Lowest-Order Flows

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

$$\simeq e_s\,n_s\,({\bf E}_\perp+{\bf V}_s\times {\bf B})- \nabla p_s,$$ 0 (2.132)

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

The previous two equations can be solved to give

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

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

where

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

$$\frac{{\bf q}_{\perp\,s}}{p_s} = \frac{5}{2}\,\frac{{\bf B}\times \nabla T_s}{e_s\,B^2}.$$ (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,

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

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

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

$$\nabla\cdot{\bf V}_s =0,$$ (2.140)

$$\nabla\cdot {\bf q}_s = 0.$$ (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]

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

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

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

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

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

where

$$V_{\psi\,s}(\psi) = -I\,\frac{d{\mit\Phi}}{d\psi} - \frac{I}{e_s\,n_s}\,\frac{dp_s}{d\psi},$$ (2.146)

$$\frac{q_{\psi\,s}(\psi)}{p_s(\psi)} = - \frac{5}{2}\,\frac{I}{e_s}\,\frac{dT_s}{d\psi}.$$ (2.147)

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