Hamiltonian from the equation of motion

Question

The equation of motion is

$$(1+e\cos f)\frac{d^2\theta}{df^2}-2e\sin f\frac{d\theta}{df}-\frac{w^2}{2}\sin 2(f-\theta)=0,$$ where $0<e<1$, $w\in const$. Taking $q=\theta$ and $p=\frac{d\theta}{df}$, with the use of the Hamilton equations

$$\frac{dq}{df}=\frac{\partial H}{\partial p},\qquad \frac{dp}{df}=-\frac{\partial H}{\partial q},$$

how can I find the Hamiltonian $H=H(p,q,f)$?

Answer 1

Expanding on a previous answer and comment: there is no trick to solving this problem.

Start from the perfectly general Lagrangian $$ L=\frac{1}{2}G(q,t)\dot{q}^2 +F(q,t)\dot q - V(q,t) $$ and obtain the equation of motion $$ G(q,t)\ddot{q}+\frac{\partial G(q,t)}{\partial t}\dot{q}+ \frac{\partial F(q,t)}{\partial t}+\frac{\partial V(q,t)}{\partial q} +\frac{1}{2}\frac{\partial G(q,t)}{\partial q}\dot{q}^2=0\, . $$ You can then compare to your form to deduce $G(q,t)$, $F(q,t)$ and $V(q,t)$. From this you can obtain the Hamiltonian $$ H=p\dot{q}-L\, . $$

Answer 2

Hints:

1. Start with a Lagrangian formulation $$ L~:=~\frac{1}{2}(1+e\cos t)^2\dot{q}^2 -V, \qquad V~:=~-\frac{w^2}{4}(1+e\cos t) \cos 2(q-t). \tag{1} $$

2. Next Legendre transform to obtain the Hamiltonian formulation.