}\displaystyle & & \displaystyle (1+\unicode[STIX]{x1D6FD})(\langle \tilde{\mathit{B}}_{\Vert }\rangle -\tilde{\mathit{B}}_{\Vert }^{(0)})+\frac{5\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD}_{\mathit{i}}}{4\unicode[STIX]{x1D6FD}}\unicode[STIX]{x1D700}^{3/2}\mathit{k}_{\bot }^{2}\unicode[STIX]{x1D70C}_{\mathit{pi}}^{2}\tilde{\mathit{B}}_{\Vert }^{(0)}\simeq \frac{\mathit{c}\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{k}_{\bot }^{2}\unicode[STIX]{x1D70C}_{\mathit{i}}}{4\mathit{v}_{\mathit{i}}}[(5\unicode[STIX]{x1D6FE}\mathit{q}^{2}\unicode[STIX]{x1D700}^{-1/2}+3)\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle -3\tilde{\unicode[STIX]{x1D6F7}}^{(0)}]\nonumber\\ \displaystyle & & \displaystyle \quad +\,\frac{\text{i}\mathit{qk}_{\bot }\unicode[STIX]{x1D6FD}}{4}\left\{5\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D700}^{1/2}\left[\langle \tilde{\mathit{A}}_{\Vert }\rangle -\tilde{\mathit{A}}_{\Vert }^{(0)}\left(1+\frac{\mathit{k}_{\bot }^{2}\mathit{c}^{2}}{\unicode[STIX]{x1D714}_{\mathit{p}}^{2}}\right)\right]+3\mathit{a}\right\}\end{eqnarray}$$

and

(5.25)$$\begin{eqnarray}\displaystyle & & \displaystyle \left(1+\frac{\unicode[STIX]{x1D70D}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D700}^{1/2}}{4}\right)\mathit{b}\simeq \unicode[STIX]{x1D700}\tilde{\mathit{B}}_{\Vert }^{(0)}\left(1-\frac{\unicode[STIX]{x1D70D}}{5}\unicode[STIX]{x1D700}^{1/2}-\frac{\unicode[STIX]{x1D70D}\unicode[STIX]{x1D6FD}_{\mathit{i}}}{4\unicode[STIX]{x1D6FD}}\unicode[STIX]{x1D700}^{1/2}\mathit{k}_{\bot }^{2}\unicode[STIX]{x1D70C}_{\mathit{pi}}^{2}\right)\nonumber\\ \displaystyle & & \displaystyle \quad -\,\frac{\text{i}\mathit{q}\unicode[STIX]{x1D6FD}\mathit{k}_{\bot }}{2}(3-\unicode[STIX]{x1D70D}\unicode[STIX]{x1D700}^{1/2})\left[\langle \tilde{\mathit{A}}_{\Vert }\rangle -\tilde{\mathit{A}}_{\Vert }^{(0)}\left(1+\frac{\mathit{k}_{\bot }^{2}\mathit{c}^{2}}{\unicode[STIX]{x1D714}_{\mathit{p}}^{2}}\right)\right]-\frac{\text{i}\unicode[STIX]{x1D712}q\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D700}^{1/2}\mathit{k}_{\bot }\mathit{a}}{4}+\frac{c\unicode[STIX]{x1D70D}q^{2}\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{k}_{\bot }^{2}\unicode[STIX]{x1D70C}_{\mathit{i}}}{4\mathit{v}_{\mathit{i}}\unicode[STIX]{x1D700}^{1/2}}\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle ,\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

where all $\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$ and $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}$ terms are corrected. The line following (5.25) and (5.26) should be deleted and replaced by the following sentence. Equations (5.22) and (5.23) are only consistent with

(5.26)$$\begin{eqnarray}\langle \tilde{\mathit{A}}_{\Vert }\rangle \simeq \tilde{\mathit{A}}_{\Vert }^{(0)}+\frac{\unicode[STIX]{x1D700}}{2}\mathit{a}\end{eqnarray}$$

in the small skin depth limit, for which field lines switch between irrational and rational when $\mathit{q}_{1}=\mathit{ik}_{\bot }\mathit{B}_{\mathit{p}}^{-1}\tilde{\mathit{A}}_{\Vert }^{(0)}\neq 0$.

