Example: Motion in a Central Potential \bgroup\color{black}$ V(r)$\egroup

Use spherical coordinates and choose the \bgroup\color{black}$ z$\egroup direction so that the motion is in the \bgroup\color{black}$ r\phi$\egroup plane. The Hamiltonian is \bgroup\color{black}$ T+U$\egroup for this simple case if we don't choose coordinates that are time dependent. Due to the spherical symmetry of the potential (and of the kinetic energy term), the momentum conjugate to \bgroup\color{black}$ \phi$\egroup is conserved.

$\displaystyle \dot{p}_i$ $\displaystyle =-{\partial H\over\partial q_i }$    
$\displaystyle \dot{p}_\phi$ $\displaystyle =-{\partial H\over\partial \phi}=0$    
$\displaystyle p_\phi$ $\displaystyle = {\partial L\over \partial \dot{\phi}}={\partial T\over \partial \dot{\phi}}$    
$\displaystyle T$ $\displaystyle ={1\over 2}mv^2={1\over 2}m(\dot{r}^2+r^2\sin^2\theta\dot{\phi}^2+r^2\dot{\theta}^2) \rightarrow {m\over 2}(\dot{r}^2+r^2\dot{\phi}^2)$    
$\displaystyle p_\theta$ $\displaystyle = 0 = const.$    
$\displaystyle p_\phi$ $\displaystyle = m r^2\dot{\phi} = const.$    
$\displaystyle p_r$ $\displaystyle = m\dot{r}$    
$\displaystyle H$ $\displaystyle = {1\over 2m}(p_r^2+{p_\phi^2\over r^2}) + V(r)$    
$\displaystyle \dot{p}_r$ $\displaystyle =-{\partial H\over\partial r} =-{\partial \over\partial r}\left({p_\phi^2\over 2m r^2} + V(r)\right)$    

Either of the last two equations can be used to solve the problem using the effective potential.

Jim Branson 2012-10-21