3.1 The ion solution

We consider the ‘collisional’ case: $\unicode[STIX]{x1D708}_{ii}>\unicode[STIX]{x1D714}$ . Introducing the proton charge, $e$ , we have

(3.1) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{Iv_{\Vert }}{R^{2}Bq}\left(\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}+\text{i}\ell q^{\prime }x\right)\,h_{i}+\boldsymbol{v}_{di}\boldsymbol{\cdot }\unicode[STIX]{x1D735}h_{i}-\text{i}\unicode[STIX]{x1D714}h_{i}-C_{i}(h_{i})-\overline{L_{i}C_{i}(g_{i}L_{i})}\nonumber\\ \displaystyle & & \displaystyle \quad =-\text{i}\frac{ef_{0i}}{T_{i}}(\unicode[STIX]{x1D714}-\unicode[STIX]{x1D714}_{\ast i}^{T})\left(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9}+\frac{v_{\bot }^{2}}{2\unicode[STIX]{x1D6FA}_{i}}\unicode[STIX]{x1D6FF}B_{\Vert }\right)-\frac{ef_{0i}}{T_{i}}\left(1-\frac{\unicode[STIX]{x1D714}_{\ast i}^{T}}{\unicode[STIX]{x1D714}}\right)\boldsymbol{v}_{di}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F9}.\end{eqnarray}$$

We solve this equation by introducing an ordering scheme in terms of a small parameter, $\unicode[STIX]{x1D700}_{i}$ , where $\unicode[STIX]{x1D714}_{dri}\sim \unicode[STIX]{x1D708}_{i}\sim \unicode[STIX]{x1D700}_{i}\unicode[STIX]{x1D714}_{bi}$ ; $\unicode[STIX]{x1D714}\sim \unicode[STIX]{x1D714}_{dr\unicode[STIX]{x1D703}}\sim \unicode[STIX]{x1D700}_{i}^{2}\unicode[STIX]{x1D714}_{bi}$ ; we also order $z_{i}\sim \unicode[STIX]{x1D700}_{i}$ . Although we shall later assume a small number of trapped particles, $f_{t}\ll 1$ , we do not order it in $\unicode[STIX]{x1D700}_{i}$ . The radial magnetic drift frequency, $\unicode[STIX]{x1D714}_{dri}$ , exceeds the azimuthal drift frequency, $\unicode[STIX]{x1D714}_{dr\unicode[STIX]{x1D703}}$ , because of the narrow radial width of the semi-collisional layer.

Writing $h_{i}=h_{0i}+\unicode[STIX]{x1D700}_{i}h_{1i}+\unicode[STIX]{x1D700}_{i}^{2}h_{2i}+\cdots \,$ , the lowest-order solution satisfies

(3.2) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}h_{0i}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}=0\quad \Rightarrow h_{0i}=h_{0i}(v,\unicode[STIX]{x1D706},x,\unicode[STIX]{x1D70E}),\end{eqnarray}$$

while in first order we obtain

(3.3) $$\begin{eqnarray}\frac{Iv_{\Vert }}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}h_{1i}=-v_{dri}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}x}h_{0i}+C_{i}(h_{0i})-\frac{ef_{0i}}{T_{i}}\left(1-\frac{\unicode[STIX]{x1D714}_{\ast i}^{T}}{\unicode[STIX]{x1D714}}\right)v_{dri}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}x}\unicode[STIX]{x1D713},\end{eqnarray}$$

where we note

(3.4) $$\begin{eqnarray}v_{drj}=\boldsymbol{v}_{drj}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}=\frac{Iv_{\Vert }}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{Iv_{\Vert }}{\unicode[STIX]{x1D6FA}_{j}}\right).\end{eqnarray}$$

Thus $h_{1i}$ for passing particles can be annihilated by applying the operation: $\langle B(\ldots )/v_{\Vert }\rangle$ :

(3.5) $$\begin{eqnarray}\langle \ldots \rangle =\left.\oint \frac{(\ldots )\,\text{d}\unicode[STIX]{x1D703}}{\boldsymbol{B}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}}\right/\oint \frac{\text{d}\unicode[STIX]{x1D703}}{\boldsymbol{B}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}}=\left.\oint (\ldots )\,R^{2}\,\text{d}\unicode[STIX]{x1D703}\right/\oint R^{2}\,\text{d}\unicode[STIX]{x1D703},\end{eqnarray}$$

since

(3.6) $$\begin{eqnarray}\frac{1}{\boldsymbol{B}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}}=\frac{q}{\boldsymbol{B}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D711}}=\frac{qR^{2}}{I}.\end{eqnarray}$$

For trapped particles, we integrate along the bounce orbit, summing over $\unicode[STIX]{x1D70E}$ in the usual way, to obtain the constraint

(3.7) $$\begin{eqnarray}\left\langle \frac{B}{v_{\Vert }}C_{i}(h_{0i})\right\rangle =0,\end{eqnarray}$$

where the integration is now between bounce points. This determines $h_{0i}$ , yielding:

(3.8) $$\begin{eqnarray}h_{0i}=\left[\frac{\hat{n}_{i}}{n_{0}}+\left(u^{2}-\frac{3}{2}\right)\frac{\hat{T}_{i}}{T_{i}}\right]f_{0i},\end{eqnarray}$$

where $u^{2}=m_{j}v^{2}/2T_{j}$ with $j=i$ .

The equation for $h_{1i}$ can then be integrated to give

(3.9) $$\begin{eqnarray}h_{1i}=-\frac{Iv_{\Vert }}{\unicode[STIX]{x1D6FA}_{i}}\left[\frac{e}{T_{i}}\left(1-\frac{\unicode[STIX]{x1D714}_{\ast i}^{T}}{\unicode[STIX]{x1D714}}\right)\unicode[STIX]{x1D6F9}^{\prime }f_{0i}+h_{0i}^{\prime }\right]+H_{1i}(v,\unicode[STIX]{x1D706},x,\unicode[STIX]{x1D70E}),\end{eqnarray}$$

where $H_{1i}$ remains to be determined. This can be rewritten as

(3.10) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}h_{1i}={\displaystyle \frac{Iv_{\Vert }}{\unicode[STIX]{x1D6FA}_{i}}}\left[A_{1}^{(i)}+\left(u^{2}-{\displaystyle \frac{5}{2}}\right)A_{2}^{(i)}\right]f_{0i}+H_{1i}(v,\unicode[STIX]{x1D706},x,\unicode[STIX]{x1D70E}),\\ A_{1}^{(i)}=-{\displaystyle \frac{e\unicode[STIX]{x1D6F9}^{\prime }}{T_{i}}}\left(1-{\displaystyle \frac{\unicode[STIX]{x1D714}_{\ast i}}{\unicode[STIX]{x1D714}}}(1+\unicode[STIX]{x1D702}_{i})\right)-\left({\displaystyle \frac{\hat{n}_{i}^{\prime }}{n_{0}}}+{\displaystyle \frac{\hat{T}_{i}^{\prime }}{T_{i}}}\right)\equiv -\left({\displaystyle \frac{\tilde{p}_{i}^{\prime }}{p_{0i}}}+{\displaystyle \frac{e\unicode[STIX]{x1D6F7}^{\prime }}{T_{i}}}\right),\\ A_{2}^{(i)}=-\left(-{\displaystyle \frac{\unicode[STIX]{x1D714}_{\ast i}\unicode[STIX]{x1D702}_{i}}{\unicode[STIX]{x1D714}}}{\displaystyle \frac{e\unicode[STIX]{x1D6F9}^{\prime }}{T_{i}}}+{\displaystyle \frac{\hat{T}_{i}^{\prime }}{T_{i}}}\right)\equiv -{\displaystyle \frac{\tilde{T}_{i}^{\prime }}{T_{i}}},\end{array}\right\}\end{eqnarray}$$

where $p_{0i}=n_{0}T_{i}$ , on using (2.13) and (2.14). Note that the quantities $\tilde{p}_{i}^{\prime }$ and $\tilde{T}_{i}^{\prime }$ thus defined, depend on $\unicode[STIX]{x1D714}$ .

The equation for $h_{2i}$ is

(3.11) $$\begin{eqnarray}\frac{v_{\Vert }I}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}h_{2i}+\frac{v_{\Vert }I}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{Iv_{\Vert }}{\unicode[STIX]{x1D6FA}_{i}}\right)\frac{\unicode[STIX]{x2202}h_{1i}}{\unicode[STIX]{x2202}x}=C_{ii}(h_{1i})+C_{ie}(h_{1i}),\end{eqnarray}$$

(where, for most purposes, we can ignore the small ion–electron collisional term, but see § 7 later), which provides the constraint

(3.12) $$\begin{eqnarray}\left\langle \frac{B}{v_{\Vert }}C_{ii}(h_{1i})\right\rangle =0.\end{eqnarray}$$

For the ions, we take the model pitch-angle scattering collision operator that conserves momentum (Rutherford et al. Reference Rutherford, Kovrizhnykh, Rosenbluth and Hinton1970):

(3.13) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}C_{ii}(h)=\unicode[STIX]{x1D708}_{ii}(u)\left(L(h)+{\displaystyle \frac{m_{i}}{T_{i}}}v_{\Vert }U_{\Vert i}^{\ast }(h)f_{0i}\right),\quad L(h)={\displaystyle \frac{2v_{\Vert }}{v^{2}B}}{\displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D706}}}\left(\unicode[STIX]{x1D706}v_{\Vert }{\displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D706}}}h\right),\\ U_{\Vert i}^{\ast }(h)=\left.\displaystyle \int \,\text{d}^{3}v\unicode[STIX]{x1D708}_{ii}(u)v_{\Vert }h\right/\displaystyle \int \,\text{d}^{3}vf_{0i}\unicode[STIX]{x1D708}_{ii}(u)m_{i}v_{\Vert }^{2}/2T_{i},\quad \unicode[STIX]{x1D708}_{ii}(u)=\unicode[STIX]{x1D708}_{0i}{\displaystyle \frac{\unicode[STIX]{x1D711}(u)}{u^{3}}},\\ \unicode[STIX]{x1D708}_{0i}={\displaystyle \frac{n_{0}\text{e}^{4}\ln \unicode[STIX]{x1D6EC}}{4\unicode[STIX]{x03C0}\unicode[STIX]{x1D700}_{0}^{2}m_{i}^{2}}}\left({\displaystyle \frac{m_{i}}{2T_{i}}}\right)^{3/2},\quad \unicode[STIX]{x1D711}(u)=\left(1-{\displaystyle \frac{1}{2u^{2}}}\right)\unicode[STIX]{x1D702}(u)+{\displaystyle \frac{\unicode[STIX]{x1D702}^{\prime }(u)}{2u}},\\ \unicode[STIX]{x1D702}(u)={\displaystyle \frac{2}{\sqrt{\unicode[STIX]{x03C0}}}}\displaystyle \int _{0}^{u}\,\text{d}t\text{e}^{-t^{2}}.\end{array}\right\}\end{eqnarray}$$

The constraint (3.12) yields an expression for $H_{1i}$ so that:

(3.14) $$\begin{eqnarray}h_{1i}=\frac{I}{\unicode[STIX]{x1D6FA}_{i}}\left(v_{\Vert }-H\hat{V}_{\Vert }\frac{B}{B_{0}}\right)\left(A_{1}^{(i)}+\left(u^{2}-\frac{5}{2}\right)A_{2}^{(i)}\right)f_{0i}+\frac{m_{i}}{T_{i}}\left\langle \frac{B}{B_{0}}U_{\Vert i}^{\ast }\right\rangle H\hat{V}_{\Vert }f_{0i},\end{eqnarray}$$

where $H(\unicode[STIX]{x1D706}-\unicode[STIX]{x1D706}_{c})$ is the Heaviside function and

(3.15) $$\begin{eqnarray}\hat{V}_{\Vert }=\frac{\unicode[STIX]{x1D70E}v^{2}B_{0}}{2}\int _{\unicode[STIX]{x1D706}}^{\unicode[STIX]{x1D706}_{c}}\frac{\text{d}\unicode[STIX]{x1D706}^{\prime }}{|\langle v_{\Vert }\rangle |},\quad \unicode[STIX]{x1D706}_{c}=\frac{B_{0}}{B_{\text{Max}}}.\end{eqnarray}$$

Using (3.14) and integrating over velocity space:

(3.16) $$\begin{eqnarray}\int (\ldots )\,\text{d}^{3}\boldsymbol{v}=\mathop{\sum }_{\unicode[STIX]{x1D70E}}\int (\ldots )\frac{B\unicode[STIX]{x03C0}v^{3}\,\text{d}v\,\text{d}\unicode[STIX]{x1D706}}{|v_{\Vert }|},\end{eqnarray}$$

we can calculate $U_{\Vert \,i}^{\ast }$ self-consistently. Thus

(3.17) $$\begin{eqnarray}U_{\Vert i}^{\ast }=\frac{I}{\unicode[STIX]{x1D6FA}_{i}}\frac{T_{i}}{m_{i}}\left(1-f_{c}\frac{B^{2}}{\langle B^{2}\rangle }\right)\left[A_{1}^{(i)}+\left(\frac{\{u^{2}\unicode[STIX]{x1D708}_{ii}\}}{\{\unicode[STIX]{x1D708}_{ii}\}}-\frac{5}{2}\right)A_{2}^{(i)}\right]+f_{c}B\frac{\langle BU_{\Vert i}^{\ast }\rangle }{\langle B^{2}\rangle },\end{eqnarray}$$

where we have defined the symbol $\left\{\ldots ..\right\}$ by

(3.18) $$\begin{eqnarray}\{w\}=\int \,\text{d}^{3}\boldsymbol{v}w\left(\frac{mv^{2}}{2T}\right)\frac{mv_{\Vert }^{2}}{T}\frac{f_{0}}{n}=\frac{8}{3\sqrt{\unicode[STIX]{x03C0}}}\int _{0}^{\infty }\,\text{d}uw(u)u^{4}\text{e}^{-u^{2}},\end{eqnarray}$$

and

(3.19) $$\begin{eqnarray}\int \,\text{d}^{3}\boldsymbol{v}w\left(\frac{mv^{2}}{2T}\right)\frac{mv_{\Vert }\hat{V}_{\Vert }}{T}\frac{B}{B_{0}}H\frac{f_{0}}{n}=f_{c}\frac{B^{2}}{B_{0}^{2}}\{w\},\end{eqnarray}$$

where $f_{c}=1-f_{t}$ , with $f_{t}$ the trapped particle fraction defined by

(3.20) $$\begin{eqnarray}f_{t}=1-\frac{3}{4}\langle B^{2}\rangle \int _{0}^{\unicode[STIX]{x1D706}_{c}}\frac{\unicode[STIX]{x1D706}\,\text{d}\unicode[STIX]{x1D706}}{\left\langle \sqrt{1-\unicode[STIX]{x1D706}B}\right\rangle }.\end{eqnarray}$$

Solving self-consistently for $\langle BU_{\Vert i}^{\ast }\rangle$ , substituting for $A_{1}^{(i)}$ and $A_{2}^{(i)}$ from (3.10) and using the values of the collisional integrals from appendix A, we obtain:

(3.21) $$\begin{eqnarray}\langle BU_{\Vert i}^{\ast }\rangle =-\frac{IT_{i}}{e}\left[\frac{\tilde{p}_{i}^{\prime }}{p_{0i}}+\frac{e\unicode[STIX]{x1D6F7}^{\prime }}{T_{i}}-1.17\frac{\tilde{T}_{i}^{\prime }}{T_{i}}\right].\end{eqnarray}$$

Then one can calculate the ion parallel flux from (3.14), to give

(3.22) $$\begin{eqnarray}U_{\Vert i}=-\frac{IT_{i}}{eB}\left(\frac{\tilde{p}_{i}^{\prime }}{p_{0i}}+\frac{e\unicode[STIX]{x1D6F7}^{\prime }}{T_{i}}-1.17f_{c}\frac{B^{2}}{\langle B^{2}\rangle }\frac{\tilde{T}_{i}^{\prime }}{T_{i}}\right).\end{eqnarray}$$

It remains to address the determination of $\hat{n}_{i}$ and $\hat{T}_{i}$ ; $\hat{n}_{i}$ is already given in terms of ${\tilde{n}}$ by (2.13), whereas $\hat{T}_{i}$ is obtained by applying the operator $\int (\ldots )(m_{i}v^{2}/2)\,\text{d}^{3}\boldsymbol{v}$ to the next-order equation,

(3.23) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{v_{\Vert }I}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}h_{3i}+\frac{v_{\Vert }I}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{Iv_{\Vert }}{\unicode[STIX]{x1D6FA}_{i}}\right)\frac{\unicode[STIX]{x2202}h_{2i}}{\unicode[STIX]{x2202}x}+\boldsymbol{v}_{d\unicode[STIX]{x1D703}i}\boldsymbol{\cdot }\unicode[STIX]{x1D735}h_{0i}-\text{i}\unicode[STIX]{x1D714}h_{0i}-C_{i}(h_{2i})-\overline{L_{i}C_{i}(g_{0i}L_{i})}\nonumber\\ \displaystyle & & \displaystyle \quad =-\frac{\text{i}e}{T_{i}}f_{0i}(\unicode[STIX]{x1D714}-\unicode[STIX]{x1D714}_{\ast i}^{T})\left(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9}+\frac{v_{\bot }^{2}}{2\unicode[STIX]{x1D6FA}_{i}}\unicode[STIX]{x1D6FF}B_{\Vert }\right),\end{eqnarray}$$

since it annihilates the first term on the left by means of the flux surface average and the collision term, $C_{i}\left(h_{2i}\right)$ , due to its conservation properties. This will be elaborated in § 4.2.

