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{\bf ADVANCED DYNAMICS --- PHY 4241/5227}\\\vspace {2mm} 
{\bf HOME AND CLASS WORK -- SET 4}\\%
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Solution for assignment 18: Legendre transformation and 
Hamilton's equations for the spherical pendulum. \bigskip

\noindent (a) Legendre transformation. With
$$ L = T- V~~{\rm with}~~
   T = \frac{1}{2}\,m\,b^2\,\dot{\theta}^2 + 
       \frac{1}{2}\,m\,b^2\,(\sin\theta)^2\,\dot{\phi}^2
   ~~{\rm and}~~ V = -m\,g\,b\,\cos\theta                $$
the generalized momenta are
$$ p_{\theta} = \frac{\partial L}{\partial\dot{\theta}} =
   m\,b^2\,\dot{\theta}~~{\rm and}~~p_{\phi} = 
   \frac{\partial L}{\partial\dot{\phi}} =
   m\,b^2\,(\sin\theta)^2\,\dot{\phi}\ . $$
Therefore,
$$ \dot{\theta}\,p_{\theta}+\dot{\phi}\,p_{\phi} = 2T~~{\rm and}~~ 
   H=\dot{\theta}\,p_{\theta}+\dot{\phi}\,p_{\phi}-L
    =\frac{1}{2}\,(\dot{\theta}\,p_{\theta}+\dot{\phi}\,p_{\phi})
    =T+V\ .$$
Eliminating 
$$ \dot{\theta} = \frac{p_{\theta}}{m\,b^2}~~{\rm and}~~
   \dot{\phi} = \frac{p_{\phi}}{m\,b^2\,(\sin\theta)^2}\ . $$
from $H$, we have
$$ H = \frac{p^2_{\theta}}{2\,m\,b^2} + \frac{p^2_{\phi}}
   {2\,m\,b^2\,(\sin\theta)^2} - m\,g\,b\,\cos\theta\ .    $$
\medskip

\noindent (b) Hamilton's equations:
\begin{eqnarray} \nonumber
  \dot{\theta} &=& \frac{\partial H}{\partial\,p_{\theta}}\ =\
  \frac{p_{\theta}}{m\,b^2}\,, \\ \nonumber
  \dot{\phi} &=& \frac{\partial H}{\partial\,p_{\phi}}\ =\
  \frac{p_{\phi}}{m\,b^2\,(\sin\theta)^2}\,, \\ \nonumber
  \dot{p_{\theta}} &=& - \frac{\partial H}{\partial\,\theta}\ =\
  \frac{p^2_{\phi}\,\cos\theta}{m\,b^2\,(\sin\theta)^3} -
   m\,g\,b\,\sin\theta\,, \\ \nonumber
  \dot{p_{\phi}} &=& - \frac{\partial H}{\partial\,\phi}\ =\ 0
  ~~\Rightarrow~~p_{\phi} = {\rm constant\ (angular\ momentum\
  conservation)}\,.
\end{eqnarray}
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