When $\tilde{\mathit{B}}_{\Vert }^{(0)}=0$, the $\unicode[STIX]{x1D714}_{\mathit{p}}^{2}/\mathit{k}_{\bot }^{2}\mathit{c}^{2}\rightarrow 0$ limits of (5.24) and (5.25) correct (5.27) and (5.28) to read

(5.27)$$\begin{eqnarray}\frac{\mathit{v}_{\mathit{i}}\langle \tilde{\mathit{B}}_{\Vert }\rangle }{\mathit{k}_{\bot }\mathit{c}}\rightarrow \frac{\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{i}}}{4}[(5\unicode[STIX]{x1D6FE}\mathit{q}^{2}\unicode[STIX]{x1D700}^{-1/2}+3)\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle -3\tilde{\unicode[STIX]{x1D6F7}}^{(0)}]-\frac{\text{i}5\unicode[STIX]{x1D6FE}q\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D700}^{1/2}\mathit{k}_{\bot }^{2}\mathit{c}^{2}}{4\unicode[STIX]{x1D714}_{\mathit{p}}^{2}}\frac{\mathit{v}_{\mathit{i}}}{\mathit{c}}\tilde{\mathit{A}}_{\Vert }^{(0)}\end{eqnarray}$$

and

(5.28)$$\begin{eqnarray}\frac{\mathit{v}_{\mathit{i}}\mathit{b}}{\mathit{k}_{\bot }\mathit{c}}\rightarrow \frac{\text{i}q\unicode[STIX]{x1D6FD}k_{\bot }^{2}\mathit{c}^{2}}{2\unicode[STIX]{x1D714}_{\mathit{p}}^{2}}(3-\unicode[STIX]{x1D70D}\unicode[STIX]{x1D700}^{1/2})\frac{\mathit{v}_{\mathit{i}}}{\mathit{c}}\tilde{\mathit{A}}_{\Vert }^{(0)}+\frac{\unicode[STIX]{x1D70D}\mathit{q}^{2}\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{i}}}{4\unicode[STIX]{x1D700}^{1/2}}\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle .\end{eqnarray}$$

The sentences that follow should then read as follows. Consequently, when $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}=0$, we see that the ratio of the $\tilde{\mathit{A}}_{\Vert }^{(0)}$ terms gives $\mathit{b}/\langle \tilde{\mathit{B}}_{\Vert }\rangle \simeq -6/5\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D700}^{1/2}$, so rather strong poloidal variation occurs in $\tilde{\mathit{B}}_{\Vert }$. For $\tilde{\mathit{A}}_{\Vert }^{(0)}=0$, the poloidal variation of $\tilde{\mathit{B}}_{\Vert }$ is strong since $\mathit{b}/\langle \tilde{\mathit{B}}_{\Vert }\rangle \simeq \unicode[STIX]{x1D70D}/2\unicode[STIX]{x1D6FE}\simeq 1.3$.

For $\tilde{\mathit{B}}_{\Vert }^{(0)}=0$ and $\unicode[STIX]{x1D6FD}\ll 1$, the small skin depth ($\unicode[STIX]{x1D714}_{\mathit{p}}^{2}/\mathit{k}_{\bot }^{2}\mathit{c}^{2}\gg 1$) limits of (5.24) and (5.25) yield the corrected (5.29) and (5.30) to be

(5.29)$$\begin{eqnarray}\frac{\mathit{v}_{\mathit{i}}\langle \tilde{\mathit{B}}_{\Vert }\rangle }{\mathit{k}_{\bot }\mathit{c}}\simeq \frac{\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{i}}}{4}[(5\unicode[STIX]{x1D6FE}\mathit{q}^{2}\unicode[STIX]{x1D700}^{-1/2}+3)\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle -3\tilde{\unicode[STIX]{x1D6F7}}^{(0)}]-\frac{\text{i}3(\unicode[STIX]{x1D700}+2\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime })q\unicode[STIX]{x1D6FD}}{4}\frac{\mathit{v}_{\mathit{i}}}{\mathit{c}}\tilde{\mathit{A}}_{\Vert }^{(0)}\end{eqnarray}$$