3.2 The electron solution

For the electrons, equation (2.10) becomes

(3.24) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{Iv_{\Vert }}{R^{2}Bq}\left(\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}+\text{i}\ell q^{\prime }x\right)h_{e}+\boldsymbol{v}_{de}\boldsymbol{\cdot }\unicode[STIX]{x1D735}h_{e}-\text{i}\unicode[STIX]{x1D714}h_{e}-C_{e}(h_{e})-\overline{L_{e}C_{e}(g_{e}L_{e})}\nonumber\\ \displaystyle & & \displaystyle \quad =\text{i}\frac{ef_{0e}}{T_{j}}(\unicode[STIX]{x1D714}-\unicode[STIX]{x1D714}_{\ast e}^{T})\left(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9}+\frac{v_{\bot }^{2}}{2\unicode[STIX]{x1D6FA}_{e}}\unicode[STIX]{x1D6FF}B_{\Vert }\right)+\frac{ef_{0e}}{T_{e}}\left(1-\frac{\unicode[STIX]{x1D714}_{\ast e}^{T}}{\unicode[STIX]{x1D714}}\right)\boldsymbol{v}_{de}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F9}.\qquad\end{eqnarray}$$

We employ a related ordering scheme to that of the ions to solve (3.24) for the electrons, introducing another small parameter, $\unicode[STIX]{x1D700}_{e}$ , where $\unicode[STIX]{x1D714}_{dre}\sim \unicode[STIX]{x1D708}_{e}\sim \unicode[STIX]{x1D700}_{e}\unicode[STIX]{x1D714}_{be}$ ; $\unicode[STIX]{x1D714}\sim \unicode[STIX]{x1D714}_{d\unicode[STIX]{x1D703}e}\sim \unicode[STIX]{x1D700}_{e}^{2}\unicode[STIX]{x1D714}_{be}$ . However, the semi-collisional electron model requires the additional ordering: $k_{\Vert }v_{\text{the}}\sim \unicode[STIX]{x1D700}_{e}^{2}v_{\text{the}}/L_{c}$ , where $k_{\Vert }=\ell q^{\prime }x/Rq$ . With $h_{e}=h_{0e}+\unicode[STIX]{x1D700}_{e}h_{1e}+\unicode[STIX]{x1D700}_{e}^{2}h_{2e}+\cdots \,$ , the lowest-order equation is

(3.25) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}h_{0e}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}=0\quad \Rightarrow \quad h_{0e}=h_{0e}(v,\unicode[STIX]{x1D706},x,\unicode[STIX]{x1D70E}),\end{eqnarray}$$

while in first order we obtain

(3.26) $$\begin{eqnarray}\frac{Iv_{\Vert }}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}h_{1e}=-v_{dre}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}x}h_{0e}+C_{e}(h_{0e})+\frac{ef_{0e}}{T_{e}}\left(1-\frac{\unicode[STIX]{x1D714}_{\ast e}^{T}}{\unicode[STIX]{x1D714}}\right)v_{dre}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}x}\unicode[STIX]{x1D6F9}.\end{eqnarray}$$

As in the case of the ions, $h_{1e}$ can be annihilated for passing particles by applying the operation: $\langle B(\ldots )/v_{\Vert }\rangle$ . For trapped particles, we again integrate along the bounce orbit, summing over $\unicode[STIX]{x1D70E}$ in the usual way. Thus, we obtain the constraint

(3.27) $$\begin{eqnarray}\left\langle \frac{B}{v_{\Vert }}C_{e}(h_{0e})\right\rangle =0,\end{eqnarray}$$

which determines $h_{0e}$ , yielding:

(3.28) $$\begin{eqnarray}h_{0e}=\left[\frac{\hat{n}_{e}}{n_{0}}+\left(u^{2}-\frac{3}{2}\right)\frac{\hat{T}_{e}}{T_{e}}\right]f_{0e},\end{eqnarray}$$

where now $u^{2}=m_{e}v^{2}/2T_{e}$ . The energy exchange term in the electron–ion collision operator is neglected (see (3.33) below), which makes the perturbation in the electron temperature independent of that in the ion temperature.

The equation for $h_{1e}$ can then be integrated to give

(3.29) $$\begin{eqnarray}h_{1e}=\frac{Iv_{\Vert }}{\unicode[STIX]{x1D6FA}_{e}}\left[\frac{e}{T_{e}}\left(1-\frac{\unicode[STIX]{x1D714}_{\ast e}^{T}}{\unicode[STIX]{x1D714}}\right)\unicode[STIX]{x1D6F9}^{\prime }f_{0e}-h_{0e}^{\prime }\right]+H_{1e}(v,\unicode[STIX]{x1D706},x,\unicode[STIX]{x1D70E}),\end{eqnarray}$$

where the function $H_{1e}$ remains to be determined. This result can be rewritten as

(3.30) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}h_{1e}={\displaystyle \frac{Iv_{\Vert }}{\unicode[STIX]{x1D6FA}_{e}}}\left[A_{1}^{(e)}+\left(u^{2}-{\displaystyle \frac{5}{2}}\right)A_{2}^{(e)}\right]f_{0e}+H_{1e}(v,\unicode[STIX]{x1D706},x,\unicode[STIX]{x1D70E}),\\ A_{1}^{(e)}={\displaystyle \frac{e\unicode[STIX]{x1D6F9}^{\prime }}{T_{e}}}\left(1-{\displaystyle \frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}}(1+\unicode[STIX]{x1D702}_{e})\right)-\left({\displaystyle \frac{\hat{n}_{e}^{\prime }}{n_{0}}}+{\displaystyle \frac{\hat{T}_{e}^{\prime }}{T_{e}}}\right)\equiv -\left({\displaystyle \frac{\tilde{p}_{e}^{\prime }}{p_{0e}}}-{\displaystyle \frac{e\unicode[STIX]{x1D6F7}^{\prime }}{T_{e}}}\right),\\ A_{2}^{(e)}=-\left({\displaystyle \frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}}{\displaystyle \frac{e\unicode[STIX]{x1D6F9}^{\prime }}{T_{e}}}+{\displaystyle \frac{\hat{T}_{e}^{\prime }}{T_{e}}}\right)\equiv -{\displaystyle \frac{\tilde{T}_{e}^{\prime }}{T_{e}}},\end{array}\right\}\end{eqnarray}$$

with $p_{0e}=n_{0}T_{e}$ , where again we have used (2.13) and (2.14). The next-order equation is

(3.31) $$\begin{eqnarray}\frac{v_{\Vert }I}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}h_{2e}+\frac{v_{\Vert }I}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{Iv_{\Vert }}{\unicode[STIX]{x1D6FA}_{e}}\right)\frac{\unicode[STIX]{x2202}h_{1e}}{\unicode[STIX]{x2202}x}+\frac{\text{i}v_{\Vert }I\ell q^{\prime }x}{qR^{2}B}h_{0e}=C_{e}(h_{1e}),\end{eqnarray}$$

leading to the constraint

(3.32) $$\begin{eqnarray}\left\langle \frac{B}{v_{\Vert }}C_{e}(h_{1e})\right\rangle =\frac{\text{i}I\ell q^{\prime }x}{q}\left\langle \frac{1}{R^{2}}\right\rangle h_{0e},\end{eqnarray}$$

which determines $H_{1e}(v,\unicode[STIX]{x1D706},x,\unicode[STIX]{x1D70E})$ .

The electron collision operator takes the form

(3.33) $$\begin{eqnarray}C_{e}(h)=C_{ee}(h)+\unicode[STIX]{x1D708}_{ei}(u)\left(L(h)+\frac{m_{e}}{T_{e}}v_{\Vert }U_{\Vert i}\,f_{0e}\right),\end{eqnarray}$$

where $C_{ee}(h)$ is the electron–electron collision operator, $U_{\Vert i}$ is the ion bulk parallel velocity, $\unicode[STIX]{x1D708}_{ei}(u)=\unicode[STIX]{x1D708}_{0e}/u^{3}$ with $\unicode[STIX]{x1D708}_{0e}=n_{0}\text{e}^{4}\ln \unicode[STIX]{x1D6EC}(m_{e}/2T_{e})^{3/2}/(4\unicode[STIX]{x03C0}\unicode[STIX]{x1D700}_{0}^{2}m_{e}^{2})$ .

At this point we introduce the two Spitzer functions (Spitzer & Harm Reference Spitzer and Harm1953): $h_{es}^{(1)}$ and $h_{es}^{(2)}$ , where

(3.34a,b ) $$\begin{eqnarray}C_{e}(h_{es}^{(1)})=v_{\Vert }\frac{\text{i}I\ell q^{\prime }x}{R^{2}Bq}\left(\frac{\hat{n}_{e}}{n_{0}}+\frac{\hat{T}_{e}}{T_{e}}\right)f_{0e};\quad C_{e}(h_{es}^{(2)})=v_{\Vert }\frac{\text{i}I\ell q^{\prime }x}{R^{2}Bq}\left(u^{2}-\frac{5}{2}\right)\frac{\hat{T}_{e}}{T_{e}}f_{0e},\end{eqnarray}$$

so that the constraint equation becomes

(3.35) $$\begin{eqnarray}\left\langle \frac{B}{v_{\Vert }}C_{e}(h_{1e}-h_{es})\right\rangle =0,\end{eqnarray}$$

with $h_{es}=h_{es}^{(1)}+h_{es}^{(2)}$ . Now that the collision operator is acting on a localised (to the trapped and barely passing region) distribution function we can use a model pitch-angle scattering operator for the electron–electron collisions (Rutherford et al. Reference Rutherford, Kovrizhnykh, Rosenbluth and Hinton1970).

(3.36) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}C_{ee}(h)=\unicode[STIX]{x1D708}_{ee}(u)\left(L(h)+{\displaystyle \frac{m_{e}}{T_{e}}}v_{\Vert }U_{\Vert e}^{\ast }(h)f_{0e}\right)\\ U_{\Vert e}^{\ast }(h)=\left.\displaystyle \int \text{d}^{3}\boldsymbol{v}\unicode[STIX]{x1D708}_{ee}(u)v_{\Vert }h\right/\displaystyle \int \text{d}^{3}vf_{0e}\unicode[STIX]{x1D708}_{ee}(u)m_{e}v_{\Vert }^{2}/2T_{e},\\ \unicode[STIX]{x1D708}_{ee}(u)=\unicode[STIX]{x1D708}_{0e}{\displaystyle \frac{\unicode[STIX]{x1D711}(u)}{u^{3}}}.\end{array}\right\}\end{eqnarray}$$

To remove the ion flow we write

(3.37a,b ) $$\begin{eqnarray}h_{es}^{(1)}=h_{s}^{(1)}+\frac{m_{e}v_{\Vert }}{T_{e}}U_{\Vert i}\,f_{0e},\quad h_{es}^{(2)}=h_{s}^{(2)}.\end{eqnarray}$$

The functions $h_{s}^{(1,2)}$ are given by Cohen et al. (Reference Cohen, Spitzer and McR. Routly1950) and Spitzer & Harm (Reference Spitzer and Harm1953):

(3.38) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}h_{s}^{(1)}={\displaystyle \frac{v_{\Vert }}{R^{2}B}}S^{(1)}{\displaystyle \frac{D^{(1)}(u)}{u}}f_{0e},\quad h_{s}^{(2)}={\displaystyle \frac{v_{\Vert }}{R^{2}B}}S^{(2)}{\displaystyle \frac{D^{(2)}(u)}{u}}f_{0e},\\ S^{(1)}={\displaystyle \frac{4\text{i}\unicode[STIX]{x1D70F}_{ei}}{3\sqrt{\unicode[STIX]{x03C0}}}}{\displaystyle \frac{I\ell q^{\prime }x}{q}}\left({\displaystyle \frac{\hat{n}_{e}}{n_{0}}}+{\displaystyle \frac{\hat{T}_{e}}{T_{e}}}\right)\equiv {\displaystyle \frac{4\text{i}\unicode[STIX]{x1D70F}_{ei}}{3\sqrt{\unicode[STIX]{x03C0}}}}{\displaystyle \frac{I\ell q^{\prime }x}{q}}\left({\displaystyle \frac{\tilde{p}_{e}}{p_{0e}}}-{\displaystyle \frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}}-{\displaystyle \frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}}{\displaystyle \frac{e\unicode[STIX]{x1D6F9}}{T_{e}}}\right),\\ S^{(2)}={\displaystyle \frac{4i\unicode[STIX]{x1D70F}_{ei}}{3\sqrt{\unicode[STIX]{x03C0}}}}{\displaystyle \frac{I\ell q^{\prime }x}{q}}{\displaystyle \frac{\hat{T}_{e}}{T_{e}}}\equiv {\displaystyle \frac{4i\unicode[STIX]{x1D70F}_{ei}}{3\sqrt{\unicode[STIX]{x03C0}}}}{\displaystyle \frac{I\ell q^{\prime }x}{q}}\left({\displaystyle \frac{\tilde{T}_{e}}{T_{e}}}-{\displaystyle \frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}}{\displaystyle \frac{e\unicode[STIX]{x1D6F9}}{T_{e}}}\right),\end{array}\right\}\end{eqnarray}$$

where $\unicode[STIX]{x1D70F}_{ei}=3\sqrt{\unicode[STIX]{x03C0}}/4\unicode[STIX]{x1D708}_{0e}$ and $D^{(1)}$ and $D^{(2)}$ are related to the normalised responses to a parallel electric field and thermal force, tabulated in Spitzer & Harm (Reference Spitzer and Harm1953) (to be precise, $D^{(1)}=-D/A$ and $D^{(2)}=2D/B$ as given in Spitzer & Harm (Reference Spitzer and Harm1953, Tables I and II)).

So, on defining $\unicode[STIX]{x1D708}_{e}=\unicode[STIX]{x1D708}_{ee}+\unicode[STIX]{x1D708}_{ei}$ , the constraint equation becomes

(3.39) $$\begin{eqnarray}\left\langle \frac{B}{v_{\Vert }}\left[\unicode[STIX]{x1D708}_{e}L(h_{1e}-h_{s})+\frac{m_{e}v_{\Vert }}{T_{e}}(\unicode[STIX]{x1D708}_{ee}U_{\Vert e}^{\ast }(h_{1e}-h_{s})+\unicode[STIX]{x1D708}_{ei}U_{\Vert i})f_{0e}\right]\right\rangle =0,\end{eqnarray}$$

where $h_{s}=h_{s}^{(1)}+h_{s}^{(2)}$ . Thus

(3.40) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D706}}\left(\unicode[STIX]{x1D706}\langle v_{\Vert }\rangle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D706}}H_{1e}\right)=\frac{IB_{0}v^{2}}{2\unicode[STIX]{x1D6FA}_{e}}\left(A_{1}^{(e)}+\left(u^{2}-\frac{5}{2}\right)A_{2}^{(e)}\right)f_{0e}\nonumber\\ \displaystyle & & \displaystyle \quad -\,\frac{m_{e}v^{2}}{2T_{e}}\left\langle \frac{B}{\unicode[STIX]{x1D708}_{e}}(\unicode[STIX]{x1D708}_{ee}U_{\Vert e}^{\ast }+\unicode[STIX]{x1D708}_{ei}U_{\Vert i})f_{0e}\right\rangle -\frac{(S^{(1)}D^{(1)}(u)+S^{(2)}D^{(2)}(u))v^{2}}{2u}\left\langle \frac{1}{R^{2}}\right\rangle f_{0e}.\qquad\end{eqnarray}$$

Integrating (3.40) in $\unicode[STIX]{x1D706}$ , we find

(3.41) $$\begin{eqnarray}\displaystyle h_{1e} & = & \displaystyle \frac{I}{\unicode[STIX]{x1D6FA}_{e}}\left(v_{\Vert }-H\hat{V}_{\Vert }\frac{B}{B_{0}}\right)\left(A_{1}^{(e)}+\left(u^{2}-\frac{5}{2}\right)A_{2}^{(e)}\right)f_{0e}\nonumber\\ \displaystyle & & \displaystyle +\,\left[\frac{m_{e}}{T_{e}}\left\langle \frac{B}{B_{0}\unicode[STIX]{x1D708}_{e}}(\unicode[STIX]{x1D708}_{ee}U_{\Vert e}^{\ast }+\unicode[STIX]{x1D708}_{ei}U_{\Vert i})\right\rangle +\frac{(S^{(1)}D^{(1)}(u)+S^{(2)}D^{(2)}(u))}{B_{0}u}\left\langle \frac{1}{R^{2}}\right\rangle \right]H\hat{V}_{\Vert }\,f_{0e}.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Using the definition of $U_{\Vert e}^{\ast }$ in (3.36) and integrating over velocity space, we can calculate $U_{\Vert e}^{\ast }(h_{1e}-h_{s})$ self-consistently. However, as argued by Helander & Sigmar (Reference Helander and Sigmar2002), $\langle BU_{\Vert e}^{\ast }\rangle =\langle BU_{\Vert i}\rangle +0(f_{t})$ , provided $f_{t}\ll 0(\unicode[STIX]{x1D708}_{ei}/\unicode[STIX]{x1D708}_{e})$ . Thus result (3.41) simplifies to

(3.42) $$\begin{eqnarray}\displaystyle h_{1e} & = & \displaystyle \frac{I}{\unicode[STIX]{x1D6FA}_{e}}\left(v_{\Vert }-H\hat{V}_{\Vert }\frac{B}{B_{0}}\right)\left(A_{1}^{(e)}+\left(u^{2}-\frac{5}{2}\right)A_{2}^{(e)}\right)f_{0e}\nonumber\\ \displaystyle & & \displaystyle +\,\left[\frac{m_{e}\langle BU_{\Vert i}\rangle }{T_{e}}+\frac{(S^{(1)}D^{(1)}(u)+S^{(2)}D^{(2)}(u))}{u}\left\langle \frac{1}{R^{2}}\right\rangle \right]\frac{H\hat{V}_{\Vert }}{B_{0}}f_{0e}.\end{eqnarray}$$

It remains to determine $\hat{n}_{e}$ and $\hat{T}_{e}$ ; these are obtained in § 4.1 by applying the operators $\langle \int (\ldots )\text{d}^{3}\boldsymbol{v}\rangle$ and $\int (\ldots )(m_{e}v^{2}/2)\text{d}^{3}\boldsymbol{v}$ to the next-order equation:

(3.43) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{v_{\Vert }I}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}h_{3e}+\frac{v_{\Vert }I}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{Iv_{\Vert }}{\unicode[STIX]{x1D6FA}_{e}}\right)\frac{\unicode[STIX]{x2202}h_{2e}}{\unicode[STIX]{x2202}x}+\frac{\text{i}I\ell q^{\prime }x}{qR^{2}B}v_{\Vert }h_{1e}+\boldsymbol{v}_{d\unicode[STIX]{x1D703}e}\boldsymbol{\cdot }\unicode[STIX]{x1D735}h_{0e}\nonumber\\ \displaystyle & & \displaystyle \quad -\,\text{i}\unicode[STIX]{x1D714}h_{0e}-C_{e}(h_{2e})-\overline{L_{e}C_{e}(g_{0e}L_{e})}=\frac{\text{i}e}{T_{e}}f_{0e}(\unicode[STIX]{x1D714}-\unicode[STIX]{x1D714}_{\ast e}^{T})\left(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9}+\frac{v_{\bot }^{2}}{2\unicode[STIX]{x1D6FA}_{e}}\unicode[STIX]{x1D6FF}B_{\Vert }\right).\qquad\end{eqnarray}$$

4.1 The perturbed electron density and temperature

As mentioned earlier, the determination of $\hat{n}_{e}$ and $\hat{T}_{e}$ is accomplished by applying the annihilators $\langle \int (\ldots )\text{d}^{3}\boldsymbol{v}\rangle$ and $\int (\ldots )(m_{e}v^{2}/2)\text{d}^{3}\boldsymbol{v}$ to the third-order equation (3.43), which both eliminate  $h_{3e}$ .

