{\bf v}_S) 
+ \Lambda d\omega .\end{aligned}$$ The energy due to the vorticity is represented by the term which is proportional to $\omega =  \nabla \times {\bf v}_S$. Differentiating Eq. (\[eq:ASAS\_ENERGY\]) with respect to time and eliminating the time derivatives using the conservation laws above we recover the conservation of the energy $$\begin{aligned}
\label{eq:ASAS_RESOLVED} 
\frac{\partial E}{\partial t} +  \nabla \cdot    Q &=& 
 R + \left(P'_{ik} - \Lambda \omega \delta_{ik} 
+ \Lambda\frac{\omega_i \omega_k}{\omega}\right)
\frac{\partial v_{Ni} }{\partial x_k}\nonumber\\
&+&\left( J -\rho{\bf v}_N+ \nabla \times \Lambda  \nu 
\right)\cdot \left\{ F + \left[({\bf v}_S - {\bf v}_N) 
\times  \omega \right] \right\} = 0,\end{aligned}$$ where $P'_{ik}$ is the part of the stress tensor associated with the vorticity; the explicit form of the energy current is not indicated since it will not be used in the following. Since the dissipative function $R$ must be positive, the remaining terms on the right-hand side of Eq. (\[eq:ASAS\_RESOLVED\]) must be quadratic forms for small deviations from equilibrium. This implies that the most general form of the mutual friction force is $$\begin{aligned}
\label{eq:AS2.4}
 F &=& -\left[\omega\times\left(\nabla\times
\Lambda\nu\right) \right]- \beta\left[\,\nu\times
\left[\,\omega\times({\bf v}_N-{\bf v}_S-
\nabla\times\Lambda\nu)\,\right]\,\right]\nonumber \\
&-&\beta'\left[\,\omega\times ({\bf v}_N -{\bf v}_S
- \nabla \times\Lambda\nu)\right] 
+\beta''\nu\cdot\left[\, \omega\cdot ({\bf v}_N -{\bf v}_S
- \nabla \times 