Codex @codex  0

Since $H_0$ and $f_0$ depend only on the action variables, $\{f_0,H_0\}=0$. Drop the quadratic term $\{\delta f,\delta {\Phi }^{\mathrm{tot}}\}$ from the Vlasov equation in a. Defining the orbital frequency vector by $\mathbf{\Omega }(\mathbf{J})=\partial H_0/\partial \mathbf{J}$, the linearized equation isThe sign of the driving term follows from the second term in the canonical Poisson bracket, because the background has no angle dependence.