The first operation results in

(4.2) $$\begin{eqnarray}\displaystyle & & \displaystyle \left\langle \int \,\text{d}^{3}\boldsymbol{v}\frac{v_{\Vert }I}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{Iv_{\Vert }}{\unicode[STIX]{x1D6FA}_{e}}\right)\frac{\unicode[STIX]{x2202}h_{2e}}{\unicode[STIX]{x2202}x}\right\rangle +\left\langle \frac{\text{i}\ell Iq^{\prime }x}{R^{2}Bq}\int \,\text{d}^{3}\boldsymbol{v}v_{\Vert }h_{1e}\right\rangle \nonumber\\ \displaystyle & & \displaystyle \qquad +\,\left\langle \int \,\text{d}^{3}\boldsymbol{v}\boldsymbol{v}_{d\unicode[STIX]{x1D703}e}\boldsymbol{\cdot }\unicode[STIX]{x1D735}h_{0e}\right\rangle -\left\langle \int \,\text{d}^{3}\boldsymbol{v}\overline{L_{e}C_{e}(g_{0e}L_{e})}\right\rangle \nonumber\\ \displaystyle & & \displaystyle \quad =\text{i}\unicode[STIX]{x1D714}\hat{n}_{e}+\text{i}n_{0}\left\langle \frac{e}{T_{e}}(\unicode[STIX]{x1D714}-\unicode[STIX]{x1D714}_{\ast e})(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})-\frac{\unicode[STIX]{x1D6FF}B_{\Vert }}{B}(\unicode[STIX]{x1D714}-\unicode[STIX]{x1D714}_{\ast e}(1+\unicode[STIX]{x1D702}_{e}))\right\rangle .\end{eqnarray}$$

Here one can recognise the first three terms on the left-hand side as representing, respectively, the surface-averaged contributions to the electron continuity equation of the divergences of the radial flux, $\unicode[STIX]{x1D6E4}_{re}$ (which we will see below is related to Pfirsch–Schlüter and neoclassical radial transport), the parallel flux, $\unicode[STIX]{x1D6E4}_{\Vert e}$ , and the ‘azimuthal flux’, $\unicode[STIX]{x1D6E4}_{e}^{Az}$ (Fitzpatrick Reference Fitzpatrick1989), while the fourth corresponds to the contribution from classical radial transport.

The second, the energy moment, takes the form

(4.3) $$\begin{eqnarray}\displaystyle & & \displaystyle \left\langle \int \,\text{d}^{3}\boldsymbol{v}\frac{v_{\Vert }I}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{Iv_{\Vert }}{\unicode[STIX]{x1D6FA}_{e}}\right)\frac{m_{e}v^{2}}{2}\frac{\unicode[STIX]{x2202}h_{2e}}{\unicode[STIX]{x2202}x}\right\rangle +\left\langle \frac{\text{i}\ell Iq^{\prime }x}{R^{2}Bq}\int \int \,\text{d}^{3}\boldsymbol{v}v_{\Vert }\frac{m_{e}v^{2}}{2}h_{1e}\right\rangle \nonumber\\ \displaystyle & & \displaystyle \qquad +\,\left\langle \int \,\text{d}^{3}\boldsymbol{v}\frac{m_{e}v^{2}}{2}\boldsymbol{v}_{d\unicode[STIX]{x1D703}e}\boldsymbol{\cdot }\unicode[STIX]{x1D735}h_{0e}\right\rangle -\left\langle \int \,\text{d}^{3}\boldsymbol{v}\frac{m_{e}v^{2}}{2}\overline{L_{e}C_{e}(g_{0e}L_{e})}\right\rangle \nonumber\\ \displaystyle & & \displaystyle \quad =\frac{3}{2}\text{i}\unicode[STIX]{x1D714}\left(\hat{n}_{e}+\frac{n_{0}\hat{T}_{e}}{T_{e}}\right)+\text{i}n_{0}\left\langle \frac{3}{2}\frac{e}{T_{e}}(\unicode[STIX]{x1D714}-\unicode[STIX]{x1D714}_{\ast e}(1+\unicode[STIX]{x1D702}_{e}))(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})\right.\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\left.\frac{5}{2}\frac{\unicode[STIX]{x1D6FF}B_{\Vert }}{B}(\unicode[STIX]{x1D714}-\unicode[STIX]{x1D714}_{\ast e}(1+2\unicode[STIX]{x1D702}_{e}))\right\rangle ,\end{eqnarray}$$

with a similar interpretation in terms of fluxes.

It is helpful to separate the electron and ion distributions into a Pfirsch–Schlüter-like, $h_{e}^{PS}$ , and a banana neoclassical part, $h_{e}^{\text{Ban}}$ , and express the quantities $\hat{n}_{j}$ and $\hat{T}_{j}$ in terms of ${\tilde{n}}_{j}$ and $\tilde{T}_{j}$ . Thus, for the electrons we write

(4.4) $$\begin{eqnarray}h_{1e}-h_{es}=h_{e}^{PS}+h_{e}^{\text{Ban}},\end{eqnarray}$$

where

(4.5) $$\begin{eqnarray}\displaystyle h_{e}^{PS} & = & \displaystyle \frac{Iv_{\Vert }m_{e}}{eB}\left(1-\frac{B^{2}}{\langle B^{2}\rangle }\right)\left[\frac{\tilde{p}^{\prime }}{p_{0e}}+\left(u^{2}-\frac{5}{2}\right)\frac{\tilde{T}_{e}^{\prime }}{T_{e}}\right]f_{0e}+\frac{Iv_{\Vert }m_{e}}{eB}\left(1-\frac{B^{2}}{\langle B^{2}\rangle }\right)\nonumber\\ \displaystyle & & \displaystyle \times \,\left[\frac{\tilde{p}_{i}^{\prime }}{p_{0e}}+\frac{e\unicode[STIX]{x1D6F7}^{\prime }}{T_{e}}\right]f_{0e}-\frac{4\text{i}\unicode[STIX]{x1D70F}_{ei}v_{\Vert }}{3\sqrt{\unicode[STIX]{x03C0}}}\frac{I\ell q^{\prime }x}{q}\left(\frac{1}{R^{2}B}-\frac{B}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \right)\nonumber\\ \displaystyle & & \displaystyle \times \,\left\{\frac{D^{(1)}(u)}{u}\left[\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}(1+\unicode[STIX]{x1D702}_{e})}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right]\right.\nonumber\\ \displaystyle & & \displaystyle +\,\left.\frac{D^{(2)}(u)}{u}\left[\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right]\right\}f_{0e}\end{eqnarray}$$

and

(4.6) $$\begin{eqnarray}\displaystyle h_{e}^{\text{Ban}} & = & \displaystyle \frac{Im_{e}}{e}\left(\frac{B}{\langle B^{2}\rangle }v_{\Vert }-\frac{H\hat{V}_{\Vert }}{B_{0}}\right)\left[\frac{\tilde{p}^{\prime }}{p_{0e}}+\left(u^{2}-\frac{5}{2}\right)\frac{\tilde{T}_{e}^{\prime }}{T_{e}}-1.17\frac{\tilde{T}_{i}^{\prime }}{T_{e}}\right]f_{0e}\nonumber\\ \displaystyle & & \displaystyle -\,\frac{4\text{i}\unicode[STIX]{x1D70F}_{ei}}{3\sqrt{\unicode[STIX]{x03C0}}}\frac{I\ell q^{\prime }x}{q}\left(\frac{B}{\langle B^{2}\rangle }v_{\Vert }-\frac{H\hat{V}_{\Vert }}{B_{0}}\right)\left\langle \frac{1}{R^{2}}\right\rangle \nonumber\\ \displaystyle & & \displaystyle \times \,\left\{\frac{D^{(1)}(u)}{u}\left[\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}(1+\unicode[STIX]{x1D702}_{e})}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right]\right.\nonumber\\ \displaystyle & & \displaystyle +\,\left.\frac{D^{(2)}(u)}{u}\left[\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right]\right\}f_{0e},\end{eqnarray}$$

where we have substituted for $U_{\Vert i}$ from (3.22).

Similarly, it is convenient to separate $h_{1i}$ in (3.14) into Pfirsch–Schlüter, $h_{i}^{PS}$ , and banana contributions, $h_{i}^{\text{Ban}}$ , writing:

(4.7) $$\begin{eqnarray}h_{1i}=h_{i}^{\text{Flow}}+h_{i}^{PS}+h_{i}^{\text{Ban}},\end{eqnarray}$$

where $h_{i}^{\text{Flow}}$ describes the mean flow:

(4.8) $$\begin{eqnarray}h_{i}^{\text{Flow}}=-\frac{Iv_{\Vert }m_{i}B}{e\langle B^{2}\rangle }\left[\frac{\tilde{p}_{i}^{\prime }}{p_{0i}}+\frac{e\unicode[STIX]{x1D6F7}^{\prime }}{T_{i}}-1.17\frac{\tilde{T}_{i}^{\prime }}{T_{i}}\right]f_{0i},\end{eqnarray}$$

while

(4.9) $$\begin{eqnarray}h_{i}^{PS}=-\frac{Iv_{\Vert }m_{i}}{eB}\left(1-\frac{B^{2}}{\langle B^{2}\rangle }\right)\left[\frac{\tilde{p}_{i}^{\prime }}{p_{0i}}+\frac{e\unicode[STIX]{x1D6F7}^{\prime }}{T_{i}}+\left(u^{2}-\frac{5}{2}\right)\frac{\tilde{T}_{i}^{\prime }}{T_{i}}\right]f_{0i}\end{eqnarray}$$

and

(4.10) $$\begin{eqnarray}h_{i}^{\text{Ban}}=-\frac{Im_{i}}{e}\left(\frac{Bv_{\Vert }}{\langle B^{2}\rangle }-\frac{H\hat{V}_{\Vert }}{B_{0}}\right)(u^{2}-1.33)\frac{\tilde{T}_{i}^{\prime }}{T_{i}}f_{0i}.\end{eqnarray}$$

The first term represents a drifting Maxwellian and therefore does not contribute to the ion–ion collision term.

4.1.1 Electron radial fluxes

First, we consider the radial fluxes arising from classical transport. Recalling the definitions of $h_{0j}$ and $L_{j}$ and using the conservation of momentum in like-particle collisions, we obtain

(4.11) $$\begin{eqnarray}\displaystyle & & \displaystyle \left\langle \int \,\text{d}^{3}\boldsymbol{v}\overline{L_{e}C_{e}(g_{0e}L_{e})}\right\rangle \equiv \frac{\text{d}}{\text{d}r}\langle \unicode[STIX]{x1D6E4}_{r,e}^{C}\rangle \nonumber\\ \displaystyle & & \displaystyle \quad =\frac{m_{e}T_{e}}{e^{2}}\frac{\text{d}^{2}}{\text{d}x^{2}}\left\langle \frac{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}{B^{2}}\right\rangle \int \,\text{d}^{3}\boldsymbol{v}\frac{m_{e}v_{r}}{T_{e}}C_{ei}\left[v_{r}\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e\unicode[STIX]{x1D6F7}}{T_{e}}+\left(u^{2}-\frac{5}{2}\right)\frac{\tilde{T}_{e}}{T_{e}}\right)\right],\qquad\end{eqnarray}$$

where $r$ represents a ‘radial-like’ coordinate labelling flux surfaces, $v_{r}$ being the corresponding radial component of the velocity. We note the gyro-correction to the scattering target ion distribution has the form

(4.12) $$\begin{eqnarray}v_{r}\frac{m_{i}}{eB}\left(\frac{\tilde{p}_{i}}{p_{0i}}+\frac{e\unicode[STIX]{x1D6F7}}{T_{i}}+\left(u^{2}-\frac{5}{2}\right)\frac{\tilde{T}_{i}}{T_{i}}\right)f_{0i}.\end{eqnarray}$$

The velocity integral can be evaluated by observing that it involves the matrix elements of the collision operator between Laguerre polynomials given by Helander & Sigmar (Reference Helander and Sigmar2002, § 4.5). The resultant contribution from (4.11) is the familiar Braginskii expression:

(4.13) $$\begin{eqnarray}\frac{\text{d}}{\text{d}r}\langle \unicode[STIX]{x1D6E4}_{r,e}^{C}\rangle =-\frac{m_{e}T_{e}}{\text{e}^{2}\unicode[STIX]{x1D70F}_{ei}}\left\langle \frac{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}{B^{2}}\right\rangle \frac{\text{d}^{2}}{\text{d}x^{2}}\left(\frac{\tilde{p}}{p_{0e}}-\frac{3}{2}\frac{\tilde{T}_{e}}{T_{e}}\right)n_{0}.\end{eqnarray}$$

A similar calculation for the classical heat flux results in

(4.14) $$\begin{eqnarray}\frac{\text{d}}{\text{d}r}\langle q_{r,e}^{C}\rangle =-\frac{m_{e}T_{e}}{\text{e}^{2}\unicode[STIX]{x1D70F}_{ei}}\left\langle \frac{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}{B^{2}}\right\rangle \frac{\text{d}^{2}}{\text{d}x^{2}}\left(-\frac{3}{2}\frac{\tilde{p}}{p_{0e}}+4.66\frac{\tilde{T}_{e}}{T_{e}}\right)n_{0}T_{e},\end{eqnarray}$$

so that the classical energy flux, $\langle Q_{r,e}^{C}\rangle =\langle q_{r,e}^{C}\rangle +(5/2)\langle \unicode[STIX]{x1D6E4}_{r,e}^{C}\rangle$ , satisfies

(4.15) $$\begin{eqnarray}\frac{\text{d}}{\text{d}r}\langle Q_{r,e}^{C}\rangle =-\frac{m_{e}T_{e}}{\text{e}^{2}\unicode[STIX]{x1D70F}_{ei}}\left\langle \frac{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}{B^{2}}\right\rangle \frac{\text{d}^{2}}{\text{d}x^{2}}\left(\frac{\tilde{p}}{p_{0e}}+0.91\frac{\tilde{T}_{e}}{T_{e}}\right)n_{0}T_{e}.\end{eqnarray}$$

Turning to the leading term in (4.2), we can eliminate $h_{2e}$ by integrating by parts in $\unicode[STIX]{x1D703}$ and using (3.31) to obtain

(4.16) $$\begin{eqnarray}\displaystyle & & \displaystyle \left\langle \int \,\text{d}^{3}\boldsymbol{v}\frac{v_{\Vert }I}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{Iv_{\Vert }}{\unicode[STIX]{x1D6FA}_{e}}\right)h_{2e}\right\rangle \equiv \langle \unicode[STIX]{x1D6E4}_{r,e}^{NC}\rangle \nonumber\\ \displaystyle & & \displaystyle \quad =\left\langle \frac{I}{\unicode[STIX]{x1D6FA}_{e}}\int \,\text{d}^{3}\boldsymbol{v}\left[\frac{v_{\Vert }^{2}I}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{I}{\unicode[STIX]{x1D6FA}_{e}}\right)\frac{\text{d}h_{1e}}{\text{d}x}+\frac{\text{i}\ell Iq^{\prime }x}{R^{2}Bq}v_{\Vert }^{2}h_{0e}-v_{\Vert }C_{e}(h_{1e})\right]\right\rangle ,\end{eqnarray}$$

where the first term vanishes if the equilibrium is up–down symmetric, which we take to be the case here. Thus, finally, we recognise this contribution as the divergence of the neoclassical radial flux

(4.17) $$\begin{eqnarray}\frac{\text{d}}{\text{d}r}\langle \unicode[STIX]{x1D6E4}_{r,e}^{NC}\rangle =-\frac{\text{d}}{\text{d}x}\left\langle \frac{I}{\unicode[STIX]{x1D6FA}_{e}}\int \,\text{d}^{3}\boldsymbol{v}v_{\Vert }C_{e}(h_{1e}-h_{es}^{(1)}-h_{es}^{(2)})\right\rangle .\end{eqnarray}$$

Likewise, for the thermal equation, we find

(4.18) $$\begin{eqnarray}\frac{\text{d}}{\text{d}r}\langle Q_{r,e}^{NC}\rangle =-\frac{\text{d}}{\text{d}x}\left\langle \frac{I}{\unicode[STIX]{x1D6FA}_{e}}\int \,\text{d}^{3}\boldsymbol{v}v_{\Vert }\frac{m_{e}v^{2}}{2}C_{e}(h_{1e}-h_{es}^{(1)}-h_{es}^{(2)})\right\rangle .\end{eqnarray}$$

To evaluate the velocity integrals in (4.17) and (4.18), we recall the result (3.19).

