The motion of a test particle around a point mass $\mu$ is governed by the Hamiltonian $$ (*)\qquad\qquad H(r,p_r,p_\phi) = \frac{p_r^2}{2} + \frac{p_\phi^2}{2r^2} - \frac{\mu}{r} $$ which has well-known solutions and action representation $$ H(J_r,J_\phi) = -\frac{\mu^2}{2(J_r+J_\phi)^2} = -\frac{\mu^2}{2J_\phi^2} \left(1 - 2\frac{J_r}{J_\phi} + 3 \frac{J_r^2}{J_\phi^2} + O(J_r^3) \right), $$ where $J_\phi=p_\phi$. Now, at fixed $p_\phi=J_\phi$, one can consider the Hamiltonian $(*)$ as that for the radial motion only and re-write it as $$ H_r(r,p_r) = \frac{p_r^2}{2} + \Phi_{\mathrm{eff}}(r) \qquad\text{with}\qquad \Phi_{\mathrm{eff}}(r) = -\frac{\mu}{r} + \frac{J_\phi^2}{2r^2}, $$ which has $J_\phi$ as a parameter. The minimum of $\Phi_{\mathrm{eff}}$ occurs at the radius $r_{\mathrm{c}}=J_\phi^2/\mu$ of the circular orbit with angular momentum $J_\phi$. $\Phi_{\mathrm{eff}}$ has the Taylor expansion $$ \Phi_{\mathrm{eff}}(r) = -\frac{\mu}{2r_{\mathrm{c}}} + \frac{\mu}{2r_{\mathrm{c}}^3} x^2 - \frac{\mu}{r_{\mathrm{c}}^4} x^3 + \frac{3\mu}{2r_{\mathrm{c}}^5} x^4 + O(x^5) $$ with $x\equiv r-r_{\mathrm{c}}(J_\phi)$. Here, the first term is the energy of the circular orbit. Now split $H_r=H_0 + H_1$ with \begin{align} H_0 &= -\frac{\mu}{2r_{\mathrm{c}}} + \frac{p_x^2}{2} + \frac{\mu}{2r_{\mathrm{c}}^3} x^2,& H_1 &= - \frac{\mu}{r_{\mathrm{c}}^4} x^3 + \frac{3\mu}{2r_{\mathrm{c}}^5} x^4. \end{align} Classicle epicycle theory (e.g. Lindblad 1926) corresponds to ignoring $H_1$ and solving $H_0$, which is simple harmonic motion with epicycle frequency $\kappa\equiv\sqrt{\mu/r_{\mathrm{c}}^3}=\mu^2/J_\phi^3$, giving \begin{align} x_0 &= \sqrt{\frac{2J_r}{\kappa}}\sin\theta, &\theta&=\vartheta+\kappa t,\\ H_0 &= -\frac{\mu^2}{2J_\phi^2} + \kappa J_r &&= -\frac{\mu^2}{2J_\phi^2} \left(1 - 2 \frac{J_r}{J_\phi}\right), \end{align} which is the first-order of the full Hamiltonian $H(J_r,J_\phi)$ above. All this is standard textbook stuff, except for $H_1(x)$ (which is correct).


Now, use canonical perturbation theory to get to the next order. According to Lichtenberg & Liebermann (Springer 1983), this amounts to averaging the perturbation $H_1(x)$ over the unperturbed orbit (note that $\langle\sin^3\theta\rangle=0$ and $\langle\sin^4\theta\rangle=3/8$): $$ \left\langle H_1\left(x=x_0(\theta)\right)\right\rangle = \frac{3\mu}{2r_{\mathrm{c}}^5} \frac{4J_r^2}{\kappa^2}\frac{3}{8} = \frac{9\mu^2}{4J_\phi^2}\frac{J_r^2}{J_\phi^2}. $$

However, from the full $H(J_r,J_\phi)$ above, we expect $$ H_1 = -\frac{3\mu^2}{2J_\phi^2}\frac{J_r^2}{J_\phi^2} $$ which differes by a factor $-3/2$.

What went wrong with my derivation?


1 Answer 1


The problem is that the $x^3$ term also contributes to the first order (in $J_r$) correction to $H$ and we must go to second-order perturbation theory. Using the Deprit perturbation series (Lichtenberg & Liebermann 1983, §2.5), we have \begin{align} H_1 &= -\frac{\mu}{r_{\mathrm{c}}^4} x^3 &&= -\frac{\mu}{r_{\mathrm{c}}^4}\left(\frac{2J_r}{\kappa}\right)^{3/2}\sin^3\theta &&= -\frac{\mu}{4r_{\mathrm{c}}^4}\left(\frac{2J_r}{\kappa}\right)^{3/2}\left(3\sin\theta-\sin3\theta\right) \\ H_2 &= \frac{3\mu}{2r_{\mathrm{c}}^5} x^4 &&= \frac{3\mu}{2r_{\mathrm{c}}^5}\left(\frac{2J_r}{\kappa}\right)^{2}\sin^4\theta &&= \frac{3\mu}{16r_{\mathrm{c}}^5}\left(\frac{2J_r}{\kappa}\right)^{2}\left(3-4\cos2\theta+\cos4\theta\right) \end{align} Then the first order correction to the Hamiltonian, $\overline{H}_1=\langle H_1\rangle=0$. For the second order, one needs \begin{align} \kappa\frac{d w_1}{d\theta} = \langle H_1\rangle - H_1 && \to && w_1 = \frac{\mu}{12r_{\mathrm{c}}^4\kappa} \left(\frac{2J_r}{\kappa}\right)^{3/2} \left(\cos3\theta-9\cos\theta\right), \end{align} when $$ \left[w_1,H_1-\langle H_1\rangle\right] = \frac{3\mu}{8r_{\mathrm{c}}^5}\left(\frac{2J_r}{\kappa}\right)^{2}\left(-5+4\cos2\theta+\cos4\theta\right) $$ and \begin{align} \overline{H}_2 &= \left\langle H_2 + \tfrac12\left[w_1,H_1-\langle H_1\rangle\right]\right\rangle && = -\frac{3J_r^2}{2r_{\mathrm{c}}^2} \end{align} as required.


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