Lowest-Order Flows

To lowest order in the small parameter $\delta_{\theta\,s}\,{\mit\Delta}_s$, Equations (2.116) and (2.118) yield

0 $\displaystyle \simeq e_s\,n_s\,({\bf E}_\perp+{\bf V}_s\times {\bf B})- \nabla p_s,$ (2.132)
0 $\displaystyle \simeq \frac{e_s}{m_s}\left[\frac{5}{2}\,p_s\,({\bf E}_\perp+{\bf...
...}_s\times {\bf B}\right]
-\nabla\left(\frac{5}{2}\,\frac{T_s\,p_s}{m_s}\right).$ (2.133)

The previous two equations can be solved to give

$\displaystyle {\bf V}_s$ $\displaystyle = V_{\parallel\,s}\,{\bf b} + {\bf V}_{\perp\,s},$ (2.134)
$\displaystyle {\bf q}_s$ $\displaystyle = q_{\parallel\,s}\,{\bf b} + {\bf q}_{\perp\,s},$ (2.135)


$\displaystyle {\bf V}_{\perp\,s}$ $\displaystyle = \frac{{\bf E}\times {\bf B}}{B^2}+\frac{{\bf B}\times \nabla p_s}{e_s\,n_s\,B^2},$ (2.136)
$\displaystyle \frac{{\bf q}_{\perp\,s}}{p_s}$ $\displaystyle = \frac{5}{2}\,\frac{{\bf B}\times \nabla T_s}{e_s\,B^2}.$ (2.137)

It can be seen that the lowest-order particle and heat flows are both confined to magnetic flux-surfaces (i.e., ${\bf V}_s\cdot\nabla \psi={\bf q}_s\cdot\nabla \psi=0$). The first term on the right-hand side of Equation (2.136) is the E-cross-B velocity,

$\displaystyle {\bf V}_E = \frac{{\bf E}\times {\bf B}}{B^2},$ (2.138)

that is common to all plasma species [18]. The second term is the diamagnetic velocity,

$\displaystyle {\bf V}_{\ast\,s} = \frac{{\bf B}\times \nabla p_s}{e_s\,n_s\,B^2},$ (2.139)

which is different for electrons and ions, and is a consequence of the rapid gyro-motions of charged particles in the presence of equilibrium pressure gradients [18]. The drift ordering (2.113) ensures that the E-cross-B and diamagnetic velocities are similar in magnitude. It is clear from Equation (2.137) that there is a diamagnetic flow of heat, as well as particles, around flux-surfaces; this heat flow is the same as that associated with the cross thermal conductivities introduced in Section 2.6.

To lowest order in the small parameter $\delta_{\theta\,s}\,{\mit\Delta}_s$, Equations (2.115) and (2.117) yield

$\displaystyle \nabla\cdot{\bf V}_s$ $\displaystyle =0,$ (2.140)
$\displaystyle \nabla\cdot {\bf q}_s$ $\displaystyle = 0.$ (2.141)

Here, use has been made of ${\bf V}_s\cdot\nabla n_s={\bf V}_s\cdot\nabla p_s= 0$. Clearly, the lowest-order particle and heat flows are both divergence free. Now, the fact that ${\bf V}_s\cdot\nabla \psi={\bf q}_s\cdot\nabla \psi=0$ implies that $V_s^\psi = q_s^\psi = 0$. Making use of these results, the previous two equations can be combined with Equations (2.123) and (2.126), as well as the fact that $\partial/\partial\varphi\equiv 0$ in an axisymmetric equilibrium, to give [34]

$\displaystyle \frac{{\bf V}_s\cdot\nabla\theta}{{\bf B}\cdot\nabla\theta}= V_{\theta\,s}(\psi),$ (2.142)
$\displaystyle \frac{{\bf q}_s\cdot\nabla\theta}{{\bf B}\cdot\nabla\theta}= q_{\theta\,s}(\psi).$ (2.143)

Taking the scalar products of Equations (2.134) and (2.135) with $\nabla\theta$, we obtain

$\displaystyle V_{\theta\,s}(\psi)$ $\displaystyle = \frac{V_{\parallel\,s}}{B} - \frac{V_{\psi\,s}(\psi)}{B^2},$ (2.144)
$\displaystyle q_{\theta\,s}(\psi)$ $\displaystyle = \frac{q_{\parallel\,s}}{B} - \frac{q_{\psi\,s}(\psi)}{B^2},$ (2.145)


$\displaystyle V_{\psi\,s}(\psi)$ $\displaystyle = -I\,\frac{d{\mit\Phi}}{d\psi} - \frac{I}{e_s\,n_s}\,\frac{dp_s}{d\psi},$ (2.146)
$\displaystyle \frac{q_{\psi\,s}(\psi)}{p_s(\psi)}$ $\displaystyle = - \frac{5}{2}\,\frac{I}{e_s}\,\frac{dT_s}{d\psi}.$ (2.147)

Here, use has been made of Equations (2.136) and (2.137).