 Solution

Since $L_z=J_{\varphi }$, Hamilton's equations giveFor the spherical unperturbed Hamiltonian, $H_0=H_0(J_r,L)$; in these particular action-angle variables the third component ${\Omega }_{\varphi }=0$ describes the fixed orientation of the orbital plane. It should not be confused with the physical azimuthal angular velocity of a particle around its orbit.

The conserved normalization of a probability distribution $\int f{d}^3\theta d^{3}J=1$ makes the expected value of specific angular momentum equal to $\int L_{z}f{d}^3\theta d^{3}J$. Its time derivative is the torque per unit halo mass:This follows either by averaging the single-particle equation or by integrating the Collisionless Boltzmann equation against $L_z$, with vanishing boundary fluxes.

The term $f_0{\partial }_{{\theta }_{\varphi }}\delta \Phi$ integrates to zero over the periodic angle variable. Thus the first nonzero torque is quadratic in the perturbation. Use the Fourier series expansions of b; the source's reference here to part (a) is a numbering slip. Integrating the product over all three angle variables yieldsReality of the potential energy per unit mass gives $\delta {\Phi }_{-\mathbf{k}}=\delta {\Phi }_{\mathbf{k}}^{\ast }$; likewise $\delta f_{-\mathbf{k}}=\delta f_{\mathbf{k}}^{\ast }$. Conjugate harmonics make the total torque real. The sum is over all integer vectors, so no extra factor of two should be inserted.