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\begin{document}

\begin{center}
{\bf ADVANCED DYNAMICS --- PHY 4241/5227}\\\vspace {2mm} 
{\bf MOMENTUM CONSERVATION}\\%
\end{center} \bigskip
\setcounter{equation}{0}

\centerline{\bf Linear Momentum} \bigskip

In accordance with Noether's theorem, we derive momentum conservation 
from translation invariance
\begin{equation}
   \delta_x\,L\ =\ 0\,,\qquad \delta_y\,L\ =\ 0\,\qquad {\rm and}
   \qquad \delta_z\,L\ =\ 0\,.
\end{equation}
Using the definitions of the variations and the Euler-Lagrange 
equations, we find for a $n$ particle system (note that the 
displacements agree for all particles)
\begin{eqnarray}
   0 = \delta_x\,L = \sum_{i=1}^n \frac{\partial L}{\partial x_i}\,
   \delta x\ \Leftrightarrow\
   0 = \sum_{i=1}^n \frac{\partial L}{\partial x_i} = 
   \sum_{i=1}^n \frac{d~}{dt} \frac{\partial L}{\partial \dot{x}_i} =
   \sum_{i=1}^n \dot{p}_i^x\ \Leftrightarrow\
   \sum_{i=1}^n p_i^x = {\rm constant}\,,\\
   0 = \delta_y\,L = \sum_{i=1}^n \frac{\partial L}{\partial y_i}\,
   \delta y\ \Leftrightarrow\
   0 = \sum_{i=1}^n \frac{\partial L}{\partial y_i} = 
   \sum_{i=1}^n \frac{d~}{dt} \frac{\partial L}{\partial \dot{y}_i} =
   \sum_{i=1}^n \dot{p}_i^y\ \Leftrightarrow\
   \sum_{i=1}^n p_i^y = {\rm constant}\,,\\
   0 = \delta_z\,L = \sum_{i=1}^n \frac{\partial L}{\partial z_i}\,
   \delta z\ \Leftrightarrow\
   0 = \sum_{i=1}^n \frac{\partial L}{\partial z_i} = 
   \sum_{i=1}^n \frac{d~}{dt} \frac{\partial L}{\partial \dot{z}_i} =
   \sum_{i=1}^n \dot{p}_i^z\ \Leftrightarrow\
   \sum_{i=1}^n p_i^z = {\rm constant}\,.
\end{eqnarray}
In vector notation the result reads
\begin{equation}
  \vec{P}\ =\ \sum_{i=1}^n \vec{p}\ =\ {\rm constant}\,.
\end{equation} \bigskip

\centerline{\bf Angular Momentum} \bigskip

The total angular momentum of a system is
\begin{equation}
  \vec{L}\ =\ \sum_j \vec{r}_j\times \vec{p}_j
\end{equation}
where the sum of over point particles. We now consider an infinitesimal
rotations by an angle $\phi$:
\begin{equation} \label{infrot}
  \delta\vec{r}_j\ =\ \delta\vec{\phi}\times\vec{r}_j\,,~~~
  \delta\dot{\vec{r}}_j\ =\ \delta\vec{\phi}\times\dot{\vec{r}}_j\,.
\end{equation}
For example, if this rotation is about the $z$ axis $|\delta\vec{r}_j|= 
|\delta\phi\,r_j\,\sin(\theta)|$ holds. The components of the rotations 
(\ref{infrot}) are
\begin{equation}
  \delta x^i_j\qquad {\rm and}\qquad \delta \dot{x}^i_j\,,
\end{equation}
where $i=1,2,3$ labels the coordinates and $j=1,\dots,n$ the particles.
Assuming symmetry of the Lagrangian under the rotation, we have
\begin{equation}
  0\ =\ \sum_j \left\{ \sum_i\frac{\partial L}{\partial x^i_j}\,
        \delta x^i_j + \sum_i\frac{\partial L}{\partial\dot{x}^i_j}\,
        \delta \dot{x}^i_j \right\}
\end{equation}
Using the definition of the genralized momentum and Euler-Lagrange
equations, this reads
\begin{equation}
  0\ =\ \sum_j \left\{ \sum_i\dot{p}^i_j\,\delta x^i_j + 
  \sum_i p^i_j\,\delta\dot{x}^i_j \right\}\ =\
  \sum_j \left\{ \dot{\vec{p}}_j\cdot\delta \vec{r}_j + 
  \vec{p}_j\cdot \delta\dot{\vec{r}}_j \right\}\,.
\end{equation}
Inserting the definitions (\ref{infrot}) 
\begin{equation}
  0\ =\ \sum_j \left\{ \dot{\vec{p}}_j\cdot(\delta\vec{\phi}
  \times\vec{r}_j) + \vec{p}_j\cdot(\delta\vec{\phi}\times
  \dot{\vec{r}}_j) \right\}
\end{equation}
and we want to pull out the $\delta\vec{\phi}$ rotations as they are
the same for all particles:
\begin{equation}
  0\ =\ \sum_j \left\{ \delta\vec{\phi}\cdot(\vec{r}_j\times\dot{p})j) 
  + \delta\vec{\phi}\cdot(\dot{\vec{r}}_j\times\vec{p}_j) \right\}\
  =\ \delta\vec{\phi}\ \frac{d~}{dt}\,\sum_j(\vec{r}_j\times\vec{p}_j)\
\end{equation}
\begin{equation}
  \Leftrightarrow\ \sum_j(\vec{r}_j\times\vec{p}_j)\ =\ \vec{L}\
  =\ {\rm Constant}\,.
\end{equation}
\end{document}