Considering the Pfirsch–Schlüter and banana contributions separately, we introduce their respective distribution functions. From momentum conservation, it is again clear that only $e{-}i$ collisions contribute to (4.17) in both cases. For the Pfirsch–Schlüter contribution the calculation has the same structure as the classical case, with the substitution $v_{r}\rightarrow v_{\Vert }$ in the distribution function. However, since we are now considering the difference $h_{1e}-h_{es}$ , the scattering ‘ion distribution’ is effectively at rest when evaluating the collisional matrix elements. Recalling (4.5), the result is

(4.19) $$\begin{eqnarray}\displaystyle \frac{1}{n_{0}}\frac{\text{d}}{\text{d}r}\langle \unicode[STIX]{x1D6E4}_{r,e}^{PS}\rangle & = & \displaystyle -\frac{m_{e}T_{e}}{\text{e}^{2}\unicode[STIX]{x1D70F}_{ei}}I^{2}\left(\left\langle \frac{1}{B^{2}}\right\rangle -\frac{1}{\langle B^{2}\rangle }\right)\frac{\text{d}^{2}}{\text{d}x^{2}}\left(\frac{\tilde{p}}{p_{0e}}-\frac{3}{2}\frac{\tilde{T}_{e}}{T_{e}}\right)\nonumber\\ \displaystyle & & \displaystyle -\,\frac{\text{i}I^{2}\ell q^{\prime }T_{e}}{eq}\left(\left\langle \frac{1}{R^{2}B^{2}}\right\rangle -\frac{1}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \right)\frac{\text{d}}{\text{d}x}\nonumber\\ \displaystyle & & \displaystyle \times \,\left[x\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e\unicode[STIX]{x1D6F7}}{T_{e}}+\left(1-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{e})\right)\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\right].\end{eqnarray}$$

For the heat flux, we obtain

(4.20) $$\begin{eqnarray}\displaystyle \frac{1}{n_{0}T_{e}}\frac{\text{d}}{\text{d}r}\langle Q_{r,e}^{PS}\rangle & = & \displaystyle -\frac{m_{e}T_{e}I^{2}}{\text{e}^{2}\unicode[STIX]{x1D70F}_{ei}}\left(\left\langle \frac{1}{B^{2}}\right\rangle -\frac{1}{\langle B^{2}\rangle }\right)\frac{\text{d}^{2}}{\text{d}x^{2}}\left(\frac{\tilde{p}}{p_{0e}}+0.91\frac{\tilde{T}_{e}}{T_{e}}\right)\nonumber\\ \displaystyle & & \displaystyle -\,\frac{\text{i}I^{2}\ell q^{\prime }T_{e}}{eq}\left(\left\langle \frac{1}{R^{2}B^{2}}\right\rangle -\frac{1}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \right)\frac{\text{d}}{\text{d}x}\nonumber\\ \displaystyle & & \displaystyle \times \,\left[\frac{5}{2}x\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e\unicode[STIX]{x1D6F7}}{T_{e}}+\frac{\tilde{T}_{e}}{T_{e}}+\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\left(1-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+2\unicode[STIX]{x1D702}_{e})\right)\right)\right].\qquad\end{eqnarray}$$

Finally, we consider the banana contribution. For the classical and Pfirsch–Schlüter contributions we used the exact collision operator, since the distributions were not localised in velocity space. However, the banana contribution to $h_{1e}-h_{ss}$ is localised and it is sufficient to use the model collision operator (3.36), which leads to:

(4.21) $$\begin{eqnarray}\langle \unicode[STIX]{x1D6E4}_{e}^{\text{Ban}}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}\rangle =-\frac{I}{e}\left\langle \frac{m_{e}}{B}\int \,\text{d}^{3}\boldsymbol{v}v_{\Vert }\unicode[STIX]{x1D708}_{e}\left(h_{1e}-h_{s}^{(1)}-h_{s}^{(2)}-\frac{m_{e}v_{\Vert }}{T_{e}\unicode[STIX]{x1D708}_{e}}(\unicode[STIX]{x1D708}_{ee}U_{\Vert e}^{\ast }+\unicode[STIX]{x1D708}_{ei}U_{\Vert i})f_{0e}\right)^{\text{Ban}}\right\rangle .\end{eqnarray}$$

Substituting expression (4.6), using result (3.22) and approximating $\langle BU_{\Vert e}^{\ast }\rangle \cong \langle BU_{\Vert i}\rangle$ as discussed below (3.41), we obtain

(4.22) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{n_{0}}\frac{\text{d}}{\text{d}r}\langle \unicode[STIX]{x1D6E4}_{r,e}^{\text{Ban}}\rangle =-f_{t}\frac{I^{2}m_{e}T_{e}\{\unicode[STIX]{x1D708}_{e}\}}{\text{e}^{2}\langle B^{2}\rangle }\frac{\text{d}^{2}}{\text{d}x^{2}}\left[\frac{\tilde{p}}{p_{0e}}+\left(\frac{\{u^{2}\unicode[STIX]{x1D708}_{e}\}}{\{\unicode[STIX]{x1D708}_{e}\}}-\frac{5}{2}\right)\frac{\tilde{T}_{e}}{T_{e}}-1.17\frac{\tilde{T}_{i}}{T_{e}}\right]\nonumber\\ \displaystyle & & \displaystyle \quad +\,\frac{4\unicode[STIX]{x1D70F}_{ei}\text{i}\ell q^{\prime }}{3\sqrt{\unicode[STIX]{x03C0}}q}\frac{f_{t}}{\langle B^{2}\rangle }\left\langle \frac{I^{2}}{R^{2}}\right\rangle \frac{T_{e}}{e}\frac{\text{d}}{\text{d}x}\!\left(x\left[\!\vphantom{\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)}\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\!\left\{\frac{\unicode[STIX]{x1D708}_{e}D^{(1)}}{u}\right\}\right.\right.\nonumber\\ \displaystyle & & \displaystyle \quad +\,\left.\left.\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\left\{\frac{\unicode[STIX]{x1D708}_{e}D^{(2)}}{u}\right\}\right]\right).\end{eqnarray}$$

Introducing the numerical values of the various collisional averages given by the integrals in appendix A, this becomes:

(4.23) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{n_{0}}\frac{\text{d}}{\text{d}r}\langle \unicode[STIX]{x1D6E4}_{r,e}^{\text{Ban}}\rangle =-1.53f_{t}\frac{I^{2}m_{e}T_{e}}{\text{e}^{2}\langle B^{2}\rangle \unicode[STIX]{x1D70F}_{ei}}\frac{\text{d}^{2}}{\text{d}x^{2}}\left[\frac{\tilde{p}}{p_{0e}}+1.39\frac{\tilde{T}_{e}}{T_{e}}-1.17\frac{\tilde{T}_{i}}{T_{e}}\right]\nonumber\\ \displaystyle & & \displaystyle \quad -\,1.67\frac{f_{t}}{\langle B^{2}\rangle }\left\langle \frac{I^{2}}{R^{2}}\right\rangle \frac{\text{i}\ell q^{\prime }T_{e}}{eq}\frac{\text{d}}{\text{d}x}\left(x\left[\vphantom{\frac{\tilde{T}_{e}}{T_{e}}}\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+1.02\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right.\right.\nonumber\\ \displaystyle & & \displaystyle \quad +\,\left.\left.0.02\frac{\tilde{T}_{e}}{T_{e}}\right]\right).\end{eqnarray}$$

Similarly, we find

(4.24) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{n_{0}T_{e}}\frac{\text{d}}{\text{d}r}\langle Q_{r,e}^{\text{Ban}}\rangle =-f_{t}\frac{I^{2}m_{e}T_{e}\{u^{2}\unicode[STIX]{x1D708}_{e}\}}{\text{e}^{2}\langle B^{2}\rangle }\frac{\text{d}^{2}}{\text{d}x^{2}}\!\!\left[\frac{\tilde{p}}{p_{0e}}+\left(\frac{\{u^{4}\unicode[STIX]{x1D708}_{e}\}}{\{u^{2}\unicode[STIX]{x1D708}_{e}\}}-\frac{5}{2}\right)\frac{\tilde{T}_{e}}{T_{e}}+\left(\frac{\{u^{2}\unicode[STIX]{x1D708}_{ii}\}}{\{\unicode[STIX]{x1D708}_{ii}\}}-\frac{5}{2}\right)\frac{\tilde{T}_{i}}{T_{e}}\right]\nonumber\\ \displaystyle & & \displaystyle \quad -\,\frac{4\unicode[STIX]{x1D70F}_{ei}\text{i}\ell q^{\prime }}{3\sqrt{\unicode[STIX]{x03C0}}q}\frac{f_{t}}{\langle B^{2}\rangle }\left\langle \frac{I^{2}}{R^{2}}\right\rangle \frac{T_{e}}{e}\frac{\text{d}}{\text{d}x}\left(x\left[\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\{u\unicode[STIX]{x1D708}_{e}D^{(1)}\}\right.\right.\nonumber\\ \displaystyle & & \displaystyle \quad +\,\left.\left.\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\{u\unicode[STIX]{x1D708}_{e}D^{(2)}\}\right]\right),\end{eqnarray}$$

leading to

(4.25) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{n_{0}T_{e}}\frac{\text{d}}{\text{d}r}\langle Q_{r,e}^{\text{Ban}}\rangle =-1.70f_{t}\frac{I^{2}m_{e}T_{e}}{\text{e}^{2}\langle B^{2}\rangle \unicode[STIX]{x1D70F}_{ei}}\frac{\text{d}^{2}}{\text{d}x^{2}}\left[\frac{\tilde{p}}{p_{0e}}-0.40\frac{\tilde{T}_{e}}{T_{e}}-1.17\frac{\tilde{T}_{i}}{T_{e}}\right]\nonumber\\ \displaystyle & & \displaystyle \quad -\,2.97\frac{f_{t}}{\langle B^{2}\rangle }\left\langle \frac{I^{2}}{R^{2}}\right\rangle \frac{i\ell q^{\prime }T_{e}}{eq}\frac{\text{d}}{\text{d}x}\left(x\left[\vphantom{\frac{\tilde{T}_{e}}{T_{e}}}\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+1.25\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right.\right.\nonumber\\ \displaystyle & & \displaystyle \quad \left.\left.+\,0.25\frac{\tilde{T}_{e}}{T_{e}}\right]\right).\end{eqnarray}$$

4.1.2 Parallel electron fluxes and plasma current

The contribution to (4.2) from the parallel flux $\unicode[STIX]{x1D6E4}_{\Vert e}$ is

(4.26) $$\begin{eqnarray}\left\langle \frac{\text{i}\ell Iq^{\prime }x}{R^{2}Bq}\int \,\text{d}^{3}\boldsymbol{v}v_{\Vert }h_{1e}\right\rangle =\left\langle \frac{\text{i}\ell Iq^{\prime }x}{R^{2}Bq}U_{\Vert e}(h_{1e})\right\rangle .\end{eqnarray}$$

The part arising from the Pfirsch–Schlüter-like part of $h_{1e}$ can be calculated directly, but for the banana contribution we take advantage of the self-adjointness of the collision operator to circumvent the fact that $h_{1e}$ is not localised in pitch angle, while still allowing us to use the model collision operator (Helander & Sigmar Reference Helander and Sigmar2002).

We first consider the parallel current, which is needed for Ampère’s equation, but also provides an expression for $U_{\Vert e}$ . We can readily calculate $j_{\Vert }^{PS}$ , the current arising from the Pfirsch–Schlüter-like contributions to the electron and ion distribution functions given in (4.5) and (4.9), obtaining:

(4.27) $$\begin{eqnarray}\displaystyle j_{\Vert }^{PS} & = & \displaystyle -I\tilde{p}^{\prime }\left(\frac{1}{B}-\frac{B}{\langle B^{2}\rangle }\right)+\frac{4\unicode[STIX]{x1D70F}_{ei}}{3\sqrt{\unicode[STIX]{x03C0}}}\frac{\text{i}I\ell q^{\prime }xp_{0e}e}{m_{e}q}\nonumber\\ \displaystyle & & \displaystyle \times \,\left(\frac{1}{R^{2}B}-\frac{B}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \right)\left[\vphantom{\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)}\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\left\{\frac{D^{(1)}}{u}\right\}\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\left\{\frac{D^{(2)}}{u}\right\}\right],\end{eqnarray}$$

which becomes

(4.28) $$\begin{eqnarray}\displaystyle j_{\Vert }^{PS} & = & \displaystyle -I\tilde{p}^{\prime }\left(\frac{1}{B}-\frac{B}{\langle B^{2}\rangle }\right)-1.97\unicode[STIX]{x1D70F}_{ei}\frac{\text{i}I\ell q^{\prime }xp_{0e}e}{m_{e}q}\nonumber\\ \displaystyle & & \displaystyle \times \left(\frac{1}{R^{2}B}-\frac{B}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \!\right)\!\left[\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}+\frac{0.34\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1\!+\!1.34\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\!\right]\!.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

To calculate the banana regime neoclassical contribution, we write

(4.29) $$\begin{eqnarray}j_{\Vert }=j_{\Vert s}+(j_{\Vert }-j_{\Vert s})=j_{\Vert s}-e\int \,\text{d}^{3}\boldsymbol{v}v_{\Vert }(h_{1e}-h_{es}),\end{eqnarray}$$

where

(4.30) $$\begin{eqnarray}j_{\Vert s}=n_{0}eU_{\Vert i}-e\int \,\text{d}^{3}\boldsymbol{v}v_{\Vert }h_{es},\end{eqnarray}$$

so that

(4.31) $$\begin{eqnarray}\displaystyle j_{\Vert s} & = & \displaystyle -\frac{4\unicode[STIX]{x1D70F}_{ei}}{3\sqrt{\unicode[STIX]{x03C0}}}n_{0}e\frac{\text{i}\ell Iq^{\prime }x}{R^{2}Bq}\frac{T_{e}}{m_{e}}\nonumber\\ \displaystyle & & \displaystyle \times \,\left[\vphantom{\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)}\left\{\frac{D^{(1)}(u)}{u}\right\}\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,\left\{\frac{D^{(2)}(u)}{u}\right\}\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\right],\end{eqnarray}$$

which reduces to

(4.32) $$\begin{eqnarray}j_{\Vert s}=1.97n_{0}e\unicode[STIX]{x1D70F}_{ei}\frac{\text{i}\ell Iq^{\prime }x}{R^{2}Bq}\frac{T_{e}}{m_{e}}\left[\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}+0.34\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+1.34\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right].\end{eqnarray}$$

Now, recalling (3.34), we can write

(4.33) $$\begin{eqnarray}j_{\Vert }-j_{\Vert s}=-e\int \,\text{d}^{3}\boldsymbol{v}v_{\Vert }(h_{1e}-h_{es}^{(1)}-h_{es}^{(2)})=-\frac{eR^{2}B}{S^{(1)}}\int \,\text{d}^{3}\boldsymbol{v}\frac{C_{e}(h_{s}^{(1)})}{f_{0e}}(h_{1e}-h_{es}^{(1)}-h_{es}^{(2)}).\end{eqnarray}$$

Using the self-adjointness property of $C_{e}(h_{s}^{(1)})$ , we obtain

(4.34) $$\begin{eqnarray}j_{\Vert }-j_{\Vert s}=-\frac{eR^{2}B}{S^{(1)}}\int \,\text{d}^{3}\boldsymbol{v}\frac{h_{s}^{(1)}}{f_{0e}}C_{e}(h_{1e}-h_{es}^{(1)}-h_{es}^{(2)}),\end{eqnarray}$$

so that

(4.35) $$\begin{eqnarray}\displaystyle (j_{\Vert }-j_{\Vert s}) & = & \displaystyle -\frac{4\unicode[STIX]{x1D70F}_{ei}}{3\sqrt{\unicode[STIX]{x03C0}}}e\int \,\text{d}^{3}\boldsymbol{v}\frac{D^{(1)}(u)}{u}v_{\Vert }C_{e}(h_{1e}-h_{es}^{(1)}-h_{es}^{(2)})\nonumber\\ \displaystyle & = & \displaystyle \frac{4\unicode[STIX]{x1D70F}_{ei}}{3\sqrt{\unicode[STIX]{x03C0}}}e\int \,\text{d}^{3}\boldsymbol{v}\frac{D^{(1)}(u)}{u}v_{\Vert }\unicode[STIX]{x1D708}_{e}\left(h_{1e}-h_{s}^{(1)}-h_{s}^{(2)}-\frac{m_{e}v_{\Vert }}{T_{e}}U_{\Vert i}\,f_{0e}\right),\end{eqnarray}$$

where we have used the definitions (3.38) and the model electron–electron collision operator (3.36), since $(h_{1e}-h_{es}^{(1)}-h_{es}^{(2)})$ is now localised. Evaluating this expression using result (4.6)

