$e$ is negative. Thus positive $v$ implies a flow of electrons from the polarising to the switching magnet. The last contribution to the torque in Eq. (\[Gamma3\]) is due to the external field $H_{\rm ext}$ with $h_{\rm ext}=H_{\rm ext}/H_{\rm
  u0}$. The external field is taken in the direction of the magnetisation of the polarising magnet, as is the case in most experimental situations.

It follows from Eq. (\[llg\]) that in a steady state ${\bm\Gamma}=0$. We shall first consider some cases of experimental importance where the steady state solutions are trivial and the important physics is concerned entirely with their stability. To discuss stability, we linearise Eq. (\[Gamma2\]), using Eq. (\[Gamma3\]), about a steady state solution ${\bf m}={\bf m}_{0}$. Thus $$\label{Gamma4}
{\bf m}={\bf m}_0+\xi{\bf e}_{\alpha}+\eta{\bf e}_{\phi},$$ where ${\bf e}_{\alpha}$, ${\bf e}_{\phi}$ are unit vectors in the direction ${\bf m}$ moves when $\alpha$ and $\phi$ are increased independently. The linearised equation may be written in the form $$\label{Gamma5}
\frac{{\rm d}\xi}{{\rm d}\tau}=A\xi+B\eta;\quad
\frac{{\rm d}\eta}{{\rm d}\tau}=C\xi+D\eta.$$ Following Sun [@sun], we have introduced the natural dimensionless time variable $\tau=tH_{\rm u0}/(1+\gamma^2)$. The conditions for the steady state to be stable are $$\label{Gamma6}
F=A+D\leqslant 0;\quad G=AD-BC\geqslant 0$$ excluding $F=G=0$ [@20]. For simplicity we give these conditions explicitly 