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{\bf ADVANCED DYNAMICS --- PHY 4241/5227}\\\vspace {2mm} 
{\bf HOME AND CLASS WORK -- SET 3}\\%
\vspace {3mm} (January 20, 2010) \\
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\noindent Solution for assignment 14: Energy conservation.
\bigskip

\noindent
The total time derivative of the Lagrangian $L(\dot{q}_j,q_j,t)$ is
$$ \frac{dL}{dt}\ =\ \sum_j \left( 
   \frac{\partial L}{\partial\dot{q}_j}\,\frac{d\dot{q}_j}{dt} +
   \frac{\partial L}{\partial q_j}\,\frac{dq_j}{dt} \right) +
   \frac{\partial L}{\partial t}\ . $$
By its definition $\delta_tL = (\partial L/\partial t)\,\delta t = 0$
implies $(\partial L/\partial t) = 0$, so that
$$ \frac{dL}{dt}\ =\ \sum_j \left( 
   \frac{\partial L}{\partial\dot{q}_j}\,\ddot{q}_j +
   \frac{\partial L}{\partial q_j}\,\dot{q}_j \right) $$
holds. Using the Euler-Lagrange equation, we find the desired result:
$$ \frac{dL}{dt}\ =\ \sum_j \left( \ddot{q}_j\,
   \frac{\partial L}{\partial\dot{q}_j} + \dot{q}_j\,
   \frac{d\ }{dt}\,\frac{\partial L}{\partial\dot{q}_j}\right) =
   \frac{d\ }{dt}\,\sum_j\dot{q}_j\,\frac{\partial L}{\partial\dot{q}_j}
   ~~\Leftrightarrow~~\frac{d}{dt}\,\left( L - \sum_j \dot{q}_j\,
   \frac{\partial L}{\partial \dot{q}_j} \right) = 0\ .  $$

\noindent For a bilinear kinetic Energy we find by differentiation and 
carrying the sum over $i$ out
\begin{eqnarray} \nonumber
   \sum_i \dot{q}_i\,\frac{\partial T}{\partial \dot{q}_i} &=&
   \sum_i \dot{q}_i\,\frac{\partial\ }{\partial \dot{q}_i}  
   \sum_{j,k} a_{jk}\,\dot{q}_j\dot{q}_k\ =\
   \sum_i \dot{q}_i\,\sum_{j,k}a_{jk}\,\left(\delta_{ij}\,\dot{q}_k 
   + \dot{q}_j\,\delta_{ik}\right) \\ \nonumber &=&
   \sum_{j,k}a_{jk}\,\left(\dot{q}_j\,\dot{q}_k + 
   \dot{q}_j\,\dot{q}_k\right)\ =\ 2\,T\ .
\end{eqnarray}
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