(4.36) $$\begin{eqnarray}\displaystyle (j_{\Vert }-j_{\Vert s})^{\text{Ban}} & = & \displaystyle \frac{4\unicode[STIX]{x1D70F}_{ei}n_{0}}{3\sqrt{\unicode[STIX]{x03C0}}}\frac{f_{t}BIT_{e}}{\langle B^{2}\rangle }\left[\frac{\tilde{p}^{\prime }}{p_{0e}}\left\{\frac{\unicode[STIX]{x1D708}_{e}D^{(1)}}{u}\right\}-\left\{\left(\frac{5}{2}-u^{2}\right)\frac{\unicode[STIX]{x1D708}_{e}D^{(1)}}{u}\right\}\frac{\tilde{T}_{e}^{\prime }}{T_{e}}\right.\nonumber\\ \displaystyle & & \displaystyle \left.-\,\left(\frac{5}{2}-\frac{\{u^{2}\unicode[STIX]{x1D708}_{ii}\}}{\{\unicode[STIX]{x1D708}_{ii}\}}\right)\left\{\frac{\unicode[STIX]{x1D708}_{e}D^{(1)}}{u}\right\}\frac{\tilde{T}_{i}^{\prime }}{T_{e}}\right]-\left(\frac{4\unicode[STIX]{x1D70F}_{ei}}{3\sqrt{\unicode[STIX]{x03C0}}}\right)^{2}\frac{f_{t}B\text{i}I\ell q^{\prime }xn_{0}}{\langle B^{2}\rangle q}\left\langle \frac{1}{R^{2}}\right\rangle \frac{eT_{e}}{m_{e}}\nonumber\\ \displaystyle & & \displaystyle \times \,\left[\vphantom{\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)}\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\left\{\frac{\unicode[STIX]{x1D708}_{e}D^{(1)2}}{u^{2}}\right\}\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\left\{\frac{\unicode[STIX]{x1D708}_{e}D^{(1)}D^{(2)}}{u^{2}}\right\}\right],\end{eqnarray}$$

leading to

(4.37) $$\begin{eqnarray}\displaystyle & & \displaystyle (j_{\Vert }-j_{\Vert s})^{\text{Ban}}=\frac{f_{t}BIp_{0e}}{\langle B^{2}\rangle }\left[-1.67\frac{\tilde{p}^{\prime }}{p_{0e}}+1.19\frac{\tilde{T}_{e}^{\prime }}{T_{e}}+1.95\frac{\tilde{T}_{i}^{\prime }}{T_{e}}\right]\nonumber\\ \displaystyle & & \displaystyle \quad -\,2.58\frac{f_{t}B\text{i}I\ell q^{\prime }xp_{0e}e\,\unicode[STIX]{x1D70F}_{ei}}{\langle B^{2}\rangle qm_{e}}\left\langle \frac{1}{R^{2}}\right\rangle \left[\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1\!+\!1.17\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}+0.17\frac{\tilde{T}_{e}}{T_{e}}\right].\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Combining expressions (4.28), (4.32) and (4.37), we obtain

(4.38) $$\begin{eqnarray}\displaystyle j_{\Vert } & = & \displaystyle -I\tilde{p}^{\prime }\left(\frac{1}{B}-\frac{B}{\langle B^{2}\rangle }\right)+1.97\unicode[STIX]{x1D70F}_{ei}\frac{\text{i}I\ell q^{\prime }xp_{0e}e}{m_{e}q}\frac{B}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \nonumber\\ \displaystyle & & \displaystyle \times \,\left[\vphantom{\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)}(1-1.31f_{t})\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)+0.34(1-0.69f_{t})\right.\nonumber\\ \displaystyle & & \displaystyle \left.\times \,\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\right]-\frac{f_{t}BIp_{0e}}{\langle B^{2}\rangle }\left[1.67\frac{\tilde{p}^{\prime }}{p_{0e}}-1.19\frac{\tilde{T}_{e}^{\prime }}{T_{e}}-1.95\frac{\tilde{T}_{i}^{\prime }}{T_{e}}\right].\end{eqnarray}$$

The first term is the usual Pfirsch–Schlüter current while the second term represents the effects of the parallel electron pressure gradient, the parallel electric field and parallel thermal force, whereas the final term is the bootstrap current.

We can now calculate $U_{\Vert e}$ from

(4.39) $$\begin{eqnarray}U_{\Vert e}=U_{\Vert i}-\frac{j_{\Vert }}{n_{0}e},\end{eqnarray}$$

with $U_{\Vert i}$ given by (3.22). The contribution to the electron continuity equation from the divergence of the parallel flux (4.26) is then:

(4.40) $$\begin{eqnarray}\displaystyle & & \displaystyle n_{0}\left\langle \frac{\text{i}\ell Iq^{\prime }x}{R^{2}Bq}U_{\Vert e}\right\rangle \nonumber\\ \displaystyle & & \displaystyle \quad =\frac{\text{i}I^{2}\ell q^{\prime }x}{eq}\tilde{p}^{\prime }\left(\left\langle \frac{1}{R^{2}B^{2}}\right\rangle -\frac{1}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \right)+\frac{\text{i}I^{2}\ell q^{\prime }xp_{0e}f_{t}}{e\langle B^{2}\rangle q}\left\langle \frac{1}{R^{2}}\right\rangle \nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\left[1.67\frac{\tilde{p}^{\prime }}{p_{0e}}-1.19\frac{\tilde{T}_{e}^{\prime }}{T_{e}}-3.12\frac{\tilde{T}_{i}^{\prime }}{T_{e}}\right]\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\frac{\text{i}I^{2}\ell q^{\prime }xp_{0e}}{eq}\left[\left\langle \frac{1}{R^{2}B^{2}}\right\rangle \left(\frac{p_{i}^{\prime }}{p_{0e}}+\frac{e\unicode[STIX]{x1D6F7}^{\prime }}{T_{e}}\right)-\frac{1.17}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \frac{\tilde{T}_{i}^{\prime }}{T_{e}}\right]\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\frac{1.97\,\unicode[STIX]{x1D70F}_{ei}(I\ell q^{\prime }x)^{2}p_{0e}}{m_{e}\langle B^{2}\rangle q^{2}}\left\langle \frac{1}{R^{2}}\right\rangle ^{2}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\left[(1-1.31f_{t})\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)+0.34(1-0.69f_{t})\right.\nonumber\\ \displaystyle & & \displaystyle \qquad \left.\times \,\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\right].\end{eqnarray}$$

The energy equation involves the parallel heat flux

(4.41) $$\begin{eqnarray}Q_{\Vert e}=\int \,\text{d}^{3}\boldsymbol{v}\frac{m_{e}v^{2}}{2}v_{\Vert }h_{1e}\equiv q_{\Vert e}+\frac{5}{2}n_{0}T_{e}U_{\Vert e};\quad q_{\Vert e}=\int \,\text{d}^{3}\boldsymbol{v}\left(\frac{m_{e}v^{2}}{2}-\frac{5}{2}T_{e}\right)v_{\Vert }h_{1e}.\end{eqnarray}$$

We can decompose $Q_{\Vert e}$ as

(4.42) $$\begin{eqnarray}Q_{\Vert e}=Q_{\Vert se}+(q_{\Vert e}-q_{||se})-\frac{5}{2n_{0}e}(j_{\Vert }-j_{||s})T_{e};\quad Q_{\Vert se}=\int \,\text{d}^{3}\boldsymbol{v}\frac{m_{e}v^{2}}{2}v_{\Vert }h_{se}.\end{eqnarray}$$

The Spitzer contribution is given as

(4.43) $$\begin{eqnarray}\displaystyle Q_{\Vert se} & = & \displaystyle \frac{5}{2}n_{0}T_{e}U_{\Vert i}+\frac{4\unicode[STIX]{x1D70F}_{ei}}{3\sqrt{\unicode[STIX]{x03C0}}}n_{0}T_{e}\frac{\text{i}\ell Iq^{\prime }x}{R^{2}Bq}\frac{T_{e}}{m_{e}}\nonumber\\ \displaystyle & & \displaystyle \times \left[\vphantom{\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)}\{uD^{(1)}(u)\}\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\right.\nonumber\\ \displaystyle & & \displaystyle \left.+\,\{uD^{(2)}(u)\}\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\right],\end{eqnarray}$$

resulting in

(4.44) $$\begin{eqnarray}\displaystyle Q_{\Vert se} & = & \displaystyle -\frac{5}{2}n_{0}T_{e}\frac{IT_{i}}{eB}\left[\left(\frac{p_{i}^{\prime }}{p_{0e}}+\frac{e\unicode[STIX]{x1D6F7}^{\prime }}{T_{e}}\right)-\frac{1.17B^{2}}{\langle B^{2}\rangle }(1-f_{t})\frac{\tilde{T}_{i}^{\prime }}{T_{e}}\right]\nonumber\\ \displaystyle & & \displaystyle -\,6.26n_{0}T_{e}\frac{\text{i}\ell Iq^{\prime }x\unicode[STIX]{x1D70F}_{ei}}{R^{2}Bq}\frac{T_{e}}{m_{e}}\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}+0.58\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+1.58\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right).\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Introducing the Pfirsch–Schlüter part of $h_{1e}$ , equation (4.5), we can calculate the corresponding Pfirsch–Schlüter contribution to $Q_{\Vert e}$ directly, obtaining:

(4.45) $$\begin{eqnarray}\displaystyle \frac{Q_{\Vert e}^{PS}}{p_{0e}} & = & \displaystyle \frac{5}{2}\frac{IT_{e}}{e}\left(\frac{1}{B}-\frac{B}{\langle B^{2}\rangle }\right)\left(\frac{\tilde{p}^{\prime }}{p_{0e}}+\frac{\tilde{T}_{e}^{\prime }}{T_{e}}\right)+6.26n_{0}T\unicode[STIX]{x1D70F}_{ei}\frac{\text{i}I\ell q^{\prime }xT_{e}}{m_{e}q}\left(\frac{1}{R^{2}B}-\frac{B}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \right)\nonumber\\ \displaystyle & & \displaystyle \times \,\left[\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}+\frac{0.58\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+1.58\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\right].\end{eqnarray}$$

We formulate the banana neoclassical contribution for $q_{\Vert e}$ by analogy to the neoclassical current, $j_{\Vert e}$ , as in (4.35), so that it can also be evaluated using the localised distribution function (4.6):

(4.46) $$\begin{eqnarray}(q_{\Vert e}-q_{\Vert se})=-\frac{4\unicode[STIX]{x1D70F}_{ei}}{3\sqrt{\unicode[STIX]{x03C0}}}T_{e}\int \,\text{d}^{3}\boldsymbol{v}\frac{D^{(2)}(u)}{u}v_{\Vert }\unicode[STIX]{x1D708}_{e}\left(h_{1e}-h_{s}^{(1)}-h_{s}^{(2)}-\frac{m_{e}v_{\Vert }}{T_{e}}U_{\Vert i}\,f_{0e}\right).\end{eqnarray}$$

Using expression (4.6), we obtain the banana contribution to $q_{\Vert e}$ :

(4.47) $$\begin{eqnarray}\displaystyle & & \displaystyle (q_{\Vert e}-q_{\Vert se})^{\text{Ban}}=\frac{f_{t}BIT_{e}p_{0e}}{e\langle B^{2}\rangle }\left[0.04\frac{\tilde{p}^{\prime }}{p_{0e}}+0.64\frac{\tilde{T}_{e}^{\prime }}{T_{e}}-0.05\frac{\tilde{T}_{i}^{\prime }}{T_{e}}\right]\nonumber\\ \displaystyle & & \displaystyle \quad +\,0.46\frac{f_{t}\text{i}I\ell q^{\prime }xT_{e}p_{0e}\unicode[STIX]{x1D70F}_{ei}}{m_{e}q}\frac{B}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \left[\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+1.94\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right.\nonumber\\ \displaystyle & & \displaystyle \quad \left.+\,0.94\frac{\tilde{T}_{e}}{T_{e}}\right].\end{eqnarray}$$

Inserting the results (4.37), (4.44), (4.45) and (4.47) into (4.42), we finally obtain

(4.48) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{n_{0}T_{e}}\left\langle \frac{\text{i}\ell Iq^{\prime }x}{R^{2}Bq}Q_{\Vert e}\right\rangle =\frac{5}{2}\frac{\text{i}\ell I^{2}q^{\prime }xT_{e}}{eq}\left(\left\langle \frac{1}{R^{2}B^{2}}\right\rangle -\frac{1}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \right)\left(\frac{\tilde{p}^{\prime }}{p_{0e}}+\frac{\tilde{T}_{e}^{\prime }}{T_{e}}\right)\nonumber\\ \displaystyle & & \displaystyle \quad +\,\frac{f_{t}}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \frac{\text{i}\ell I^{2}q^{\prime }xT_{e}}{eq}\left[4.18\frac{\tilde{p}^{\prime }}{p_{0e}}-2.36\frac{\tilde{T}_{e}^{\prime }}{T_{e}}-7.86\frac{\tilde{T}_{i}^{\prime }}{T_{e}}\right]-\frac{5}{2}\frac{\text{i}\ell I^{2}q^{\prime }xT_{i}}{eq}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\left[\left\langle \frac{1}{R^{2}B^{2}}\right\rangle \left(\frac{p_{i}^{\prime }}{p_{0e}}+\frac{e\unicode[STIX]{x1D6F7}^{\prime }}{T_{e}}\right)-\frac{1.17}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \frac{\tilde{T}_{i}^{\prime }}{T_{e}}\right]+6.26\frac{(\ell Iq^{\prime }x)^{2}}{\langle B^{2}\rangle q^{2}}\left\langle \frac{1}{R^{2}}\right\rangle ^{2}\frac{T_{e}\unicode[STIX]{x1D70F}_{ei}}{m_{e}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\left[\vphantom{\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D702}_{e}\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)}(1-1.11f_{t})\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\right.\nonumber\\ \displaystyle & & \displaystyle \quad +\,\left.0.58(1-0.31f_{t})\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D702}_{e}\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\right].\end{eqnarray}$$

4.1.3 Azimuthal electron fluxes

Finally, we require the contributions from the azimuthal drift terms that appear in (4.2) and (4.3).

(4.49) $$\begin{eqnarray}\displaystyle \left\langle \int \,\text{d}^{3}\boldsymbol{v}\boldsymbol{v}_{de}\boldsymbol{\cdot }\unicode[STIX]{x1D735}h_{0e}\right\rangle & = & \displaystyle -\text{i}\left\langle \int \int \,\text{d}^{3}\boldsymbol{v}\boldsymbol{v}_{de}\boldsymbol{\cdot }(m\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}-\ell \unicode[STIX]{x1D735}\unicode[STIX]{x1D711})h_{0e}\right\rangle \nonumber\\ \displaystyle & = & \displaystyle -\text{i}\left\langle \frac{\boldsymbol{n}}{\unicode[STIX]{x1D6FA}_{e}}\times (\unicode[STIX]{x1D735}\ln B+\unicode[STIX]{x1D73F})\boldsymbol{\cdot }(m\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}-\ell \unicode[STIX]{x1D735}\unicode[STIX]{x1D711})\right\rangle \frac{T_{e}}{m_{e}}\left(\frac{\hat{n}_{e}}{n_{0}}+\frac{\hat{T}_{e}}{T_{e}}\right)n_{0}\nonumber\\ \displaystyle & = & \displaystyle -\text{i}\left\langle \frac{\boldsymbol{n}}{\unicode[STIX]{x1D6FA}_{e}}\times \frac{\unicode[STIX]{x1D735}(\unicode[STIX]{x1D707}_{0}p_{0}+B^{2})}{B^{2}}\boldsymbol{\cdot }(m\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}-\ell \unicode[STIX]{x1D735}\unicode[STIX]{x1D711})\right\rangle \frac{T_{e}}{m_{e}}\left(\frac{\hat{n}_{e}}{n_{0}}+\frac{\hat{T}_{e}}{T_{e}}\right)n_{0}.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

This can be expressed as

(4.50) $$\begin{eqnarray}\displaystyle & & \displaystyle \left\langle \int \,\text{d}^{3}\boldsymbol{v}\boldsymbol{v}_{de}\boldsymbol{\cdot }\unicode[STIX]{x1D735}h_{0e}\right\rangle =\text{i}\ell \frac{T_{e}}{e}\left\langle \left(\frac{1}{B^{2}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D707}_{0}p_{0}+B^{2})+\left(\frac{1}{RB^{2}}\right)^{2}\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}\frac{\unicode[STIX]{x2202}B^{2}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)\right\rangle \nonumber\\ \displaystyle & & \displaystyle \quad \times \,\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)n_{0}.\end{eqnarray}$$

Similarly, we also obtain

(4.51) $$\begin{eqnarray}\displaystyle & & \displaystyle \left\langle \int \,\text{d}^{3}\boldsymbol{v}\frac{m_{e}v^{2}}{2}\boldsymbol{v}_{de}\boldsymbol{\cdot }\unicode[STIX]{x1D735}h_{0e}\right\rangle \nonumber\\ \displaystyle & & \displaystyle \quad =\frac{5}{2}\text{i}\ell \frac{T_{e}}{e}\left\langle \left(\frac{1}{B^{2}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D707}_{0}p_{0}+B^{2})+\left(\frac{1}{RB^{2}}\right)^{2}\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}\frac{\unicode[STIX]{x2202}B^{2}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)\right\rangle \nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\left(\frac{\tilde{p}_{e}}{p_{0e}}+\frac{\tilde{T}_{e}}{T_{e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+2\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)n_{0}T_{e}.\end{eqnarray}$$

4.2 The perturbed ion temperature

In the case of the ions there is no need to obtain the ion density equation for $\hat{n}_{i}$ as it is determined by quasi-neutrality and given in (2.13). To obtain the equation for $\hat{T}_{i}$ we apply the operator $\int (\ldots )(m_{i}v^{2}/2)\,\text{d}^{3}\boldsymbol{v}$ to (3.23).