and

(5.30)$$\begin{eqnarray}\frac{\mathit{v}_{\mathit{i}}\mathit{b}}{\mathit{k}_{\bot }\mathit{c}}\simeq \frac{\text{i}q\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D700}^{1/2}(\unicode[STIX]{x1D700}+2\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime })}{4}[\unicode[STIX]{x1D712}+(3-\unicode[STIX]{x1D70D}\unicode[STIX]{x1D700}^{1/2})\unicode[STIX]{x1D700}^{1/2}]\frac{\mathit{v}_{\mathit{i}}}{\mathit{c}}\tilde{\mathit{A}}_{\Vert }^{(0)}+\frac{\unicode[STIX]{x1D70D}\mathit{q}^{2}\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{i}}}{4\unicode[STIX]{x1D700}^{1/2}}\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle ,\end{eqnarray}$$

where we assume $\unicode[STIX]{x1D700}^{3/2}\gg \mathit{k}_{\bot }^{2}\mathit{c}^{2}/\unicode[STIX]{x1D714}_{\mathit{p}}^{2}\gg \unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime }\unicode[STIX]{x1D700}^{5/2}$. The three sentences that follow should be replaced by the following two sentences. When $\tilde{\mathit{A}}_{\Vert }^{(0)}=0$, the poloidal variation of $\tilde{\mathit{B}}_{\Vert }$ is strong with $\mathit{b}/\langle \tilde{\mathit{B}}_{\Vert }\rangle \simeq \unicode[STIX]{x1D70D}/2\unicode[STIX]{x1D6FE}\,\simeq 1.33$. For $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}=0$, it is also rather strong, since $\mathit{b}/\langle \tilde{\mathit{B}}_{\Vert }\rangle \simeq \unicode[STIX]{x1D712}\unicode[STIX]{x1D700}^{1/2}/3$.

The sentence including (5.31)–(5.32) should be corrected to read as follows. Keeping $\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle$ terms, the $\unicode[STIX]{x1D714}_{\mathit{p}}^{2}/\mathit{k}_{\bot }^{2}\mathit{c}^{2}\rightarrow 0$ limit gives

(5.31)$$\begin{eqnarray}\mathit{k}_{\bot }\mathit{a}\rightarrow \text{i}2\mathit{q}\tilde{\mathit{B}}_{\Vert }^{(0)}+\text{i}\frac{5\unicode[STIX]{x1D6FE}}{4}\unicode[STIX]{x1D700}^{1/2}\mathit{q}^{2}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}\frac{\mathit{c}\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{k}_{\bot }\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle }{\mathit{v}_{\mathit{i}}}\end{eqnarray}$$

and

(5.32)$$\begin{eqnarray}\mathit{k}_{\bot }\langle \tilde{\mathit{A}}_{\Vert }\rangle \rightarrow \text{i}\unicode[STIX]{x1D6FE}q\unicode[STIX]{x1D700}^{1/2}\tilde{\mathit{B}}_{\Vert }^{(0)}+\text{i}\frac{\unicode[STIX]{x1D70E}}{4}\unicode[STIX]{x1D700}^{1/2}\mathit{q}^{2}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}\frac{\mathit{c}\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{k}_{\bot }\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle }{\mathit{v}_{\mathit{i}}}.\end{eqnarray}$$

Evaluating (5.12) and (5.14) without inserting (5.9), (5.33) and (5.34) should read as follows:

(5.33)$$\begin{eqnarray}\displaystyle & & \displaystyle \langle \tilde{\mathit{B}}_{\Vert }\rangle \rightarrow \left[1-\frac{3}{2}\mathit{q}^{2}\unicode[STIX]{x1D6FD}-\frac{5\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FD}_{\mathit{i}}}{4\unicode[STIX]{x1D6FD}}\unicode[STIX]{x1D700}^{3/2}\mathit{k}_{\bot }^{2}\unicode[STIX]{x1D70C}_{\mathit{pi}}^{2}\right]\tilde{\mathit{B}}_{\Vert }^{(0)}+\frac{5\unicode[STIX]{x1D6FE}}{4}\mathit{q}\unicode[STIX]{x1D700}^{1/2}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}\frac{\mathit{c}\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{k}_{\bot }\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle }{\mathit{v}_{\mathit{i}}}\nonumber\\ \displaystyle & & \displaystyle \quad =\left[1-\frac{3}{2}\mathit{q}^{2}\unicode[STIX]{x1D6FD}\right]\tilde{\mathit{B}}_{\Vert }^{(0)}\end{eqnarray}$$

