Generalized thermostat on a Riemannian surface ($F=X+\lambda V,D_t\dot{\gamma }=\lambda (\gamma ,\dot{\gamma })i\dot{\gamma }$)

A smooth real function $\lambda$ on the unit tangent bundle determines this unit-speed smooth flow. Its base velocity is $v$, so its generator never vanishes. In the canonical frame its divergence in Liouville volume of a surface geodesic flow is $V\lambda$. The special case where $\lambda$ depends only on the base is a magnetic flow; when $\lambda (x,v)=⟨E(x),iv⟩$ it is a Gaussian thermostat.

Write a tangent variation of a surface thermostat as $aF+yH+zV$ in the canonical frame. The commutators $[F,H]=-\lambda F+(K-H\lambda +{\lambda }^2)V$ and $[F,V]=-H-(V\lambda )V$ give $\dot{a}=\lambda y$ and the stated transverse equation. Near the vertical projective direction, $r=y/z$ obeys $\dot{r}=1-(V\lambda )r+(K-H\lambda +{\lambda }^2)r^2$.

At unit speed on a Riemannian surface, the right-hand side is $⟨E,iv⟩iv$. This removes the velocity-parallel component of the forcing, so differentiating $|\dot{\gamma }{|}^2$ gives zero. The dynamics need not preserve the ordinary Liouville volume of a surface geodesic flow.

For $F=X+\lambda V$, the canonical fields $X,V$ preserve Liouville volume of a surface geodesic flow, so the divergence is $V\lambda$. For a Gaussian thermostat, $\lambda =⟨E,iv⟩$ and $V(iv)=-v$, giving the stated sign. If $E=\nabla f$, then the density $e^{f\circ \pi }$ makes this divergence zero.