(4.52) $$\begin{eqnarray}\displaystyle & & \displaystyle \left\langle \int \,\text{d}^{3}\boldsymbol{v}\frac{v_{\Vert }I}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{Iv_{\Vert }}{\unicode[STIX]{x1D6FA}_{e}}\right)\frac{m_{i}v^{2}}{2}\frac{\unicode[STIX]{x2202}h_{2i}}{\unicode[STIX]{x2202}x}\right\rangle +\left\langle \int \,\text{d}^{3}\boldsymbol{v}\frac{m_{i}v^{2}}{2}\boldsymbol{v}_{d\unicode[STIX]{x1D703}i}\boldsymbol{\cdot }\unicode[STIX]{x1D735}h_{0i}\right\rangle \nonumber\\ \displaystyle & & \displaystyle \quad -\,\left\langle \int \,\text{d}^{3}\boldsymbol{v}\frac{m_{i}v^{2}}{2}\overline{L_{i}C_{ii}(g_{i}L_{i})}\right\rangle =\frac{3}{2}\text{i}\unicode[STIX]{x1D714}\left(\frac{\tilde{p}_{i}}{p_{0i}}+\frac{e\unicode[STIX]{x1D6F7}}{T_{i}}-\frac{e\unicode[STIX]{x1D6F9}}{T_{i}}\left(1+\frac{T_{i}}{T_{e}}\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{i})\right)\right)\nonumber\\ \displaystyle & & \displaystyle \quad -\,\text{i}n_{0}\left\langle \frac{3}{2}\frac{e}{T_{i}}\left(\unicode[STIX]{x1D714}+\frac{T_{i}}{T_{e}}\unicode[STIX]{x1D714}_{\ast e}(1+\unicode[STIX]{x1D702}_{i})\right)(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})+\frac{5}{2}\frac{\unicode[STIX]{x1D6FF}B_{\Vert }}{B}\left(\unicode[STIX]{x1D714}+\frac{T_{i}}{T_{e}}\unicode[STIX]{x1D714}_{\ast e}(1+2\unicode[STIX]{x1D702}_{i})\right)\right\rangle .\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Integrating the first term on the left-hand side by parts and substituting for $h_{2i}$ from (3.11), it can be written as the divergence of the neoclassical ion heat flux

(4.53) $$\begin{eqnarray}\displaystyle \frac{\text{d}}{\text{d}r}\langle Q_{^{i}}^{NC}\rangle \cong \frac{\text{d}}{\text{d}x}\left\langle \int \,\text{d}^{3}\boldsymbol{v}\frac{m_{i}v^{2}}{2}\frac{Iv_{\Vert }}{\unicode[STIX]{x1D6FA}_{i}}C_{ii}(h_{1i})\right\rangle . & & \displaystyle\end{eqnarray}$$

The second term is the ion azimuthal flux and the third is the ion classical heat flux:

(4.54) $$\begin{eqnarray}\frac{\text{d}}{\text{d}r}\langle Q_{ri}^{C}\rangle =\left\langle \int \,\text{d}^{3}\boldsymbol{v}\frac{m_{i}v^{2}}{2}\overline{L_{i}C_{ii}(g_{0_{i}}L_{i})}\right\rangle .\end{eqnarray}$$

This can be evaluated in a similar manner to the electron case, resulting in

(4.55) $$\begin{eqnarray}\frac{\text{d}}{\text{d}x}\langle Q_{ri}^{C}\rangle =-2\frac{m_{i}T_{i}}{\text{e}^{2}\unicode[STIX]{x1D70F}_{i}}\left\langle \frac{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|}{B^{2}}^{2}\right\rangle n_{0}\frac{\text{d}^{2}}{\text{d}x^{2}}\tilde{T}_{i},\end{eqnarray}$$

where $\unicode[STIX]{x1D70F}_{i}=(3\sqrt{\unicode[STIX]{x03C0}}/4)4\unicode[STIX]{x03C0}\unicode[STIX]{x1D700}_{0}^{2}m_{i}^{1/2}T_{i}^{3/2}/n_{0}\text{e}^{4}\ln \unicode[STIX]{x1D6EC}$ .

The calculation of the Pfirsch–Schlüter heat flux proceeds similarly to the electron one and yields

(4.56) $$\begin{eqnarray}\frac{\text{d}}{\text{d}r}\langle Q_{ri}^{PS}\rangle =-\frac{2m_{i}T_{i}}{\text{e}^{2}\unicode[STIX]{x1D70F}_{i}}I^{2}\left(\left\langle \frac{1}{B^{2}}\right\rangle -\frac{1}{\langle B^{2}\rangle }\right)n_{0}\frac{\text{d}^{2}}{\text{d}x^{2}}\tilde{T}_{i},\end{eqnarray}$$

while the banana regime contribution is

(4.57) $$\begin{eqnarray}\frac{\text{d}}{\text{d}r}\langle Q_{ri}^{\text{Ban}}\rangle =-\frac{m_{i}T_{i}I^{2}f_{t}}{\text{e}^{2}\langle B^{2}\rangle }\left(\{u^{4}\unicode[STIX]{x1D708}_{ii}\}-\frac{\{u^{2}\unicode[STIX]{x1D708}_{ii}\}^{2}}{\{\unicode[STIX]{x1D708}_{ii}\}}\right)n_{0}\frac{\text{d}^{2}}{\text{d}x^{2}}\tilde{T}_{i}=-0.92\frac{m_{i}T_{i}I^{2}f_{t}}{\text{e}^{2}\langle B^{2}\rangle \unicode[STIX]{x1D70F}_{i}}n_{0}\frac{\text{d}^{2}}{\text{d}x^{2}}\tilde{T}_{i}.\end{eqnarray}$$

Finally, the azimuthal drift contribution is

(4.58) $$\begin{eqnarray}\displaystyle \left\langle \int \,\text{d}^{3}\boldsymbol{v}\frac{m_{i}v^{2}}{2}\boldsymbol{v}_{di}\boldsymbol{\cdot }\unicode[STIX]{x1D735}h_{0i}\right\rangle & = & \displaystyle -\frac{5}{2}i\ell \frac{T_{i}}{e}\left\langle \!\left(\frac{1}{B^{2}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D707}_{0}p_{0}+B^{2})+\left(\frac{1}{RB^{2}}\right)^{2}\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}\frac{\unicode[STIX]{x2202}B^{2}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\!\right)\!\!\right\rangle \nonumber\\ \displaystyle & & \displaystyle \times \,\left(\frac{\tilde{p}_{i}}{p_{0i}}+\frac{\tilde{T}_{i}}{T_{i}}+\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{i}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+2\unicode[STIX]{x1D702}_{i})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)n_{0}T_{i}.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

4.3 Summary

Here we collect together the above results to obtain the final form of the equations determining the electron density and temperature perturbations in terms of the perturbed fields $\unicode[STIX]{x1D6F7}$ and $\unicode[STIX]{x1D6F9}$ . The first, equation (4.2) can be written in the form

(4.59) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{\text{d}}{\text{d}r}(\langle \unicode[STIX]{x1D6E4}_{r,e}^{C}\rangle +\langle \unicode[STIX]{x1D6E4}_{r,e}^{PS}\rangle +\langle \unicode[STIX]{x1D6E4}_{r,e}^{\text{Ban}}\rangle )+\text{i}k_{\Vert 0}\langle \unicode[STIX]{x1D6E4}_{\Vert ,e}\rangle +\text{i}k_{\unicode[STIX]{x1D703}}\langle \unicode[STIX]{x1D6E4}_{e}^{Az}\rangle \nonumber\\ \displaystyle & & \displaystyle \quad =\text{i}\unicode[STIX]{x1D714}\left[{\tilde{n}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F7}}{T_{e}}n_{0}+\left\langle \frac{\unicode[STIX]{x1D707}_{0}\tilde{p}}{B^{2}}\right\rangle \left(1-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{e})\right)n_{0}\right],\end{eqnarray}$$

where $k_{\Vert 0}=\ell Iq^{\prime }x\langle 1/R^{2}B\rangle /q$ , $k_{\unicode[STIX]{x1D703}}=\ell q\langle |\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}|\rangle$ and we have substituted (4.1) for $\unicode[STIX]{x1D6FF}B_{\Vert }$ . From (4.13) we have:

(4.13a ) $$\begin{eqnarray}\frac{\text{d}}{\text{d}r}\langle \unicode[STIX]{x1D6E4}_{r,e}^{C}\rangle =-\frac{m_{e}T_{e}}{\text{e}^{2}\unicode[STIX]{x1D70F}_{ei}}\frac{\text{d}^{2}}{\text{d}x^{2}}\left\langle \frac{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}{B^{2}}\right\rangle \left(\frac{\tilde{p}}{p_{0e}}-\frac{3}{2}\frac{\tilde{T}_{e}}{T_{e}}\right)n_{0};\end{eqnarray}$$

from (4.19):

(4.19a ) $$\begin{eqnarray}\displaystyle \frac{1}{n_{0}}\frac{\text{d}}{\text{d}r}\langle \unicode[STIX]{x1D6E4}_{r,e}^{PS}\rangle & = & \displaystyle -\frac{m_{e}T_{e}}{\text{e}^{2}\unicode[STIX]{x1D70F}_{ei}}I^{2}\left(\left\langle \frac{1}{B^{2}}\right\rangle -\frac{1}{\langle B^{2}\rangle }\right)\frac{\text{d}^{2}}{\text{d}x^{2}}\left(\frac{\tilde{p}}{p_{0e}}-\frac{3}{2}\frac{\tilde{T}_{e}}{T_{e}}\right)\nonumber\\ \displaystyle & & \displaystyle -\,\frac{\text{i}I^{2}\ell q^{\prime }T_{e}}{eq}\left(\left\langle \frac{1}{R^{2}B^{2}}\right\rangle -\frac{1}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \right)\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{\text{d}}{\text{d}x}\left[x\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e\unicode[STIX]{x1D6F7}}{T_{e}}+\left(1-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{e})\right)\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\right];\end{eqnarray}$$

from (4.23):

(4.23a ) $$\begin{eqnarray}\displaystyle \frac{1}{n_{0}}\frac{\text{d}}{\text{d}r}\langle \unicode[STIX]{x1D6E4}_{r,e}^{\text{Ban}}\rangle & = & \displaystyle -1.53f_{t}\frac{I^{2}m_{e}T_{e}}{\text{e}^{2}\langle B^{2}\rangle \unicode[STIX]{x1D70F}_{ei}}\frac{\text{d}^{2}}{\text{d}x^{2}}\left[\frac{\tilde{p}}{p_{0e}}+1.39\frac{\tilde{T}_{e}}{T_{e}}-1.17\frac{\tilde{T}_{i}}{T_{e}}\right]\nonumber\\ \displaystyle & & \displaystyle -\,1.67\frac{f_{t}}{\langle B^{2}\rangle }\left\langle \frac{I^{2}}{R^{2}}\right\rangle \frac{\text{i}\ell q^{\prime }T_{e}}{eq}\frac{\text{d}}{\text{d}x}\left(x\left[\vphantom{\frac{\tilde{T}_{e}}{T_{e}}}\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}\right.\right.\nonumber\\ \displaystyle & & \displaystyle \left.\left.-\,\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+1.02\,\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}+0.02\frac{\tilde{T}_{e}}{T_{e}}\right]\right).\end{eqnarray}$$

From (4.40), we have defined

(4.40a ) $$\begin{eqnarray}\displaystyle & & \displaystyle \left\langle \frac{1}{R^{2}B}\right\rangle \langle \unicode[STIX]{x1D6E4}_{\Vert ,e}\rangle \nonumber\\ \displaystyle & & \displaystyle \quad =\frac{I}{e}\tilde{p}^{\prime }\left(\left\langle \frac{1}{R^{2}B^{2}}\right\rangle -\frac{1}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \right)+\frac{Ip_{0e}f_{t}}{e\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \left[1.67\frac{\tilde{p}^{\prime }}{p_{0e}}-1.19\frac{\tilde{T}_{e}^{\prime }}{T_{e}}-3.12\frac{\tilde{T}_{i}^{\prime }}{T_{e}}\right]\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\frac{Ip_{0e}}{e}\left[\left\langle \frac{1}{R^{2}B^{2}}\right\rangle \left(\frac{p_{i}^{\prime }}{p_{0e}}+\frac{e\unicode[STIX]{x1D6F7}^{\prime }}{T_{e}}\right)-\frac{1.17}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \frac{\tilde{T}_{i}^{\prime }}{T_{e}}\right]-\text{i}\frac{1.97\unicode[STIX]{x1D70F}_{ei}(I\ell q^{\prime }x)p_{0e}}{m_{e}\langle B^{2}\rangle q}\left\langle \frac{1}{R^{2}}\right\rangle ^{2}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\left[\vphantom{\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)}(1-1.31f_{t})\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\right.\nonumber\\ \displaystyle & & \displaystyle \qquad \left.+\,0.34\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)(1-0.69f_{t})\right],\end{eqnarray}$$

while from (4.50) we have defined

(4.50a ) $$\begin{eqnarray}\displaystyle \langle \unicode[STIX]{x1D6E4}_{e}^{Az}\rangle & = & \displaystyle \frac{T_{e}}{eq\langle |\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}|\rangle }\left\langle \left(\frac{1}{B^{2}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D707}_{0}p_{0}+B^{2})+\left(\frac{1}{RB^{2}}\right)^{2}\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}\frac{\unicode[STIX]{x2202}B^{2}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)\right\rangle \nonumber\\ \displaystyle & & \displaystyle \times \,\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)n_{0}.\end{eqnarray}$$

Similarly, for (4.3) we obtain

(4.60) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{\text{d}}{\text{d}r}(\langle Q_{r,e}^{C}\rangle +\langle Q_{r,e}^{PS}\rangle +\langle Q_{r,e}^{\text{Ban}}\rangle )+\text{i}k_{\Vert 0}\langle Q_{\Vert e}\rangle +\text{i}k_{\unicode[STIX]{x1D703}}\langle Q_{e}^{Az}\rangle \nonumber\\ \displaystyle & & \displaystyle \quad =\frac{3}{2}\text{i}\unicode[STIX]{x1D714}\left[\tilde{p}_{e}-\frac{\unicode[STIX]{x1D714}_{\ast e}(1+\unicode[STIX]{x1D702}_{e})}{\unicode[STIX]{x1D714}}e\unicode[STIX]{x1D6F7}_{0}+\frac{5}{3}\left\langle \frac{\unicode[STIX]{x1D707}_{0}\tilde{p}}{B^{2}}\right\rangle \left(1-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+2\unicode[STIX]{x1D702}_{e})\right)p_{0e}\right],\end{eqnarray}$$

where, from (4.15),

(4.15a ) $$\begin{eqnarray}\frac{\text{d}}{\text{d}r}\langle Q_{r,e}^{C}\rangle =-\frac{m_{e}T_{e}}{\text{e}^{2}\unicode[STIX]{x1D70F}_{ei}}\left\langle \frac{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}{B^{2}}\right\rangle \frac{\text{d}^{2}}{\text{d}x^{2}}\left(\frac{\tilde{p}}{p_{0e}}+0.91\frac{\tilde{T}_{e}}{T_{e}}\right)n_{0}T_{e},\end{eqnarray}$$

from (4.20),

(4.20a ) $$\begin{eqnarray}\displaystyle \frac{1}{n_{0}T_{e}}\frac{\text{d}}{\text{d}r}\langle Q_{r,e}^{PS}\rangle & = & \displaystyle -\frac{m_{e}T_{e}I^{2}}{\text{e}^{2}\unicode[STIX]{x1D70F}_{ei}}\left(\left\langle \frac{1}{B^{2}}\right\rangle -\frac{1}{\langle B^{2}\rangle }\right)\frac{\text{d}^{2}}{\text{d}x^{2}}\left(\frac{\tilde{p}}{p_{0e}}+0.91\frac{\tilde{T}_{e}}{T_{e}}\right)\nonumber\\ \displaystyle & & \displaystyle -\,\frac{\text{i}I^{2}\ell q^{\prime }T_{e}}{eq}\left(\left\langle \frac{1}{R^{2}B^{2}}\right\rangle -\frac{1}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \right)\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{\text{d}}{\text{d}x}\left[\frac{5}{2}x\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e\unicode[STIX]{x1D6F7}}{T_{e}}+\frac{\tilde{T}_{e}}{T_{e}}+\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\left(1-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+2\unicode[STIX]{x1D702}_{e})\right)\right)\right]\qquad\end{eqnarray}$$

and from (4.25),

(4.25a ) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{n_{0}T_{e}}\frac{\text{d}}{\text{d}r}\langle Q_{r,e}^{\text{Ban}}\rangle =-1.70f_{t}\frac{I^{2}m_{e}T_{e}}{\text{e}^{2}\langle B^{2}\rangle \unicode[STIX]{x1D70F}_{ei}}\frac{\text{d}^{2}}{\text{d}x^{2}}\left[\frac{\tilde{p}}{p_{0e}}-0.40\frac{\tilde{T}_{e}}{T_{e}}-1.17\frac{\tilde{T}_{i}}{T_{e}}\right]\nonumber\\ \displaystyle & & \displaystyle \quad -\,2.97\frac{f_{t}}{\langle B^{2}\rangle }\left\langle \frac{I^{2}}{R^{2}}\right\rangle \frac{\text{i}\ell q^{\prime }T_{e}}{eq}\frac{\text{d}}{\text{d}x}\left(x\left[\vphantom{\frac{\tilde{T}_{e}}{T_{e}}}\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+1.25\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right.\right.\nonumber\\ \displaystyle & & \displaystyle \quad \left.\left.+\,0.25\frac{\tilde{T}_{e}}{T_{e}}\right]\right),\end{eqnarray}$$

while, from (4.48),

(4.48a ) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{n_{0}T_{e}}\left\langle \frac{1}{R^{2}B}\right\rangle \langle Q_{\Vert e}\rangle =\frac{5}{2}\frac{IT_{e}}{e}\left(\left\langle \frac{1}{R^{2}B^{2}}\right\rangle -\frac{1}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \right)\left(\frac{\tilde{p}^{\prime }}{p_{0e}}+\frac{\tilde{T}_{e}^{\prime }}{T_{e}}\right)\nonumber\\ \displaystyle & & \displaystyle \quad +\,\frac{f_{t}\langle R^{2}B\rangle }{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \frac{IT_{e}}{e}\left[4.18\frac{\tilde{p}^{\prime }}{p_{0e}}-2.36\frac{\tilde{T}_{e}^{\prime }}{T_{e}}-7.86\frac{\tilde{T}_{i}^{\prime }}{T_{e}}\right]\nonumber\\ \displaystyle & & \displaystyle \quad -\,\frac{5}{2}\frac{\text{i}\ell I^{2}q^{\prime }xT_{i}}{eq}\left[\left\langle \frac{1}{R^{2}B^{2}}\right\rangle \left(\frac{p_{i}^{\prime }}{p_{0e}}+\frac{e\unicode[STIX]{x1D6F7}^{\prime }}{T_{e}}\right)-\frac{1.17}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \frac{\tilde{T}_{i}^{\prime }}{T_{e}}\right]\nonumber\\ \displaystyle & & \displaystyle \quad -\,6.26\frac{\text{i}\ell Iq^{\prime }x}{\langle B^{2}\rangle q}\left\langle \frac{1}{R^{2}}\right\rangle ^{2}\frac{T_{e}\unicode[STIX]{x1D70F}_{ei}}{m_{e}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\left[(1-1.11f_{t})\left(\frac{\tilde{p}_{e}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\right.\nonumber\\ \displaystyle & & \displaystyle \quad \left.+\,0.58(1-0.31f_{t})\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D702}_{e}\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\right]\end{eqnarray}$$

and, from (4.51),

(4.51a ) $$\begin{eqnarray}\displaystyle \langle Q_{e}^{Az}\rangle & = & \displaystyle \frac{5}{2}\frac{T_{e}}{qe\langle |\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}|\rangle }\left\langle \left(\frac{1}{B^{2}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D707}_{0}p_{0}+B^{2})+\left(\frac{1}{RB^{2}}\right)^{2}\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}\frac{\unicode[STIX]{x2202}B^{2}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)\right\rangle \nonumber\\ \displaystyle & & \displaystyle \times \,\left(\frac{\tilde{p}_{e}}{p_{0e}}+\frac{\tilde{T}_{e}}{T_{e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+2\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)n_{0}T_{e}.\end{eqnarray}$$