and

(5.34)$$\begin{eqnarray}\displaystyle & & \displaystyle \left(1+\frac{\unicode[STIX]{x1D70D}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D700}^{1/2}}{4}\right)\mathit{b}\rightarrow \left[\unicode[STIX]{x1D700}\left(1-\frac{\unicode[STIX]{x1D70D}}{5}\unicode[STIX]{x1D700}^{1/2}-\frac{\unicode[STIX]{x1D70D}\unicode[STIX]{x1D6FD}_{\mathit{i}}}{4\unicode[STIX]{x1D6FD}}\unicode[STIX]{x1D700}^{1/2}\mathit{k}_{\bot }^{2}\unicode[STIX]{x1D70C}_{\mathit{pi}}^{2}\right)+\frac{\unicode[STIX]{x1D712}+3\unicode[STIX]{x1D6FE}}{2}\mathit{q}^{2}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D700}^{1/2}\right]\tilde{\mathit{B}}_{\Vert }^{(0)}\nonumber\\ \displaystyle & & \displaystyle \quad +\,\frac{\unicode[STIX]{x1D70D}}{4}\mathit{q}\unicode[STIX]{x1D700}^{1/2}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}\frac{\mathit{c}\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{k}_{\bot }\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle }{\mathit{v}_{\mathit{i}}}=\left[\unicode[STIX]{x1D700}\left(1-\frac{\unicode[STIX]{x1D70D}}{5}\unicode[STIX]{x1D700}^{1/2}\right)+\frac{\unicode[STIX]{x1D712}+3\unicode[STIX]{x1D6FE}}{2}\mathit{q}^{2}\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D700}^{1/2}\right]\tilde{\mathit{B}}_{\Vert }^{(0)},\end{eqnarray}$$

where $c\unicode[STIX]{x1D6FD}k_{\bot }\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle /\mathit{v}_{\mathit{i}}\tilde{\mathit{B}}_{\Vert }^{(0)}\simeq \mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{i}}$ is inserted as this result will be determined from quasineutrality in the next subsection, and the $\unicode[STIX]{x1D6FD}$ correction on the left side is small in the final form.

Correcting the next paragraph, it should read as follows. In the small skin depth limit $\unicode[STIX]{x1D700}^{3/2}\unicode[STIX]{x1D714}_{\mathit{p}}^{2}/\mathit{k}_{\bot }^{2}\mathit{c}^{2}\gtrsim 1\gg \unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime }\unicode[STIX]{x1D700}^{5/2}\unicode[STIX]{x1D714}_{\mathit{p}}^{2}/\mathit{k}_{\bot }^{2}\mathit{c}^{2}$ for finite $\unicode[STIX]{x1D6FD}\ll 1$ and $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}=0=\tilde{\mathit{A}}_{\Vert }^{(0)}$, we use (5.22) and (5.23) along with the quasineutrality result $c\unicode[STIX]{x1D6FD}k_{\bot }\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle /\mathit{v}_{\mathit{i}}\tilde{\mathit{B}}_{\Vert }^{(0)}\simeq \mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{i}}$ to find $\unicode[STIX]{x1D700}a\,\simeq 2\langle \tilde{\mathit{A}}_{\Vert }\rangle$. Then (5.22)–(5.25) give

(5.35)$$\begin{eqnarray}\displaystyle & \mathit{k}_{\bot }\mathit{a}\simeq \text{i}2\mathit{q}\tilde{\mathit{B}}_{\Vert }^{(0)}, & \displaystyle\end{eqnarray}$$

(5.36)$$\begin{eqnarray}\displaystyle & \mathit{k}_{\bot }\langle \tilde{\mathit{A}}_{\Vert }\rangle \simeq \text{i}q\unicode[STIX]{x1D700}\tilde{\mathit{B}}_{\Vert }^{(0)}, & \displaystyle\end{eqnarray}$$

