Chew-Goldberger-Low Forms

As is well known, the smallness of the toroidal magnetization parameter, $\delta _s$, ensures that the dominant parallel viscosity tensors, $\pi$$_{\parallel\,s}$ and $\Theta$$_{\parallel\,s}$, take the so-called Chew-Goldberger-Low forms [11,29]:

$$\mbox{\boldmath$\pi$} _{\parallel\,s} = {\mit\Delta}\pi_{\parallel\,s}\left({\bf b}\,{\bf b}- \frac{1}{3}\,{\bf I}\right),$$ (2.152)

$$\mbox{\boldmath$\Theta$} _{\parallel\,s} = {\mit\Delta}\Theta_{\parallel\,s}\left({\bf b}\,{\bf b}- \frac{1}{3}\,{\bf I}\right),$$ (2.153)

where $\langle {\mit\Delta} \pi_{\parallel\,s}\rangle =\langle {\mit\Delta} \Theta_{\parallel\,s}\rangle = 0$. Incidentally, it is clear from Equation (2.60) that the parallel viscosity tensor in the classical closure scheme does indeed take this form.

Now, if $\pi$$_{\parallel\,s}$ takes the Chew-Goldberger-Low form then

$$\nabla\psi\cdot \mbox{\boldmath$\pi$} _{\parallel\,s}\cdot {\bf b} =0,$$ (2.154)

$$\nabla\psi\cdot \mbox{\boldmath$\pi$} _{\parallel\,s}\cdot\nabla\varphi = 0.$$ (2.155)

These results follow because ${\bf b}\cdot\nabla \psi= \nabla\varphi\cdot\nabla\psi=0$.

Furthermore,

$$R^2\,\nabla\varphi \cdot\nabla\cdot \mbox{\boldmath$\pi$} _{\parallel\,s} = \nabla\cdot\left(R^2\,\nabla\varphi\cdot\mbox{\... ...dmath$\pi$}_{\parallel\,s} : \widetilde{\nabla}\left(R^2\,\nabla\varphi\right),$$ (2.156)

where, in Cartesian coordinates,

$$\left(\widetilde{\bf A}\right)_{jk} = \frac{1}{2}\left(\frac{\partial A_j}{\partial r_k} + \frac{\partial A_k}{\partial r_j}\right),$$ (2.157)

and use has been made of the fact that $\pi$$_{\parallel\,s}$ is a symmetric tensor. In fact, it can be shown that [19]

$$\widetilde{\nabla}\left(R^2\,\nabla\varphi\right) = {\bf0}.$$ (2.158)

Hence, we deduce that

$$R^2\,\nabla\varphi \cdot\nabla\cdot \mbox{\boldmath$\pi$} _{\parallel\,s} = \nabla\cdot\left(R^2\,\nabla\varphi\cdot\mbox{\boldmath$\pi$}_{\parallel\,s}\right).$$ (2.159)

Making use of Equation (2.150), we obtain

$$\left\langle R^2\,\nabla\varphi \cdot\nabla\cdot\mbox{\boldmath$\... ...rphi\cdot\mbox{\boldmath$\pi$}_{\parallel\,s}\cdot\nabla {\cal V}\right\rangle.$$ (2.160)

However, if $\pi$$_{\parallel\,s}$ takes the Chew-Goldberger-Low form then $\nabla\varphi\cdot$$\pi$$_{\parallel\,s}\cdot\nabla {\cal V}=0$, because ${\bf b}\cdot\nabla{\cal V}= \nabla\varphi\cdot\nabla{\cal V} = 0$. It follows that [32,34]

$$\left\langle R^2\,\nabla\varphi \cdot\nabla\cdot\mbox{\boldmath$\pi$}_{\parallel\,s} \right\rangle =0.$$ (2.161)

Likewise, if $\Theta$$_{\parallel\,s}$ takes the Chew-Goldberger-Low form then

$$\left\langle R^2\,\nabla\varphi \cdot\nabla\cdot\mbox{\boldmath$\Theta$}_{\parallel\,s} \right\rangle =0.$$ (2.162)