I'm working on a problem in Kibble's Classical Mechanics (Chapter 11, Problem 12) and came upon a method for exploring the stability of an equilibrium point that I wasn't familiar with. The question basically says to explore the stability of one of the stable Lagrangian points in the restricted three-body problem by assuming a solution where $x - x_0$, $y - y_0$, $p_x + m \omega y_0$ and $p_y - m \omega x_0$ are all small quantities proportional to $e^{pt}$, with $p$ constant. The question then asks to show that the possible values of $p$ are given by

$$p^4 + w^2p^2 + \frac{27M_1M_2 \omega^4}{4(M_1 + M_2)^2} = 0$$

I was wondering if someone could help me better understand this type of small displacement from equilibrium. Would the small quantities all have the same constant of proportionality, or would each have a different one? Is the p in the exponent related to the $p_x$, $p_y$ for momentum or is it just an unrelated constant?

I'm posting some of my work below in case it's helpful! Thank you for any help!

The problem initially asks you to solve for the Hamiltonian of the restricted three-body problem. (Based on an earlier question, where they ask you to find the Lagrangian in a frame rotating at $\omega$, the rate at which the larger masses, $M_1$, $M_2$ are revolving around their CM. So, if $M_1$ is located at $(-a_1, 0)$ and $M_2$ is located at $(a_2, 0)$, then in NON-rotating coordinates, the Lagrangian is given by:

$$L = \frac{1}{2} m \dot{r'}^2 + \frac{G m M_1}{r_1} + \frac{G m M_2}{r_2}$$

In the rotating frame, $\dot{r'} = \dot{r} + \omega x r$, and we get the Lagrangian in the rotating frame...

$$L = \frac{1}{2} m \dot{x}^2 - m \omega \dot{x} y + \frac{1}{2} m \dot{y}^2 + m \omega \dot{y} x + \frac{1}{2} m \omega^2 (x^2 + y^2) + \frac{G m M_1}{r_1} + \frac{G m M_2}{r_2}$$

And from this you get the Hamiltonian...

$$H = \frac{p_x^2}{2m} + \omega y p_x + \frac{p_y^2}{2m} - \omega x p_y - \frac{G m M_1}{r_1} - \frac{G m M_2}{r_2}$$

From here, I've tried using Hamilton's equations and plugging in as a solution e.g. $x = x_0 + Ae^{pt}$ and $\ddot{x} = p^2 A e^{pt}$ and using the fact that, at equilibrium, the accelerations and changes in momentum must be $0$, but I'm not having any luck. Any push in the right direction would be much appreciated!

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    $\begingroup$ A proof is given in Why are $L_4$ and $L_5$ lagrangian points stable? $\endgroup$ – Qmechanic Jul 26 '20 at 3:56
  • $\begingroup$ Yes, that post examines the same area, but I'm more curious about when and why you assume a small displacement from equilibrium proportional to an exponential. I guess if p is imaginary, you'd get oscillatory behavior, and the equilibrium would be stable. $\endgroup$ – TKT Jul 26 '20 at 18:41
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    $\begingroup$ That's because the linearized EOM around a stationary point is a 2nd-order homogeneous ODE with constant coefficients. Its solutions are precisely on the aforementioned form. $\endgroup$ – Qmechanic Jul 26 '20 at 19:38
  • $\begingroup$ Okay, thank you! This makes sense now! $\endgroup$ – TKT Jul 27 '20 at 2:13

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