(5.37)$$\begin{eqnarray}\displaystyle & \langle \tilde{\mathit{B}}_{\Vert }\rangle \simeq \tilde{\mathit{B}}_{\Vert }^{(0)}\left(1-\frac{3}{2}\mathit{q}^{2}\unicode[STIX]{x1D6FD}\right) & \displaystyle\end{eqnarray}$$

and

(5.38)$$\begin{eqnarray}\mathit{b}\simeq \tilde{\mathit{B}}_{\Vert }^{(0)}\left[\unicode[STIX]{x1D700}\left(1-\frac{\unicode[STIX]{x1D70D}}{5}\unicode[STIX]{x1D700}^{1/2}\right)+\frac{\unicode[STIX]{x1D712}}{2}\mathit{q}^{2}\unicode[STIX]{x1D700}^{1/2}\unicode[STIX]{x1D6FD}\right]\end{eqnarray}$$

for $\unicode[STIX]{x1D6FD}\sim \unicode[STIX]{x1D700}\mathit{k}_{\bot }^{2}\unicode[STIX]{x1D70C}_{\mathit{pi}}^{2}$. As a result, for this $\tilde{\mathit{B}}_{\Vert }^{(0)}\neq 0$ case with $\tilde{\unicode[STIX]{x1D6F7}}^{(0)}=0$ and $\tilde{\mathit{A}}_{\Vert }^{(0)}=0$, the poloidal variation of $\tilde{\mathit{A}}_{\Vert }$ is strong with $\langle \tilde{\mathit{A}}_{\Vert }\rangle /\mathit{a}\simeq \unicode[STIX]{x1D700}/2$, while the poloidal variation of $\tilde{\mathit{B}}_{\Vert }$ is weak with $\mathit{b}/\langle \tilde{\mathit{B}}_{\Vert }\rangle \simeq \unicode[STIX]{x1D700}$.

In § 5.4, there are sign errors in the $\tilde{\mathit{A}}_{\Vert }$ terms of (5.41)–(5.45). The signs can be corrected by changing the signs of all $\tilde{\mathit{A}}_{\Vert }$, $\tilde{\mathit{A}}_{\Vert }^{(0)}$, $\langle \tilde{\mathit{A}}_{\Vert }\rangle$, and ‘$a$’ terms. In addition, $5\unicode[STIX]{x1D6FE}/16$ replaces $(\unicode[STIX]{x1D70E}+5\unicode[STIX]{x1D6FE})/16$ in (5.45) and $\unicode[STIX]{x1D700}\rightarrow \unicode[STIX]{x1D700}+2\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime }$ in the $\tilde{\mathit{A}}_{\Vert }^{(0)}$ term, so it becomes

(5.45)$$\begin{eqnarray}\displaystyle \left(\frac{\unicode[STIX]{x1D6FE}\mathit{q}^{2}}{\unicode[STIX]{x1D700}^{1/2}}+1\right)\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle & = & \displaystyle \tilde{\unicode[STIX]{x1D6F7}}^{(0)}\left[1+\frac{5\unicode[STIX]{x1D6FE}}{16}\mathit{q}^{2}\unicode[STIX]{x1D6FD}_{\mathit{i}}\unicode[STIX]{x1D700}^{3/2}\mathit{k}_{\bot }^{2}\unicode[STIX]{x1D70C}_{\mathit{pi}}^{2}\right]-\text{i}\frac{5\unicode[STIX]{x1D6FE}\mathit{v}_{\mathit{i}}}{8\mathit{c}}\mathit{q}^{2}\unicode[STIX]{x1D700}^{1/2}(\unicode[STIX]{x1D700}+2\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime })\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}\tilde{\mathit{A}}_{\Vert }^{(0)}\nonumber\\ \displaystyle & & \displaystyle +\,\frac{\unicode[STIX]{x1D6FE}\mathit{v}_{\mathit{i}}}{\mathit{c}\unicode[STIX]{x1D6FD}}\mathit{q}\unicode[STIX]{x1D700}^{1/2}\unicode[STIX]{x1D70C}_{\mathit{pi}}\tilde{\mathit{B}}_{\Vert }^{(0)}.\end{eqnarray}$$

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

Section 6 summarizes the small skin depth zonal flow residual responses for the various initial perturbations. The complete summary of the corrected results is as follows.