Summarising for the ions, we can rewrite (4.52) as:

(4.61) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{\text{d}}{\text{d}r}(\langle Q_{r,i}^{C}\rangle +\langle Q_{r,i}^{PS}\rangle +\langle Q_{r,i}^{\text{Ban}}\rangle )+\text{i}k_{\unicode[STIX]{x1D703}}\langle Q_{i}^{Az}\rangle \nonumber\\ \displaystyle & & \displaystyle \quad =\frac{3}{2}\text{i}\unicode[STIX]{x1D714}\left[\tilde{p}_{i}-\frac{\unicode[STIX]{x1D714}_{\ast e}(1+\unicode[STIX]{x1D702}_{i})}{\unicode[STIX]{x1D714}}\frac{T_{i}}{T_{e}}e\unicode[STIX]{x1D6F7}n_{0}+\frac{5}{3}p_{0i}\left\langle \frac{\unicode[STIX]{x1D707}_{0}\tilde{p}}{B^{2}}\right\rangle \left(1+\frac{T_{i}}{T_{e}}\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+2\unicode[STIX]{x1D702}_{i})\right)\right],\qquad\end{eqnarray}$$

where, from (4.55),

(4.55a ) $$\begin{eqnarray}\frac{\text{d}}{\text{d}x}\langle Q_{ri}^{C}\rangle =-2\frac{m_{i}T_{i}}{\text{e}^{2}\unicode[STIX]{x1D70F}_{i}}\left\langle \frac{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|}{B^{2}}^{2}\right\rangle n_{0}\frac{\text{d}^{2}}{\text{d}x^{2}}\tilde{T}_{i},\end{eqnarray}$$

from (4.56)

(4.56a ) $$\begin{eqnarray}\frac{\text{d}}{\text{d}r}\langle Q_{ri}^{PS}\rangle =-\frac{2m_{i}T_{i}}{\text{e}^{2}\unicode[STIX]{x1D70F}_{i}}I^{2}\left(\left\langle \frac{1}{B^{2}}\right\rangle -\frac{1}{\langle B^{2}\rangle }\right)n_{0}\frac{\text{d}^{2}}{\text{d}x^{2}}\tilde{T}_{i},\end{eqnarray}$$

from (4.57)

(4.57a ) $$\begin{eqnarray}\frac{\text{d}}{\text{d}r}\langle Q_{ri}^{\text{Ban}}\rangle =-0.92\frac{m_{i}T_{i}I^{2}f_{t}}{\text{e}^{2}\langle B^{2}\rangle \unicode[STIX]{x1D70F}_{i}}n_{0}\frac{\text{d}^{2}}{\text{d}x^{2}}\tilde{T}_{i}\end{eqnarray}$$

and, finally, from (4.58):

(4.58a ) $$\begin{eqnarray}\displaystyle \langle Q_{i}^{Az}\rangle & = & \displaystyle -\frac{5}{2}\frac{T_{i}}{eq\langle |\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}|\rangle }\left\langle \left(\frac{1}{B^{2}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D707}_{0}p_{0}+B^{2})+\left(\frac{1}{RB^{2}}\right)^{2}\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}\frac{\unicode[STIX]{x2202}B^{2}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right)\right\rangle \nonumber\\ \displaystyle & & \displaystyle \times \,\left(\frac{\tilde{p}_{i}}{p_{0i}}+\frac{\tilde{T}_{i}}{T_{i}}+\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{i}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+2\unicode[STIX]{x1D702}_{i})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)n_{0}T_{i}.\end{eqnarray}$$

5.1 The parallel Ampère’s law

The parallel Ampère’s law states that

(5.1) $$\begin{eqnarray}|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}\frac{\text{d}^{2}A_{\Vert }}{\text{d}x^{2}}=\unicode[STIX]{x1D707}_{0}\,j_{\Vert }.\end{eqnarray}$$

We expand (5.1) in the localisation, $x$ , expressing $A_{\Vert }$ in terms of $\unicode[STIX]{x1D6F9}$ as defined above (2.9), which we expand in the form $\unicode[STIX]{x1D6F9}=\unicode[STIX]{x1D6F9}^{(0)}+\unicode[STIX]{x1D6F9}^{(1)}+\cdots \,$ . In leading order we have

(5.2) $$\begin{eqnarray}\frac{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}{\text{i}\unicode[STIX]{x1D714}}\frac{I}{R^{2}Bq}\frac{\text{d}^{2}}{\text{d}x^{2}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\unicode[STIX]{x1D6F9}^{(0)}=0\quad \Rightarrow \quad \unicode[STIX]{x1D6F9}^{(0)}=\unicode[STIX]{x1D6F9}^{(0)}(x),\end{eqnarray}$$

while in next order

(5.3) $$\begin{eqnarray}-\frac{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}{\text{i}\unicode[STIX]{x1D714}}\frac{I}{R^{2}Bq}\frac{\text{d}^{2}}{\text{d}x^{2}}\left(\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\unicode[STIX]{x1D6F9}^{(1)}+\text{i}\ell q^{\prime }x\unicode[STIX]{x1D6F9}^{(0)}(x)\right)=\unicode[STIX]{x1D707}_{0}j_{\Vert },\end{eqnarray}$$

leading to the solubility condition

(5.4) $$\begin{eqnarray}\frac{\ell q^{\prime }I}{\unicode[STIX]{x1D714}q}\left\langle \frac{1}{R^{2}}\right\rangle \frac{\text{d}^{2}}{\text{d}x^{2}}(x\unicode[STIX]{x1D6F9}^{(0)})=-\unicode[STIX]{x1D707}_{0}\left\langle \frac{j_{\Vert }B}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle ,\end{eqnarray}$$

where, from (4.38),

(5.5) $$\begin{eqnarray}\displaystyle & & \displaystyle \left\langle \frac{j_{\Vert }B}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle =-I\tilde{p}^{\prime }\left(\left\langle \frac{1}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle -\left\langle \frac{B^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle \frac{1}{\langle B^{2}\rangle }\right)\nonumber\\ \displaystyle & & \displaystyle \quad +\,1.97\unicode[STIX]{x1D70F}_{ei}\frac{\text{i}I\ell q^{\prime }xp_{0e}e}{m_{e}q}\left\langle \frac{B^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle \frac{1}{\langle B^{2}\rangle }\left\langle \frac{1}{R^{2}}\right\rangle \nonumber\\ \displaystyle & & \displaystyle \quad \times \,\left[\vphantom{\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)}(1-1.31f_{t})\left(\frac{\tilde{p}}{p_{0e}}-\frac{e(\unicode[STIX]{x1D6F7}-\unicode[STIX]{x1D6F9})}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{e})\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\right.\nonumber\\ \displaystyle & & \displaystyle \quad \left.+\,0.34(1-0.69f_{t})\left(\frac{\tilde{T}_{e}}{T_{e}}-\frac{\unicode[STIX]{x1D714}_{\ast e}\unicode[STIX]{x1D702}_{e}}{\unicode[STIX]{x1D714}}\frac{e\unicode[STIX]{x1D6F9}}{T_{e}}\right)\right]\nonumber\\ \displaystyle & & \displaystyle \quad -\,\frac{f_{t}Ip_{0e}}{\langle B^{2}\rangle }\left\langle \frac{B^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle \left[1.67\frac{\tilde{p}^{\prime }}{p_{0e}}-1.19\frac{\tilde{T}_{e}^{\prime }}{T_{e}}-1.95\frac{\tilde{T}_{i}^{\prime }}{T_{e}}\right].\end{eqnarray}$$

5.2 The vorticity equation

The vorticity equation was obtained in the previous publication (Connor et al. Reference Connor, Hastie and Helander2009). Here we list the key steps in its derivation and quote the result. The procedure was to add the velocity moments of the gyro-kinetic equations for the electrons and ions, take the long wavelength limit: $k_{\bot }^{2}\unicode[STIX]{x1D70C}_{i}^{2}\ll 1$ and exploit the lowest-order quasi-neutrality condition: ${\tilde{n}}_{e}={\tilde{n}}_{i}$ , to obtain

(5.6) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{\boldsymbol{B}\boldsymbol{\cdot }\unicode[STIX]{x1D735}}{\unicode[STIX]{x1D714}^{2}}\left(\frac{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}{\unicode[STIX]{x1D707}_{0}B^{2}}\frac{\text{d}^{2}}{\text{d}x^{2}}\boldsymbol{B}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F9}\right)\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{j}\frac{e_{j}^{2}}{T_{j}}\int \,\text{d}^{3}vf_{0j}\left\{\frac{\unicode[STIX]{x1D714}_{\ast j}^{T}}{\unicode[STIX]{x1D714}}\unicode[STIX]{x1D6F7}-\left(1-\frac{\unicode[STIX]{x1D714}_{\ast j}^{T}}{\unicode[STIX]{x1D714}}\right)\left[\frac{m_{j}^{2}v_{\bot }^{2}|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}{2e_{j}^{2}B^{2}}\frac{\text{d}^{2}}{\text{d}x^{2}}\unicode[STIX]{x1D6F7}+\frac{m_{j}v_{\bot }^{2}}{2e_{j}B}\unicode[STIX]{x1D6FF}B_{\Vert }\right]\right\}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\text{i}\mathop{\sum }_{j}e_{j}\int \,\text{d}^{3}\boldsymbol{v}\frac{\boldsymbol{v}_{dj}\boldsymbol{\cdot }\unicode[STIX]{x1D735}g_{j}}{\unicode[STIX]{x1D714}}+\frac{1}{4\unicode[STIX]{x1D714}}\mathop{\sum }_{j}e_{j}\int \,\text{d}^{3}\boldsymbol{v}\overline{L_{j}^{2}C_{j}(g_{j}L_{j}^{2})}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\frac{\text{i}}{\unicode[STIX]{x1D714}}\mathop{\sum }_{j}\frac{m_{j}}{e_{j}}\int \,\text{d}^{3}\boldsymbol{v}\unicode[STIX]{x1D70E}|v_{\Vert }|\frac{\boldsymbol{B}\boldsymbol{\cdot }\unicode[STIX]{x1D735}}{B}\left(\frac{m_{j}v_{\bot }^{2}|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}{2B^{2}}\right)\frac{\text{d}^{2}}{\text{d}x^{2}}g_{j},\end{eqnarray}$$

where the divergence of the parallel current has been expressed in terms of parallel gradients of $A_{\Vert }$ , i.e. $\unicode[STIX]{x1D6F9}$ , through Ampère’s law, equation (5.1). The distribution functions $g_{j}$ in the velocity space integrations over the magnetic drift terms are expressed in terms of the quantities $h_{j}$ and we note the final term vanishes for up–down symmetric equilibria. The expansions $\unicode[STIX]{x1D6F9}=\unicode[STIX]{x1D6F9}^{(0)}+\unicode[STIX]{x1D6F9}^{(1)}+\unicode[STIX]{x1D6F9}^{(2)}\ldots$ ; $h_{j}=h_{0j}+h_{1j}+h_{2j}\ldots$ are introduced and the equation for $\unicode[STIX]{x1D6F9}$ solved order by order.

In leading order one finds $\unicode[STIX]{x1D6F9}^{(0)}$ is independent of $\unicode[STIX]{x1D703}$ , while the equation for $\unicode[STIX]{x1D6F9}^{(1)}$ :

(5.7) $$\begin{eqnarray}\frac{I^{2}}{\unicode[STIX]{x1D714}^{2}R^{2}q^{2}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left[\frac{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}{\unicode[STIX]{x1D707}_{0}R^{2}B^{2}}\frac{\unicode[STIX]{x2202}^{2}}{\unicode[STIX]{x2202}x^{2}}\left(\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\unicode[STIX]{x1D6F9}^{(1)}+\text{i}\ell q^{\prime }x\unicode[STIX]{x1D6F9}^{(0)}\right)\right]=\text{i}\frac{I^{2}}{\unicode[STIX]{x1D714}R^{2}q}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{\tilde{p}^{\prime }}{B^{2}}\right)\end{eqnarray}$$

can be integrated, introducing a constant of integration which can in turn be determined through a periodicity condition in $\unicode[STIX]{x1D703}$ on $\unicode[STIX]{x1D6F9}^{(1)}$ . Applying the same periodicity condition on $\unicode[STIX]{x1D6F9}^{(2)}$ in second order provides the required equation for $\unicode[STIX]{x1D6F9}^{(0)}(x)$ :

(5.8) $$\begin{eqnarray}\displaystyle & & \displaystyle \frac{I^{2}}{\unicode[STIX]{x1D714}^{2}}\left[\frac{\ell ^{2}{q^{\prime }}^{2}}{\unicode[STIX]{x1D707}_{0}q^{2}}\left\langle \frac{1}{R^{2}}\right\rangle ^{2}x\frac{\text{d}^{2}(x\unicode[STIX]{x1D6F9}^{(0)})}{\text{d}x^{2}}+\frac{\ell q^{\prime }x\unicode[STIX]{x1D714}}{q}\tilde{p}^{\prime }\left(\left\langle \frac{1}{B^{2}R^{2}}\right\rangle \!\left\langle \frac{B^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle -\left\langle \frac{1}{R^{2}}\right\rangle \!\left\langle \frac{1}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle \right)\right]\nonumber\\ \displaystyle & & \displaystyle \quad =m_{i}n_{0}\left\langle \frac{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}{B^{2}}\right\rangle \left\langle \frac{B^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle \left[1-\frac{\unicode[STIX]{x1D714}_{\ast i}}{\unicode[STIX]{x1D714}}(1+\unicode[STIX]{x1D702}_{i})\right]\frac{\text{d}^{2}\unicode[STIX]{x1D6F7}}{\text{d}x^{2}}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,2\frac{\ell }{\unicode[STIX]{x1D714}}\tilde{p}\left\langle \frac{B^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle \left\langle \frac{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}{R^{2}B^{2}}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D712}}\left(\unicode[STIX]{x1D707}_{0}p_{0}+\frac{B^{2}}{2}\right)+\frac{\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D703}}{R^{2}B^{2}}\frac{\unicode[STIX]{x2202}B}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\right\rangle \nonumber\\ \displaystyle & & \displaystyle \qquad -\,\text{i}\frac{I}{\unicode[STIX]{x1D714}}\left\langle \frac{B^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle \left\langle \mathop{\sum }_{j}\int \,\text{d}^{3}\boldsymbol{v}\frac{m_{j}v_{\Vert }}{B}\frac{I}{R^{2}q}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{v_{\Vert }}{B}\right){h^{\prime }}_{j}\right\rangle \nonumber\\ \displaystyle & & \displaystyle \qquad -\,\frac{\ell {p^{\prime }}_{0}I^{2}}{\unicode[STIX]{x1D714}}\tilde{p}\left[\left\langle \frac{1}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}B^{2}}\right\rangle \left\langle \frac{B^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle -\left\langle \frac{1}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle ^{2}\right]\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\frac{\ell ^{2}q^{\prime }{p^{\prime }}_{0}I^{2}}{\unicode[STIX]{x1D714}^{2}q}(x\,\unicode[STIX]{x1D6F9}^{(0)})^{\prime }\left[\left\langle \frac{1}{R^{2}}\right\rangle \left\langle \frac{1}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle -\left\langle \frac{1}{B^{2}R^{2}}\right\rangle \left\langle \frac{B^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle \right]\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{1}{4\unicode[STIX]{x1D714}}\left\langle \frac{B^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle \left\langle \mathop{\sum }_{j}e_{j}\int \,\text{d}^{3}\boldsymbol{v}\overline{L_{j}^{2}C_{j}(g_{0j}L_{j}^{2})}\right\rangle ,\end{eqnarray}$$

where we have substituted for $\unicode[STIX]{x1D6FF}B_{\Vert }$ from (4.1) (which has the effect of replacing the $\unicode[STIX]{x1D735}B$ drift by the curvature drift) and substituted for $\unicode[STIX]{x1D6F9}^{(1)}$ from the solution of (5.7). The term involving $\langle \sum _{j}e_{j}\int \,\text{d}^{3}\boldsymbol{v}\overline{L_{j}^{2}C_{j}(g_{0j}L_{j}^{2})}\rangle$ represents the contribution from the classical radial transport of toroidal momentum, calculated in (B 2) of appendix B.

