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\begin{document}

\noindent Solution for assignment 13: \bigskip

\begin{eqnarray}
  \left(d\vec{r}\right)^2 &=& (dx)^2+(dy)^2+(dz)^2\\
  &=& (d\rho)^2 + \rho^2 (d\phi)^2 + (dz)^2\\
  &=& (dr)^2 + r^2 \sin^2\theta\,(d\phi)^2 + r^2 (d\theta)^2\\
  \left(\vec{v}\right)^2 &=& (\dot x)^2+(\dot y)^2+(\dot z)^2\\
  &=& (\dot\rho)^2 + \rho^2 (\dot\phi)^2 + (\dot z)^2\\
  &=& (\dot r)^2 + r^2 \sin^2\theta\,(\dot\phi)^2 + r^2 (\dot\theta)^2
\end{eqnarray}
Locally orthonormal unit vectors are given by
\begin{eqnarray}
  \hat{\rho} &=& \cos(\phi)\,\hat{x} + \sin(\phi)\,\hat{y}\\
  \hat{\phi} &=&-\sin(\phi)\,\hat{x} + \cos(\phi)\,\hat{y}\\
  \hat{r} &=& \sin(\theta)\,\hat{\rho} + \cos(\theta)\,\hat{z}\ 
  =\ \sin(\theta)\,\cos(\phi)\,\hat{x} + 
     \sin(\theta)\,\sin(\phi)\,\hat{y} + \cos(\theta)\,\hat{z}\\ 
  \hat{\theta} &=& \cos(\theta)\,\hat{\rho} - \sin(\theta)\,\hat{z}\ 
  =\ \cos(\theta)\,\cos(\phi)\,\hat{x} + 
     \cos(\theta)\,\sin(\phi)\,\hat{y} - \sin(\theta)\,\hat{z}
\end{eqnarray}
Usefull is table~\ref{tab_dot} of dot products between the Cartesian
and the new, local orthonormal unit vectors.

\begin{table}[th]  
\centering % \bigskip 
\begin{tabular}{|c|c|c|c|} \hline
              & $\hat{x}$          & $\hat{y}$          & $\hat{z}$   \\ \hline
$\hat{\rho}$  & $\cos\phi$         & $\sin\phi$         &     0       \\ \hline
$\hat{\phi}$  &$-\sin\phi$         & $\cos\phi$         &     0       \\ \hline
$\hat{r}$     &$\sin\theta\cos\phi$&$\sin\theta\sin\phi$&$ \cos\theta$\\ \hline
$\hat{\theta}$&$\cos\theta\cos\phi$&$\cos\theta\sin\phi$&$-\sin\theta$\\ \hline
\end{tabular}
\caption{Dot products between Cartesian and local orthonormal  unit 
vectors.} \label{tab_dot} \end{table}

The velocity vector is in Cartesian, cylindrical and spherical coordinates
\begin{eqnarray}
  \vec{v} &=& \dot x\,\hat{x} + \dot y\,\hat{y} +\dot z\,\hat{z}\\
          &=& \dot\rho\,\hat{\rho}+\rho\,\dot\phi\,\hat{\phi}+\dot z\,\hat{z}\\
          &=& \dot r\,\hat{r}+r\,\sin(\theta)\,\dot\phi\,\hat{\phi}+
              r\,\dot\theta\,\hat{\theta}
\end{eqnarray}
\vfil \eject

{\bf Application (Landau-Lifshitz, p.21, Problems 1 and 2):} Cartesian 
angular momentum components and angular momentum squared in cylindrical 
and spherical coordinates.