6.1 $\mathit{A}_{\Vert }^{(0)}=0=\mathit{B}_{\Vert }^{(0)}$

(6.2)$$\begin{eqnarray}\displaystyle & \displaystyle \left(\frac{\unicode[STIX]{x1D6FE}\mathit{q}^{2}}{\unicode[STIX]{x1D700}^{1/2}}+1\right)\frac{\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle }{\tilde{\unicode[STIX]{x1D6F7}}^{(0)}}=1+\frac{5\unicode[STIX]{x1D6FE}}{16}\mathit{q}^{2}\unicode[STIX]{x1D6FD}_{\mathit{i}}\unicode[STIX]{x1D700}^{3/2}\mathit{k}_{\bot }^{2}\unicode[STIX]{x1D70C}_{\mathit{pi}}^{2}, & \displaystyle\end{eqnarray}$$

(6.3)$$\begin{eqnarray}\displaystyle & \displaystyle \frac{\mathit{v}_{\mathit{i}}\langle \tilde{\mathit{B}}_{\Vert }\rangle }{\mathit{k}_{\bot }\mathit{c}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}}=\frac{\unicode[STIX]{x1D6FD}_{\mathit{i}}\unicode[STIX]{x1D700}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}}{2\mathit{q}}, & \displaystyle\end{eqnarray}$$

(6.4)$$\begin{eqnarray}\displaystyle & \displaystyle \frac{\mathit{v}_{\mathit{i}}\mathit{b}}{\mathit{k}_{\bot }\mathit{c}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}}=\frac{\unicode[STIX]{x1D70D}\unicode[STIX]{x1D700}\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}}{4\unicode[STIX]{x1D6FE}\mathit{q}}, & \displaystyle\end{eqnarray}$$

(6.5)$$\begin{eqnarray}\displaystyle & \displaystyle \frac{\mathit{v}_{\mathit{i}}\langle \tilde{\mathit{A}}_{\Vert }\rangle }{\text{i}\mathit{c}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}}=\frac{1}{4}\unicode[STIX]{x1D700}^{2}\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}\left(1-\frac{\unicode[STIX]{x1D700}^{1/2}}{2\unicode[STIX]{x1D6FE}\mathit{q}^{2}}\right)\simeq \frac{1}{4}\unicode[STIX]{x1D700}^{2}\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}, & \displaystyle\end{eqnarray}$$

(6.6)$$\begin{eqnarray}\displaystyle & \displaystyle \frac{\mathit{v}_{\mathit{i}}\mathit{a}}{\text{i}\mathit{c}\tilde{\unicode[STIX]{x1D6F7}}^{(0)}}=\frac{1}{2}\unicode[STIX]{x1D700}\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}\left(1-\frac{\unicode[STIX]{x1D700}^{1/2}}{2\unicode[STIX]{x1D6FE}\mathit{q}^{2}}\right)\simeq \frac{1}{2}\unicode[STIX]{x1D700}\unicode[STIX]{x1D6FD}_{\mathit{i}}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}. & \displaystyle\end{eqnarray}$$

6.2 $\unicode[STIX]{x1D6F7}^{(0)}=0=\mathit{B}_{\Vert }^{(0)}$

(6.7)$$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{i}\mathit{c}\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle }{\mathit{v}_{\mathit{i}}\tilde{\mathit{A}}_{\Vert }^{(0)}}=\frac{5\unicode[STIX]{x1D700}(\unicode[STIX]{x1D700}+2\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime })\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}}{8(1+\unicode[STIX]{x1D700}^{1/2}/\unicode[STIX]{x1D6FE}\mathit{q}^{2})}\simeq \frac{5}{8}\unicode[STIX]{x1D700}(\unicode[STIX]{x1D700}+2\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime })\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}, & \displaystyle\end{eqnarray}$$

(6.8)$$\begin{eqnarray}\displaystyle & \displaystyle \frac{\langle \tilde{\mathit{A}}_{\Vert }\rangle }{\tilde{\mathit{A}}_{\Vert }^{(0)}}=1-\frac{1}{2}\unicode[STIX]{x1D700}(\unicode[STIX]{x1D700}+2\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime }), & \displaystyle\end{eqnarray}$$