It remains to evaluate the term involving the $h_{j}$ , which we do by repeated application of the gyro-kinetic equations for $h_{0j}$ , $h_{1j}$ , $h_{2j}$ and $h_{3j}$ with integration by parts in $\unicode[STIX]{x1D703}$ and noting momentum conservation in ion–ion collisions. The result can be expressed as

(5.9) $$\begin{eqnarray}\displaystyle & & \displaystyle \text{i}\frac{I}{\unicode[STIX]{x1D714}}\left\langle \mathop{\sum }_{j}\int \,\text{d}^{3}\boldsymbol{v}\frac{m_{j}v_{\Vert }}{B}\frac{I}{R^{2}q}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{v_{\Vert }}{B}\right){h^{\prime }}_{j}\right\rangle \nonumber\\ \displaystyle & & \displaystyle \quad =\text{i}\frac{I}{\unicode[STIX]{x1D714}}\left\langle \mathop{\sum }_{j}\int \,\text{d}^{3}\boldsymbol{v}\frac{Im_{j}v_{\Vert }}{B}\left\{-\frac{Iv_{\Vert }}{R^{2}Bq}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D703}}\left(\frac{v_{\Vert }}{\unicode[STIX]{x1D6FA}_{i}}\right){h^{\prime \prime }}_{j}+C_{j}({h^{\prime }}_{j})\right.\right.\nonumber\\ \displaystyle & & \displaystyle \qquad \left.\left.+\,\text{i}{\unicode[STIX]{x1D714}h^{\prime }}_{j}-\text{i}\ell q^{\prime }\frac{Iv_{\Vert }}{R^{2}Bq}({xh^{\prime }}_{j})\vphantom{\mathop{\sum }_{j}}\right\}\right\rangle .\end{eqnarray}$$

The first term can be recognised as involving the neoclassical radial transport of toroidal angular momentum:

(5.10) $$\begin{eqnarray}\langle \unicode[STIX]{x1D6F1}^{NC}\rangle =\left\langle \mathop{\sum }_{j}\int \,\text{d}^{3}\boldsymbol{v}vm_{j}\frac{I}{B}v_{\Vert }v_{drj}h_{j}\right\rangle ,\end{eqnarray}$$

where the first non-vanishing contribution is from $h_{3_{j}}$ . The right-hand side of expression (5.9) thus reduces to

(5.11) $$\begin{eqnarray}-\frac{\text{i}}{\unicode[STIX]{x1D714}}\langle \unicode[STIX]{x1D6F1}^{NC}\rangle ^{\prime \prime }-Im_{i}n_{0}\left\langle \frac{U_{\Vert i}}{B}\right\rangle ^{\prime }+\frac{\ell q^{\prime }I^{2}}{\unicode[STIX]{x1D714}q}\left\langle \frac{1}{R^{2}B^{2}}\right\rangle \left[x\left(\tilde{p}-\frac{\ell }{\unicode[STIX]{x1D714}}p_{0}^{\prime }\unicode[STIX]{x1D6F9}^{(0)}\right)^{\prime }\right].\end{eqnarray}$$

Further use of the gyro-kinetic equation and integrations by parts in $\unicode[STIX]{x1D703}$ implies

(5.12) $$\begin{eqnarray}-\frac{\text{i}}{\unicode[STIX]{x1D714}}\langle \unicode[STIX]{x1D6F1}^{NC}\rangle ^{\prime \prime }=\text{i}\frac{I}{2\unicode[STIX]{x1D714}}\left\langle \mathop{\sum }_{j}\int \,\text{d}^{3}\boldsymbol{v}\frac{m_{j}v_{\Vert }}{B}\frac{Iv_{\Vert }}{\unicode[STIX]{x1D6FA}_{j}}\left[C_{j}(h_{2j}^{\prime \prime })+\frac{Iv_{\Vert }}{3\unicode[STIX]{x1D6FA}_{j}}C_{j}(h_{1j}^{\prime \prime })\right]\right\rangle .\end{eqnarray}$$

Assembling all the contributions to (5.8) and substituting for $U_{\Vert i}$ from (3.22),

(5.13) $$\begin{eqnarray}\displaystyle & & \displaystyle x\frac{\text{d}^{2}}{\text{d}x^{2}}(x\unicode[STIX]{x1D6F9}^{(0)})+\left(D_{I}+\frac{1}{4}\right)\frac{\unicode[STIX]{x1D714}}{\ell p_{0}^{\prime }}\tilde{p}-(L+H)x\frac{\text{d}}{\text{d}x}\left(\frac{\unicode[STIX]{x1D714}}{\ell p_{0}^{\prime }}\tilde{p}-\unicode[STIX]{x1D6F9}^{(0)}\right)\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{\unicode[STIX]{x1D707}_{0}m_{i}n_{0}}{\ell ^{2}}\left(\frac{q}{Iq^{\prime }\langle 1/R^{2}\rangle }\right)^{2}\left\langle \frac{B^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle \left\langle \frac{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}{B^{2}}\right\rangle \unicode[STIX]{x1D714}(\unicode[STIX]{x1D714}-\unicode[STIX]{x1D714}_{\ast i}(1+\unicode[STIX]{x1D702}_{i}))\frac{\text{d}^{2}}{\text{d}x^{2}}\unicode[STIX]{x1D6F7}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{\unicode[STIX]{x1D707}_{0}m_{i}n_{0}}{\ell ^{2}}\left(\frac{q}{q^{\prime }\langle 1/R^{2}\rangle }\right)^{2}\frac{\unicode[STIX]{x1D714}^{2}T_{i}}{e}\left\langle \frac{B^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle \frac{\text{d}^{2}}{\text{d}x^{2}}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\left\{\left\langle \frac{1}{B^{2}}\right\rangle \left(\frac{{\tilde{n}}}{n_{0}}+\frac{e\unicode[STIX]{x1D6F7}}{T_{i}}\right)+\left(\left\langle \frac{1}{B^{2}}\right\rangle -\frac{1.17}{\langle B^{2}\rangle }(1-f_{t})\right)\frac{\tilde{T}_{i}}{T_{i}}\right\}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\frac{\text{i}\unicode[STIX]{x1D714}\,\unicode[STIX]{x1D707}_{0}}{\ell ^{2}}\left(\frac{q}{Iq^{\prime }\langle 1/R^{2}\rangle }\right)^{2}\left\langle \frac{B^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle \frac{\text{d}^{2}}{\text{d}x^{2}}\langle \unicode[STIX]{x1D6F1}^{C}+\unicode[STIX]{x1D6F1}^{NC}\rangle .\end{eqnarray}$$

Here we have introduced the quantity $D_{I}$ in the notation of Glasser et al. (Reference Glasser, Greene and Johnson1975), where

(5.14) $$\begin{eqnarray}\displaystyle D_{I} & = & \displaystyle -\frac{1}{4}+p^{\prime }\frac{q}{q^{\prime }}\left\langle \frac{R^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle -\left(p^{\prime }\frac{q}{q^{\prime }}\right)^{2}\left\langle \frac{R^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle ^{2}\nonumber\\ \displaystyle & & \displaystyle +\,\left(\frac{p^{\prime }}{I}\frac{q}{q^{\prime }}\right)^{2}\left\langle \frac{R^{4}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle \left\langle \frac{B^{2}R^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle \nonumber\\ \displaystyle & & \displaystyle -\,p^{\prime }\left(\frac{1}{I}\frac{q}{q^{\prime }}\right)^{2}\left\langle \frac{B^{2}R^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle \left\langle \frac{\unicode[STIX]{x2202}R^{2}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D712}}-\frac{R^{2}q}{I}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D712}}\left(\frac{I}{q}\right)\right\rangle ,\end{eqnarray}$$

and $L$ , appearing in the work of Hahm (Reference Hahm1988), where

(5.15) $$\begin{eqnarray}L=\left(\frac{\unicode[STIX]{x1D707}_{0}p_{0}^{\prime }q}{q^{\prime }\langle 1/R^{2}\rangle \langle B^{2}\rangle }\right)\left\langle \frac{B^{2}}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle .\end{eqnarray}$$

The quantity $D_{I}$ plays a role in the Mercier stability criterion: $D_{I}+1/4>0$ (Mercier Reference Mercier1960) while the combination $L+H$ in (5.13) is given by

(5.16) $$\begin{eqnarray}L+H=\left(\frac{\unicode[STIX]{x1D707}_{0}p_{0}^{\prime }q}{q^{\prime }\langle 1/R^{2}\rangle }\right)\left\langle \frac{1}{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{2}}\right\rangle .\end{eqnarray}$$

An expression for $\langle \unicode[STIX]{x1D6F1}^{NC}\rangle$ has been given by Wong & Chan (Reference Wong and Chan2005): $\langle \unicode[STIX]{x1D6F1}^{NC}\rangle =0.19(n_{0}m_{i}^{2}T_{i}\,f_{t}^{3}/\unicode[STIX]{x1D70F}_{ii}e^{3})\,\text{d}^{2}\tilde{T}_{i}/\text{d}x^{2}$ , where $\unicode[STIX]{x1D70F}_{ii}=\unicode[STIX]{x1D70F}_{i}/\sqrt{2}$ , but we generalise this in appendix B for arbitrary axisymmetric toroidal geometry, to obtain the total collisional toroidal angular viscosity:

(5.17) $$\begin{eqnarray}\langle \unicode[STIX]{x1D6F1}\rangle =\langle \unicode[STIX]{x1D6F1}^{C}\rangle +\langle \unicode[STIX]{x1D6F1}^{NC}\rangle ,\end{eqnarray}$$

where

(5.18) $$\begin{eqnarray}\langle \unicode[STIX]{x1D6F1}^{C}\rangle =-\frac{n_{0}e}{5\unicode[STIX]{x1D70F}_{i}}\left(\frac{2m_{i}T_{i}}{\text{e}^{2}}\right)^{2}\left\langle \frac{|\unicode[STIX]{x1D735}\unicode[STIX]{x1D712}|^{4}}{B^{4}}\right\rangle \frac{\text{d}^{2}}{\text{d}x^{2}}\left\{\frac{\tilde{p}_{i}}{p_{0i}}+\frac{e\unicode[STIX]{x1D6F7}}{T_{i}}+3\frac{\tilde{T}_{i}}{T_{i}}\right\},\end{eqnarray}$$

and

(5.19) $$\begin{eqnarray}\displaystyle & & \displaystyle \langle \unicode[STIX]{x1D6F1}^{NC}\rangle =0.80\frac{n_{0}e}{\unicode[STIX]{x1D70F}_{i}}\left(\frac{I^{2}m_{i}T_{i}}{\text{e}^{2}}\right)^{2}\left[\left\langle \frac{1}{B^{4}}\right\rangle +\frac{(G^{(3)}-G^{(2)})}{\langle B^{2}\rangle ^{2}}\right]\frac{\text{d}^{2}}{\text{d}x^{2}}\left(\frac{\tilde{p}_{i}}{p_{0i}}+\frac{e\unicode[STIX]{x1D6F7}}{T_{i}}\right)\nonumber\\ \displaystyle & & \displaystyle \quad +\,\frac{n_{0}e}{\unicode[STIX]{x1D70F}_{i}\langle B^{2}\rangle }\left(\frac{I^{2}m_{i}T_{i}}{e^{2}}\right)^{2}\left[0.37\left(\frac{1}{\langle B^{2}\rangle }\left\langle \frac{1}{B^{2}}\right\rangle -\left\langle \frac{1}{B}\right\rangle \left\langle \frac{1}{B^{3}}\right\rangle \right)+0.54\left\langle \frac{1}{B^{4}}\right\rangle \right]\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\frac{\text{d}^{2}}{\text{d}x^{2}}\left(\frac{\tilde{T}_{i}}{T_{i}}\right)\nonumber\\ \displaystyle & & \displaystyle \quad +\,\frac{n_{0}e}{\unicode[STIX]{x1D70F}_{i}\langle B^{2}\rangle ^{2}}\left(\frac{I^{2}m_{i}T_{i}}{e^{2}}\right)^{2}(0.37G^{(0)}-0.62G^{[1]}+0.94G^{(2)}-0.2G^{(3)}-0.74G^{(4)})\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\frac{\text{d}^{2}}{\text{d}x^{2}}\left(\frac{\tilde{T}_{i}}{T_{i}}\right),\end{eqnarray}$$

with

(5.20) $$\begin{eqnarray}G^{(0)}=\frac{15\langle B^{2}\rangle ^{2}}{4}\left\langle \frac{1}{B^{2}}\int _{0}^{1/B_{\text{Max}}}\,\text{d}\unicode[STIX]{x1D706}\frac{\unicode[STIX]{x1D706}(1-\unicode[STIX]{x1D706}B)}{\left\langle \sqrt{1-\unicode[STIX]{x1D706}B}\right\rangle }\right\rangle ,\end{eqnarray}$$

(5.21) $$\begin{eqnarray}\displaystyle \frac{G^{(1)}}{3\langle B^{2}\rangle } & = & \displaystyle \frac{15}{4}\int _{0}^{1/B_{\text{Min}}}\,\text{d}\unicode[STIX]{x1D706}\unicode[STIX]{x1D706}\frac{\left\langle B\sqrt{1-\unicode[STIX]{x1D706}B}\right\rangle }{\left\langle \sqrt{1-\unicode[STIX]{x1D706}B}\right\rangle }\left\langle \frac{\sqrt{1-\unicode[STIX]{x1D706}B}}{B}\right\rangle \nonumber\\ \displaystyle & & \displaystyle +\,\frac{15}{16}\langle B^{2}\rangle \left\langle \frac{1}{B}\right\rangle \int _{0}^{1/B_{\text{Max}}}\frac{\text{d}\unicode[STIX]{x1D706}\unicode[STIX]{x1D706}^{2}}{\left\langle \sqrt{1-\unicode[STIX]{x1D706}B}\right\rangle }\nonumber\\ \displaystyle & & \displaystyle -\,\frac{15}{8}\langle B^{2}\rangle \int _{0}^{1/B_{\text{Max}}}\,\text{d}\unicode[STIX]{x1D706}\unicode[STIX]{x1D706}\frac{1}{\left\langle \sqrt{1-\unicode[STIX]{x1D706}B}\right\rangle }\left\langle \frac{\sqrt{1-\unicode[STIX]{x1D706}B}}{B}\right\rangle \int _{\unicode[STIX]{x1D706}}^{1/B_{\text{Max}}}\frac{\text{d}\unicode[STIX]{x1D706}^{\prime }}{\left\langle \sqrt{1-\unicode[STIX]{x1D706}^{\prime }B}\right\rangle },\qquad\end{eqnarray}$$

(5.22) $$\begin{eqnarray}G^{(2)}=\frac{15\langle B^{2}\rangle ^{2}}{4}\left\langle \frac{1}{B^{2}}\int _{0}^{1/B_{\text{Max}}}\unicode[STIX]{x1D706}\,\text{d}\unicode[STIX]{x1D706}\frac{\left\langle B\sqrt{1-\unicode[STIX]{x1D706}B}\right\rangle }{\left\langle \sqrt{1-\unicode[STIX]{x1D706}B}\right\rangle }\left\langle \frac{\sqrt{1-\unicode[STIX]{x1D706}B}}{B}\right\rangle \right\rangle ,\end{eqnarray}$$

(5.23) $$\begin{eqnarray}G^{(3)}=\frac{15\langle B^{2}\rangle ^{2}}{4}\int _{0}^{1/B_{\text{Min}}}\,\text{d}\unicode[STIX]{x1D706}\unicode[STIX]{x1D706}\frac{1}{\left\langle \sqrt{1-\unicode[STIX]{x1D706}B}\right\rangle }\left\langle \frac{\sqrt{1-\unicode[STIX]{x1D706}B}}{B}\right\rangle \left(\frac{\left\langle B\sqrt{1-\unicode[STIX]{x1D706}B}\right\rangle }{\langle B^{2}\rangle }-\left\langle \frac{\sqrt{1-\unicode[STIX]{x1D706}B}}{B}\right\rangle \right)\end{eqnarray}$$

and

(5.24) $$\begin{eqnarray}\displaystyle G^{(4)} & = & \displaystyle \frac{15\langle B^{2}\rangle ^{2}}{4}\int _{0}^{1/B_{\text{Max}}}\,\text{d}\unicode[STIX]{x1D706}\unicode[STIX]{x1D706}\frac{1}{\left\langle \sqrt{1-\unicode[STIX]{x1D706}B}\right\rangle }\nonumber\\ \displaystyle & & \displaystyle \times \,\left(\frac{1}{\left\langle \sqrt{1-\unicode[STIX]{x1D706}B}\right\rangle }\left\langle \frac{\sqrt{1-\unicode[STIX]{x1D706}B}}{B}\right\rangle \left\langle \frac{1-\unicode[STIX]{x1D706}B}{B}\right\rangle -\left\langle \frac{1-\unicode[STIX]{x1D706}B}{B^{2}}\right\rangle \right).\end{eqnarray}$$

Insertion of the results (5.18) and (5.19) completes the form of the vorticity equation (5.13). Analytic evaluations of the coefficients in (5.19) in the large aspect ratio limit are presented in appendix B and are consistent with those given by Wong & Chan (Reference Wong and Chan2005).