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{\bf SOLUTIONS MIDTERM PHY 4936 (Fall 2011)}\\ 
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\centerline{\bf PROBLEM 3}\bigskip

\noindent 3.1 Principle of least action: Every mechanical system
is characterized by a definite function $L=L(q_1,\dots,q_s,\,\dot q_1,
\dots,\dot q_s,\,t)$ and the motion of the system is such that the
system moves between two fixed positions at different times $t_1$ 
and $t_2$ in a way that for sufficiently short time differences the 
integral
$$ S\ =\ \int_{t_1}^{t_2} L\,dt $$
takes the least possible value. \medskip

\noindent 3.2 Deviation of the Euler-Lagrange equations from the least 
action principle in general coordinates $q_i,\dot{q}_i$, $i=1,\dots s$:
\begin{eqnarray} \nonumber
  0 &=& \delta \int_{t_1}^{t_2} dt\, L(q_i,\dot{q}_i,t)\ 
  =\ \int_{t_1}^{t_2} dt\, \left\{ L(q_i+\delta q_i,\dot{q}_i+
  \delta\dot{q}_i,t)- L(q_i,\dot{q}_i,t)\right\} \\ \nonumber
  &=&  \int_{t_1}^{t_2} dt\,\left\{L(q_i,\dot{q}_i,t) + \sum_i
  \frac{\partial L}{\partial q_i}\,\delta q_i + \sum_i
  \frac{\partial L}{\partial \dot{q}_i}\,\delta\dot{q}_i 
  - L(q_i,\dot{q}_i,t)\right\} \\ \nonumber
  &=& \int_{t_1}^{t_2} dt\, \left\{\sum_i \frac{\partial L}
  {\partial q_i}\,\delta q_i + \sum_i\frac{\partial L}
  {\partial\dot{q}_i}\,\frac{d\ }{dt}\delta q_i \right\} \\ \nonumber
  &=& \int_{t_1}^{t_2} dt\,\sum_i\left\{ \frac{\partial L}
  {\partial q_i} - \frac{d\ }{dt}\,\frac{\partial L}
  {\partial\dot{q}_i}\right\}\,\delta q_i + \left[
  \frac{\partial L}{\partial\dot{q}_i}\,\delta q_i\right]_{t_1}^{t_2}
  \ =\ \int_{t_1}^{t_2} dt\,\sum_i\left\{ \frac{\partial L}
  {\partial q_i} - \frac{d\ }{dt}\,\frac{\partial L}
  {\partial\dot{q}_i}\right\}\,\delta q_i\ . 
\end{eqnarray}
The last equality holds because of $\delta q_i(t_1)=\delta q_i(t_2)=0$.
As the variations are independent, the finally obtained relation is 
equivalent to
$$ \frac{\partial L}{\partial q_i} - \frac{d\ }{dt}\,\frac{\partial L}
  {\partial\dot{q}_i}\ =\ 0~~{\rm for}~~i=1,\dots,s\ .$$
\smallskip

\noindent 3.3 $L(q_k+\epsilon_k,\dot{q}_k,t)=L(q_k,\dot{q}_k,t)$ 
implies 
$$ 0 = \frac{\partial L}{\partial q_k} = \frac{d~}{dt}\,
   \frac{\partial L}{\partial \dot{q}_k}\ \Rightarrow\  
   p_k = \frac{\partial L}{\partial \dot{q}_k}\ {\rm conserved\
   (generalized\ momentum).} $$
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