(6.9)$$\begin{eqnarray}\displaystyle & \displaystyle \frac{\mathit{a}}{\tilde{\mathit{A}}_{\Vert }^{(0)}}=-(\unicode[STIX]{x1D700}+2\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime }), & \displaystyle\end{eqnarray}$$

(6.10)$$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{i}\langle \tilde{\mathit{B}}_{\Vert }\rangle }{\mathit{k}_{\bot }\tilde{\mathit{A}}_{\Vert }^{(0)}}=\frac{3}{4}\mathit{q}\unicode[STIX]{x1D6FD}(\unicode[STIX]{x1D700}+2\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime }), & \displaystyle\end{eqnarray}$$

(6.11)$$\begin{eqnarray}\displaystyle & \displaystyle \frac{\mathit{b}}{\text{i}\mathit{k}_{\bot }\tilde{\mathit{A}}_{\Vert }^{(0)}}=\frac{\unicode[STIX]{x1D712}}{4}q\unicode[STIX]{x1D6FD}\unicode[STIX]{x1D700}^{1/2}(\unicode[STIX]{x1D700}+2\unicode[STIX]{x1D6E5}_{\mathit{s}}^{\prime }). & \displaystyle\end{eqnarray}$$

6.3 $\unicode[STIX]{x1D6F7}^{(0)}=0=\mathit{A}_{\Vert }^{(0)}$

(6.12)$$\begin{eqnarray}\displaystyle & \displaystyle \frac{c\unicode[STIX]{x1D6FD}k_{\bot }\langle \tilde{\unicode[STIX]{x1D6F7}}\rangle }{\mathit{v}_{\mathit{i}}\tilde{\mathit{B}}_{\Vert }^{(0)}}=\frac{\unicode[STIX]{x1D700}k_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}}{\mathit{q}(1+\unicode[STIX]{x1D700}^{1/2}/\unicode[STIX]{x1D6FE}\mathit{q}^{2})}\simeq \frac{\unicode[STIX]{x1D700}\mathit{k}_{\bot }\unicode[STIX]{x1D70C}_{\mathit{pi}}}{\mathit{q}}, & \displaystyle\end{eqnarray}$$

(6.13)$$\begin{eqnarray}\displaystyle & \displaystyle \frac{\mathit{k}_{\bot }\langle \tilde{\mathit{A}}_{\Vert }\rangle }{\text{i}\tilde{\mathit{B}}_{\Vert }^{(0)}}=\mathit{q}\unicode[STIX]{x1D700}, & \displaystyle\end{eqnarray}$$

(6.14)$$\begin{eqnarray}\displaystyle & \displaystyle \frac{\mathit{k}_{\bot }\mathit{a}}{\text{i}\tilde{\mathit{B}}_{\Vert }^{(0)}}=2\mathit{q}, & \displaystyle\end{eqnarray}$$

(6.15)$$\begin{eqnarray}\displaystyle & \displaystyle \frac{\langle \tilde{\mathit{B}}_{\Vert }\rangle }{\tilde{\mathit{B}}_{\Vert }^{(0)}}=1-\frac{3}{2}\mathit{q}^{2}\unicode[STIX]{x1D6FD}, & \displaystyle\end{eqnarray}$$

(6.16)$$\begin{eqnarray}\displaystyle & \displaystyle \frac{\mathit{b}}{\tilde{\mathit{B}}_{\Vert }^{(0)}}=\unicode[STIX]{x1D700}\left(1-\frac{\unicode[STIX]{x1D70D}}{5}\unicode[STIX]{x1D700}^{1/2}\right)+\frac{\unicode[STIX]{x1D712}}{2}\mathit{q}^{2}\unicode[STIX]{x1D700}^{1/2}\unicode[STIX]{x1D6FD}. & \displaystyle\end{eqnarray}$$

All small skin depth results are consistent with $\langle \tilde{\mathit{A}}_{\Vert }\rangle \simeq \tilde{\mathit{A}}_{\Vert }^{(0)}+\unicode[STIX]{x1D700}\mathit{a}/2$, but field lines can only switch between irrational and rational when $\mathit{q}_{1}=\text{i}\mathit{k}_{\bot }\mathit{B}_{\mathit{p}}^{-1}\tilde{\mathit{A}}_{\Vert }^{(0)}\neq 0$.