Vector products between cylindrical unit vectors:
\begin{eqnarray}
  \hat{\rho}\times\hat{\phi}\ =\ \hat{z}\,,~~~
  \hat{\phi}\times\hat{z}\ =\ \hat{\rho}\,,~~~
  \hat{z}\times\hat{\rho}\ =\ \hat{\phi}
\end{eqnarray}
Angular momentum in cylindrical coordinates:
\begin{eqnarray}
  \vec{M} = m\,\vec{r}\times\vec{v} &=& m\,\left(\rho\,\hat{\rho}+
  z\,\hat{z}\right) \times \left(\dot\rho\,\hat{\rho}+
  \rho\,\dot\phi\,\hat{\phi} +\dot z\,\hat{z}\right) \\ &=&
  m\,\left(\rho^2\dot\phi\,\hat{z} - \rho\,\dot{z}\,\hat{\phi} +
  z\,\dot\rho\,\hat{\phi} - z\,\rho\,\dot\phi\,\hat{\rho}\right)\\
  &=& m\,\left(-z\,\rho\,\dot\phi\,\hat{\rho} + \left(z\,\dot\rho
   -\rho\,\dot{z}\right)\,\hat{\phi}+\rho^2\dot\phi\,\hat{z}\right)
\end{eqnarray}
Therefore,
\begin{eqnarray}
  \vec{M}^{\,2} = M^2 &=& m^2\rho^2\dot\phi^2\left(\rho^2+z^2\right)
  +  m^2 \left(z\,\dot\rho -\rho\,\dot{z}\right)^2
\end{eqnarray}
Vector products between spherical unit vectors:
\begin{eqnarray}
  \hat{\phi}\times\hat{r}\ =\ \hat{\theta}\,,~~~
  \hat{r}\times\hat{\theta}\ =\ \hat{\phi}\,,~~~
  \hat{\theta}\times\hat{\phi}\ =\ \hat{r}
\end{eqnarray}
Angular momentum in spherical coordinates:
\begin{eqnarray}
  \vec{M} = m\,\vec{r}\times\vec{v} &=& m\,r\,\hat{r}\times \left(
  \dot r\,\hat{r}+r\,\sin(\theta)\,\dot\phi\,\hat{\phi}+
  r\,\dot\theta\,\hat{\theta} \right) \\ &=& m\,r^2 \left(
  \dot\theta\,\hat{\phi}-\sin(\theta)\,\dot\phi\,\hat{\theta}\right)
\end{eqnarray}
Therefore,
\begin{eqnarray}
  \vec{M}^{\,2} = M^2 &=& m^2\,r^4\left( \dot\theta^2 +
  {\dot\phi}^2 \sin^2\theta \right)
\end{eqnarray}

Now use $M_x=\vec{M}\cdot\hat{x}$ and so on. \medskip

Cylindrical coordinates:
\begin{eqnarray}
  M_x\ =\ \vec{M}\cdot\hat{x} &=& m\,\left(\rho\,\dot{z}-
  z\,\dot{\rho}\right)\,\sin\phi - m\,\rho\,z\,\dot\phi\,\cos\phi\\
  M_y\ =\ \vec{M}\cdot\hat{y} &=& - m\,\left(\rho\,\dot{z}-
  z\,\dot{\rho}\right)\,\cos\phi - m\,\rho\,z\,\dot\phi\,\sin\phi\\
  M_z\ =\ \vec{M}\cdot\hat{z} &=& m\,\rho^2\,\dot\phi
\end{eqnarray}

Spherical coordinates:
\begin{eqnarray}
  M_x\ =\ \vec{M}\cdot\hat{x} &=& -m\,r^2 \left(\dot\theta\,\sin\phi
       + \dot\phi\,\sin\theta\,\cos\theta\,cos\phi\right)\\
  M_y\ =\ \vec{M}\cdot\hat{y} &=& m\,r^2 \left(\dot\theta\,\cos\phi
       - \dot\phi\,\sin\theta\,\cos\theta\,sin\phi\right)\\
  M_z\ =\ \vec{M}\cdot\hat{z} &=& m\,r^2 \left(\dot\theta^2
       + \dot\phi^2\sin^2\theta\right)
\end{eqnarray}

\end{document}
 \vfil\eject
% \bigskip 

Another, less elegant possibility: \medskip

Cartesian velocity components in cylindrical coordinates:
\begin{eqnarray}
   v_x = \dot x = \vec{v}\cdot\hat{x} &=& 
         \dot\rho\,\cos\phi - \rho\,\dot\phi\,\sin\phi \\
   v_y = \dot y = \vec{v}\cdot\hat{y} &=& 
         \dot\rho\,\sin\phi + \rho\,\dot\phi\,\cos\phi \\
   v_z = \dot z = \vec{v}\cdot\hat{z} &=& \dot z\,\hat z
\end{eqnarray}
Cartesian angular momentum components in cylindrical coordinates:
\begin{eqnarray}
   M_x = m\,(y\,\dot z - z\,\dot y) &=& m\,[\rho\,\sin(\phi)\,\dot z
       - z\,\dot\rho\,\sin(\phi)-z\,\rho\,\dot\phi\,\cos(\phi)] \\
       &=& m\,(\rho\,\dot z-z\,\dot\rho)\,\sin(\phi) - 
           m\,\rho\,z\,\dot\phi\,\cos(\phi)
\end{eqnarray}
\end{document}
